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Students get through the MP Board Class 11th Physics Important Questions Chapter 4 Motion in a Plane which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 4 Motion in a Plane

Motion in a Plane Class 11 Important Questions Very Short Answer Type

Question 1.
State for each of the following physical quantities, if it is a scalar or a vector : Volume, mass, speed, acceleration, density, number of moles, velocity, angular
frequency, displacement, angular velocity.
Answer:
Scalar quantities : Volume, mass, speed, density, number of moles, angular frequency.
Vector quantities : Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path, length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Impulse. As impulse = force × time = change in momentum . As force and momentum are vector quantities, so it is a vector.

Question 3.
What do you mean by magnitude of a vector quantity?
Answer:
The positive value of the modulus of a vector quantity is called magnitude of
vector quantity. It is represented by \(|\vec{a}|\)

Question 4.
Whether addition of two vectors depends upon the order of placement of the vectors.
Answer:
No, according to the commutative law of vector addition \(\vec{a}+\vec{b}\) = \(\vec{b}+\vec{a}\).

Question 5.
Can two vectors of different dimension can be added?
Answer:
No, we can add two vectors of same dimension.

Question 6.
How can a vector and scalar be added?
Answer:
A vector and a scalar cannot be added.

Question 7.
How can three vectors be added so that its resultant is zero?
Answer:
If three vectors can be represented completely by the three sides of a triangle taken in the same order, then their resultant is zero.

Question 8.
What do you mean by unit vector and zero vector?
Answer:
If the magnitude of a vector is one then it is called as unit vector and if its
magnitude is zero and direction is uncertain is called zero vector, unit vector
\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\)

Question 9.
What do you mean by resolution of a vector?
Answer:
When a vector is divided into two or more components such that their addition is the same vector then it is called resolution of vector.

Question 10.
Can the magnitude of component of vector more than its original vector?
Answer:
No, it cannot be.

Question 11.
Can a vector be multiplied with a scalar? Illustrate with example.
Answer:
Yes, \(\vec{F}=m \cdot \vec{a}\)

Question 12.
When the scalar product of two vectors is maximum?
Answer:
When both the vectors are parallel, then angle between them is 0° or 180°. Then scalar product will be maximum.

Question 13.
Area of a surface is vector or scalar quantity?
Answer:
It is vector quantity because its direction is perpendicular to its surface.

Question 14.
Which type of quantity is modulus of a vector?
Answer:
Modulus of a vector is scalar quantity.

Question 15.
What do you mean by displacement vector?
Answer:
Displacement vector is those vector which specify that how much and in which direction an object has displaced in time intervel t₁ and t₂.

Question 16.
What do you mean by equal vectors?
Answer:
Two vectors whose magnitude and directions are same are called equal vectors.

Question 17.
Write the triangle law of vector addition.
Answer:
Triangle law of vector addition : If two vectors are represented by the adjacent sides of a triangle in a definite order. Then the
resultant will be given by the third side taken in the opposite manner.
i.e., \(\vec{R}=\overrightarrow{O A}+\overrightarrow{A B}\)

Question 18.
Can a vector of IS dyne be added in a vector of 10 Newton?
Answer:
Yes, it can be added because dimensions of both physical quantities are same.

Question 19.
Why displacement vector and velocity vector cannot be added?
Answer:
They cannot be added because their dimensions are different.

Question 20.
What will happen if a vector is extended in its parallel direction?
Answer:
No changes will occur in vector.

Question 21.
When the addition of two vectors are maximum and minimum?
Answer:
When both the vectors are parallel the addition is maximum and when they are opposite their addition is minimum.

Question 22.
Can the resultant of two different vectors be zero?
Answer:
No, because the minimum magnitude of \(\vec{A}+\vec{B}\) is A – B.

Question 23.
Whether addition of vectors depends on order they placed?
Answer:
No.

Question 24.
If P and Q are two vector quantities then PQ will be which quantity?
Answer:
It is a scalar quantity.

Question 25.
When vector product of two vectors is maximum?
Answer:
When two vectors are perpendicular to each other, then their vector product is maximum.

Question 26.
If a component of a vector is zero and other component of same vector is non-zero, can the magnitude of that vector be zero?
Answer:
No.

Question 27.
If the angle between two vectors are changed keeping magnitude to be same; then what will be the effect on resultant vector?
Answer:
Magnitude and direction of resultant vector will get changed.

Question 28.
When addition and subtraction of two vectors are equal?
Answer:
When the magnitude of both vectors are same and mutually perpendicular to each other then addition and subtraction of two vectors will be equal.

Question 29.
If \(\vec{A} \cdot \vec{C}=\vec{B} \cdot \vec{C}\) whether \(\vec{A}\) and \(\overrightarrow{\boldsymbol{B}}\) will be always equal?
Answer:
No.

Question 30.
What is projectile motion? When does the trajectory is parabola?
Answer:
When a body is projected at an angle other than vertical, the body moves under gravity and falls at different points on the earth. The path traced out by the body is called trajectory and its motion is called projectile.
Example : Motion of cannon ball, the object released from an aeroplane flying horizontally. Trajectory is parabola when air of friction is negligible.

Question 31.
At what angle the horizontal range is maximum?
Answer:
When θ = 45°, = R _max = \(\frac{u^{2}}{g}\)

Question 32.
From top of a tower two bullets are shot with different velocities in horizontal direction. Which one will reach on the ground first?
Answer:
Both bullets will reach the ground at the same time because initial velocity in vertical direction is same for them.

Motion in a Plane Class 11 Important Questions Short Answer Type

Question 1.
Establish the following vector inequalities geometrically or otherwise :
(a) \(|a+b| \leq|a|+|b|\)
(b) \(|\boldsymbol{a}+\boldsymbol{b}| \geq\|\boldsymbol{a}|-| \boldsymbol{b}\|\).
When does the equality sign above apply?
Answer:
Let the two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the sides OP and PQ of the ΔOPQ taken in the same order, then their resultant is represented by the side OQ of the triangle such that
OQ = a + b as shown in fig.
Thus, OP = a,PQ = b
∴ \(|O P|=|a|,|P Q|=|b|\)
and \(|O Q|=|a+b|\)
(a) \(|a+b| \leq|a|+|b|\)

Proof:
We know from the property of a triangle that its one side is always less than the sum of the lengths of its two other sides.
Now in Δ OPQ,
OQ<OP+PQ
or \(|a+b|<|a|+|b|\) …(1)
If the two vectors a and b are acting along the same straight line i.e., are collinear and in the same direction, then
θ (angle between them) = 0
∴ \(|a+b|=\sqrt{a^{2}+b^{2}+2 a b \cos \theta}=\sqrt{(a+b)^{2}}\)
= a + b = \(|a|+|b|\) ..(2)
From eqns. (1) and (2), we get
\(|a+b| \leq|a|+|b|\)
The equality sign applies if a and b are collinear and act in same direction.
(b) \(\text { (b) }|a+b| \geq \| a|-| b||\)

Proof:
One side is more than the difference of two other sides i.e., in ΔOPQ in fig.
OQ >\(|O P-P Q|\)
The modulus of (OP – PQ) has been taken because L.H.S. is + ve and R.H.S. may be – ve if OP < PQ.
or \(|a+b|>\| a|-| b||\) ….(3)
If a and b are acting along the same straight line in the opposite direction, then
\(|a+b|>\| a|-| b||\)
From eqns. (3) and (4), we get
\(|a+b| \geq \| a|-| b||\)
Equality sign holds for as stated in eqn. (4).

Question 2.
Explain scalar product of two vectors?
Answer:
Let the angle between two vectors \(\vec{A}\) and \(\vec{B}\) is θ
∴ Scalar product of \(\vec{A}\) and \(\vec{B}\) will be
\(\vec{A} \cdot \vec{B}\) = AB cosθ

Example:
Let a force\(\vec{F}\) is acting on a particle and is displaced through \(\vec{r}\) at an angle θ then
\(\vec{F} \cdot \vec{r}\) = Frcosθ
Now, \(\vec{A} \cdot \vec{B}\) = AB cosθ
Hence, \(\vec{A} \cdot \vec{B}\) = 0 if either \(|\vec{A}|\) = 0 or \(|\vec{B}|\)= 0 or θ= 90°.

Question 3.
What type of quantity is work?
Answer:
Since work = Force × Displacement
i.e., W = \(\vec{F} \cdot \vec{d}\)
Since scalar product of two vectors are scalar, therefore work is a scalar quantity.

Question 4.
What type of quantity is kinetic energy?
Answer:
K.E =\(\frac{1}{2}\)mv² = \(\frac{1}{2}\)m \((\vec{v} \cdot \vec{v})\)
Since, \(\vec{v} \cdot \vec{v}\) will be a scalar quantity, so kinetic energy will be a scalar quantity.

Question 5.
Prove that addition of vectors obeys commutative law.
Or
State and prove the commutative law of vector addition.
Answer:
Consider two vectors \(\vec{a}\) and \(\vec{b}\) which are in the same plane \(\overrightarrow{O A}=\vec{a}\) and \(\overrightarrow{A B}=\vec{b}\).
Join \(\overrightarrow{O B}\).
By triangle law of addition,
\(\vec{a}+\vec{b}=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\) …(1)
Complete the parallelogram OABC. In a parallelogram, op¬posite sides are parallel and equal. Hence,
\(\overrightarrow{A B}=\overrightarrow{O C}=\vec{b} \text { and } \overrightarrow{O A}=\overrightarrow{C B}=\vec{a}\)
By triangle law of addition,
\(\vec{b}+\vec{a}=\overrightarrow{O C}+\overrightarrow{C B}=\overrightarrow{O B}\) …..(2)
From eqns. (1) and (2), \(\vec{a}+\vec{b}=\vec{b}+\vec{a}\)
Hence, the addition of vectors obey commutative law.

Question 6.
Show that addition of vectors obey the associative law.
Answer:
In a plane choose origin O and construct

Question 7.
Explain the law of parallelogram of vector’s addition.
Answer:
If the two vectors acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, then their resultant vector is represented in magnitude and direction by the diagonal passing through their common origin.
If two vectors \(\vec{A}\) and \(\vec{B}\) are represented by two adjacent sides of a parallelogram OACB, the diagonal OC represents the resultant of these two vectors.
In parallelogram OACB,
OB= AC
∴ \(\overrightarrow{O B}=\overrightarrow{A C}=\vec{B}\)
In ? OAC, as per triangle law of vector addition,
\(\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A C}\)
∴ \(\vec{R}=\vec{A}+\vec{B}\)

Question 8.
Write the expression of position vector, velocity vector and acceleration vector for a particle moving in a plane.
Answer:

Question 9.
What is the effect of air on the time of flight and range of projectile?
Answer:
Time of flight and range will be decreased.

Question 10.
A food packet is released from an aeroplane flying horizontally. What will be the path of packet:
(i) Seen from the plane and
(ii) Seen from the earth?
Answer:
(i) In a straight line vertically downward.
(ii) Parabolic.

Question 11.
An aeroplane is moving horizontally with uniform velocity. A bomb is dropped from the plane. What will be the position of plane, when the bomb strikes the ground?
Answer:
When the friction of medium is zero, the plane will be just above the point where the bomb strike the ground.
When the friction is not zero, the plane will be little ahead above the point where the bomb will strike the ground.

Question 12.
What will be the change in centripetal force when speed of an object and radius of path are doubled in circular motion.
Answer:
Centripetal force F = \(\frac{m v^{2}}{r}\)
When v₁ = 2v and r₁ = 2r
Then, F₁ = \(\frac{m v_{1}^{2}}{r_{1}}\)
or F₁ = \(\frac{m(2 v)^{2}}{2 r}=\frac{4 m v^{2}}{2 r}\)
or F₁ = \(\frac{2 m v^{2}}{r}\)
or F₁= 2F is centripetal force will get doubled.

Question 13.
What do you mean by uniform circular motion? Give two examples.
Answer:
When a body moves in a circular path with constant speed, then its motion is called uniform circular motion.
Example: Motion of the tip of the blade of a fan, motion of the earth around the sun is nearly a uniform circular motion.

Question 14.
What do you mean by angular displacement?
Answer:
The angle suspended by the particle on the circular path in given time interval is called an angular displacement of the particle its unit is radian.

Question 15.
If the velocity of projectile motion is doubled then what will be the effect on its horizontal range?
Answer:
since R = \(\frac{u^{2} \sin ^{2} \theta}{g}\)
i.e., R ∝ u²
When velocity of projectile motion is doubled then its horizontal range get four times increased.

Question 16.
What do you understand by time-period and frequency?
Answer:
Time-period is defined as the time taken by an object to complete one rotation. Its unit is second. If the angular velocity of an object is a> and angle suspended at the centre during one complete rotation is 2π, then
Time-period T = \(\frac{2 \pi}{\omega}\)
or ω = \(\frac{2 \pi}{T}\)
Frequency is defined as the number of rotations completed by the body in one second. Its unit is hertz.
If time-period is T, then frequency n = \(\frac{1}{T}\)
or n = \(\frac{1}{2 \pi / \omega}=\frac{\omega}{2 \pi}\)
or ω = 2πn

Question 17.
Define linear velocity and angular velocity of a particle in circular motion. Establish the relation between them.
Answer:
Angular velocity :
The rate of change of angle subtended by a body moving in a circular path, at the centre of the path per second is called angular velocity of the body.
Linear velocity :
The distance travelled by the particle on the circular path in one second is called linear velocity.
Its SI unit is ms^-1.

Relation between linear velocity and angular velocity:
Let a particle is moving in a circular path of radius r and centre O. Its linear velocity is v and angular velocity is ω
If in small time Δt, the particle travels distance Δs on the circle,
∴ ∠POQ = Δθ
Hence’ ω = \(\frac{\Delta \theta}{\Delta t}\) ….(1)
and ν = \(\frac{\Delta s}{\Delta t}\) ….(2)
But, Angle = \(\frac{\text { Arc }}{\text { Radius }}\)
∴ Δθ = \(\frac{\Delta s}{r}\)
∴ From eqn. (1), we get
ω = \(\frac{\Delta s}{r \Delta t}\)
or \(\frac{\Delta s}{\Delta t}\) = ωr
∴ ν = ωr

Hence, Linear velocity = Angular velocity × Radius.

Question 18.
Define centripetal acceleration and derive its expression. Also deduce its direction.
Answer:
Centripetal acceleration :
It is defined as the acceleration which always acts towards the centre along the radius of the circular path.
“Centripetal” comes from a Greek term which means “centre seeking”.
Expression for centripetal acceleration:
Let a body be moving with constant angular velocity? and constant speed? in a circular path of radius R and centre O.

Let the body be at points P and Q at time t and t + Δt respectively.
Also let Δθ= ∠POQ be the angular displacement in a small time interval Δt, thus
ω = \(\frac{\Delta \theta}{\Delta t}\) = or Δθ = ωΔt …(1)
If ν and \(\vec{v}+\Delta \vec{v}\)be the velocity vectors of the body at points P and Q acting along PL and QM respectively. As the body is moving with uniform speed, so lengths PL and PM are equal i.e., PL = PM. The change in velocity \(\Delta \vec{v}\) from P to Q is due to the change in the direction of velocity vector.
Hence magnitude of acceleration of the body is given by

From point’a’draw \(\overrightarrow{a b}\) parallel and equal to \(\overrightarrow{P L}\) and \(\overrightarrow{a c}\) parallel and equal to \(\overrightarrow{Q M}\),
thus
\(\overrightarrow{a b}=\vec{v}, \overrightarrow{a c}=\vec{v}+\Delta \vec{v}\)
∴ \(\overrightarrow{b c}=\Delta \vec{v}\)
Clearly the angle between \(\overrightarrow{a b}\) and \(\overrightarrow{a c}\) is Δθ . As Δt is quite small so a and c lie very close to each other, thus be can be taken to be an arc of \(\widehat{b c}\) a circle of radius
ab = \(|\vec{v}|\)

Also when Δt → 0, then L.H.S. of eqn. (4) represents acceleration. Thus eqn. (4) can be written as

Eqn. (6) gives the magnitude of the acceleration produced in the body.
Direction of acceleration :
Acceleration always acts in the direction of Δν which acts from b to c. When Δt is decreased, Δθ also decreases. If Δt → 0, then Δθ → 0, so Δν tends to be perpendicular in ab. As ab || PL, so Δν tends to act perpendicular to PL i.e., along PO is along the radius towards the centre of circular path.
Since ν and R are always constant, so magnitude of centripetal acceleration is also constant. But the direction changes pointing always towards the centre. So centripetal acceleration is not a constant vector.

Motion in a Plane Class 11 Important Questions Long Answer Type

Question 1.
State parallelogram law of vector addition of two vectors and derive an expression for their resultant vector.
Ans. Parallelogram law of vector addition : If the two vectors acting at a point are represented in magnitude and direction by the adjacent sides of a parallelogram, then their resultant vector is represented in magnitude and direction by the diagonal passing through their common origin.
If two vectors \(\vec{A}\) and \(\vec{B}\) are represented by two adjacent sides of a parallelogram OACB, the diagonal OC represents the resultant of these two vectors.
In parallelogram OACB,
OB= AC
∴ \(\overrightarrow{O B}=\overrightarrow{A C}=\vec{B}\)
In Δ OAC, as per triangle law of vector addition,
\(\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A C}\)
∴ \(\vec{R}=\vec{A}+\vec{B}\)

Two vectors \(\vec{a}\) and \(\vec{b}\) are completely represented by the two sides OA and OB of a parallelogram respectively as shown in fig.

The angle between two vectors \(\vec{a}\) and \(\vec{b}\) is θ. The diagonal
OC of the paralellogram gives the resultant \(\vec{R}\),
Such that \(\vec{R}\) = \(\vec{R}=\vec{a}+\vec{b}\)
The resultant \(\vec{R}\) makes a angle with direction of \(\vec{a}\) From point C draw a perpendicular CN on extended OA.
In ΔANC,
sinθ = \(\frac{C N}{A C}\) ⇒ CN= AC sinθ
⇒ CN = b sinθ ….(1)
Also, cosθ = \(\frac{A N}{A C}\)
⇒ AN = AC cosθ = b cosθ ….(2)
∴ ON = OA + AN =a+b cosθ ….(3)
In ΔONC, OC² = ON² + CN²
⇒ R² = (a +b cosθ)² +(bsin?)²
or R² = a² +b² cos² θ+2abcosθ+b² sin²θ
or R² = a² +b²(sin²θ+ cos² θ) + 2abcosθ
or R² = a² +b² +2abcosθ
or R = \(\sqrt{a^{2}+b^{2}+2 a b \cos \theta}\) …(4)
The above eqn. (4) is the required expression.
Direction of R is given by
tanα = \(\frac{b \sin \theta}{a+b \cos \theta}\) or a = tan^-1\(\left(\frac{b \sin \theta}{a+b \cos \theta}\right)\)

Question 2.
What do you mean by two-dimensional motion? Establish the relation between displacement, velocity and acceleration.
Answer:
TWo-dimensional motion: When a body moves in a plane, its motion is called two-dimensional motion.
Example: Motion in circular path, motion of the bullet fired from the gun.
Let a body is moving with uniform acceleration \(\vec{a}\). At time t = 0 its velocity is \(\vec{?}\) (0) and at t sec velocity is ν(t). 1f the position vectors are \(\vec{r}\) (0) and \(\vec{r}\) (t) respectively.

Question 3.
What do you understand by projectile motion? Prove that the path of projectile is parabolic.
Answer:
Projectile motion :
When a body is projected at an angle other than vertical, the body moves under gravity and falls at different points on the earth. The path traced out by the body is called trajectory and its motion is called projectile.
Example : Motion of cannon ball, the object released from an aeroplane flying horizontally. Trajectory is parabola when air of friction is negligible.

Let a body is projected with velocity u, at an angleθ with the horizontal. Thus, we have two components of u.
i.e., Horizontal component along X-axis = u cosθ
and Vertical component along T-axis = u sinθ
∵ The horizontal velocity is free from gravitational acceleration,
∴ u_x = v_x(0) = u cosθ and a_x = 0
If x (t) is the position of the particle of any instant of time, then the displacement will be x(t) – λ(0),
But, λ(0) = 0
∴ x(t) – λ(0) = λ(t) = x
Hence, x = v_x(0)t + \(\frac{1}{2}\) a_xt²
or x = u cosθ t
or t = \(\frac{x}{u \cos \theta}\) …(1)
Now, for the vertical component,
v_y = u sinθ and a_y = -g, (∵ the particle moves in upward direction) If y(t) is the position of the particle at any instant of time, then the displacement will be y(t) -y(0)
But, y(0) = 0
∴ y(t) – y(0)=y(t) = y

or y = x tanθ –\(\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\) …(3)
Since, u, g and θ are constant, therefore the eqn. (3) is in the form of
y = ax – bx²
Where, a = tanθ and b = \(\frac{g}{2 u^{2} \cos ^{2} \theta}\)
Which is the equation of a parabola. Since, it is a quadratic equation in x. Proved.

Question 4.
Derive the expression for time of flight, horizontal range and maximum height attain for a projectile motion.
Answer:
Expression for time of flight: Let a body is projected with initial velocity u at an angle θ with the horizontal.
∴ The components of u are:
(i) Horizontal component, u_x = u cosθ and
(ii) Vertical component, u_y = u Sinθ
The time taken by the projectile to travel from point O to A, is called its time of flight.
Now, initial velocity along Y-axis is
u_y=ν_y(0)=u sinθ
Now,at highest point
ν_y(t) = 0 and a_y = -g
If the time taken by the projectile from O to P is t, then
ν_y(t)= ν_y(0)+at
or
0=u sinθ-gt
or t = \(\frac{u \sin \theta}{g}\)
But Time of flight, T = 2t
∴ T = \(\frac{2 u \sin \theta}{g}\)

Horizontal Range:
The horizontal distance travelled in the time of flight is called range.
Now, Range=OA=R
∴ R = Horizontal velocity × Time of flight
or R= u cosθ × \(\frac{2 u \sin \theta}{g}\)
or R= \(\frac{u^{2} \cdot 2 \sin \theta \cos \theta}{g}\)
∴ R = \(\frac{u^{2} \sin 2 \theta}{g}\) (∵ 2sinθ cosθ = sin2θ)

For maximum range, sin2θ = 1
or sin2θ= sin9o°
or 2θ = 90°
∴ θ =45°
For maximum range the angle of projection should be 45°.
∴ R_max = \(\frac{u^{2}}{g} .\)

Maximum height attain: Now, we have two resolved parts of initial velocity u, i,e.,
(i) Horizontal component u_x = u cosθ and
(ii) Vertical component u_y = u Sinθ.
Suppose, the maximum height of projectile is Hand it takes r sec. to reach the highest point.
Now, for vertical motion,
ν_y(0) = u sinθ and a_y = -g
At highest point, ν_y(t) = 0
:. By the equation, ν_y(t)=ν_y(0)+a_yt
or 0 = u sinθ – gt

Motion in a Plane Class 11 Important Numerical Questions

Question 1.
Find modulus and unit vector of the vector
\(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\)
Solution:

Question 2.
If \(\vec{a}=\hat{i}+3 \hat{j}-4 \hat{k}\) and \(\vec{b}=2 \hat{i}-3 \hat{j}+4 \hat{k}\) then find out \(\vec{a}+\vec{b}\),
\(\vec{a} \cdot \vec{b}\) and \(\vec{a} \times \vec{b}\)
Solution:

Question 3.
If\(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then prove that \(\text { a }\) and \(\text { b }\) are mutually perpendicular to each other.
Solution:

Question 4.
When the resultant of two equal vectors will be A.
Solution:
R² = A² + B² + 2ABcosθ
When \(\vec{B}=\vec{A}\), then
R² = A² + A² + 2A.Acosθ
= 2A² +2A²cosθ
If costθ = – \(\frac{1}{2}\) = cos 120°
Then R² = A²
i.e., R = A
i.e., θ = 120°.

Question 5.
Prove that \(3 \hat{i}+2 \hat{j}-\hat{k}\) and \(2 \hat{i}-2 \hat{j}+2 \hat{k}\) are perpendicular to each other.
Solution:

Question 6.
If A = \(\vec{A}=2 \hat{i}+2 \hat{j}+p \hat{k}\) and \(\vec{B}=2 \hat{i}-\hat{j}+\hat{k}\)are perpendicular, then find the value of p.
Solution:
Given : \(\vec{A}\) and \(\vec{B}\) are perpendicular
∴ \(\vec{A} \cdot \vec{B}\) = 0
or \((2 \hat{i}+2 \hat{j}+p \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 0
∴ 2.2 + 2.(-1) + p.(1) = 0
or 4 -2 + p = 0
or P = -2.

Question 7.
Rain is falling vertically with a speed of 30 ms *. A woman rides a bicycle
with a speed of 10 ms 1 in the north to south direction. What is the direction in which she should hold her umbrella?

Solution:
Let the rain be falling vertically downward with the speed of 30 ms^-1 along OA. The woman is moving along OS with a speed of 10 ms^-1.
To protect herself from rain, the woman should hold her umbrella in the direction of relative velocity of the rain with respect to the woman i.e., ν_rw. To determine the ν_rw the woman must be brought to rest by imposing an equal and opposite velocity of the observer both on woman as well as on the rain. Rain thus has two velocities :
(i) Its own velocity of 30 ms ^-1 along \(\overrightarrow{O A}\)
(ii) Imposed velocity of 10 ms^-1 along \(\overrightarrow{O B}\)
The relative velocity is the resultant of these two velocities.

tan α = tan18°26′
α = 18° 26′
So woman should hold umbrella at an angle of 18°26′ with the vertical towards south.

Question 8.
A cyclist is riding with a speed of 27 kmh^-1. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.5 ms^-1 every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer:

Speed of cyclist ν = 27kmh^-1
= 27 x \(\frac{5}{18}\) ms^-1
= \(\frac{15}{2}\) ms^-1
Radius of circular turn, r = 80 m
Here the cyclist will be acted upon by two accelerations a_c and a_t Centripetal acceleration (a_c) is given by

Let P be the point at which the cyclist applies brakes, then the tangential acceleration a_t, (which will be negative) will act opposite to velocity of cyclist and is given by
a_t = \(\frac{\text { Change in velocity }}{\text { Time }}=\frac{0 \cdot 5}{1}\) ms^-2
= 0.5 ms^-2
As both a_c and a_t, act along the radius of circle and tangent to the circle respectively, so ac and a, are mutually perpendicular, thus the magnitude of the net acceleration ‘a’
(i.e., resultant acceleration) is given by

Direction of a: Let θ be the angle made by the net acceleration with the velocity of cyclist, then
tanθ \(\frac{a_{c}}{a_{t}}\) = \(\frac{0.703}{0.5}\)
= 1.406
or tanθ = tan54°33′
θ = 54° 33′ = 54-44°.

Question 9.
A ball is projected at an angle of 30° with a velocity of 20 ms^-1. Find the time of flight.
Solution:
u = 20 ms^-1, θ= 30°.

Question 10.
A cricketor can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketor throw the same ball?
Solution:
Given:
R = 100 m
From the formula R = \(\frac{u^{2}}{g}\), we get g
100 = \(\frac{u^{2}}{g}\)
and H = \(\frac{u^{2}}{2 g}=\frac{R}{2}=\frac{100}{2}\) = 50m

Question 11.
A ball is projected at an angle of 60° with a velocity of 40 ms^-1. What will be the maximum height and range?
Solution:

Question 12.
A body is projected at an angle of 15° and its range is 50 m. If it is projected at an angle of 45°, then find the range.
Solution:

Question 13.
A particle is executing 3 revolutions per second in a circular path of radius 15 cm. Find out angular velocity and linear velocity.
Solution:
Given r = 15 cm, frequency n = 3 revolutions per second
From ω = 2πn, we get
ω = 2 × π × 3 = 6π radian/sec.
Again, from ν = rω, we get
ν = 15 × 6π = 90πcm/sec.

Question 14.
In a projectile motion of the horizontal range and maximum height attain of an object is same, then find out the angle of projection of the object.
Solution:
Given, R = H

Question 15.
The coordinate of a particle moving in a plane at any time f is given by JC =
ct2 and y = bt² Determine the velocity of the particle.
Solution:

Question 16.
For any projectile motion the horizontal range is four times of maximum height attain. Find out the angle of projection.
Solution:
Given R = 4 H

Question 17.
A bullet is fired with a velocity of 1,000 m/s from a rifle at an angle of 30° w.r.t horizontal, find out:
(i) Time taken to reach highest point from the rifle.
(ii) Total time of flight
(iii) What will be the velocity of bullet when it strike ground surface?
(iv) At what height the buUet will raise ?
(v) At what distance from the rifle the bullet will strike the ground surface?
Solution:

Question 18.
A football player kicks the football with a velocity of 20 m/s making an angle of 30° from the horizontal. Find out
(i) Time of flight
(ii) Maximum height attain by the football
(iii) Velocity of the football and its direction after striking the ground.
(iv) Horizontal range, (given g = 10 m/s²).
Solution:

Question 19.
An aeroplane is flying at a height of 1960 m with a velocity of 600 km/hr. When it is just above a point A on the earth, a packet is dropped, which falls at point B. Calculate AB.
Solution:
u = 600 km/hr =600 × \(\frac{5}{18}\) m/s
= \(\frac{500}{3}\) m/s
For motion of a packet in downward direction,
y = \(\frac{1}{2}\) gt²
or 1960 =\(\frac{1}{2}\) × 9.8 × t²
or t² =400t = 20 sec.
∴ AB = ut = \(\frac{500}{3}\) × 20
Horizontal distance covered by packet i.e.,AB
AB = ut = \(\frac{500}{3}\) × 20
= 3333.3m
= 3.33km.

Motion in a Plane Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
1. Torque is a :
(a) Vector quantity
(b) Scalar quantity
(c) Both
(d) None of these
Answer:
(b) Scalar quantity

Question 2.
If \(\vec{A}=2 \hat{i}+4 \hat{j}\)and \(\vec{B}=2 \hat{i}-5 \hat{j}\), then \(\vec{A}+\vec{B}\) will be :
(a) \(5 \hat{i}-\hat{j}\)
(b) \(3 \hat{i}-\hat{j}\)
(c) \(4 \hat{i}-\hat{j}\)
(d) \(3 \hat{i}-3 \hat{j}\)
Answer:
(c) \(4 \hat{i}-\hat{j}\)

Question 3.
Unit vector in direction of vector \(\vec{A}\) will be :
(a) \(\hat{\mathrm{A}}=\frac{A}{\vec{A}}\)
(b) \(\hat{\mathrm{A}}=\frac{\vec{A}}{A}\)
(c) \(\text { (c) } \hat{\mathrm{A}}=A \cdot \vec{A}\)
(d) \(\text { (d) } \hat{\mathrm{A}}=A+\vec{A}\)
Answer:
(b) \(\hat{\mathrm{A}}=\frac{\vec{A}}{A}\)

Question 4.
Angle between velocity and acceleration of maximum height of a projectile is :
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(d) 90°

Question 5.
Path of a projectile is :
(a) Circular
(b) Straight line
(c) Parabolic
(d) Uncertain.
Answer:
(c) Parabolic

Question 6.
For maximum range of a projectile, the angle must be :
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 7.
Time of flight of a projectile is tx and time taken to reach maximum height is t2, then the relation between them will be :
(a) t₁ =2t₂
(b) t₂ =2t₁
(c) t₂ =3t₁
(d) t ₁=t₂
Answer:
(a) t₁ =2t₂

Question 8.
If the velocity of projectile is doubled, then its range will be :
(a) Same
(b) Become double
(c) Become half
(d) Become four times.
Answer:
(b) Become double

Question 9.
At maximum height of projectile, velocity is :
(a) Zero
(b) Minimum
(c) Equal to initial velocity
(d) None of these.
Answer:
(b) Minimum

Question 10.
Rate of change of angular velocity is called :
(a) Angular displacement
(b) Angular speed
(c) Angular acceleration
(d) Angular momentum.
Answer:
(c) Angular acceleration

2. Fill in the blanks:

1. Angular momentum is a ………………….quantity.
Answer:
Vector

2. Magnitude of a vector \(2 \hat{i}-\hat{j}\) is …………………….
Answer:
√5

3. If an object is thrown horizontally from a height, then its path is ………………..
Answer:
Parabolic

4. At maximum height of a projectile motion, velocity of an object is ………………..
Answer:
Minimum

5. The maximum horizontal distance covered by a projectile in its time of flight is called ………………………
Answer:
Horizontal range

6. ……………………….. is constant in uniform circular motion.
Answer:
Speed

7. At maximum height the direction of acceleration in projectile motion is ………………….
Answer:
Vertically downwards

8. In uniform circular motion there is ……………………. changes in angular momentum of a particle.
Answer:
Zero

9. Angle between velocity and acceleration in uniform circular motion is …………………..
Answer:
90°

10. SI unit of angular momentum is …………………
Answer:
Radian/second

3. Match the following:

I.

Answer:
1. (c) \(\frac{u^{2} \sin 2 \theta}{2 g}\)
2. (d) \(\frac{2 u \sin \theta}{g}\)
3. (e) \(\frac{u^{2} \sin 2 \theta}{g}\)
4. (a) u cosθ
5. (b) u sinθ

II.

Answer:
1. (c) Changes
2. (e) Toward centre
3. (b) ν = rω
4. (a) Tangential
5. (d) Constant

 4. Write true or false:

1. If \(\vec{A} \cdot \vec{B}\) = AB, then angle between \(\vec{A}\) and \(\vec{B}\) is zero.
Answer:
True

2. If a unit vector is added or subtracted from a given vector, we always obtain given vector.
Answer:
True

3. A scalar quantity can be added with vector quantity.
Answer:
False

4. Resultant of three coplaner vector can be zero.
Answer:
True

5. Path of motion is a straight line in two-dimensional motion moving with uniform velocity.
Answer:
True

6. Acceleration is constant in projectile motion.
Answer:
True

7. Horizontal velocity is constant in projectile motion.
Answer:
True

8. If an object is dropped from running train, then its path will be parabolic.
Answer:
True

9. Kinetic energy is zero at maximum height of a projectile motion.
Answer:
False

10. In a given time interval, the ratio of change in velocity to the time-interval gives average acceleration.
Answer:
True

5. Answer in one word:

1. A projectile is thrown upwards. At the topmost point, the acceleration is in which direction.
Answer:
Vertically downwards

2. When an object is projected from earth surface, then which quantity remains constant?
Answer:
Horizontal component of velocity

3. Twp bullets are fired horizontally with different velocities from the same height. Which bullet would reach the earth’s surface earlier.
Answer:
Both bullets reach the earth’s surface in the same time

4. A particle is moving in a circular path. What is angle between its velocity and acceleration?
Answer:
90°

5. A particle has uniform circular motion. What is the change in its angular velocity after one half cycle.
Answer:
No change in angular velocity

Students get through the MP Board Class 11th Physics Important Questions Chapter 13 Kinetic Theory  which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 13 Kinetic Theory

Kinetic Theory Class 11 Important Questions Very Short Answer Type

Question 1.
State Boyle’s law.
Answer:
According to Boyle’s law “At constant temperature the volume of a given mass of a gas inversely proportional to its pressure, i.e.,
V∝\( \frac{1}{\mathrm{P}} \)
or
V=K.\( \frac{1}{\mathrm{P}} \)
or
PV = K (constant).

Question 2.
Write the condition at which Boyle’s law function.
Answer:
Boyle’s law function at low pressure and high temperature.

Question 3.
Write down Charle’s law.
Answer:
According to Charle’s law “ The volume of given mass of a gas at constant pressure increases by \( \frac{1}{273} \) of its volume at 0°C, for each 1°C rise in temperature.

Question 4.
Prove that with help of Charle’s law that at -273°C, volume of a gas is zero.
Answer:
According to Charle’s law.
V = V₀( 1 + \( \frac{1}{273} \) t)
At t =-273° C
or
V = V₀(1+ \( \frac{-273}{273} \) )
or
V = V₀ (1-1) = 0
i.e., at -273°C volume of a gas is zero.

Question 5.
What are ideal gases?
Answer:
The gases which obey Boyle’s law and Charle’s law completely are called ideal gases.

Question 6.
State Avogadro’s law.
Answer:
According to Avogadro’s law, “Equal volume of all the gases under similar con-ditions of temperature and pressure contain the same number of molecules.”

Question 7.
Write down Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gases which do not interact in any way is equal to the sum of their individual pressures.

Question 8.
What is Grahm’ law of diffusion?
Answer:
According to Grahm’s law of diffusion, “The rate of diffusion of a gas is directly proportional to the square root of its density.”

Question 9.
What is meant by absolute scale of temperature?
Answer:
The scientist Kelvin had developed a temperature scale by considering – 273°C as zero of the scale. This scale is called absolute scale of temperature. The value of its each division is 1°C. It is also called as Kelvin scale.

Question 10.
What do you mean by absolute zero?
Answer:
The absolute zero is that temperature at which the volume of a gas becomes zero.

Question 11.
If a gas is suddenly compressed, then its temperature increased. Why?
Answer:
The temperature of a gas is directly proportional to its mean kinetic energy. When a gas is suddenly compressed, then the mean kinetic energy of gas is increased. Therefore, the temperature of gas is increased suddenly.

Question 12.
What is Boltzmann constant? Write its value.
Answer:
The ratio of universal gas constant (R) and Avogadro’s number (N) is known as Boltzmann’s constant
K= \( \frac{\mathrm{R}}{\mathrm{N}} \)
Its value is 1.38 x 10^-23 joule/kelvin.

Question 13.
Write down relation between universal gas constant and specific gas constant.
Answer:
The relation between universal gas constant and specific gas constant is given below:
Specific gas constant r = \( \frac{\text { Universal gas constant }(\mathrm{R})}{\text { Molecular weight of } operatorname{gas}(\mathrm{M})} \) .

Question 14.
Write the ideal gas equation.
Answer:
PV = nRT.

Question 15.
Find out dimensional formula for R.
Answer:
From PV = RT
R = \( \frac{\mathrm{PV}}{\mathrm{T}} \)
or
[R] = \( \frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{3}\right]}{[\theta]} \)
= \( \left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right] \) .

Question 16.
At equal temperature (T) and pressure (P), two gases of same volume (V) are mixed together. If the temperature and volume of the mixure is T and V, then what will be its pressure?
Answer:
According to Dalton’s law of partial pressure it will be 2P.

Question 17.
At NTP, 1cm³ H₂ and 1cm³ O₂ are given. In which gas number of molecules will be more and why?
Answer:
According to Avogadro’s law number of molecules of 1 cm³ H₂ and 1 cm³ O₂ will be equal.

Question 18.
Oxygen and hydrogen gases are filled in a porous pot in equal amount, which one will diffuse soon?
Answer:
By the kinetic theory of gas,
ν rms ∝\( \frac{1}{\sqrt{\mathrm{M}}} \)
Since M_H < M₀ hence ν_H > ν₀
Thus, the hydrogen will diffuse soon.

Kinetic Theory Class 11 Important Questions Short Answer Type 

Question 1.
What is gas equation? Establish it.
Or
For a gas equation prove that PV = RT.
Or
Establish ideal gas equation.
Answer:
Let the initial pressure of one gram mole of a gas of molecular weight M is P₁ its volume is V₁ and absolute temperature is T₁. After a thermodynamic process the pressure of gas become P₂, volume and temperature becomes V₂ and T₂ respectively. The change occurs in the state of gas can be considered to be a combination of two processes.
(i) Keeping the temperature T₁ of gas constant, its pressure is varied from P₁ to P₂, so that as shown in fig. the volume of gas becomes V’. Hence, as per Boyle’s law :
P ₁V₁ = P₂V’ …(1)

(ii) Keeping the pressure P2 constant, the temperature of gas is varied from T₁ to T₂ so that its volume changes from V’ to V₂ as shown in fig. then according to Charle’s law :

\( \frac{\mathrm{V}^{\prime}}{\mathrm{T}_{\mathrm{l}}} \) = \( \frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}} \)
or
\( V’ = \frac{\mathrm{V}_{2} \mathrm{~T}_{1}}{\mathrm{~T}_{2}} \) ………. (2)
Substituting the value V’ from eqn. (2) in eqn.(1)
P₁V ₁ = \( \frac{\mathrm{P}_{2} \mathrm{~V}_{2} \mathrm{~T}_{1}}{\mathrm{~T}_{2}} \)
or
P₁V ₁ = \( \frac{P_{2} V_{2}}{T_{2}} \)
\( \frac{\mathrm{PV}}{\mathrm{T}} \) = Constant …….. (3)
The value of this constant is same (= 8.314 joule per mole-kelvin) for 1 mole of all the gases. It is denoted by R. Hence,
\( \frac{\mathrm{PV}}{\mathrm{T}} \) = R
or
PV = RT …(4)
Where, R is called universal gas constant. Eqn. (4) is called gas equation for one mole of gas.

Question 2.
Write down postulates of kinetic theory of gases.
Answer:
The postulates of kinetic theory of gas are given below :

• Every gas is composed of minute particles called molecules.
• The size of these molecules is negligible as compared to their intermolecular dis¬tance. t
• The molecules of gas are spherical uniform in all aspects rigid and perfectly elastic.
• The molecules are always in state of random motion and move in all possible direction with all possible velocities.
• Due to their continuous motion, these molecules collide with each other and also with the walls of container during their random motion. Due to these collision, there is no change in density of gas, i.e., the number of molecules per unit volume of the gas remains unchanged.
• After collision the direction of velocity of molecules changes. The collision is instantaneous, i.e., the time taken in collision is negligible as compared with the time taken between two consecutive collisions.
• The collisions are perfectly elastic, i.e., kinetic energy of molecules remains con-served.
• Between the two successive collisions a molecule travels in a straight line with uniform velocity. The distance covered by a molecule between two consecutive collisions is called ‘Free path’, The average distance travelled by a molecule between successive collisions is called mean-free-path.
• The molecules of gas collide with the walls of container and hence they exert the force on the walls. The force acting per unit area on the wall is called pressure of gas.
• The mass of molecules of gas is negligible and velocity is high. Therefore, there is no effect of gravity on the motion of molecules.

Question 3.
Prove that: P = \( \frac{1}{3} \) ρc^-2
Answer:
We know P = \( \frac{1}{3} \) \( \frac{m N c^{2}}{V} \)
Here mN = M
∴ P = \( \frac{1}{3} \) \( \frac{M \bar{c}^{2}}{V} \)
But \( \frac{M}{V} \) = ρ (density)
∴ P = \( \frac{1}{3} \) ρc^-2

Question 4.
On the basis of kinetic theory of gases, prove that p = \( \frac{2}{3} \) E where symbols have their meanings,
Answer:
According to kinetic theory of gases, the pressure exerted by the gas is gi ven by
P = \( \frac{m N \bar{c}^{2}}{3 V} \) ………….. (1)

Where, N is the number of molecules, c ^-2 mean square velocity and V is volume. Let m be the mass of one molecule, then the mass of gas be mN
∴Density of gas ρ = \(\frac{m N}{V}\) …………. (2)
From eqns. (1) and (2),
P = \( \frac{1}{3} \) ρc^-2
= \( \frac{2}{3} \) .\( \frac{1}{2} \) ρc^-2 = \( \frac{2}{3} \) E
Where, E is kinetic energy of per unit volume of gas.

Question 5.
Explain kinetic energy of a gas on the basis of kinetic theory of gases.
Or
Prove that : P = \( \frac{2}{3} \) E
Answer:
Kinetic energy of gas: Let the mass of one molecule of gas is m and number of molecules is N. Hence, the mass of gas will be mN.
∴Density of gas ρ = \(\frac{m N}{V}\) …………. (1)
According to Kinetic theory of gas the pressure exerted by gas is
P= \( \frac{m N \bar{c}^{2}}{3 V} \) ……………. (2)
From equns. (1) and (2)
P = \( \frac{1}{3} \) ρc^-2 = \( \frac{2}{3} \) . \( \frac{1}{3} \) ρc^-2
P= \( \frac{2}{3} \) E …….. (3)

Where, E is kinetic energy of a gas per unit volume. From eqn. (3),
or
E = \( \frac{3}{2} \) P
or
P = \( \frac{2}{3} \) E.

Question 6.
On the basis of kinetic theory of gases prove that the mean kinetic energy of molecules of gas is directly proportional to the absolute temperature of gas.
Or
Prove that the mean kinetic energy of gas is E = \( \frac{3}{2} \) KT.
Answer:
According to kinetic theory of gases, the pressure exerted by the gas is given by :
P = \( \frac{m N \bar{c}^{2}}{3 V} \)

Where, m is mass of one molecule of gas, N is the number of molecules in the volume V,c^-2 is mean square velocity of gas and V is volume. In the above equation if V is volume of one mole of gas, then N be the Avogadro’s number, mN be the gram molecular weight (M) hence from the above equation :
PV=\( \frac{1}{3} \)mNc^-2
= \( \frac{1}{3} \)Mc^-2 …(1)

But by the gas equation,
PV = RT ……………. (2)
From eqns. (1) and (2), we get
\( \frac{1}{3} \)Mc^-2 =RT
or
c^-2 = \( \frac{3 R T}{M} \) …… (3)
From eqn. (3), it is clear that c^-2 ∝T
Thus, the absolute temperature of a gas is directly proportional to the mean square velocity of the molecules of a gas.
Again from eqn. (3),

Where, K =\( \frac{R}{N} \) is called Boltzmann’s constant. Its value is 1.38 × 10^-23 joule per kelvin. This is formula of mean kinetic energy of molecules. From this equation, it is clear that
E∝T.

Question 7.
Explain absolute zero on basis of kinetic theory of gas.
Answer:
We know E =\( \frac{3}{2} \) KT
, If T=0, then E = 0.
Therefore, absolute zero temperature is that temperature at which average kinetic energy of the gaseous molecules is zero.

Question 8.
Prove that\( \overline{\boldsymbol{c}} \) =\( \sqrt{\frac{3 K T}{M}} \) where K is Boltzmann’s constant, T is absolute temperature and M is mass of one molecule of gas.
Answer:
According to kinetic theory of gases,
\( \overline{\boldsymbol{c}} \) = \( \sqrt{\frac{3 R T}{M}} \) =\( \sqrt{\frac{3 R T}{mN}} \), [∵M= mn]
= \( \sqrt{\frac{3 K T}{m}} \), [∵ K = \( \frac{R}{N} \) ]
or
\( \overline{\boldsymbol{c}} \) ∝ \( \sqrt{T} \)
i.e., r.m.s. value of velocity of gaseous molecule is directly proportional to square root of absolute temperature.

Question 9.
If the number of molecules in a box is halved, then what will be the effect on its pressure?
Answer:
According to kinetic theory of gas, the pressure of gas :
P= \(\frac{1}{3} \frac{m N \bar{c}^{2}}{V}\)
If the number of molecules is halved ,then the pressure of gas :
P’=\(\frac{1}{3} \frac{m}{V} \frac{N}{2} \bar{c}^{2}\)
P’ = \( \frac{1}{2}\left[\frac{m N \bar{c}^{2}}{3 V}\right]\)
or
\(\frac{P^{\prime}}{P}\) = \(\frac{1}{2}\)
or
P’ = \(\frac{1}{2}\)P
Thus, the pressure will Also be halved.

Question 10.
Derive Boyle’s law on basis of kinetic theory of gases.
Answer:
Derivation of Boyle’s law: According to kinetic theory of gases the pressure exerted by gas is :
P = \(\frac{1}{3}\) \( \frac{m N \bar{c}^{2}}{V} \)
or
PV = \( \frac{1}{3} \) \( m N \bar{c}^{2} \)
= \( \frac{1}{3} \) . \( m N \bar{c}^{2} \)
= \( \frac{2}{3} \).N .\(\frac{1}{2} \) \( m \bar{c}^{2}\)
= \( \frac{2}{3} \) NE, [ ∵ E = \( \frac{1}{2} \) \( m \bar{c}^{2}\)]
= \( \frac{2}{3} \) NKT, (∵E = KT ) …….. (1)
If the mass and absolute temperature of gas are constant,then
= \( \frac{2}{3} \) NKT be also constant
∵ PV = constant
This is Boyle’s law.

Question 11.
Derive Ctiarie’s law on basis of kinetic theory of gases.
Answer:
Derivation of Charle’s law : According to kinetic theory of gases, the pressurue exertd by gas is
P = \( \frac{1}{3} \) \( \frac{m N \bar{c}^{2}}{V} \) …….. (1)
or
PV = \( \frac{1}{3}\) \( m N \bar{c}^{2} \)
or
PV = \( \frac{1}{3} \) \( M \bar{c}^{2} \) ……… (2)

Where, M = mN is mass of gas which remains constant .
If the pressure of gas remains constant,then from the above equation.
V= \( \frac{M}{3 P} \bar{c}^{2} \)
or
V ∝\( \bar{c}^{2} \)
or
Since, \( \bar{c}^{2} \) ∝T therefore
V ∝T
This is Charle’s law.

Question 12.
Derive Dalton’s law of partial pressure on basis of kinetic theory of gases.
Answer:
Dalton’s law of partial pressure: According to Dalton’s law of partial pressure, the total pressure exerted by a mixture of gases which do not interact in any way is equal to the sum of their individual pressure.
Let us consider about a vessel of volume V₁ having number of gases mixed together.
Let the gases one, two, three ………….contain N₁ molecules of mass muN₂ molecules of mass
m₂, N₃ molecules of mass m₃ ……………respectively. Let their root mean square velocities are
\( \bar{c}_{1}, \bar{c}_{2}, \bar{c}_{3}\)…………… respectively.

Then the pressure due to first gas
P₁ = \( \frac{1}{3} \) m₁N₁c^-2
Similarly, the pressure due to second, third,………….gases are

If all the gases are mixed at same temperature, then mean kinetic energy óf the molcules of each gas will be the same i.e.,


This is Delton’s law of partial pressure.

Question 13.
A vessel is filled up with mixture of two different gases. Explain with reason :
(i) Is Average kinetic energy per unit molecule are same?
(ii) Is root-mean-square value of velocity are same?
(iii) Is pressure same?
Answer:
(i) Yes, as E = \(\frac{3}{2} \)KT, it depends on absolute temperature only.

(ii) No, because \( \bar{c} \) = \( \sqrt{\frac{3 R T}{M}} \) and it depends on molecular weight M and temperature T.

(iii) Nothing cannot be said about pressure as mass is not known.

Question 14.
Write the law of equipartition of energy.
Answer:
According to this law, for any dynamical system in thermal equilibrium, the total energy distributed equally amongst all the degree of freedom, and the energy associated
with each molecule per degree of freedom is\( \frac{1}{2} \) KT, where K is Boltzmann’s constant and T is temperature of the system.

Question 15.
Explain degree of freedom.
Answer:
The number of degrees of freedom of a dynamical system is defined as the total number of coordinates or independent quantities required to describe completely the position and configuration of the system.
For Example:
(i) When a particle moves along a straight line, say along X-axis, its position can be specified by its displacement along the X-axis. Therefore, such a particle has one-translational degree of freedom.

(ii) If the particle is moving in a plane, its position at any instant can be determined by knowing the displacements of the particle along the X-axis and Y-axis. Therefore it has two- translational degrees of freedom.

(iii) If the particle is moving in space, its position at any instant can be determined by knowing the displacement of the particle along X-axis, Y-axis and Z-axis. Therefore, such a particle has three-translational degrees of freedom.
For example, a bob of an oscillating simple pendulum has one degree of freedom, an insect moving on a horizontal floor has two degrees of freedom and a buzzing bee has three degrees of freedom.

Question 16.
Find out ratio of specific heats for monoatomic gas.
Answer:
For monoatomic gas like He or Ar, whose degree of freedom is 3 but according to law of equipartition of energy.
Energy =3 × \( \frac{1}{2} \) KT = \( \frac{3}{2} \)kT
But energy associated with one mole of gas
U =\( \frac{3}{2} \) nkT
Where n is number of gas molecule for 1 mole of gas.
But Boltzmann’s constant k = \( \frac{R}{n} \)
or
nk = R
Putting value in eqn. (1), we get
U = \( \frac{3}{2} \)RT
But C_v = \( \frac{d U}{d T} \) = \( \frac{d}{d T} \) \( \left(\frac{3}{2} R T\right) \)
or
C_v =\( \frac{3}{2} \) R

Question 17.
Find out ratio of specific heat for diatomic gas.
Answer:
For diatomic gas like Hydrogen, Oxygen etc. degree of freedom is 5.
Therefore energy associated with one mole of gas is
U = \(\frac{5}{2} \)nkT
or
U= \(\frac{5}{2} \)RT
But, nk =R

since C_v = \( \frac{d U}{d T} \) = \( \frac{d}{d T} \) \( \left(\frac{5}{2} R T\right) \) =\(\frac{5}{2} \) R
∴ From C_p – C_v = R
C_p = R +C_v
or
C_p = R +\(\frac{5}{2} \) R = \(\frac{7}{2} \) R
∴ γ = \( \frac{C_{P}}{C_{V}} \) = \( \frac{\frac{7}{2} R}{\frac{5}{2} R} \) = \(/frac{7}{5} \) = 1.40

Question 18.
Find out ratio of specific heat for triatomic gas.
Answer:
For triatomic gas like C02, H2S degree of freedom is 6. Energy associated with one mole of gas is
U =\(\frac{6}{2} \)nKT
or
U = \(\frac{6}{2} \)RT = 3RT

But Cv=\( \frac{d U}{d T} \) = \( \frac{d}{d T} \) (3RT) = 3R
∴ From C_p-C_v = R
C_p =R + C_v =R + 3 R = 4R
∴ γ = \( \frac{C_{P}}{C_{V}} \) = \( \frac{4_{R}}{3_{R}} \) = \(\frac{4}{3} \) = 1.33

Kinetic Theory Class 11 Important Questions Long Answer Type

Question 1.
Establish formula for pressure of a gas on the basis of kinetic theory of gases.
Answer:
When a gas is filled in a closed vessel, then molecules of gas are in state of continuous random motion. They collide with one another and also with the walls of con-tainer. Due to these collisions, the molecules of gas exert the force on the walls of the vessel. The force acting per unit area of the walls is called pressure of gas.
Let us consider a hollow cubical vessel of sides of length ‘l’ as depicted in fig. An ideal gas is filled in it. Let the mass of one molecule is m. Consider a molecule of gas moving with velocity C. The components of velocity C along X, Y and Z direction are, u, v and w respectively.
∴ C² = u²+ v² + w² ……….. (1)
Let us consider about two faces A and B, perpendicular to the direction of X-axis.

If the molecule P collides with the face A with velocity u. It will returned with a velocity -u after collision.
∴ Linear momentum of molecule before collision = mu Linear momentum of molecule after collision = – mu.
∴ Change in linear momentum due to collision = mu – (-mu) = 2mu
The molecule rebounded from A collides with the opposite face B, rebounds and again strikes with face A. Thus, the total distance travelled by the molecule will be 2l.

∴ Time taken by the molecule in travelling the distance 2l be
= \( \frac{2_{l}}{{u}} \), [ ∵ T ime = \( \frac{\text { Distance }}{\text { Velocity }} \) ]
Thus, after covering the distance 21 i.e., after every interval \( \frac{2l}{u} \) , the molecule P will
again strike with the face A.
Number of collision with face A per second be = \( \frac{u}{2 l} \)
The momentum transferred by molecule to the face A per second i.e., rate of change ofmomentum
= 2mu ×\( \frac{u}{2 l} \) = \( \frac{m u^{2}}{l} \)

But according to Newton’s second law of motion, the rate of change of momentum is equal to the force exerted on that face.
∴The force exerted by molecule at face Abe = \( \frac{m u^{2}}{l} \)
∴ Pressure exerted by the molecule on the face A be
= \( \frac{\mathrm{mu}^{2}}{\mathrm{l}} \) = \( \frac{n u^{2}}{l^{3}} \)

Let N be the number of molecules in gas and their velocity components along X direction are u₁, u₂, u₃, un respectively, then pressure exerted by these molecules on the face A

Where, V = l³ = Volume of vessel i.e., volume of gas.
Similarly, the pressure exerted by N molecule along Y-axis and Z-axis be
P_y = \( \frac{m}{V} \) \( \left[v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+\ldots \ldots+v_{n}^{2}\right] \) ……… (3)
P_z = \( \frac{m}{V} \) \( \left[w_{1}^{2}+w_{2}^{2}+w_{3}^{2}+\ldots \ldots+w_{n}^{2}\right] \) ………… (4)
But a gas exerts the same pressure in all directions i.e., P_x=P_y=P_z=P (say)
∴ P = \( \frac{P_{x}+P_{y}+P_{z}}{3} \) …………….. (5)
Hence, from eqns. (2), (3), (4) and (5),
P = \( \frac{P_{x}+P_{y}+P_{z}}{3} \)

Let \( \bar{c} \) = \( \sqrt{\frac{c_{1}^{2}+c_{2}^{2}+\ldots . .+c_{n}^{2}}{N}} \) where \( \bar{c} \) is root mean square velocity , then
N\( \bar{c}^{2}\) = \( c_{1}^{2}+c_{2}^{2}+\ldots \ldots+c_{n}^{2} \) ………….. (7)

From eqns. (6) and (7),
P = \( \frac{m}{3 V} \)N\( \bar{c}^{2}\)
or
P = \( \frac{m N \bar{c}^{2}}{3 V} \)
This is expression of pressure exerted by gas.

Kinetic Theory  Class 11 Important Numerical Questions

Question 1.
Up to what temperature should an ideal gas initially at 27°C be heated so that its volume becomes doubled at constant pressure?
Solution:
Given:T₁ = 27°C = 273+27 = 300K
V₁ = V,V₂ = 2V
∴ According to Charle’s law,
\( \frac{V_{1}}{T_{1}} \) = \( \frac{V_{2}}{T_{2}} \)
or
T₂ = \( \frac{V_{2}}{V_{1}} \) T₁= \(\frac{2 V}{V} \) × 300
= 600 K
= 600 – 273 = 327°C

Question 2.
A gas is filled in a vessel at 127°C at 4 atm. pressure. If the temperature of gas increased up to 527°C, then what would be the pressure of gas?
Answer:
T₁ = 127°C = 127+273 = 400K
T₂ = 527°C = 527 + 273 = 800 K
P₁ = 4 atm. pressure
∴ By the pressure law,

\( \frac{P_{1}}{P_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
or
P₂ = P₁ \( \frac{T_{2}}{T_{1}} \) = 4 × \( \frac{800}{400} \) = 8
= 8 atm.pressure

Question 3.
Number of molecule per cubic centimetre ¡n a space is five and temperature is 3 K. What will be the pressure, there? (R 1.38 × 10^-23 joule/m/K)
solution:
PV = nRT
P = \( \frac{n R T}{V} \)
Given: V = 1 cm³ = 10 ^-6, R = 1.38 ×10^-23 J/m /K, n =5 , T = 3K
∴ P = \( \frac{5 \times 1 \cdot 38 \times 10^{-23} \times 3}{1} \) = 2.07 ×10^-22N/m²

Question 4.
Temperature of any gas is -68°C. By what temperature it must be heated so that
(i) kinetic energy between the molecules become double
(ii) value of velocity of molecule become double.
Solution :
(i) From E = \(\frac{3}{2} \)KT,
E ∝ T
or
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
Given: E ₂ = 2E₁,=T₁ = 273-68 = 205 K
∴ \( \frac{E_{1}}{2 E_{1}} \) = \( \frac{205}{T_{2}} \)
or
T ₂ = 205 ×2 = 410K = 410 – 273° =137°C

(ii) From C^-2 = \( \frac{3 R T}{\gamma} \) we get
\( \bar{c}^{2}\) ∝ \( \sqrt{T} \)
or
\(\frac{\bar{c}_{1}}{\bar{c}_{2}}\) = \( \sqrt{\frac{T_{1}}{T_{2}}} \)
Given: \(\bar{c}_{2}\) = 2\( \bar{c}_{1} \) ,T ₁ = 205K
∴ \( \frac{\bar{c}_{1}}{\bar{2c}_{1}} \) = \( \sqrt{\frac{205}{T_{2}}} \)
or
\( \frac{1}{4} \) =\( \frac{205}{T_{2}} \)
or
T ₂ = 205 × 4 = 820
or
T ₂ = 820 – 273 = 574°C

Question 5.
At 30°C temperature, mixture of Helium and Hydrogen gas is filled in a vessel. Find out the ratio of r.m.s. value of velocity of the molecule at this temperature.
Solution:
From C^-2 = \( \frac{3 R T}{M} \)
\( \bar{c} \) = \( \frac{1}{\sqrt{M}} \)
or
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{M_{2}}{M_{1}}} \)

Given, molecular weight of Helium M₁ = 4
and Molecular weight of Hydrogen M₂ =2
∴ \( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{2}{4}} \) = \( \frac{1}{\sqrt{2}} \) = 1:\( \sqrt{2} \)
or
\( \bar{c}_{1} \): \( \bar{c}_{2} \) = 1:\( \sqrt{2} \)

Question 6.
If the absolute temperature of gas is done four times then by how much times their r.m.s. velocity of molecule increases? Also by how much times their kinetic energy and pressure also increases?
Solution: From \( \bar{c} \) = \( \frac{3 R T}{M} \) , where R and M are constant
∴ \(\bar{c} \) ∝ \( \sqrt{T}\)
or
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{T_{1}}{T_{2}}} \)
or
\( \frac{\bar{c}_{1}}{\bar{c}_{2}} \) = \( \sqrt{\frac{T_{1}}{4 T_{1}}} \) = \( \sqrt{\frac{1}{4}} \) = \( \frac{1}{2} \)
or
\( \bar{c}_{2} \) = 2 \( \bar{c}_{1} \)
i.e., r.m.s velocity will increase by two Times

From formula E = \( \frac{3}{2} \) KT
E ∝ T or \( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \)
Putting T₂ = 4T₁
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{4 T_{1}} \) = \( \frac{1}{4} \)
or
E₂ = 4E ₁
\( \frac{P_{1}}{P_{2}} \) = \( \frac{T_{\mathrm{l}}}{4 T_{1}} \) = \( \frac{1}{4} \)
or
P₂ = 4P₁
∴ Pressure will become four Times.

Question 7.
If the temperature of a gas increased from 77°C to 227°C, then what will be the ratio of kinetic energy of then molecules?
Solution:
Given:T₁ = (273+ 77)K = 350K and T₂ = (273 + 227)K = 500K.
From formula E ∝ T
\( \frac{E_{1}}{E_{2}} \) = \( \frac{T_{1}}{T_{2}} \) = \( \frac{350}{500}\) = \( \frac{7}{10} \)
E₁: E ₂ = 7 : 10

Question 8.
Volume of vessel is two times the volume of vessel B and same gas is filled in both vessels. If the temperature and pressure of the vessel A is double w.r.t. vessel B, then what will be the ratio of molecules of the gas of vessel A and B?
Solution:
From PV = nRT

Question 9.
The velocities of four molecules of a gas are 2, 4, 6 and 8 km/sec.
Calculate : (i) Average velocity, (ii) root-mean-square velocity.
Solution:
Given : c₁ = 2 km/sec, c₂ = 4 km/ sec, c₃ = 6 km/sec and c4 = 8 km/sec.
(i) Average velocity:
c = \( \frac{c_{1}+c_{2}+c_{3}+c_{4}}{4} \)
= \( \frac{2+4+6+8}{4} \) = 5 km/sec

(ii) Root-mean-square velocity:

Question 10.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at N.T.P. Take the diameter of oxygen molecule to be roughly 3Å. (NCERT)
Solution:
Given, diameter of one molecule of oxygen = 3 Å
∴ Radius r = \( \frac{3}{2} \) = 1.5 Å = 1.5 × 10 ^-10m
∴ Volume of one molecule of oxygen = \( \frac{4}{3} \) πr³
= \( \frac{4}{3} \) ×3.14 (1.5 × 10 ^-10)³
= 14.13 × 10 ^-30 m³
∴ Volume of 1 mole of oxygen = 6.02 × 10 ²³ × 14.13 ×^-30
= 85.06 × 10 ^-7 m³

The volume of 1 mole of oxygen at STP = 22.4 litre
= 22.4 × 10^-3m³
∴ Fraction of the molecular volume of the actual volume = \( \frac{85 \cdot 06 \times 10^{-7}}{22 \cdot 4 \times 10^{-3}} \)
= 3.797 × 10 ^-4
= 3.8 ×10^-4 ≈ 4 × 10^-4

Question 11.
Molar volume is the volume occupied by 1 mole of any ideal gas at standard temperature and pressure (STP: 1 atm pressure, 0°C). Show that it is 22.4 litre. (NCERT)
Solution:
At S.T.P., T = 0°C = 0 + 273 =273K
Pressure P = 1 atm = 1-013 x 10⁵Nm^-2 ; R = 8.3 J mol^-1‘K^-1 By gas equation
∴ By gas equation for 1 mole PV =RT
V = \( \frac{R T}{P} \) = \( \frac{8 \cdot 3 \times 273}{1 \cdot 013 \times 10^{5}} \)
= 2236.82 ×10 ^-5
= 22.3682 ×10 ^-3 m ³
= 22.4 ×10 ^-3m ³ = 22.4 litre.

Question 12.
An air bubble of volume l-0cm3 rises from the bottom of a lake 40m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface which is at a temperature of 35°C? (NCERT)
Solution:
Given, Initial volume of bubble
V₁ = 10cm³ = 1.0 × 10^-6m³
Initial temperature T₁ = 273 +12 = 285 K
Initial pressure on bubble P₁ = Atmospheric pressure + Pressure of 40 m high water column
= 1.013×10⁵+hdg
= 1.013×10⁵ + 40×10³ × 9.8
= 4.933 × 10⁵ Pa

Final pressure on bubble P₂ = 1 atm = 1.013 ×10⁵ Pa
Fina temperature T₂ = 273+ 35=308 K
From the equation \( \frac{P_{V_{1}}}{T_{1}} \) = \( \frac{P_{2} V_{2}}{T_{2}} \)
or
V₂ = \( \frac{P_{V_{1}}}{T_{1}} \) ×\( \frac{T_{2}}{P_{2}} \) = \( \frac{4 \cdot 933 \times 10^{5} \times 1 \cdot 0 \times 10^{-6} \times 308}{285 \times 1 \cdot 013 \times 10^{5}} \)
= 5.26 ×10^-6 m³ ≈ 5.3 × 10^-6 m³

Question 13.
At what temperature is the root-mean-square speed of an atom is an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at – 20°C? (Atomic mass of Ar = 39.9 u, of He = 4.0 u). (NCERT)
Solution:
Given, atomic mass of He M₁=4.0u
Atomic mass of Ar M₂ =39.3u
Temperature of He gas T ₁ = -20+273 = 253K

Kinetic Theory  Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
There are N molecules in a vessel, if the number of molecules are doubled then the pressure of gas will be :
(a) Double
(b) Remain same
(c) Become four times
(d) Become one fourth.
Answer:
(a) Double

Question 2.
Motion of gaseous molecule at absolute zero temperature :
(a) Become less
(b) Increases
(c) Become zero
(d) None of these.
Answer:
(c) Become zero

Question 3.
At -273°C, molecules of gas moves with :
(a) Maximum velocity
(b) Minimum velocity
(c) Zero velocity
(d) None of these.
Answer:
(c) Zero velocity

Question 4.
Reason for deviating from gaseous law of ideal gas at less temperature is :
(a) Maximum collision become inelastic
(b) Volume of molecules cannot be negligible
(c) Force acting between molecules become less
(d) Molecular velocity become less.
Answer:
(b) Volume of molecules cannot be negligible

Question 5.
Root-mean-square velocity of ideal gas molecules at constant temperature :
(a) Remain same
(b) Inversely proportional to square root of molecular weight
(c) Proportional to square root of molecular weight
(d) Inversely proportional to molecular weight.
Answer:
(b) Inversely proportional to square root of molecular weight

Question 6.
Graph between PV and P of a gas which obey Boyle’s law will be :
(a) Hyperbola
(b) Parallel line with PV axis
(c) Parallel line with P axis
(d) None of these.
Answer:
(c) Parallel line with P axis

Question 7.
False statement regarding kinetic theory of gas is :
(a) Collision between two molecules are perfectly elastic
(b) Kinetic energy between molecules is proportional to absolute temperature
(c) Absolute temperature of gas is inversely proportional to root-mean-square velocity.
(d) At absolute temperature average kinetic energy of molecules is zero.
Answer:
(c) Absolute temperature of gas is inversely proportional to root-mean-square velocity.

Question 8.
Reason for pressure exerted by gaseous molecules on the wall of vessel is :
(a) Losses its own kinetic energy
(b) Get stick with the walls of vessel
(c) Due to collision with wall of vessel its momentum get change
(d) Get accelerated toward wall.
Answer:
(c) Due to collision with wall of vessel its momentum get change

Question 9.
There is no atmosphere in moon, because :
(a) It is closer to earth
(b) It revolve around the earth
(c) It obtain light from sun
(d) Escape velocity is less than root-mean-square velocity.
Answer:
(d) Escape velocity is less than root-mean-square velocity.

Question 10.
The temperature of an ideal gas is raised from 27°C to 927°C the root-mean-square velocity of its molecules will become :
(a) Two times
(b) Half
(c) Four times
(d) One fourth.
Answer:
(a) Two times

Question 11.
Every gas behaves as ideal gas at:
(a) Low pressure and high temperature
(b) High pressure and low temperature
(c) At equal pressure and temperature
(d) High pressure and high temperature.
Answer:
(a) Low pressure and high temperature

Question 12.
Unit of universal gas constant is :
(a) joule/mole-kelvin
(b) mole/joule-kelvin
(c) joule-mole-kelvin
(d) kelvin/joule/mole.
Answer:
(a) joule/mole-kelvin

Question 13.
If the temperature of gas remain constant and pressure become half then the volume will get.
(a) Half
(b) Double
(c) Unchanged
(d) Four times.
Answer:
(b) Double

Question 14.
In gas equation PV = RT, V is volume of:
(a) Gas
(b)1 gram gas
(c) 1 litre gas
(d) 1 mole gas.
Answer:
(d) 1 mole gas.

Question 15.
Pressure of gas filled in closed vessel is due to :
(a) Large number of molecules
(b) Attraction between wall and molecules
(c) Collision of molecules with wall
(d) None of these.
Answer:
(c) Collision of molecules with wall

Question 16.
The average kinetic energy associated with each degree of freedom is :
(a) \( \frac{3}{2} \) KT
(b) KT
(c) \( \frac{1}{2} \) KT
(d) \( \frac{3}{2} \) RT.
Answer:
(c) \( \frac{1}{2} \) KT

Question 17.
The mean kinetic energy of the molecule of gas depends upon :
(a) Nature of gas
(b) Absolute temperature
(c) Volume of gas
(d) None of these.
Answer:
(b) Absolute temperature

Question 18.
Root-mean-square velocity of gas is :
(a) Directly proportional to its specific molecular weight
(b) Directly proportional to its square of molecular weight
(c) Directly proportional to its molar weight
(d) Directly proportional to its absolute temperature.
Answer:
(c) Directly proportional to its molar weight

Question 19.
Magnitude of R for 1 gram mole of gas is :
(a) 8.31 erg
(b) 8.31 MKS unit
(c) 4.2 joule
(d) 4.2 calorie.
Answer:
(b) 8.31 MKS unit

Question 20.
If the r.nus. velocity of a gas is doubled, then its pressure will:
(a) Increase
(b) Decrease
(c) Remain same
(d) None of these.
Answer:
(a) Increase

2. Fill in the blanks:

1. Momentum applied per unit area on the wall of a vessel by the molecule of gas is equal to ………………………. .
Answer:
Pressure

2. At absolute temperature ………………………. of gas becomes zero.
Answer:
Kinetic energy

3. For a diatomic gas, the degree of freedom is ………………………. .
Answer:
Five

4. Kinetic energy associated with each degree of freedom is ………………………. .
Answer:
\( \frac{1}{2} \)KT

5. The value of 0°C on the kelvin scale is ………………………. .
Answer:
273 K.

3. Match the following:

 Answer:
1. (e) \( \frac{1}{3} \) \( \frac{m N}{V} \)c^-2
2. (d) ∝ c^-2
3. (a) ∝ T
4. (b) 3
5. (c) 5.

Students get through the MP Board Class 11th Physics Important Questions Chapter 14 Oscillations  which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 14 Oscillations

Oscillations Class 11 Important Questions Very Short Answer Type

Question 1.
Define simple harmonic motion with example.
Answer:
When a body moves to and fro about a point in a straight line such that its acceleration is always directly proportional to the displacement and directed towards the mean position, then its motion is called simple harmonic motion.
Example: Motion of a body suspended by a spring, motion of simple pendulum.

Question 2.
What is oscillatory motion?
Answer:
When a body moves periodically to and fro or back and forth about a definite point, then its motion is called oscillatory motion.
Motion of simple pendulum, vibration of tuning fork etc. are the examples of oscillatory motion.

Question 3.
For what type of motion of a body acceleration is directly proportional to displacement?
Answer:
It is simple harmonic motion.

Question 4.
When the velocity and acceleration will be maximum in S.H.M.?
Answer:
The velocity is maximum at the mean position and acceleration is maximum at the extreme position.

Question 5.
Write the formula for kinetic energy and potential energy for a body executing S.H. M.?
Answer:
Kinetic Energy = \(\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)\)
Potential Energy = \(\frac{1}{2} m \omega^{2} y^{2}\)
Where m – particle of mass,
ω – angular frequency,
a – amplitude and
y = displace-ment of the particle form its mean position.

Question 6.
At which position kinetic energy and potential energy is zero for a body executing S.H.M.?
Answer:
At maximum displacement position (y = ±a) kinetic energy is zero and at mean position (y = 0) potential energy is zero.

Question 7.
Define simple pendulum.
Answer:
When a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string by a rigid support, then it is called simple pendulum.

Question 8.
Which of the following relationships between acceleration ‘a’ and the dis-placement x: of a particle involve simple harmonic motion ? (NCERT)
(a) a = 0.7x,
(b) a = 200x²,
(c) a = -10x²
(d) a = 100x³.
Answer:
In a S.H.M. the relation between acceleration and displacement is
a=-ω²x
(a) a = 0.7 x
Comparing with eqn. (1),
-ω² = 0.7
or’
ω = \(\sqrt{-0 \cdot 7}\)
∵ Value of a is not real therefore the motion is not S.H.M.

(b) a = -200x²
This motion is not a S.H.M.

(c) a = -10x
Comparing-with eqn. (1),
ω² =10
or
ω = \(\sqrt{10}\)
∵ ω has a real value therefore, the relation shows a S.H.M.

(d) a = 100 x³
This motion is not a S.H.M.

Question 9.
The length of a spring is increased, how the time-period be affected?
Answer:
The time-period will increase.

Question 10.
On which conservation law S.H.M. is based?
Answer:
S.H.M. is based on law of conservation of energy.

Question 11.
What is second pendulum? What is its length?
Answer:
The pendulum whose periodic time is two seconds is known as second pendulum. The length of second pendulum is about 100 cm.

Question 12.
Write the characteristics of S.H.M.
Answer:
Characteristics of S.H.M.:

• The motion is periodic.
• The particle moves to and fro about a point in a straight line.
• The acceleration is directly proportional to the displacement.
• The direction of acceleration is always towards the mean position.

Question 13.
Define time-period, frequency and amplitude.
Answer:
Time period: The time taken by an oscillating body to complete one oscillation is called time-period.
Frequency: The number of vibrations or oscillations in one second is called frequency.
Amplitude: The maximum displacement of a particle about its mean position is called amplitude. It is denoted by a or A.

Question 14.
What is the relation between S.H.M. and Circular motion?
Answer:
When any particle is performing uniform motion on the circumference of a circle, then the perpendicular dropped on the diameter of the circle from any point of rotation execute S.H.M. along the diameter of circle.

Question 15.
Define spring constant and write its S.I. unit and dimensional formula.
Answer:
Spring constant: The force required to elongate or compress a spring through unit length is called spring constant.
Unit: SI unit of spring constant is Nm1.
Dimensional formula : Dimensional formula is [ML°T^-2].

Question 16.
Write the formula for the time-period of a body suspended by a spring.
Answer:
Where, T= Time-period, m = Mass suspended and k = Force constant.

Question 17.
What is meant by the effective length of a simple pendulum?
Answer:
The length of pendulum from the point of suspension to the centre of bob is called effective length.

Question 18.
Why the length of simple pendulum is measured up to the centre of the bob?
Answer:
The centre of gravity point of spherical bob lies at its centre. The concept of heavy point mass is assumed that the weight of the body acts at the C.G. point. Therefore the effective length is measured up to centre of the bob.

Question 19.
How time period of a simple pendulum change with effective length of a simple pendulum.
Answer:
We know \(T=2 \pi \sqrt{\frac{l}{g}}\)
or
T ∝ \(\sqrt{l}\)
When effective length is increased, time period also get increased.

Question 20.
Can the pendulum clock be used in a satellite?
Answer:
No, because g = 0 inside satellite, therefore T = ∞.
So, in place of pendulum clock, spring clock can be used inside it.

Question 21.
The pendulum clock cannot be used in artificial satellite. Why?
Answer:
In an artificial satellite, the bodies are in the state of weightlessness. Hence, g becomes zero. Thus, the time-period becomes infinite.Therefore, it cannot be used in the satellite.

Question 22.
A boy is swinging, sitting on a swing. If another boy sits beside him, what will happen to its time-period?
Answer:
The time period is independent of mass, hence there will be no effect.

Question 23.
A boy is swinging, standing on a swing. If he sits on it, what happen to the time-period?
Answer:
When the boy will sit on the swing, his C.G. point will shift downwards. Hence, the effective length will increase therefore the time-period will also increase.

Question 24.
What is the reason that the frequency of oscillation clocks depend on the rise and fall of mercury level in the thermometer?
Answer:
If the mercury level increases, it shows that temperature has increased and the decreases in mercury level shows a decrease in temperature. Due to increase in temperature, the length of pendulum increases and \(T \propto \sqrt{l}\) so time period also increases. Hence clock runs slow, i.e., the frequency of oscillation decreases. But with decrease in temperature, length decreases and so time-period decreases. Hence clock runs fast i.e., frequency of oscillation increases.

Question 25.
The pendulum clock becomes fast when it is taken to pole from equator. Why?
Answer:
We have, T = \(2 \pi \sqrt{\frac{l}{g}}\)
If l is constant ,then T ∝ \(\frac{1}{\sqrt{g}}\)
Hence, time-period is inversely proportional to the square root of acceleration due to gravity. At poles the value of g is greatest, hence the time-period will decrease at poles, hence the clock will be fast.

Question 26.
Why pendulum clock does not show time at satellite?
Answer:
At satellite g = 0, therefore, i.e., time period of pendulum become infinity, therefore it does not show time of satellite.

Question 27.
What is damped oscillation?
Answer:
When the amplitude of an oscillating body decrease gradually by some external opposing force, then its oscillation is called damped oscillation.

Question 28.
Define forced vibration.
Answer:
When an oscillating body oscillates under the influence of some external force, then it oscillates with the frequency of external force, then its oscillation is called forced vibration.

Question 29.
What do you mean by forced oscillation?
Answer:
When the body oscillate under the influence of an external periodic force, with a constant amplitude and a frequency equal to the periodic force, the oscillation of the body is called forced oscillation.

Question 30.
Define resonance.
Answer:
When a body vibrates under the influence of a periodic force and its frequency becomes equal to that of periodic force then the amplitude of the body increases. This phenomenon is known as resonance.

Question 31.
Why marching troops are asked to break their steps while crossing the bridge?
Answer:
The footsteps of marching troops produces a periodic force on the bridge. If the frequency of foot-steps becomes equal to the natural frequency of bridge, the amplitude of bridge will increase and hence the bridge may break.

Question 32.
What is the effect of forced oscillation on amplitude and frequency?
Answer:
Due to forced oscillation amplitude of oscillating body go on reducing and frequency become less than the natural frequency.

Question 33.
When forced oscillation become resonance oscillation?
Answer:
When the natural frequency of the body becomes equal to the frequency due to periodic force then forced oscillation is converted into resonance oscillation.

Question 34.
A simple pendulum is suspended in a lift, whose time-period is T. What will aeration ‘a’ (i) upward, (ii) downward.
Answer:
(i) When the lift accelerate with acceleration ‘a’ upward then the time-period will decrease
( T = \(2 \pi \sqrt{\frac{l}{g+a}}\))
(ii) On acceleration with acceleration ‘a’ downward then its timeperiod will increase
( T = \(2 \pi \sqrt{\frac{l}{g-a}}\) )

Question 35.
What will be the time-period of effective length infinity?
Answer:
84.6 minute.

Question 36.
What will be the time period of pendulum at the centre of earth? Why?
Answer:
At the centre of earth the time-period of pendulum will be infinity as ‘g’ is zero at the centre.

Question 37.
A spring of spring constant k is cut into three equal pieces. What will be the time period of each?
Answer:
The spring constant of each piece will be 3k.

Oscillations Class 11 Important Questions Short Answer Type 

Question 1.
The bob of simple pendulum is a ball filled with water, if a small hole is made at the bottom of the ball, how will its time period change as water is drains out of it?
Answer:
When the ball is completely filled with water, its centre of gravity be at its geometric centre. So, it will oscillate with definite time period. But as the water drains out of it through the hole, the centre of gravity is displaced gradually downward. Hence the effective length also increases gradually. Therefore the time period of the pendulum also increases gradually. But when whole of the water is removed then its centre of gravity again shifted to its geometric centre and its time-period become same as was in initial position.

Question 2.
What is simple harmonic motion. Write its four characteristics.
Answer:
When a body moves to and fro about a point in a straight line such that its acceleration is always directly proportional to the displacement and directed towards the mean position, then its motion is called simple harmonic motion.
Example: Motion of a body suspended by a spring, motion of simple pendulum.
Characteristics of S.H.M.:

• The motion is periodic.
• The particle moves to and fro about a point in a straight line.
• The acceleration is directly proportional to the displacement.
• The direction of acceleration is always towards the mean position.

Question 3.
Write the relation between acceleration and displacement of a particle ex-ecuting1 S.H.M. Also deduce expression for time-period from it?
Answer:
Acceleration of the particle with displacement y is α = -ω²y
Where ω is angular velocity of particle here (-) sign indicate the direction of accel-eration only.
Hence, ω² = \(\frac{\alpha}{y}\) = \(\frac{\text { Acceleration }}{\text { Displacement }}\)
or
ω = \(\sqrt{\frac{\alpha}{y}}\)
But T =\(\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{\frac{\alpha}{y}}}=2 \pi \sqrt{\frac{y}{\alpha}}\)
T = \(2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}\)

Question 4.
Write the expression for displacement, velocity and acceleration of a particle executing S.H.M. and say :
(i) When the velocity of the particle will be maximum and zero.
(ii) When the acceleration of the particle will be maximum and zero.
Answer:
Displacement y = a sin ωt
Velocity ν = ω \(\sqrt{a^{2}-y^{2}}\)
Acceleration a = – ω²y
Where a is amplitude and ω is angular acceleration.

(i) Velocity is maximum at mean position (y = 0)
νmax = ω.a
Velocity is zero at maximum displacement position ν = 0 , when y = ±a.

(ii) Acceleration is maximum when displacement is maximum (y = ±a)
α = ± ω² a, when y = ±a
Acceleration is zero at mean position α = 0, when y = 0.

Question 5.
What is meant by simple pendulum? When the bob is displaced from its mean position it starts oscillations. Why?
Answer:
If a heavy point mass be suspended by a weightless, perfectly flexible and inextensible string from a rigid support and an arrangement is made for its frictionless oscillations, then this arrangement is called simple pendulum. As shown in fig. when bob is displaced from its mean position A to a position B, then the bob raised up to height ‘h\ so that its potential energy increases.

Now, if the bob is released, then its centre of gravity falls down to obtain the stable equilibrium and when it reaches to the mean position its potential energy is converted into kinetic energy. At points, it will not stay, but due to inertia moves forward and hence the bob starts to execute simple harmonic motion to and fro about its mean position.

Question 6.
Write down the formula of time-period of simple pendulum and state the factors on which it does depend.
Answer:
The time-period of simple pendulum is given by :
T = 2π \(\sqrt{\frac{l}{g}}\)
From the above formula, it is clear that the time period of simple pendulum depends on its effective length ‘l’ and acceleration due to gravity ‘g’.
(i) Dependence on length f: The time period of simple pendulum is directly proportional to the square root of effective length of pendulum, i.e., T α \(\sqrt{l}\)
Example: When a boy swinging on a swing stand up suddenly, then his centre of gravity rises up and so the effective length of the swing reduces which results an increase in time period.

(ii) Dependence on acceleration due to gravity ‘g’: The time-period T of simple pendulum is inversely proportional to the square root of acceleration due to gravity at that place i.e., T α \(\frac{1}{\sqrt{g}}\)
Example: When a pendulum is taken up to hills or down in a mine get slow because time-period T increases due to decrease in the value of ‘g’.

Question 7.
Write down the laws of simple pendulum. Give practical application of each law.
Answer:
The laws of simple pendulum and their practical applications are given below :
(i) Law of length: If the value of ‘g’ remains constant, then the time period of simple pendulum is directly proportional to the square root of effecti ve length of simple pendulum.
i-e., T α \(\sqrt{l}\)
This law is used to repair the pendulum clocks when they get slow or becomes fast while taken up to hills or down in a mine.

(ii) Law of acceleration due to gravity : If the effective length of simple pendulum remains constant, then the time period T of the simple pendulum is inversely proportional to the acceleration due to gravity, i.e., T α \(\frac{1}{\sqrt{g}}\)
Due to this reason the pendulum clock get slow while taken up to hills or down in mines.

(iii) Law of mass: The time period of simple pendulum does not depend upon the mass of bob or that of thread.
Hence, either bob is heavy or lighter, if the value of 7’ and ‘g’ are constant, then there is no effect in its time-period. .

(iv) Law of isochronism: If the amplitude of oscillations of simple pendulum are very small, then the time-period of pendulum does not depend upon the amplitude of oscillations.
Because of this, the angular displacement of the bob is kept small in the experiment of simple pendulum.

Question 8.
Establish the relation between time-period and frequency.
Answer:
Let the time-period of a particle is T and its frequency is υ.
∴ The particle complete one vibration in T sec.
Hence, in 1 sec the number of vibration, υ = \(\frac{1}{T}\)
or
υT = 1.

Question 9.
Derive an expression for the displacement of a particle executing S.H.M.
Or
Find an expression for the displacement of S.H.M.
Answer:
Let XYX’Y’ is a circle whose centre is O and radius a. A particle is moving with uniform angular speed ω on the circle.
Let at time t = 0, the particle is at X and in time t, it is at P. A perpendicular PN is drawn on YOY’ from P and N is foot of the perpendicular.

The particle takes t second to subtend an ∠POX = θ.
∴ ω = \(\frac{\theta}{t}\)
or
θ = ωt
Let the displacement ON = y
∴ In ∆ NPO, sin NPO = \(\frac{O N}{O P}\)
But, ∠NPO = ∠POX = θ = ωt
∴ sin ωt = \(\frac{y}{a}\)
or
y = a sin ωt
This is the required equation for the displacement of S.H.M.

Question 10.
Derive the expression for the velocity of a particle executing S.H.M.
Or
Derive an expression for velocity of a particle executing S.H.M. When is the velocity maximum and minimum?
Answer:
Expression for velocity: Let XYX’Y’ is a circle of centre O and radius a. A particle is moving on the circle with angular velocity ω.
Let at t = 0, the particle is at X and after t sec, it is at P.
Let at point P, the velocity of the particle is ν along the tangent at P.

Now, velocity v is resolved into two parts :
(i) ν sin θ, parallel to PN and
(ii) ν cos θ, perpendicular to PN.
Since,ν cos θ is parallel to the direction of motion of foot of the perpendicular N.
∴ Velocity of N, υ = ν cosθ
υ = ν cos ωt.
or
υ = ν \(\sqrt{1-\sin ^{2} \omega t}\)
( ∴ sin² ωt + cos² = 1)

This is the required expression.

Case 1. Ify = 0, then from eqn. (1), we get
ν =ω \(\sqrt{a^{2}-0}\) = ωa
∴ Velocity will be maximum at mean position.
Case 2. If y = a, then from eqn. (1), we get
ν =ω \(\sqrt{a^{2}-a^{2}}=\) = 0
∴ Velocity will be minimum at the extreme position.

Question 11.
Derive the expression for the acceleration of a particle executing S.H.M. Find where the acceleration is maximum and minimum.
Answer:
We have by displacement equation,
y = a sin(ωt + Φ) …(1)
If the acceleration is α, then we have


This is the required expression.
Case 1. If y = 0, then by eqn. (3),
α =0
i.e., at the mean position the acceleration will be zero i.e., minimum.
Case 2. If y = a, then by eqn. (3),
α = – ω²a
i.e., at extreme position the acceleration is maximum.

Question 12.
Derive the expression for the time>period and frequency of a particle ex-ecuting S.H.M.
Answer:
Now, we have the magnitude of acceleration is given by

Again, we know that frequency,
υ = \(\frac{1}{T}\)
∴ υ = \(\frac{1}{2 \pi} \sqrt{\frac{\text { Acceleration }}{\text { Displacement }}}\)

Question 13.
Mass m is suspended by an ideal spring. It oscillates up and down. If the force constant of spring Is k, then prove that the time-period,
T =2 π \(\sqrt{\frac{m}{k}}\)
or
A mass is suspended by a spring. It Is pulled downwards and then left. Prove that it executes S.H.M. Find the expression for time-period.
Answer:
Let a spring is suspended by a rigid support and mass ni is suspended at its lower end.
It is displaced downwards at a distance y and left.
The restoring force F = -ky …( 1)
Where, k is force constant.
Also, by Newton’s second law of motion,
F=mα …………. (2)
Where, α is acceleration.
∴ From eqns. (1)and (2), we get
mα =-ky ……….. (3)
or
α =\(-\frac{k}{m}\) .y
Also, we have, for S.H.M.
α = – ω² y ……..(4)
Since, the L.H.S. of eqns. (3) and (4) are equal, hence the R.H.S. must be equal.

Question 14.
Derive the expression for K.E. and P.E. of a particle executing S.H.M. and also prove that total energy of a particle remains constant.
Answer:
K.E. of the particle : We know that velocity of a particle executing S.H.M. is given by,
ν = ω\( \sqrt{a^{2}-y^{2}} \)

K.E.= \( \frac{1}{2}\)mv²
=\( \frac{1}{2}\) m{ ω\( \sqrt{a^{2}-y^{2}} \)²
=\( \frac{1}{2}\)mω²(a²-y²) …(1)
P.E. of the particle: Now, the acceleration of a particle executing S.H.M. is given by α = -ω²y

Also, by Newton’s second law of motion,
Force = Mass x Acceleration
∴ F = mα
or
F = -mω²y
Now initially the displacement was zero and force was zero. The force was gradually increases and hence the displacement also increases to y.

∴Mean force =\(\frac{0+F}{2} \) = mω²y
∵ Work W= Mean force x Displacement
∴ W = \(\frac{1}{2} \) mω²y ×y
or
W = \(\frac{1}{2} \) mω²y²

By equns.(1) and (2), we get
K.E +P.E = \(\frac{1}{2} \) mω² (a²-y²)+ \(\frac{1}{2} \) mω² y²
or
E = \(\frac{1}{2} \) mω²a² –\(\frac{1}{2}\)mω²y² +\(\frac{1}{2}\)mω²y²
or
E = \(\frac{1}{2}\)mω²a²
As ω and a are constant, therefore the total energy remains constant.

Question 15.
What is second’s pendulum? Calculate its effective length.
Answer:
Second’s pendulum: A simple pendulum whose time-period is 2 seconds, is called second’s pendulum.

Oscillations Class 11 Important Questions Long Answer Type

Question 1.
Deduce the expression for time-period of simple pendulum T = 2π \(\sqrt{\frac{l}{g}} \)
Answer:
Motion of a simple pendulum: Suppose that m is the mass of the bob and l is the effective length of a simple pendulum. S is a point of suspension and O is mean position of the bob. Suppose at any instant during oscillation, the bob is in the position P. Then displacement OP = y and ∠ OSP = θ. In this position of the bob, two forces act upon it:
(i) Tension T is the thread, in the direction PS upward along the thread and

(ii) The weight of the bob, mg, vertically downward. The weight mg can be resolved into two components :
(a) mg cos θ, in the direction SP downward along the thread.
(b) mg sin θ, perpendicular to the thread SP.
The component mg cos θ balances the tension Tin the thread. The component mg sin θ tends to bring the bob back to its mean position. This is known as restorting force on the bob. Thus, the restorting force acting on the bob is given as F= -mg sinθ

Where negative sign shows that the direction of force is opposite to the direction which displacement increases, i.e., towards the mean position O.
1f the angular displacement of the bob be small, then sin θ = 0 (for example, if θ = 5⁰, then sin θ= 0.0872 and θ=0.0873 radian). Thus,
F= -mgθ
But,
θ= \( \frac{\mathrm{Arc}}{\mathrm{Radius}} \) = \( \frac{O P}{S P} \) = \( \frac{y}{l} \)
F= -mg.\( \frac{y}{l} \) …….. (1)

If α be the acceleration produced in the bob due to restoring force, then by Newton’s law of motion
F = mα ……. (2)
From eqns. (1) and (2), we get
mα = -mg \( \frac{y}{l} \)
or
α= –\( \frac{g}{l} \).y …(3)

At a given place g is constant and for a given pendulum is constant.
Therefore,
α ∞ -y …….. (4)
Thus, the acceleration of the bob is directly proportional to its displacement. Hence, for small amplitude, the motion of a simple pendulum is simple harmonic motion. Its periodic time is

Oscillations Class 11 Important Numerical Questions

Question 1.
The time-period of a particle executing S.H.M. is 2 sec. Calculate the time after t =0, that its amplitude behalf of the displacement.
Solution:
Given: T= 2 sec and y = \(\frac{a}{2}\)
∵ y = a sin ωt
Where ω =\( \frac{2 \pi}{T} \)
∴ y = a sin \( \frac{2 \pi}{T} \) t ⇒ \( \frac{a}{2}= \) = a sin \( \frac{2 \pi}{T} \) t

Question 2.
The length of a simple pendulum ¡s 39.2 / π² metre. If g = 9.8 ms^-2, then calculate the time-period of simple pendulum.
Solution:
Given:l = \( \frac{39 \cdot 2}{\pi^{2}} \), g = 9.8 ms^-2
Formula:

Question 3.
The mass of a particle executing S.H.M. is 0.4 kg and amplitude and time period are 0.5 m and π/ 2 respectively. Calculate the velocity and ICE. of particle at a displacement 0.3 m.
Solution :
Given:m = 0.4kg , a = 0.5 m, T= π/2 sec, y = 0.3 m.
Npw we have

Again,

Question 4.
A simple pendulum execute 60 oscillation per minute. Find its effective length ? (g = 981 cm/ s²)
Solution:
Given: Time-period T = \( \frac{\text { Time }}{\text { No.of oscillations }} \) = \frac{1 \mathrm{~min}}{60} = 1 sec

Question 5.
The value of acceleration due to gravity on a planet is \( \frac{1}{4} \) th of that of earth.
If the time-period of simple pendulum on earth is 2 sec, then find the time-period on the planet.
Solution:
g₂= \( \frac{1}{4}\) g₁

Question 6.
What will be the time.period of a simple pendulum of length \( \frac{9 \cdot 8}{\pi^{2}} \) m? Name such pendulum.
Solution:
Given: l= \( \frac{9 \cdot 8}{\pi^{2}} \)m g = 9.8 ms^-2
Formula:
T =2π \( \sqrt{\frac{l}{g}} \)

Question 7.
Two pendulum of length 100 cm and 110*25 cm start oscillating at same time. After how much oscillation again they will oscillate at same time.
Solution:
Let the time-period of the 100 cm pendulum be T₁
T₁ = 2 π\( \sqrt{\frac{100}{g}} \)

and if T₂is the time-period for 110.25 cm pendulum, then
T₂ = 2 π\( \sqrt{\frac{110 \cdot 25}{g}} \)
T₁ <T₂
∴ Both the pendulum to oscillate at same time if the big pendulum perform n oscillaion and small pendulum, perform (n + 1) oscillation.
i.e.,(n+1)T₁ = nT ₂

i.e., Big pendulum after 20 oscillation or small pendulum after 21 oscillation will start oscillating together.

Question 8.
Find the length of second pendulum at (i) Surface of earth (g = 9-8 m/s2), (ii) Surface of moon (g = 1.65 m/s2).
Solution:
Time-period for second pendulum is T= 2 sec.
∴From T = 2 π\( \sqrt{\frac{l}{g}} \)

(i) Length at earth surface
2 = 2π² \( \sqrt {\frac{l}{9.8}} \)
or
l = π² \( \ {\frac{l}{9.8}} \)
or
π².l = 9.8
or
l =\( \frac{9 \cdot 8}{(\pi)^{2}} \) = \( \frac{9 \cdot 8}{(3 \cdot 14)^{2}} \) = 0.99m

(ii) Length of moon
l = \( \frac{1 \cdot 65}{\pi^{2}} \)
or
l = \( \frac{1 \cdot 65}{(3 \cdot 14)^{2}} \) = 0.167 m.

Question 9.
A mass of 98 kg suspended in a spring is oscillating whose spring constant is 200 N/m. Find its time-period.
Solution:
We know,
T = 2π\( \sqrt{\frac{m}{k}} \)
or
T = 2 ×3.14 \( \sqrt{\frac{98}{200}} \)
T= 4-396 second.

Question 10.
The total energy of a particle executing S.H.M. is E. What will be the K.E. and P.E. of particle at the displacement half of the amplitude?
Solution:
We know that total energy is
E = \( \frac{1}{2} \) mω² a²
Also, K.E = \( \frac{1}{2} \) mω²( a² – y ²)

Now, at y = \( a / 2 \),
K.E = \( \frac{1}{2} \) mω²\( \left(a^{2}-\frac{a^{2}}{4}\right) \)
= \( \frac{1}{2} \) mω² . \( \frac{3 a^{2}}{4} \)
= \( \frac{3}{8} \) mω²a² = \( \frac{3}{4} \) E.

Again, P.E = \( \frac{1}{2} \) mω² y ²
Now, at y = \( a / 2 \),
P.E = \( \frac{1}{2} \) mω² \( \frac{a^{2}}{4} \)
P.E = \( \frac{1}{8} \) mω² a² = \( \frac{1}{4} \) E.

Question 11.
The amplitude of a particle executing S.H.M. is 0*01 m and frequency is 60 Hz. What will be the maximum acceleration of the particle?
Solution:
Given : a = 0.01m, υ = 60 Hz Now, maximum acceleration is given by
α = ω² a
= (2π υ ) ²a
= 4π²υ ²a
= 4×3.14 ×3.14×60×60×0.01
= 1419.78ms^-2

Question 12.
At a place a body travels 125 m in 5 sec during the free fall under gravity. What will be the time period of a simple pendulum of length 2-5 m at that place?
Solution:
Given : S= 125 m, υ = 0, t = 5 sec.
Now, s = ut + \( \frac{1}{2} \) gt²
or
125 = 0 +\( \frac{1}{2} \)g×25
or
g = 10 ms^-2
Again,
= 2π\( \sqrt{\frac{l}{g}} \)
= 2π \( \sqrt{\frac{2.5}{10}} \)
= 2π ×0.5 = π sec.

Question 13.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0-6 s. What is the weight of the body ? (NCERT)
Solution:
Given :m = 50 kg, y = 20 cm = 0.2 m, T- 0.6 s
From F=ky
∴ mg = ky
or
K = \( \frac{\mathrm{mg}}{\mathrm{y}} \) = [ latex] \frac{50 \times 9 \cdot 8}{0 \cdot 2} [/latex] = 2450Nm^-1

Weight of the body
W = m’g
= 22.36×9.8 = 219.17 N
≈ 219N.

Question 14.
A spring having spring constant 1200 Nm’ is mounted on a horizontal table as shown in fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled side ways to a distance of 2.0 cm and released. (NCERT)
Determine
(j) The frequency of oscillations.
(ii) Maximum acceleration of the mass and
(iii) The maximum speed of the mass.

Solution:
Given:k=1200Nm^-1,m= 3kg, amplitude a =2 cm =2 × 10²m
(i) Frequency υ =\( \frac{1}{2 \pi} \) \( \sqrt{\frac{k}{m}} \) = \( \frac{1}{2 \times 3 \cdot 14} \sqrt{\frac{1200}{3}} \)
= \( \frac{1}{6 \cdot 28} \) ×\( \sqrt{400}\) = \( \) \frac{20}{6 \cdot 28}
= 3.18 s^-1 ≈3.2 ^-1

(ii) α = -ω ² y
∴α _max = ω ²a = \( \frac{k}{m} \) .a = \( \frac{1200}{3} \) ×2 ×10^-2
= 8.0 ms ^-2

(iii) ∴ν _max=ωa= \( \sqrt{\frac{k}{m}} \) ×a =\( \sqrt{\frac{1200}{3}} \) ×2 ×10^-2
= 20 × 2 × 10^-2 = 0.4 ms ^-1

Oscillations Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Which of the following equation does not represent simple harmonic, motion:
(a) x = asin(ωt+δ)
(b) x = bcos(ωt+Φ)
(c) x = atan(ωt + Φ)
(d) x = asin(ωt cosωt.
Answer:
(c) x=atan(ωt + Φ)

Question 2.
Time-period and amplitude of a particle executing simple harmonic motion is Tand a. The minimum time taken to reach distance will be:
(a) T
(b) \(\frac{T}{4}\)
(c) \(\frac{T}{8}\)
(d) \(\frac{T}{16}\)
Answer:
(c) \(\frac{T}{8}\)

Question 3.
Time-period and amplitude of a particle executing simple harmonic motion is T and a. Its maximum velocity is :
(a) \(\frac{4 a}{T}\)
(b) \(\frac{4 a}{T}\)
(c ) \(2 \pi \sqrt{\frac{a}{T}}\)
(d) \(\frac{2 \pi a}{T}\)
Answer:
(d) \(\frac{2 \pi a}{T}\)

Question 4.
Acceleration in S.H.M. is :
(a) Maximum at amplitude position
(b) Maximum at mean position
(c) Remain constant
(d) None of these.
Answer:
(a) Maximum at amplitude position

Question 5.
The ratio of acceleration to displacement of a particle executing S.H.M. is measurement of:
(a) Spring constant
(b) Angular acceleration
(c) (Angular acceleration)²
(d) Restoring force.
Answer:
(c) (Angular acceleration)²

Question 6.
When displacement is half of amplitude, then ratio of potential energy to the total energy is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) 1
(d) \(\frac{1}{8}\)
Answer:
(b) \(\frac{1}{4}\)

Question 7.
What remain constant in S.H.M.:
(a) Restoring force
(b) Kinetic energy
(c) Potential energy
(d) Time-period.
Answer:
(d) Time-period.

Question 8.
Graph between ‘f and ‘F obtained as :
(a) Hyperbola
(b) Parabola
(c) Straight line
(d) None of these.
Answer:
(b) Parabola

Question 9.
The time-period of a seconds pendulum is :
(a) 1 second
(b) 2 second
(c) 3 second
(d) 4 second.
Answer:
(b) 2 second

Question 10.
Resonance is example of:
(a) Tuning fork
(b) Forced oscillation
(c) Free oscillation
(d) Damped oscillation.
Answer:
(b) Forced oscillation

2. Fill in the blanks:

1. A periodic motion of constant amplitude and same frequency is called ………… .
Answer:
Simple harmonic motion

2. In simple harmonic motion its total energy is ………. .
Answer:
Constant

3. The restoring force setup per unit extension in the spring is called ………… .
Answer:
Spring constant (Force constant)

4. When a body oscillates on both sides of its mean position in a straight line, then this bind of motion is called ………… .
Answer:
Simple harmonic motion

5. The maximum displacement of a body in oscillatory motion is called ………… .
Answer:
Amplitude

6. In the presence of damping forces, the amplitude of oscillations of a body ………… .
Answer:
Decreases

7. The time-period of second pendulum is ………… second.
Answer:
Two

8. The periodic time of simple pendulum does not depends upon
Answer:
Mass.

3. Match the following:
I.

Answer:
1. (c) At mean position
2. (d) At maximum displacement
3. (b) Remain constant
4. (e) Directly proportional to displacement.
5.  (a) 2π\( \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}} \).

II.

Answer:
1. (c) Periodic motion
2. (a) Forced oscillation
3. (e) S.H.M.
4. (b) Conservation of energy and momentum
5. ((d) Main frequency.

4. Write true or false:

1. Every S.H.M. is a periodic motion.
Answer:
True

2. Every periodic motion is S.H.M.
Answer:
False

3. In S.H.M. total energy is directly proportional to square of amplitude.
Answer:
True

4. Acceleration is always zero for a particle executing S.H.M.
Answer:
False

5. Time-period of simple pendulum is directly proportion to square root of its length.
Answer:
True

6. Time-period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.
Answer:
True

7. Time-period of simple pendulum depends on mass, material and size of the pendulum.
Answer:
False

8. In S.H.M., velocity is maximum when acceleration is minimum.
Answer:
True

9. The motion of moon around the earth is S.H.M.
Answer:
False

10. Time-period of hard spring is less than soft spring.
Answer:
True

11. Time-period of a simple pendulum cannot be one day.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 3 Motion in a Straight Line which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 3 Motion in a Straight Line

Motion in a Straight Line Class 11 Important Questions Very Short Answer Type

Question 1.
When do an object of definite size is considered as a point object, in motion?
Answer:
When the distance travelled by the object is very large in comparison to the size of object.

Question 2.
What is one-dimensional motion? Give three examples.
Answer:
When an object moves in a straight line, then its motion is called one-dimensional motion.
Examples :

• A body falling under gravity.
• Train running on straight rails.
• A man moving on a straight road.

Question 3.
Define displacement and distance.
Answer:
Displacement : The change in position in a particular direction is called displacement.
Distance : The length of a path covered by a particle in a given time interval is called its distance.

Question 4.
Direction of motion of an object is determined by velocity or acceleration.
Answer:
Velocity.

Question 5.
Define velocity. Write its SI unit.
Answer:
Displacement per unit time of a moving body is called its velocity.
Or
In a particular direction, the distance travelled in unit time is called velocity.
Formula: Velocity = \(\frac{\text { Displacement }}{\text { Time }}\) Time
The SI unit of velocity is m/s. It is a vector quantity.

Question 6.
Define speed. Is it a scalar or vector quantity?
Answer:
The rate of change of position of a moving body is called its speed.
Or
The distance travelled in unit time by a moving body is called its speed. It is a scalar quantity.
speed = \(\frac{\text { Distance }}{\text { Time }}\)

Question 7.
What do you understand by relative velocity?
Answer:
Relative velocity : The relative velocity of a body w.r.t. another body is the rate of change of position of that body relative to another body.
Let the velocity of bodies A and B are v₁ and v₂ respectively, then
Velocity of B w.r.t. A = v₂ – v₁
and Velocity of A w.r.t. B= v₁ – v₂.

Question 8.
Define acceleration and write its SI unit.
Answer:
Acceleration is defined as rate of change in velocity. Its SI unit is metre/second².

Question 9.
If the velocity of a body is constant, then what will be its acceleration?
Answer:
Zero.

Question 10.
What will be the displacement of a bullet fired vertically upward when it returns to its original point?
Answer:
Displacement of bullet will be zero.

Question 11.
Explain with reason, can a moving body have acceleration when :
(1) It is moving with uniform speed.
Answer:
Yes, because when speed is uniform its direction may be changed.

(2) It is moving with uniform velocity.
Answer:
No, as acceleration is defined as change in velocity.

Question 12.
Can a moving particle has constant speed but variable velocity?
Answer:
Yes, in uniform circular motion.

Question 13.
Can a body moving with constant speed has acceleration?
Answer:
Yes, in uniform circular motion.

Question 14.
Can uniform acceleration change the direction of velocity?
Answer:
Yes, when a body thrown vertically upwards, at its highest point the direction of velocity changes, while the direction of g is always acting downwards.

Question 15.
Can the speed be zero in one-dimensional motion and acceleration be not zero?
Answer:
Yes, at the highest point, when the body is thrown vertically upwards.

Question 16
What does the area enclosed between the velocity-time curve and time axis show?
Answer:
Distance travelled by the body.

Question 17.
(i) Draw velocity-time graph of a body moving uniformly in a straight line,
(ii) Draw position-time graph of a body moving uniformly in a straight line.
Answer:

Question 18.
Draw the graphs of positive and negative accelerated motion of velocity-time graph.
Answer:

Question 19.
What does the slope represent in position-time graph?
Answer:
The slope of a position-time graph represents velocity of an object.

Question 20.
Which quantity is represented by the area under the velocity-time graphs?
Answer:
It represents displacement of an object.

Motion in a Straight Line Class 11 Important Questions Short Answer Type

Question 1.
Differentiate displacement and distance.
Answer:
Difference between displacement and distance :

Question 2.
Differentiate velocity and speed.
Answer:
Difference between velocity and speed :

Question 3.
Define uniform motion and uniform velocity.
Answer:
Uniform motion :
When a body travels equal distance in equal interval of time, in the same direction, then its motion is called uniform motion.
Uniform velocity :
When the displacement of a moving body is equal in equal interval of time, then its velocity is called uniform velocity.

Question 4.
In which of the following examples of motion can the body be considered approximately a point object:
(a) A railway carriage moving without jerks between two stations.
(b) A monkey sitting on the top of a man cycling smoothly on a circular track.
(c) A spinning cricket ball that turns sharply on hitting the ground.
(d) A tumbling beaker that has slipped off the edge of a table?
Answer:
In (a) and (b), we can consider the body approximately as a point object because the motion of the object involves changes of position by distance much greater than its size.

Question 5.
A player throws a ball upwards with an initial speed of 29 m/s :
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c)Choose the x = 0 m and t₀ = 0 s to be the location and time of the ball at its highest point, vertically downward direction to the positive direction of X-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hand? (Take g = 9.8 ms^-2 and neglect air resistance)
Answer:
(a) Since the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.

(b) When the ball is at the highest point of its motion, its velocity becomes zero and the acceleration is equal to the acceleration due to gravity = 9.8 ms^-2in vertically downward direction.

(c) When the highest point is chosen as the location for x = 0 and t = 0 and vertically downward direction to be the positive direction of X-axis.
During upward motion : Sign of position is negative, sign of velocity negative and the sign of acceleration positive i.e., v < 0, a > 0.
During downward motion : Sign of position is positive, sign of velocity is positive and the sign of acceleration is also positive i.e:, v > 0, a > 0.

(d) Let, t = Time taken by the ball to reach the highest point.
H = Height of the highest point from the ground.
Taking vertically upward motion of the ball.
∴ Initial velocity, u =-29.4 ms^-1, a =g = 9-8 ms^-2,
Final velocity v = 0, s = H = ?, t =?
Using the relation, v²-u² = 2as, we get
0² – (29.4)² = 2 × 9.8 H
or H = \(\frac{-29 \cdot 4 \times 29 \cdot 4}{2 \times 9 \cdot 8}\) = -44.1 m
Where -ve sign shows that the distance is covered in upward direction. Using equations v=u + at, we get
0 = -29.4 + 9.8 × t
t = \(\frac{29 \cdot 4}{9 \cdot 8}\) = 3s
i.e., Time of ascent = 3s.
Also we know that when the object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
∴ Total time after which the ball returns to the player’s hand = 2t = 2 × 3 = 6s.

Question 6.
Derive graphically the first equation of motion.
Answer:
Velocity-time graph for an uniformly accelerated motion is shown in the figure. It is clear from the graph that at t = 0 the initial velocity of a particle is u and after t sec the final velocity is v.
From ΔEBC

∴ Slope of the curve = Acceleration
or a = \(\frac{v-u}{t}\)
or at = v-u or v=u + at.

Question 7.
What is relative velocity? Derive the expression for it.
Answer:
The relative velocity with respect to a body at rest or in motion is defined the rate of change of position of a body with respect to another body.
Let A and B are two bodies moving with velocities v₁and v₂ and at time t their position coordinate are x₁(t) and x₂(t) respectively.
x₁(t) = x₁1(0) + v₁t …(1)
∴ x₂(t) = x₂(0) + v₂t …(2)
Where x₁(0) and x₂ (0) are position coordinates at t = 0.
Subtracting eqn. (1) from eqn. (2),
∴ x₂(t) – x₁(t) = x₂(0) – x₁(0) + (v₂ -v₁)t …(3)
In the eqn. (3), x₂(t)- x₁(t) represents the position of B w.r.t. A at time t.
∴ x₂(0) – x₁(0) represents the relative position or distance of B w.r.t. A,
Putting t = 0 in eqn. (3) then the relative displacement in 1 sec. will be v₂-v₁
∴ Relative velocity of B w.r.t. A = v₂-v₁
∴ By eqn. (3),

Question 8.
Derive graphically the second equation of motion.
Answer:
From the given graph displacement of an object = Area under the curve OACE
or
s = Area of rectangle OABE + Area of ∆EBC
or
s = OA × AB + \(\frac{1}{2}\)EB × BC
or
s = ut+\(\frac{1}{2}\)t(v – u)
v – u=at
s = ut + \(\frac{1}{2}\)t ×at
s = ut + \(\frac{1}{2}\)at ²

Question 9.
Derive graphically the third equation of motion.
Answer:
From the above v-t graph
Displacement of an object = Area of trapezium OACE

Motion in a Straight Line Class 11 Important Questions Long Answer Type

Question 1.
Derive the equation of motion by Integration method.
Answer:
First equation of motion : v = u + at :Let at any instant t, the velocity of a particle is v.
∴ Acceleration, a = \(\frac{d v}{d t}\)
or dv = adt …(1)
Let at t = 0, v = u
∴ Integrating eqn. (1), we get

∴ v-u = a(t- 0)
or v – u- at or v = u + at.
Second equation of motion : Let a particle is moving with velocity u along X-axis and at any instant t, its displacement is x.
∴ v = \(\frac{d x}{d t}\)
or
dx = vdt
Also, we know that,
v = u + at
∴ dx = (u + at)dt ….(1)
Now, at t = 0, x = x₀
∴ Integrating eqn. (1), we get

Third equation of motion :
Accordrng to definition of acceleration

Question 2.
Derive graphically the second and third equation of motion.
Answer:
From the given graph displacement of an object = Area under the curve OACE
or
s = Area of rectangle OABE + Area of ∆EBC
or
s = OA × AB + \(\frac{1}{2}\)EB × BC
or
s = ut+\(\frac{1}{2}\)t(v – u)
v – u=at
s = ut + \(\frac{1}{2}\)t ×at
s = ut + \(\frac{1}{2}\)at ²

From the above v-t graph
Displacement of an object = Area of trapezium OACE

Motion in a Straight Line Class 11 Important Numerical Questions

Question 1.
A car travels from a city A to city B with a speed of 40 km/h and returns back to city A, with the speed of 50 km/h. What is the average speed of car? What will be its average velocity?
Solution:
Let the distance between A and B is x.

Question 2.
A body obtain a velocity of 10m/s in 3 second from rest. Find its acceleration.
Solution:
Given, u = 0, v = 10m/s and t = 3sec.
From Ist equation of motion,
v = u + at
a = \(\frac{v-u}{t}\)
Putting the values in above equation, we get
a = \(\frac{10-0}{3}\) = \(\frac{10}{3}\) m/s²
or
a = 3.33 m/s²

Question 3.
A body describes 45 m in 8^th second with uniform acceleration from rest. Find its acceleration.
Solution:
Given :u-0,t=8s,s = 45m

Question 4.
A car moving with a velocity of 36 km/h. When brakes are applied it stop after 10m. Calculate acceleration and time taken to stop.
Solution:

Question 5.
A body describe 15m in sixth second and 23m in tenth second. Find initial velocity and acceleration of the body.
Solution:

or 2u+19a=46
Solving eqns. (1) and (2),…..(2)
u = 4 m/s and a = 2m/s²

Question 6.
Two trains are running parallely in two tracks in the same direction with a speed of 10 m/s and 15 m/s. Find relative velocity.
Solution:
Given, V₁ = 10 m/s, v₂ = 15 m/s
Relative velocity = v₂ – v₁
= 15-10 = 5 m/s.

Question 7.
A car is running on a straight road with a velocity of 126 km/hr. It stops after travelling a distance of 200 m. Find the retardation of the car. How much time does it take to stop?
Solution:
Given, initial velocity u = 126km/h
= 126 × \(\frac{5}{18}\) = 35m/s.
Distance curved s = 200 m
Final velocity v = 0.
Using third equation of motion,
v² = u² + 2as
or 0 = 35 × 35 + 2 × a × 200
a = \(-\frac{35 \times 35}{2 \times 200}\) = 3.06m/s²
Using first equation of motion,
v = u + at
or 0=35 – 3.06t
or t= \(\frac{35}{3 \cdot 06}\)
=11.43sec.

Question 8.
A car travels first \(\frac{1}{3}\) part of total distance at a speed of 10km/hr, second\(\frac{1}{3}\) part at 20km/hr and last \(\frac{1}{3}\) part at 60 km/hr. Find out average speed of the car.
Solution:
Let the distance travelled by car = x km
Given, v₁ = 10 km/hr, v₂ = 20km/hr, v₃ 60 km/hr

Question 9.
A ball is thrown vertically upward reaches the roof of a house 100m in height. At same instant a second ball is dropped downward from the roof of the house will zero initial velocity. At what height both the ball will cross each other?
Solution:
Given, h = 100m
Let, both the ball cross each other at a distance of x m after t sec.
For vertically upward motion,
u = ? v = 0 a = -g and h = 100 m
v² = u² +2as, we get
0 = u² – 2 × g × 100
or u² = 200g = 200 × 9.8 = 1960
or
for vertically downward motion, .
u = 0, a = g, s = (100 -x)
From s = ut + \(\frac{1}{2}\)at², weget
100 – x = 0 +\(\frac{1}{2}\) × 9.8 × t² =4.9t²
or 100 – x = 4.9 t²
When the ball is thrown upward, distance covered by it
s = ut + \(\frac{1}{2}\) at² ….

or x = 100 – 25 = 75m

Question 10.
A jet plane is moving with a velocity of 500km/hr. Velocity of gas ejecting out with respect to jet plane is 1500 km/hr. find out the velocity of gas ejecting out w.r.t. a man standing on ground surface.
Solution:
Velocity of gas ejecting out from the jet plane w.r.t. man standing on ground surface
= v_g-v_j =1500-500
= 1000 km/hr.

Question 11.
A woman starts from her home at 9-00 am., walks with a speed of 5 kmh^-1 on a straight road up to her Office 2-5 km away, stays at the office up to 5-00 p.m. and returns home by an auto with a speed of 25 kmh^-1. Choose suitable scales and plot the x – t graph of her motion.
Solution:
x -t graph of the motion of woman is shown in fig. V₁ = speed of woman while walking at 5kmh^-1.
x = Distance covered by her = 2-5 km If t¹ = time taken to reach office, then it can be calculated by using
x₁ = v₁t₁
or t₁ = \(\frac{x}{v_{1}}\)
∴ t₁ = \(\frac{2 \cdot 5}{5}\) = \(\frac{1}{2}\) h = 30 minutes

If O is regraded as origin for both time and distance, then at t = 9.00 am., x = 0 at t = 9.30 am., x = 2.5 km and she reaches in her office. So OA represents x – t graph of the motion, when the woman walks from her home to office.
When she stays at her office from 9.30 a.m to 5.00 p.m., then she is stationary and hence her stay is represented by the straight line AB in the graph.
On return, speed of auto, v₂ = 25 km/h
∴ If t₂ = time taken by her i.e., by auto from office to her home, then
t₂ \(\frac{x}{v_{2}}\) = \(\frac{2 \cdot 5}{25}\) = \(\frac{1}{10}\) = 6minutes
Thus, she reaches back to her home at 5.06 p.m.
Her motion on the return journey in shown by BC part of the graph.
Scale chosen :
Time on X-axis, 1 division = 1 hour.
Distance on Y-axis, 1 division = 0.5 km.

Question 12.
A body travels half of its total path in the last second of its fall from rest. Find the time and height of its fall.
Solution:
Let the body falls from a height h and takes time t to fall.
From Second equation of motion, fall from/I to B is given by

Motion in a Straight Line Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
If the velocity of an object is doubled then among the following which will be doubled:
(a) Acceleration
(b) Kinetic energy
(c) Momentum
(d) Weight.
Answer:
(c) Momentum

Question 2.
An object dropped from peak of a tower takes 2 second to cross half its height. The height of the tower will be : (If g = 9-8 m/s²)
(a) 9.8m
(b) 19.6m
(c) 20m
(d) 39.2m.
Answer:
(d) 39.2m.

Question 3.
Ratio of instantaneous velocity and instantaneous speed is ……………….
(a) Less than 1
(b) More than 1
(c) Zero (d) Equal to 1.
(d) Equal to 1.
Answer:
(d) Equal to 1.

Question 4.
If acceleration-time graphs, the area represents :
(a) Displacement
(b) Velocity
(c) Change in velocity
(d) Distance covered.
Answer:
(c) Change in velocity

Question 5.
An object is falling freely under the gravitation. The ratio of the distance covered by it in first, second and third seconds will be :
(a) 1 : 3 : 5
(b) 1 : 2 : 3
(c) 1 : 4 : 9
(d) 1 : 5: 6.
Answer:
(a) 1 : 3 : 5

Question 6.
The motion of a particle of a gas are :
(a) One-dimensional
(b) Two-dimensional
(c) Three-dimensional
(d) Vertically upward and downward.
Answer:
(c) Three-dimensional

Question 7.
Which of the following quantity may be zero for any particle in motion :
(a) Displacement
(b) Distance
(c) Speed
(d) None of these.
Answer:
(a) Displacement

Question 8.
If a player throws a ball vertically upward with initial velocity 29 m/s. The velocity of the ball at its peak of motion will be :
(a) 29 m/s
(b) -29 m/s
(c) Zero
(d) None of these.
Answer:
(c) Zero

Question 9.
A car covers half a distance between two points with a speed of 40 km/hr and next half with a speed of 60 km/hr. The average speed will be :
(a) 40 km/hr
(b) 48 km/hr
(c) 50 km/hr
(d) 60 km/hr.
Answer:
(b) 48 km/hr

Question 10.
The position vector of a particle is represented by the equation x = (3t³ + 7t²+14t + 8)m, the acceleration of the particle at t = 1 sec will be :
(a) 10 m/s²
(b) 32 m/s²
(c) 23 m/s²
(d) 16 m/s².
Answer:
(b) 32 m/s²

2. Fill in the blanks:

1. A device used to measure instantaneous velocity is called ………………
Answer:
Speedometer

2. The motion of the Billiard ball is …………….. motion.
Answer:
Two-dimensional

3. The change in position of a body with respect to time is called ………………….
Answer:
Motion

4. When a body moves in a straight line, then its motion is ……………………
Answer:
One-dimensional

5. The distance travelled by a body in a definite direction is called ………………….
Answer:
Displacement

6. The rate of change of position of a body in motion is called its ………………..
Answer:
Velocity

3. Match the following:

 4. Write true or false:

1. From a velocity-time graph, the distance travelled can be measured.
Answer:
True

2. In two dimensional motion, the path of uniform velocity is a straight line.
Answer:
True

3. The displacement can be zero, negative as well as positive.
Answer:
True

4. The unit of acceleration in SI system is not metre/sec².
Answer:
False

5. Displacement is a vector quantity.
Answer:
True

6. The slope of velocity time graph represents acceleration.
Answer:
True

5. Answer in one word:

1. A body is moving with constant velocity. What is the value of its acceleration?
Answer:
Zero

2. What does the slope of position-time graph give?
Answer:
Velocity

3. What does the slope of velocity-time graph give?
Answer:
Acceleration

4. Who had derived the equations of motion?
Answer:
Gallileo,

5. What is the value of acceleration due to gravity on earth’s surface?
Answer:
9.81 m/s²

Students get through the MP Board Class 11th Physics Important Questions Chapter 2 Units and Measurements which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 2 Units and Measurements

Units and Measurements Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by measurement?
Answer:
In physics, the measurement means comparison of the given physical quantity with standard quantity of same nature.

Question 2.
What are the different type of measurement?
Answer:
The different type of measurement are :

1. F.P.S. (Foot Pound Second)
2. C.G.S. (Centimetre Gram Second)
3. M.K.S. (Metre Kilogram Second).

Question 3.
Define unit.
Answer:
The standard value of any physical quantity is known as its unit.

Question 4.
What are fundamental physical quantities?
Answer:
The physical quantities which do not depend upon other quantities are called fundamental.
Example: Mass, Length, Time etc.

Question 5.
What are derived physical quantities?
Answer:
The physical quantities which are derived from fundamental quantities are called derived quantities e.g.,
Volume = length × breadth × height = (length)³
Similarly Force, Momentum, Pressure, Power etc. are derived quantities.

Question 6.
Define metre.
Answer:
The distance travelled by light ray in \(\frac{1}{29,97,92,458}\) seconds in vaccum is called metre.

Question 7.
What are the different units to measure astronomical distance?
Answer:

• Light year: Distance travelled by light rays in one year in vacuum is called one light year.
  1 light year = 3 × 10⁸ × 365 × 24 × 60 × 60 metre = 9.46 × 10¹⁵ m.
• Parsec : It is the unit to measure the distance of stars from the earth.
  1 Parsec = 3.26 light year= 3.08 × 10¹³ km.
• Astronomical unit: Average distance between sun and earth is called astronomical units. It is denoted by A.U.
  1A.U.= 1.496 × 10¹¹metre.

Question 8.
What do you mean by one kilogram?
Answer:
The mass of platinum-iridium cylinder, kept at International Bureau of weights and measure at severes near Paris in France is defined as one kilogram.

Question 9.
Define one second.
Answer:
One second is the time-interval in which Cs – 133 atom execute 9 19 26 31 770 vibrations.

Question 10.
Define the dimensions of a physical quantity with example.
Answer:
The suitable powers, by which a derived unit is expressed in terms of fundamental units are called dimensions of that physical quantity.
Examples :
(i) Area = Length × Breadth
= [L] × [L] = [L]² = [M⁰L²T⁰].

(ii) Velocity = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\begin{array}{l}
{[\mathrm{L}]} \\
\hline[\mathrm{T}]
\end{array}\) = [LT^-1] = [M⁰LT^-1
Thus, dimension of area in length is 2 and that of velocity are, in length 1 and time -1.

Question 11.
What is dimensional equation? Explain with examples.
Answer:
The equation which represents the relation of a derived unit with the fundamental units is called dimensional equation. In mechanics dimensional equation is represented as
[X] = [M^aL^bT^c]
Examples : Velocity = [M⁰LT^-1 ]
and Force = [MLT^-2].

Question 12.
Which of the three physical quantities have the dimensional formula [ML^-1T^-2]?
Answer:
Pressure, Stress and Young’s modulus.
∵ [ML^-1T^-2]=\(\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}\) = \(\frac{\text { Force }}{\text { Area }}\)
[L2] Area

Question 13.
What are dimensionless quantities? Give three examples.
Answer:
The quantities which are having no dimension are called dimensionless quantities e.g., relative density, strain, relative humidity.

Question 14.
What is significant figure? What is the significant figure in 7.000 × 10³ m?
Answer:
Significant figure :
The number of digits used to express accurately the measured value of the physical quantity is called the significant figures.
The significant figure of 7.000 × 10³ m is 3.

Question 15.
How many significant figures are in the following :
(i) 6.320 joule,
(ii) 0.2370 gm,
(iii) 0.007 kg,
(iv) 2.64 × 10^-5m.
Answer:
(i)  4,
(ii) 4,
(iii) 1,
(iv) 3.

Question 16.
Candela is unit of which physical quantity?
Answer:
Candela is unit of Luminious intensity.

Question 17.
What do you mean by Armstrong unit?
Answer:
It is the unit to measure micro distance 1 Å = 10^-10 metre.

Question 18.
What are the merits of SI system?
Answer:
The main features of SI system are :

• It is a coherent system of units.
• In this system a physical quantity has only one unit, e.g., the unit of energy is joule, whether it is mechanical energy, heat energy or electrical energy.
• It is a decimal system. Smallest and biggest quantities are expressed in the power of 10. e.g.,
  1 kilometre = 10³ m; 1 mm = 10^-3m etc.
• It is closely related to C.G.S. and M.K.S. system.
• All the derived units in this system are accepted on international level.

 Units and Measurements Class 11 Important Questions Short Answer Type

Question 1.
Which of the following is the most precise device for measuring length :
(a) A vernier callipers with 20 divisions on the sliding scale.
(b) A screw gauge of pitch 1mm and 100 divisions on the circular scale.
(c) An optical instrument that can measure length to within a wavelength of light.
Answer:
The most precise device is that whose least count is minimum.
Now, (a) Least count of vernier callipers = 1 MSD – 1 VSD
= 1MSD – \(-\frac{19}{20}\) MSD = \(\frac{1}{20}\) MSD
= \(\frac{1}{20}\) mm =\(\frac{1}{200}\) cm
= 0.005 cm
(b) Least count of screw guage
= 0.001cm
(c) Wavelength of light, λ ≈ 10 ^-5 cm = 0.00001 cm
∴ Least count of optical instrument = 0.00001 cm Thus, clearly the optical instrument is the most precise.

Question 2.
What are the uses of dimensional analysis?
Answer:
Uses :

• To find the unit of a physical quantity.
• To change one system of units into another system.
• To check the correctness of a physical relation.
• To establish relation between different physical quantities.

Question 3.
Distinguish between fundamental units and derived units.
Answer:
Difference between fundamental and derived units :

Question 4.
Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diametre of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
Answer:
(a) The diameter of a thread is so small that is cannot be measured using a metre scale. We wind the thread on in close turns on a pencil or a cylindrical glass rod so that the turns touch one another and are tight. Count the number of turns (n) and find the average length (l) of the coiled thread using the metre scale. Divide this average length by the number of turns to get the diameter i.e.,
Diameter = \(\frac{\text { Average length }}{\text { Number of turns }}\) = \(\frac{l}{n}\)

(b) Yes, the accuracy of the gauge increases by increasing the number of divisions of the circular scale because the least count is given by :
Least count = \(\frac{\text { Pitch }}{\text { No. of divisions on the circular scale }}\)
So, theoretically speaking, least count decreases on increasing the number of divisions on the circular scale.

Question 5.
Let us choose such a unit accordance to which velocity of light be one unit. On the basis of this new unit, what will be the distance between Sun and Earth if light takes 8 minute and 20 second to reach earth?
Answer:
Velocity of light = 1 (as per new unit)
Time taken by light to reach earth surface = (60 × 8 + 20) = 500 sec.
∴ Distance between Sun and Earth = Velocity of Light × Time
= 1 × 500 = 500 Unit of length.

Question 6.
Check the correctness of formula, v² = u² +2las by dimensional method.
Answer:
v²=u²+las
Now, dimensions ofu and v = [LT^-1], dimensions of a =[LT^-2]
Dimensions of s = [L]
and 2 is dimensionless
∴ [LT^-1]² =[LT^-1]²+[LT^-2L]
or [ L²T^-2 ] = [ L²T^-2] + [ L²T^-2]
Since, the dimensions of each term is same, hence the formula is true.

Question 7.
What do you mean by principle of homogeneity of dimensions?
Answer:
According to this principle the dimensions of all terms on both the sides of a dimensional equation are same.

Units and Measurements Class 11 Important Questions Long Answer Type

Question 1.
The velocity of sound waves in a medium depends upon the coefficient of elasticity of the medium (E) and its density (d). Establish the formula for the velocity of wave by dimensional method.
Answer:
Let the velocity of sound in the air depends upon the coefficient of elasticity and density of medium, then
v ∝ E^ad^b
or v = k ,E^ad^b ……..(1)
Where v = velocity of sound, E = coefficient of elasticity, d – density and a and b are numbers and k is dimensionless constant.
Now, the dimensional formulae of quantities are :
v=[LT^-1]
E = [ML^-1T^-2]
d = [ML^-3]
By eqn. (1) we get
or [LT^-1] =[ML^-1T^-2]^a[ML^-3]^b
or [LT^-1] =[M^aL^-a’T^-2aM^bL^-3b]
or [LT^-1] =[M^a+b + L^-a-3bT^-2a]
Now,by the principle of homogeneity, the dimension of both the sides will be equal.
a + b = 0 …(2)
∴ -a-3b=1 …(3)
-2a = -1 …(4)
Hence, a=\(\frac{1}{2}\)
Putting the value of a in eqn. (1), we get
\(\frac{1}{2}\)+ b = 0
or
b = – \(\frac{1}{2}\)

Putting the values of a and b in eqn. (1), we get
v = k.E^1/2d-d^-1/2
or

This is the required relation.

Question 2.
How much dyne is in one Newton?
Answer:
Dimensional formula for force is [MLT^-2].
Here, a=1,b=1 and c = -2

Putting the value, we get
n₂= n₁[10³] [10² ] [1]
or
n₂ = n₁ × 10n⁵
If n₁ = 1 dyne
Then n₂ = 1 Newton
∴ 1 Newton = 10⁵ dyne.

Question 3.
According to van der Wall relation between pressure P, volume V and temperature T is given by

Where a and b are constant. Find the dimensions of a and b.
Answer:
According to homogineity of dimensional equation, a
Dimension of \(\frac{a}{V^{2}}\)= Dimension of P.
∴ Dimension of a = Dimension of P ×(Dimension of V)²
= [M¹L^-1T^-2][L³]²
∴ Dimension of a =[M¹L⁵T^-2] and Dimension of b = Dimension of V = [L³]
or Dimension of b = [M⁰L³T⁰].

Question 4.
Find the fundamental unit of force and work using dimensional equation.
Answer:
(1) Force = Mass × Acceleration

(2) Work Force Displacement
= Mass x Acceleration x Displacement

∴ Dimensional formula of work = [ML²T^-2]
SI unit = kg-metre²/sec²
C.G.S. unit = g-cm²/sec².

Question 5.
What are the limitations of dimensional equations?
Explain the limitations of dimensional analysis.
Answer:
Limitations :

• The dimensional formulae of only those quantities can be ob- :d which can be expressed in terms of fundamental quantities.
• It is not applicable to trigonometrical, exponential or logarithmic functions.
• The value of constants cannot be determined by this system.
• If an equation has a constant having dimensions e.g., G, Then relation cannot be blished.

Question 6.
The periodic time of a simple pendulum T, depends upon its effective length (l) acceleration due to gravity (g). Derive the formula for time-period by dimensional
Or
Find out an expression for periodic time of a simple pendulum using dimensional sis.
Answer:
Let the formula for periodic time (T) be,
T ∝ l^a g^b
T = K. l^a g^b
Putting the dimensions of periodic time (7), effective length (l) and acceleration due to gravity in eqn. (1).
[M⁰L⁰T] = [L]^a x [LT^-2]^b .
or
[M⁰L⁰T] = [M⁰L^a+bT^2b]
By the principle of homogeneity, comparing the powers of length and time, we get
a + b = 0 …(2)
and – 2b = 1
⇒ b = – \(\frac{1}{2}\)
Putting in eqn. (2), we get
a- \(\frac{1}{2}\) = 0
∴ a = \(\frac{1}{2}\)
Putting the values of a and b in eqn. (1), we get
T = kl^1/2g^-1/2
or

This is the required formula.

Question 7.
What are the different type of errors in measurement? Explain.
Answer:
Random error :
If an observer measures a physical quantity and each time he finds different readings, then such an error is called random error.
Example : Suppose the diameter of a wire is determined by a screw gauge and each time the reading is different because the wire is not uniform.
To remove this error, so many readings are taken and then mean is calculated.
If a physical quantity is measured n times and its readings are a₁,a₂, a₃ …..a_n

Absolute error :
The difference between the observed value and actual value of a physical quantity is called absolute error. It is always positive.
The absolute error of a quantity x is denoted by | Δx |
Let the observed values of a physical quantity are x₁,x₂, x₃ …..x_nand the average value is x.

Relative error :
The ratio of absolute error and actual value is called relative error.
∴ Relative error = \(\frac{\text { Average absolute error }}{\text { Actual value }}\)
= \(\frac{\text { Actual value }-\text { Measured value }}{\text { Actual value }}\)
or

Percentage error: When the relative error is expressed in percentage, then it is called percentage error.
∴ Percentage error = \(\frac{\text { Average absolute error }}{\text { Actual value }}\) ×100

Question 8.
What do you understand by significant figure ? How do you find significant figure of a quantity?
Answer:
The number of digits used to express accurately, the measured value of a physical quantity is called significant figure.

Counting of significant figures :

• By changing the decimal point, the number of significant figures does not change.
  For example; 1.234, 12.34 and 123.4 the number of significant figure is 4.
• All the zeros between two non-zero numbers are significant.
  For example; Significant figures in 3100-05 is 6.
• The zeros before the first non-zero digit are ignored while all the zeros to the right are counted.
  For example; Significant figure in 0 007 is 1 and that of 6-320 is 4,
• The zeros in a whole number at the end are not counted.
  For example; Significant figure in 6320 is 3 and in 13200 is 3.
• The significant figure does not change with the change of units i.e.,
  0.0432m = 4.32 cm, both have the significant figure 3.
• When a number is expressed in the power of 10, then the power is not counted, i.e., 10.52 × 10⁵ has significant figure 4.

Units and Measurements Class 11 Important Numerical Questions

Question 1.
Convert 0.53 Å into metre.
Solution:
We know that, 1Å = 10^-10 m
0.53 Å = 0.53 ×10^-10m
= 5.3 ×10^-11m

Question 2.
A student measures the thickness of a human hair by looking at it through
a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 35mm. What is theestimate on the thickness of hair?
Solution:
Estimate on the thickness of hair is given by:

Question 3.
State the number of significant figures in the following :
(a) 0.007m²,
(b) 2.64 × 10²⁴kg,
(c) 0.2370 g cm^-3,
(d) 6.320 J,
(e) 6.032 Nm^-2,
(f) 0.0006032 m².
Answer:
The number of significant figures is as given below :
(a) 1,
(b) 3,
(c) 4,
(d) 4,
(e) 4,
(f) 4.

Question 4.
The length, breadth and thickness of a rectangular sheet of metal are 4.234m, l-005m and 2.01cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
Here, Length, l = 4.234m
Breadth, b = 1.005m
Thickness, h = 2.01cm = 0 0201m
Area of the sheet = 2(lb + bh + hl)
= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)
= 8.7209468m²
As the least number of significant figures in thickness is 3,
∴ Area = 8.72m²
Volume = l × b × h
= 4.234 × 1.005 × 0.0201m³
= 0-0855m³

Question 5.
A physical quantity P is related to four observables a, b, c and d as follows :

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P ? If the value of P calcu-lated using the above relation turns out to be 3.763, to what value should you round off the result?
Solution:

The calculation of error clearly shows that the number of significant figures is 2, so the result of P may be rounded off to two significant digits i.e.,
P = 3.763 = 3.8.

Question 6.
The percentage errors in the measurement of length and mass are 2% and 5% respectively. What will be the percentage error in the measurement of density of a cube?
Solution:

Maximum % error in the measurement of density will be 11%.

Question 7.
Given that the period T of oscillation of a gas bubble from an explosion under water depends on P, d and E, where the symbols are pressure, density and total energy of explosion. Find dimensionally a relation for T.
Solution:
Let T∝P^ad^bE^c
or T =kPsup>ad^bE^c, (where His a dimensionless constant) …(1)
Writing dimensional formula of each physical quantity on both sides of eqn. (1), we
[M⁰L⁰T¹] = [ML^-1T^-2 ]^a [ML^-3 ]^b [ML²T^-2 ]^c
= [M^a+b+c][L^-a-3b+2c][T^-2a-2c]
Comparing power of,
M, 0 = a + b + c ……(2)
L, 0 = – a – 3b + 2c ……(3)
T, 1 = -2a – 2c
or a + c = –\(\frac{1}{2}\)
∴ From eqns. (2) and (4), we get
b = – (a + c)
= \(-\left(-\frac{1}{2}\right)\) = \(-\left(\frac{1}{2}\right)\)
Putting the value of b in eqn. (3),
– a + 2c = 3b
=> -a + 2c =3 x \(\frac{1}{2}\)= \(\frac{3}{2}\)
Adding eqns. (4) and (5),

Question 8.
An object of mass ‘m’ is executing simple harmonic motion suspended at one end of a spring. If its spring constant is K and time-period is T, then prove with the help of dimensional method that equation T=2π\(\frac{m}{\mathbf{K}}\) incorrect. Also find its correct
equation.
Solution:
Given, T = 2π\(\frac{m}{\mathbf{K}}\)
Writing dimensional formula of the above equation
[T] = \(\frac{[\mathrm{M}]}{\left[\mathrm{MT}^{-2}\right]}\) = [T²]
Since, the dimension of T on both sides are not same, therefore the given equation is incorrect.
Let us assume that time-period T, a^th power of mass (m) and b^th power of spring constant (K), then
T∝m^aK^b
or T = K.m^aK^b
Let K = 2π
∴ T = 2πm^aK^b
or [T] = 2π[M]^a [MT^-2]^b
= [M] ^(a+b)[T]^-2b
Comparing both sides, we get
a + b = 0
and -2b = 1
or b = –\(\frac{1}{2}\)
∴ a+ \(-\left(-\frac{1}{2}\right)\)=0
or
a = \(\frac{1}{2}\)
Putting the values of a and b in the above equation,
T = 2πm\(\frac{1}{2}\). K-\(\frac{1}{2}\)

This is the correct form of the equation.

Question 9.
Equation of a particle of velocity (v) and time (t) is given by v = A + Bt +\(\frac{C}{D+t}\) find out the dimension of A, B, C and D.
Solution:
Each form of R.H.S. must have dimension of v.
∴ Dimensional formula of A = Dimensional formula of velocity
v = [LT^-1].
Dimensional formula of B.t = Dimensional formula of v = [LT^-1]
∴ Dimensional formula of B =\(\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}\) [LT^-2]
∴Dimensional formula off D = Dimensional formula off t = [T].
Dimensional formula of\(\frac{C}{D+t}\) = Dimensional formula of v= [LT^-1]
and Dimensional formula of C = [LT^-1] x [Dimensions of D + t] – [L¹].

Question 10.
Assuming energy E, length L and time T as the fundamental unit, what would be dimensions of mass ? If energy E is being replaced by force F then what would be the dimensions of mass ?
Solution:

∴ Dimensions of mass will be = [E¹L^-2T²].
Replacing energy E by force F, we get
Force F = Mass x Acceleration

Question 11.
If m and c represent mass and velocity, then find with the help of dimensional method the quantity mc² is equivalent to which form?
Solution:
Dimensions of mc² – Dimensions of mass x (Dimensions of velocity)²
= M[LT^-1]²
= [M¹L²T^-2 ] = Dimensions of energy.

Question 12.
The centripetal force acting on a particle revolving in a circular path depends upon velocity of the particle v and radius of circle (r). Using dimensional method establish the formula for the centripetal acceleration (a).
Solution:
According to given question
a ∝ v^a r^b
or a = kv^a r^b …(1)
Writing the dimensional formula for the quantities
[M⁰LT^-2] = [M⁰LT⁰]^b
=[M⁰L^a+bT^-a]
Comparing the powers of M, L and T both sides,
a + b= 1 …(2)
and – a = -2 or a = 2
Putting the value of a in eqn. (2), we get
2 + b=1
or b = 1 – 2 = – 1
Hence, from eqn. (1), we get
a = k.v² r^-1 or a = \(\frac{k v^{2}}{r}\)
If k= 1, then a= \(\frac{v^{2}}{r}\)

Question 13.
Diameter of a sphere is 2.348 cm. Find out volume and surface area of it up to required significant number.
Solution:
Given,
2R = 2.348 cm
∴ R= 1.174 cm
Volume of sphere V =\(\frac{4}{3}\)πR³ = \(\frac{4}{3}\) × 3.14 ×(1.174)³
or
V = 6.774 cm³
and Surface area of sphere S = 4πR²
=4 × 3.14 × (1.174)²
= 17.311 = 17.31cm².

Motion in a Straight Line Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Light year is unit of:
(a) Time
(b) Distance
(c) Speed
(d) Intensity of light.
Answer:
(b) Distance

Question 2.
Fermi is unit of:
(a) Energy
(b) Momentum
(c) Impulse
(d) Length.
Answer:
(d) Length.

Question 3.
The dimensional formula of gravitational constant is :
(a) [ML²T^-2]
(b) [M^-1L³T^-2]
(c) [M^-1L²T^-3]
(d) [ML³T²]
Answer:
(b) [M^-1L³T^-2]

Question 4.
Choose a pair of physical quantities having the same dimensional formula:
(a) Pressure and stress
(b) Stress and strain .
(c) Pressure and force
(d) Power and force.
Answer:
(a) Pressure and stress

Question 5.
Units of energy is :
(a) watt
(b) joule
(c) Electron volt
(d) Calorie.
Answer:
(a) watt

Question 6.
Relation between v and t, velocity of sound and time is
v = at+\(\frac{b}{t+c}\) dimensional formula for a, b and c will be:
(a) [L²], [T], [LT^-2]
(b) [LT^-2], [L], [T]
(c) [LT⁺²], [LT], [L]
(d) [L], [LT], [T²].
Answer:
(b) [LT^-2], [L], [T]

Question 7.
Order of 4-12 × 10⁵ is :
(a) 4
(b) 6
(c) 5
(d) None of these.
Answer:
(b) 6

Question 8.
Which among the following have significant number four :
(a) 0.630
(b) 0.0024
(c) 6.023
(d) 0.004.
Answer:
(c) 6.023

Question 9.
Which among the following is not a fundamental unit:
(a) Metre
(b) Ampere
(c) Kelvin
(d) Litre.
Answer:
(a) Metre

Question 10.
Which among the following physical quantities do not have the dimensional
formula [ML^-1T^-2 ] :
(a) Youngs modulus
(b) Stress
(c) Strain
(d) Pressure.
Answer:
(c) Strain

Question 11.
Light year is equal to :
(a) 9.46 × 10¹⁰ km
(b) 9.46 × 10¹² km
(c) 9.46 × 10¹² metre
(d) 9.46 × 10^-12 km.
Answer:
(b) 9.46 × 10¹² km

Question 12.
If the length of a simple pendulum is increased, then its percentage increase in its time-period will be:
(a) 0.5%
(b) 2.0%
(c) 1%
(d) 4%.
Answer:
(b) 2.0%

2. Fill in the blanks:

1. The volume of cube of side 1cm is equal to ……………….. m³.
Answer:
10^-6

2. A vehicle moving with speed of 18 km/hr covers ……………….. m in 1 second.
Answer:
5

3. 3.0 m/s² = ………………….. km/h².
Answer:
3.9 × 10^-4

4. Number of significant figure in 2.64 × 10²⁴ kg is ………………….
Answer:
3

5. Number of fundamental unit in SI system is …………………..
Answer:
7

6. 1 micron = ………………… metre.
Answer:
10^-6

7. 1Å = …………………….. metre.
Answer:
10^-10

8. The error due to observer measurement is called ………………..
Answer:
Individual error

9. 1 newton = ……………………. dyne.
Answer:
10⁵

10. Dimensional formula of charge is ……………………
Answer:
[AT]

3. Match the following:
I.

Answer:
1. (c) [M⁰LT⁰]
2. (e) [M⁰L⁰T^-1].
3. (b) Dimensional of planck constant
4. (a) Dimension of momentum
5. (d) [ML²T^-2]

II.

Answer:
1. (c) Pressure
2. (e) Force.
3. (d) Candela
4. (b) Newton × second
5. (a) Light year

4. Write true or false:

1. Dimensional formula for heat and work are same.
Answer:
True

2. The equation v = u + at, can be derived from dimensional method.
Answer:
False

3. Dimensional formula of impulse and momentum are different.
Answer:
False

4. Dimensional formula for pressure, coefficient of Young modulus and stress are [ML^-1T^-2].
Answer:
True

5. Candela is units of Luminious intensity.
Answer:
True

6. Significant figure changes with change in its units of a physical quantity.
Answer:
False

7. It is not necessary that the physical quantity will be the same, if their dimensional formula and units are same.
Answer:
True

8. The value of constant can be determined by dimensional method.
Answer:
False

9. On knowing one dimensions of a physical quantity, its unit can be determined.
Answer:
True

10. The physical quantity is more significant which have more significant figure.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 15 Waves which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 15 Waves

Waves Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by wave motion?
Answer:
It is a method of energy transformation in which the disturbance produced in the medium propagates without the flow of medium.

Question 2.
What is progressive wave. Write the formula for displacement of it?
Answer:
When disturbance is created continuously in a medium, then the particle of medium vibrates to and fro about their mean position without leaving their place. The distur¬bance move forward. The wave thus produced are called progressive wave.
Its displacement equation is :
\(y=a \sin \frac{2 \pi}{\lambda}(v t-x)\)

Question 3.
Write the principle of superposition of waves.
Answer:
When any number of waves meet simultaneously at a point in a medium, the net displacement at a given time is the algebraic sum of the displacement due to each wave at that time.

Question 4.
Write the name of the phenomenon obtained due to superposition of waves.
Answer:
Interference,

• Beats,
• Stationary waves.

Question 5.
What are stationary waves?
Answer:
When two identical waves of equal frequency travel along the same line but in opposite direction superimpose, then the resultant wave thus formed is called stationary waves.

Question 6.
How many types of stationary waves are there? Write one example of them.
Answer:
There are two types of stationary wave :

1. Transverse wave,
2. Longitudinal wave.

1.Transverse wave: If during the propagation of mechanical wave in a medium, the particles of the medium vibrate about their mean position in a direction perpendicular to the direction of wave propagation then the wave is called transverse wave

2. Longitudinal wave: If during the propagation of mechanical wave in a medium, the particles of the medium vibrate about their mean position along the direction of wave propagation then this type of wave is called longitudinal wave.

Question 7.
What are beats?
Answer:
When two waves with slight difference in frequencies travel in a medium simul¬taneously in the same direction then as a result of their superposition, the intensity of sound at any point increase and decreases alternately. This successive rise and fall in the intensity of sound is called beats.

Question 8.
What are the conditions of beats?
Answer:

• Both the waves should travel along the same straight line, in the same medium, with same velocity.
• These should be slight difference in their frequencies.
• The amplitudes of both the waves should be nearly equal.

Question 9.
If two tunning fork of frequency n₁ and n₂ are hammered together. Then how many beats will be obtained per second?
Answer:
Per second number of beat = n₁~ n₂.

Question 10.
Write three applications of beats?
Answer:

• Determination of frequency of a tuning fork.
• Matching the frequency of sound instruments.
• To enquire about poisonous gas in the mines.

Question 11.
On which factors the velocity of transverse wave produced in stretched string depends on?
Answer:

• Tension of string,
• Mass per unit length of string.

Question 12.
Write the formula for frequency of a stretched string
Answer:
\(n=\frac{1}{2 l} \sqrt{\frac{T}{m}}\)
Where, l = Length of stretched string, T = Time-period, m = Mass per unit length of string.

Question 13.
Explain beats period and frequency of beats.
Answer:
One rise and one fall of sound form one beat. The time interval between two consecutive intense sounds is called ‘beat period’.
The number of times the intensity of sound rises and falls in one second is called ‘beat frequency’.
Let the frequencies are υ₁ and υ₂
Beat frequency = \(v_{1}-v_{2}\)
and Beat period = \(\frac{1}{v_{1}-v_{2}}\)

Question 14.
What is Doppler’s effect?
Answer:
When, there is a relative motion between a sound source and a listener, then the change in the frequency of the source appears to the listener. This effect is called Doppler’s effect.

Question 15.
What are the limitations of Doppler’s effect?
Answer:
Limitations :

• The effect expresses the apparent change in the frequency, not the intensity of sound. ,
• The velocity of source or listener should not be greater than the velocity of sound.

Question 16.
What is difference between the Doppler’s effect on sound and in light waves?
Answer:
Doppler’s effect on sound waves depends upon the relative velocity and velocity of source and listener. While in light waves Doppler’s effect depends upon the relative velocity only.

Question 17.
Explain one application of Doppler’s effect.
Answer:
To estimate the speeds of aeroplanes or submarine: When radio waves transmitted from the Radar, come back after reflection from aeroplane, then their wavelength is changed. If an aeroplane is approaching then its wavelength decreases or frequency increases and if the aeroplane receding from the station, the wavelength increases. Thus, by the change of wavelength or frequency, the speed of aeroplane can be calculated.

Question 18.
What do you mean by organ pipe?
Answer:
The pipe in which sound is produced by vibrating the molecules of air or gas-filled in it.

Question 19.
The open organ pipe produces more melody sound than dosed organ pipe. Why?
Answer:
In open organ pipe the harmonics produced are odd and even both while in closed organ pipe only odd harmonics are produced, therefore harmonics produced by open organ pipe are melody.

Question 20.
A sound-producing whistle is whirled fast in a horizontal circle. What change in the frequency will be observed by an observer standing (i) at the centre of the circle, (ii) outside of the circle?
Answer:

• Since, the whistle is always constant distance from the observer, hence no change in frequency will be observed.
• For the observer standing outside the circle, the whistle will approach and move away from him in every half rotation. Hence, the frequency will be observed increasing and decreasing.

Question 21.
Why a flute has several holes in it?
Answer:
A flute is an open organ pipe, melodious sound of different frequencies can be produced by changing the length of air column by putting fingers on the holes.

Question 22.
Why sound travels faster in moist air than in dry air?
Answer:
Presence of water vapour decreases the density of air. Since velocity of sound is inversely proportional to the density. Hence, sound travels faster in moist air than in dry air.

Question 23.
Why the sound heard is more in carbon dioxide than in air?
Answer:
Velocity of sound in a gas is inversely proportional to the square root of the density of gas. Since carbon dioxide is lighter than air. Thus, velocity of sound is greater in carbon dioxide. Hence, sound heard is more in carbon dioxide than in air.

Question 24.
Sound travels faster on a hot day than on a cold day. Why?
Answer:
Velocity of sound is directly proportional to the square root of temperature. Therefore, sound travels faster on a hot day than on a cold day.

Question 25.
Why are transverse waves not produced in gases?
Answer:
Because the gases are not rigid.

Question 26.
Which property is common to all types of mechanical waves?
Answer:
When the waves propagate through a medium, the particles of a medium do not advance along with waves.

Question 27.
What will be the effect on the speed of sound waves in a gas if the absolute temperature of gas is increased to 4 times of its previous value?
Answer:
According to the formula \(v \propto \sqrt{T}\), speed of sound waves will be doubled.

Question 28.
The speed of sound in air is 330m/s. What will be the effect on the speed of sound waves if pressure is doubled, keeping the temperature constant?
Answer:
No effect, because speed of sound does not depend upon pressure.

Question 29.
Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases? (NCERT)
Answer:
For the propagation of transverse waves through a medium, the medium should posses rigidity. Gases do not have property of rigidity, so the transverse waves cannot propagate through them. Due to rigidity of solids both, transverse as well as longitudinal wave can propagate through them.

Question 30.
The shape of pulse gets distorted during propagation in a dispersive medium. (NCERT)
Answer:
A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions i.e., with different velocities. So the shape of the pulse gets distorted i.e., a plane wavefront in a non-dispersive medium does not remain a plane wavefront in a dispersive medium.

Question 31.
A sitar, a tambour, and a harmonium all the three are made in unison at the same frequency, even then their voice can be recognised separately. Why is it so?
Answer:
Because the sounds of different qualities are produced by them.

Waves Class 11 Important Questions Short Answer Type 

Question 1.
What are progressive waves? Derive its equation.
Answer:
When the disturbance produced from the body executing simple harmonic motion propagates in the medium, then the particles of medium vibrates to and fro about their mean position and the disturbance advances in the medium. The waves thus produced are called progressive waves.

Consider a harmonic wave propagat- g O ing in a medium along the positive direction of x-axis as shown in fig. The speed ofpropagation of wave is v. Its source is at origin ‘O’. As the source is executing
S.H.M., then the displacement of the particle situated at ‘O’ at any instant ‘f be \(y=a \sin \omega t\) ………(1)
Where, ‘a’ is amplitude of vibration and ‘ co ‘ is angular frequency. Now, \(\omega=2 \pi v=\frac{2 \pi}{T}\)

Where, u is frequency and T is time-period. Since the speed of wave is v hence the disturbance will reach the point ‘P’ at a distance V from the origin ‘O’ in \(\frac{x}{v}\) sec. Thus, at any instant ‘f the displacement of the particle situated at P will be \(y=a \sin \omega\left(t-\frac{x}{v}\right)\) …….(2)

The particle of medium situated at ‘P’ will begin to vibrate after \(\frac{x}{v}\) sec. from the particle situated at ‘O’. Thus, in time ‘t’ the displacement of particle at P will be the same as that the particle at O, before \(\frac{x}{v}\) sec.Therefore, instead of ‘t’ in the eqn. (1) we should write \(\left(t-\frac{x}{v}\right)\) Eqn. (2) shows a progressive wave which is propagating along positive direction of X- axis with velocity v. If a progressive wave is travelling along negative direction of A-axis, then its equation will be :
\(y=a \sin \omega\left(t+\frac{x}{v}\right)\) …………. (3)
As ω = 2πυ
∴ From eqn.(2)
\(y=a \sin 2 \pi v\left(t-\frac{x}{v}\right)\) ……………. (4)
But ν = υλ

\(v=\frac{v}{\lambda}\)
∴ \(y=a \sin 2 \pi \frac{v}{\lambda}\left(t-\frac{x}{v}\right)\)
or
\(y=a \sin \frac{2 \pi}{\lambda}(v t-x)\)
This is the equation of a progressive wave.

Question 2.
Explain Newton’s formula for velocity of sound in air.
Answer:
From purely theoretical considerations, Newton gave an empirical relation to calculate the velocity of sound in a gas.
\(v=\sqrt{\frac{B}{\rho}}\) …………. (1)
Where B is bulk modulus and P is density of the gas.

Sound travels through a gas in the form of compressions and rarefactions. Newton assumed that the changes in pressure and volume of a gas, when sound waves are propagated through it, are isothermal. The amount of heat produced during compression is lost to the surroundings and similarly, the amount of heat lost during rarefaction is gained from the surroundings, so as to keep the temperature constant. Using

coefficient of isothermal elasticity, i.e., Bi in Eqn. (1), Newton’s formula becomes: …(2)
\(v=\sqrt{\frac{B_{i}}{\rho}}\)
Calculation of B_i:
Consider a certain mass of the gas.
Let P = initial pressure of the gas, V= initial volume of the gas. Under isothermal conditions PV= constant
Differentiating both sides, we get
PdV + VdP = 0
PdV = -VdP
\(P=-\frac{V d P}{d V}=-\frac{d P}{\frac{d V}{V}}=B_{i}\) (by definition)
Substituting this value in eqn. (2), we obtain
\(v=\sqrt{\frac{P}{\rho}}\)

Question 3.
Write the characteristics of progressive waves.
Answer:
Characteristics of progressive wave are as follows :

• A harmonic (progressive) wave is a disturbance which travels through a medium with a definite velocity, without change in shape.
• Energy is transferred along the direction of propagation of wave, but there is no net transport of the material of the medium.
• The velocity of an oscillating particle is different on different states of oscillation, being maximum at the mean position and momentarily zero at the extreme positions. The velocity of propagation of the wave is however, constant, depending on the nature of the medium.
• At any instant, the phase of the oscillation varies from particle to particle, because each particle starts oscillating a little later than the previous particle.
• For particles separated by an integral multiple of wavelength λ, the displacement, velocity and acceleration of the particle have the same values at any instant.

Question 4.
Write characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

• These waves do not advance in the medium but remain steady at its place. This is why these waves are called stationary waves.
• No energy is transferred by these waves because the flow of energy at any point of the medium in one direction due to incident wave is the same as the flow of energy at that point in opposite direction due to reflected wave.
• To produce these waves, the presence of bounded medium is necessary because the wave travelling in such a medium, after being reflected from the boundary, produce an identical wave moving in opposite direction. As a result of superposition of incident and reflected wave, the stationary wave is produced.
  The stationary waves cannot be produced in vacuum or in unlimited medium.
• In the waves, certain points in the medium are always at rest, i.e., their displacement remain zero. These points are called nodes. These points are situated at equal distances. In case of longitudinal stationary waves, change in pressure and in density is maximum at nodes as compared to other points.
• The displacement of the point in between two consecutive nodes is maximum as compared to other points. These points are called antinodes. In longitudinal stationary waves, there is no change in pressure and in density at antinodes.
• The distance between two consecutive nodes or between two consecutive antinodes is λ / 2. The distance between a node and its neighbouring antinode is λ / 4.
• Except at nodes, all the particles of the medium vibrates but the amplitude of vibration of the particles are different. The amplitude of the vibration is zero at the nodes and maximum at the antinodes.
• All the particles between two consecutive nodes vibrate in the same phase, i.e., they reach simultaneously through their equilibrium positions.
• At any instant, the particles of both the sides of a node are in mutually opposite phases but the particles of both the sides of an antinode are in the same phase.
• All the particles of medium pass through their mean positions simultaneously twice in each period.

Question 5.
Give difference between progressive waves and stationary waves.
Answer:
Comparison of Progressive and Stationary waves

Question 6.
What are beats? What are the necessary conditions to obtain beats?
Answer:
Beats: When two waves with slight difference in frequencies travel in a medium simultaneously in the same direction then as a result of their superposition, the intensity of sound at any point increase and decreases alternately. This successive rise and fall in the intensity of sound is called beats.

Conditions:

• Both the waves should travel along the same straight line, in the same medium, with same velocity.
• These should be slight difference in their frequencies.
• The amplitudes of both the waves should be nearly equal.

Question 7.
Derive Newton’s formula for velocity of sound. Point our error in Newton’s formula and hence discuss Laplace correction.
Answer:
Newton assumed that sound travels through air under isothermal conditions and for elastic medium velocity of longitudinal waves,
\(v=\sqrt{\frac{E}{\rho}}\)
∴ PV = constant under isothermal condition.
On differentiating, PdV+ VdP = 0
or
PdV = -VdP
\(P=\frac{-d P}{d V / V}=\frac{\text { stress }}{\text { strain }}=E\)
∴ \(v=\sqrt{\frac{P}{\rho}}\) [from eqn. (1)]
This is Newton’s formula for the velocity of sound in air, which given 280 m/s whereas actual value is 332 m/s.
Laplace correction : Laplace suggested that sound waves travel under adiabatic conditions because:
(i) When sound waves travels in air, changes in volume and pressure takes place rapidly.
(ii) ( Air is a bad conductor of heat. Thus, PVγ = constant
On differentiation Pd(Vγ) + VγdP = 0
or
PγVγ-1 dV + VγdP = 0
or
\(\gamma P=\frac{-V d P}{d V}=\frac{-d P}{d V / V}\)

Hence, according to Laplace, the velocity of sound is \(v=\sqrt{\frac{\gamma P}{\rho}}\) (for air, γ =1.4)
∴ \(v=\sqrt{1 \cdot 4} \times 280=331 \cdot 3 \mathrm{~m} / \mathrm{s}\)

Question 8.
Derive the equation of stationary waves when it is reflected by rigid boundary.
Answer:
When a progressive wave is reflected by a rigid surface, then phase difference of n is produced.
Let the wave is travelling in +ve direction along the Y-axis,
Equation of wave will be
\(y_{1}=a \sin \frac{2 \pi}{\lambda}(v t-x)\) ……….. (1)
Where, a = Amplitude, y =Velocity and λ Wavelength.

When this wave is reflected back, then it will travel along -ve direction of X-axis. ……….. (2)
\(y_{2}=-a \sin \frac{2 \pi}{\lambda}(v t+x)\)
Let the two waves superimpose and stationary wave is formed.
∴By the principle of superposition.
y = y₁ + y₂
or

This is the required equation.

Question 9.
Derive the equation of stationary wave, when reflected by free boundry.
Answer:
Let the incident wave is
\(y_{1}=a \sin \frac{2 \pi}{\lambda}(v t-x)\)
∴ Reflected wave, as the phase does not change
\(y_{2}=a \sin \frac{2 \pi}{\lambda}(v t+x)\)
∴ By the principle of superposition,

Question 10.
Explain beats period and frequency of beats.
Answer:
One rise and one fall of sound form one beat. The time interval between two consecutive intense sounds is called ‘beat period’.
The number of times the intensity of sound rises and falls in one second is called ‘beat frequency’.
Let the frequencies are υ1 and υ2
∴ Beat frequency = υ1 – υ2
and Beat period = \(\frac{1}{v_{1}-v_{2}}\).

Question 11.
Differentiate interference and beats.
Answer:
Distinction between Interference and Beats :

Question 12.
Use the formula \(v=\sqrt{\frac{\gamma P}{\rho}}\)
to explain why the speed of sound in air:
(a) Is independent of pressure,
(b) Increases with temperature,
(c) Increases with humidity.
Answer:
Given : \(v=\sqrt{\frac{\gamma P}{\rho}}\)
By gas equation PV = RT
or
\(P=\frac{R T}{V}\)
∴ \(v=\sqrt{\frac{\gamma R T}{V \cdot \rho}}\)
But V, p = M (molecular weight of gas)
\(v=\sqrt{\frac{\gamma R T}{M}}\)
(a) In the eqn. (1) P does not appear so the speed is independent of pressure at constant temperature.

(b) If y, R and M are constant, then,
\(v \propto \sqrt{T}\)
i.e., the velocity (speed) of sound is directly proportional to the square root of absolute temperature.
∴ With increase in temperature the speed of sound increases.

(c) From the formula \(v=\sqrt{\frac{\gamma P}{\rho}}\) , it is clear that
\(v \propto \frac{1}{\sqrt{\rho}}\)
The density of humid air is less than that of dry air therefore with increase in humidity the speed of sound increases.

Question 13.
What are harmonics? Which harmonics can be produced in a stretched string?
Answer:
The harmonics are those tones, whose frequency is integral multiple of that of fundamental tone. The harmonics whose frequencies are even multiple of that of fundamental tone, are called even harmonics and the harmonics whose frequencies are odd multiple of that of fundamental tone are called odd harmonics.

In a stretched string even and odd both harmonics can be produced. In Fig. the first, second and third harmonics are shown.

Question 14.
How do you determine the frequency of a tuning fork with the help of beats?
Answer:
Determine frequencies of a tuning fork: Suppose that we have to determine the frequency of a tuning fork. For this purpose, we take such a fork of known frequency
υ₂ whose frequency υ₂ differ slightly from unknown frequency υ₁. Now, both the tuning forks are sounded together and the beats are heard. Let the number of beats heard per second be x. Then,
υ₁ = υ₂ +x or υx = υ₂-x

To find which value is correct out of two values, a little wax is attached to the prong of the tuning fork of unknown frequency υ₁.
Again, the two forks are sounded together and beats are heard.
If number of beats per second increase, then the frequency of the first tuning fork will be (υ₂ – ×) because according to
(υ₁ – υ₂) = ×, × increases as υ₁ decreases.
If number of beats per second decreases, then the frequency of the first tuning fork will be (υ₂ + x) because according to
(υ₁ – υ₂) = -x, x decreases as υ₁ decreases.

Question 15.
Prove that the frequency of fundamental frequency of open organ pipe is double of the fundamental frequency of closed organ pipe of same length.
Or
If the frequency of fundamental tone of an open organ pipe is «, and that of closed organ pipe is n₂, then prove that n₁ = 2n₂.
Answer:
Let the length of organ pipe is l. For the fundamental frequency of open organ pipe.

\(l=\frac{\lambda_{1}}{2}\)
Where, λ1 is the wavelength
λ1 = 2l
If the fundmental frequency is n₁,then

Waves Class 11 Important Questions Long Answer Type

Question 1.
Prove that beats obtained by two sound sources per second is equal to the differences of frequencies of the sound source.
Answer:
When two sources of sound of nearly equal frequencies are sounded together, the amplitude of resultant wave due to superposition of waves at a point is in space. Some¬times the amplitude is maximum and sometimes, it is minimum.

As intensity is directly proportional to the square of amplitude, the intensity also varies periodically with time from maximum to minimum and to maximum again. Let two waves be,
\(y_{1}=a \sin 2 \pi v_{1} t\) and \(y_{2}=a \sin 2 \pi v_{2} t \) are superimposed. On superposition, we get a resultant wave as,
y = y₁+y₂

Question 2.
Prove that even and odd harmonics are produced by a stretched string.
Answer:
When a stretched string is plucked, then transverse waves are produced. The wave reflected from both- the ends and produced stationary waves.
∴The wave is reflected from the rigid end, hence the equation of stationazy wave will be
\(y=-2 a \sin \frac{2 \pi x}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\) ……………… (1)

Let one end of string is at x = 0 and another at x =l. where l is the length of string.
Now, at x = 0, from eqn. (1), we get
\(y=-2 a \sin 0^{\circ} \cdot \cos \left(\frac{2 \pi v t}{\lambda}\right)\)
or
y=0
Again, at x = l, from eqn. (1),
\(y=-2 a \sin \frac{2 \pi l}{\lambda} \cos \frac{2 \pi v t}{\lambda}\)
Now, at x =l, y = 0

Now, the mode of vibration will be obtained by putting k = 1, 2, 3,….
First mode of vibration: If k = 1 and the wavelength of stationary wave is λ₁, then from eqn. (2),
λ₁ = 2l
\(l=\frac{\lambda_{1}}{2}\)
Hence, the string vibrates in one segment. Now, if the velocity is and frequency n₁, then
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{2 l}\) ………… (3)

This frequency is called fundamental frequency or first harmonics.

Second mode of vibratioñ : If k =2 and wavelength ‘λ₂, then from eqn. (2),
λ₂ = l
Also, if the frequency is n₂,then
\(n_{2}=\frac{v}{\lambda_{2}}\)
or
\(n_{2}=\frac{v}{l}=\frac{2 v}{2 l}=2 n_{1}\)

Question 3.
Prove that in a dosed organ pipe odd harmonics are produced.
Or
In a closed organ pipe, describe the mode of vibrations and prove that:
(i) The frequency of fundamental note is inversely proportional to the length of the organ pipe.
(ii) Only odd harmonics are produced.
Answer:
Organ pipe is a long cylindrical tube of uniform diameter and made of wood or metal. Sound is produced by this pipe as a result of vibrations in air column inside this. If one end of this tube is closed then it is called closed organ pipe and if both ends are open then it is called open organ pipe.

When we place a source of sound near the open end of a closed organ pipe, a wave travels towards the closed end. This wave, after being reflected by the closed end, returns. A stationary longitudinal wave is produced inside the closed organ pipe as a result of superposition of incident wave and reflected wave. An antinode is always formed at the open end because the air particles are free to vibrate there. A node is always formed at the closed end because at the closed end the air particles are not free to vibrate.

Suppose that the closed end of the closed organ pipe is at x = 0 and open end at x = L, where L is the length of the organ pipe.
\(y=-2 a \sin \frac{2 \pi x}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\)
Case (i): At x = 0, we have y = o
Thus, the wave function vanishes at x = 0.

Case (ii) : At x = L, we have
\(y=-2 a \sin \frac{2 \pi L}{\lambda} \cdot \cos \frac{2 \pi v t}{\lambda}\)
At x =L, the displacement y has its maximum value ,y will be maximum when \(\sin \frac{2 \pi L}{\lambda}\) maximum i.e,

Where, k = 0, 1,2, 3,4,….
\(\lambda=\frac{4 L}{(2 k+1)}\)
Here, k = 0, 1, 2,… correspond to first, second, third,…. mode of vibration.

First mode of vibration: If λ₁ be the wavelength of the wave produced in air column corresponding k = 0, then from eqn. (2), we have
λ₁ = 4L
If n₁ be the frequency of the wave set up, then
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{4 L}\)

It is the tone of minimum frequency which produces in closed end organ pipe. It is called fundamental tone or first harmonic.
Also,
\(n_{1} \propto \frac{1}{L}\)
i.e., Fundamental frequency is inversely proportional to the length of the organ pipe.

Second mode of vibration :
If λ₂ be the wavelength of the wave set up in air column corresponding to k = 1, then from eqn. (2), we have
\(\lambda_{2}=\frac{4 L}{3}\)
If n₂ be the frequency of the wave set up, then

Thus, in this case, the frequency of the tone produced is three times the frequency of fundamental tone. It is called third harmonic or first overtone.

Third mode of vibration : If k = 2 and corresponding wavelength is λ₃, then from eqn. (2), we get
\(\lambda_{3}=\frac{4 L}{5}\)
If n₃ be the frequency of the wave set up, then
\(n_{3}=\frac{v}{\lambda_{3}}=\frac{v}{4 L / 5}=5 \cdot \frac{v}{4 L}\)
or
n₃ = 5n₁
Hence, the fifth harmonics of second overtone is five times of fundamental frequency.
Hence, the harmonics are n₁, 3n₁ 5n₁,….
Thus, odd harmonics are produced.

Question 4.
Prove that in open organ pipe both odd and even harmonics are produced. Ans. When a source of sound is kept at one end of an open organ pipe, the compressions and rarefactions are set up in the pipe. At the open ends antinodes are formed, hence rarefactions will be produced at the open ends and it returns in the form of compression. Thus, incident wave and reflected wave superimpose and stationary wave is formed.
Since, the reflection is, from free boundary,
\(y=2 a \cos \frac{2 \pi x}{\lambda} \cdot \sin \frac{2 \pi v t}{\lambda}\)

Now, the one end of organ pipe is at x = 0 and another end is at x = l.
Putting x = 0, we get
\(y=2 a \cos 0 \cdot \sin \frac{2 \pi v t}{\lambda}=2 a \sin \frac{2 \pi v t}{\lambda}\)
Again, putting x = l, we get
\(y=2 a \cos \frac{2 \pi l}{\lambda} \cdot \sin \frac{2 \pi v t}{\lambda}\)
For y is maximum,
\(\left|\cos \frac{2 \pi l}{\lambda}\right|=1 t\)
or
\(\frac{2 \pi l}{\lambda}=k \pi \) …………. (2)
Where,k= 1,2,3………

First mode of vibration: If k = I and wavelength is λ₁, then from eqn. (2), we get
λ₁ = 2l
Also
\(n_{1}=\frac{v}{\lambda_{1}}=\frac{v}{2 l} \)

n₁ s the least frequency called fundamental frequency or first harmonics.

Second mode of vibration : 1f k = 2 and corresponding wavelength is λ₂.
then
\(\lambda_{2}=\frac{2 l}{2}=l \)
∴ Frequency ,n₂= \(\frac{v}{\lambda_{2}}=\frac{v}{l}=\frac{2 v}{2 l}\)
n₂ = 2n₁
Hence, second harmonic is obtained at double of fundamental frequency.

Third mode of vibration: If k = 3 and corresponding wavelength is λ₃ and frequency n₃, then

Hence, third harmonic is 3 times of fundamental frequency.
Thus, the harmonics are both odd and even multiple of fundamental frequency.

Question 5.
Derive the expression for the apparent frequency, when the source of sound is moving towards a stationary listener.
Answer:
Suppose S is a source moving towards a stationary listener O. Let υ be the real frequency of the source and υ is the velocity of sound wave. Now, the first wave emitted by the source covers a distance υ in 1 sec. If the source is stationary then these υ waves will be
spread in the distance υ. Thus, the length of one wave i.e., the wavelength will be \(\lambda=\frac{u}{v}\)

Now, suppose the sound-source S is moving with velocity us towards the observer. Then the source covers a distance us in one second towards the listener. Therefore, the v waves emitted by the source in 1 second will spread in a distance (υ-υ_s) only. Thus, the wavelength will be shortened. Let the changed wavelength be λ’.
Then, \(\lambda^{\prime}=\frac{u-u_{s}}{v}\) ………….(1)

Thus, the listener will receive waves of wavelength λ’. Hence, the frequency of sound will appear to him changed. If the apparent frequency be υ’, then
\(v^{\prime}=\frac{u}{\lambda^{\prime}}\) (∴ Velocity of sound υ is steady )
Substituting the value of λ’ in eqn. (1), we have
\(\dot{v^{\prime}}=\frac{u}{\left(u-u_{s}\right) / v}\)
or
\(v^{\prime}=\left(\frac{u}{u-u_{s}}\right) v\) …………. (2)
As (υ-υ_s)< υ,therefore υ’ > υ i.e., the apparent frequency is greater than the real frequency.

If the source S is going away from the listener O, then the wavelength will be in-creased and its value will be given by
\(v^{\prime}=\left(\frac{u}{u+u_{s}}\right) v\)

As (υ + υ_s) > υ_s, therefore, υ‘ < υ i.e., the apparent frequency is less than the real frequency.

Question 6.
Derive the expression for the apparent frequency, when the listener is moving toward stationary source.
Answer:
Suppose, S is a sound source of frequency u and u is the velocity of sound. Again suppose that O is a listener who is coming towards the source with the velocity υ₀. If the listener were stationary then he would have received u waves in 1 second. Therefore, the real wavelength is given by
λ = \(\frac{u}{v}\)

But, the listener himself is moving towards the source with a velocity υ₀ that is, he covers a distance υ₀ toward the source in 1 second. Hence the listener, in addition to υ waves, also receives υ₀/ λ waves contained in the distance υ₀. Therefore, the total waves received by the listener in 1 second i.e., the apparent frequency of the sound is

As (υ+ υ₀) > u, hence υ‘ > υ i.e., apparent frequency will be greater than the actual frequency.

Question 7.
Derive the expression for the apparent frequency heared by a listener, when the source and listener both are moving in the same direction.
Answer:
Let the source and listener are moving with the velocities υ_s and υ₀ respectively in the same direction as shown in the figure.
If the source is moving only.
∴ The frequency heard by the listener will be \(n_{1}=\left(\frac{u}{u-u_{s}}\right) n\)
But, listener is moving away from the source

∴ Apparent frequency, \(n^{\prime}=\left(\frac{u-u_{o}}{u}\right) n_{1}\)

Putting the value of n₁, we get
\(n^{\prime}=\left(\frac{u-u_{o}}{u}\right) \times \frac{u}{u-u_{s}} \cdot n\)
or
\(n^{\prime}=\left(\frac{u-u_{o}}{u-u_{s}}\right) n\)
This is the required expression.

Waves Class 11 Important Numerical Questions

Question 1.
The speed of a wave ¡n a medium is 960m/s. If 3600 waves are passing through a point in the medium in 1 minute, then calculate the wavelength.
Solution:
Given: Speed of wave ν = 960 m/s
Frequency of wave n = \(\frac{3600}{60}\) =60/ second
Wavelength λ = \(\frac{v}{n}=\frac{960}{60}\)
or
λ = 16m.

Question 2.
The speed of sound in air at 14°C is 340 m/s. What is the speed of sound in air at 157.5°C when pressure becomes two times of its initial value.
Solution:
The change in pressure has no effect on speed of sound.
Given: t₁= 14°C,ν₁= 340m/s,t₂ = 1575°C,ν₂ =?

Question 3.
The distance between two consecutive nodes of a stationary wave is 25 cm. If the velocity ¡s 300 mc’, then calculate the frequency.
Solution:
Given: λ /2=25cm
or
λ =50cm = 50 ×\(10^{-2}\)
or
ν = 300 ms^-1
ν = nλ
or
300 = n ×50 ×\(10^{-2}\)
or
\(n=\frac{300}{50 \times 10^{-2}}=6 \times 10^{2}\) = Hz

Question 4.
The frequency of a whistle is 256 Hz. It is moving towards a stationary listener with a velocity of \(\frac{v}{3}\) , where v is the velocity of sound waves, then calculate the apparent frequency.
Solution :
Given : n = 256 Hz, \(v_{s}=\frac{v}{3}\)

Question 5.
An observer is moving towards a stationary source of frequency n. The frequency heard by the observer is 2n. If the velocity of sound is 332 ms^-1, then find the velocity of observer.
Solution:
Given : n’ = 2n, ν = 332 ms^-1.

Question 6.
The frequencies of two sources are 512 Hz and 516 Hz. Find the time inter-val between two consecutive beats.
Solution:
Given : n₁ =512 Hz, n₂ =516 Hz.
∴ Beat period = \(\frac{1}{n_{2}-n_{1}}\)
\(=\frac{1}{516-512}=\frac{1}{4}=0 \cdot 25 \mathrm{sec}\).

Question 7.
Two sound waves of wavelength 1 metre and 1.01 metre produce 40 beats in 10 seconds in a gas. Find the speed of sound in gas.
Solution:
Beats per second \(=\frac{40}{10}=4\)
If ν is the velocity of sound in gas, then
Frequency of first wave υ₁ = \(\frac{v}{1}\)
Frequency of second wave υ₂ \(\frac{v}{1.01}\)

Question 8.
Two tuning forks produce 6 beats when they are sounded together. The frequency of one tuning fork is 288 Hz. Now, the second tuning fork is loaded by little wax, then the beats are not heard. Find the frequency of another tuning fork.
Solution:
Let the frequency of second tuning fork is n.
∴ n = 288 + 6 or 288 – 6
or
n = 294 or 282 Hz
Since, second fork is loaded hence its frequency will decrease.
Since, no beat is heard.
Hence, the frequency of second tuning fork = 294 Hz.

Question 9.
Two sources of sound of equal frequencies are kept at a distance 100 m apart. A listener is moving between the sources and hears 4 beats. If the distance between the sources becomes 400 m, what will be the beats heard ?
Answer:
Spl. The frequency of beats depends upon the frequencies of the sources, not the distance. Hence, 4 beats will be heard.

Question 10.
Frequency of open end organ pipe is 256 Hz. What will be the frequency of same length closed end organ pipe ?
Solution:
Frequency of l length open end organ pipe
\(n=\frac{v}{2 l}\)
Frequency of closed ended organ pipe
\(n^{\prime}=\frac{v}{4 l}\)
or
\(\frac{n}{n^{\prime}}=\frac{v / 2 l}{v / 4 l}=\frac{2}{1}\)
Given, n = 256
∴ \(\frac{256}{n^{\prime}}=\frac{2}{1}\)
or
\(n^{\prime}=\frac{256}{2}=128 \mathrm{~Hz}\).

Question 11.
A string of mass 2-50 kg is under a tension of 200 N. The length of the stretched string is 20-0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end ? (NCERT)
Solution:
Given : M= 2.50 kg, l = 20m, T= 200N
Mass per unit length m = \(\frac{M}{l}=\frac{2 \cdot 5}{20}\) = 0.125kg/m
∴ \(v=\sqrt{\frac{T}{m}}\)
or
\(v=\sqrt{\frac{200}{0 \cdot 125}}=\sqrt{1600}\) = 40 m/sec.
Time taken by distance (jerk) be
\(t=\frac{l}{v}\)
or
(∴ Velocity = \(\frac{\text { displacement }}{\text { time }}\) )
\(t=\frac{20}{40}\) = 0.5 sec.

Question 12.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the spHsh heard at the top given that the speed of sound in air is 340 ms^-1? (g = 9.8 ms 2) (NCERT)
Solution:
Given : Height of tower h = 300 m Speed of sound v = 340 ms^-1, initial velocity of stone u = 0 Let the time taken by stone to base of tower is t₁ Then by equation of motion
\(h=u t+\frac{1}{2} g t^{2}\)
300 = \(0 \times t_{1}+\frac{1}{2} \times 9 \cdot 8 t_{1}^{2}\)
or
\(4 \cdot 9 t_{1}^{2}=300\)
or
\(t_{1}=\sqrt{\frac{300}{4 \cdot 9}}=\) = 7.82 sec.

If the time taken to reach the sound up to top of tower is t₂, then \(t_{2}=\frac{h}{v}=\frac{300}{340}\) = 0.885
∴ Total time t = t₁+t₂.
= 7.82+0.885=8.705 = 8.7 sec .

Question 13.
A bat emits ultrasonic sound of frequency 1,000 kHz in air. 1f the sound meets a water surface. What is the wavelength of (a) the reflected sound (b) the transmiffed sound? Speed of sound in air is 340 ms^-1 and in water 1486 ms. (NCERT)
Solution:
Given :
n = 1,000 kHz =I000x10³Hz =10⁶Hz
Speed of sound in air ν_a = 340ms^-1 and that in water νw = 1486ms^-1 
(a) Reflected sound propagate in air

(b) Transmitted sound propagate in water

Waves Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Sound of which frequency can be heard by human beings :
(a) 5 vib /sec
(b) 500 vib /sec
(c) 27000 vib / sec
(d) 50, 000 vib /sec.
Answer:
(b) 500 vib /sec

Question 2.
Astronaut cannot hear sound of his partner in moon, because :
(a) Frequency produced is more than audible frequency
(b) No medium is there for propagation of sound
(c) Temperature at night is very low
(d) There are lots of reason in moon.
Answer:
(b) No medium is there for propagation of sound

Question 3.
Two sources of frequency f₁ and f₂ will have beat frequency :
(a) \(\text { (a) } \sqrt{f_{2}}-\sqrt{f_{1}}\)
(b) f₁-f₂
(C) \(2\left(f_{1}-f_{2}\right)\)
(d) \(\sqrt{\left(f_{1}^{2}-f_{2}^{2}\right)}\)
Answer:
(b) f₁-f₂

Question 4.
The law relating sound between source and observer is :
(a) Doppler’s law
(b) Huygens law
(c) Newton’s law
(d) Galilean’s law.
Answer:
(a) Doppler’s law

Question 5.
The effect of beats due to two waves is because of:
(a) Refraction
(b) Reflection
(c) Dispersion
(d) Interference.
Answer:
(d) Interference.

Question 6.
As a utensil is filled with water, its frequency :
(a) Increases
(b) Decreases
(c) Remains same
(d) None of the above.
Answer:
(a) Increases

Question 7.
Waves transport from one end to another :
(a) Energy
(b) Amplitude
(c) Wavelength
(d) Matter.
Answer:
(a) Energy

Question 8.
Energy is not carried by :
(a) Transverse progressive waves
(b) Longitudinal progressive waves
(c) Stationary waves
(d) Electromagnetic waves.
Answer:
(c) Stationary waves

Question 9.
Velocity of sound in air is 350 m/s, frequency in open organ pipe of 50 cm length is:
(a) 175 Hz
(b) 350 Hz
(c) 700 Hz
(d) 50 Hz.
Answer:
(b) 350 Hz

Question 10.
y = 0.15 sin 5x cos 3001 represent an equation of stationary wave. Its wavelength will be :
(a) Zero
(b) 1.25m
(c) 2.512 m
(d) None of these.
Answer:
(b) 1.25m

2. Fill in the blanks:

1. The velocity of sound at 0°C is …………….. .
Answer:
332 m/s

2. …………….. waves require a physical medium for their propagation.
Answer:
Mechanical

3. …………….. waves do not require any medium for their propagation.
Answer:
Electromagnetic

4. In transverse waves, the distance between any two consecutive crest or trough is called …………….. .
Answer:
Wavelength

5. After one …………….. the phase of a particle is same as in the initial moment.
Answer:
Time period

6. Sound waves cannot propagate through …………….. .
Answer:
Vacuum.

.

3. Match the following:

Answer:
1. (d) Frequency × Wavelength
2. (a) \(\sqrt{\frac{\gamma P}{D}}\)
3. (b) \(\frac{2 \pi}{\lambda}\) × path difference,
4.  (e)\(\vec{y}=\overrightarrow{y_{1}}+\overrightarrow{y_{2}}+\ldots+\overrightarrow{y_{n}}\)
5.  (c) Sound navigation and Ranging

4. Write true or false:

1. When two sound sources travelling perpendicularly. Doppler effect will implement.
Answer:
False

2. Propagation of mechanical wave in medium is possible due to elasticity and inertia.
Answer:
True

3. Propagation of wave in air is a isothermal changes.
Answer:
False

4. If the velocity of listener is more than velocity of sound then Doppler’s effect will not implement.
Answer:
True

5. Velocity of sound in moisture is less than velocity of sound in air.
Answer:
False

6. A person is standing at a point and a car is coming toward him blowing horn. The frequency of horn will be less than original frequency.
Answer:
False

7. Stationary wave does not carry energy.
Answer:
True

8. Beats is produced when frequency of two sound source are same.
Answer:
True

9. In a resonance air column, always transverse non- progressive wave is produced.
Answer:
False

10. Expansion of universe is observed by Doppler’s effect. ‘
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 1 Physical World which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 1 Physical World

Physical World Class 11 Important Questions Very Short Answer Type

Question 1.
Define science.
Answer:
The systematic knowledge based on observation, experiment and scientific logic, is called science.

Question 2.
Define physics.
Answer:
Physics is that branch of science in which matter, energy and their interactions are studied.

Question 3.
What are the different branches in which physics is divided?
Answer:

1. Classical physics : All the discoveries before 1900 A.D. is included in it. It deals with mechanics, light, heat, sound, electricity and magnetism.
2. Modern physics: All the discoveries after 1900 A.D. is included in it. It deals with electron, molecule, nucleus, quantum mechanics etc.

Question 4.
Name any four physicists and their discoveries.
Answer:

Question 5
Name two devices used in daily life and their working principle.
Answer:

Question 6.
What are the different branches of science?
Answer:
The different branches of science are Physics, Chemistry, Biology, Botony, Maths, Medical science etc.

Question 7.
What do you mean by matter?
Answer:
The substance which occupy space is called matter. It has mass for example : Stone, Iron etc.

Question 8.
What do you mean by energy?
Answer:
Capacity of doing work is called energy. Without energy no work can be perform.

Question 9.
What are different forms of energy?
Answer:
The different forms of energy are: Mechanical energy, Heat energy, Light energy, Electrical energy, Chemical energy, Nuclear energy etc.

Question 10.
Name any three recent inventions in physics.
Answer:
The recent inventions are LASER, Computer and Super conductivity.

Question 11.
Which scientist discovered Mass-energy relation?
Answer:
Mass-energy relation E = mc² was discovered by Einstein.

Question 12.
Write the name of the force which is always attractive?
Answer:
Gravitational force.

Question 13.
Write the following forces in order of their increasing magnitude. Electromagnetic force, Nuclear force, Gravitational force.
Answer:
Gravitational force, Electromagnetic force and Nuclear force.

Question 14.
Write the name of that strong force which is responsible for stability of nucleus.
Answer:
Nuclear force.

Question 15.
What is the range of weakest force?
Answer:
10^-16m.

Question 16.
Write the name of the field-produce due to :
(i) Static charge,
(ii) Moving charge.
Answer:
(i) Static charge : Only electrical field.
(ii) Moving charge : Magnetic field.

Physical World Class 11 Important Questions Short Answer Type

Question 1.
What do you mean by systematic observation?
Answer:
Systematic observations: To work out on a problem, the scientists collected all relevant data concern to the problem, by systematic experiments and observations.

Question 2.
How hypothesis is formulated?
Answer:
Formulation of hypothesis: On the basis of observations working law is formed called hypothesis.

Question 3.
How hypothesis is tested?
Answer:
Testing of hypothesis: On the basis of hypothesis and predictions, conclusions are drawn and it is verified by new experiments. This is called testing of hypothesis.

Question 4.
How theory is established?
Answer:
Final theory: If the conclusions and predictions are verified by the experiments, then the hypothesis is accepted as final theory, otherwise corrections are made and verified again. The phenomena are repeated till the final theory is achieved.

Question 5.
What is theory? When does it need modification? Explain it with one example.
Answer:
The hypothesis, tested by experiments is called theory.
Sometimes a new measurement shows difference between the existing theory and observations then it needs to be modified.
Example: Huygens thought that light waves are longitudinal in nature. But polarization of light could not be explained on the basis of this thought. Therefore, Fresnel assumed that the light waves are transverse in nature.

Question 6.
Why physics is called as basic science?
Answer:
Physics is called basic science because its seek out and understand the basic laws of nature upon which all physical phenomenon is depended. It has brought to us deeper and deeper levels of understanding nature.

Question 7.
What is scientific method? Explain the steps of scientific method.
Answer:
Scientific method: The methods by which a scientist works to establish theories
and laws are called scientific method.
The steps of scientific method are :

• Systematic observations,
• Formulation of hypothesis,
• Testing of hypothesis,
• Final theory.

Physical World Class 11 Important Questions Long Answer Type

Question 1.
What is the relation of physics to other branches of science?
Answer:
The different branches of science in relation to physics are given below :
1. Physics in relation to Mathematics:
The knowledge of mathematics is necessary to understand the laws and theories of physics which is a powerful tool. Without mathematics expansion of physics is impossible. Radio, television, computer, satellite, telecommunication etc. are the result of combined applications of physics and mathematics.

2. Physics in relation to chemistry:
Physics plays an important role in the development of chemistry. The structure of atoms described by physics, the arrangement of atoms in periodic table, the nature of valence electrons and chemical bonds. The X-rays and neutron diffractions are the results of research in physics.

3. Physics in relation to biological science :
The technology of physics are used in Biology. The optical microscopes developed by physics are extensively used in biological science. The electron microscopes which is nearly a million times powerful than optical microscopes used in the study of the structure of the cells with the help of X-rays position of bone fracture is found and also the presence of foreign materials hidden in the body. X-rays are also used in the treatment of cancer.

Question 2.
Explain the influence of Physics on the society.
Answer:
The Physics has influenced the society in two ways:
(i) Living standard :
Physics has developed the technology and hence the inventions of machines, devices and instruments give comfort and protection to human life.
By the television and radio transmission, has made possible to see the happenings live from any part of the world. Computer can calculate very complicated calculations within no time. Thus, the efficiency of men have increased.

(ii) Effect on the thinking of human beings: The contribution of Physics in the field of education, medical sciences etc., developed the scientific aspect and therefore the narrow thinking of caste and creed, religion and superstitions are removed from the society.

Question 3.
Name few scientists and their contribution ¡n physics.
Answer:
Scientists and their contribution in physics:

Physical World Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Which among the following is called basic science :
(a) Physics
(b) Chemistry
(c) Maths
(d) Biology
Answer:
(a) Physics

Question 2.
The oldest branch of physics is :
(a)Thermodynamics
(b) Optics
(c) Mechanics
(d) Sound.
Answer:
(c) Mechanics

Question 3.
The weakest force of the nature is :
(a) Electrical force
(b) Magnetic force
(c) Nuclear force
(d) Gravitational force.
Answer:
(d) Gravitational force.

Question 4.
Rocket propulsion is based on the law :
(a) Faraday’s law
(b) Newton’s law
(c) Galileo’s law
(d) Bernoulli ‘s theorem.
Answer:
(b) Newton’s law

Question 5.
Working principle of aeroplane is based on …………………….:
(a) Laws-of thermodynamics
(b) Faraday’s laws of electromagnetic induction
(c) Bernoulli’s theorem
(d) Newton’s laws of motion.
Answer:
(c) Bernoulli’s theorem

Question 6.
Quantum mechanics was discovered by the scientist:
(a) Michael Faraday
(b) Chandrashekhar Venkat Raman
(c) Satyendranath Bose
(d) Somnath Saha.
Answer:
(c) Satyendranath Bose

Question 7.
Planetory laws of motion was propounded by :
(a) Max-plank
(b) Newton
(c) Kepler
(d) Faraday.
Answer:
(c) Kepler

Question 8.
Who gave the electromagnetic theory of waves :
(a) Niel-Bohr
(b) James dark Maxwell
(c) Michael Faraday
(d) Johans kepler.
Answer:
(c) Michael Faraday

2. Fill in the blanks:

1. Theory of relativity was given by ……………..
Answer:
Albert Einstein

2. Neutron was discovered by…………………..
Answer:
James Chadwick

3. Abdus Salam is from the country ………………….
Answer:
Pakistan

4. Computer is based on the principle of ………………….
Answer:
Digital logic of electronic circuit

5. Sir Issac Newton belonged to the country …………………
Answer:
England

3. Match the following:
I.

Answer:
1. (e) Laws of gravitation.
2. (f) Explanation of photoelectric effect
3 (a) Principle of electromagnetic waves
4 (c) Dual nature of light
5 (d) Scattering of light
6. (b) Cosmic ray

II.

Answer:
1. (c) Conversion of gravitational energy into electrical energy
2. (f) Bernoulli’s principle.
3. (e) Faraday’s laws of electromagnetic induction
4. (a) Laws of motion
5. (b) Propogation of electromagnetic waves
6. (d) Digital logic of electronic circuits

III.

Answer:
1. (e) England
2. (d) Italy
3. (b) Germany
4. (a) Scotland
5. (f) Greece.
6. (c) India

4. Write true or false:

1. Nuclear reactor is based on the concept that a slow moving neutron splits uranium.
Answer:
True

2. Mesons were discovered by Clark David Anderson.
Answer:
False

3. Newton had given the corposcular theory of light.
Answer:
True

4. Tn 1921, Albert Einstein got the noble prize for theory of relativity.
Answer:
False

5. Fast moving particles are studied using the theory of relativity.
Answer:
True

5. Answer in one word:

1. What is Einsteins mass energy equivalence equation?
Answer:
E = mc²

2. Substance which has mass is called?
Answer:
Matter

3. Rocket propulsion is based on which law?
Answer:
Newton’s third law of motion

4. The laws of photoelectric effect is explained by which equation?
Answer:
Einsteins photoelectric of equation

5. Who gave the principle of uncertainty?
Answer:
Heisenberg (Germany)

Students get through the MP Board Class 11th Biology Important Questions Chapter 11 Transport in Plants which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 11 Transport in Plants

Transport in Plants Class 11 Important Questions Very Short Answer Type

Question 1.
Which factors effect rate of diffusion?
Answer:
Factors affecting the rate of diffusion are :

• Concentration of the medium,
• Solubility of solutes,
• Temperature,
• Size and mass of diffusing particles,
• Permeability of the separating membrane.

Question 2.
What is Porin? Give its role in diffusion.
Answer:
Porin is a type of protein, which forms pore in the outer membrane of plastids, mitochondria and bacteria, so that protein molecules can pass through them.

Question 3.
Which part of the roots absorb water from the soil?
Answer:
Root hair zone of the roots absorb water from the soil.

Question 4.
Which soil water is absorbed by the roots?
Answer:
Roots absorb capillary water.

Question 5.
What is Guttation?
Answer:
The process of loss of water in the form of water droplets through the hydath- odes of herb plants is called as Guttation.

Question 6.
What is semipermeable membrane?
Answer:
Membranes which allow only the solvent to pass through them not the solute particles are called as semipermeable membrane

Question 7.
Name the tissue helps for conduction of water in plants.
Answer:
Xylem helps for conduction of water and minerals in plants.

Question 8.
What is Transpiration pull?
Answer:
Due to continuous loss of water vapour through stomata during transpiration process tension is created on unbroken water column in the Xylem, which remain continuous due to cohesive force of water molecules, thus water ascend in the plants in upward direction through Xylem. This is called as Transpiration pull.

Question 9.
What is Wilting?
Answer:
When rate of transpiration is very high and rate of absorption fails to keep balance with the loss of water, a kind of water deficit takes place in plants. This condition is called as Wilting. Plant cells become flaccid in this condition.

Question 10.
What do you mean by antitranspiration substance?
Answer:
Substance which reduces rate of transpiration without effecting gaseous exchange in plants is called as Antitranspiration substance e.g. Phenyl mercuric acetate.

Question 11.
What is ascent of sap?
Answer:
The upward movement of water from the root towards the top of the plant is known as ascent of sap. Sap is water with dissolved ingredients (Minerals). Ascent of sap takes place through tracheary elements of Xylem.

Question 12.
Define DPD.
Answer:
The amount by which the diffusion pressure of the solution is lower than that of its solvents at the same temperature and atmospheric pressure is called as diffusion pressure deficit (DPD). DPD is the index of absorbing power of a solution, it is also called as suction pressure.

Question 13.
Define osmotic pressure.
Answer:
A pressure which is developed in a solution when it is separated from its solvent by a semipermeable membrane due to presence of dissolved solutes in it. This is called osmotic pressure (OP).

Transport in Plants Class 11 Important Questions Short Answer Type

Question 1.
Explain role of protein pump during active transport in plants.
Answer:
Absorption of substances which occurs with expenditure of energy through plasma membrane is called as active absorption.

Cell is surrounded by a lipo-protein membrane, which is impermeable to free ions. But some carrier compounds (protein in nature) acts as carrier to carry ions by binding with ions. Carrier combines with ions to form carrier-ion complex, which can move across the membrane with the expenditure of metabolic energy (ATP) of respiration. On the inner surface of the membrane this complex breaks releasing ions into the cell while the carrier goes back to the outer surface to pick fresh ion.

Question 2.
Why water potential of pure water is highest? Explain.
Answer:
Kinetic energy of water molecule is more. If concentration of water is more in any system then its kinetic energy will be more, thus its water potential or osmotic potential is high. Addition of solvent to pure water lowers its osmotic potential. It is represented by Greek symbol psi (Ψ) and its value is equal to osmotic pressure. Its unit is Pascal (Pa).

Question 3.
Explain Water Potential. What are the various factors which effect it? Explain relationship between water potential, solute potential and pressure potential.
Answer:
Water travels from high energy area to low energy area. Kinetic energy of pure water is highest. Water move from one place to other place according to concentration gradient or potential gradient. Water potential of pure water is highest, its value is equal to zero. Addition of solute to pure water lowers its free energy therefore reduces value of water potential, thus its value become negative (-ve).
The differences between the free energy of water molecule is pure and any other system is called as water potential. It is represented by (Ψ_w).
The amount of solute present in water is called solute potential (Ψ_s).
Pressure potential is equal to osmotic pressure, when cell is in turgid condition. Pressure potential is represented by sign Ψ_p.
The water potential (Ψ_w) is actually the sum of other two potentials.
Ψ_w = Ψ_s + Ψ_p

Question 4.
What will happen if pressure more than atmospheric pressure is applied to the pure water or solvent?
Answer:
If pressure more than atmospheric pressure is applied in pure water then its water potential increases. Its value is equal to the amount of water pumped from one place to the other place. When water enters into a cell, it apply pressure on the cell wall. Cell become turgid. It increases pressure potential. Generally, water potential is positive. Water potential is represented by sign Ψ_p.

Question 5.
How does mycorrhiza helpful to plant in absorption of water and minerals?
Answer:
There are fungi which live in intimate association with some other unlike living organisms, both being beneficial to each other, the process being called symbiosis. A symbiotic, non-pathogenic or weakly pathogenic association of various fungi and bryophytes, pteridophytes or flowering plants is known as mycorrhiza. Most of these associations are not specific. A fungus can associate with many plants and a plant can associate with many fungi. The fungus may grow within the root surface or may grow around the root surface.

The fungal partners are usually those that can breakdown the plant cell wall only in a limited way. The fungus takes its nutrition from the damaged portion and grow, but the growth of the fungus within the plant does not damage it. The fungus remains in a restricted area. The plants are benefitted as the fungi spread in soil, absorbs water, nitrogen and other minerals from the soil. The fungi also produce various growth promoting substances and antimicrobial substances. The fungi use carbohydrates produced by other plant partner.

Question 6.
Give role of endodermis in absorption of minerals in plants.
Answer:
Many transport proteins are found embedded in the plasma membrane of endodermal cells of the roots of the plants, which allow only few particles to pass through them.
Transport protein of endodermal cells of roots of the plants can bind to selected solute particles only based on the shape and utility of the mineral ions for the plants. They also control quantity and type of ions to be absorbed.

Question 7.
Derive relationship between osmotic pressure, turgor pressure and suction pressure or diffusion pressure deficit.
Answer:
To understand relationship between OP, TP, and DPD, we must see first what are these.
Osmotic pressure (OP) :
A pressure which is developed in a solution, when it is separated from its solvent by a semipermeable membrane.

Turgor pressure (TP):
When a cell is placed in distilled water, it absorbs water by the process of osmosis and become turgid and the contents of cell exert a pressure against cell wall which is known as turgor pressure.

Diffusion pressure deficit (DPD) :
The amount by which the diffusion pressure of the solution is lower than that of its solvent at the same temperature and atmospheric pressure is called as Diffusion pressure deficit or Suction pressure.
Relationship between Diffusion Pressure Deficit, Osmotic Pressure and Turgor Pressure : When a cell absorbs water the turgor pressure of cell is increased but cell wall also applies an equal but opposite pressure called wall pressure (WP) against the turgor pressure (TP). Therefore, actual force responsible for entry of water into the cell will be OP-TP (or WP). The relationship between DPD, OP, TP and WP can be expressed as follows :
DPD(SP) =OP-WP
or
DPD = OP – TP, (∵ TP = WP)
Due to endosmosis of water the OP of the cell sap, decreases while WP is increased, so that in fully turgid cell TP is equal to OP.
OP = TP (in fully turgid cell)
Hence, DPD (SP) = 0 (zero)

Thus, there will be no absorption of water by the cell in fully turgid cell.
On the other hand, the removal of water from cell sap (exosmosis) results in the increase of its OP and decrease of TP so much so that cell becomes flaccid (fully plasmo- lyzed) and the value of turgor pressure becomes zero.
TP = 0 (in fully flaccid cell)
Hence, DPD (SP) = OP
Consider that a plant cell with OP = 10 atm is immersed in pure water. In beginning, TP inside the cell will be zero, i.e., DPD = OP = 10 atm.

Question 8.
What is Osmosis? Give impotance of osmosis process in plants.
Answer:
Osmosis :
The process of movement of solvent from higher concentration to lower concentration through a semipermeable membrane is called as Osmosis.
It is of following types :

1. Endosmosis :
It is the movement of the solvent molecules from outside to inside the cell through semipermeable membrane (SPM). e.g., Some raisins when kept in water for few hours, they swell as they contain sugar solution of higher concentration.

2. Exosmosis :
It is the movement of solvent molecules from inside to outside the cell through semipermeable membrane (SPM). e.g.,

(i) Swollen raisins shrink if kept in a sugar solution of higher concentration.
(ii) If erythrocytes are placed in 2% NaCl solution, they quickly lose water, shrink up and assume the wrinkled appearance.
Significance of Osmosis :

•  It helps in the absorption of water and minerals from the soil by roots.
• Osmosis helps in the movement of water from one cell to another.
• The phenomenon of plasmolysis is dependent on osmosis.
• It develops turgidity and turgor pressure in plant cell and hence the shape or form of organs is maintained.
• Opening- and closing of stomata is regulated by the osmotic entry and exit of water in guard cells.
• The resistance of plants to drought and frost increases with increase in osmotic pressure of their cells.

Question 9.
What is the role of root pressure in transport of water in plants?
Answer:
According to Priestley (1916), upward flow of water is due to development of a hydrostatic pressure, which is developed in root system. The hydrostatic pressure developed due to accumulation of water absorbed by root, is called root pressure. This pressure is developed in the tracheal elements of the Xylem as a result of metabolic activities of roots. This led to the view that root pressure can raise water in tall trees.

Objections to root pressure theory :

• Root pressure has not been found in all plants.
• No or little root pressure has been observed in gymnosperms which have some of the tallest trees of the world.
• Water continues to rise upwards even in the absence of roots.
• It was observed that, root pressure by itself cannot be held responsible for the ascent of sap in tall trees.Because its maximum value does not normally exceed two atmospheric pressure, hence it can raise the sap up to 15-30 m.

Experiment 1.

Demonstration of root pressure in plants.
Requirements :
Potted Tomato, Balsam or Biyophyllum plant, rubber tube, knife, narrow gas tube, coloured water, petridish, non-drinking oil.
Method:
Take a previously well-watered potted plant of Tomato, Balsam or Bryophyllum. Cut their stem 5-8 cm above the soil level particularly in the morning. Fix a narrow glass tube to the cut end of stem with the help of rubber tube and fill it with coloured water. Cover the glass tube with a small petridish in order to prevent evaporation. A drop of non-drying oil can also be used over the surface of water to prevent evaporation.

Question 10.
Describe the factors affecting Ascent of sap through Xylem in plants.
Answer:
Ascent of sap through Xylem in plants depends on following physical factors :

• Cohesion: Water molecules has tendency to link with each other by force of attraction.
• Adhesion: An attraction exists also between water molecules and the elements of the Xylem wall called as adhesion.
• Surface tension : Molecules of water has more attraction in liquid state as compared to gaseous state.

Water molecules are in a continuous movement strongly attracted to each other to form a continuous column of water in Xylem elements, which help them to move in upward direction through Xylem.

Question 11.
Give role of transpiration pull in transport of water. What are the factors affecting transpiration ? Which factors are useful for plants?
Answer:
Transpiration pull or Tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water vapours move out of the plant through stomata. The DPD of mesophyll cells increases.

With the increase in DPD, these cells absorb water from adjoining cells, ultimately the water is absorbed from Xylem elements of vascular bundles of leaf. Since the xylem elements are filled with continuous water column this tension of pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Thus, water ascends in the plants because of transpiration pull and the column of water remains continuous because of cohesive force of water molecules. This theory is the most accepted theory at the present time.
Factors affecting transpiration : Transpiration is influenced by a number of factors both external and internal. A brief account of them are given below :

[A] External factors :
1. Humidity :
If the plant is surrounded by a saturated atmosphere, there can be no escape of water vapour from the intercellular spaces of the leaf to the outside atmosphere. On the other hand, if the atmosphere is less humid, transpiration goes on at a rapid rate.

2. Temperature :
Rise in temperature reduces the relative humidity of the air as a result, the rate of transpiration increases.

3. Wind :
Wind has a direct influence on humidity. Dry air lowers the amount of air moisture and increases the rate of transpiration, whereas humid air decreases the rate of transpiration.

4. Light:
The rate of transpiration increases markedly in light and decreases in the dark. Light affects transpiration into two ways : Firstly, it increases transpiration by raising the temperature of leaves. Secondly, there is a close relationship between the opening of stomata and light as the stomata opens in day and gets closed in night.

5. Available soil water :
If available soil water is such that the rate of absorption of water is slowed down, the rate of transpiration is correspondingly decreased. Higher con-centration of salts in the soil water reduces the absorption rate and thereby the rate of transpiration.

6. Atmospheric pressure:
At lower atmospheric pressure at the altitudes, the rate of transpiration is increased but the decrease in rate occurs due to the prevailing low tempera-ture at these heights.

[B] Internal factors:
The histological and internal physiological conditions influence the rate of transpira-tion to a great extent. The following structures of the leaves influence the rate of transpiration :

• Presence of thick cuticle and deposition of wax on epidermis, decreases the rate of transpiration.
• Presence of hair on the outer surface of stomata also reduces the rate of transpiration.
• The rate of transpiration depends upon size, position and distribution of the leaves.
• The rate of transpiration is directly proportional to the water contents of the mesophyll tissue.

Question 12.
Xylem transport is unidirectional and phloem transport is bidirectional. Explain.
Answer:
In Xylem transport of water and minerals is unidirectional, i.e. from root to top part of the plant (upward) because suberin layer found in the endodermis of roots causes active transport of ions in one direction.

Whereas in phloem transport of food occurs bidirectional. Generally, food is prepared in the green parts of the plants which contain chlorophyll and then it is collected in sink, which is to be distributed to all parts of the plant. Young buds act as sink. In plants, source and sink for transport of food is changeable as it changes with season thus phloem shows bidirectional movement of food.

Question 13.
Give reason for opening and closing of guard cells during transpiration.
Answer:
Opening and closing of stomata :
Opening and closing of stomata depends upon the activity of guard cells. Guard cells are crescent like or kidney shaped. Their concave wall are thicker than convex which face each other. The opening and closing of stomata depend upon the turgidity of the stomata, when guard cells become turgid, stomata open while guard cells become flaccid, stomata closed. Guard cells contain chloroplast hence they synthesize food by photosynthesis, using CO₂ during day time.

Thus, due to lack of CO₂ the pH value of the cell increases (7.5 p^H). At this p^H starch produced during photosynthesis reacts with inorganic phosphate in presence of phosphorylase enzyme to form glucose-1-phosphate, which is soluble in water thus concentration of the cell sap of the guard cell increases. This lead to absorption of water by the guard cells from neighbouring cells by osmosis, thus increasing turgor pressure due to which the outer thinner wall of the guard cells stretches out and stomata open for diffusion of water vapour.

In dark the reverse reactions occur in the guard cells. During this period photosynthesis does not occur and respiration releases CO₂, thus reducing the pH of guard cell towards acidic side (p^H 5.0).
Now soluble carbohydrates are converted into insoluble starch by the enzyme phosphorylase. This brings about rapid fall of the osmotic concentration of cell sap of guard cell which become flaccid, thus closing the stomatal aperture.

Transport in Plants Class 11 Important Questions Long Answer Type

Question 1.
(a) Describe the process of plasmolysis with labelled diagram.
(b) What will happen if a plant cell is kept in high water potential solution?
AnsWER:
(a) Plasmolysis :
When a living cell is placed in a hypertonic solution then exosmosis of water takes place hence the water of cell comes out into outer solution. Due to exosmosis of water the protoplasm shrinks away from the cell wall and an irregular mass at the centre. This shrinkage of protoplasm is known as plasmolysis.

When plasmolysed cell is placed in water, the water enters into the cell sap, the cell becomes turgid and the protoplasm again assumes its normal sap and position. This phenomenon is called as deplasmolysis.

(b) When a plant cell is kept in high water potential solution, i.e. hypotonic solution then endosmosis occurs. Cell become turgid. Protoplasm apply pressure on the cell wall called as Turgor pressure. Pressure of the protoplasm against cell wall is called as pressure potential (Ψ_P). This turgor pressure is responsible for increased size of the cell.

Question 2.
Explain hypothesis of translocation of organic substances in plants.
Answer:

Munch mass flow hypothesis :
This theory was proposed by Munch (1930). According to this theory, translocation occurs on mass along a gradient of turgor pressure from supply end to the consumption end and which can be established by the following arrangement.

Two bulbs ‘A’ and ‘B’ with semipermeable membrane walls are – connected with a tube ‘C’ containing water to form a closed system. Bulb ‘A’ contains more concentrated solution than bulb ‘B’ and are dipped in water. Now due to higher osmotic pressure of the concentrated sugar solution in ‘A’, water enters into it by endomosis, increasing turgor pressure. This will result in mass flow of sugar solution to bulb ‘B’ through the tube ‘C’ till the concentration of sugar solution in both become equal. If in the above, it could be possible to maintain continuous supply of sugar in ‘A’ and its utilization in ‘B’, the process will go on indefinitely.

According to Munch, similar arrangement exists in plants. Due to photosynthesis, the mesophyll cells of the leaves have higher concentration of sugar solution in soluble form corresponding to bulb ‘A’ and is utilized by different parts (bulb ‘B’). The sieve tubes of the phloem act as tube ‘C’.

In plants mesophyll cells draw water from the xylem of leaf, due to high osmotic and suction pressure or DPD of their cell sap, increasing their turgor pressure. The turgor pressure in the cells of stem and roots are comparatively low hence, the soluble organic solute begins to flow from mesophyll through phloem down to the cells of stem and roots under the gradient of turgor pressure. These organic materials are either consumed or stored and water returns back to the xylem.

Question 3.
Write differences between following :
(a) Diffusion and Osmosis,
(b) Transpiration and Evaporation,
(c) Osmotic pre-ssure and Osmotic potential,
(d) Diffusion and Imbibition,
(e) Apoplast and Symplast pathway,
(f) Guttation and Transportation,
(g) Transpiration and Guttation,
(h) Endosmosis and Exosmosis.
Answer:
(a) Differences between Diffusion and Osmosis

(b) Differences between Transpiration and guttation

(c) Differences between Osmotic Pressure and Osmotic Potential

(d) Differences between Diffusion and Imbibition

(e) Differences between Apoplast and Symplast Pathway

(f) Differences between Guttation and Transportation

(g) Differences between Transpiration and Guttation

(h) Differences between Endosmosis and Exosmosis

Question 4.
What is ascent of sap? Explain the view of Dixon and Jolly regarding the ascent of sap.
Or,
Explain the cohesion theory of ascent of sap.
Answer:
Ascent of sap :
The mechanism of upward translocation of water from the roots to the top of the plant is still a mystery. The upward movement of water from the root towards the top of the plant is known as ascent of sap. Sap is water with dissolved ingredients (minerals). The ascent of sap takes place through the tracheary elements of xylem. The mechanism involved in the transport of water from the roots to the leaves of plants, some of them being more than 200 metre high (e.g., Sequoia, Gigantia, Eucalyptus), is still unresolved.

Transpiration pull and cohesion of water theory :
This theory was proposed by Dixon and Jolly in 1894. According to this theory, water molecules are in a continuous movement strongly attracted to each other to form a continuous column of water in xylem elements. The process of transpiration pulls up the water. This theory has the following three essential features:

(A) Strong cohesion force or tensile strength of water :
The principle is mairily based on the facts that:

• Water molecules have strong mutual attraction (cohesion) with each other and cannot be easily separated from one another.
• An attraction exists also between water molecules and the elements of the xylem wall (cohesion). These cohesive and adhesive attractions together maintain water column in xylem tracheids.

(B) Continuity of water column in the plant:
A strong objection against the theory has been the evidence for the occurrence of air bubbles in the water column of xylem elements. They disturb the continuity of the water column.

(C) Transpiration pull or tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water va¬pours move out of the plant through stomata. The DPD of mesophyll cell increases. With the increase in DPD These cells absorb water from adjoining cells, ultimately the water is absorbed from xylem elements of vascular bundles of leaf. Since, the xylem elements are filled with continuous water column. This tension or pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Question 5.
By giving significance of transpiration explain that “Transpiration is a necessary harm”. Give factors affecting transpiration.
Answer:
Significance of Transpiration :
It is one of the most controversial issues for the physiologists. Some consider it a very’ significant and beneficial process but many other consider it as entirely harmful and useless as plants. Spend energy to absorb water continuously but 90% of absorbed water is lost by transpiration process. The significant role of transpiration are as follow:

• Excess amount of water is removed from the plant tissues.
• Bring coolness for the foliage and thus help them in maintaining their vigour in sunlight.
• Causes transpiration pull, through which water is conducted to all parts of the plant.
• Helpful in the absorption of water and minerals.
• Helpful to maintain the concentration of cell sap.
• Saturate the atmosphere with water vapour and regulate the water cycle of the atmosphere.
• Certain plants liberate certain hygroscopic salts with transpired water, which accumulates on the leaf surface. These salts absorb water from the atmosphere and this saves the plant from wilting.

Thus, transpiration seems to be of much significance in the life of plants. However, excessive transpiration may cause permanent wilting and death of the plant. Therefore, “transpiration is a necessary harm” for the plants.

Factors affecting Transpiration :
Transpiration pull or Tension on the unbroken water column :
As a result of transpiration, water is lost from mesophyll cells to the intercellular spaces. The water vapours move out of the plant through stomata. The DPD of mesophyll cells increases.

With the increase in DPD, these cells absorb water from adjoining cells, ultimately the water is absorbed from Xylem elements of vascular bundles of leaf. Since the xylem elements are filled with continuous water column this tension of pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Thus, water ascends in the plants because of transpiration pull and the column of water remains continuous because of cohesive force of water molecules. This theory is the most accepted theory at the present time.
Factors affecting transpiration : Transpiration is influenced by a number of factors both external and internal. A brief account of them are given below :

[A] External factors :
1. Humidity :
If the plant is surrounded by a saturated atmosphere, there can be no escape of water vapour from the intercellular spaces of the leaf to the outside atmosphere. On the other hand, if the atmosphere is less humid, transpiration goes on at a rapid rate.

2. Temperature :
Rise in temperature reduces the relative humidity of the air as a result, the rate of transpiration increases.

3. Wind :
Wind has a direct influence on humidity. Dry air lowers the amount of air moisture and increases the rate of transpiration, whereas humid air decreases the rate of transpiration.

4. Light:
The rate of transpiration increases markedly in light and decreases in the dark. Light affects transpiration into two ways : Firstly, it increases transpiration by raising the temperature of leaves. Secondly, there is a close relationship between the opening of stomata and light as the stomata opens in day and gets closed in night.

5. Available soil water :
If available soil water is such that the rate of absorption of water is slowed down, the rate of transpiration is correspondingly decreased. Higher con-centration of salts in the soil water reduces the absorption rate and thereby the rate of transpiration.

6. Atmospheric pressure:
At lower atmospheric pressure at the altitudes, the rate of transpiration is increased but the decrease in rate occurs due to the prevailing low tempera-ture at these heights.

[B] Internal factors:
The histological and internal physiological conditions influence the rate of transpira-tion to a great extent. The following structures of the leaves influence the rate of transpiration :

• Presence of thick cuticle and deposition of wax on epidermis, decreases the rate of transpiration.
• Presence of hair on the outer surface of stomata also reduces the rate of transpiration.
• The rate of transpiration depends upon size, position and distribution of the leaves.
• The rate of transpiration is directly proportional to the water contents of the mesophyll tissue.

Question 6.
Explain the pathway of water absorption in plants with the help of diagram.
Answer:
In plants, water is absorbed from root hairs. These root hairs absorb the capillary water of the soil by the process of osmosis. Each root hair possessing a vacuole which is filled with cell sap. The cytoplasm of the root hairs functions like that of a semipermeable membrane. These root hairs are associated with the cortical cells. Cortical cells are elaborated up to endodermis. The absorbed water is passed into xylem through root hairs to cortical cells. Cortical cells to pericycle and pericycle to vascular tissues of the stem.

Question 7.
What is Transpiration? What are its types?
Or,
Write three types of transpiration process.
Answer:
Transpiration :
The process of loss of water in the form of water vapour from aerial parts of the plant is known as transpiration. There are three types of transpiration based on plant surface :

1. Stomatal or foliar transpiration :
The loss of water which takes place through specialized apertures which are present in leaves is called foliar transpiration. Most of the (80-90%) foliar transpiration takes place through stomata is called stomatal transpi-ration.

2. Cuticuiar transpiration :
Transpiration through cuticle is called the cuticular transpiration. Cuticular transpiration is only 3-10% of the total transpiration. It is continuous throughout day and night.

3. Lenticular transpiration :
Loss of water through the lenticels of fruits and woody stem is called the lenticular transpiration. The lenticular transpiration is only 0-1 % of the total transpiration. It however, continues day and night because lenticels have no mechanism of closure.

Question 8.
Describe Starch ⇌ Sugar interconversion hypothesis of opening and closing of stomata.
Answer:
Starch ⇌ Sugar interconversion hypothesis :
(i) This theory was proposed by Lloyd in 1908. According to him, the turgidity of guard cells is controlled by changes in OP, caused by interconversion of starch and sugar. He found that the amount of starch in guard cells increased by night and decreased by day.

During day time the guard cells of the stomata contain sugar synthesized by their chloroplasts by the use of light. The sugar is soluble and increases the concentration of the sap of guard cells. Due to higher concentra¬tion of the cytoplasm of guard cells, the water comes to them from the neighbouring cells by osmosis and they become turgid with the result the stomata remain open.

(ii) According to Sayre (1926), stomata open at pH 4-2 to 4-4 when starch contents is very low. The closure of stomata is always associated with an increase in starch contents. Sayre thought that utilization of C02 during photosynthesis during day time will cause the increase of pH resulting in the conversion of starch into sugar.

(iii) According to Hanes, enzyme phosphorylase catalyses the interconversion of starch and sugar.

(iv) According to this hypothesis, CO₂ liberated from respiration is used in photosynthesis by mesophyll cells in daytime. This leads to the lowering of acidity (H⁺ ion concen-tration) of the guard cells. As the H⁺ ion concentration (acidity) decreases the pH increases and the enzymatic interconversion of starch into sugar is favoured.

During night (darkness) CO₂ released from respiration, accumulates in the intercellular spaces increasing the H⁺ ion concentration (lowering the p^H) this favours the conversion of sugar into starch.

(v) According to Steward (1964), OP of guard cells is not affected unless glucose-1- phosphate is further converted into glucose and inorganic phosphate.
According to Steward :
(A) At higher p^H stomata opens in the following manner :

(B) At low p^H closing of the stomata will be take place in the following manner:

Transport in Plants Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
Transpiration occurs through :
(a) Stomata
(b) Lenticel
(c) Cuticle
(d) All of these.
Answer:
(d) All of these.

Question 2.
Stomata opens during day time because the guard cells have :
(a) Outer membrane is thin
(b) Kidney shaped
(c) Chloroplast is present
(d) Large nucleus is found.
Answer:
(a) Outer membrane is thin

Question 3.
Rate of transpiration increases due to :
(a) Drought
(b) High temperature
(c) Moist soil
(d) Slow speed of wind.
Answer:
(c) Moist soil

Question 4.
Rate of transpiration is measured by :
(a) Potometer
(b) Porimeter
(c) Photometer
(d) None of these.
Answer:
(a) Potometer

Question 5.
When temperature rises, rate of transpiration :
(a) Increases
(b) Decreases
(c) Transpiration stops
(d) None of these.
Answer:
(a) Increases

Question 6.
Stomata closes due to lack of water. This condition is created due to :
(a) Formation of cytokinin
(b) Formation of Auxin
(c) Formation of ethylene
(d) Formation of Abscisic acid.
Answer:
(b) Formation of Auxin

Question 7.
Which of the following reason decreases rate of transpiration :
(a) Wind
(b) Increase in temperature
(c) Decrease in light intensity
(d) Absorption of water by plants.
Answer:
(c) Decrease in light intensity

Question 8.
Water reaches the upper part of the plants by :
(a) Root pressure
(b) Cell pressure
(c) Evaporation
(d) Diffusion.
Answer:
(c) Evaporation

Question 9.
Maximum water is absorbed by plants through :
(a) Root cap
(b) Root hair
(c) Root hair area
(d) Division area.
Answer:
(c) Root hair area

Question 10.
Plants only utilize this water present in the soil:
(a) Runaway water
(b) Gravitational water
(c) Vapourised water
(d) Capillary water.
Answer:
(d) Capillary water.

Question 11.
When the cell is completely turgid which of the following pressure becomes zero :
(a) Turgor pressure
(b) Cellular wall pressure
(c) Vapour pressure
(d) Osmotic pressure.
Answer:
(b) Cellular wall pressure

Question 12.
Stomata are open in day time in most of the plants, such stomata are called :
(a) Photoactive type
(b) Potato type
(c) Barley type
(d) Dark active type.
Answer:
(a) Photoactive type

Question 13.
Which type of stomata is found in the lower surface of leaves :
(a) Appie type
(b) Potato type
(c) Oat type
(d) None of these.
Answer:
(b) Potato type

Question 14.
In which type of plants Sunken stomata is found :
(a) Xerophytes
(b) Hydrophytes
(c) Mesophytes
(d) Sporophytes.
Answre:
(a) Xerophytes

Question 15.
What will happen if a plant cell is kept in concentrated salt solution :
(a) Plasmolysis
(b) Turgid
(c) Contraction
(d) Deplasmolysis.
Answer:
(a) Plasmolysis

Question 16.
Reason for Root pressure is :
(a) Passive absorption
(b) Active absorption
(c) Increase in transpiration
(d) Decrease in rate of photosynthesis.
Answer:
(b) Active absorption

Question 17.
Potometer is used for measurement of:
(a) Rate of absorption
(b) Rate of transpiration
(c) Rate of photosynthesis
(d) Phototropism.
Answer:
(b) Rate of transpiration

Question 18.
In guttation process loss of water occurs through :
(a) Stomata
(b) Hydathodes
(c) Wound
(d) Lenticels
Answer:
(b) Hydathodes

Question 19.
Which wall of the guard cells is thick :
(a) Outer
(b) Inner
(c) Lateral
(d) All of these.
Answer:
(b) Inner

Question 20.
When a cell is at equilibrium then :
(a) DPD = 0
(b) DPD = TP
(c) OP = TP
(d) DPD=OP.
Answer:
(c) OP = TP

Question 21.
What will happen if a plant cell is kept in pure water :
(a) Turgid
(b) Flaccid
(c) Plasmolysis
(d) Impermeable.
Answer:
(a) Turgid

Question 22.
Water potential of cell become positive in plant cell due to :
(a) Transpiration
(b) Low transpiration
(c) High absorption
(d) Guttation.
Answer:
(c) High absorption

Question 23.
Meaning of Osmosis is :
(a) Diffusion of dissolved particles from higher to lower concentration
(b) Diffusion of dissolved particles from lower to higher concentration
(c) Diffusion of water from higher to lower concentration
(d) Diffusion of water from lower to higher concentration.
Answer:
(c) Diffusion of water from higher to lower concentration

Question 24.
Guttation occurs due to :
(a) Transpiration
(b) Osmosis DPD
(c) Root pressure
(d) Osmotic pressure.
Answer:
(c) Root pressure

Question 25.
Transfer of water from one cell to other cell and direction of water flow depends on:
(a) WP
(b) TP
(c) DPD
(d) Primary plasmolysis.
Answer:
(c) DPD

2. Fill in the blanks:

1. The cavity and loose tissues present below the hydathodes is called ………………….
Answer:
Epithem

2. …………………. gave the name protoplasm.
Answer:
Purkinje

3. …………………. gave the K+ exchange process for opening and closing of stomata.
Answer:
Levitt

4. The guard cells for dicot stomata are ………………….
Answer:
Kidney shaped

5. Osmoscope is used to measure ………………..
Answer:
Osmosis

6. Parchment paper, is an example of …………………… membrane.
Answer:
Semipermeable

7. DPD = OP – …………………..
Answer:
TP

8. The cavity present below hydathodes is called ………………..
Answer:
Epithem

9. Loss of water from the margin of leaves in the form of water droplets is called………………….
Answer:
Guttation

10. In osmosis, diffusion of solvent takes place through ………………….
Answer:
Semipermeable membrane

11. The DPD of a fully turgid cell is always …………………..
Answer:
0

12. The DPD of a solution is ……………….. to its concentration.
Answer:
Proportional

13. ………………… is an instrument which is used to measure the rate of transpiration.
Answer:
(b) Potometer

14. The transport of water and mineral salts takes place through …………………
Answer:
Xylem

15. Absorption of water with use of energy is known as …………………..
Answer:
Active absorption

16. In plants, rate of transpiration is measured with the help of ……………….
Answer:
Potometer

17. Loss of water in the form droplets from the margin of leaves is called ……………….
Answer:
Guttation

3. Match the following:

(A)

Answer:
1. (b) Stomata
2. (c) Hydathode
3. (d) Injured part
4. (e) Epithem
5. (a) Semipermeable membrane

(B)

Answer:
1. (e) J. C. Bose,
2. (d) Atkins and Priestley
3. (a) Pnestley
4. (b) Dixon and Jolly
5. (c) Godlewski

4. Answer in one word:

1. Which part of the root take part in water absorption?
Answer:
Root hairs

2. What type of water is absorbed by roots?
Answer:
Capillary water

3. From which tissue of the plant body, translocation of water takes place?
Answer:
Xylem

4. Name the pressure responsible for stomatal opening.
Answer:
Turgor Pressure

5. “Transpiration is a necessary evil”, who stated it?
Answer:
Kartis

6. The pressure exerted in plant cells due to the absorption of water.
Answer:
Osmotic pressure

7. What will be the DPD of a fully turgid cells?
Answer:
Zero

8. Name the water absorbing part of the roots.
Answer:
Root hair

9. Name the process of loss of water from the aerial part of the plant body in the form of water vapour.
Answer:
Transpiration

10. Name the enzyme which converts starch into glucose, 1-phosphate.
Answer:
Phosphorylase

Students get through the MP Board Class 11th Biology Important Questions Chapter 12 Mineral Nutrition which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Chapter 12 Mineral Nutrition

Mineral Nutrition Class 11 Important Questions Very Short Answer Type

Question 1.
Write the name of symbiotic bacteria present in the roots of leguminous plant.
Answer:
Rhizobium bacteria are present in the roots of leguminous plant which absorb atmospheric nitrogen.

Question 2.
Write the name of two micronutrients.
Answer:
The two micronutrients present in plants are Manganese (Mn) and Molybdenum (Mo).

Question 3.
Write the deficiency symptoms of potassium.
Answer:
Deficiency of potassium inhibits protein synthesis. Scorching of older leaves is the most important symptoms of trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 4.
Why nitrogen is essential for growth of plant?
Answer:
Protein is essential for growth of plant and for development of protein nitrogen is essential, so for the proper development of plant nitrogen is very much essential.

Question 5.
What are the drawbacks of soilless culture?
Answer:
Drawbacks :

• It will be very expensive.
• Limited production will be obtained.
• Crops will never be grown by this method.

Question 6.
What are micronutrients? Give two examples.
Answer:
The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

Question 7.
Describe the deficiency symptoms of nitrogen in plants.
Answer:
Deficiency symptoms of nitrogen:

• Yellowing i.e., Chlorosis of leaves, older leaves become yellow first.
• Acceleration in the rate of formation of purple pigment anthocyanin in the leaves which become purple in colour.
• Flowering is delayed or is completely suppressed.

Question 8.
Which pigment is present in the roots of leguminous plants?
Answer:
Leghaemoglobin.

Question 9.
Write the importance of molybdenum in plants.
Answer:
Molybdenum act as an activator for the enzyme nitrate reductase involved in nitrogen metabolism.

Question 10.
In plants, the organic solutes are transported through which tissue?
Answer:
In plants, the organic solutes are transported through phloem tissue.

Question 11.
What is Mineral nutrition?
Answer:
The utilization of various kinds of absorbed ions by a plant for its growth and development is called as Mineral nutrition.

Question 12.
What are nutrient elements?
Answer:
Elements which are essential for plant growth are called as nutrient elements.

Question 13.
In study of water culture purity of water and mineral nutrient is essential. Why? (NCERT)
Answer:
In study of water culture purity of water and pure mineral nutrients are essential because plants can absorb only pure water and pure minerals and their essentiality for the plants can be determined correctly. Impure mineral nutrient cannot help to determine essentiality of the minerals for plants.

Question 14.
Explain following With examples: Macronutrients, micronutrients, useful elements, essential elements, toxic elements.
Answer:
1. Macronutrients: Essential elements which are required in large quantity for normal growth and development of plants are called as macronutrients.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, etc.

2. Micronutrients: The mineral elements which are required in very low concentration for the plant growth are known as micronutrient or microelements.
e.g., Zn, Mo, Cl, Mn, B, Cu, etc.

3. Useful elements: Some essential elements which are essential for higher plants called as useful elements.
e.g., Na, Si, Ca and Se.

4. Essential elements: Out of approximately 30 elements universally present in the plants, 16 elements are essential for plants, i.e., they require them.
e.g. C, H, O, N, P, K, Ca, S, Mg, Fe, Mn, Zn, B, Cu, Cl, Al.

5. Toxic element: Concentration of any mineral ion which decreases dry weight of plant by 10% is called as Toxic element. It is different for different plants.
e.g. As, Cu, Hg etc.

Question 15.
Give four general importance of mineral salt.
Answer:
Importance of Mineral salt:

1. Nutrition of plants: Some minerals like N, P, S, etc. help to produce protein.
2. As catalyst: Some minerals act as catalyst in biochemical reactions. e.g. Mn.
3. Bodybuilding: Some elements help to produce structural part, such as cell wall of the cell.
4. Balancing work: Some mineral elements stop toxic effect of some other elements. e.g., Calcium, Magnesium.

Question 16.
Write difference between Active and Passive absorption.
Answer:
Differences between Active and Passive absorptions

Mineral Nutrition Class 11 Important Questions Short Answer Type 

Question 1.
Mention the name of any five macroelements and their deficiency symptoms. (NCERT)
Answer:
Macronutrients :

1. Nitrogen
2. Phosphorus,
3. Sulphur,
4. Magnesium and
5. Calcium.

1.Deficiency symptoms of nitrogen:

• Stunted growth with yellowish-green leaves in young plants.
• Yellowing and drying of older leaves.
• Flowering reduced.

2. Deficiency symptoms of phosphorus :

• Stunted growth.
• The leaves become purple coloured due to excessive anthocyanin.
• Premature leaf falls and necrotic spots on the fruits.
• Vascular tissues less developed.
• Low rate of protein synthesis.

3. Deficiency symptoms of sulphur :

• Plants stunted in growth and flowering delayed.
• Chloroplast becomes pale green in colour.
• Chlorosis of leaves.
• Hard woody stem.

4. Deficiency symptoms of magnesium :

• Interveinal chlorosis or molted chlorosis with green veins.
• Brittleness of the leaves and necrosis.
• Wilting, drying off and dropping of the tips of plant.
• Necrotic patches on the leaves.

5. Deficiency symptoms of calcium :

• Calcium deficiency causes malformation of the younger leaves.
• Chlorosis also occurs along the margins of younger leaves.

Question 2.
Describe essentiality of potassium in plants and give its deficiency symptoms.
Answer:
Potassium: It is essential for activating enzymes concerned in the synthesis of polypeptides from amino acids and is also essential for the process of photosynthesis, and respiration.

Deficiency symptoms: Deficiency of potassium inhibits protein synthesis. Scorch-ing of older leaves is the most important symptom on trees. The leaf tips and margins show general chlorosis. Shortening of internodes also occurs.

Question 3.
All elements are not essential for plants. Explain it. (NCERT)
Answer:
All elements absorbed by the plants are not essential.
Criteria for Essentiality: According to Arnon and Stout, 1939, the criteria for essentiality of mineral elements are as follows :

• The plant is unable to grow normally and complete its life cycle, in the absence of the elements.
• The element is specific and cannot be replaced by another element, however close it may be in the periodic table.
• The element plays a direct role in the metabolism of the plant.

Question 4.
Give atleast essential elements. Describe them and give their deficiency symptoms. (NCERT)
Answer:
1. Sulphur: It is absorbed in the form of sulphate ions from the soil. It is important constituent of some amino acids. It is also essential for the synthesis of sulphur con¬taining vitamins like biotin, thiamine and coenzyme A. Disulphide bridge (S-S) plays an important role in determining protein structure and sulphydryl groups (SH) are necessary for the activity of many enzymes.

Deficiency symptoms :

• Reduced leaf, stunted growth, a general chlorosis, followed by the production of anthocyanin pigments in some species. The younger leaves are affected first.
• Inward rolling of leaf margins and tips.
• Due to development of sclerenchymatous tissue, stem becomes hard.

2. Phosphorus: It is an important constituent of phospholipids, nucleic acids and various coenzymes like NAD, NADP and ATP. It is absorbed from soil in the form of phosphate. Phospholipids are the constituents of membranes. NAD, NADP and ATP are required for oxidation-reduction reactions of photosynthesis, respiration and fat metabo-lism. ATP is the chief source of energy.

Deficiency symptoms :

• Premature leaf fall.
• Formation of necrotic areas on leaves and fruits.
• Leaves become dark blue in colour.
• The growth of roots and shoots is completely restricted.

3. Calcium: It is a constituent of the middle lamella, in which it is present in the form of calcium pectate. It helps to stabilize the structure of the chromosomes.
Deficiency symptoms: Calcium deficiency causes malformation of the younger leaves. Chlorosis also occurs along the margins of the younger leaves.

4. Magnesium: It is one and only constituent of the chlorophyll molecule. It acts as activator for many enzymes in phosphate transfer reactions particularly in carbohydrate metabolism and nucleic acid synthesis. It is also believed to be an important binding agent in ribosomes, where protein synthesis takes place.
Deficiency symptoms: Magnesium deficiency results in extensive interveinal chlorosis of leaves. The older leaves are affected first. Ultimately, leaves develop necrotic spots.

5. Iron: It is essential for chlorophyll synthesis. It is important constituent of ferredoxin, flavoprotein, iron-porphyrin, catalase, peroxidase and cytochromes. Iron is primarily concerned in the formation of chloroplast protein in the leaves.
Deficiency symptoms: Pronounced interveinal chlorosis occurs due to deficiency of iron.

Question 5.
Which statement is right out of following if it is wrong, give reason : (NCERT)
(a) Deficiency of boron causes formation of sclerenchymatous axis.
(b) AH elements present in the cell are essential.
(c) Nitrogen is not an essential element for plants.
(d) It is very easy to determine essentiality of micronutrients in plants because it is taken in very less quantity.
Answer:
(a) Wrong, because deficiency of Boron effect on activity of membrane, pollen germination, cell elongation, cell differentiation and transport of carbohydrates, (b), (c) and (d) are right statements.

Question 6.
If plant shows deficiency symptoms of two elements how will you determine deficient element practically? (NCERT)
Answer:
Deficiency symptoms of different elements are different. By providing particu¬lar deficient element if deficiency symptoms eradicates then deficient element can be confirmed.

Question 7.
Why deficiency symptoms are observed in newly formed parts of some plants whereas in some other plants it is observed in mature part? (NCERT)
Answer:
Deficiency symptoms in plant parts depends on mobility of elements. Deficiency symptoms are observed in old tissue first as compared to new tissues of young part. For example, symptoms of deficiency of nitrogen, potassium and magnesium are observed in old leaves first of all.

Biomolecules of old leaves decomposes and element move towards the new leaves.
When elements are non-mobile then they cannot reach to young parts thus deficiency symptoms are observed in new leaves.

Question 8.
Describe passive absorption of minerals in plants. (NCERT)
Answer:
Passive Absorption: In this method, the movement of minerals into the cells or tissue takes place without the expenditure of energy on the part of absorbing cells. This is a purely physical phenomenon and takes place in many ways.

1. Diffusion: It can be observed by first placing a plant cell or tissue in a medium of high concentration of salt and then transferring it into lower salt concentration. During this transfer, some of the ions taken up will diffuse into the lower salt concentration from the higher salt concentration, until the concentration of both the solutions get balanced. The whole process is unaffected by temperature or metabolic inhibitors.

2. Mass flow: Plants which have a higher rate of transpiration contain comparatively high concentrations of minerals. It is well-known that increase in transpiration is accompanied by increased absorption of ions from the soil. In this way the ions are absorbed by the plants in a mass flow due to the higher rates of transpiration.

3. Donnan equilibrium: According to Donnan, cells contain some fixed ions which do not exert any diffusion pressure. Therefore, ions similar in charge to that of fixed ions, move inwardly through diffusion. Oppositely charged ions also pass inwardly to balance them. Such a diffusion continues till the multiple of cations and anions inside cells becomes equal to that of ions present in outside medium.

Question 9.
Describe process of formation of Root nodules. (NCERT)
Answer:
Formation of Root nodule: It is a combined activity of roots of nutrient plants (Leguminous plants) and Rhizobium bacterias.

Steps of formation of Root nodules are as follows :
First of all Rhizobium bacteria multiply and are collected surrounding the roots of leguminous plants and get connected to root hair cells.

Root hair turns and bacteria attack them. An infected thread appears which carry Rhizobium bacteria to the cortex, where they begin to produce nodule. Now bacterias are released from the thread and enters into the cells which help for differentiation of nitrogen fixation cells. In this way, root nodules are formed. Leguminous plants provide nutrients to bacteria and Rhizobium bacteria provides nitrogenous compounds to plant by nitrogen fixation.

Mineral Nutrition Class 11 Important Questions Long Answer Type 

Question 1.
What are the materials required for atmospheric nitrogen fixation? Give their role in Nitrogen fixation. (NCERT)
Answer:
Rhizobium bacteria shows symbiotic relationship with the roots of leguminous plants like pea, moong, gram, beans etc. Symbiosis occurs in the root nodules. Rhizobium bacteria obtains food from the leguminous plants and Rhizobium bacteria help for nitrogen fixation and provides nitrogenous compounds to the plants.
Symbiotic Nitrogen fixation: The process of nitrogen fixation is done by nitrogen-fixing bacterias like Rhizobium, Cyanobacteria, nitrogen-fixing fungi etc.
Mechanism of Nitrogen fixation: Following enzymes and materials are required for nitrogen fixation:

• Nitrogenase and Hydrogenase enzymes.
• Leghaemoglobin: It protects the enzyme from oxygen.
• One non-haem iron protein: e.g. Ferredoxin: It is an electron carrier.
• Hydrogen donor substances like: NADPH₂, FMNH₂, Pyruvate, Sucrose, Glucose etc.
• Energy compound ATP.
• Cofactor like TPP (Thiamine pyrophosphate), CoA, inorganic phosphate, Mg++ etc.
• Co and Mo.
• A compound to fix ammonia produced by reduction of nitrogen.
  Following are the steps of natural nitrogen fixation:
  1. Reduction of molecular nitrogen:
  
  

Question 2.
What is Hydroponics? Give its uses.
Answer:
Hydroponics (Soilless culture): Large- scale cultivation of plants in water culture or soilless culture is known as hydroponics. It was discovered by Prof. W. F. Gariak in 1929. The plants are grown in large, shallow pots which are full of nutrient solutions. The tubs or tanks are covered with wire netting to provide support to the seedlings. The solution is aerated at regular intervals by means of an inlet tube.
There are certain advantages and disadvantages of this technique.

Uses:

• Possibility to provide desirable nutrient environment.
• Regulation of pH of the nutrient solution conductive to specific crop.
• It is possible to replace the nutrient, so more healthy plant as compared to soil culture can be obtained.
• On the roof of multistoried buildings of metropolitan cities, vegetables and fruits may be produced in large scale.
• Ripening time of fruit is less in soilless culture.

Question 3.
Give an experiment to demonstrate the essentiality of minerals required by plants.
Answer:
Essentiality of minerals required by plants can be demonstrated by using a normal culture medium such as Sach’s culture solution. Components of this solution are as follows:
Sach’s culture solution :

This experiment is performed to ascertain the role of individual elements in the normal growth and development of plants and the effect produced by their deficiencies. In water culture (solution culture) experimental plants are grown in a medium, the chemical composition of which is known and kept under control.

Method: Take seven wide-mouthed bottles of neutral glass of the same size. Clean and thoroughly sterilize them with nitric acid and distilled water. Fit each bottle with a cork containing two holes; one in the centre for introducing a seedling and the other at the side for a bent glass tube for aerating the culture solution. Make a slit running from the central hole to the rim of the cork for readily taking out the seedling.

Now, prepare a normal culture solution and pour it into the first bottle. Then, prepare six other solutions from which omit by turn Mg, Ca, Fe, K, P and N, and designate them as Mg, Ca, Fe, K, P and N, and fill the remaining seven jars with these solutions. Take seed¬lings of same kind and more or less same size. Introduce a seedling in each bottle through split cork. Wrap the bottles with black paper and expose them to light. Make arrangement

for proper aeration of roots. Renew the culture solution fortnightly. Observe carefully the symptoms of plants appeared due to the deficiency of various elements.

Question 4.
Name any five micronutrients and give their specific role and deficiency symptoms.
Answer:
Nutrient elements which are required in traces for normal growth and development of the plants are called as Micronutrients.
e.g. Manganese, Boron, Copper, Zinc, Molybdenum and Chlorine. Their specific role and deficiency symptoms are as follows :

Mineral Nutrition Class 11 Important Questions Objective Type

1. Choose the correct answers:

Question 1.
In Nitrogen cycle nitrifying bacteria convert:
(a) NH₃ into N₂
(b) Nitrogen fixation
(c) NH₃ into NO₂
(d) Amino acid into NHr
Answer:
(c) NH₃ into NO₂

Question 2.
Function of Zinc is :
(a) Production of Chlorophyll
(b) Production of IAA
(c) Closing of stomata
(d) Oxidation of Carbohydrates.
Answer:
(b) Production of IAA

Question 3.
Symptoms of destruction of Chlorophyll is called as :
(a) Necrosis
(b) Chlorosis
(c) Stunted shoot
(d) Rotting.
Answer:
(c) Stunted shoot

Question 4.
Plants which can prepare their own food are called as :
(a) Heterotrophs
(b) Autotrophs
(c) Mixotrophs
(d) None of these.
Answer:
(b) Autotrophs

Question 5.
Falling of premature leaves, flowers and fruits is called as :
(a) Abscission
(b) Dieback
(c) Stunted
(d) Hypertrophy.
Answer:
(a) Abscission

Question 6.
In which form ‘C’ is available to plants :
(a) ‘C’ element
(b) CO₂
(c) CO₃
(d) Amino acid.
Answer:
(b) CO₂

Question 7.
Which element is required by the plants in large quantity :
(a) N
(b) P
(c) Ca
(d) S.
Answer:
(a) N

Question 8.
Which element out of following is not macroelement:
(a) P
(b) K
(c) Mg
(d) Mo.
Answer:
(d) Mo.

Question 9.
Hydroponics name was given by :
(a) Knop and Sach
(b) Hoagland
(c) Gariak
(d) Sach.
Answer:
(c) Gariak

Question 10.
Soil element available to plant is called as :
(a) Mineral salt
(b) Micro salt
(c) Nutrient substance
(d) None of these.
Answer:
(a) Mineral salt

Question 11.
How much percentage of dry weight of plants is formed by Carbon, Hydrogen and Oxygen
(a) 10-15%
(b) 15-25%
(c) 25-35%
(d) 85-95%.
Answer:
(d) 85-95%.

Question 12.
Element required for synthesis of IAA :
(a) B
(b) Cu
(c) Zn
(d) Mo.
Answer:
(c) Zn

Question 13.
Active absorption of minerals in plants depends on :
(a) Availability of O₂
(b) Light
(c) Temperature
(d) Availability of CO₂.
Answer:
(a) Availability of O₂

Question 14.
Element which is not essential for the plants out of the following is :
(a) Iron
(b) Magnesium
(c) Lead
(d) Phosphorus.
Answer:
(c) Lead

Question 15.
Trace element out of following is :
(a) Ca
(b) Mg
(c) C
(d) Cu.
Answer:
(d) Cu.

2. Fill in the blanks:

1. Monotropa is an example of ………………………………… plant.
Answer:
Saprophyte

2. Orobanche is an example of ………………………………… plant.
Answer:
Total root parasite

3. Zn, Cu, Mn, B, Mo and Cl are ………………………………… elements.
Answer:
Micro

4. Extensive interveinal Chlorosis is caused due to deficiency of ………………………………… .
Answer:
Mg

5. Element act as activator for the enzyme nitrate reductase is ………………………………… .
Answer:
Mo

6. Rhizobium is a ………………………………… nitrogen fixing bacteria.
Answer:
Symbiotic

7.
Answer:
2HNO₂

8. ……………………………….. element participates in various metabolic activities of plant.
Answer:
Essential

9. Tetra-pyrrol porphyrin ring of chlorophyll has ………………………………… element at the centre.
Answer:
Mg

10. Lack of ………………………………… causes the absence of glands in the roots of leguminous plants.
Answer:
Boron

11. Rhizobium is found in the glands of the leguminous plants situated at ………………………………… .
Answer:
Roots

12. The element which are required for the growth of plants are called ………………………………… element.
Answer:
Essential

13.‘Whiptail disease’ in cabbage is due to deficiency of ………………………………… .
Answer:
Molybdenum.

3. Match the following:

(A)

Answer:
1. (e) Insectivorous.
2. (d) Plant ash
3. (a) Minor trace element
4. (b) Hydroponics
5. (c) Critical element

(B)

Answer:
1. (d) Stomatal movement
2.  (e) Chlorophyll.
3.  (a) Nitrogen metabolism
4. (b) Photolysis of water
5.  (c) Amino acid.

(C)

Answer:
1.  (d) Total stem parasite
2. (e) Partial stem parasite.
3. (b) Total root parasite
4. (c) Insectivorous
5. (a) Symbiotic

4. Answer in one word:

1. Write two mineral element essential for production of chloroplast.
Answer:
Mg and Fe

2. Name the pigment found in the root nodule of leguminous plant.
Answer:
Leghaemoglobin

3. Full form of NPK is.
Answer:
Nitrogen, Phosphorus, Potassium

4. Name the enzyme induce Nitrogen fixation.
Answer:
Nitrogenase

5. Name one free-floating aquatic insectivorous plant.
Answer:
Utricularia

6. Name the process in which roots of the plants are kept in nutrient solution for plant growth.
Answer:
Hydroponics

7. Deficiency of which element causes abscission of premature leaves.
Answer:
Phosphorus

8. Name the root of the Cuscuta plant which penetrate into the body of host plant.
Answer:
Haustorial root

9. Name the macroelement which forms structural component of plant body.
Answer:
Carbon

10. From where plants obtain hydrogen.
Answer:
Water

Students get through the MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division which are most likely to be asked in the exam.

MP Board Class 11th Biology Important Questions Chapter 10 Cell Cycle and Cell Division

Cell Cycle and Cell Division Class 11 Important Questions Very Short Answer Type

Question 1.
What is the average cell cycle duration of cells of mammalia?
Answer:
Average cell cycle duration of cells of mammals is 24-25 hours.

Question 2.
What is the difference between cytokinesis and karyokinesis?
Answer:
Division of the cytoplasm is called as cytokinesis, whereas division of the nucleus is called as karyokinesis.

Question 3.
What is G₀ phase of cell cycle?
Answer:
Cells which do not divide further and reaches to inactive phase, is called as G₀ phase of cell cycle. At this stage cell shows high rate of (active) metabolism but does not divide. They divide depending upon requirement of the organism.

Question 4.
Why Mitosis cell division is called as similar division?
Answer:
At the end of this cell division daughter cells have same number of chromosomes like parent cell thus Mitosis division is called as similar cell division.

Question 5.
Where daughter cells formed by meiosis cell division are same and different?
Answer:
Four daughter cells formed by meiosis division in the testis are the spermatids which are same in size whereas in the ovary of female four daughter cells formed by meiosis are different in size, i.e. one ovum which is larger in size and three polar bodies, which are smaller in size.

Question 6.
Is Mitosis division possible without DNA replication in ‘S’ phase of cell cycle?
Answer:
No, Mitosis division is not possible without DNA replication in ‘S’ phase of cell cycle, because duplication of DNA is essential for Mitosis division.

Question 7.
Name the stage of cell division which involves crossing-over.
Answer:
Crossing-over occurs during pachytene stage of Prophase-I of meiosis cell division.

Question 8.
Name the stain used for staining chromosomes of the cells for study.
Answer:
Acetocarmine stain is used for staining chromosomes of the cell for study.

Question 9.
Name the cells exhibiting mitosis cell division.
Answer:
Mitosis is the characteristic feature of somatic cells.

Question 10.
Name the cells exhibiting meiosis cell division.
Answer:
Meiosis takes place in generative cells to produce gametes.

Question11.
What is chiasmata?
Answer:
The place of chromosomes where they come into contact with each other during crossing-over is called as chiasmata.

Question 12.
List the various phases of the first meiotic division.
Answer:
The following stages are found in the first meiotic division :

1. 1. Prophase-I: It has following substages :
   • Leptotene,
   • Zygotene,
   • Pachytene,
   • Diplotene and
   • Diakinesis.
2. Metaphase-I,
3. Anaphase-I,
4. Telophase-I.

Question 13.
What is synapsis?
Answer:
Pairing of homologous chromosomes in zygotene stage of Prophase-I of meiosis cell division is known as synapsis. The paired structure formed by synapsis is called as bivalent.

Question 14.
What do you mean by Mitotic poison?
Answer:
Chemical substances which inhibit the mitotic division are called as Mitotic poison.
e.g.,

1. Cholchicine : It stops formation of spindles during Metaphase.
2. Ribonuclease : It stops Prophase stage.
3. Mustard gas : It breaks chromosomes.

Question 15.
Name the parts of a flowering plant where meiosis cell division occurs.
Answer:
In the pollen sac of androecium and ovary of gynoecium meiosis division occurs to produce male and female gametes respectively.

Cell Cycle and Cell Division Class 11 Important Questions Short Answer Type

Question 1.
Describe the phenomenon occurs during interphase of cell cycle.
Answer:
Interphase is the period between two successive cell divisions. It has following three stages:

• G_1-phase : RNA and protein synthesis occurs during this stage.
• S-phase : DNA synthesis occurs during this stage.
• G₂-phase : Synthesis of protein and RNA continue in this stage.

Question 2.
What is the significance of Meiosis division?
Answer:
Significance of meiosis : The significance of meiosis are as follows :

• The number of chromosomes is stable generation to generation due to the meiotic division.
• The number of chromosomes is reduced to half in the gametes.
• The characters of parents are transmitted into offspring by this process.
• Crossing-over takes place during this stage which can produce new characters in the organisms.
• Variations are also occurred which help in the study of evolution.
• There are four haploid cells formed from one diploid cell.

Question 3.
Name the stage of cell cycle when following phenomenon occurs :
(i) Movement of chromosomes towards the equator of the cell.
(ii) Breaking of centromere and separation of chromosomal half.
(iii) Pairing of homologous chromosomes.
(iv) Exchange between homologous chromosomes.
Answer:
(i) Metaphase,
(ii) Anaphase,
(iii) Zygotene stage of Prophase-I of Meiosis-I,
(iv) Pachytene stage of Prophase-I of Meiosis-I.

Question 4.
Explain the following :
(i) Synapsis,
(ii) Bivalent,
(iii) Chiasmata.
Answer:
(i) Synapsis :
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

(ii) Bivalent:
The pairs of homologous chromosome formed during zygotene stage of meiosis-I. The chromosomes contract and become short and thick.

(iii) Chiasmata :
In Pachytene stage of Prophase-I of Meiosis-I division, non-sister chromatids of two chromosomes, which come in contact of each other and fixes themselves tightly for crossing-over. These fixed parts are called as Chiasmata.

Question 5.
Write differences between cytokinesis in plant and animal cell.
Answer:
Cytokinesis : Division of cytoplasm after Karyokinesis to produce daughter cells is called as Cytokinesis.

(a) Cytokinesis in plant cell:
It immediately follows karyokinesis. In plant cells, a cell plate of dense cistemae is laid down at the equatorial plate. It grows centrifugally (external laterally) towards plasma membrane until it divides the cell into two parts. The space in between the two membranes of cistemae is filled by the deposition of pectates of calcium and magnesium to become middle lamella.

(b) Cytokinesis in animal cell:
In animal cell, cytokinesis occurs by an invagination of cell membrane almost in the middle of the cell. The furrow gradually depens and ultimately divides the cell into two parts.

Question 6.
Write differences between Anaphase of Mitosis and Meiosis cell division.
Answer:
Differences between Anaphase of Mitosis and Meiosis cell division

Question 7.
Analyse each phase of cell division and explain how does following changes occurs in the cell:
(i) Number of chromosomes (N) in each cell.
(ii) Quantity of DNA in each cell.
Answer:
(i) In all cells during prophase, metaphase and anaphase stage of cell division, number of chromosomes become double and in telophase stage number of chromosomes become half again, i.e., (N).
(ii) During cell division quantity of DNA changes in all stages. During prophase, metaphase and anaphase number of DNA increases, it becomes double, but in telophase stage quantity of DNA becomes half.

Question 8.
Discuss following with your teacher :
(i) When cell division occurs in haploid seed and lower category of plants ?
(ii) Where cell division does not occur in the haploid cells of higher category of plant?
Answer:
(i) Haploid spores of lower category of plant divides by mitosis division to form gametophyte, whereas zygote develops to form sporophyte and meiosis division occurs for formation of haploid spores, e.g., moss, fern etc.
(ii) Synergids and antipodal cells found in the ovules of angiosperms are haploid. They do not show cell division. After completion of life span they dies.

Question 9.
What is crossing-over? Throw light on its significance.
Answer:
Crossing-over : Crossing-over is the exchange of chromosomal segments or genes between homologous chromosomes during diplotene stage of meiosis-I.

Significance of crossing-over :

• New characters are originated by this process,
• Variations occur in organisms,
• It plays an important role in organic evolution,
• It results in the production of adaptation in organisms.

Question 10.
Write the significance of mitosis.
Answer:
Significance of mitosis:

• Growth and development takes place due to mitosis,
• Some micro-organisms use this method for asexual reproduction,
• Dead cells are substituted by new cells by mitotic division,
• General repairing process is result of mitotic division,
• Genetic information are transferred from parents to offspring by this process.

Question 11.
What is the difference between anaphase-1 of meiosis and anaphase of mitosis ? What is its effect on the whole process?
Answer:
In meiotic anaphase-I, the two chromosomes of each bivalent separate from each other and move towards opposite poles of the spindle. But the two chromatids of each chromosome still continue to remain joined with one another with centromere. Thus, only one chromosome of each homologous chromosome pair reaches each pole and the number of chromosomes are reduced to half in daughter nuclei.

In mitotic anaphase-I, centromere of each chromosomes divides and permits the separation of two sister chromatids, each chromatid after separation is called daughter chromosomes. The daughter chromosomes move apart and migrate towards opposite poles. The centromere is pulled towards poles of the spindle and the arms of chromosomes are dragged behind.

Effect of whole process:
Anaphase-I of meiosis resulting in the formation of daughter nuclei having half of the chromosomes to their parents. Thus, the number of chromosomes is constant from generation to generation.

Question 12.
Write a note on necessity of two types of cell divisions in multicellular organisms.
Answer:
Necessity of two types of cell divisions :
There is complexity in structure and physiology of multicellular organisms. Thus, twc types of cell divisions are required in them. The process of meiosis cell division help in the maintenance of chromosome number in the species. Diploid (2x) organism produce haploid (x) gametes, which on fusion form diploid zygote, whereas zygote divides by mitosis division to form diploid organism. Healing of wound in multicellular organisms occurs by mitosis cell division.

Question 13.
Describe the prophase and anaphase of mitosis along with diagram.
Answer:
1. Prophase : (Pro = before, phasis – stage).

• In this stage chromosomes are found in the form of long threads which are called as chromatin net.

• Chromosomes become short and thick and each of them splits lengthwise forming two chromatids.
• At the ending of prophase nuclear membrane and nucleolus start to disappear.

2. Anaphase :

• During this stage the centromere of each chromosome gets divided into two which make the chromatids free from each other.
• Every chromatid with its centromere moves towards its poles.
• Termination of anaphase movement occurs when the chromosomes form a densely packed group at the two poles.

Question 14.
Draw labelled diagram of different stages of Mitosis cell division.
Answer:

Question 15.
Describe the metaphase of mitosis along with diagram.
Answer:

Metaphase : The following events occur during mitotic metaphase: ‘

• Nuclear membrane and nucleolus disappear during this stage.
• Chromosomes move towards equatorial plane of the cell.
• Chromatids of chromosomes become dense and separated from each other except centromere.
• Spindle fibres are formed.
• The centromeres of the chromosomes are arranged on equator and chromatid towards the opposite poles. Fig Metaphase

Question 16.
Write differences between following :
(i) Mitosis and Meiosis cell division,
(ii) Chromatin and Chromatid,
(iii) Centromere and centriole,
(iv) Centromere and Chromomere,
(v) Metaphase-I and Metaphase-II,
(vi) Zygotene and Pachytene,
(vii) Cell furrow and Cell plate.
Answer:
(i) Differences between Mitosis and Meiosis cell division :

(ii) Differences between Chromatin and Chromatid

(iii) Differences between Centromere and Centriole

(iv) Differences between Centromere and Chromomere

(v) Differences between Metaphase-I and Metaphase-II

(vi) Differences between Zygotene and Pachytene

(vii) Differences between Cell furrow and Cell plate

Question 17.
Define the following :
1. Homologous chromosomes,
2. Synapsis,
3. G₂alent,
5. Tetrad.
Answer:
1. Homologous chromosomes :
Homologous chromosomes are paired during zygotene and then form chiasmata at some places where crossing over takes place between them and finally resulting in the production of new characters and possibilities of organic evolution.

2. Synapsis:
Pairing of homologous chromosomes in zygotene substage of prophase-I of meiosis cell division is known as synapsis. The paired structure formed as result of synapsis is called as bivalent.

3. G₂-Phase :
In this phase, RNA and protein continues to be synthesised. The cell is now ready to enter the mitotic phase. The cells of this stage contain double amount of DNA.

4. Bivalent:
The pairs of homologous chromosome formed during meiosis-I. The chromosomes contract and become short and thick.

5. Tetrad:
In Pachytene stage of Prophase-I of Meiosis-I, Chromosomes split lengthwise but still remain attached at the point of centromere to form four chromatids called as Tetrad.

Question 18.
Which of the following statement is associated with :
(a) Prophase,
(b) Metaphase,
(c) Anaphase,
(d) Telophase,
(e) Interphase of mito-sis.
1. The nuclear membrane reappears.
2. Chromosomes are thickest and shortest.
3. Chromosomes begin to coil.
4. Centromere divides into two.
5. Nucleus is active, but chromosomes are not distinct.
6. Followed by cytokinesis.
7. Each chromosome consists of two chromatids.
Answer:
1. (d) Telophase
2. (b) Metaphase
3. (c) Anaphase
4. (c) Anaphase
5. (e) Interphase of mito-sis
6. (d) Telophase
7. (c) Anaphase

Cell Cycle and Cell Division Class 11 Important Questions Long Answer Type

Question 1.
Write important differences between Mitosis and Meiosis cell division.
Answer:
Differences between Mitosis and Meiosis