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Students get through the MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 9 Mechanical Properties of Solids

Mechanical Properties of Solids Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by deforming force?
Answer:
External force applied on a body is known as deforming force.

Question 2.
What is deformation?
Answer:
If an application of deforming force, a change in the shape or size of a body takes place, then it is said to be deformed.

Question 3.
What is meant by elasticity?
Answer:
It is the property of a body by virtue of which, it regains its original length, volume and shape when the deforming force is removed.

Question 4.
What is stress and its unit?
Answer:
The internal force of restitution per unit area of a deformed body is called stress.
Stress = \(\frac{\text { Force }}{\text { Area }}\)
Unit: SI unit of stress is Nm^-2.

Question 5.
Define plasticity.
Answer:
Plasticity is the property of remaining deformed even after the removal of deforming forces.

Question 6.
What do you mean by elastic limit?
Answer:
The maximum value of deforming force, by which if the force is applied on a body, the body looses its elastic property, is called elastic limit.

Question 7.
What do you mean by brittleness?
Answer:
Some substances are neither elastic nor plastic but they break easily into pieces, this property of a body is called brittleness and the bodies are called brittle substances.

Question 8.
What is difference between stress and pressure?
Answer:
Stress is internal force while pressure is external force.

Question 9.
How many types of stress are there?
Answer:
There are three types of stress :

1. Tensile stress or longitudinal stress,
2. Normal stress,
3. Tangential or shearing stress

Question 10.
What is strain? Write its unit.
Answer:
The ratio of change in length, volume or shape and its original length, volume or shape is called strain.
As it is the ratio of same quantities, therefore it has no unit.

Question 11.
What is longitudinal stress?
Answer:
It is the restoring force developed per unit cross-sectional area of a body when the length of the body increases in the direction of deforming force.

Question 12.
Define tangential stress.
Answer:
When the deforming force is applied along the plane of the area of cross-section of the body then the stress produced is known as tangential stress.

Question 13.
State the types of strain.
Answer:
There are three types of strain :

1. Longitudinal strain,
2. Shearing strain and
3. Volume strain.

Question 14.
What is longitudinal strain?
Answer:
Within elastic limit, it is the ratio of change in length to original length.
i.e., Longitudinal strain = \(\frac{\text { Change in length }}{\text { Initial length }}\)

Question 15.
Define volume strain.
Answer:
The change in unit volume due to deforming force is called volume strain.
Or
Within elastic limit, the ratio of change in volume to the initial volume is called volume strain, i.e.,
Volume strain = \(\frac{\text { Change in volume }}{\text { Initial volume }}\)

Question 16.
Define shearing strain.
Answer:
When a deforming force is applied on a body along its surface in such a way that the volume of the body remains constant, but the shape changes, then the body is said to be sheared and the strain produced is called shearing strain. Within elastic limit, it is measured by the ratio of the relative displacement of one plane to its distance from fixed plane, i.e.,
Shearing strain, θ = \(\frac{\Delta l}{l}\)
Δ l is relative displacement between two layers and l is the distance between the layers.

Question 17.
State Hooke’s law.
Answer:
Within elastic limit the stress is directly proportional to the strain. i.e.,
Stress ∝ Strain.

Question 18.
What are the types of modulus of elasticity?
Answer:
There are three types of modulus of elasticity :

1. Young’s modulus,
2. Bulk modulus,
3. Modulus of rigidity.

Question 19.
Define Young’s modulus and give its unit.
Answer:
Within elastic limit the ratio of longitudinal stress to the longitudinal strain is called Young’s moduluSIe.,
Young s modulus = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
Unit: Its SI unit is Nm^-2.

Question 20.
What is compressibility?
Answer:
‘The reciprocal of bulk modulus is called compressibility.

Question 21.
Define coefficient of rigidity.
Answer:
Within elastic limit, the ratio of tangential stress to the shearing strain is called coefficient of regidity.

Question 22.
Write the following in form of increasing elasticity of copper, iron, glass and rubber.
Answer:
Rubber < Glass < Copper < Iron.

Question 23.
If an ivory ball and an iron ball are dropped from the same height on a hard floor. Which one will rebounce higher?
Answer:
The ivory ball will rebounce higher because its elasticity is greater than that of iron ball.

Question 24.
Springs are made of steel and not of copper. Why?
Answer:
The Young modulus of steel is more than that of copper. If the same deforming force is applied on the spring of steel and spring of copper then the steel regain its initial position faster than that of copper spring. So the spring are made of steel.

Question 25.
What do you mean by restoring force?
Answer:
The force due to which the body regain its original shape and size is called restoring force.

Question 26.
State the condition for deforming force and restoring force to be equal and opposite.
Answer:
When the strain produced in the body is within the elastic limit then the deform¬ing force and restoring force are equal and opposite to each other.

Question 27.
On increasing temperature, Young modulus of elasticity increases or de¬creases.
Answer:
Decreases.

Question 28.
Why the work is done in stretching a wire?
Answer:
When a wire is stretched, the inter reaction force is developed within the wire and hence work is done against this force.

Question 29.
What happens with work to stretch a wire?
Answer:
The work done to stretch a wire is stored as elastic potential energy within wire.

Question 30.
If the length of the wire is halved, then what will be the effect in its Young modulus of elasticity?
Answer:
No effect on Young modulus of elasticity as it depends on the mass of the mate¬rial of the body not in its length.

Question 31.
If the length of wire is cut to its half then :
(i) What will be the effect on increase in length?
Answer:
Increase in length will get halved with respect to initial length.

(ii) Effect on maximum weight it can resist?
Answer:
No effect.

Question 32.
If a wire is replaced by another wire of same length and material but of double diameter, then :
(i) What will be increase in length for same given weight?
Answer:
Increase in length will remain one-fourth.

(ii) Effect on maximum weight that it can resist?
Answer:
Maximum weight it can resist will become four times.

Question 33.
Why the suspension wires are made up of quartz or phospher bronze?
Answer:
This is because the elastic after effect is negligible in quartz and phospher bronze.

Question 34.
In Searl’s apparatus, two similar wires of same material are taken, why?
Answer:
The temperature of room changes, hence the effect of temperature will be same on both the wires. Thus, the change in temperature does not cause any error.

Question 35.
In Searl’s experiment the readings are taken both by loading and unload¬ing the weights, why?
Answer:
There are two reasons :

1. It ensures that the experiment is performed within elastic limit.
2. The errors produced due to elastic fatigue, torsion and backlash ‘are removed.

Question 36.
Why the readings are taken after some time in the experiments of elasticity?
Answer:
To remove the error caused by elastic after effect.

Question 37.
What is elastic hysteresis?
Answer:
On applying deforming force on a body the strain is produced in it on removal of deforming force some time body do not regain its original position completely. Some strain remains left in that body. It is called elastic hysteresis.

Question 38.
Write down the effect of impurities on elasticity.
Answer:
In the presence of impurities the elasticity of the bodies is increased.

Question 39.
What is the effect of temperature on elasticity?
Answer:
With increase in temperature, usually the elasticity of the bodies decreases and with decrease in temperature, the elasticity increases. But the effect of temperature is negligible in the elasticity of’Invar steel’.

Mechanical Properties of Solids Class 11 Important Questions Short Answer Type

Question 1.
State the difference between elastic and plastic bodies.
Answer:
Difference between Elastic and Plastic bodies :

Question 2.
Steel is more elastic than rubber. Explain with reason.
or
Why steel is more elastic than rubber?
Answer:
Let two wires of steel and rubber of equal length L and equal area of cross-section A are taken. Both are subjected by a force F, so that the elongation produced in steel is l_s and in rubber l_r.
∴ Modulus of elasticity of steel, Y_s = \(\frac{F \cdot L}{A \cdot l_{s}}\) …(1)
and Modulus of elasticity of rubber, Y_r = \(\frac{F \cdot L}{A \cdot l_{r}}\) …(2)
Now, dividing eqn. (1) by eqn. (2),
\(\frac{Y_{s}}{Y_{r}}=\frac{l_{r}}{l_{s}}\)
∵ l_r > l_s
∴ Y_s > Y_r
Hence, it is clear that steel is more elastic than rubber.

Question 3.
What do you mean by Young’s modulus of elasticity. Write its unit and dimensional formula. Derive the formula for Young’s modulus and hence define it.
Answer:
Within elastic limit, the ratio of longitudinal stress to the longitudinal strain is called Young’s modulus.
i.e., Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)

SI unit is Nm^-2 and dimensional formula is [ML^-1T^-2].
Formula: Let a mass M is suspended by a wire of length L and hence the elongation produced is l. If the radius of the wire is r, then
Stress = \(\frac{\text { Force }}{\text { Area }}=\frac{M g}{\pi r^{2}}\)
and strain = \(\frac{l}{L}\)
But Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
∴ Y = \(\frac{M g / \pi r^{2}}{l / L}\)
Or
Y = \(=\frac{M g L}{\pi r^{2} l}\)
Within elastic limit, Young’s modulus of elasticity is numerically equal to that force, which extends the length of wire of unit cross-sectional area by the original length i.e., the final length becomes double the initial length.

Question 4.
What do you mean by bulk modulus? Establish the formula for bulk modulus.
Answer:

Bulk modulus : Within elastic limit, the normal stress is directly proportional to the volume strain.
∴ K = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
Formula : Let initial volume of a gas is V and pressure P. If the pressure is increased to P + p, so that its volume becomes V- v, hence the decrease in volume is v.
Since, force acting per unit area is called pressure. Hence,
Normal stress = Increase in pressure = (P + p)- P = p
Volume strain = – \(\frac{v}{V}\)
By Hooke’s law, within elastic limit,
Bulk modulus of elasticity = \(\frac{\text { Normal stress }}{\text { Volume strain }}\)
K = \(\frac{p}{-\frac{v}{V}}=\frac{-p V}{v}\)
-ve sign shows that, with increase in pressure, volume decreases.

Question 5.
What do you mean by coefficient of rigidity? Derive formula for it.
Answer:
Coefficient of rigidity : Within elastic limit the ratio of tangential stress and shearing strain is called modulus of rigidity.
i.e., η = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)

Formula derivation : Let ABCDEFGH be a cuboid (of side l). A tangential force F is applied on the surface ADEF, If the area of surface is A and angular displacement due to force F is θ, then
Tangential stress = \(\frac{F}{A}\)
If the relative displacement between the layers is
AA’ = X, then
Shearingstrain, θ = \(\frac{A A^{\prime}}{A B}=\frac{x}{l}\)
Modulus of rigidity = \(\frac{\text { Tangential stress }}{\text { Shearing strain }}\)
∴ η = \(\frac{F / A}{x / l}=\frac{F l}{x A}\)

Question 6.
Why are girders I shaped?
Answer:

Girders of iron are made of I shapped.
Let l be the length of a rod, b be the breadth and d be its thickness. If weight W be suspended at its middle point the bending (δ) is given by
δ = \(\frac{\mathrm{W} l^{3}}{b d^{3} \mathrm{Y}}\) …(1)
For the strength of girder, should be smaller, δ will be smaller if
(a) Y is large.
(b) From eqn. (1) it is clear that δ∝\(\frac{1}{b}\) and δ∝\(\frac{1}{d^{3}}\) Therefore depth d should be greater than breadth b.
Therefore, girders of made of I shape.

Question 7.
Define Poisson’s ratio. What is its unit? What is its value?
Answer:
When a deforming force F is applied to a wire, then its length increases, but its radius or breadth decreases. This strain which occurs along the width of the body is called lateral strain. Within elastic limit, the lateral strain is proportional to longitudinal strain.
i.e., Lateral strain ∝ Longitudinal strain
or
Lateral strain = σ × Longitudinal strain
or
Poisson’s ratio,
Where D = Initial diameter of wire, d = Change in diameter, L = Original length of wire, l = Change in length.
It has no unit. Theoretical values lies from – 1 to 0.5.

Question 8.
Derive an equation for the force acting on the clamps, when the wire gets cooled. f’
Answer:

Let a wire of length L and area of cross-section A be clamped between the two ends. If it is being cooled, then its temperature decreases by Δ t°C, as a result it gets strinked and so it tries to contract the clamps. This force has to be found out. If Y is the Young’s moduliis of elasticity and a is the coefficient of linear expansion, then
α = \(\frac{\text { Change in length }(l)}{\text { Original length }(L) \times \text { Change in temp. }(\Delta t)}\)
Hence, αLΔt = l or \(\frac{l}{L}\) = αΔt
∴ Strain = αΔt …(1)
But, Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Stress = Y × Strain= YαΔt …(2)
Moreover, Stress = \(\frac{\text { Force }}{\text { Area }}\)
or
Force = Stress × Area = YαΔt × A
∴ F =YAαΔt
This is the required relation.

Question 9.
Figure shows the strain-stress curve for a given material. What are:
(a) Young’s modulus and
(b) Approximate yield strength for his material?

Answer:
From the given graph for a stress of 150 × 10⁶ Nm^-2, the strain is 0.002.
(a) Young’s modulus of the material (Y) is given by
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{150 \times 10^{6}}{0.002}=\frac{150 \times 10^{6}}{2 \times 10^{-3}}\)
= 75 ×10⁹ Nm^-2
= 7.5 × 10¹⁰Nm^-2.

(b) Yield strength of a material is defined as the maximum stress it can sustain.
∴ From graph, the approximate yield strength of the given material = 300 × 10⁶ Nm2 = 3 × 10⁸ Nm^-2. Actually, it is slightly less than 3 × 10⁹ Nm^-2.

Question 10.
The stress-strain graphs for materials A and if are shown in figure (a) and (b):

The graphs are drawn to the same scale :
(a) Which of the materials has greater Young’s modulus?
(b) Which of the two is the stronger material?
Answer:
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence Young’s modulus = \(\left(\frac{\text { Stress }}{\text { Strain }}\right)\) is greater forA than that of B.

(b) Strength of a material is determined by the amount of stress (load) required to cause breaking or fracture of material corresponding the breaking point.
∴ Material A is stronger than B as it can withstand more load without breaking than the material B corresponding to point D.

Question 11.
Prove that energy stored per unit volume in a stretched string is, given by:
U = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × Young’s modulus × (Strain )²
Derive the formula for the potential energy of a stretched string.
Answer:
Consider a wire of length L and it is extended by a force F due to which its extension takes place by l.
If A is the area of cross-section of the wire and Fis the Young’s modulus of elasticity,
then
Y = \(\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\) = \(\frac{F / A}{l / L}\) = \(\frac{F L}{A l} .\)
Or
YAl = Fl or F = \(\frac{Y A l}{L}\) …(1)

When, the wire is stretched by a force F, then work is done by the deforming force
which is stored in the wire in the form of potential energy.
Small amount of work done to stretch by dl i.e.,
dW = Fdl= \(\frac{Y A l}{L}\)dl
∴ The total amount of work done will be

Where, \(\frac{l}{L}\)is the longitudinal strain
Work done per unit volume is called elastic potential energy per unit volume (U).
∴ U= \(\frac{1}{2}\) × Y × (Strain)², [from eqn. (3)]
But, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ U= \(\frac{1}{2}\) × \(\frac{\text { Stress }}{\text { Strain }}\) × (Strain)²
or
U= \(\frac{1}{2}\) × Stress × Strain. Proved.

Mechanical Properties of Solids Class 11 Important Numerical Questions

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10^-5 m²stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0 × 10^-5
m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Here, for steel wire,
Length of wire, l₁ = 4.7 m
Area of cross-section, a₁ = 3.0 × 10^-5 m²
Stretching, Δl₁, = Δl(say)
Stretching force on steel, F₁ = F
For copper wire,
Length of wire,l₂ = 3.5 m
Area of cross-section, a₂ = 4.0 × 10^-5 m²
Stretching, Δl₂ = Δl(given);
Stretching force on copper, F₂ = F
Let Y₁ and Y₂ be the Young’s modulus of steel and copper wire respectively.

Dividing eqn. (1) by eqn. (2), we get

= \(\frac{18 \cdot 8}{10 \cdot 5}\) = 1.79 = 1.8:1

Question 2.
Two wires of diameter 0.25 cm. One made of steel and the other made of brass are loaded as shown in fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. If Young’s modulus of steel and brass are 2.0 × 10¹¹ Pa and 0.91 × 10¹¹ Pa, then compare the elongation of the steel and the brass wires.
Solution:

Using, Y = \(\frac{M g L}{\pi r^{2} l}\)
l = \(\frac{M g L}{\pi r^{2} Y}\)
For brass wire,
M= 6 kg, L = 1.0 m, Y = 0.91 × 10¹¹N/m²
and 2r = 0.25 cm = 0.25 × 10^-2m
r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{6 \times 9 \cdot 8 \times 1 \cdot 0}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 0 \cdot 91 \times 10^{1}}\)
= 1.3 × 10^-4 m.
For steel wire,
M = 6 + 4 = 10 kg,L = 1.5 m, Y = 2.0 × 10¹¹ N/m²and 2r = 0.25cm
∴ r = 0.125 × 10^-2m
Putting the values in formula,
l = \(\frac{10 \times 9 \cdot 8 \times 1 \cdot 5}{3 \cdot 14 \times\left(0 \cdot 125 \times 10^{-2}\right)^{2} \times 2 \cdot 0 \times 10^{1}}\)
= 1.5 × 10^-4m

Question 3.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N. force, producing only elastic deformation. Calculate the resulting strain. Y for copper = 1.1 × 10¹¹ Nm^-2.
Solution:
Here, Y = 1.1 × 10¹¹Nm^-2
A = Area of cross-section
= 15.2 mm × 19.1 mm
= 15.2 × 10^-3m × 19.1 × 10^-3 m
Force, F = 44500 N
Resulting Strain = Longitudinal Strain = ?
∴ Y = \(\frac{\text { Stress }}{\text { Strain }}\)
or
Strain \(=\frac{\text { Stress }}{Y}=\frac{F}{A Y}\)
or
Longitudinal Strain = \(\frac{44500}{15 \cdot 2 \times 19 \cdot 1 \times 10^{-6} \times 1 \cdot 1 \times 10^{1}}\)
= 139.34 × 10^-3 m = 0.139.

Question 4.
A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress is not to exceed 10⁸ Nm^-2, what is the maximum load the cable can support?
Solution:
Here, radius of steel cable r = 1.5cm = 1.5 × 10^-2 m
Max. Stress = 10⁸ Nm^-2
∴ Area of cross-section of cable A = πr^-2 = π(1.5 × 10^-2)²
Maximum load the cable can withstand = Maximum force ? We know that,
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross-section }}\)
or
Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π × (1.5 × 10^-2)²
= 3.142 × 2.25 × 10⁸ × 10^-4 N
or
Maximum load the cable can withstand = 7.1 × 10⁴N.

Question 5.
A 4.0 metre long copper wire of area of cross-section 1.2 cm² is stretched by a force of 4.8 × 10³N. If the Young’s modulus of copper is Y = 1.2 × 10¹¹ N/m², then calculate:
(i) Stress,
(ii) Strain,
(iii) Increase in length of wire.
Solution:
Given : L=4.0 metre, A = 1.2 cm² = 1.2 × 10^-4m², F = 4.8 × 10³ N, Y= 1.2 × 10¹¹ N/m²

(i) Stress : The stress on wire will be = \(\frac{F}{A}=\frac{4 \cdot 8 \times 10^{3}}{1 \cdot 2 \times 10^{-4}}\)
∴ Stress = 4 × 10⁷ N/m².

(ii) Strain: The Young’s modulus of wire is Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Strain = \(\frac{\text { Stress }}{\mathrm{Y}}=\frac{4 \times 10^{7}}{1 \cdot 2 \times 10^{11}}\)
∴ Strain = 3.3 × 10^-4

(iii) Increase in Length of wire : From the formula of Young’s modulus
Y = \(\frac{\mathrm{FL}}{\mathrm{A} . l}\)
∴ l = \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
\(=\frac{4 \cdot 8 \times 10^{3} \times 4}{1 \cdot 2 \times 10^{11} \times 1 \cdot 2 \times 10^{-4}}\)
= 13.2 × 10^-4 = 1.32 mm.

Question 6.
Two wires made of same material, length of first wire is half the length of second wire and its diameter is double the diameter of second wire. If same weight is suspended in both wires then find the ratio of increase in length.
Solution:

Question 7.
On stretching a gold wire, Its length Increases by 1%. Find out linear strain of it, if its coefficient of Young modulus is 8 × 10¹⁰ N/m². Find out the stress produced.
Solution:
According to question,
l = 1%, Y = 8 × 10¹⁰ N/m²
∴ Linear strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
= \(\frac{1}{100}\)
= 0.01
Stress = Y × Strain = (8 × 10¹⁰) × (0.01)
=8 ×10⁸N/m².

Question 8.
2 kg weight elongates a wire of 3 m length by 1 mm. The diameter of wire is 1 mm. Determine Young’s modulus of the material of the wire.
Solution:
Given : M = 2 kg, L = 3 m, l = 1 mm = 10^-3 m, r= \(\frac{1}{2}\)mm= 0.5 x 10^-3m, g = 9.8 ms^-2. ,
We have, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Y = \(\frac{2 \times 9 \cdot 8 \times 3}{3 \cdot 14 \times\left(0 \cdot 5 \times 10^{-3}\right)^{2} \times 10^{-3}}\)
=  \(\frac{58 \cdot 8 \times 10^{6} \times 10^{3}}{3 \cdot 14 \times 0 \cdot 25}\)
∴ Y = \(\frac{588 \times 10^{12}}{314 \times 25}\)
= 7.49 × 10¹⁰Nm^-2

Question 9.
The diameter of a wire is \(\frac{2}{\sqrt{\pi}}\) cm. The force required to double the length is 2 × 10¹²dyne. Calculate Young’s modulus of the material of the wire.
Solution:
Given: r = \(\frac{1}{2}\) × \(\frac{2}{\sqrt{\pi}}\)cm = \(\frac{1}{\sqrt{\pi}}\) 10^-2 m, L = l
and Mg = 2 × 10¹²dyne = 2 × 10¹² × 10^-5N = 2 × 10⁷N
Now, we have Y = \(\frac{M g L}{\pi r^{2} l}\)

∴ Y = \(\frac{2 \times 10^{7}}{10^{-4}}\) = 2 × 10¹¹ Nm^-2

Question 10.
Calculate the force required to elongate 1 m long the 2 mm thick wire by 1 mm. (Y = 2 x 10¹¹ Nm^-2)
Solution:
Given: L = 1m,r = \(\frac{2}{2}\) = 1mm =10^-3m = 10^-3m, l = 1mm 10^-3m Y = 2 × 10¹¹Nm^-2
Frmula, Y = \(\frac{M g L}{\pi r^{2} l}\)
or
Mg = \(\frac{Y \pi r^{2} l}{L}\)
\(=\frac{2 \times 10^{11} \times 3 \cdot 14 \times\left(10^{-3}\right)^{2} \times 10^{-3}}{1}\)
= 6.28 × 10¹¹ × 10^-9
= 6.28 × 10² = 628 N.

Question 11.
Calculate the work done in extending a wire of length lm and of cross-section 1mm2 through 2mm. (Given y = 2 × 10¹¹Nm^-2)
Solution:
Given: L= 1m, A = nr² = 1mm² = (1 × 10^-3m)²= 10^-6m²
l = 2mm = 2 × 10^-3m and Y = 2 × 10^-11 Nm^-2
Strain = \(\frac{l}{L}=\frac{2 \times 10^{-3}}{1}\) = 2 × 10^-3
∴ Volume of wire = A.L = 10^-6 × 1 = 10^-6m³
Work done = \(\frac{1}{2}\) × Y × (Strain)² × Volume
= \(\frac{1}{2}\) × 2 × 10¹¹ × (2 ×10^-3)¹¹× 10^-6
= 0.4 J.

Question12.
A rubber ball while taking to the bottom of a lake 200 m deep its volume de-creases by 0.1%. Calculate the Bulk modulus of rubber. Density of water is 10³ kg/ m³ and g = 10 m/sec².
Solution:
Given : Depth h = 200m, Density of water = 10³ kg/m³ Acceleration due to gravity g = 10 m/sec²,
Volume strain = 0.1% = –\(\frac{0 \cdot 1}{100}\)
∴ Bulk modulus K = \(\frac{\frac{-p}{\Delta v}}{\frac{V}{V}}\)
or
K = \(\frac{-200 \times 10^{3} \times 10}{\frac{-0 \cdot 1}{100}}\)
= 2 × 10⁹N/m²

Question 13.
A weight of 20 kg is suspended in a wire. Area of cross-section of it is 1mm². The length of the wire in extended state is 6 m. On withdrawing the weight its length become 5.995 m. Find out Young modulus of elasticity.
Solution:
Given : M = 20 kg, A = 1mm² = 10^-6m², L = 6m,
Change in length = 6 – 5.995 = 0.005 m
∴ Y = \(\frac{M g L}{A l}\)
or
Y \(=\frac{20 \times 9 \cdot 8 \times 6}{10^{-6} \times 0-005}\)
= 2.35 × 10¹¹N/m².

Question 14.
On iron wire of radius 0.5 mm is heated up to 300°C and then it is fixed between two clamp. When the temperature reduces to 30°. What will be the force acting upon the clamp. (α = 1.2 × 10^-5/°C and Y= 1.4 × 10¹² dyne/cm²)
Solution:
Given : r = 0.5mm = 0.05 cm, α = 1.2 × 10×^-5/°C, Y = 1.4 × 10¹²dyne/cm²
ΔT = 300 – 30 = 270° C
From F = YΔ α. AT, we get
F = Y πr² α ΔT
∴ F = 1.4 × 10¹² × 314 × (0.05)² × 1.2 × 10^-5 × 270
F = 3.56 × 10⁷ dyne.

Mechanical Properties of Solids Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
A wire of radius r and length L on which a mass M is suspended so that its increase in length is T. Young modulus will be:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)
(b) \(\frac{\mathrm{Mgl}}{\pi r^{2} \mathrm{~L}}\)
(c) \(\frac{\pi r^{2} L}{M g l}\)
(d) \(\frac{\pi r^{2} l}{\mathrm{MgL}}\)
Answer:
(a) \(\frac{\mathrm{MgL}}{\pi r^{2} l}\)

Question 2.
Which substance is more elastic:
(a) Glass
(b) Steel
(c) Plastic
(d) Rubber.
Answer:
(b) Steel

Question 3.
Magnitude of Poisson’s ratio is in between :
(a) – 1 and \(\frac{1}{2}\)
(b) –\(\frac{3}{4}\) and –\(\frac{1}{2}\)
(c) \(\frac{1}{2}\) and 1
(d) 1 and 2
Answer:
(a) – 1 and \(\frac{1}{2}\)

Question 4.
Work done per unit volume for a wire is :
(a) Stress × Strain
(b) \(\frac{1}{2}\)× Stress × Strain
(e) 2(Stress × Strain)
(d) Stress/Strain.
Answer:
(b) \(\frac{1}{2}\)× Stress × Strain

Question 5.
Four wire made of same material, equal weight is suspended on them, increase in length will be maximum for wire :
(a) Length 50 cm and diameter 0.5 mm
(b) Length 100 cm and diameter 1 mm
(c) Length 200 cm and diameter 2 mm
(d) Length 300 cm and diameter 3 mm.
Answer:
(a) Length 50 cm and diameter 0.5 mm

Question 6.
10 newton force is needed to break a wire of radius one millimetre. What is force needed to break a wire of radius 3 mm :
(a) 90N
(b) \(\frac{10}{3}\)N
(c) \(\frac{10}{9}\)N
(d) 30N
Answer:
(a) 90N

Question 7.
In general modulus of rigidity is than modulus of elasticity:
(a) Less
(b) More
(c) Equal
(d) None of these.
Answer:
(a) Less

Question 8.
A wire of length L, Area of cross-section A, Young modulus of elasticity Y and coefficient of linear expansion a is heated up to t°C. Force experience on wire will be :
(a) γAαt
(b) γAαLt
(c) tALα
(d) \(\frac{t \alpha \mathrm{L}}{\mathrm{A}}\)
Answer:
(a) γAαt

2. Fill in the blanks:

1. SI unit of stress is …………………
Answer:
Newton/metre²

2. On increasing temperature, coefficient of elasticity ………………….
Answer:
decreases

3. Within elastic limit ……………….. is directly proportional to strain.
Answer:
stress

4. Rubber is ………………. elastic than steel.
Answer:
less

5. For an ideal rigid body Young modulus is ………………..
Answer:
infinity

3. Match the following:

Answer:
1. (d) Applicable within elastic limit
2. (e) Volume strain
3. (a) Proportional to area of cross-section
4. (b) Zero
5. (c) Shearing strain

4. Write true or false:

1. Hooke’s law is applicable within elastic limit.
Answer:
True

2. Diamond can be considered as rigid body.
Answer:
True

3. Product of strain and stress is equal to stored energy.
Answer:
False

4. Young modulus of elasticity is dimensionless.
Answer:
False

5. Breaking stress depends on body of material.
Answer:
True

6. Young modulus of elasticity is valid for solid only.
Answer:
True

7. Rubber is more elastic than steel.
Answer:
False

8. Hookes law is defined within elastic limit.
Answer:
False

9. Quartz does not exhibit elastic after effect practically.
Answer:
True

10. It is difficult to twist a small rod in comparison to long rod.
Answer:
True

11. A hollow rod of same length and mass is more strong than a solid rod.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 10 Mechanical Properties of Fluids which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 10 Mechanical Properties of Fluids

Mechanical Properties of Fluids Class 11 Important Questions Very Short Answer Type

Question 1.
What do you mean by pressure? Is it scalar or vector? Write its SI unit.
Answer:
The force acting normally on unit area of the surface is known as pressure. It is a scalar quantity. The SI unit of pressure is N/m².

Question 2.
What is thrust? Write its SI unit.
Answer:
The total force acting normally on a surface, by the liquid pressure is called thrust.
Unit: Its SI unit is N.

Question 3.
What is Pascal’s law? State its two applications.
Answer:
The pressure in a fluid in equilibrium is the same everywhere if the effect of gravity can be neglected.
Or
Pressure applied to any part of an enclosed fluid at rest is transmitted equally to every portion of the fluid and the walls of the containing vessel.
Applications :

• Hydraulic lift,
• Hydraulic brakes.

Question 4.
Why is the dam of water reservoir thick at the bottom?
Answer:
It is so because the pressure of water in the reservoir increases with depth.

Question 5.
What do you understand by atmospheric pressure? What are its various units?
Answer:
The force exerted by the air of atmosphere, per unit area on the surface of earth is called atmospheric pressure.
latm = 1.013 × 10⁵Nm^-2 = 1 Pascal.

Question 6.
Explain 1 bar and 1 tor.
Answer:
1 bar: It is the unit of pressure.
1 bar = 1.0 × 10⁵Nm^-2 =10⁵Pa.
1 tor: It is also unit of pressure named after Torricelli’s.

Question 7.
What is the value of atmospheric pressure?
Answer:
1.013 ×10⁵ Newton/metre².

Question 8.
“Atmospheric pressure is 760 mm of Hg”. What it means?
Answer:
It means that it equilibrium the 760 mm height of mercury column pressure.

Question 9.
What is known as Torricellian vacuum?
Answer:
The vacuum is the simple barometer above the mercury column is known as Torricellian vacuum.

Question 10.
What will be the effect on mercury level of barometer if a small hole is made in the upper portion?
Answer:
The level of mercury will come down.

Question 11.
What will be the effect on the level of mercury in a barometer if a water droplet is inserted inside it.
Answer:
The level of mercury will fall due to the water vapour formation when a water droplet is inserted inside the barometer.

Question 12.
What will be the effect on level of mercury of the barometer is baught to hill?
Answer:
The level of mercury will fall as atmospheric pressure at hill is less.

Question 13.
How atmospheric pressure is change with altitude?
Answer:
Atmospheric pressure will decrease with altitude.

Question 14.
Which liquid is used in barometer?
Answer:
Mercury is used in barometer.

Question 15.
The surface tension of a liquid does not depend upon the area of the surface. Why?
Answer:
Because the surface tension is defined as the force acting on unit length of the line drawn on the liquid surface.

Question 16.
What is difference between cohesive force and adhesive force?
Or
What do you mean by cohesive force and adhesive force?
Answer:
Cohesive force: The force of attraction between the molecules of same sub-stance is called cohesive force.
Adhesive force: The force of attraction between the molecules of different sub-stances is called adhesive force.

Question 17.
What is surface tension? Give its SI unit.
Answer:
In equilibrium, the force acting per unit length on an imaginary line drawn on the liquid surface, perpendicular to the line and tangential to the surface is called surface tension.
Its SI unit is Nm^-1.

Question 18.
How can be the surface tension of a liquid be reduced?
Answer:
Surface tension of a liquid can be reduced by increasing temperature.

Question 19.
Soaps and detergents are helpful in cleaning the clothes. Why?
Answer:
The soap or detergent reduces the surface tension of water. Therefore, the soap solution can penetrate deep into the pores, where water cannot normally reach and thus dirt and dust are removed.

Question 20.
Why hot soup tastes better than cold soup?
Answer:
Soup is a liquid. Whenever a liquid is heated, its surface tension decreases and it gets more spread up in the tongue and it reaches to the sensitive layers rather than cold soup. Hence, hot soup tastes better than cold soup.

Question 21.
At which surface the pressure is excess, on what factors it depends?
Answer:
The excess of pressure is at concave surface. This is because of surface tension. The excess pressure depends upon :

• Surface tension of liquid and
• Radius of curvature of liquid surface.

Question 22.
What is angle of contact? How do the angles of contact differ for the liquids which (i) wet the solid surface and (ii) does not wet the solid surface?
Answer:
The angle that the tangent to the liquid surface at the point of contact makes with the solid surface inside the liquid is called the angle of contact. The angle of contact is acute for the liquids which wet the solid surface and is obtuse which do not wet.

Question 23.
What is capillarity? Give example.
Answer:
The phenomenon of ascending and descending of liquid in a capillary is called capillarity. In kerosene lamp oil rises in the wick due to capillarity.

Question 24.
Towels are used to wipe and dry the body after bathe. Why?
Answer:
A towel has number of pores, which act as capillary, hence it soaks the water of the body.

Question 25.
If wax is polished on clothes, they become water proof. Why?
Answer:
When wax is polished, then the pores of the threads, which act as capillaries get closed and it cannot suck water. Hence it becomes water proof.

Question 26.
Why the blood pressure in humans is greater at the feet than at the brain? (NCERT)
Answer:
The height of the blood column is quite large at the feet than at the brain. There fore the blood pressure in humans is greater at the feet than at the brain.

Question 27.
What are the factor that effect air pressure at a place?
Answer:

1. Height of atmosphere
2. Density of atmosphere
3. Acceleration due to gravity.

Question 28.
On which principle hydraulic lift work?
Answer:
Hydraulic lift work on Pascal’s law.

Question 29.
Why the droplets of liquid are spherical?
Answer:
The free surface of liquid has tendency to achieve to least surface area, because of its potential energy. The sphere has the least surface area therefore droplets are spherical.

Question 30.
The oil spreads over the water surface but when the water is put into oil it becomes spherical droplets. Why?
Answer:
This is because the adhesive force between the molecules of water and oil is greater than the cohesive force of oil.

Question 31.
When a glass rod is heated at one end, that end becomes spherical, why?
Answer:
When the end of a glass rod is heated, it melts and changes into liquid, hence due to the surface tension the liquid tries to achieve the least surface area, thus it becomes spherical.

Question 32.
In which material of capillary the water will descent not ascent? What will happen if a capillary made up of silver is dipped into water?
Answer:
The water will descent in the capillary made up of paraffin wax. The water will neither rise nor descent in silver capillary.

Question 33.
When two lead pieces are pressured against each other they stick together, why?
Answer:
When two pieces of lead are pressured against each other, then the molecules come into contact and the cohesive force acts between the molecules and hence they stick together.

Question 34.
The small needle of steel floats in the water but it sinks into water with little soap solution. Why?
Answer:
The needle floats in the water because the surface tension force balance the weight of the needle. When the soap solution is added into water its surface tension decrease and hence the force which balances the needle decreases and needle sinks.

Question 35.
Why the farmers plough the fields after rain?
Answer:
When the fields are ploughed the capillaries of earth broken and the under- ground water remains there which is used by the plants.

Question 36.
What is molecular range?
Answer:
The maximum distance, within which the molecules of a liquid experience cohesive force.
Unit: It is of the order 10^-9 m.

Question 37.
What is meant by surface film of a liquid?
Answer:
Surface film is a thin film of liquid near its surface and having thickness equal to the molecular range for that liquid.

Question 38.
What is viscosity ?
Answer:
The inherent property of the fluids (liquids or gases) by virtue of which they oppose the relative motion between their layers is called viscosity.

Question 39.
What do you mean by ideal liquid?
Answer:
The liquid whose viscosity and compressibility both are zero called an ideal liquid.

Question 40.
What do you understand by streamline flow?
Answer:
When a fluid flows in such a way, that at any point the velocity of all the particles of fluid remain constant, then its flow is called streamline flow.

Question 41.
What do you mean by turbulent flow?
Answer:
The flow of the fluid in which the motion of the particles are irregular and zig-zag is called turbulent flow.
Fast-flowing rivers during the flood, storms are the examples of turbulent flow.

Question 42.
Two streamline flow do not intersect each other, why?
Answer:
If two streamline flow intersects at a point, then at that point, there will be two directions of velocity of the particles of fluid, which is not possible, hence they do not intersect.

Question 43.
State Stoke’s law.
Answer:
Stoke’s law states that, if a sphere of radius r moving with uniform velocity v, through a homogeneous incompressible fluid of viscosity p, the retarding force Fis given by
F = 6 πηrν.

Question 44.
What is terminal velocity?
Answer:
When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.

Question 45.
Define coefficient of viscosity.
Answer:
The coefficient of viscosity of a liquid is the viscous force acting tangentially per unit area of a liquid layer having unit velocity gradient in a direction perpendicular to the direction of flow of liquid.

Question 46.
What is critical velocity?
Answer:
The maximum velocity of a fluid, below which the flow remains streamline flow, is called critical velocity.

Question 47.
State the principle of continuity.
Answer:
When a non-viscous and incompressible liquid flows through a tube of non-uniform bore and its flow is streamlined, then the product of velocity of flow and the area of cross-section of the tube at every transverse cross-section remains constant.
i.e., A.v = a constant
or A₁v₁=A₂v₂.

Question 48.
How many types of energy does a flowing liquid possess?
Answer:
The flowing liquid possesses three types of energies :

1. Potential energy,
2. Kinetic energy and
3. Pressure energy.

Question 49.
Two small light balls are hanging near to each other by two threads. If air is blown between them, they come near to each other, why?
Answer:
When air is blown the speed of air increases, hence the pressure between the balls decreases. Thus due to outer pressures, the balls come closer.

Question 50.
Air is blown below the pan of a balance in equilibrium, what will happen?
Answer:
When the air is blown below the pan, the velocity of air is increased and hence the pressure below the pan will decrease. Hence the air above the pan, presses the pan downward, thus that pan will be lowered down.

Question 51.
When a fast-moving train passes through a platform, all the dust particle and others attract toward train, why?
Answer:
When a fast-moving train passes through a platform according, Bernoulli’s theorem pressure of that place get dropped at once, so the dust particle and other’s get attracted toward the train.

Mechanical Properties of Fluids Class 11 Important Questions Short Answer Type 

Question 1.
Explain why :
(a) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
(b) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. (NCERT)
Answer:
(a) Atmospheric pressure at height of 6km
h = 6km = 6× 10₃m, d = 1.30kg m^-3, g = 9.8 ms^-2
∴ P = hdg = 6 × 10³ × 1.30 × 9.8 = 0.7644 × 10⁵Pa
Atmospheric pressure at sea level = 1.013 × 10⁵Pa
Thus, it is clear that the atmospheric pressure at height of 6 km is nearly half the atmospheric pressure. But the atmosphere is extended nearly up to 100km from the surface of earth. It is so because the density of air decreases as we go up from the surface of earth.

(b) When the force is exerted at a point on the liquid, then the pressure is equally transmitted in all directions. Thus, the pressure is not concerned with direction. Hence, the hydrostatic pressure is a scalar quantity.

Question 2.
Explain why :
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) (NCERT)
Answer:
(a) When a drop of liquid is put on the solid surface then three interfaces are formed:

1. Solid-liquid interface,
2. Solid-vapour interface,
3. Liquid-vapour interface.

Let the surface tension in these interfaces are T_SLT_SV and T_LV respectively. Then at point 0 for the equilibrium of liquid
T_SV = T_SL+T_LV cos θ,
where θ is angle of contact.
∴ cos θ = \(\frac{T_{S V}-T_{S L}}{T_{L V}}\)
For mercury – glass pair Tsv < TSL ∴ cos θ is negative i.e., θ is obtuse angle.
For water-glass Tsv >T_SL
cos θ is positive i.e., θ is acute angle.

(b) The angle of contact for water glass pair is acute angle. Hence, the water spreads over the surface of glass. The angle of contact for mercury-glass pair the angle of contact is obtuse angle. Hence, the mercury tries to contract and acquire the shape of drop so that the angle of contact becomes obtuse angle.

In the other words, the adhesive force between molecules of water and glass is greater than the cohesive forces between the molecules of water. Hence, water wets the glass. While the adhesive force between the molecules of mercury and glass is less than cohesive forces between the molecules of mercury, hence the mercury does not wet the glass.

Question 3.
Explain why :
(a) Surface tension of a liquid is independent of the area of the surface.
(b) A drop of liquid under no external forces is always spherical in shape. (NCERT)
Answer:
(a) The force acting per unit length on either side of an imaginary line at right angles on the surface of liquid is called surface tension. This force does not depend on the area of surface.

(b) In the absence of external force, only surface tension acts on the liquid surface. The surface tension tries to minimize the surface area of liquid. For the given volume the surface area of sphere is least. Hence, in the absence of external forces the liquid drop acquires the spherical shape.

Question 4.
Explain why :
(a) To keep a piece of paper horizontal, you should blow over, not under it.
(b) When we try to close a water tap with our Angers, fast jets of water gush through the openings between our Angers. (NCERT)
Answer:
(a) When a piece of paper (paper strip) is blown over, then the velocity of air in that region increases, hence the kinetic energy of air increases, then according to Bernoulli’s theorem the pressure in that region decreases than the pressure in below region. Hence, due to decrease in pressure in upper region, the paper strip becomes horizontal.

(b) When we try to close the water tap with our fingers. The area for the flow of liquid decreases.
According to principle of continuity Av = constant.

Obviously if A decreases then v increases. Thus, the velocity of flow of liquid increases and fast jets of water gush through the opening between our finger.

Question 5.
Explain the construction and working of hydraulic lift.
Answer:
Construction: The construction of a hydraulic lift is shown in the Fig. In it there are two cylinders C₁ and C₂ of different area of cross-sections. These are interconnected by means of a narrow tube. Two frictionless and watertight pistons A and B are fitted in the cylinders C₁ and C₂ respectively. The area of cross-section of the piston A is small while that of the piston B is quite large. The body to be lifted up is placed on the platform attached to the piston B. The force is applied on the piston A. The two cylinders are filled with water.

Principle: The hydraulic lift consists of two cylinders C₁ and C₂ and they are connected by a horizontal tube T. There are two air resistance pistons P₁ and P₂ and their area of cross-section is A₁ and A₂ respectively. Such that A₁ solid A₂ i. e., area of container C₁ is less than that of C₂ On piston P₁ if F, is the force applied, then applied pressure will be
P = \(\frac{F_{1}}{A_{1}}\) …….. (1)
This pressure now gets transmitted to the container C₂ and so the liquid gives an upward force to the piston P₂. By Pascal’s law,
Pressure on piston P₂ = Pressure on piston P1 = \(\frac{F_{1}}{A_{1}}\)
Force F₂ on piston P₂ = Pressure × Area
or F₂ = \(\frac{F_{1}}{A_{1}} \) × A₂ ……….. (2)
As A₁ < A₂, hence F₂>F₁ So by applying lesser force on small area, we get large force on larger areas. This is the principle of hydraulic lift.

Question 6.
Show that a body when immersed in a liquid looses its weight equal to the weight of the liquid displaced by the body.
Answer:
According to this principle, when a body is immersed wholly or partially in a liquid at rest, it loses some of its weight which is equal to the weight of liquid displaced by immersed part of the body.

Consider a rectangular body of mass M and height h is completely immersed in a liquid of density p. Let A be the area of the lower and upper face of the body and l is the depth of the upper face of the body from the free surface of liquid.

Volume of liquid displaced = Volume of the body ( V) = Ah.
Mass of the liquid displaced, m = Vρ= Ahρ ………… (1)
Pressure acting on the upper face of the body due to
liquid, P₁ = lρg ……………… (2)
and Pressure acting on the lower face of body, P₂ = (l+ h)ρg ………….. (3)
Downward force on the upper face of the body is Fx = P1A = IρgA …………… (4)
Upward force on the lower face of the body is F₂ = P₂A = A(l + h)ρg …………… (5)
Net upward thrust on the body
F = F₂-F₁ =(l + h)ρgA – lρgA
or
F = Ahρg …………….. (6)
From eqns. (1) and (6), we get
F = mg = Weight of liquid displaced Actual weight of the body W = Mg
Upward thrust F = mg
Hence, the apparent weight of the body in liquid = W-F= Mg – mg.

Question 7.
What is Pascal’s law? State and prove it.
Answer:
In a closed liquid the pressure applied at any part is equally transmitted in all direction and in the same amount.
Or
If a liquid is in equilibrium, then the pressure in every part of it is equal.
Deduction of Pascal’s law: Consider two points O₁ and O₂ inside a liquid. Now, imagine a right circular cylinder with O₁, O₂ as the axis. Its end faces will be circular with O₁ and O₂ as centres.

The liquid inside the cylinder is in equilibrium, therefore the forces exerted by the outside liquid on the inner liquid will be perpendicular to the surface.
In accordance with it the forces acting on the end faces of the cylinder with O₁ and O₂ as centre will be perpendicular to them.

If the force acting on the face with O₁ as centre is F₁ then
Pressure at O₁ P = \(\frac{F_{1}}{A}\)
Where A is the area of the end face.
∴ F₁ = P₁A
Similarly, if F₂ is the force acting on the end face with O₂ as centre, then
Pressure at O₂ is P₂ = \(\frac{F_{2}}{A}\)
or F₂ = P₂ A
As the liquid is in equilibrium.
∴ F₁=F₂
P₁ A = P₂A
or P₁ = P₂
Therefore, the pressure on O₁ and O₂ are equal.

Question 8.
State Pascal’s law. Explain the effect of gravity on Pascal’s law.
Answer:
Pascal’s law:

In a closed liquid the pressure applied at any part is equally transmitted in all direction and in the same amount.
Or
If a liquid is in equilibrium, then the pressure in every part of it is equal.
Deduction of Pascal’s law: Consider two points O₁ and O₂ inside a liquid. Now, imagine a right circular cylinder with O₁, O₂ as the axis. Its end faces will be circular with O₁ and O₂ as centres.

The liquid inside the cylinder is in equilibrium, therefore the forces exerted by the outside liquid on the inner liquid will be perpendicular to the surface.
In accordance with it the forces acting on the end faces of the cylinder with O₁ and O₂ as centre will be perpendicular to them.

If the force acting on the face with O₁ as centre is F₁ then
Pressure at O₁ P = \(\frac{F_{1}}{A}\)
Where A is the area of the end face.
∴ F₁ = P₁A
Similarly, if F₂ is the force acting on the end face with O₂ as centre, then
Pressure at O₂ is P₂ = \(\frac{F_{2}}{A}\)
or F₂ = P₂ A
As the liquid is in equilibrium.
∴ F₁=F₂
P₁ A = P₂A
or P₁ = P₂
Therefore, the pressure on O₁ and O₂ are equal.

Suppose a trough is filled with a liquid of density ρ. A and B are two points on the same vertical line and separated by a distance of h. A is above the point B shown in Fig.
Imagine a right cylinder of height h and area of cross-section A with the vertical line AB as axis, Weight of the cylinder = Mass × g
= Volume × Density × g
= Ahpg, (∵ Volume=Area × Height)

This weight will be acting vertically downwards. If the pressures at the points A and B are P₁ and P₂ respectively, then the force acting on the upper face of the cylinder in the downward direction is
F₁ = Pressure x Area
or
F₁ = P₁A
Similarly, the force acting on the lower face of the cylinder in the upward direction is F₂ = P₂A
But, the liquid is in equilibrium, therefore F₁ + Ahρg – F₂ = 0
∴ P₁A + Ahρg- P₂A = 0
or (P₂ – P₁ )A = Ahρg
or P₂ – P₁ = hρg …………… (1)
From eqn. (1), it is clear that there is a difference of pressure between any two points in the same vertical line in a liquid.

Question 9.
Explain why: The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection. (NCERT)
Answer:
The area of needle of a syringe is very small. According to principle of continuity Aν = constant. Since area is small therefore ν will be large. The doctor exerts pressure P with his thumb. According to Bernoulli’s theorem.
P + \(\frac{1}{2}\) dv²+ dgh = Constant

In this expression, the power of P is 1 and that of ν is 2. Obviously, ν is more effective than P. This is reason due to which the size of needle of syringe controls flow rate better than the thumb pressure exerted by doctor while administering an injection.

Question 10.
The farmer plough their field after rain. Why?
Answer:
In a dry clay, number of pores are there, which acts as capillaries. After rain if farmers plough their field, then the capillaries of earth broken and the underground water remain there, which is used by the plant.

Question 11.
Explain why: A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel. (NCERT)
Answer:
The velocity of liquid emerging out of a small hole is high (∵ Aν = constant). Hence, its linear momentum will be large. Since no external force is acting on the system, so the linear momentum of the system remains conserved. Thus, it is clear that a reaction will act on the vessel and it will experience a thrust.

Question 12.
Explain why: A spinning cricket ball in air does not follow a parabolic trajectory. (NCERT)
Answer:
When a spinning cricket ball is thrown in air then at one side of ball the resultant velocity of air increases and at other side, the velocity decreases. According to Bernoulli’s theorem, a pressure difference is created between the two sides and a resultant force acts on the ball. Hence, the ball follows a curved path, it is called Magnus effect. Due to this effect, the ball does not follow a parabolic trajectory.

Question 13.
Describe Torrecelli’s experiment of atmospheric pressure. How does it measure the pressure?
Answer:
Torricelli’s Experiment: Torricelli performed an experiment to measure the atmospheric pressure. This experiment is called the
Torricelli’s experiment. The device used for the mea¬surement of atmospheric pressure is called a Barometer.
A barometer consists of a one-metre long glass tube of nearly 1.5 cm internal diameter.

One end of this tube is closed. This tube is completely filled with mercury. Now, closing the open end with thumb, we invert the tube and place the open end in a trough containing mercury such that the tube is vertical and the open end is well inside the mercury. Care is taken that no air bubble enters the tube. On removing the thumb the level of mercury in the tube starts falling and at a certain definite height the fall in level stops.

In this state the height of the level of mercury in the tube above the (fee surface of mercury in the trough i. e., the height of the mercury column is nearly 76 cm.
In the space inside the tube above the mercury, column is fully vacuumed, this empty space is called the Torricellian vacuum.

Explanation: Torricelli explained the above phenomenon as follows: The column of mercury in the tube tries to fall due to its own weight while the atmospheric pressure tries to push it upwards. When the pressure of the mercury column equals the atmospheric pressure an equilibrium is achieved and the fall of mercury level stops. In this way by measuring the height of mercury column in the tube the atmospheric pressure can be measured.

Question 14.
What are the factors which effect surface tension? How they effect surface tension?
Answer:
Factor affecting surface tension are :

• Substance dissolved in the liquid: The substance which are highly soluble in the liquid increases the surface tension, while the insoluble substance decreases surface tension.
• Contamination of the surface: Dust particles, oil, grease etc. decrease the surface tension of liquid.
• Temperature of the liquid: Generally on increasing the temperature, the surface tension decreases. According to the formula,
  T = T₀ (1 – αt)
  Where, T= S.T. at t°C, TQ = S.T. at 0°C and α = Temperature coefficient of surface tension.

Question 15.
Explain the phenomenon of capillary rise.
Answer:
Capillary rise: Let a capillary be dipped vertically in a liquid which wets the glass surface. The me¬niscus of the liquid in the capillary will be concave. Suppose the liquid does not rise in the capillary [figure (a)]. Imagine two points A and B at the same horizontal level below the surface of the liquid. Another point C is above the meniscus of the liquid

The pressure on the concave side of the meniscus is more than that on the convex side. This excess of pressure on the concave side is 2T/R, where, T is the surface tension of the liquid and R is the radius of the meniscus. The pressure on the point C is equal to the atmospheric pressure P. Therefore, the net pressure at the point
B = P – \(\frac{2 T}{R}\)

At point A the pressure is equal to the atmospheric pressure P. Points A and B are in the same horizontal level. Therefore, the pressure on both these points must be equal. Clearly, the liquid will rise in the capillary to a height such that the pressure of the liquid column at the point B equals 2 T/R. Thus, the total pressure at the point
B = P – \(\frac{2 T}{R}+\frac{2 T}{R}\) = P.

Question 16.
In strong storm, the roofs of houses fly away, High-velocity Low pressure why?
Answer: At the time of strong wind or storms the air above the roof moves with very high velocity, creating low pressure above the roof. Hence P₂ becomes greater than P₁. Due to the pressure difference P₁ -P₂, a large upthrust acts on the roof and therefore it fly away.

Question 17.
What is critical velocity? What is its relationship with Reynold’s number?
Answer:
The maximum velocity of a fluid, till the flow is streamline is called critical velocity. If the critical velocity is V_c, coefficient of viscosity is η, density of liquid is ρ and diameter of tube is D, then
V_c = \(\frac{K \eta}{\rho D}\)
Where, K is called Reynold’s number.
K is a dimensionless quantity.

Question 18.
Differentiate between the streamlined flow and turbulent flow of liquids.
Answer:
Difference between Streamlined flow and Ttirbulent flow :

Question 19.
State Newton’s law of viscosity. Define coefficient of viscosity and write its unit and dimensional formula.
Answer:
Newton’s law of viscosity: According to this law, the friction force F between the two layers is :
(i) Directly proportional to the area A of the layers, i.e.,
F ∝ A

(ii) Directly proportional to the velocity gradient between the layers, i.e.,
F ∝\(\frac{d v}{d x}\)
Combining both the laws, we get
F ∝ A .\(\frac{d v}{d x} \)
F = – η A. \(\frac{d v}{d x}\)

Where, η is a constant, called coefficient of viscosity.
Negative sign shows that viscous force acts in the opposite direction of flow.
Now, if A = 1 and \(\frac{d v}{d x}\) = 1, then F = – η
Thus, the coefficient of viscosity of a liquid is the tangential force required to maintain a unit velocity gradient between two layers of unit area separated by unit distance. Unit: MKS unit of η is N-s/m².

Question 20.
Find out dimensional formula of coefficient of viscosity.
Answer:
We know that fricition force
F = – ηA \(\frac{d v}{d x}\)
∴ Coefficient of viscosity η = – \(\frac{F}{A \frac{d v}{d x}}\)
or [η] = \(\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]\left[\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right]}\) = [ML^-1T^-1].

Question 21.
What is terminal velocity? Calculate the terminal velocity of a sphere falling through a viscous liquid.
Answer:
Terminal velocity: When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.
Formula for terminal velocity: Let a sphere of radius r, density d is falling through a liquid of density ρ and coefficient of viscosity η.
Now, the forces acting on falling body are:

• Weight of the sphere acting vertically downwards.
• Force of buoyancy liquid, acting vertically upward.
• Viscous force acting vertically upwards.

Now, Weight of sphere = Volume × Density × g
= \(\frac{4}{3}\) πr³dg,
Where, r is the radius of ball.
Force of buoyancy = Weight of liquid displaced
= \(\frac{4}{3}\) πr³ρg
∴ Resultant weight of body =\(\frac{4}{3}\) πr³ dg – \(\frac{4}{3}\) πr³ρg
= \(\frac{4}{3}\) πr³ (d – ρ) g

Again, Viscous force = 6 πηrν
When the body is falling with terminal velocity, then Viscous force = Resultant weight
∴ 6 πηrν = \(\frac{4}{3}\) πr³(d – ρ) g
or ν = \(\frac{2}{9} \frac{r^{2}(d-\rho) g}{\eta}\).

Question 22.
What is terminal velocity? On what factors does it depend?
Answer:
Terminal velocity:

Terminal velocity: When a ball falls through a liquid then initially the ball is accelerated due to gravity but soon it achieves a constant velocity called terminal velocity.
Formula for terminal velocity: Let a sphere of radius r, density d is falling through a liquid of density ρ and coefficient of viscosity η.
Now, the forces acting on falling body are:

• Weight of the sphere acting vertically downwards.
• Force of buoyancy liquid, acting vertically upward.
• Viscous force acting vertically upwards.

Now, Weight of sphere = Volume × Density × g
= \(\frac{4}{3}\) πr³dg,
Where, r is the radius of ball.
Force of buoyancy = Weight of liquid displaced
= \(\frac{4}{3}\) πr³ρg
∴ Resultant weight of body =\(\frac{4}{3}\) πr³ dg – \(\frac{4}{3}\) πr³ρg
= \(\frac{4}{3}\) πr³ (d – ρ) g

Again, Viscous force = 6 πηrν
When the body is falling with terminal velocity, then Viscous force = Resultant weight
∴ 6 πηrν = \(\frac{4}{3}\) πr³(d – ρ) g
or ν = \(\frac{2}{9} \frac{r^{2}(d-\rho) g}{\eta}\).

Factors on which it depends :

• It is directly proportional to the square of the radius of the body.
• Directly proportional to the difference of the densities of the liquid and body.
• Inversely proportional to the coefficient of viscosity of the liquid.
• Directly proportional to acceleration due.to gravity.

Question 23.
State and prove principle of continuity.
Answer:
Principle of continuity: When a non-viscous and incompressible liquid flows through a tube of non-uniform bore and its flow is streamline, then the product of velocity of flow and the area of cross-section of the tube at every transverse cross-section remains constant.
i. e., A.ν = a constant
or A₁ν₁ = A₂V₂ .
Proof: Let an incompressible and non-viscous liquid be flowing through a tube XY of non-uniform bore.
A and B are two transverse sections of the tube having areas A₁ and A₂ respectively. Also let the velocity of flow of the liquid at these transverse sections be V₁, and v₂ respectively. Let the density of the liquid bed.
The liquid flowing through the transverse section of A travels a distance v₁ in 1 second, therefore the volume of the liquid flowing per second through section at A is A₁v₁. Hence, the mass of the liquid flowing per second through this section is A₁V₁d.

Similarly, the mass of the liquid flowing through the transverse section at B is A₂v₂d. Since the liquid is incompressible, it cannot accumulate in any portion of the tube. Therefore, whatever amount of liquid enters the segment AB of the tube at section A, the same amount of liquid will flow out of the section at B.
A₁V₁d = A₂v₂d
or A₁v₁ = A₂V₂
or A.v = Constant
This is the principle of continuity.

Question 24.
How many types of energy are there in a flowing liquid? Write the expressions for them.
Answer:
A non-compressible flowing liquid possesses three types of energies :
1. Kinetic energy: If the mass of liquid flowing is m, then its kinetic energy,
K.E = \(\frac{1}{2}\) mv²
If the volume of liquid be V, then K.E. of unit volume of liquid = \(\frac{1}{2}\) \(\frac{m v^{2}}{V}\)
But, \(\frac{m}{V}\) = d. (density)
∴ K.E. of unit volume of liquid =\(\frac{1}{2}\) dv².

2. Potential energy : If a liquid of mass m is at a height h, then its
P.E.= mgh
Where, g is acceleration due to gravity.
∴ P.E of unit volume of liquid = \(\frac{m g h}{V}\) = dgh, (∵ \(\frac{m}{V}\) = d)

3. Pressure energy: For the flow of liquid pressure is applied, with result in pressure energy.
Let the pressure P is applied on a liquid of area of cross-section A, so that liquid flows through a distance of x.
∴ Pressure energy = Work done
= Force × Distance
= Pressure × Area × Distance
= P × A × x =PV, (∵ Ax = V)
∴ Pressure energy per unit volume =\(\frac{P V}{V}\) = P
and pressure energy per unit mass = \(\frac{P V}{V}\) = \(\frac{P}{\frac{m}{V}}\) = \(\frac{P}{d}\).

Question 25.
What is Stoke’s law? What are the conditions of Stoke’s law?
Answer:
According to Stoke, if a spherical body moves through a homogeneous viscous liquid, the viscous force F is :
(i) Directly proportional to the coefficient of viscosity
i.e., F ∝η
(ii) Directly proportional to the radius of body
i.e., F ∝ r
(iii) Directly proportional to the velocity of the body
i.e., F ∝ v
Combining all the laws, we get
F ∝ ηrv
or
F = Kηrv
Where K is constant of proportionality. For spherical ball, K is equal to 6π.
∴ F=6πηrv

Conditions of Stoke’s law :

• The sphere should be very small.
• The sphere should be rigid and smooth.
• The liquid should be extended infinitely.
• The liquid should be viscous and homogeneous.

Mechanical Properties of Fluids Class 11 Important Questions Long Answer Type

Question 1.
State-Bernoulli’s theorem and prove it.
Or
State Bernoulli’s theorem and prove that:
P +\(\frac{1}{2}\) ρv₂ + ρgh = A constant.
Answer:
Bernoulli’s theorem: When a non-viscous and incompressible liquid flows in streamlined condition through a tube then the total energy of its unit mass or unit volume remains constant, i.e.,
For unit volume,
P +\(\frac{1}{2}\) ρv + ρgh = a constant
and For unit mass,
\(\frac{P}{\rho}+\frac{1}{2}\)v₂ +gh = a constant
Where, P – Pressure, p = Density, v = Velocity of liquid flow, g = Acceleration due to gravity, h – Height of liquid above ground level.

Proof: Suppose an incompressible and non-viscous liquid is flowing through a tube XY of non-uniform bore. Its flow is streamlined [See fig.].
A and B are two transverse sections of the tube. The areas of these sections are A₁ and A₂ respectively. At the section A let the velocity of flow of the liquid be v, and its pressure P₁. At the section B let the corresponding quantities be v₂ and P₂ respectively.

Also let the heights of A and B above the ground be h₁ and h₂ respectively and p the density of the liquid.
The force, acting on the liquid at the section A is
F₁=P₁A₁
Therefore, the work done on the liquid entering the cross-section at A in each second
W₁ = F₁ × V₁ = P₁A₁v₁
Similarly, the amount of work done by the liquid coming out of the section at B in each second
W₂ = P₂A₂V₂
∴ Net amount of work done on the liquid
W = W₁ – W₂
= P × A × v₁ – P₂A₂V₂ ……………… (1)
But, from the principle of continuity,
A₁ V₁ = A₂V₂

Where, A₁v₁, or A₂V₂ is the volume of the liquid flowing in each second through any cross-section.
∴ Mass of the liquid flowing any cross-section in one-second
m = A₁v₁ρ= A₂v₂ρ
or
A₁v₁ = \(\frac{m}{\rho}\)
Putting this value in eqn. (1), we get
W = P₁ \(\frac{m}{\rho}\) – P₂\(\frac{m}{\rho}\)
= (P₁ – P₂)\(\frac{m}{\rho}\)
The potential energies of the liquid flowing per second at A and B are mgh₁ and mgh₂ respectively.
∴ Increases in the potential energy of the liquid
= mgh₂ – mgh₁ = mg(h₂-h₁)

The K.E. of the liquid flowing per second at A is \(\frac{1}{2}\)mv₁ ² and at B is \(\frac{1}{2}\) mv₂ ².
∴ Increase in the kinetic energy of the liquid
= \(\frac{1}{2}\) mv₂ ² – \(\frac{1}{2}\)mv₁ ²
= \(\frac{1}{2}\) m(v₂ ² – v₁ ²)

From the principle of conservation of energy,
Work done on the liquid = Increase in potential energy of the liquid + Increase in the kinetic energy of the liquid
∴ (P₁ – P₂)\(\frac{m}{\rho}\) = mg (h₂ – h₁) + \(\frac{1}{2}\)m (v₂² – v₁²)
or
P₁ – P₂ = ρg (h₂ – h₁) + \(\frac{1}{2}\)ρ(v₂² – v₁²)
or
P₁ – P₂ = ρgh₂ – ρgh₁ + \(\frac{1}{2}\)ρ(v₂² – \(\frac{1}{2}\)ρ(v₁²
or
P₁ + ρgh₁+\(\frac{1}{2}\)ρ(v₁² = P₂ + ρgh₂ + \(\frac{1}{2}\)ρ(v₂² ………… (2)
or
P + ρgh + \(\frac{1}{2}\)ρv² ……………… (3)

∴ Therefore, the total energy of unit volume of the liquid remains constant at every point during flow.
From eqn. (3),
\(\frac{P}{\rho}\) + \(\frac{1}{2}\)v² + gh = constant
The total energy of unit mass of the liquid remains constant. From eqn. (2),
P₁ + \(\frac{1}{2}\)ρv₁²+ dgh₁ = P₂ + \(\frac{1}{2}\)ρv₂² + dgh₂
For horizontal tubes : h1 = h2.
∴ P₁ + \(\frac{1}{2}\)ρv₁² = P₂ + \(\frac{1}{2}\)ρv₂²
P + \(\frac{1}{2}\)ρv² = Constant.

Question 2.
In venturi meter the area of cross-sections at two points are A₁ and A₂ and pressure difference is h. Derive an expression for the volume of liquid flowing per second.
Or
Establish the formula for the flow of liquid by venturi meter.
Or
Explain an application of Bernoulli’s theorem.
Answer:
Venturimeter: It is a device based upon the Bernoulli’s theorem. It is used to measure the rate of flow of a liquid through a given pipe.
The construction of a venturi meter is shown in the fig., AB is a horizontal pipe, the middle portion R of it, is constricted. Two vertical tubes M and N are attached to it as shown in the figure. They constitute the manometer arrangement.

When the liquid (say, water) flows through the pipe AB, then in accordance with the principle of continuity, the velocity of flow at the points is smaller and that at/? it is larger. Therefore, according to the Bernoulli’s theorem, the pressure at A will be larger than the pressure at R. This causes the liquid to rise more in-tube Ethan in the tube N. By measuring the difference in the heights of liquid column M and N we can calculate the rate of flow of liquid through pipe.

Let at A the area of cross-section of the pipe be A₁, the velocity of flow V₁ and pressure P₁. At R the area of cross-section of the tube is A₂, the velocity of flow v₂ and the pressure is P₂.
As the pipe is horizontal, according to Bernoulli’s theorem,
P₁ + \(\frac{1}{2}\)ρv₁² = P₂\(\frac{1}{2}\)ρv₂²
or
P₁ – P₂ = \(\frac{1}{2}\) ρ (v₂² – v₁²) ……………… (1)
But, from the principle of continuity,
A₁v₁=A₂V₂ =Q

Where, Q is the volume of the liquid flowing per second i.e., the rate of flow of the liquid.
Therefore,
v₁ = \(\frac{Q}{A_{1}}\) and V₂ =\(\frac{Q}{A_{2}}\)
Putting these values in the eqn. (1), we have
or
P₁ – P₂ = \(\frac{1}{2}\)ρ (\(\frac{Q^{2}}{A_{2}^{2}}-\frac{Q^{2}}{A_{1}^{2}}\) )
or
P₁ – P₂ = \(\frac{1}{2}\) \(\frac{\rho Q^{2}}{A_{1}^{2} A_{2}^{2}}\) (A₁² – A ₂²)

If the difference in levels of the liquid in tubes M and N is h, then
P₁ -P₂ = hpg
Therefore, from eqn. (2),
hpg = \(\frac{1}{2}\) [/latex] \(\frac{\rho Q^{2}}{A_{1}^{2} A_{2}^{2}}\) (A₁² – A ₂²).

Question 3.
State and prove Torricelli’s theorem.
Answer:
Torricelli’s theorem: The velocity of efflux is equal to the velocity that a body acquires when allowed to fall freely from rest through a height equal to the depth of the hole below the free surface of liquid.
Let liquid is filled in a jar, up to height H and from the free surface there is a hole below depth h.
The velocity by the liquid emerges from the hole is called the velocity of efflux.

Now, the K.E. of free surface is zero and P.E. is maximum.
∴ Energy of free surface, per unit volume is
= P+0+ ρgH = P+ ρgH
Where, P is atmospheric pressure and ρ is the density of liquid.

Similarly, the energy per unit volume of liquid at the hole
= P + \(\frac{1}{2}\)ρv²+ ρg (H -h)
Again, by the Bernoulli’s theorem,
P +ρgH=P +\(\frac{1}{2}\)ρv² +pg(H-h)
or
pgH = \(\frac{1}{2}\)ρv² +pgH-pgh
or
\(\frac{1}{2}\)v² =gh
or
v² = 2 gh,
or
v = \(\sqrt{2 g h}\)

Again, if a body is allowed to fall from a height h freely, then u = 0.
∴ v² = u² + 2 gs
or
v² = 0+2gh = 2gh
or v = \(\sqrt{2 g h}\).

Question 4.
What is meant by surface energy of a liquid? Prove that surface energy per unit area is equal to the surface tension.
Or
What is surface energy? Derive an expression for it.
Answer:
Surface energy: The potential energy stored in unit area of free surface of liquid is called its surface energy.
Let PQRS be a rectangular frame of wire. AB is another straight wire which can slide over the arms PQ and RS of the frame as shown in the Fig.

Let the frame be dipped in soap solution and taken out so that a soap film ABRQ is formed. Due to the force of surface tension, the wire AB starts sliding towards QR. If the wire is to be prevented from sliding, a force F must be applied on the wire AB. In equilibrium this force will be equal and opposite to the force of surface tension on the wire AB.
If the length of the wire AB is l and the surface tension of the soap film is T, then F = 2Tl

The factor of 2 appears because soap film has two surfaces.
If the wire AB is displaced through a distance x against the force of surface tension to the position A ‘B’, this will increase the area of the film.
Work done in increasing the area of the film,
W=Fx = 2Tlx ,
or W=T2lx
But, 2lx= A, the total increase in the area of the two surfaces of the film.
∴ W=TA
or
T = \(\frac{W}{A}\)

In the above expression if A = 1, then T= W
Hence, the surface tension of a liquid is numerically equal to the amount of work done in increasing the surface area of the film by unity under isothermal conditions. Thus, surface tension is equal to the surface energy per unit area.

Question 5.
Derive an expression for the excess of pressure inside a drop.
Answer:
Excess of pressure inside a drop: A liquid drop is spherical in shape and its outer surface is convex. Therefore, due to surface tension a net inward force acts on every molecule situated on the surface of the drop. As a result the pressure inside the drop is more than that outside. Let the radius of the drop be R and the excess of pressure inside it be P. Due to this excess of pressure an outward force acts on the surface of the drop.

Suppose due to this excess of pressure, the radius of the drop increases from R to R + ΔR.

Therefore, the work done by the excess of pressure in expansion of the drop,
W = Force × Displacement
= Pressure × Area × Displacement
= P × 4πR² × A R
Increase in the surface area of the drop = Final surface area – Initial surface area
= 4π(R + ΔR)² – 4πR² =8 πRΔR
∵ 4π( ΔR)² will be very small and can be neglected.
∴ Increase in the surface energy
= Surface tension x Increase in the surface area = T ×8 πRΔR
This increase in surface energy is due to the excess of pressure.
∴ P × 4πR² × ΔR = T × 8 πR.ΔR
or P = \(\frac{2 T}{R}\).

Question 6.
Derive an expression for excess pressure inside a bubble.
Answer:
Excess of pressure inside a bubble: Due to air inside it, the bubble has two free surface. One inside and another outside. Let the radius of the bubble be R. Due to excess of pressure P in it, an outward force acts on the surface of the bubble. This is balanced by the force due to the surface tension.
Suppose due to excess of pressure P, let the radius of the bubble increase from R to R +ΔR.
∴ Work done by the excess of pressure
= Force × Distance = Pressure × Area × Distance
= P × 4πR² × ΔR
Since, there are two surfaces in a bubble, the increase in the surface area of the bubble
= 2[4π(R + ΔR)² – 4 πR² ]
= 16πR.ΔR
∴ Increase in surface energy = T×16πR.ΔR
The increase in surface energy is due to the excess of pressure.
∴ P × 4πR² × ΔR = T × 16 πR.ΔR
or P =\(\frac{2 T}{R}\).

Question 7.
What is meant by capillarity? Give two examples of capillarity in daily life. Establish the formula to determine surface tension of water by capillary rise.
Or
Derive the formula to determine the surface tension in the laboratory.
Answer:
Capillarity: The phenomenon of ascenting and descenting of liquid in a capillary is called capillarity. In kerosene lamp oil rises in the wick due to capillarity.
Formula derivation: Let a uniform capillary tube of radius r be immersed into water and water rises up to height h. The meniscus is concave in the tube and its ra¬dius is nearly equal to the radius of capillary. The surface tension of liquid acts at the point of contact of meniscus along the direction of tangent, making an angle θ with the wall of tube.

Now, according to Newton’s third law, reaction force R is equal to T.
Resolving R we get horizontal components R sin θ = T sin θ acting diametrically opposite, around the meniscus, hence their resultant will be zeros.
And vertical component R cos θ= T cos θ acting upwards around the circumference of meniscus, due to which the water rises.
The total force which rises the water
= Tcosθ × 2 πr …………. (1)
(2 πr has been multiplied because it is the total circumference on which T cosθ acts)

Now, the upward force is balanced by the weight of the water.
Now, volume of water raised = Volume of water to height h + Volume of water in meniscus.
Where, volume of water in meniscus = Volume of cylinder of radius and height r -Volume of hemisphere filled with air
= πr²r – \(\frac{2}{3}\) πr³
∴ Volume of water raised = πr² h + (πr² r – \(\frac{2}{3}\) πr³)
= πr²h + \(\frac{1}{3}\)πr³

If the density of water is ρ then weight of water acting vertically downwards.
W = mg – Volume × Density × g
=( πr²h + \(\frac{1}{3}\)πr³)ρg
= πr²(h +\(\frac{1}{3}\)r) ρg

Now, in equilibrium position,
Total force upwards = Total force downwards
T cosθ.2πr = πr²(h+ \(\frac{1}{3}\)r)ρg
or T = \(\frac{r\left(h+\frac{1}{3} r\right) \rho g}{2 \cos \theta}\)
Since,\(\frac{r}{3}\) is negligible in comparison to h.
∴T = \(\frac{r h \rho g}{2 \cos \theta}\)

But, for pure water and clean glass θ = 0°
∴ cos θ = cos 0° = 1
∴ T = \(\frac{r h \rho g}{2}\)
This is the required formula.

Mechanical Properties of Fluids Class 11 Important Numerical Questions

Question 1.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor? (NCERT)
Solution:
Given, m = 50kg; 2r = 1.0cm ⇒ r = 0.5 cm = 0.5 × 10^-2 m
F = mg = 50 × 9.8 = 490N.
Now pressure,
P =\(\frac{F}{A}\) = \(\frac{F}{\pi r^{2}}\) = \(\frac{490}{3 \cdot 14\left(0 \cdot 5 \times 10^{-2}\right)^{2}} \)
= \(\frac{490 \times 10^{4}}{0 \cdot 785}\)
= 624.20 × 10⁴
= 6.24 × 10⁶ N/m².

Question 2.
Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kgm^-3. Determine the height of the wine column for normal atmospheric pressure.(NCERT)
Solution:
Given, d = 984kg/m³; g = 9.8ms ^-2, P = 1.013 × 10⁵Pa
Formula, P = hdg
∴ h = \(\frac{P}{d g}\) = \(\frac{1 \cdot 013 \times 10^{5}}{984 \times 9 \cdot 8}\) =10.5m.

Question 3.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m^-3. (NCERT)
Solution:
Given, ν₁ = 70ms^-1 ;ν₂ = 63ms^-1;A = 2.5m² , d = 1.3 kgm^-3
According to Bernoulli’s theorem,
\(\frac{1}{2} d v_{1}^{2}\) + P₁ = \(\frac{1}{2} d v_{2}^{2}\) +P₂
or
P₁ – P₂ = \(\frac{1}{2} \)d (ν₂² – ν₁²)A
∴ Applied thrust force will be
F = (P₁ – P₂)A
= \(\frac{1}{2}\)d (ν₂² – ν₁²)A
= \(\frac{1}{2}\) ×1.3 [ (63)² – (70)²] × 2.5
= 0.5 × 1.3 [3969 – 4900] × 2.5
= -1.5 × 10³N.

Question 4.
What is the pressure inside the drop of mercury of radius 3-00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10^-1 Nm^-1. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop. (NCERT)
Solution:
Given, R = 3.00mm = 3 × 10³m, T = 4.65 × 10^-1Nm –¹
Atmospheric pressure = 1.01 × 10⁵Pa
Formula: ρ = \(\frac{2 T}{R}\)
Or ρ = \(\frac{2 \times 4 \cdot 65 \times 10^{-1}}{3 \times 10^{-3}}\) = 310Pa.

Total pressure inside the drop = Atmospheric pressure + Excess pressure
= 1.01 × 10⁵ +310
= 1.01 × 10⁵ +0.00310 × 10⁵
= 1.01310 × 10⁵Pa
= 1.01 × 10⁵ Pa (UP to three significant digits).

Question 5.
At what depth in the water the pressure will be double the atmospheric pressure ?
Solution:
Let the depth = hm
∴ Pressure of Am depth water + Atmospheric pressure
= 2 × Atmospheric pressure
or Pressure of water at h m depth = Atmospheric pressure
or h × 1 × 10³ × 9.8 = 1 × 10⁵(∵ P = hρg)
or h = \(\frac{10^{5}}{10^{3} \times 9 \cdot 8}\) = 10.20m.

Question 6.
The weight of a piece of lead in water is 100 g. Find out its weight in air if relative density of lead is 11.3.
Solution:
Let die weight of lead in air be x
The weight of lead in air = 100 g.
The loss in its weight in water = x – 100

Relative density of lead = \(\frac{\text { The weight of lead in air }}{\text { The loss in its weight in water }} \)
∴ \(\frac{11 \cdot 3}{1}\) = \(\frac{x}{x-100}\)
or 11.3x – 11.30 = x
or 10.3x = 1130
x = 109.7 gm.

Question 7.
The surface tension of a liquid is 70 × 10.3 Nm^-1. A needle of 5 cm is floating on it. Calculate the lateral force acting on the pin.
Solution:
Given : T = 70 × 10³Nm^-1, l = 5 cm = 5 x 10^-2m.
Now we have T = \(\frac{F}{l}\)
F=T.l
= 70 × 10^-3 × 5 × 10 ^-2
= 3.5 × 10^-3N.

Question 8.
The radius of a soap bubble is 0.5 cm. If the surface tension of soap solution is 0.2 Nm^-1, then find the surface energy of the bubble.
Solution:
Given : R = 0.5cm = 0.5 × 10^-2m, T = 0.2Nm ^-1.
A = 4πR² = 4 × 3.14 × (0.5 × 10^-2)²
= 12.56 × 0.25 × 10^-4m²
Now, Surface energy,
W=2TA,
= 2 × 0.2 × 12.56 × 0.25 × 10^-4 = 1.256 × 10^-4 joule.

Question 9.
Calculate the work done in blowing a bubble of soap solution of radius 5cm. Surface tension of soap solution is = 3 × 10 ^-2Nm^-1
Solution:
Given : r = 5 cm =\(\frac{5}{100}\) = \(\frac{1}{20}\)m,
T = 3 × 10^-2 Nm^-1
Since, the bubble has two surfaces.
∴ A = 2 × 4πr²
or
A = 8π\(\left(\frac{1}{20}\right)^{2}\) = 628 × 10^-4m²
Now,
W = T.A
= 3 ×10^-2 × 628 x 10^-4
= 1884 x 10^-6 = 0.001884 joule.

Question 10.
Radius of a soap bubble is r and its surface tension is T. At constant temperature the radius of bubble is blown doubled, calculate the required force.
Solution:
ΔA = 2 [4πr²₂ – 4πr²₂ ]
= 8π (r²₂ – r²₂ )
Given:
r₁=r and r₂ = 2r.
ΔA = 8π (4r² -r²)
= 24πr²
∵ W=T.ΔA
∴ W=T.24 πr² = 24 πr². T.

Question 11.
A drop of diameter 2.8 mm is broken into 125 droplets of equal radii. If the surface tension of liquid is 75 dyne × cm^-1, then calculate the change in energy.
Solution:
Given :
2R = 2.8 mm
or
R = 1.4 mm
= 1.4 × 10^-3m
and, T = 75 dyne × cm^-1
= 75 × 10^-5 N × (10^-2 m)^-1
∴ T = 75 × 10^-3 Nm-1

Now, Volume of big drop = Volume of 125 droplets.
∴ \(\frac{4}{3} \pi R^{3}\) = 125 × \(\frac{4}{3} \) πr³
or R3 = 125r³
or
R = 5r
or
r = \(\frac{R}{5}\)
∴ Surface area of bigger drop = 4πR²
and Surface area of 125 droplets = 125 × 4πr²
= 125 × 4π × \(\left(\frac{R}{5}\right)^{2}\)
∴ Δ A = 20πR² – 4πR² = 16 πR²
∴ Change in energy,
ΔW = T.Δ A
= 75 × 10^-3 × 16 × πR²
= 75 × 10^-3 × 16 × \(\frac{22}{7}\) × (1.4 × 10 ^-3)²
= 7392 × 10^-9
= 7.392 × 10^-9 joule.

Question 12.
The surface tension of soap solution is l-9xl0-2 Nm”1. Calculate the excess of pressure inside the bubble of diameter 2 cm.
Solution:
Given : T = 1.9 × 10^-2 Nm^-1
R = \(\frac{2}{2}\) cm = 1cm = 1 × 10^-2m.
Now, ρ = \(\frac{4 T}{R}\)
or
ρ = \(\frac{4 \times 1 \cdot 9 \times 10^{-2}}{10^{-2}}\) = 7.6 Nm^-2.

Question 13.
Water is flowing through a tube of non-uniform cross-section. The velocity at a point A is 4 times that of another point B. Compare the diameters of tube at A and B.
Solution:
Given: ν₁ = 4v₂
We have, by the principle of continuity,

Question 14.
The diameter of a spherical ball A is half of that of B. What will be the ratio of their terminal velocities in water?
Solution:
Given: a1 = \(\frac{a_{2}}{2}\)
We have,
ν ∝ a²
or
\(\frac{v_{1}}{v_{2}}\) = \(\frac{a_{1}^{2}}{a_{2}^{2}}\)
∴ \(\frac{v_{1}}{v_{2}}\) = \(\frac{\left(a_{2} / 2\right)^{2}}{a_{2}^{2}}\) = \(\frac{1}{4}\)
∴ ν₁ :ν₂ = 1:4

Question 15.
Water is flowing through a horizontal tube of non-uniform area of cross-section. At a place where the pressure is 20 cm (of water) the velocity of flow of water is 60 cm/s. Find the velocity of flow at a place where the pressure is 15 cm (of water).
Solution:
Given, P₁ = 20 cm(of water) =20 × 1 × 980 dyne /cm², v, = 60 cm/s,
P₂ = 15 × 1 × 980 dyne/cm² and ρ= 1 gm/cm3.

By the Bernoulli’s theorem,
\(\frac{1}{2}\) ρ (ν₂² – ν₁²) = P₁ – P₂
⇒ \(\frac{1}{2}\) × 1 × (ν₂² – 60²) = 20 × 1 × 1980 – 15 × 1 × 1980
⇒ \(\frac{1}{2}\) (ν₂² – 60²) = 5 × 980
⇒ ν₂² = 9800 + 3600 = 13400
∴ ν₂ = 115.76 cm/s.

Question 16.
Calculate the terminal velocity of a spherical ball of diameter 2mm, in a viscous liquid. The density of liquid 1.0 × 10³ kg m^-3 and density of ball is 8.0 × 10³kg m³ and coefficient of viscosity of liquid is 1.0 kg m^-1s^-1.
Solution:
Given : 2a = 2mm or a = 1mm = 1 x 10^-3m,
ρ = 1.0 × 10 ³kg m^-3 ,d = 8.0 × 10³ kg m^-3, η = 1.0 kg m^-1s^-1

Now, ν = \(\frac{2}{9} \cdot \frac{a^{2}(d-\rho) g}{\eta}\)
= \(\frac{2}{9}\) . \(\frac{\left(1 \times 10^{-3}\right)^{2}\left(8 \cdot 0 \times 10^{3}-1 \cdot 0 \times 10^{3}\right) 9 \cdot 8}{1 \cdot 0}\)
= \(\frac{2}{9}\) × 10^-6 × 7 × 10³ × 9.8
= 15.2 ×10^-3 ms^-1

Question 17.
A drop of water radius 0.0015 mm is falling in the air. If the coefficient of viscosity of air is 1.8 × 10 ^-6 kgm^-1s^-1 then calculate the terminal velocity of drop. The density of air is negligible.
Solution:
Given : a = 0.0015 mm = 0.0015 × 10^-3m = 1.5 × 10⁶m,
d = 1 × 10³ kgm^-3, ρ = 1.0 × 10³ kg m^-3, η = 8.0 × 10^-6 kg m^-1 s^-1

Formula,
ν = \(\frac{2}{9} \cdot \frac{a^{2}(d-\rho) g}{\eta}\)
= \(\frac{2}{9}\) .\(\frac{\left(1 \cdot 5 \times 10^{-6}\right)^{2} \times 1 \times 10^{3} \times 9 \cdot 8}{1 \cdot 8 \times 10^{-6}}\)
= \(\frac{2}{9}\) \(\frac{2 \cdot 25 \times 10^{-12} \times 9 \cdot 8 \times 10^{3}}{1 \cdot 8 \times 10^{-6}}\)
= 2.72 × 10^-3ms^-1.

Question 18.
Two horizontal pipes of different diameters are joined together, water is flowing through them. The velocity and pressure through one pipe are 4 ms^-1 and 2.0 × 10^-3Nm^-2 respectively. The diameters of pipes are 3 cm and 6 cm. Calculate the velocity and pressure in second pipe.
Solution:
Given :
v₁ = 4 ms^-1 , P ¹ = 2.0 × 10⁴ Nm^-1
r₁cm = 1.5 × 10^-2m, r² = \(\frac{6}{2}\) = 3cm = 3 × 10^-2
Now we have
A₁ v₁ = A₂v₂
or πr₁ ²v₁ = πr₁ ²v₂
or
r₁ ²v₁ = r₁ ²v₂

∴ ν₂ = \(\frac{r_{1}^{2} \cdot v_{1}}{r_{2}^{2}} \)
= \(\frac{\left(1 \cdot 5 \times 10^{-2}\right)^{2} \times 4}{\left(3 \times 10^{-2}\right)^{2}}\)
= 1 ms^-1
Again, P₂ – P₁ = \(\frac{1}{2}\) ρ (ν₁² – ν₂²)
or P₂ – 2 × 10⁴ = \(\frac{1}{2}\) × 10³ (4² – 1²)
or
P₂ = \(\frac{1}{15}\) × 10³ + 2 × 10⁴
= 2.75 × 10 ⁴ Nm^-2

Question 19.
What will be the maximum velocity of water for streamline flow through a tube of diameter 2 cm? (η= 10³ kgm^-1s^-1).
Solution:
Given: D = 2cm = 2 × 10²m. η= 10³ kgm^-1s^-1
K=2000,
ρ= 10 kgm^-3
Now, ν_c = \(\frac{K \eta}{\rho D}\)
or ν_c = \(\frac{2000 \times 10^{-3}}{10^{3} \times 2 \times 10^{-2}}\) = 0.1ms^-1.

Mechanical Properties of Fluids Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
The sudden fall of atmospheric pressure indicates :
(a) Storm
(b) Rainfall
(c) Clear weather
(d) Coldwave.
Answer:
(a) Storm

Question 2.
On which principle the Hydraulic lift work :
(a) Pascal’s law
(b) Archimedis principle
(c) Boyle’s law
(d) Bernoulli’s theorem.
Answer:
(a) Pascal’s law

Question 3.
Systolic and diastolic blood pressure of a healthy human being is :
(a) 140 mm/80 mm of Hg
(b) 120 mm/80 mm of Hg
(c) 160 mm/90 mm of Hg
(d) 80 mm/ 120 mm of Hg.
Answer:
(b) 120 mm/80 mm of Hg

Question 4.
On which condition liquid flowing through a tube :
(a) When high density and more viscous liquid flows through less radius tube
(b) When low density and more viscous liquid flows through less radius tube
(c) When high density and less viscous liquid flows through more radius tube
(d) When less density and less viscous liquid flows through more radius tube.
Answer:
(b) When low density and more viscous liquid flows through less radius tube

Question 5.
On increasing temperature viscosity of liquid :
(a) Increases
(b) Becomes zero
(c) Decreases
(d) No effect.
Answer:
(c) Decreases

Question 6.
Bernoulli’s equation is based on which law :
(a) Conservation of momentum
(b) Conservation of energy
(c) Conservation of mass
(d) No law.
Answer:
(b) Conservation of energy

Question 7.
Rain droplet is spherical in shape, its reason is :
(a) Gravitational force
(b) Viscosity
(c) Atmospheric pressure
(d) Surface tension.
Answer:
(d) Surface tension.

Question 8.
On increasing temperature, surface tension of liquid :
(a) Increases
(b) Becomes zero
(c) Decreases
(d) No effect.
Answer:
(c) Decreases

Question 9.
At critical temperature, the surface tension of the liquid is :
(a) Zero
(b) Infinity
(c) Same as that any other temperature
(d) Cannot be determined.
Answer:
(a) Zero

Question 10.
The shape of meniscus of mercury in capillary tube is :
(a) Convex
(b) Cancave
(c) Flat
(d) Not definite.
Answer:
(a) Convex

Question 11.
Magnitude of angle of contact due to which surface of glass become wet is :
(a) 0°
(b) 90°
(c) More than 90°
(d) Less than 90°.
Answer:
(d) Less than 90°

Question 12.
On dipping glass capillary inside a mercury :
(a) Level of mercury in capillary tube increases
(b) Level of mercury in capillary tube increases and overflow
(c) Level of mercury in capillary tube falls
(d) Level of mercury in capillary tube falls and does not rise.
Answer:
(c) Level of mercury in capillary tube falls

Question 13.
Which phenomenon is not related to surface tension :
(a) Dancing of piece of camphor in water surface
(b) Spherical shape of small mercury drop
(c) The liquid become stable after jerking in a vessel
(d) None of these.
Answer:
(c) The liquid become stable after jerking in a vessel

2. Fill in the blanks:

1. Pressure exerted by one-millimeter column of mercury is called ……………………. .
Answer:
one tor

2. Blood pressure at leg of a human is ……………………. in comparison to his brain.
Answer:
more

3. In generally surface tension of a liquid ……………………. on increasing temperature.
Answer:
decreases

4. Viscosity of gas ……………………. and viscosity of liquid ……………………. on increasing temperature.
Answer:
increases, decreases

5. On mixing detergent in water, its ……………………. decreases.
Answer:
surface tension

6. When the magnitude of Reynold’s number is between 0 to 2000, the flow of liquid is ……………………. .
Answer:
streamline

7. ……………………. type of energy is associated with flowing liquid.
Answer:
three

8. Poisson’s is C.G. S. unit of ……………………. .
Answer:
coefficient of viscosity

9. The liquid which is incompressible is known as ……………………. liquid.
Answer:
ideal

10. Mercury does not wet glass because cohesive force is ……………………. than adhesive force.
Answer:
less.

3. Match the following:

I.

Answer:
1.  (c) Capillarity
2. (a) Use of detergent,
3 (b) Pressure is less
4. (e) Pressure is more
5. (d) Cohesive force.

II.

Answer:
1. (d) \( \eta A \frac{d v}{d x} \)
2. (a) \( \frac{K \cdot \eta}{\rho \cdot D} \),
3. (e) \( \frac{2}{9} \cdot \frac{r^{2}(d-\sigma) g}{\eta} \)
4. (b) A.v = constant
5. (c) P+ \( \frac{1}{2} d v^{2} \)+ dgh = constant.

4. state true or False:

1. Hydrostatic pressure is vector quantity.
Answer:
False

2. Surface tension of a liquid does not depend upon area of cross-section of the surface.
Answer:
True

3. Angle of contact of mercury with glass is more.
Answer:
True

4. The man who jump with parachute, its velocity increases initially and then become constant.
Answer:
True

5. Viscosity of gas decreases on increasing temperature.
Answer:
False

6. Surface tension is numerically equal to surface energy.
Answer:
True

7. Greece is more viscous than honey.
Answer:
False

8. In flowing liquid, streamline of layer cannot intersect each other.
Answer:
False.

MP Board Class 10th Hindi Navneet Solutions पद्य Chapter 5 प्रकृति चित्रण

प्रकृति चित्रण अभ्यास

बोध प्रश्न

प्रकृति चित्रण अति लघु उत्तरीय प्रश्न

प्रश्न 1.
कवि के अनुसार ‘वनों और बागों में किसका विस्तार है?
उत्तर:
कवि के अनुसार ‘वनों और बागों’ में बसन्त ऋतु का विस्तार है।

प्रश्न 2.
तन, मन और वन में परिवर्तन किसके प्रभाव से दिखाई दे रहा है?
उत्तर:
तन, मन और वन में परिवर्तन बसन्त ऋतु के प्रभाव से दिखाई दे रहा है।

प्रश्न 3.
कन्हैया के मुकुट की शोभा क्यों बढ़ गई है?
उत्तर:
शरद ऋतु की चाँदनी की छवि पाकर कन्हैया के मुकुट की शोभा बढ़ गई है।

प्रश्न 4.
कवि ने हिमालय की झीलों में किसको तैरते हुए देखा है?
उत्तर:
कवि ने हिमालय की झीलों में कमल नाल को खोजने वाले तैरते हुए हंसों को देखा है।

प्रश्न 5.
कवि ने बादलों को कहाँ घिरते देखा है?
उत्तर:
कवि ने मानसरोवर झील के समीप हिमालय की ऊँची चोटियों पर बादलों को घिरते देखा है।

प्रश्न 6.
कौन अपनी अलख नाभि से उठने वाले परिमल के पीछे-पीछे दौड़ता है?
उत्तर:
कस्तूरी मृग अपनी अलख नाभि से उठने वाले परिमल के पीछे-पीछे दौड़ता है।

प्रकृति चित्रण लघु उत्तरीय प्रश्न

प्रश्न 1.
बसन्त ऋतु के आगमन पर प्रकृति में कौन-कौन से परिवर्तन होते हैं?
उत्तर:
बसन्त ऋतु के आगमन पर कुंजों में भौरे गुंजार करने लगते हैं, उनके झुण्ड के झुण्ड आम के बौरों पर चक्कर लगाने लगते हैं। विहग (पक्षी) समाज में तरह-तरह की आवाजों के द्वारा बसन्त की खुशी का वर्णन होने लगता है। ऋतुराज के आने से प्रकृति में भाँति-भाँति के राग-रंग बिखरने लगते हैं।

प्रश्न 2.
यमुना तट पर किसी छटा बिखरी हुई है?
उत्तर:
यमुना तट पर अखण्ड रासमण्डल की छटा बिखरी

प्रश्न 3.
कवि के अनुसार ‘घनेरी घटाएँ’ क्या कर रही
उत्तर:
कवि के अनुसार घनेरी घटाएँ घुमड़-घुमड़ कर गर्जना करने लग जाती हैं और फिर उनकी गर्जना थमने का नाम नहीं लेती।

प्रश्न 4.
‘निशाकाल से चिर अभिशापित’ किसे कहा गया है?
उत्तर:
निशाकाल से चिर अभिशापित चकवा-चकवी को कहा गया है। कवियों की मान्यता है कि चकवा-चकवी का रात के समय वियोग हो जाता है। यह वियोग उन्हें किसी शाप के द्वारा प्राप्त हुआ है।

प्रश्न 5.
कवि ने भीषण जाड़ों में किससे ‘गरज-गरज कर भिड़ते देखा है?
उत्तर:
कवि ने भीषण जाड़ों में महादेव को झंझावात से गरज-गरज कर भिड़ते देखा है।

प्रकृति चित्रण दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
पद्माकर ने वर्षा ऋतु में प्रकृति का चित्रण किस तरह किया है? उल्लेख कीजिए।
उत्तर:
पद्माकर ने वर्षा ऋतु में चारों ओर चमकने वाली बिजली और शीतल, मन्द सुगन्धित वायु के बहने की तथा घनेरी घटाओं के घुमड़-घुमड़ कर गर्जन करने का वर्णन किया है।

प्रश्न 2.
कवि के अनुसार बसन्त का प्रभाव कहाँ-कहाँ दिखाई दे रहा है?
उत्तर:
कवि के अनुसार बसन्त का प्रभाव कूलों में, केलि में, कछारों में, कुंजों में, क्यारियों में और सुन्दर किलकती हुई कलियों में, पत्तों में, कोयल में, पलासों में, समस्त द्वीपों में और ब्रज के बाग-बगीचों तथा युवतियों में देखा जा सकता है।

प्रश्न 3.
नागार्जुन ने कवि कल्पित’ सन्दर्भ किसे माना है? स्पष्ट कीजिए।
उत्तर:
नागार्जुन ने कवि कल्पित सन्दर्भ मेघदूत को माना है। अलकापुरी से जब यक्ष को निर्वासित कर दिया गया तो वह अपनी प्रेमिका के वियोग में दुःखी रहने लगा। वर्षा ऋतु के आने पर यक्ष ने बादल को दूत बनाकर अर्थात् मेघदूत के माध्यम से ही अपनी प्रेमिका को प्रेम-सन्देश भेजा है।

प्रश्न 4.
नागार्जुन के अनुसार बसन्त ऋतु के उषाकाल का वर्णन कीजिए।
उत्तर:
नागार्जुन के अनुसार बसन्त ऋतु के उषाकाल का वर्णन इस प्रकार है-
बसन्त ऋतु का सुप्रभात था, उस समय मन्द-मन्द गति से हवा चल रही थी, सूर्य की प्रातःकालीन कोमल किरणें पर्वत शिखरों पर स्वर्णिम आभा में गिर रही थीं। रात में चकवा-चकवी का वियोग क्रन्दन सुनाई पड़ता था तथा सरवर के किनारे काई की हरी दरी पर प्रात:काल में उनको प्रणय आलाप करते देखा है।

प्रश्न 5.
निम्नलिखित काव्यांशों की सप्रसंग व्याख्या कीजिए
(अ) कौन बताए वह छायामय ………… गरज-गरज भिड़ते देखा है।
उत्तर:
कविवर नागार्जुन कहते हैं कि इस कैलाश शिखर पर धनपति कुबेर की अलका नगरी का पुराणों में उल्लेख मिलता है लेकिन आज तो इसका यहाँ कोई नामोनिशान भी नहीं है, न ही हमें कालिदास के आकाश में विचरण करने वाले उस मेघदूत का कहीं पता मिलता है। हमने बहुत प्रयत्न कर लिए पर वह ढूँढने से भी नहीं मिला। ऐसा हो सकता है कि वह छायामय मेघदूत यहीं कहीं जाकर बरस गया होगा। फिर कवि कहता है कि इन बातों को छोड़ो, मेघदूत तो कवि कालिदास की कल्पना थी।

मैंने तो भयंकर कड़कड़ाते जाड़ों में आकाश को चूमने वाले कैलाश के शिखर पर घनघोर झंझावातों में महादेव जी को उनसे गरज-गरज कर युद्ध करते देखा है। बादल को घिरते देखा है।

(ब) और भाँति कुंजन ………… और बन गए।
उत्तर:
कविवर पद्माकर कहते हैं कि बसन्त ऋतु में कुंजों में और ही प्रकार की सुन्दरता छा गई है तथा भरों की भीड़ में अनौखी गुंजार सुनाई दे रही है। डालियों, झुण्डों तथा आम के बौरों में नयी आभा आ गई है। गलियों में और ही भाँति की शोभा छा गयी है पक्षियों के समाज में भी और ही प्रकार की बोलियाँ सुनाई दे रही हैं। ऐसे श्रेष्ठ ऋतुराज बसन्त के अभी दो दिन भी नहीं चुके हैं फिर भी सर्वत्र और ही प्रकार का रस, और ही प्रकार की रीति, और ही प्रकार के राग तथा और ही प्रकार के रंग सर्वत्र छाये हुए हैं। इस ऋतु में मनुष्यों के शरीर और ही प्रकार के हो गए हैं, और ही प्रकार के मन तथा और ही प्रकार के वन और बाग हो गए हैं।

प्रकृति चित्रण महत्त्वपूर्ण वस्तुनिष्ठ प्रश्न

प्रकृति चित्रण बहु-विकल्पीय प्रश्न

प्रश्न 1.
‘कहाँ गया धनपति कुबेर वह’ पंक्ति पाठ्य-पुस्तक की किस कविता से ली गयी है? (2009)
(क) नीति-अष्टक
(ख) श्रद्धा
(ग) पथ की पहचान
(घ) बादल को घिरते देखा है।
उत्तर:
(घ) बादल को घिरते देखा है।

प्रश्न 2.
कन्हैया के मुकुट की शोभा किसके द्वारा बढ़ गई है?
(क) नगों द्वारा
(ख) शरद की चाँदनी से
(ग) तारों की आभा से
(घ) सोने का होने के कारण
उत्तर:
(ख) शरद की चाँदनी से

प्रश्न 3.
कवि ने बादलों को कहाँ घिरते देखा है? (2010, 17)
(क) नवल धवल गिरि पर
(ख) तालाबों में
(ग) चट्टानों में
(घ) बागों में।
उत्तर:
(क) नवल धवल गिरि पर

प्रश्न 4.
निशाकाल से चिर अभिशापित किसे कहा गया है?
(क) मृगों को
(ख) नायक-नायिका को
(ग) चकवा-चकवी को
(घ) भौंरों को।
उत्तर:
(ग) चकवा-चकवी को

प्रश्न 5.
हिमालय की झीलों में कवि नागार्जुन ने तैरते हुए देखा है (2015)
(क) मछली को
(ख) मृग को
(ग) हंसों को
(घ) नारी को।
उत्तर:
(ग) हंसों को

रिक्त स्थानों की पूर्ति

1. अमल धवल गिरि के शिखरों पर ……… को घिरते देखा है।
2. वर्षा ऋतु का वर्णन पद्माकर ने ….. को बढ़ाने वाली ऋतु के रूप में किया है।
3. “बीथिन में ब्रज में नवेलिन में बेलनि में बगन में बागन में बगर्यो …….. है।”
4. ……… की हरी दरी पर प्रणय-कलह छिड़ते देखा है।
5. हिमालय की झीलों में कवि नागार्जुन ने ………. को तैरते देखा है। (2011)

उत्तर:

1. बादल
2. विरह
3. बसन्त
4. शैवालों
5. हंसों।

सत्य/असत्य

1. ताप तीन प्रकार के होते हैं – दैहिक, दैविक और भौतिक।
2. ‘ऋतु वर्णन’ कविता में कविवर पद्माकर ने वर्षा ऋतु का सुन्दर वर्णन किया है।
3. हिमालय पर बादलों की स्थिति सदैव एक-सी रहती है।
4. ‘बादल को घिरते देखा है’ कविता के कवि तुलसीदास हैं। (2009)

उत्तर:

1. सत्य
2. असत्य
3. असत्य
4. असत्य

सही जोड़ी मिलाइए

उत्तर:
1. → (ख)
2. → (ग)
3. → (क)
4. → (घ)

एक शब्द/वाक्य में उत्तर

1. पद्माकर कवि को किस काल का कवि माना जाता है?
2. ‘ऋतु वर्णन’ में कवि ने प्रमुख रूप से इन पदों में किस ऋतु का वर्णन किया है।
3. पद्माकर कवि ने विरह को बढ़ाने वाली ऋतु कौन-सी मानी है?
4. चकवा और चकवी कब बेबस हो जाते हैं?
5. कवि नागार्जुन ने हिमालय की झीलों में किसको तैरते देखा? (2012)

उत्तर:

1. रीतिकाल
2. शरद ऋतु,बसन्तु ऋतु तथा वर्षा ऋतु
3. वर्षा ऋतु
4. रात्रि में
5. हंसों को।

ऋतु वर्णन पाठ सारांश

ऋतु वर्णन में पद्माकर कवि ने वसन्त ऋतु की महिमा का गुणगान किया है। यमुना के तट पर,गोपियों की रासलीला का सुन्दर वर्णन किया है। वसन्त ऋतु में कोयल की कूक का मधुर स्वर गूंजता सुनायी पड़ता है। विभिन्न प्रकार के पुष्प खिले रहते हैं तथा उन पुष्पों की सुगन्ध दूर-दूर तक फैली रहती है। बागों में पुष्पों की सुगन्ध से आकर्षित होकर भौरे गुंजार करते हैं।

आम के वृक्ष मंजरियों से भर गये हैं। प्रकृति के हरा-भरा होने से पक्षी भी मस्त होकर कलरव कर रहे हैं। कवि ने शरद ऋतु की अपूर्व शोभा का सुन्दर,वर्णन किया है। शरद ऋतु की चाँदनी चारों ओर व्याप्त है। लताओं पर तथा तमाल के वृक्षों पर चारों ओर शरद ऋतु का सौन्दर्य छाया है। शरद ऋतु में चाँदनी दिग्दिगन्त को अपनी आभा से मंडित कर रही है।

ऋतु वर्णन संदर्भ-प्रसंगसहित व्याख्या

(1) कुलन में केलि में कछारन में कंजन में,
क्यारिन में कलित कलीन किलकत है।
कहैं पद्माकर परागन में पौन में,
पातन में पिक में पलासन पतंग है।
द्वार में दिसान में दुनी में देस देसन में,
देखौ दीप दीपन में दीपत दिगंत है।
बीथिन में ब्रज में नवेलिन में बेलनि में,
बनन में बागन में बगर्यो बसंत है।

शब्दार्थ :
कूलन = नदी के किनारों में। कलित = सुन्दर। केलि क्रीड़ा में। कलीन-कलियों में। किलकंत=किलकारी मारता है। पौन = पवन, हवा। पातन = पत्तों। पिक = कोयल। दुनी = दुनिया। दीप = द्वीप। दीपत = दीप्त। दिगंत = दिशाओं में। बीथिन = गलियों में। नवेलिन = नवयुवतियों में। बगर्यो = फैला हुआ है।

सन्दर्भ :
प्रस्तुत छन्द ‘प्रकृति चित्रण’ के अन्तर्गत ‘ऋतु-वर्णन’ शीर्षक से लिया गया है। इसके रचयिता श्री पद्माकर हैं।

प्रसंग:
इस छन्द में कवि ने बसन्त की चारों ओर फैली हुई मादकता का वर्णन किया है।

व्याख्या :
कविवर पद्माकर कहते हैं कि नदी के कूलों में क्रीड़ा स्थलों में, कछारों में, कुंजों में, क्यारियों में, सुन्दर कलियों में बसन्त किलकारी मार रहा है। पुष्पों के पराग में, पवन में, पत्तों में, कोयल में और पलाशों में बसन्त पगा हुआ है। घर के द्वारों में दिशाओं में, दुनिया में, देशों में, द्वीपों में और दिशाओं में बसन्त दीप्तमान हो रहा है। गलियों में, ब्रजमण्डल में, नवयुवतियों में बेलों में, वनों में, बागों में चारों ओर बसन्त की बहार छाई हुई है।

विशेष:

1. कवि ने बसन्त की मादक छटा का सम्पूर्ण संसार में प्रसार दिखाया है।
2. अनुप्रास अलंकार की सुन्दर छटा।
3. भाषा-ब्रज।

(2) और भाँति कुंजन में गुजरत भीरे भौंर,
और डौर झौरन पैं बोरन के वै गए।
कहै पद्माकर सु और भाँति गलियान,
छलिया छबीले छैल औरे छ्वै छ्वै गए।
औरे भाँति बिहग समाज में अवाज होति,
ऐसे रितुराज के न आज दिन द्वै गए।
और रस औरे रीति और राग औरे रंग,
औरे तन और मन और बन है गए॥

शब्दार्थ :
भीरे = भीड़। बौरन = आम के बौर में। छैल = छवि। छ्दै छ्वै = छा-छा गए। विहग = पक्षी।

सन्दर्भ एवं प्रसंग :
पूर्ववत्।

व्याख्या :
कविवर पद्माकर कहते हैं कि बसन्त ऋतु में कुंजों में और ही प्रकार की सुन्दरता छा गई है तथा भरों की भीड़ में अनौखी गुंजार सुनाई दे रही है। डालियों, झुण्डों तथा आम के बौरों में नयी आभा आ गई है। गलियों में और ही भाँति की शोभा छा गयी है पक्षियों के समाज में भी और ही प्रकार की बोलियाँ सुनाई दे रही हैं। ऐसे श्रेष्ठ ऋतुराज बसन्त के अभी दो दिन भी नहीं चुके हैं फिर भी सर्वत्र और ही प्रकार का रस, और ही प्रकार की रीति, और ही प्रकार के राग तथा और ही प्रकार के रंग सर्वत्र छाये हुए हैं। इस ऋतु में मनुष्यों के शरीर और ही प्रकार के हो गए हैं, और ही प्रकार के मन तथा और ही प्रकार के वन और बाग हो गए हैं।

विशेष :

1. बसन्त ऋतु को इन्हीं सब विशेषताओं के कारण ऋतुराज की पदवी दी गयी है।
2. बसन्त में प्रकृति तथा पुरुष सभी में एक विचित्र प्रकार की नवीनता आ जाती है।
3. अनुप्रास की छटा।

(3) चंचला चलाकै चहूँ ओरन तें चाह भरी,
चरजि गई तो फेरि चरजन लागी री।
कहै पद्माकर लवंगन की लोनी लता,
लरजि गई तो फेरि लरजन लागी री।
कैसे धरौं धीर वीर त्रिविधि समीरै तन,
तरजि गई तो फेरि तरजन लागी री।
घुमड़ि-घुमड़ि घटा घन की घनैरों अबै,
गरजिं गई तो फेरि गरजन लागी री॥

शब्दार्थ :
चंचला = बिजली। चलाकै = चमककर। त्रिविध समीरें = शीतल, मन्द, सुगन्ध वाली हवा। घनेरी = घनी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने पावस ऋतु का सुन्दर एवं मनोहारी वर्णन किया है।

व्याख्या :
कविवर पद्माकर कहते हैं कि पावस ऋतु में बिजली चारों ओर बड़ी चाहना से चमक रही है। एक बार वह बिजली आकाश में छा गयी तो सभी को आश्चर्यजनक लगने लगी। कवि पद्माकर कहते हैं कि लौंग (लवंग) की सुन्दर लता इस ऋतु को बहुत ही अच्छी लग रही है। एक बार यदि वह लवंग लता झुक गई तो फिर वह निरन्तर झुकती ही जाती है। वे वीर! तू ही बता ऐसे सुहावने समय में जबकि शीतल, मन्द और सुगन्धित वायु बह रही हो, तो मैं तेरे बिना कैसे धैर्य धारण करूँ। यह सुगन्धित वायु एक बार यदि तन को सुवासित करने लगी, तो फिर वह निरन्तर ही सुवासित करती रहेगी। इस ऋतु। में बादलों की घनी घटाएँ घुमड़-घुमड़ कर बार-बार आ रही हैं। यदि वे घटाएँ एक बार गर्जन करने लगी, तो फिर वे निरन्तर गर्जन करने लगेंगी।

विशेष :

1. वर्षा ऋतु का बड़ा ही मनमोहक वर्णन हुआ
2. अनुप्रास की छटा।
3. ब्रजभाषा का प्रयोग।

(4) तालने पै ताल पै तमालन पै मालन पै,
वृन्दावन बीथिन बहार बंसीवट पै। 
कहै पद्माकर अखंड रासमंडल पै, 
मण्डित उमंड महा कालिन्दी के तट पै। 
छिति पर छान पर छाजत छतान पर, 
ललित लतान पर लाड़िली की लट पै। 
छाई भली छाई यह सदर जुन्हाई जिहि, 
पाई छबि आजु ही कन्हाई के मुकुट पै।

शब्दार्थ :
तमालन = तमाल वृक्ष पर। बंसीवट = वह वट वृक्ष जहाँ श्रीकृष्ण बंसी बजाते थे। कालिन्दी = यमुना। छिति = पृथ्वी। छान = छाजन। छतान = छतों पर। लाड़िली = राधा जी। जुन्हाई = चाँदनी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने शरद ऋतु की सुन्दर शोभा का वर्णन किया है।

व्याख्या :
कवि पद्माकर कहते हैं कि तालाबों पर, ताड़ पर, तमाल वृक्षों पर और मालाओं पर, वृन्दावन की गलियों एवं बंसीवट पर, सभी ओर शरद ऋतु की बहार छाई हुई है। कविवर पद्माकर कहते हैं कि सम्पूर्ण रासमण्डली पर तथा कालिन्दी के तट पर शरद ऋतु की महान् छटाएँ उमड़ रही हैं। पृथ्वी पर, छप्परों पर और छतों पर शरद ऋतु छाई हुई है। सुन्दर लताओं पर, लाड़िली अर्थात् राधाजी की लटों पर शरद ऋतु की यह चाँदनी भली-भाँति छाई हुई है और यही शोभा आज श्रीकृष्ण के मुकुट पर भी छाई हुई है।

विशेष :

1. शरद ऋतु की सुन्दरता का वर्णन।
2. अनुप्रास की छटा।
3. ब्रजभाषा का प्रयोग।

बादल को घिरते देखा है भाव सारांश

प्रस्तुत कविता में कवि ने हिमालय की ऊँची-ऊँची चोटियों का तथा उमड़ने-घुमड़ने वाली घटाओं का चित्ताकर्षक वर्णन किया है। हिमालय की चोटियों के नीचे अनेक प्रकार की झीलें बहती रहती हैं। उन झीलों के समीप बरसात की उमस से व्याकुल होकर हंस कमलदण्डों को खोजते हुए इधर-उधर घूम रहे हैं।

उन जल-क्रीड़ा करते हुए हंसों को देखकर मेरा मन प्रफुल्लित हो उठता है। कवि कहता है कि हिमालय पर्वत की बर्फ से ढकी अगम्य घाटियों में कस्तूरी मृग विचरण करते हुए देखे जा सकते हैं। उन मृगों की नाभि से निकलने वाली कस्तूरी की सुगन्ध को खोजते हुए वे अज्ञानतावश इधर-उधर भटक रहे हैं। मृग कस्तूरी न मिलने के कारण स्वयं से दुखी प्रतीत होते हैं।

कवि बादलों का वर्णन करते हुए कहता है कि प्राचीनकाल में कुबेर के शाप से अलकापुरी से निष्कासित यक्ष ने बादलों को दूत बनाकर अपनी प्रेमिका के समीप भेजा था। इसमें कितनी सत्यता है कौन जानता है? मैंने शीत के दिनों में बर्फ से ढकी कैलाश पर्वत की ऊँची-ऊँची चोटियों पर काले-काले बादलों को घिरते देखा है।

बादल को घिरते देखा है संदर्भ-प्रसंगसहित व्याख्या

(1) अमल धवल गिरि के शिखरों पर,
बादल को घिरते देखा है।
छोटे-छोटे मोती जैसे
उसके शीतल तुहिन कणों को
मानसरोवर के उन स्वर्णिम
कमलों पर गिरते देखा है,
बादल को घिरते देखा है।

शब्दार्थ :
अमल = निर्मल। धवल = सफेद। गिरि = पर्वत। तुहिन = ओस।

सन्दर्भ :
प्रस्तुत छन्द ‘बादल को घिरते देखा है’ शीर्षक कविता से लिया गया है। इसके कवि ‘नागार्जुन’ हैं।

प्रसंग :
घिरते हुए बादलों को देखकर कवि की कल्पना स्फुटित हुई है, उसी का यहाँ वर्णन है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि निर्मल तथा श्वेत पर्वत के शिखरों पर मैंने बादल को घिरते हुए देखा है। छोटे-छोटे मोती जैसे उसके ठण्डे बर्फ के कणों को मानसरोवर के उन सुनहरे कमलों पर गिरते देखा है। बादल को घिरते देखा है।

विशेष :

1. कवि का प्रकृति चित्रण बड़ा ही मोहक है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(2) तुंग हिमालय के कन्धों पर
छोटी-बड़ी कई झीलें हैं,
उनके श्यामल नील सलिल में
समतल देशों से आ-आकर
पावस की ऊमस से आकुल
तिक्त-मधुर विसतंतु खोजते
हंसों को तिरते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
तुंग = ऊँचे। नील सलिल में = नीले जल में। समतल = मैदानी भागों से। पावस= वर्षा। आकुल= व्याकुल। तिक्त = तीखे। मधुर =मीठे। विसतंतु = कमल नाल।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने मानसरोवर झील में तैरने वाले हंसों का वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि ऊँचे हिमालय पर्वत की तलहटी में छोटी और बड़ी कई झीलें हैं। उनके साँवले और नीले जल में मैदानी भागों से उड़-उड़कर आने वाले तथा
वर्षा की उमस से व्याकुल हुए एवं तीखे तथा मीठे कमल नाल को खोजते-फिरते हंसों को मैंने तैरते देखा है। बादल को घिरते देखा है।

विशेष :

1. मानसरोवर की प्राकृतिक सुषमा तथा उसमें तैरने वाले सुन्दर हंसों का मनमोहक वर्णन है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(3) ऋतु बसंत का सुप्रभात था
मंद-मंद था अनिल बह रहा
बालारुण की मृदु किरणें थीं
अगल-बगल स्वर्णाभ शिखर थे
एक-दूसरे से विरहित हो
अलग-अलग रहकर ही जिनको
सारी रात बितानी होती,
निशा काल से चिर-अभिशापित
बेबस उन चकवा-चकई का
बंद हुआ क्रन्दन फिर उनमें
उस महान सरवर के तीरे
शैवालों की हरी दरी पर
प्रणय-कलह छिड़ते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
अनिल = वायु। बालारुण = प्रात:कालीन सूर्य। मृदु = कोमल। स्वर्णिम = सुनहरे। विरहति = अलग। होकर, वियोग की दशा में। निशाकाल = रात के समय। चिर = दीर्घ काल से। अभिशापित = अभिशाप पाये हुए। सरवर = तालाब। तीरे = किनारे। शैवालों = काई। प्रणय-कलय = प्रेम का झगड़ा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत छन्द में कवि ने बसन्त ऋतु के सुप्रभात की मनोहर झाँकी का वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि बसन्त ऋतु का सुप्रभात था। उस समय वायु मंद-मंद गति से बह रही थी तथा प्रात:कालीन सूर्य की कोमल किरणे अगल-बगल के सुनहरे शिखरों पर बिखर रही थीं। उस समय एक-दूसरे से अलग होकर! सारी रात वियोग में बिता देने वाले, अनन्त काल से अलग रहने का अभिशाप पाये हुए, उन बेवस चकवा-चकवी का वियोग। दुःख से उत्पन्न क्रन्दन यकायक बन्द हो गया। प्रात:काल होने पर उस महान सरोवर के किनारे काई की हरी पट्टी पर उन चकवा-चकवी को परस्पर प्रेमालाप में झगड़ते हुए मैंने देखा है। बादल को घिरते देखा है।

विशेष :

1. चकवा-चकवी रात में एक-दूसरे से अलग होकर विरह जन्य क्रन्दन करते रहते हैं। प्रात:काल दोनों का संयोग होने पर प्रणय क्रीड़ाएँ करने लगते हैं।
2. बसन्त ऋतु की मादकता का वर्णन है।
3. भाषा-खड़ी बोली।

(4) दुर्गम बर्फानी घाटी में
शत-सहस्त्र फुट ऊँचाई पर
अलख-नाभि से उठने वाले
निज के ही उन्मादक परिमल
के पीछे धावित हो होकर
तरल तरुण कस्तूरी मृग को
अपने पर चिढ़ते देखा है,
बादल को घिरते देखा है।

शब्दार्थ :
दुर्गम = जहाँ जाना सरल न हो। बर्फानी = बर्फ से ढकी हुई। शत-सहस्र = सैकड़ों-हजारों। अलख नाभि = अदृश्य नाभि। उन्मादकं = नशीले। परिमल = पराग, सुगन्ध। धावित = दौड़कर।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने दुर्गम बर्फीली घाटियों में दौड़ते हुए कस्तूरी मृगों की चेष्टाओं का सुन्दर वर्णन किया है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि दुर्गम बर्फ से ढकी हुई चोटियों में सैकड़ों-हजारों फुट की ऊँचाई पर, नाभि से उठने वाली स्वयं की ही अदृश्य उन्मादक गन्ध को ढूँढ़ते हुए तथा उसकी खोज में भटकते हुए चंचल एवं तरुण कस्तूरी मृगों को स्वयं अपने ऊपर चिढ़ते हुए मैंने देखा है, बादल को घिरते देखा है।

विशेष :

1. कस्तूरी मृग की नाभि में अदृश्य रूप से छिपी रहती है, पर कस्तूरी मृग अपने अन्दर छिपी हुई कस्तूरी को न जानकर उसकी खोज में इधर-उधर भटकता फिरता है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

(5) कहाँ गया धनपति कुबेर वह
कहाँ गई उसकी वह अलका
नहीं ठिकाना कालिदास के
व्योम-प्रवाही गंगाजल का,
ढूँढा बहुत परन्तु लगा क्या
मेघदूत का पता कहीं पर,
कौन बताए वह छायामय
बरस पड़ा होगा न यहीं पर,
जाने दो, वह कवि कल्पित था,
मैंने तो भीषण जाड़ों में
नभ-चंबी कैलाश शीर्ष पर
महादेव को झंझानिल से
गरज-गरज भिड़ते देखा है।
बादल को घिरते देखा है।

शब्दार्थ :
अलका = कुबेर की राजधानी। व्योम-प्रवाही। = आकाश में बहने वाली। नभ-चुंबी = आकाश को चूमने वाले। झंझानिल = झंझावात।।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस छन्द में कवि ने कैलाश पर्वत की सुन्दरता का वर्णन करते हुए पौराणिक आख्यानों की चर्चा की है।

व्याख्या :
कविवर नागार्जुन कहते हैं कि इस कैलाश शिखर पर धनपति कुबेर की अलका नगरी का पुराणों में उल्लेख मिलता है लेकिन आज तो इसका यहाँ कोई नामोनिशान भी नहीं है, न ही हमें कालिदास के आकाश में विचरण करने वाले उस मेघदूत का कहीं पता मिलता है। हमने बहुत प्रयत्न कर लिए पर वह ढूँढने से भी नहीं मिला। ऐसा हो सकता है कि वह छायामय मेघदूत यहीं कहीं जाकर बरस गया होगा। फिर कवि कहता है कि इन बातों को छोड़ो, मेघदूत तो कवि कालिदास की कल्पना थी।

मैंने तो भयंकर कड़कड़ाते जाड़ों में आकाश को चूमने वाले कैलाश के शिखर पर घनघोर झंझावातों में महादेव जी को उनसे गरज-गरज कर युद्ध करते देखा है। बादल को घिरते देखा है।

विशेष :

1. इस छन्द में कवि ने गगनचुम्बी कैलाश शिखर की प्राकृतिक छटा का सुन्दर वर्णन किया है।
2. अनुप्रास की छटा।
3. भाषा-खड़ी बोली।

MP Board Class 10th Hindi Solutions

Students get through the MP Board Class 11th Physics Important Questions Chapter 8 Gravitation which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 8 Gravitation

Gravitation Class 11 Important Questions Very Short Answer Type

Question 1.
Write the difference between gravitation and gravity.
Answer:
Gravitation : The force of attraction between any two material bodies in the universe is called force of gravitation.
Gravity: If one of the attracting bodies is earth or some other planet or natural satel¬lite, then the force of attraction is called force of gravity.

Question 2.
Explain Newton’s universal gravitational laws.
Answer:
Every body of the universe attract each other. The force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them and it acts along the line joining the two bodies.

\(\vec{F}\)AB → Force given by B on A, \(\vec{F}\)BA → Force given by A on B. Let two bodies of mass m₁ and m₂ are kept d distance apart.
∴ (i) F ∝ m₁m₂and
(ii) F ∝ \(\frac{1}{d^{2}}\)
Combining both the laws,
F ∝\(\frac{m_{1} m_{2}}{d^{2}}\)
F = G\(\frac{m_{1} m_{2}}{d^{2}}\)
Where, G is constant called universal gravitational constant.

Question 3.
Define universal gravitational constant.
Answer:
We know that, F = G\(\frac{m_{1} m_{2}}{d^{2}}\)
If m₁ = m₂= 1 and d= 1, then F= G
Hence, the universal gravitational constant is numerically equal to the force of attraction between the two bodies each of mass unity, kept at unit distance apart.

Question 4.
Find out the unit and dimensional formula of universal gravitational constant. Also, write the value.
Answer:
Since, F = \(\frac{G m_{1} m_{2}}{d^{2}}\)
or
G = \(\frac{F d^{2}}{m_{1} m_{2}}\)
Unit: S.I. unit of G will \(\frac{\mathrm{Nm}^{2}}{\mathrm{~kg} \mathrm{~kg}}\) = Nm² kg^-2 .
Dimensional formula : [G] = \(\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{M}][\mathrm{M}]}\) = [M^-1L³T^-2]
Value : Value of G = 6.67 × 10^-11 Nm² kg ^-2.

Question 5.
The value of G = 6.67 × 10^-11 Nm² kg ^-2. What does it means? Why G is called universal constant?
Answer:
When two bodies, each of mass 1 kg are kept 1 m apart, then the force of attraction between them is 6.67 × 10^-11 N.
The value of G does not depend upon nature, medium, time, temperature etc. There-fore, it is called universal constant.

Question 6.
What is acceleration due to gravity ? What is its standard value?
Answer:
The acceleration produced in a body due to force of gravity is called acceleration due to gravity.
or
The rate of change of velocity in a body falling freely is called acceleration due to gravity.
It is denoted by g and its standard value is 9.8 m × s^-2.

Question 7.
Write the relation between ‘g’ and ‘G’.
Answer:
g = \(\frac{\dot{G} M}{R^{2}}\)

Question 8.
What is the value of ‘g’ at the centre of earth?
Answer:
The value of ‘g’ at the centre of earth is zero.

Question 9.
On what factors the weight of a body depends upon a planet?
Answer:

• Size of planet,
• Daily rotation of planet,
• Position of body in planet.

Question 10.
How the magnitude of ‘g’ changes on the surface of earth?
Answer:
From equator to pole value of ‘g’ increases. It is minimum at equator and maxi¬mum at pole.

Question 11.
At which place value of ‘g’ is minimum and maximum?
Answer:
Value of ‘g’ is minimum at centre of earth and maximum at pole.

Question 12.
A body is taken to moon from the centre of the earth. What change will occur in the weight and the mass of the body?
Answer:
The weight of a body at the centre of the earth is zero. When the body is moved towards the moon, the weight of the body will increase. On the surface of the earth the weight will be maximum. Further the weight will decrease and on the moon it will be \(\frac{1}{6}\) th of the weight on the surface of earth. The mass will remain same.

Question 13.
What is the value of g at the centre of the earth? What will be its weight?
Answer:
Zero, zero.

Question 14.
How does the value of g change on the surface of earth ? Where it is maximum and minimum?
Answer:
The value of g increases when the body is taken from equator to poles. The value of g is maximum at poles and minimum at the equator.

Question 15.
Can the gravity of earth be zero, at some height?
Answer:
No as F ∝ \(\frac{1}{r^{2}}\)

Question 16.
If the earth stops rotation about its axis, what will happen to the weight of the body? What change will occur in the body placed at the poles?
Answer:
If the earth stops rotation the weight of the bodies will increase. The weight of the body placed at poles will remain same.

Question 17.
A man can jump six times more on the moon, than on the earth. Why?
Answer:
Because the value of acceleration due to gravity is six times less than that of earth.

Question 18.
The gravitational mass on earth is \(\frac{g}{6}\). If a body is taken to moon from the earth, then what will happen to weight, inertial mass and gravitational mass ?
Answer:
The weight will be \(\frac{1}{6}\) time that of earth. The inertial mass and gravitational mass will not change.

Question 19.
The gravitational potential energy of a body on the surface of earth is 6.4 × 10⁶ joule. Explain.
Answer:
The work done required to throw a body away from gravitational field of earth will be 6.4 × 10⁶ joule.

Question 20.
Weight of a body at equatorial line is less than its weight at pole. Why?
Answer:
Because value of ‘g’ at equator is less than pole and since weight = Mass × ‘g’ Weight at equator is less than pole.

Question 21.
What is escape velocity? On what factors does it depend?
Answer:
It is the least velocity with which a body must be thrown vertically upwards so that it just escapes the gravitational pull of earth.
It depends upon

• Radius of planet and
• Acceleration due to gravity of the planet.

Question 22.
What are the value of escape velocities of earth and moon?
Answer:
Escape velocity of earth, = 11.2 kms^-1
and that of moon = 2.38 kms^-1

Question 23.
What is an artificial satellite?
Answer:
The man made satellite which revolves around the earth are called artificial satellite.

Question 24.
What is geo-stationary satellite or synchronous satellite ? What is its height above the earth?
Answer:
A satellite revolving around the earth in west to east direction; in the equatorial plane perpendicular to the axis of rotation with time-period equal to that of earth i.e., 24 hours, is called geo-stationary satellite.
Its height from the surface of earth is nearly 36000 km.

Question 25.
What is parking orbit? What is its radius?
Answer:
The orbit in which the geo-stationary satellite is launched is called parking orbit. Its radius is nearly 42309 km.

Question 26.
Why the orbit of artificial satellite is set out of the atmosphere?
Answer:
The orbit of satellite is set out of the atmosphere so that the satellite may not bum due to the friction with air.

Question 27.
What do you mean by weightlessness in the satellite?
Answer:
The necessary centripetal force to revolve in a circular orbit is provided by the gravity of earth. Hence, the net force on the satellite and the bodies inside it will be zero. Therefore, the weight of the body becomes zero.

Question 28.
Weightlessness is not felt on the moon, why?
Answer:
Due to gravity of moon, the bodies have weights on the moon, of course it is a satellite of earth.

Question 29.
Can a satellite be launched in an orbit which is not in the plane, passing through the centre of earth?
Answer:
No, because under this condition, the horizontal component of the force of at-traction between earth and satellite cannot balance the centripetal force. As a result, the satellite will be attracted towards the equatorial plane and its orbit will not be stable.

Question 30.
Can a satellite be launched in an orbit, so that it can be seen always above
Delhi?
Answer:
No, this is because Delhi is not on equator line.

Question 31.
What is the relation between orbital velocity and escape velocity?
Answer:
Escape velocity = √2 × Orbital velocity.

Question 32.
What is the direction of revolution of a geo-stationary satellite?
Answer:
The direction of revolution of geo-stationary satellite is same as the direction of rotation of earth about its own axis, i.e., west to east.

Question 33.
What is the angle between the plane of revolution of satellite and axis of rotation ?
Answer:
90°.

Question 34.
In an artificial satellite a bob is hanging with a string, what will be the tension in the string ?
Answer:
Zero.

Question 35.
Can the velocity of a body be equal to the velocity of light?
Answer:
No, as the mass of a body increases with the velocity. When its velocity becomes equal to that of light, the mass of the body will be infinite, which is not possible.

Question 36.
If the height of the parking orbit of a satellite in space is increased then its orbital velocity decreases, why?
Answer:
Since orbital velocity ν₀ = R\(\sqrt{\frac{g}{R+h}}\)
Here, R and g are constant.
∴ ν₀∝ \(\frac{1}{\sqrt{h}}\)
i.e., if h increases then ν₀ decreases.

Question 37.
To find out time in a satellite pendulum clock is used or spring clock.
Answer:
Spring clock is used to determine time at satellite because pendulum clock will get stop there.

Question 38.
In an artificial satellite a mass of 2 kg is kept at spring balance. What will be the reading of spring balance ?
Answer:
Zero.

Question 39.
What will happen if a glass full of water is tilt in an artificial satellite?
Answer:
Water will start swinging as a droplet inside an artificial satellite due to weightlessness.

Question 40.
Inside an artificial satellite a body of mass 1 kg is hanged with a string. What will be the tension in the string ?
Answer:
Zero.

Question 41.
Escape velocity of 5 gram object in earth is 11.2 km/s. What will be the escape velocity for 10 gram object?
Answer:
Escape velocity of 10 gram object will be 11.2 km/s because escape velocity does not depend upon the mass of the object.

Question 42.
From where a satellite obtain the centripetal force required to revolve around a planet?
Answer:
Any satellite gets the required centripetal force to revolve around the planet from the gravitation force of the planet.

Question 43.
Sun attracts earth toward itself with its gravitational force, but earth does not move toward sun, explain with reason.
Answer:
Earth is not stationary, it revolves around the sun in a circular orbit. Centripetal force required to revolve around the sun is obtained by earth from the gravitational force of sun. In this way earth remains in equilibrium.

Question 44.
On which factor the period of revolution of geo-stationary satellite depends?
Answer:
Period of revolution of geo-stationary satellite depends on distance of it from earth surface. If the distance from earth surface increases, its period of revolution also increases.

Question 45.
How the orbital velocity of a satellite depends upon the distance of its from the earth surface?
Answer:
When the distance of satellite from earth surface increases, its orbital velocity decreases.

Question 46.
Why total energy of a satellite is negative?
Answer:
Since, energy is required for satellite to place it at orbit in infinity. So total energy of a satellite is negative.

Question 47.
Gravitational potential energy of any object at earth surface is 6.4 × 10⁶ joule. Explain its meaning.
Answer:
It means energy required to send an object out of gravitational field is 6.4 × 10⁶ joule.

Gravitation Class 11 Important Questions Short Answer Type

Question 1.
Establish the relation between g and G.
Answer:
Let the mass of the earth be M and its radius be R. A body of mass m is placed on it. Therefore, the force of attraction by the earth on the body :
By Newton’s laws of gravitation,
F = \(\frac{G M m}{R^{2}}\) …(1)
But, the force with which a body is attracted towards the earth gives the weight of body.
F = mg …(2)
Equating eqns. (1) and (2), we get
mg= \(\frac{G M m}{R^{2}}\)
or
g = \(\frac{G M}{R^{2}}\)
It is the required relation.

Question 2.
Explain with mathematical calculation, how the value of g changes with altitude?
or
How does the value of ‘g’ change above the surface of the earth ? Find out the expression and explain.
Answer:
The value of g decreases with the increase of altitude.
Let the mass of earth is M and its radius is R.
∴ g = \(\frac{G M}{R^{2}}\) …(1)
If at point P at height h from the surface of earth is g’, then
g’ = \(\frac{G M}{(R+h)^{2}}\) …(2)
∴ By eqns. (1) and (2), we get

Hence, g’ < g.
Thus as height increases the value of g decreases.

Question 3.
How the value of g changes with the increase of depth? Find the value of g
at the centre of earth.
or
Prove that the value of g at the centre of earth is zero.
Answer:
Assuming the earth to be a homogeneous sphere of mass M, radius R and acceleration due to gravity on the surface is g. Let ρ be the mean density of earth.
We know that, g = \(\frac{G M}{R^{2}}\) …(1)
∵ M = Volume × Density
or
M = \(\frac{4}{3}\)πR³ × ρ …(2)
Putting the value of Min eqn. (1),
g = \(\frac{G \frac{4}{3} \pi R^{3} \times \rho}{R^{2}}\)
or
g = \(\frac{4}{3}\)πRGρ …(3)

Now, let the body be taken to depth h, below the free sur¬face hence the distance of the body from the centre of earth will be (R-h). Hence, the force of attraction on the body will be acted by the sphere of radius (R-h). Mass of earth of radius (R-h) will be M’.
∴ M’ = \(\frac{4}{3}\)π(R – h)³ ρ
If g_d be the acceleration due to gravity at depth h, then using eqn. (1),

Dividing eqn. (4) by eqn. (3), we have

Hence, from eqn. (5), the value of g decreases with the increase of depth. At the centre
ofearth, h=R
∴ g_d = g\(\left(1-\frac{R}{R}\right)\)
or
g_d= g(1 – 1)g × 0
∴ g_d= 0
Hence, value of g at the centre of earth is zero.

Question 4.
How does the value of ‘g’ change due to shape of the earth?
Answer:
The earth is not perfectly spherical in shape. It is slightly flattened at the poies
and is bulged at the equators
Let R_p and R_e be polar and equational radius respectively.,
Hence, g_p =\(\frac{G M}{R_{p}^{2}}\) and g_e =\(\frac{G M}{R_{p}^{2}}\)
Where g_p and g_e are the acceleration due to gravity at the poles and equator respectively.

Now \(\frac{g_{e}}{g_{p}}=\frac{G M}{R_{e}^{2}} \times \frac{R_{p}^{2}}{G M}=\left(\frac{R_{p}}{R_{e}}\right)^{2}\)
∵ R_p < R_e
∴ \(\left(\frac{R_{p}}{R_{e}}\right)^{2}\) < 1, hense \(\frac{g_{e}}{g_{p}}\) < 1
or
g_e < g_p

Question 5.
Show that the gravitational potential energy of an object near the earth
surface is ΔU = mgh.
Answer:
If a particle is raised from the surface of the earth to a height h above, then r_{1 =} R (earth radius) and r₂ = R + h (see figure).
So that, the change in potential energy )

Using Binomial theorem and neglecting the higher powers of h/R, we get

or
ΔU = G \(\frac{M m h}{R^{2}}\)
But, \(\frac{G M}{R_{p}^{2}}\)= g, (acceleration due to gravity at the surface of the earth).
Hence, ΔU = mgh
Thus, when a body of mass m is taken h distance above the earth’s surface, its potential energy increases by mgh.

Question 6.
What do you mean by gravitational field intensity? Establish relation be-tween intensity of gravitational field and acceleration due to gravity.
Answer:
Gravitational field :
The space around a material body in which its gravita-tional pull can be experienced is called its gravitational field.
Gravitational field intensity:
The intensity of gravitational field of a body at a point in a field is defined as the force experienced by a body of unit mass placed at that point provided the presence of unit mass does not disturb the original gravitational field.
i.e.. E = \(\frac{F}{m}\)

Let an object of mass M is at point O and at a distance of r from it at point P, we have to find out the intensity.
Let us consider a unit mass (m = 1) is placed at point P, therefore force of attraction applied by mass M on unit mass will be
F = \(\frac{G M \cdot 1}{r^{2}}=\frac{G M}{r^{2}}\)
Intensity at point P
E = \(=\frac{F}{m}=\frac{G M}{r^{2}} / 1=\frac{G M}{r^{2}}\) …(1)
and we know, acceleration due to gravity
g = \(\frac{G M}{r^{2}}\) …(2)
From eqns. (1) and (2), we get E = g
i.e., Intensity of gravitational field is equal to acceleration due to gravity of any point.

Question 7.
What is orbital velocity? Derive an expression for the orbital velocity. Obtain an expression if satellite is near to the surface of earth.
Answer:
Orbital velocity:
Orbital velocity is the velocity which is given to an artificial satellite, a few hundred kilometre above the earth’s surface so that it may start revolving round the earth.
Let the mass of satellite is m and the radius of orbit from the centre of earth is r, revolving with a velocity of ν_o
∴ Centripetal force = \(\frac{m v_{o}^{2}}{r}\) …(1)
Also, by Newton’s law of gravitation,
F = \(\sqrt{\frac{G M m}{r^{2}}}\) …(2)
Since, centripetal force is provided by gravitation force, then

\(\frac{m v_{o}^{2}}{r}=\frac{G M m}{r^{2}}\)
or
ν_o² = \(\frac{G M}{r}\)
∴ ν_o = \(\sqrt{\frac{G M}{r}}\) …(3)
If the radius of earth is R and height of satellite from the surface of earth is h, then
r = R + h
But g = \(\frac{G M}{R^{2}}\)
or
GM = gR²
Putting the values of GM = gR² and r = R + h in eqn. (3), we get
ν_o = \(\sqrt{\frac{g R^{2}}{R+h}}=R \sqrt{\frac{g}{R+h}}\)
If the satellite is near to the surface of earth,then h<<R, then R + h ≈ R
∴ ν_o = R\(\sqrt{\frac{g}{R}}=\sqrt{g R}\)

Question 8.
What is meant by the period of revolution ? Derive an expression for It.
Answer:
Period of revolution :
The time taken to complete one revolution is called period of revolution of satellite.
Let the period of revolution is Tand the radius of orbital is r.
∴ Orbital velocity = \(\frac{\text { Total path travelled }}{\text { Total time taken }}\)
ν_o = \(\frac{2 \pi r}{T}\)
or
T = \(\frac{2 \pi r}{v_{o}}\)
But r = R + h
∴ T = \(\frac{2 \pi(R+h)}{v_{o}}\)
But we have
ν_o = R \(\sqrt{\frac{g}{R+h}}\)
∴ T = \(\frac{2 \pi(R+h)}{R \sqrt{\frac{g}{R+h}}}\)
or
T = 2π(R + h)\(\frac{\sqrt{(R+h)}}{R \sqrt{g}}\)
∴ T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}}\)

Question 9.
Determine the revolution time of a satellite, near to earth.
Answer:
We have T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}}\)
If h<<R, then
T = \(\frac{2 \pi}{R} \sqrt{\frac{R^{3}}{g}}\)
= \(2 \pi \sqrt{\frac{R}{g}}\)
Now
R = 6.38 × 10⁶ m,g = 9.8ms^-2
∴ T = \(2 \times 3 \cdot 14 \sqrt{\frac{6 \cdot 38 \times 10^{6}}{9 \cdot 8}}\)
= 84.4 min.(approx.).

Question 10.
Planet A is heavier than planet B. Which will have the greater escape velocity?
Answer:
We have, ν = \(\sqrt{\frac{2 G M}{R}}\)
If R is constant, then
ν ∝ √M
∴ \(\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{A}}{M_{B}}}\)
∵ M_A>M_B
∴ ν _A> ν_B
Hence, escape velocity of planet A will be greater than that of planet B.

Question 11.
Establish the relation between escape velocity and orbital velocity.
Answer:
Escape velocity : ν_e = \(\sqrt{2 g R}\) …(1)
Orbital velocity : ν_o = \(\sqrt{g R}\) …(2)
Dividing eqn. (1) by eqn. (2), we get
\(\frac{v_{e}}{v_{o}}=\frac{\sqrt{2 g R}}{\sqrt{g R}}=\sqrt{2}\)
∴ ν_e = \(\sqrt{2} v_{o}\)

This is the required relation between escape velocity and orbital velocity.

Question 12.
What are the characteristics of geo-stationary satellite?
Answer:

• The geo-stationary satellite revolves around the earth in equatorial plane, perpendicular to the axis of rotation of earth.
• Its direction is from west to east.
• Its time-period of revolution is 24 hours.

Question 13.
What is geo-stationary satellite ? Prove that its height from the surface of earth is 36000 km.
Answer:
Geo-stationary satellite:
A satellite revolving around the earth in west to east direction; in the equatorial plane perpendicular to the axis of rotation with time-period equal to that of earth i.e., 24 hours, is called geo-stationary satellite.
Its height from the surface of earth is nearly 36000 km.
Now, we have

= 42.22 × 10⁶ m = 42220 km (Approx.)
∴ h=(R + h) – R
= 42.22 × 10⁶ – 6.38 × 10⁶ = 35840 km
or
h = 36000 km (Approx.) Proved.

Question 14.
Why the moon has no atmosphere?
Or
Why the moon has no atmosphere while Jupiter and Saturn have dense atmosphere?
Answer:
The escape velocity for moon is 2.38 kms^-1, while the r.m.s. speed of molecules of gases at 500 K is 2.5 kms^-1, hence the molecules of gases escape from the gravity of moon. The escape velocities of Jupiter and Saturn are greater, therefore, they have atmosphere.

Question 15.
What conclusion did Newton obtain from Kepler’s law of planetary motion?
Answer:
Conclusions obtained by Newton : On the basis of Kepler’s law, Newton pro-pounded the following:
(i) The path of the planets round the sun can be approximately taken as circular, for which the necessary centripetal force can be obtained from the gravitational force of attraction between the planet and sun. The force acting on the planet is always towards the sun.

(ii) By the third law :
T² ∝ r³
For the circular path of planet, the centripetal force acting will be
F = \(\frac{m v^{2}}{r}=\frac{m(r \omega)^{2}}{r}\) = mrω²
or
F = mr \(\left(\frac{2 \pi}{T}\right)^{2}=\frac{4 \pi^{2}}{T^{2}}\) mr …(1)
Since T² ∝ r³ ⇒ T² = Kr³
Putting the value of T²in eqn (1),we get
F = \(\frac{4 \pi^{2}}{K r^{3}}\) mr = \(\frac{4 \pi^{2} m}{K r^{2}}\)
But,\(\frac{4 \pi^{2} m}{K}\) is a Constant.
∴ F = \(\frac{\text { Constant }}{r^{2}}\)
or
F ∝ \(\frac{1}{r^{2}}\)
Hfence, the force acting on the planet is inversely proportional to the square of dis-tance between them.

(iii) The force acting on the planet is directly proportional to its mass.
ie., F∝m.

Gravitation Class 11 Important Questions Long Answer Type

Question 1.
Derive an expression for the escape velocity. Does it depend upon mass?
Answer:
Escape velocity : It is the least velocity with which a body must be thrown vertically upwards so that it just escapes the gravitational pull of earth.
It depends upon

• Radius of planet and
• Acceleration due to gravity of the planet.

Let the mass of earth be M and its radius is R .At any time let the body is at a distance x from the centre of earth.
∴ Force on the body due to earth, F = \(\frac{G M m}{x^{2}}\)
∴ Work done in moving the body through dx distance, against gravity,
dW = Fdx = \(\frac{G M m}{x^{2}}\) .dx
∴ Work done in moving up to infinity,

∴ W = \(\frac{G M m}{R}\)
If the escape velocity is ν_e, then

This is the required expression.
Since, the formula is free from the mass of the body, hence the escape velocity does not deppnd upon the mass of the body.

Gravitation Class 11 Important Numerical Questions

Question 1.
Two masses, each of 1 gm, are kept 1 cm apart. Calculate the force of at-traction between them in kg wt.
Solution:
Given : m₁, = m₂ = 1 gm = 10^-3 kg, r = 1 cm = 10^-2 m and
G = 6.67 × 10^-11 Nm²kg^-2
Now F = \(\frac{G m_{1} m_{2}}{r^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{-11} \times 10^{-3} \times 10^{-3}}{\left(10^{-2}\right)^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{-17}}{10^{-4}}\)
6.67 × 10^-13 N
\(=\frac{6 \cdot 67 \times 10^{-13}}{10}\) kg wt
∴ F = 6.67 × 10^-14kg wt

Question 2.
The mass of a planet is 10¹⁸ kg and its radius is 10³ m. Calculate the acceleration due to gravity on that planet.
Solution:
Given : M = 10¹⁸kg, R = 10³m and G = 6.67 × 10^-11 Nm²kg^-2.
Now, g = \(\frac{G M}{R^{2}}\)
∴ g = \(\frac{6 \cdot 67 \times 10^{-11} \times 10^{18}}{\left(10^{3}\right)^{2}}\)
\(=\frac{6 \cdot 67 \times 10^{7}}{10^{6}}\)
or
g = 6.67 × 10= 66.7ms^-2.

Question 3.
Calculate the value of g at 600 km height above the surface of earth. Given the radius of earth R = 6400 km and g = 9.8 ms^-2.
Solution:
Given : R = 6400 km = 6400 × 10³ m, g = 9.8 ms ^-2,
h = 600 km = 600 × 10³ m
Formula \(\frac{g^{\prime}}{g}=\frac{R^{2}}{(R+h)^{2}}\)
0r
\(\frac{g^{\prime}}{9 \cdot 8}=\left(\frac{6400 \times 10^{3}}{(6400+600) \times 10^{3}}\right)^{2}=\left(\frac{6400}{7000}\right)^{2}\)
∴ g’ = 9.8 × \(\left(\frac{64}{70}\right)^{2}\)
= 8.192ms^-2

Question 4.
At what height the weight of a body be \(\frac{1}{3}\) times that of the surface of earth
Solution:
Given:


∴ h = R(1.732 – 1) = 0.732Rm

Question 5.
What would be the angular velocity of earth so that the weight of a body at equator becomes zero?
Solution:
Now, we have at equator,
g’ = g – Rω²
But, g’ = 0
∴ g – Rω² = 0
or
-Rω² = -g
or
ω²=g|R
or
ω = \(\sqrt{\frac{g}{R}} \text { . }\)

Question 6.
Calculate the potential energy of a body of mass 100 kg placed at a height 600 km above the surface of earth (M = 6 × 10²⁴ kg, R = 6400 km)-
Solution:
Given : M = 6 × 10²⁴ kg,R = 6400 km = 6400 × 10³m,
h = 600 km = 600 × 10³m,m = 100 kg.
Formula U = \(-\frac{G M m}{r}\)
\(=-\frac{6 \cdot 67 \times 10^{-11} \times 6 \times 10^{24} \times 100}{(6400+600) \times 10^{3}}\) (r = R +h)
\(=-\frac{6 \cdot 67 \times 6 \times 10^{15}}{7000 \times 10^{3}}\)
U = – 5.717 10⁹ joule.

Question 7.
The gravitational field intensity at a point 8000 km away from the earth surface is 6 N/kg. What is gravitational potential at that point?
Solution:
Given : E = 6 N/kg, d – 8000 km = 8000 × 10³m
∴ Formula V = -E × d
Gravitational potential V = -6 × 8000 × 10³
= -48 × 10⁶
= -4.8 × 10⁷Nm/kg.

Question 8.
An artificial satellite is revolving around the earth at height 3600 km above the surface of earth. Calculate
(i) Orbital velocity,
(ii) Period of revolution.
Solution:
Given : h = 3600 km = 3600 × 10³m, g = 9.8 m/s² and R = 6.38 × 10⁶m.
Formula: ν_o = \(\sqrt{\frac{g R^{2}}{R+h}}\)

= 10⁴ × 0.6322m/s
= 6.322 × 10³m/s = 6.322m/s

= 626744 × 1.58 = 9913.9 sec
= 2.734 hours

Question 9.
The mass of planet A Is 4 times that of planet I? and radius of A, ¡s 2 times
that of B. Compare their escape velocities.
Solution:
Given : M₁= 4M₂and R₁ = 2R₂

Question 10.
The distance of planet A is three times that of planet B. If the time of revolution of B is 6 months, then find the time of revolution of A.
Solution:By kepler’s law,
T² ∝ r³

Question 11.
The time of revolution ofaplanet revolving in a circular orbit of radius R is T. Find the time of revolution of a planet revolving in an orbit of radius 4R.
Solution:
Given: = T₁ = T, r₁ = R,r₂ = 4R.
By Kepler’s law,
\(\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
or
\(\left(\frac{T}{T_{2}}\right)^{2}=\left(\frac{R}{4 R}\right)^{3}=\frac{1}{64}\)
or
\(\frac{T}{T_{2}}=\frac{1}{8}\)
∴ T₂ = 8T

Question 12.
If the radius of the moon is 1.74 × 10⁶ m and its mass is 7.36 × 10²² kg, then calculate the escape velocity from the surface of the moon.
Solution:
Given: R = 1.74 × 10⁶ m, M = 7.36 × 10²² kg,
G = 6.67 × 10^-11Nm²/kg²
∵ Escape velocity, ν_e = \(\sqrt{\frac{2 G M}{R}}\)
∴ ν_e = \(\sqrt{\frac{2 \times 6 \cdot 67 \times 10^{-11} \times 7 \cdot 36 \times 10^{22}}{1 \cdot 74 \times 10^{6}}}\)
= 2.38 × 10³ m/sec = 2.38 km/sec.

Question 13.
The escape velocity of earth is 11 kms1. If the radius of a planet is double of earth’s radius and mean density are same, then find the escape velocity of that planet.
Solution:
Escape velocity,
ν = \(\sqrt{\frac{2 G M}{R}}\)

or
ν₂ = 22 km/s.

Question 14.
The mass of a planet is 3 × 10²⁴ kg and radius is 5 × 10⁶ m. What will be the energy required to move, 5 kg body to infinity ?
Solution:
Given : M = 3 × 10²⁴kg,R = 5 × 10⁶m, m = 5kg,
G = 6.67 × 10¹¹ Nm²kg^-2.
Formula U = \(\frac{G M m}{2 R}\)
\(=\frac{6 \cdot 67 \times 10^{-11} \times 3 \times 10^{24} \times 5}{2 \times 5 \times 10^{6}}\)
U = 100.05 × 10⁶ = 1010⁸ joule.

Gravitation Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Newton’s laws of gravitation is true for :
(a) All bodies
(b) Only for heavenly bodies
(c) Only for small bodies
(d) Only for charged objects.
Answer:
(a) All bodies

Question 2.
Value of ‘g’ at moon with comparison to earth is :
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{6}\)
Answer:
(d) \(\frac{1}{6}\)

Question 3.
value of universal gravitational constant G is :
(a) Same at all places
(b) Unequal at all places
(c) Zero
(d) Infinity.
Answer:
(b) Unequal at all places

Question 4.
Relation between g and G is :
(a) g = GM
(b) gR² = GM
(c) gM = GR²
(d) gM/GR².
Answer:
(b) gR² = GM

Question 5.
Magnitude of g outside the gravitational field is:
(a) Infinity
(b) Zero
(c) 9.8 m/s²
(d) 980 newton.
Answer:
(b) Zero

Question 6.
Magnitude of g at the centre of earth is:
(a) Zero
(b) Infinity
(c) 9.8 newton
(d) 980 dyne.
Answer:
(a) Zero

Question 7.
Magnitude of g at pole is:
(a) Less
(b) More
(c) Zero
(d) None of these.
Answer:
(b) More

Question 8.
Magnitude of g at equator:
(a) More
(b) Less
(c) Infinity
(d) Zero
Answer:
(b) Less

Question 9.
Formula for gravitational potential energy is:
(a) \(\frac{G M m}{r}\)
(b) \(\frac{G m}{r}\)
(c) \(\frac{R}{G M}\)
(d) gR² = GM.
Answer:
(a) \(\frac{G M m}{r}\)

Question 10.
Formula for escape velocity is:
(a) gR
(b) MR .
(c) \(\sqrt{2 g R}\)
(d) \(\text { (d) } \sqrt{g R} \text { . }\)
Answer:
(c) \(\sqrt{2 g R}\)

Question 11.
There is no atmosphere at moon, because :
(a) Moon is closer to earth
(b) It revolves round the earth
(c) It gets light from sun
(d) Escape velocity of gas molecules is less than their root mean-square velocity.
Answer:
(d) Escape velocity of gas molecules is less than their root mean-square velocity.

Question 12.
If a body is projected at angle of 50° then its escape velocity will be :
(a) 11.6 km/s
(b) 11.2km/s
(c) 12.8 km/s
(d) 16.2 km/s.
Answer:
(b) 11.2km/s

Question 13.
Distance of geo-stationary satellite from earth surface :
(a) 6 R
(b) 7 R
(c) 5 R
(d) 3 R.
Answer:
(a) 6 R

Question 14.
Kepler’s second law is based on :
(a) Newton’s first law
(b) Theory of Relativity
(c) Newton’s second law
(d) Laws of conservation of angular momentum.
Answer:
(d) Laws of conservation of angular momentum.

Question 15.
Period of revolution of geo-stationary satellite is :
(a) 20 hours
(b) 22 hours
(c) 24 hours
(d) 25 hours.
Answer:
(c) 24 hours

Question 16.
Formula for orbital velocity is :
(a) gR
(b) g² R²
(c) g³R³
(d) \(\sqrt{g R}\)
Answer:
(c) g³R³

Question 17.
Who invented first the laws of planetary motion :
(a) Newton
(b) Kepler
(c) Galileo
(d) Aryabhatt.
Answer:
(b) Kepler

Question 18.
Radius of parking orbit is :
(a) 36000 km
(b) 42309 km
(c) 6400 km
(d) 80000 km.
Answer:
(b) 42309 km

2. Fill in the blanks:

1. The acceleration of body due to gravity is called …………………..
Answer:
acceleration due to gravity

2. Gravitational force in nature is a …………………….. force.
Answer:
weak

3. Acceleration due to gravity does not depend upon ………………. of the body.
Answer:
mass

4. At height the value of ‘g’ is ………………………
Answer:
less

5. At centre of earth the value of ‘g’ is ………………………
Answer:
Zero

6. G is a ……………… quantity.
Answer:
scalar

7. Gravitational mass and inertial mass are …………………… to each other.
Answer:
equal

8. Gravitational intensity is equal to ……………………..
Answer:
acceleration due to gravity

9. Gravitational potential energy at infinity is ……………………..
Answer:
zero

10. Magnitude of ‘g’ at equatorial position is …………………..
Answer:
minimum

11. Magnitude of ‘g’ at pole is ……………..
Answer:
maximum

12. Orbital velocity does not depend upon ………………… of a body.
Answer:
mass

13. Escape velocity does not depend upon …………….. of the body.
Answer:
mass

14. Magnitude of escape velocity is ……………… km/s.
Answer:
11.2

15. The time-period of geo-stationary satellite is ………………. hours.
Answer:
24

3. Match the following:
I.

Answer:
1. (c) Planetary motion
2. (a) Zero
3. (e) \(\sqrt{2 g R}\)
4. (b) \(-\frac{G M m}{2 r}\)
5. (d) Time-period of revolution is 24 hours

II.

Answer:
1. (e) Vector.
2. (c) Variable
3. (b) S calar quantity
4. (a) Weak force
5. (d) E = mc²

4. Write true or false:

1. Newton’s laws is universal.
Answer:
True

2. Magnitude of G is different at different places.
Answer:
False

3. Magnitude of ‘g’ is equal at any point on surface of earth.
Answer:
False

4. Ratio of gravitational mass and inertial mass is one.
Answer:
True

5. Gravitational potential is negative at every point on earth surface.
Answer:
True

6. Principle of superposition is not applicable for gravitation.
Answer:
False

7. Value of escape velocity of a body depends upon its mass.
Answer:
False

8. Height of polar satellite from earth surface is 36000 km.
Answer:
False

9. Magnitude of gravitation potential energy at infinity is zero.
Answer:
True

10. Gravitational force is a conservative force.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 7 System of Particles and Rotational Motion which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 7 System of Particles and Rotational Motion

System of Particles and Rotational Motion Class 11 Important Questions Very Short Answer Type

Question 1.
Define the rotational motion.
Answer:
When a force is applied on a rigid body then it revolves about its own axis. This motion of the body is called rotational motion.

Question 2.
Define the term rigid body.
Answer:
A body is said to be a rigid body if an applying external force or during its motion the distance between constituent particle remain unchanged.

Question 3.
Differentiate rotatory motion and circular motion.
Answer:
In rotatory motion the axis of motion lies within the body while in circular motion the axis lies outside the body.

Question 4.
What is torque? Write its unit and dimensional formula.
Answer:
The turning effect of a force, to rotate a body about an axis is called torque. The torque is equal to product of force and perpendicular distance of force from the axis of rotation.
Torque = Force × Moment arm
τ = F × d
Unit: Its S.I. unit is.N × m.
Dimensional formula: [ML²T^-2].

Question 5.
What do you mean by angular velocity and angular acceleration. Write its S.I. units and dimensional formula.
Answer:
Angular velocity : Rate of Change of angular displacement per unit time.
Its unit is radian/second and dimensional formula is [M⁰L⁰T^-1].
Angular acceleration : The rate of change in angular velocity of the body about the given axis of rotaion.
Its unit is radian/second² and dimensional formula is [M⁰L⁰T^-2]

Question 6.
What do you mean by angular momentum?
Answer:
It is defined on the product of linear momentum and perpendicular distance of line of action of linear vector from the axis of rotation.
Angular momentum = Linear momentum × perpendicular distance of linear momentum from the axis of rotation.
Its SI unit is J × s. and dimensional formula is [ML²T^-1 ].

Question 7.
Define moment of inertia and give its unit and dimensional formula.
Answer:
Moment of inertia of a rotating body is the property by virtue of which, it opposes the torque applied to change its position of rest or rotational motion.
If the mass of a body is m and its distance from the axis of rotation is r, then moment of inertia, I = mr²
Unit: Its S.I. unit is kg × m².
Dimensional formula: [ML²T⁰].

Question 8.
Define radius of gyration and give its unit.
Answer:
The radius of gyration of a body is equal to the perpendicular distance from the axis of rotation of a point at which if the mass of a body is assumed to be concentrated the moment of inertia is the same as the actual ‘moment of inertia of the body about that axis. It is represented by k.
UnIt: Its S.I. unit is metre.

Question 9.
On what factors does the moment of inertia of a body depend?
Answer:
(a) On mass of body (2) Distribution of mass from the axis of rotation.

Question 10.
What will be the effect on moment of inertia if the direction of motion of the body is changed?
Answer:
There will be no effect on moment of inertia.

Question 11.
Is it essential to apply moment of couple to a rotating body? Explain with reason?
Answer:
No, it is not essential to apply moment of couple to a rotating body because moment of couple only create angular acceleration.

Question 12.
It is easy to rotate a stone tied with a small string, than that of a longer. Why?
Answer:
We know that, τ = Iω. If the string is long its moment of inertia will be greater and therefore greater torque will be required to rotate the stone. But, for small string less torque is required. Therefore, it is easy to rotate the stone with smaller string.

Question 13.
Write the relation between torque and angular momentum.
Answer:
τ = \(\frac{d L}{d t}\)

Question 14.
Write the relation between angular momentum and moment of inertia.
Answer:
Angular momentum = Moment of inertia × Angular velocity.

Question 15.
Why the spokes are used in a wheel?
Answer:
The spokes increases the distance of rim from the axis of rotation, hence its moment of inertia increases, which makes the rotation of wheel uniform and maintains its rotation.

Question 16.
The handle of a door is fixed away from the hinge. Why?
Answer:
The handle is fixed away from the hinge because, it increases the perpendicular distance between the line of action of force and hinge. Hence, the torque increases.

Question 17.
The handle of a hand pump is made long, why?
Answer:
Long handle increases the distance between the piston and the force, thus the torque increases, so less force is required to pump the water.

Question 18.
What will be the velocity of a point on a rotational axis in rotational motion?
Answer:
The velocity of a point situated in rotation axis in rotational motion is zero.

Question 19.
Differentiate inertia and moment of inertia.
Answer:
Difference between Inertia and Moment of inertia :

Question 20.
What is the physical importance of moment of inertia?
Answer:
To rotate a body about an axis from the rest or to change angular velocity or angular acceleration of rotating body, a torque is applied, more the moment of inertia more will be the torque. This is the physical importance of moment of inertia.

Question 21.
What do you mean by mechanical equilibrium of a system?
Answer:
When the vector addition of total forces and vector addition of total moment of inertia is’ zero then the system is said to be in mechanical equilibrium.

Question 22.
Write the principle of moments.
Answer:
According to this principle anti clockwise moments are taken positive and clockwise moments are taken negative when the body is in rotational equilibrium.

Question 23.
What is moment of couple? Write its SI unit and dimensional formula. Write the factor on which it depends.
Answer:

The product of a force and the perpendicular distance between the line of action of forces is called moment of couple.
Moment of couple = Force × Couple arm
Unit: Its S.I. unit is N × m.
The factor it depends are :

• Magnitude of force.
• Perpendicular distance between the line of action of forces.

Question 24.
What is the physical importance of angular momentum?
Answer:
We know that,
Torque = Force × Perpendicular distance of line of action from the axis and
Angular momentum = Linear momentum × Perpendicular distance of particle from the
axis.
Hence, as the torque measures the tinning effect, similarly angular momentum measures the rotational motion.

Question 25.
What is geometrical meaning of angular momentum?
Answer:
The angular momentum is equal to twice of the product of mass and areal velocity of a particle.
i.e., Angular momentum = 2 × Mass × Areal velocity

Question 26.
A dancer is rotating with stretched hands, when the closes her hands, her speed of rotation increases. Why?
Answer:
We know that in the absence of external force the angular momentum of a rotating body is conserved and is equal to Iω. When the dancer closes her hands, her moment of inertia I decreases hence ω i.e., angular velocity increases.

Question 27.
When a rotating stone tied with a string is stoped, the string winds up in the hand fastly. Why?
Answer:
When the string is started wind up in the hand, the length of string decreases and hence the moment of inertia of stone decreases. But, we know that angular momentum Iω is constant. As I decreases ω increases, therefore the string winds up very fast.

Question 28.
Which factor depends on moment of inertia of the body that moves around a body.
Answer:
We know I = Σmr²
Therefore moment of inertia,

• Is directly proportional to the mass.
• Is directly proportional to the square of distance from the axis of rotation.

Question 29.
Is centre of mass a reality?
Answer:
No, it is only a mathematical concept.

Question 30.
It is difficult to open the door by pushing it or pulling it at the hinge. Why?
Answer:
Because at the hinge the distance of the point of application of the force from the axis of rotation is negligible, so the torque is very small.

Question 31.
Why a helicopter must necessarily have two propellors?
Answer:
If there were only one propellor in the helicopter, then it would have to turn in the opposite direction to conserve its angular momentum.

System of Particles and Rotational Class 11 Important Questions Short Answer Type

Question 1.
Define vector product of two vectors and give one example of vector product. Also write its properties.
Answer:

Let the angle between two vectors \(\vec{A} \text { and } \vec{B}\) is θ, then their cross product is defined as
\(\vec{A} \times \vec{B}=A B \sin \theta \hat{n}\)
Where, \(\hat{n}\) is unit vector perpendicular to the plane consisting vectors \(\vec{A} \text { and } \vec{B}\) .
The direction of \(\vec{A} \times \overrightarrow{\vec{B}}\) follows the screw rule, as shown in the figure.

Example : Let a force \(\vec{F}\) is acting at a point whose position vector is \(\vec{r}\) , then the torque \(\vec{τ}\) is given by the cross product of \(\vec{F}\) and \(\vec{r}\)
i.e.. \(\vec{\tau}=\vec{r} \times \vec{F}\)
Properties: I. Vector product of two vector is always a vector quantity.
2. It does not obey commutative law
\(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)
3. It does not obey associative law
\(\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c}\)
4. \(\vec{a} \times \vec{a}\) = 0 (zero vector)

Question 2.
Show that the area of the triangle contained between the vectors  \(\vec{a} \text { and } \vec{b}\) is one half of the magnitude of a× b.
Answer:

Let the vectors \(\vec{a} \text { and } \vec{b}\) represent the two adjacent sides of ΔAOB such that
OA =b, OB = a
Let angle between a and b be θ.
Such that∠AOB= θ
and also let h be the height of the triangle such that
h=AC
Now in right angled ΔOCA,
sinθ = \(\frac{A C}{O A}\)
or
AC=OA sinθ
or
h = b sinθ … (1)
We know that the area of the triangle AOB, is given b
= \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × OB x AC
= \(\frac{1}{2}\) × a × h
= \(\frac{1}{2}\) × a × b sinθ
= \(\frac{1}{2}\)absinθ …(2)
Also by the definition of cross-product of two vectors, we know that
a × b = ab sinθ\(\hat{n}\)
or
\(|a \times b|=|a b \sin \theta \hat{n}|\)
= ab sinθ\(|\hat{n}|\)
= ab sinθ (\(|\hat{n}|\)
= absinθ, (\(|\hat{n}|\)
= absinθ, (∵ (\(|\hat{n}|\) = 1) …(3)
∴ From eqns. (2) and (3), we get
Area of ΔAOB = \(\frac{1}{2} \mid a \times b\)
= \(\frac{1}{2}\) [magnitude of a × b].

Question 3.
Give the location of the centre of mass of a
(i) sphere
(ii) cylinder
(iii) ring and
(iv) cube each of uniform mass density. Does the centre of mass of a body neccessarily lie inside the body?
Answer:
C.M. lies at the :
(i) Centre of sphere.
(ii) Mid-point of axis of symmetry of the cylinder i.e., its geometrical centre.
(iii) Centre of ring.
(iv) Point of intersection of diagonals i.e., at its geometrical centre.

No, in some cases, C.M. of a body like ring, hollow cylinder, hollow sphere and hollow cube etc. may lie outside.

Question 4.
Define work in rotational motion.
Answer:
As work is define as product of force and displacement in direction of force in linear motion, same like that work done by a torque in rotational motion is define as product of torque and angular displacement.
Work W = τ.d.θ

Question 5.
Derive an expression for the angular momentum of a body in rotational motion.
Or
Prove that J = Iω
Answer:
Consider a body which is made up number of particles of masses m₁,m₂,m₃,…. situated at the distances r₁,r₂, r₃ from the axis of rotation. If the linear velocities are
ν₁, ν₂, ν₃,…. respectively, then
Linear momentum of m₁ = m₁ν₁ = m₁r₁ω , (∵ ν = r ω)
and Angular momentum = m₁r₁ω x r₁ = m₁r₁² ω
Similarly, angular momentum of m₂,m₃,…. will be m₂r₂² ω,m₃r₃² ω,…..
∴ Angular momentum of whole body = m₁r₁²ω + m₂r₂² ω, + m₃r₃² ω, +….
or
J = ω (m₁r₁² +m₂r₂² +m₃r₃² +….)
= ωΣ mr²
or
J = ωI
∴ Angular momentum = Angular velocity × Moment of inertia.

Question 6.
What do you understand by angular momentum? Establish the relation between angular momentum and rotational K.E.
Answer:
Angular momentum : If a particle is moving about an axis, then the moment of linear momentum is called its angular momentum about that axis.
∴ Angular momentum = Linear momentum × Perpendicular distance of particle from the axis of rotation

Relation between angular momentum and rotational K.E.: We know that rotational kinetic energy (E_k) is given by the formula
E_k = \(\frac{1}{2}\) Iω² …(1)
But, Angular momentum, J = Iω
or
ω = \(\frac{J}{I}\)
Putting the value of ω in eqn. (1), we get
E_k = \(\frac{1}{2}\) I\(\left(\frac{J}{I}\right)^{2}\) = \(\frac{1}{2}\) \(\frac{J^{2}}{I}\)
or
J² =2I E_k
or
J = \(\sqrt{2 I E_{k}}\) …(2)
Eqn. (2) is the required relation.
If angular momentum remains conserved, then \(\sqrt{2 I E_{k}}\) lEk will be constant
or
2I E_k = Constant
or
E_k = \(\frac{\text { Constant }}{2 I}\)
or
E_k ∝ \(\frac{1}{I}\)

Question 7.
What is the law of conservation of angular momentum? Prove it.
Answer:
Law of conservation of angular momentum: In the absence of external torque, the angular momentum of a system remains constant.
i.e., J = a constant
Then, Iω = a constant
Now, we know that rate of change of angular momentum is equal to the external torque
applied.
i.e., τ = \(\frac{d J}{d t}\)
If τ = 0
Then, \(\frac{d J}{d t}\) = 0
or
J = a constant
or
J = Iω
If in two different conditions the moment of inertia of a body are I₁ and I₂and angular velocities are ω₁and ω₂respectively, then
I₁ω₁ = I₂ω₂

Question 8.
Prove that the rate of change of angular momentum is equal to the torque applied.
Or
Prove that τ = \(\frac{d J}{d t}\)
Answer:
Let a torque τ is applied on a body of moment of inertia I, so that its angular
acceleration becomes α, then
τ = I.α ….(1)
But, α = \(\frac{d \omega}{d t}\)
∴ τ = I.\(\frac{d \omega}{d t}\) …(2)
Also we have J = Iω
∴ \(\frac{d J}{d t}=\frac{d}{d t}\) (Iω) = \(\frac{d \omega}{d t}\)
From eqn. (2), we get
Or
\(\frac{d J}{d t}\) = Iω = τ [ from eqn .(1)]
Hence, rate of change of angular momentum is equal to the torque applied.

Question 9.
Derive an expression for the moment of inertia of a rigid body.
Or
Prove that I = Σ mr².
Answer:

Let a body rotating about an axis XY supposing that the body is made up of number of particles of masses m₁,m₂,m₃,…. situated at the distances r₁,r₂,r₃, from the axis of rotation.
∴ The moment of inertia of the particles will be m₁r₁² , m₂r₂² , m₃r₃²,
Hence, the moment of inertia of whole body,
I = m₁r₁² + m₂r₂² + m₃r₃² +
or
I = Σmr²
Where, Σ is the sign of summation.

Question 10.
Compare the different derivatives in linear motion and rotational motion.
Answer:
Different derivatives in linear and rotational motion :

Question 11.
A planet is revolving around a sun in an elliptical orbit. How angular velocity and linear velocity get changes in the orbit?
Answer:
(1) Since angular momentum L = mr²ω is constant. Therefore when planet go
far from sun r increases and hence co decreases and when planet go near to sun r decreases
and hence co increases.
(2) Since L = mνr constant. Therefore when planet go far from sun r increases and
hence ν decreases and when planet is close to sun r decreases and hence ν increases.
[∵ ν ∝\(\frac{1}{r}\)

Question 12.
A ladder is at rest against a wall. Is the ladder more likely to slip when a person climb up why?
Answer:
It is more likely to slip when a man climb upward in the ladder because due to the fact that the weight of the man will provide an extra torque for the slipping of the ladder.

Question 13.
What do you understand by angular momentum? Derive its’s expression.
Answer:
If a particle is moving about an axis, then the moment of linear momentum is called its angular momentum about that axis.
Angular momentum = Linear momentum × Perpendicular distance of particle from the axis of rotation.
Let’ m’ be the mass of a particle P whose position vector be V with respect to origin
O and its linear momentum be \(\vec{p}(=m \vec{v})\) and angular momentum be \(\vec{J}=(\vec{p} \times \overrightarrow{O N})\) = pr sinθ
In right angled Δ OPN,

ON = OP sinθ =r sinθ
or
J=rp sinθ
Where θ is angle between \(\vec{r} \text { and } \vec{p}\)
or
\(\vec{J}=\vec{r} \times \vec{p}\)
or
\(\vec{J}=\vec{r} \times(m \cdot \vec{v})\)
or
\(\vec{J}=m(\vec{r} \times \vec{v})\)

Question 14.
What do you understand by angular acceleration? Prove that :a = r × α.
Answer:
Angular acceleration : Rate of change in angular velocity is defined as angular acceleration.
Angular acceleration = \([latex]\frac{\text { Angular velocity }}{\text { Time interval }}\)[/latex]
If Δω be the change in angular velocity in time Δt, then average angular acceleration = \(\frac{\Delta \omega}{\Delta t}\)
or instantaneous angular acceleration α = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t}=\frac{d \omega}{d t}\)
But ν = rω .where r is the distance of particle, from axis of rotation,
∴ Integration both side, we get
\(\frac{d ν}{d t}\) = r \(\frac{d ω }{d t}\) = r.α

Question 15.
Prove that the radius of gyration of a body is equal to the root mean square of>the distances of particles from the axis of rotation.
Answer:
Expression for the radius of gyration : Let the mass of a body is M and its radius of gyration is K,
Moment of inertia of the body
∴ I = MK²
Also I = Σmr²
∴ MK² = Σmr²
or
MK² = m₁ r₁² + m₂r₂² + m₃r₃ ² +….
If m₁ = m₂ = m₃ =….= m (let), then
MK² = mr₁² +mr₂² +mr₃² + ……
MK² = m(r₁² + r₂² + r₃ ² + …..)
But M = m × N(where N is number of particles)
∴ mNK² = m(r₁² + r₂² + r₃ ² + …..)
or
K² = \(\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+\ldots}{N}\)
∴ K = \(\sqrt{\frac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+\ldots}{N}}\)
Hence, Radius of gyration = r.m.s. of distance of particle from the axis of rotation.

Question 16.
Derive an expression for the K. E. of a rotation.
Or
Prove that the kinetic energy of rotation of a rigid body =\(\frac{1}{2}\)ω², where I is moment of inertia and co is angular velocity.
Or
Define the rotational kinetic energy and derive its formula.
Answer:
Let a body is rotating about an axis XY, with an angular velocity ω. The body consists of number of particle and angular velocity of each particel will be same. But the distances of all the particles are different, therefore each particle will have different linear velocity.
Let the distances of the particles of masses m₁m₂m₃are r₁, r₂, r₃,…. and their
linear velocities are v₁ v₂, v₃,…. respectively.
Then, K.E. of particles will be

System of Particles and Rotational Class 11 Important Questions Long Answer Type

Question 1.
State perpendicular axes theorem and prove it. Is it applied to every body?
Or
State theorem of perpendicular axis of moment of inertia and prove it.
Answer:
Perpendicular axes theorem :
According to this theorem. The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moment of inertia of the body about two mutually perpendicular axes lying in the plane of the body and intersecting at the point through which the perpendicular axis passes.

Proof:
Let P be a particle on a lamina, at distances x,y and r from the axes of X, Y and Z and OP = r.
If the mass of the particle is m, then
M.I. about X – axis, I_x = Σ my²
M.I. about y – axis, I_y = Σ mx²
and M.I. about Z-axis, I_z = Σ mr²
From right angle Δ PMO,
r² = x² + y²
∴ mr² = mx² + my²
or
Σmr² = Σmx² + Σmy²
or
I_z = I_x + I_y

This threorem is true for all types of bodies. The solid body is divided into laminas and the theorem can be applied.

Question 2.
State theorem of parallel axes and prove it. Is it applicable for all the bddies?
Answer:
Parallel axes theorem :
According to this theorem, the moment of inertia of a body about the given axis is equal to the sum of the moment of inertia of the body about an axis through the centre of mass and parallel to the given axis and the product of the mass of the body and the square of distance between the two axes.

Proof:
Let CD is the axis passing through centre of mass of the body and AB is the axes parallel to CD at a distance a.
Consider a particle P of mass m at a distance x from CD.
∴ M.I. of P about axis AB = m(a + x)²
∴ M.I. of whole body about AB
I =Σm(a+ x)²
or
I = Σm(a² + 2ax + x²)
or
I = Σma² + Σ2amx + Σmx²
or
I = a²Σm + 2.a Σmx + Σmx²
But, Σmx² = I_cm and Σm = M.
∴ Now, Σx is the moment of body about the axis passing through centre of mass. .•.
∴ Σmx = 0
Hence, by eqn. (1)
I = a²M + 0 + I_cm
or
I = I_cm + Ma²
This theorem is applicable for all the bodies.

Question 3.
Write the equations of rotational motion and derive them.
Answer:
If the initial angular velocity is ω₀, angular acceleration is α and after t sec. the displacement is θ and angular velocity becomes ω, then equations are :
(i) ω = ω₀ + αt
(ii) θ = ω₀t + \(\frac{1}{2}\)αt²
(iii) ω² = ω₀² +2αθ

(i) Derivation of first equation :ω = ω₀ + αt
By the definition of angular acceleration,
α = \(=\frac{\omega-\omega_{0}}{t}\)
or
ω – ₀ = αt
∴ ω = ω₀ + αt

(ii) Derivation of second equation: θ = ω₀t + \(\frac{1}{2}\)αt²
The body is rotating with uniform angular acceleration, therefore
Average angular velocity = \(\frac{\omega+\omega_{0}}{2}\)
But, from the first eqn., ω = ω₀ + αt
∴ Average angular velocity = \(\frac{\omega_{0}+\alpha t+\omega_{0}}{2}\)
= ω₀ + \(\frac{1}{2}\)αt
Now, Angular displacement = Average angular velocity × Time
θ = \(\left(\omega_{0}+\frac{1}{2} \alpha t\right)\) × t
or
θ = ω₀t +\(\frac{1}{2}\) αt²

(iii) Derivation of third equation : ω² = ω₀² + 2αθ
Now, by first equation, we have
ω = ω₀ + αt
Squaring both the sides, we get
ω² = ( ω₀ + αt)²
or
ω² = ω₀² + 2ω₀αt + α²t²
or
ω² = ω₀² + 2α\(\left(\omega_{0} t+\frac{1}{2} \alpha t^{2}\right)\)
But, from second eqn., we have
ω₀t + \(\frac{1}{2}\)at² = θ
∴ ω² = ω₀²+ 2 α θ

Question 4.
Derive the expression of kinetic energy in rolling motion.
Answer:
When the body is rolling over any surface, its kinetic energy is the sum of translatory kinetic energy and rotatory kinetic energy.

The above formula is applicable for cylinder, disc and sphere.

System of Particles and Rotational Class 11 Important Numerical Questions

Question 1.
Two circular disc made of different metals having same mass and diameter dA and dB (dA > dB). If the moment of inertia about the axis perpendicular to its plane and passing through its centre is I_A and I_B. Find which is more I_A or I_B ?
Solution:
Let the mass of each disc b e‘m’ and radius be r_Aand r_B.

If the thickness of the disc is ‘t then

Since, d_A > d_B
There fore I_A < I_B

Question 2.
A solid sphere of mass M and radius R is rolling over a horizontal surface with velocity ν. What will be its kinetic energy?
Solution:
Total kinetic energy = Rotatory kinetic energy + Linear kinetic energy

Question 3.
Three particles of massess 2, 3, 4 kilogram are kept at three vertices of a triangle of side 1 metre. Find out centre of mass of the system.
Solution:
Let the vertices A be at origin (0, 0) and B be at X-axis of a ΔABC, then the
coordinates will be : A (0,0), B (1,0) and C\(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)
∴ Centre of mass of the system

Question 4.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2MR²/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to the MR²/4. Find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Solution:
(a) The moment of inertia of sphere about its diameter be
I_cm = \(\frac{2}{5}\)MR²
By the theorem of parallel axis
I = I_cm + Ma²
= \(\frac{2}{5}\)MR² + MR², [∵ a = R]
= \(\frac{7}{5}\)MR²

(b) The moment of inertia of disc about its diameter

I_d = \(\frac{1}{4}\)MR²
By the theorem of perpendicular axis the moment of inertia of disc about an axis AB passing through its centre, perpendicular to its plane be
I_cm = \(\frac{1}{4}\)MR² + \(\frac{1}{4}\)MR² [∵I_z=I_x+I_y]
= \(\frac{1}{2}\)MR²
By the theorem of parallel axis the moment of inertia about an axis passing through a point on its edge and perpendicular to the disc be
I = I_cm + Ma² = \(\frac{1}{2}\)MR² + MR² , [∵ a=R]
= \(\frac{3}{2}\)MR²

Question 5.
A solid sphere of mass 0.5 kg and radius 1 metre is rolling over a smooth surface with a velocity of 5 m/s. Find its total energy. What is the percentage of rotational kinetic energy in total energy ?
Solution:
Given : m = 0.5 kg, 2r = 1 m, r = 0.5 m, ν =5 m/s.

Question 6.
If = \(\vec{a}=3 \hat{i}-4 \hat{j}+\hat{k}\) and \(\vec{r}=5 \hat{i}-6 \hat{j}+6 \hat{k}\) then find out the magnitude of linear velocity.
Solution:
\(\vec{a}=3 \hat{i}-4 \hat{j}+\hat{k}, \vec{r}=5 \hat{i}-6 \hat{j}+6 \hat{k}\)
From \(\vec{v}=\vec{a} \times \vec{r}\)
\(=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -4 & 1 \\
5 & -6 & 6
\end{array}\right|\)
or
\(\vec{v}=\hat{i}(-24+6)-\hat{j}(18-5)+\hat{k}(-18+20)\)
\(=-18 \hat{i}-13 \hat{j}+2 \hat{k}\)

Question 7.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 gm are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at
45.0 cm. What is the mass of the metre stick?
Solution:
Given : AB = 1 m = 100 cm
AG = BG = 50 cm AC = 12 cm;

For the rotational equilibrium about point O
10g × OC = Mg × OG
or
10g (45 -12)= Mg (50 -45)
or
10 × 33 =M × 5
∴ M = \(\frac{10 \times 33}{5}\) = 66 gm.

Question 8.
Moment of inertia of a flywheel is 4 kg/m². Find out the angular acceleration of it if the torque acting on it is 10 newtonmetre.
Solution:
Given : I = 4 kg/m², τ = 10 newton-metre
From torque τ = Iα, we get
α = \(\frac{\tau}{I}=\frac{10}{4}\) = 2.5 rad/s²

Question 9.
The oxygen molecule has a mass of 5.30 × 10^-26 kg and a moment of inertia
1.94 × 10^-46kgm² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that if kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average anglular velocity of the molecule.
Solution:
Given, M = 5.30 × 10^-26kg; I = 1.94 × 10^-46kgm² and ν = 500 m/s
Rotational K.E. = \(\frac{2}{3}\) × Translatory K.E.

= 6.75 x 10¹²rad/s.

Question 10.
Find out the moment of inertia of a circular disc of mass 2 kg and radius 0.5 metre along:
(i) Its geometrical axis,
(ii) Its diameter,
(iii) Tangential direction on its circum-ference.
Solution:
Given : M= 2 kg, R = 0.5 m
(i) M.I. along its geometrical axis
I = MR² = 2 × (0.5)²
= 2 × (0.25)
= 0.5 kg m².

(ii) M.I. along diameter
I_d = \(\frac{1}{2}\)I = \(\frac{1}{2}\)MR²
= \(\frac{1}{2}\).2 × (0.5 )²
= (0.5)2 = 0.25 kgm²

(iii) M.I. in tangential direction on its circumference
I_T =I_d + Ma²
= \(\frac{M R^{2}}{2}\) + MR² = \(\frac{3}{2}\) MR²
= \(\frac{3}{2}\).2 × (0.5)²
= 3 × 0.25 =0.75 kgm².

Question 11.
The mass of a circular ring is 2 kg and its radius is 2m. The speed of its centre of mass is 20 cm × s^-1 Calculate the K.E. of rotation of ring. (M. I. of ring = MR²)
Solution:
Given : r = 2 m,M = 2kg,ν = 20 cm × s^-1 = 20 × 10 ms^-1.
Total K.E. = \(\frac{1}{2} I \omega^{2}+\frac{1}{2} M v^{2}\)
\(=\frac{1}{2} M r^{2} \frac{v^{2}}{r^{2}}+\frac{1}{2} M v^{2}\)
\(=\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}\) = Mν²
= 2 × (20 × 10^-2 )² = 8 × 10^-2 joule.

Question 12.
The wheel of a train revolve 6 revolution per second. On applying brakes it get stopped after 12 sec. Find out angular acceleration produced by brake.
Solution:
Given : Frequency n = 6 revolution/sec, t = 12 sec, ω = 0
and ω₀ = 2πn = 12π rad/sec
By the formula ω = ω₀ + α.t, we get
0 = 12π – α × 12
or
α = \(\frac{12 \pi}{12}\) = π
or
α = 3.14 rad/sec ²

Question 13.
A man is standing on a rotating table with hands stretched, the moment of inertia is 50 kg m2 and angular velocity is 3 rad/s. Now, the man closes his hands thus the moment of inertia becomes 30 kg m². Calculate the angular velocity.
Soution:
Given: I₁=50kgm², ω₁ =3rad × s^-1, I₂=30kgm².
We have, I₁ ω₁ = I₂ ω₂
∴ ω₂ = \(\frac{I_{1} \omega_{1}}{I_{2}}\)
or
ω₂ = \(\frac{50 \times 3}{30}\) = 5rad s^-1

Question 14.
If the diameter of the earth reduces to half of it, then what will be the duration of the day?
Solution:
We know I₁ ω₁ = I₂ ω₂
or
\(\frac{2}{5} M R^{2} \times \frac{2 \pi}{T_{1}}=\frac{2}{5}\left(\frac{R}{2}\right)^{2} M \times \frac{2 \pi}{T_{2}}\)
∴ \(\frac{1}{T_{1}}=\frac{1}{4 T_{2}}\)
or
T₂ = \(\frac{T_{1}}{4}=\frac{24}{4}\) = 6 hr

Question 16.
If the radius of revolving disc is made half keeping mass same, then what will be its angular velocity?
Solution:
From = I₁ ω₁ = I₂ ω₂
Or
MR²ω₁ = \(\left(\frac{R}{2}\right)^{2}\)M ω₂
∴ ω₂ = 4ω₁
i.e., Angular velocity will become 4 times of the earlier.

Question 17.
Angular momentum of a body changes from 220 kg m2 to 340 kg m2 in 10 sec. Find out the torque acting on it.
Solution:
Since τ = \(\frac{d J}{d t}\)
∴ τ = \(\frac{340-220}{10}=\frac{120}{10}\)
τ = 12 N-m.

Question 18.
The mass of a fly-wheel is 25 kg and its radius is 0.2 m. How much force
will be applied in tangential direction to produce angular acceleration of 2 rad × s^-2?
Solution:
Given : M = 25 kg, R = 0.2 m, α = 2 rad × s^-2, d = 0.2m.
Now,
I = MR²
= 25 × (0.2 )² = 1 kgm
Again, τ =I.α
= 1 × 2=2 N-m
Also, τ=F × d
∴ F = \(\frac{\tau}{d}\)
= \(\frac{2}{0 \cdot 2}\)
= 10N

Question 19.
A ball of mass 1 kg is tied at one end of a 20 cm long string and it is rotated in horizontal phase in such a way that it revolves 8 revolution per second. Find out the following :
(i) Ahgular velocity,
(ii) Linear velocity,
(iii) Moment of inertia along the finger of hand,
(iv) Angular momentum.
Solution:
Given : r = 20cm =20 × 10^-2m, m= 1 kg, υ =8rev/sec

(i) Angular velocity ω = 2πυ
or
ω = 2π × 8 = 16π rad/sec.

(ii) Linear velocity v = r.ω
or
ν = 20 × 10^-2 × 16π
ν = 3.2π m/s.

(iii) Moment of inertia I = mr²
or
I = 1 × (20 × 10^-2)²
or
I = 0.04 kg m².

(iv) Angular momentum J = Iω
or
J = 0.04 × 16π
= 0.64π kg m²/sec.

Question 20.
A ball tied with a string takes 4 second for one revolution rotating in cicular path. If the radius of the circular path is made half, then what will be the time taken by the ball to complete one revolution.
Solution:
By the laws of conservation of momentum
= I₁ ω₁ = I₂ ω₂
Let the mass of the ball be ‘wi’ and radius be ‘r’.
Then, I₁ = mr²

System of Particles and Rotational Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
In rotational motion, moment of inertia has the same importance as the physical quantity in linear motion :
(a) Velocity
(b) Acceleration
(c) Mass
(d) Force
Answer:
(c) Mass

Question 2.
Centre of mass of moving object:
(a) Remain stationary
(b) Does not get influenced by internal forces
(c) Does not get influenced by external forces
(d) None of these.
Answer:
(b) Does not get influenced by internal forces

Question 3.
The moment of linear momentum is called :
(a) Couple
(b) Torque
(c) Impulse
(d) Angular momentum.
Answer:
(d) Angular momentum.

Question 4.
Which relation is incorrect:
(a) L = Iω
(b )I = MK²
(c )E = \(\frac{1}{2}\)Iω
(d) τ = \(\frac{d L}{d t}\)
Answer:
(c )E = \(\frac{1}{2}\)Iω

Question 5.
Moment of inertia depends upon :
(a) Angular velocity
(b) Mass
(c) Distribution of mass
(d) On axes of rotation and mass distribution.
Answer:
(d) On axes of rotation and mass distribution.

Question 6.
What is the formula for rotational kinetic energy of an object:
Moment of inertia Angular Acceleration
(a) R.K.E. = \(\frac{1}{2}\) × Moment of inertia × (Angular velocity)²
(b) R.K.E. = \(\frac{\text { Moment of inertia }}{\text { Angular Acceleration }}\)
(c) R.K.E. = \(\frac{\text { Force }}{\text { Time }}\)
(d) R.K.E. = \(\frac{\text { Time }}{\text { Displacement }}\)
Answer:
(a) R.K.E. = \(\frac{1}{2}\) × Moment of inertia × (Angular velocity)²

Question 7.
Rate of change in angular velocity is known as :
(a) Angular displacement
(b) Angular speed
(c) Angular acceleration
(d) Angular momentum.
Answer:
(c) Angular acceleration

Question 8.
Moment of inertia of a solid cylinder of mass M and radius R along geometrical axis is:
(a) \(\frac{2}{3}\)MR²
(b) \(\frac{4}{3}\)MR²
(c) \(\frac{5}{4}\)MR²
(d) \(\frac{1}{2}\)MR².
Answer:
(d) \(\frac{1}{2}\)MR²

Question 9.
In a merry-go-around, a boy suddenly comes and sits down which physical quantity will be conserved :
(a) Linear momentum
(b) Kinetic energy
(c) Angular momentum
(d) None of the above.
Answer:
(c) Angular momentum

Question 10.
The unit of moment of inertia is :
(a) kg/m²
(b)kg/m
(c)kg² × m
(d)kg × m²
Answer:
(d)kg × m²

2. Fill in the blanks:

1. Formula for angular momentum is …………………….
Answer:
Moment of inertia × Angular velocity

2. If the axis of rotation changes, then its ………………… also changes.
Answer:
Moment of inertia

3. The rate of change of angular velocity is called ………………..
Answer:
Angular acceleration

4. Angular momentum gives a measure of ………………… of body.
Answer:
Rotational motion

5. The angular momentum of a body is …………….. the product of its mass and areal
velocity.
Answer:
Twice

6. ‘In rotational motion, work = moment of force × …………….
Answer:
Perpendicular distance from rotational axis.

3. Match the following:
I.

Answer:
1. (b) \(\frac{m_{1} \vec{r}_{1}+m_{2} \vec{r}_{2}}{m_{1}+m_{2}}\)
2. (a) \(\frac{d \overrightarrow{\mathrm{R}}}{d t}\)
3. (e) \(\vec{r} \times \overrightarrow{\mathrm{F}}\)
4. (c) Σmr²
5. (d) \(\frac{1}{2}\)Iω²

II.

Answer:
1. (d) MR²
2. (c) F × r
3. (e) \(\sqrt{\frac{I}{M}}\)
4. (b) \(\frac{5}{2}\)MR²
5. (a) m/sec²

4. Write true or false:
1. If two particles have equal mass, then their position of center of mass is at the center of the line joining them.
Answer:
True

2. If internal force acting on a system is zero, then its momentum will be zero.
Answer:
False

3. If no entemal force acts on a system, then the velocity of its center of mass will remain constant.
Answer:
True

4. During motion, the volume or shape of a rigid body may change.
Answer:
False

5. The center of mass always lies inside the body.
Answer:
False

6. According to theorem of parallel axis for moment of inertia I = I_o + Ma² where symbols used have their, usual meanings.
Answer:
True

Students get through the MP Board Class 11th Physics Important Questions Chapter 11 Thermal Properties of Matter which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 11 Thermal Properties of Matter

Thermal Properties of Matter Class 11 Important Questions Very Short Answer Type

Question 1.
Ice is cold, why?
Answer:
When the ice is touched by hand, the ice takes heat from the hand. Thus, heat is lost from the hand and the temperature of hand decreases and therefore we feel cold.

Question 2.
What is temperature?
Answer:
It is that physical quantity which determines the direction of flow of heat when two objects are placed in contact with each other.

Question 3.
Define calorie.
Answer:
Amount of heat required to raise the temperature of one gram pure water from 14.5°C to 15.5°C is known as one calorie.

Question 4.
What are the different scale to measure temperature? Write relation be¬tween them.
Answer:

• Centigrade or celcius scale.
• Farenheit scale.
• Kelvin scale.
• Reumar scale.

Relation between them is
\(\frac{C}{5}\) = \(\frac{F-32}{9}\) = \(\frac{R}{4}\) = \(\frac{K-273}{5}\).

Question 5.
What is ideal gas thermometer?
Answer:
Ideal gas thermometer is that thermometer whose zero is that temperature at which pressure remain zero and 1 degree is equal to 1°C.

Question 6.
What is absolute zero ? Write its value.
Answer:
It is that temperature at which pressure and volume of a gas becomes zero, known as absolute zero. Its value is 273.15°C.

Question 7.
In absolute temperature scale, what is the melting point and boiling point of pure water?
Answer:
In absolute temperature scale melting point of pure water is 273K and boiling point is 373K.

Question 8.
Can the temperature of an object have a negative value on Kelvin’s scale?
Answer:
No.

Question 9.
Why H₂ – thermometer is called standard thermometer? What is its range?
Answer:
The H₂– thermometer is called standard thermometer because it is used for the calibration of other thermometers. It can measure the temperature from – 200°C to 500°C.

Question 10.
Define coefficient of linear expansion and write its unit.
Answer:
It is defined by the change in length per unit original length per degree rise of temperature.
Let the initial length of a rod be l and the length increases by Δl for Δt change in temperature, then
Coefficient of linear expansion, α = \(\frac{\Delta l}{l \Delta t}\)
Unit: Its unit is °C ^-1 or K ^-1.

Question 11.
What are the two types of specific heat? Why C_p is greater than C_v?
Answer:
The two specific heats of gases are specific heat at constant pressure (C_p) and specific heat at constant volume (C_v).
C_p is greater than C_v because same part of heat given at constant pressure is utilized to increase the volume, (hence work done) against the applied pressure.

Question 12.
Define coefficient of superficial expansion and give its unit.
Answer:
It is defined by the change in area per unit original area per degree rise in temperature.
Let initial area of a body be A and the area change by ΔA, with the rise of temperature Δt.

∴ Coefficient of superficial expansion, β = \(\frac{\Delta A}{A \Delta t}\)
Unit: Its unit is °C^-1 or K ^-1.

Question 13.
Define coefficient of volume expansion and give its unit.
Answer:
It is defined by the change in volume per unit original volume per degree rise in temperature.
Let initial volume of a substance is V and change in volume with change in temperature ΔT is ΔV.
∴ Coefficient of volume expansion, γ = \(\frac{\Delta V}{V \Delta T}\)
Unit: Its unit is °C ^-1 or K ^-1.

Question 14.
Write the relations between α,β and γ.
Answer:
β = 2α and γ = 3α
∴ α : β : γ = 1:2:3.

Question 15.
Define specific heat and give the unit of specific heat.
Answer:
We know that,
Q = ms Δt
or
s = \(\frac{Q}{m \Delta t}\)
If m = 1 gm, Δt = 1°C
Then, s = Q
The amount of heat required to raise the temperature of 1 gm substance through 1°C is called specific heat of the substance.
Unit: Unit of specific heat is calorie × gm^-1 × °C^-1.

Question 16.
Define latent heat of fusion.
Answer:
Latent heat of fusion: The amount of heat required to change unit mass of a solid into liquid at its melting point is called latent heat of fusion.
If the latent heat of fusion is L, then Q = mL
Latent heat of ice is 80 cal × gm^-1.

Question 17.
Define latent heat of vaporization.
Answer:
Latent heat of vaporization: The amount of heat required to change a liquid of unit mass into vapour at its boiling point, is called latent heat of evaporation.
Latent heat of steam is 536 cal × gm^-1.
Unit: The practical unit of latent heat is cal × gm^-1.

Question 18.
Write the principle of thermometry.
Answer:
The properties of any object which depends on temperature and can be used to measure temperature.

Question 19.
Which is more sensitive, liquid thermometer or gas thermometer?
Answer:
Gas thermometer is more sensitive than liquid thermometer because expansion of gas is more than liquid for a given temperature difference.

Question 20.
Among metal and alloy, whose coefficient of thermal expansions is more?
Answer:
Thermal expansion of metal is more than that of alloy.

Question 21.
Can the value of coefficient of thermal expansion be always positive?
Answer:
No.

Question 22.
There is a hole in a metal plate. What will happen on heating it?
Answer:
The size of hole will increase on heating the metal plate.

Question 23.
A gap is left between two rails, why?
Answer:
The temperature of the atmosphere is always changing. With the rise of temperature the rails expands, therefore the gaps are left between them. If gap is not kept, then with the rise in temperature the rails will bend and will cause serious accidents.

Question 24.
What do you mean by the anomalous expansion of water?
Answer:
Generally, the volume of liquids increases with the rise of temperature. But, the volume of water decreases when it is heated from 0°C to 4°C and then after 4°C the volume increases. Similarly, when the water is cooled its volume decreases up to 4°C and then increases from 4°C to 0°C. Such expansion of water is called anomalous expansion.

Question 25.
The bulb of one thermometer is spherical while that of the other is cylindrical. Both have equal amounts of mercury. Which one will response quickly to temperature change?
Answer:
For a given volume,the surface area of the cylinder is large than that of sphere. Therefore heat will be quickly conducted through the cylindrical bulb. Thus, the response of cylindrical bulb thermometer will be quicker.

Question 26.
Pendulum invar of a pendulum clock is made of alloy. Why?
Answer:
Coefficient of thermal expansion of alloy is very less, thus when temperature changes there is no effect on pendulum invar, so it is made of alloy.

Question 27.
During change of state, heat given to a body is lost in which form?
Answer:
During change of state, heat given to the body is lost to increase internal molecu¬lar potential energy of the body.

Question 28.
Hot water is used for formentation. Why?
Answer:
Due to high specific heat of water, it gives more heat in comparison to other liquids. Therefore, hot water is used for formentation.

Question 29.
Ice at 0°C appear to be more cold than water at 0°C. Why?
Answer:
When the ice melt, amount of heat required for melting is absorbed from our body, therefore ice appear more cold than water at 0°C.

Question 30.
Burning due to steam is more severe than that of boiling water, why?
Answer:
When steam vapour gets contact with the skin, then first it gets converted to water of 100°C. In this process, it liberates 536 calories of heat. This extra heat is not present in boiling water. Since steam has more latent heat, hence, it produces severe bums.

Question 31.
The sand near the bank of river becomes colder than water, why?
Answer:
The specific heat of sand is very low, therefore it gains heat soon and radiate heat soon. While the specific heat of water is very high. Hence, the sand becomes colder than water.

Question 32.
Atmospheric temperature reduces after icefall. Why?
Answer:
Due to high value of latent heat, when the ice start melting it takes heat from the atmosphere, hence atmospheric temperature reduces after icefall.

Question 33.
Why the specific heat of the coolent used in nuclear reactor is high?
Answer:
Due to high specific heats coolent is able to absorb more heat in nuclear reactor.

Question 34.
At boiling point of a liquid, how much is the specific heat ?
Answer:
We know,
ΔQ = m.sΔT
or
s = \(\frac{\Delta Q}{m \cdot \Delta T}\)
At boling point, change in temperature
ΔT = o
s = \(\frac{\Delta Q}{m .0}\) = \( \frac{\Delta Q}{0}\) = ∞
∴ At boiling point, specific heat of a liquid is infinity.

Question 35.
Name the methods of transmission of heat.
Answer:
There are three methods of transmission of heat:

1. Conduction,
2. Convection and
3. Radiation.

Question 36.
What do you understand by conduction?
Answer:
In this process the particle which is heated first gives a part of its energy to its adjacent particles. The particles undergo simple harmonic motion about their mean position and the heat is transmitted from one particle to the next. Solids are heated by this process only.

Question 37.
What do you mean by convection?
Answer:
In this mode of flow the particles of substance absorb heat and leave their place or get displaced. Liquids and gases are heated by this method.

Question 38.
What is radiation?
Answer:
In this method heat is transmitted from one body to another body without any medium or without affecting the medium. Heat from sun reaches to earth by radiation.

Question 39.
What is meant by variable state in conduction?
Answer:
When one end of a rod of a conductor is heated then temperature of the different parts of the rod go on increasing slowly. This state is called variable state of temperature.

Question 40.
What is meant by steady-state?
Answer:
When one end of a conductor is heated then initially the temperature of its different parts go on increasing slowly but a stage comes when the temperature of each part of the conductor becomes constant, though the temperature of different parts are different. This state of temperature is called steady state.

Question 41.
What is temperature gradient? What type of quantity is it? Write the unit.
Answer:
Temperature gradient: Fall of temperature per unit length is called temperature gradient.
Let temperature of two surfaces of a conductor at distances x and x + Δx are θ and θ- Δ θ.
∴ Temperature gradient = \( \frac{(\theta-\Delta \theta)-\theta}{(x+\Delta x)-x}\)
Negative sign shows that as the distance increases the temperature decreases. Temperature gradient is a vector quantity and its unit is °C × m ^-1.

Question 42.
Define coefficient of thermal conductivity. Write its unit and find out its dimensional formula.
Or
Define the coefficient of thermal conductivity and obtain expression for it.
Answer:
The coefficient of thermal conductivity is defined by the amount of heat, flowing per second in steady state through the rod of length 1 long and area of cross-section 1 m², with a unit temperature difference between the opposite faces.

The quantity of heat, flowing per second through a unit cube having unit temperature difference between opposite faces, during steady-state.
i.e., Q = \(\frac{K A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
K = \(\frac{Q d}{A\left(\theta_{1}-\theta_{2}\right) t}\)
Unit: Its SI unit is \(\frac{\text { joule } \times \mathrm{m}}{\mathrm{m}^{2} \mathrm{~K} \mathrm{sec}}\) = J m^-1K ^-1sec ^-1
But, joule /sec is watt.
Other unit is watt × m^-1 × K^-1.

Dimensional formula :\( \frac{\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right][\theta][\mathrm{T}]} \) = \( \left[\mathrm{MLT}^{-3} \theta^{-1}\right] \).

Question 43.
What do you mean by thermal resistance? Write its unit and find out its dimensional formula.
Answer:
The resistance offered by the body in the flow of heat is called thermal resistance,
Thermal resistance = \( \frac{d}{\mathrm{~K} A} \) = \( \frac{\left(\theta_{1}-\theta_{2}\right) t}{Q}\)
Unit: Its SI unit is \( \frac{\mathrm{K} \mathrm{sec}}{\mathrm{J}}\) = K sec J^-1 = K watt^-1

Dimensional Formula: [Thermal resistance] = \(\frac{[\theta][\mathrm{T}]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}\) = \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{3} \theta\right] \) .

Question 44.
What is meant by thermal diffusivity?
Answer:
Thermal diffusivity: The ratio of thermal conductivity and thermal capacity of unit volume of a substance is called thermal diffusivity of the substance.
Thermal diffusivity = \( \frac{K}{\rho S} \).

Question 45.
The ice blocks are covered with saw dust. Why?
Answer:
Sawdust is bad conductor of heat. Hence, the heat of atmosphere does not conduct through sawdust and thus ice does not melt soon.

Question 46.
Woollen clothes are used in winter. Why?
Answer:
Woollen clothes have fine pores which are filled with air. Air and woollen clothes are bad conductor of heat. There it does not allow the heat to escape from the body to atmosphere. Thus, it keeps body warm.

Question 47.
A woollen blanket keeps the body warm and also prevent the ice from melting. Why?
Answer:
Air is insulator which is filled between the fibres of blanket. Therefore, blanket does not allow the heat of body to radiate into the atmosphere as well as the heat of atmosphere could not reach to ice. Thus, it keeps the body warm and protect the ice from melting.

Question 48.
State two properties of thermal radiation.
Answer:

1. Thermal radiations are electromagnetic waves.
2. They do not affect the medium through they pass.

Question 49.
Differentiate between light and heat radiations.
Answer:
Difference between Light and Heat radiations :

Question 50.
What is emissive power? Write its SI unit.
Answer:
Emissive power: The amount of heat radiated per unit area per unit time, is called emissive power of the surface.
If int sec, through area A, amount of heat radiated is Q, then
Emissive power, e = \(\frac{Q}{A \times t}\)
SI unit: joule × m^-2 × s^-1 or watt × m^-2.

Question 51.
What is absorptive power of surface?
Answer:
The ratio of amount of heat radiation absorbed by a surface in a given time and total amount of heat incident is called absorptive power of that surface.
i.e., a = \(\frac{q}{Q}\)
Where, q = Amount of heat radiation absorbed and Q = Amount of heat radiation incident.

Question 52.
Define perfectly black body.
Or
What do you mean by perfectly black body?
Answer:
A perfectly black body is one which absorbs completely all the thermal radiation incident on it and does not reflect or transmit any fraction of it. Emissivity for this body is 1.
No such body exists in nature, but in practice we consider lamp black and platinum black as perfectly black body.

Question 53.
State Stefan’s law of black body radiation.
Answer:
Stefan’s law : The total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature.
If the radiant energy per second per unit area is E and absolute temperature of the surface is T, then
E∝T⁴
or E = σT⁴
Where σ is called Stefan’s constant.
If the area of a black body is A and its temperature is T, then the radiant energy per second will be
E= σ AT⁴.

Question 54.
State Newton’s law of cooling.
Answer:
The rate of cooling of a body is proportional to the temperature difference of the average temperature of the body and surroundings provided the temperature difference between the source and surrounding is very less.

Question 55.
State Kirchhoff’s law of radiation.
Answer:
According to Kirchhoff’s law, at a given temperature the ratio of emi¬ssive power and absorptive power for a given wavelength is equal to the emissive power of black body for the same wavelength.
If the emissive power of a surface for wavelength λ is e_λ and absorptive power is a_λ, then
\(\frac{e_{\lambda}}{a_{\lambda}}\) = E_λ
Where, E_λ is emissive power of black body.

Question 56.
If a red glass is heated up to high temperature and placed in a dark room, then it appears green. Why?
Answer:
A body appears red because the body absorbs all the colours but red. It reflects red only thus it appears red. The emissive power of a red body is maximum for the green colour. Thus, it appears green, when heated up to a high temperature.

Question 57.
The cloudy nights are warmer than the day. Why?
Answer:
The clouds reflect the heat radiation waves, which are radiated from the earth. Thus, the temperature of earth does not fall. Hence, the nights become hot.

Question 58.
Deserts are very hot during the day and very cold during the night Why?
Answer:
We know that good absorber are good radiators. Sand is good absorber, during the day time the sand absorbs heat and becomes very hot. But, during the nights it radiates heat soon and becomes cold.

Question 59.
State three uses of heat radiation.
Answer:

• In the cold country, the hot water is flown through the metallic pipes to warm the rooms.
• The walls and roof of greenhouse are made up of glass.
• To decrease the temperature of hot substance, it is spread over, thus it becomes cold soon.

Question 60.
In a chilled weather, animals curl themselves. Why?
Answer:
The rate of radiation of heat is directly proportional to the area of the surface. The sphere has the minimum surface area for a given volume. Thus, by curling, the animals reduce the area through which the heat is radiated and they keep the body warmer.

Question 61.
State two applications of Newton’s law of cooling.
Answer:

1. The hot water cools soon than the warm water.
2. When milk is added to tea, the rate of cooling decreases.

Question 62.
Hot water cools soon than the warm water. Why?
Answer:
The rate of cooling is directly proportional to the difference of mean temperature of the body and the temperature of the surrounding. The difference of temperature of hot water and surrounding is greater than that of warm water, therefore hot water cools soon.

Question 63.
On what factors does the loss of heat energy per second depend?
Answer:

• Nature of the surface,
• Area of the surface,
• Temperature difference between the temperature of the surface and surrounding.

Question 64.
The walls and roofs of greenhouse are made of glass, why?
Answer:
Glass has the property, that it transmits the heat radiation of high temperature but absorbs the heat radiations of low temperature. Hence, the glass allows the radiations coming from the sun, but it does not allow the radiations waves coming from the plants. Thus, the room remains warm.

Question 65.
In summer the hills remains cold, why?
Answer:
The reasons are :

• On the hill the air is thin and dust particles are less, thus radiations are absorbed less.
• The sun rays are reflected due to inclined surface.

Question 66.
The outer surface of cooking utensils are made black and inner bright Why?
Answer:
The black surfaces are good absorber of heat. Thus, more heat is absorbed and bright surface reflects the heat into the cooking pot. Thus, more heat is given to the food, which cooks the food soon.

Question 67.
The stars X and Y are radiating yellow and blue rays respectively. Which has greater temperature?
Answer:
By Wien’s displacement law, λ_m ∝\(\frac{1}{T}\). As the wavelength of blue rays is lesser
than that of yellow rays, hence, the temperature of the Y-star will be greater than X-star.

Question 68.
Whenever an object is heated, why does it appear red first?
Ans. According to Wien’s law, that λ_m T = constant, the maximum wavelength is for the red light in the visible spectrum. Hence, for ordinary temperatures i.e., for less T, λ_m must be more. Hence, an object appears red. As temperature increases, wavelength decreases and the emitted colour will shift towards violet.

Question 69.
A teacup has shining surface whereas the bottom of cooking pot is blackened, why?
Answer:
The shining surface is a good reflector of heat radiation, whereas the blackened surface is good absorber. The shining surface reflects the heat into the cup and maintains the tea hot. for long time, outer surface of cup does not radiate heat. Black surface of cooking pot absorbs more heat and makes the cooking fast.

Question 70.
The steady-state is necessary for the different substances to compare their conductivity. Why?
Answer:
In the steady-state the heat is not absorbed by any part of the body. Hence, the amount of heat given to one end transmit through the body completely. Therefore, the substance are brought in the steady-state.

Question 71.
In Searl’s apparatus to find the conductivity of a rod, the holes are filled with some mercury then thermometer are fixed. Why?
Answer:
The mercury is good conductor of heat therefore, it gives good contact between the bulb of thermometer and the surface of rod.

Question 72.
Eskimos built double-walled houses of snow. Why?
Answer:
The air is filled between the walls, which is bad conductor of heat. Ice itself is bad conductor of heat. Hence, the heat of the room does not transmit to the atmosphere.

Question 73.
In summer, in the evening when the temperature of outside the room decreases, the temperature of the inner surface of room remains at higher temperature. Why?
Answer:
After sunset, the temperature of the atmosphere decreases fast. The wall is bad conductor of heat, hence the heat of room conduct slowly and the room temperature remains higher than the atmosphere.

Question 74.
When a tumbler of thick glass is filled with hot tea it breaks down. Why?
Answer:
Glass is a bad conductor of heat. When hot tea is poured into the tumbler, the temperature of inner surface increases and the surface expands. While the outer surface remains at same position. Hence, the tumbler breaks down.

Thermal Properties of Matter Class 11 Important Questions Short Answer Type 

Question 1.
Establish the relation between coefficient of linear expansion and coefficient of superficial expansion.
Or
Prove that coefficient of superficial expansion is double of coefficient of linear expansion.
Answer:
Let one side of a square lamina be l.Its temperature is increased through Δt so that each side increases by Δl.
∴ Initial area = l²
and Final area = (l + Δl)²
∴ Change in area, ΔA = (l + Δl)² -l²
or =l² +2l.Δl + Δl² -l²
or = 2lΔl+ Δl²

Since Δl is a small quantity, hence Δl² can be neglected.
∴ ΔA=2l.Δl.
Now, coefficient of superficial expansion,
β = \(\frac{\Delta A}{A \times \Delta t}\)
or
β = \(\frac{2 l \Delta l}{l^{2} \Delta t}\) = 2. \(\frac{\Delta l}{l \Delta t}\)
or
β = 2α (∵ α = \(\frac{\Delta l}{l \Delta t}\) )

Question 2.
Establish the relation between coefficient of linear expansion and coefficient of volume expansion.
Or
Prove that coefficient of volume expansion is three times of coefficient of linear expansion.
Answer:
Consider a cube of side of length l. The temperature of cube is increased by, Δt so that the length of the cube becomes (l + Δl).

Now, Initial volume = l³
and Final volume = (l + Δl)³
∴ Change in volume, ΔV = (l + Δl)³ — l³
= l³ +3l²Δl + 3l.Δl² + Δl³ -l³ = 3l²Δl + 3l Δl2 + Δl3
Since, Δl is a small quantity, therefore powers of Δl can be neglected.
∴ ΔV = 3l²Δl .

Now, coefficient of volume expansion,
γ = \(\frac{\Delta V}{V \times \Delta t}\)
= \(\frac{3 l^{2} \Delta l}{l^{3} \Delta t}\)
γ = 3. \(\frac{\Delta l}{l \Delta t}\) = 3α,
( ∵ α = \(\frac{\Delta l}{l \Delta t}\)).

Question 3.
Distinguish between specific heat and thermal capacity.
Answer:
Difference between Specific heat and Thermal capacity :

Question 4.
Give an application of anomalous expansion of water.
Answer:
The volume of water decreases when it is heated from 0°C,to 4°C and then the volume increases, above 4°C. Hence, the density of water at 4°C is maximum. In the cold countries when the temperature of the atmosphere decreases, the temperature of the water of the lake also decreases.

Now, weigh the calorimeter with stirrer. Fill it with water about 2/3 and weigh it again. Find the initial temperature of water.

When the temperature reaches to 4°C, the density of the water becomes maximum. Then the ice begins to float on the surface, preventing the heat of water to go out into the atmosphere. Hence, water remains at 4°C at bottom. Thus, the animals survive in this water of 4°C.

Question 5.
Describe the method to find out the specific heat of a solid.
Answer:
(i) Method: Take the solid and measure its mass by physical balance and place it inside the steam chamber with the help of a thread. Fix a thermometer so that the bulb of thermometer touches the solid. Then start flowing the steam through the steam chamber.

Now, weigh the calorimeter with stirrer. Fill it with water about 2/3 and weigh it again. Find the initial temperature of water. When the temperature of the solid becomes constant for sometimes, then remove the wooden screen and move the calorimeter below the steam chamber. Then drop the solid into calorimeter carefully. Close the lid and stirr it slowly. Measure the temperature of the mixture when it becomes constant.

(ii) Observation:

•  Mass of solid = m gm
• Mass of calorimeter + stirrer = m₁ gm
• Mass of calorimeter + stirrer + water = m₂ gm
• Initial temperature of water = t₁°C
• Temperature of solid = t₂°C
• Temperature of mixture = t₃°C
• Specific heat of calorimeter = S₁ (known).

(iii) Calculation : Let the Specific heat of solid = s
Mass of water = (m₂ – m₁) gm
∴ Heat given by the solid = Mass × Specific heat × Temperature difference = m.s.(t2-t3)
Heat taken by calorimeter =m₁s₁ (t₃ – t₁)
and Heat taken by water = (m₂ – m₁ ).(t₃– t₁)

Now, by the principle of calorimetry, we have
Heat given = Heat taken
∴ ms(t₂ -t₃) = m₁s₁(t₃ -t₁)+(m₂ – m₁ ) (t₃-t₁)
∴ S = \(\frac{m_{1} s_{1}\left(t_{3}-t_{1}\right)+\left(m_{2}-m_{1}\right)\left(t_{3}-t_{1}\right)}{m\left(t_{2}-t_{3}\right)} \)
Thus, specific heat can be calculated.

Question 6.
Write difference between heat and temperature.
Answer:
Difference between Heat and Temperature :

Question 7.
On what factors does the amount of heat-conducting through a rod in steady state depend?
Or
In steady stae, find the expression for the amount of heat flowing through a rod.
Answer:
In steady-state the amount of heat Q flowing through one face to another is :
(i) Directly proportional to the area of cross-section A i.e.,
Q ∝ A
(ii) Directly proportional to the temperature difference ( θ₁ – θ₂) of the faces i.e.,
Q ∝ θ₁ -θ₂
(iii) Directly proportional to the time t of flowing heat i.e.,
Q ∝ t
(iv) Inversely proportional to the distance between the faces i.e.,
Q ∝ \(\frac{1}{d} \)
Combining all these, we get
Q ∝\(\frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
∴ Q = K \(\frac{A\left(\theta_{1}-\theta_{2}\right) t}{d}\)

Where, K is a constant called thermal conductivity or coefficient of thermal conductivity of the substance.

Question 8.
Differentiate between steady-state and variable state.
Answer:
Difference between Steady state and Variable state :

Question 9.
State the properties of heat radiation.
Answer:
Properties :

• Heat radiations are electromagnetic waves.
• Heat radiations travel in straight line and can travel through vacuum also.
• Thermal radiations travel with the speed of light i.e., 3 × 10¹⁰ cm × s^-1 and transmit in the form of radiant energy.
• Like light, they follow the inverse-square law.
• They follow the laws of reflection, refraction, polarization etc.
• They do not affect the mediums through they pass,but the medium which stops the radiation, are heated up.

Question 10.
State Newton’s law of cooling? What are the limitations of this law? Establish the relation between the temperature difference and rate of cooling. What is cooling curve.
Or
Describe the Newton’s law of cooling, deduce the formula and write its limitations.
Answer:
Newton’s law of cooling: According to this law, the rate of cooling is directly proportional to the difference of mean temperature of the body and the temperature of surroundings.

Limitations:

• The temperature difference between the body and surrounding should not be more than 30°C.
• The loss of heat should be only by radiation.
• The temperature of the surrounding and nature of body should not be changed. Let a substance be cooled from θ₁ °C to θ₂ °C in t sec and temperature of surrounding is θ .

∴ Rate of cooling = \(\frac{\theta_{1}-\theta_{2}}{t}\)
And Mean temperature of the substance = \(\frac{\theta_{1}+\theta_{2}}{2}\)
∴ Temperature difference = \(\frac{\theta_{1}+\theta_{2}}{2} \) – θ
Hence by Newtons law \(\frac{\theta_{1}-\theta_{2}}{t}\) ∝( \(\frac{\theta_{1}+\theta_{2}}{2} \) – θ)
∴ \(\frac{\theta_{1}-\theta_{2}}{t}\) = K ( \(\frac{\theta_{1}+\theta_{2}}{2} \) – θ)
Where, K is constant.
Cooling curve: The graph plotted between rate of cooling and temperature difference is called cooling curve. It is a straight line.

Question 11.
Derive the Newton’s law of cooling by Stefan’s-Boltzmann law.
Answer:
Let the temperature of a black body is T and temperature of surrounding is T₀. Now by Stefan’s-Boltzmann law, we have
Radiant energy per second per unit area,
E =σ (T₄ – T₀⁴)
If the temperature difference is ΔT, then

T₀ and σ are constant.
E ∝ ΔT
Hence, the rate of radiation of heat energy is directly proportional to the temperature difference.

Question 12.
Define coefficient of thermal conductivity. Write its unit and find out its dimensional formula.
Or
Define the coefficient of thermal conductivity and obtain expression for it.
Answer:
The coefficient of thermal conductivity is defined by the amount of heat, flowing per second in steady-state through the rod of length 1 m long and area of cross-section lm2, with a unit temperature difference between the opposite faces.

The quantity of heat, flowing per second through a unit cube having unit temperature difference between opposite faces, during steady state.
i.e., Q = \(\frac{K A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
∴ K = \(\frac{Q d}{A\left(\theta_{1}-\theta_{2}\right) t}\)
Unit: Its SI unit is \(\frac{\text { joule } \times \mathrm{m}}{\mathrm{m}^{2} \mathrm{~K} \mathrm{sec}}\) = Jm^-1K^-1sec^-1
But, joule/sec is Watt
∴ Other unit is Watt × m^-1×K^-1

Dimensional Formula :[K] = \(\frac{\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right][\mathrm{L}]}{\left[\mathrm{L}^{2}\right][\theta][\mathrm{T}]} \) = [MLT³θ^-1].

Question 13.
What do you mean by thermal resistance? Write its unit and find out its dimensional formula.
Answer:
The resistance offered by the body in the flow of heat is called thermal resistance,
Thermal resistance = \(\frac{d}{K A}\) = \(\frac{\left(\theta_{1}-\theta_{2}\right) t}{Q}\)
Unit: Its SI unit is =\(\frac{\mathrm{K} \mathrm{sec} .}{\mathrm{J}}\) = K sec^-1 = K watt^-1

Dimensional formula:
[Thermal resistance] = \(\frac{[\theta][T]}{\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]}\) = [M^-1L^-2T³θ].

Thermal Properties of Matter Class 11 Important Questions Long Answer Type

Question 1.
Prove that for thermal expansion in solid :
α : β : γ= 1 : 2 : 3.
Answer:
Let one side of a square lamina be l.Its temperature is increased through Δt so that each side increases by Δl.
∴ Initial area = l²
and Final area = (l + Δl)²
∴ Change in area, ΔA = (l + Δl)² -l²
or =l² +2l.Δl + Δl² -l²
or = 2lΔl+ Δl²

Since Δl is a small quantity, hence Δl² can be neglected.
∴ ΔA=2l.Δl.
Now, coefficient of superficial expansion,
β = \(\frac{\Delta A}{A \times \Delta t}\)
or
β = \(\frac{2 l \Delta l}{l^{2} \Delta t}\) = 2. \(\frac{\Delta l}{l \Delta t}\)
or
β = 2α (∵ α = \(\frac{\Delta l}{l \Delta t}\) )

Consider a cube of side of length l. The temperature of cube is increased by, Δt so that the length of the cube becomes (l + Δl).

Now, weigh the calorimeter with stirrer. Fill it with water about 2/3 and weigh it again. Find the initial temperature of water. When the temperature of the solid becomes constant for some times, then remove the wooden screen and move the calorimeter below the steam chamber. Then drop the solid into calorimeter carefully. Close the lid and stirred it slowly. Measure the temperature of the mixture when it becomes constant.

(ii) Observation :

• Mass of solid =m gm
• Mass of calorimeter + stirrer = m₁ gm
• Mass of calorimeter + stirrer + water = m₂ gm
• Initial temperature of water = t₁°C
• Temperature of solid = t₂°C
• Temperature of mixture = t₃°C
• Specific heat of calorimeter = S₁ (known).

(iii) Calculation : Let the Specific heat of solid = s
Mass of water = (m₂ – m₁) gm
∴ Heat given by the solid = Mass × Specific heat × Temperature difference
= m.s.(t₂-t₃)
Heat taken by calorimeter = m₁S₁(t₃ – t₁)
and Heat taken by water = (m₂ – m₁).(t₃– t₁)
Now, by the principle of calorimetry, we have
Heat given = Heat taken
∴ ms(t₂ – t₃) =m₁s₁(t₃ – t₁) +(m₂ – m₃) (t₃ – t₁)
∴ S = \(\frac{m_{1} s_{1}\left(t_{3}-t_{1}\right)+\left(m_{2}-m_{1}\right)\left(t_{3}-t_{1}\right)}{m\left(t_{2}-t_{3}\right)} \)
Thus, specific heat can be calculated.

Question 2.
Describe Newton’s experiment to prove the law of cooling on following points:
(i) Labelled diagram of apparatus, (ii) Method and (iii) Cooling curve.
Answer:
(i) Apparatus : C = Calorimeter, S = Stirrer, T= Thermometer, V= Vessel.
(ii) Method: Take some water in a beaker and heat it. Then put hot water into calorimeter about 2/3 of it. When the temperature difference is nearly 30°C, then stir the water slowly and after every half minute record the temperature of the water, with the help of stop watch.

Then, draw graph between temperature and time. This curve is called cooling curve.
On the cooling curve take some points A, B, C, D,… Now, for each pair of points find the rate of cooling and temperature difference. For example, let the temperature of A and B points are θ1 and θ2°C.
∴ Rate of cooling = \(\frac{\theta_{1}-\theta_{2}}{t}\)
and Temp. difference = \(\frac{\theta_{1}-\theta_{2}}{2}\) – θ
Where, θ is the temperature of surroundings.

Hence, the graph between rate of cooling and temperature difference is a straight line. Thus, rate of cooling is directly proportional to temperature difference.

Question 3.
Describe the experiment to find conductivity of a rod by Searl’s apparatus on following points:
(i) Labelled diagram of apparatus, (ii) Procedure, (iii) Observation, (iv) Calculation and (v) Precautions.
Answer:
(i) Searl’s apparatus :

(ii) Procedure: First of all water is regulated through the copper tube, slow and continuous. Then steam is passed through the steam chamber. When the temperature of all the four thermometers become constant, steady state is achieved. Note the reading of ther mometer and with the help of stop watch collect the water in a weighed beaker for a particular time.

(iii) Observation :

• Area of cross-section of rod = A cm²
• Distance between the holes on rod = d cm
• Temperature of first thermometer = θ₁ °C
• Temperature of second thermometer = θ₂ °C
• Initial temperature of water = θ₃ °C
• Final temperature of water = θ₄°C
• Mass of water collected in t sec = m gm

(iv) Calculation: We know that,
Q = \(\frac{K A\left(\theta_{1}-\theta_{2}\right) t}{d}\)
or
K = \(\frac{Q \cdot d}{A\left(\theta_{1}-\theta_{2}\right) t}\)
Now, Q = Heat taken by water
= m × 1 × (θ₄ – θ₃) = m(θ₄ – θ₃)
∴ K = \(\frac{m\left(\theta_{4}-\theta_{3}\right) d}{A\left(\theta_{1}-\theta_{2}\right) t}\)
Hence, K can be calculated.

(v) Precautions :

• The steam should be passed after regulating the flow of water.
• The flow of water should be slow and continuous.
• When the temperature of all the four thermometers becomes constant, then only the readings should be taken.

Question 4.
Write the differences between conduction, convection and radiation.
Answer:
Difference between Conduction, Convection and Radiation :

Thermal Properties of Matter  Class 11 Important Numerical Questions

Question 1.
The triple point of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperature on the Celsius and Fahrenheit scales. (NCERT)
Solution:
Here, Triple point of neon,
C = K-273
For neon,
C = 24-57 – 273 = -248-43°C
For C02,
C = 216-55-273=-56-45°C Relation between Kelvin scale and Fahrenheit scale:
\(\frac{\mathrm{K}-273}{5}\) = \(\frac{\mathrm{F}-32}{9}\)
or
F =\(\frac{9}{5}\) (K – 273) + 32
For neon,
F =\(\frac{9}{5}\) (24-57-273)+ 32
⇒ F = \(\frac{9}{5}\) × (-248.43) + 32
⇒ F =-447-2+32 = -415-2°
For CO₂,
F = \(\frac{9}{5}\) (K-273) + 32
⇒ F = \(\frac{9}{5}\) (216.55-273)+ 32
⇒ F = \(\frac{9}{5}\) × (-56.45) + 32
∴ F = -101.61 + 32 = -69.61° .

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B ? (NCERT)
Solution:
Here, triple point of water on absolute scale A = 200 A and triple point of water on obsolute scale B = 350 B
Triple point of water T = 273.16 K
According to question 200 A = 350 B = 273.16 K
or
1A = \(\frac{273 \cdot 16}{200}\) K and 1B = \(\frac{273 \cdot 16}{350}\) K
As T_A and T_B are at the same temperature,
\(\frac{273 \cdot 16}{200}\) T_A = \(\frac{273 \cdot 16}{350}\) T_B
or T_B = \(=\frac{350}{200} \) T_A
⇒ T_B = \(\frac{7}{5}\) T_A.

Question 3.
A steel tape 1 m long is correctly calibrated for a temperature of 27°C. The length of steel rod measured by this tape is found to be 63’0 cm on a hot day when the temperature is 45°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when temperature is 27°C? Coefficient of linear expansion of steel a= 1.2 × 10^-5 K-1. (NCERT)
Solution:
The steel tape is calibrated at 27°C is correct. That is the length of 100 cm steel tape has correct magnitude 100 cm.
When the temperature raises from 27°C to 45°C then length of steel tape.
l_t =l₀+ l₀ αt
l_t = 100 + 100× 1.2× 10^-5 (45-27)
l_t =100 + 100 × 1.2 × 10^-5 × 18
l_t =100 + 0.0216 = 100.0216 cm.

At 45°C the measure of 63 cm. steel tape.
= \(\frac{100 \cdot 0216}{100}\) × 63
= 63.0136 cm.
If the tape is measured at 27°C then its measure is 63 cm.

Question 4.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min, when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass 109 Js^-11m^-1°C^-1 . (Heat of vaporization of water = 2256 × 10³ J/kg.) (NCERT)
Solution:
Let the temperature of flame of source be T°C .
Given: K = 109 J/s.m°C = 109 × 10^-2 J/s.cm°C
A = 0.15 m²; d= 1 cm = 10^-2m
T₁-T₂ =(T-100)° C ,
t= 1 minute = 60 sec
m = 6 kg, L = 2256 J/gm
Q = \(\frac{109 \times 0 \cdot 15 \times(T-100) \times 60}{1 \times 10^{-2}} \)
= 98100(T -100)
Now, Q = mL = 6 ×2256 × 10³
∴ 98100(T – 100) = 6 × 2256 × 10³
or
T = \(\frac{6 \times 10^{3} \times 2256}{98100} \) + 100
= 138+ 100 = 238° C.

Question 5.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surrounding is 20°C. (NCERT)
Solution:
According to Newton’s law,
\(\frac{\theta_{1}-\theta_{2}}{t}\) = K (\(\frac{\theta_{1}+\theta_{2}}{2}\) – θ

Question 6.
At what temperature °C and F are equal to each other.
Solution:
Let at x degree temperature of both °C and °F are equal.
∴ \(\frac{\mathrm{C}}{5} \) = \(\frac{\mathrm{F}-32}{9}\)
\(\frac{x}{5}\) = \(\frac{x-32}{9}\)
∴9x = 5x – 160
or 9x – 5x = -160
4x = -160
∴ x =\(\frac{-160}{4}\) = -40
i.e., -40°C = -40°F.

Question 7.
Volume of a brass ball at 0°C is 50 cm³ and at 100°C is 50.25 cm³. Find out coefficient of linear expansion.
Solution:
Given, V₀ = 50 cm3, V₁₀₀ = 50.25 cm³
ΔV = 50.25-50, =0.25 cm³ and Change in temperature ΔT = 100° C – 0° C = 100° C
∴ γ = \(\frac{\Delta V}{V \cdot \Delta T}\)
or
γ = \(\frac{0 \cdot 25}{50 \times 100}\) = 5 × 10^-5
Also, γ = 3α
∴ α = \(\frac{\gamma}{3}\) = \(\frac{5 \times 10^{-5}}{3}\)
or = 1.7 × 10^-5/°C.

f

Question 8.
Ratio of density of two metal A and B is 1 : 3. On giving equal amount of heat on equal volume of A and B, ratio of change in temperature is 2 :1. Find out ratio of specific heats.
Solution:
Given, d₁ : d₂ = 1 : 3, V₁ = V₂, Q₁ = Q₂, θ₁ : θ₂ = 2:1
Heat given to metal A, Q₁ = Mass × Specific heat × Change in temperature.
Q₁ =V₁d₁s₁θ₁
Heat given to metal B,Q₂ =Mass × Specific heat × Change in temperature.
Q₂ = V₂d₂s₂θ₂
Since Q₁ = Q₂
∴ V₁d₁s₁θ₁ = V₂d₂s₂θ₂
or
\(\frac{s_{1}}{s_{2}}\) = \(\frac{V_{2}}{\mathrm{~V}_{1}} \times \frac{d_{2}}{d_{1}} \times \frac{\theta_{2}}{\theta_{1}} \)
= 1 ×\(\frac{3}{1} \times \frac{1}{2}\) = \(\frac{3}{2} \)
i.e., s₁:s₂ = 3:2

Question 9.
Temperature of 1 kg of water is 60°C. It is mixed in a water of 1 kg at 40°C. What will be the temperature of mixture ? Specific heat of water = 1 calorie/gram °C.
Solution:
Let the temperature of the mixture be t°C.
Heat given by hot water =ms Δt = 1000× 1 × (60-t)calorie.
Heat taken by cold water = ms Δt = 1000 × 1 × (t – 40) calorie.
According to principle of mixture.
1000 × 1 × (60-t)= 1000 × 1 × (t-40)
∴ 60-t = t-40
or
2t = 100
or
t= 50°C.

Question 10.
What will be the resultant temperature when 5-gram ice at 0°C is mixed in 50 gram water at 30°C. Letent heat of ice = 80 calorie/gm.
Solution:
Let the resultant temperature be t°C.
Heat given by ice = Heat taken by water 5 × 80 + 5 × 1×(t-0) = 50× 1 ×(30-t)
or
400 + 5t = 1500-50t
or
55t = 1500-400= 1100
t =\(\frac{1100}{55}\) = 20°C.

Question 11.
How much heat is required to change 0°C of 10 gm ice into 100°C steam? The latent heat of ice and steam are 80 and 540 cal/gm respectively.
Solution:
Heat required to change 0°C ice to 0°C water = mL = 10×80 = 800 cal
Heat required to change 0°C water to 100°C water = ms ΔT
= 10× 1 × (100-0) = 1000 cal
and Heat required to change 100°C water to 100°C steam = mL = 10 × 540 = 5400cal
Total amount of heat = 800+1000 + 5400 = 7200cal.

Question 12.
Calculate the heat required to convert 15 gm ice at -15°C into steam of 100°C. Given, specific heat of ice is 0.5 cal × gm^-1 × °C^-1. Latent heat of ice is 80 cal × gm^-1 and latent heat of steam is 536 cal × gm^-1.
Solution:
Heat required to convert 15 gm ice at -15°C into 0°C
= mst = 15 × 0.5 × (0+15) = 112.5cal
Heat required to convert 0°C ice into 0°C water
= mL =15 × 80= 1200cal
Heat required to convert 0°C water into 100°C water
= mst = 15 × 1 × (100 – 0) = 1500 cal
Heat required to convert 100°C water into 100°C steam = mL = 15 × 536 = 8040 cal
∴ Total heat = 112.5 + 1200 + 1500 + 8040 = 10852.5 cal.

Question 13.
A piece of metal of mass 50 gm is heated up to 200°C and then put into 400 gm water kept into a beaker. The initial and final temperatures of water are 20°C and 22.4°C respectively. Calculate the specific heat of metal.
Solution:
Heat given by metal = Heat taken by water
or ( 50 × s × (200 – 22.4) = 400 × 1 × (22.4 – 20)
or
8880s = 960
or
s = \(\frac{960}{8880}\) = 0.108
∴ s = 0.108 cal ×gm^-1 × °C^-1

Question 14.
An iron piece of 100 gm is heated up to 10°C. How much water will be riased through 1°C, by the same amount of heat? specific heat of iron is 01 unit.
Solution:
Heat required to raise the temperature of iron piece
= 100× 0.1 × 10 = 100 cal.
Let the temperature of m gm water is raised through 1°C.
∴ Q = msΔT
or 100 = m × 1×1
or m = 100gm.

Question 15.
A wall has two faces of thickness 3 cm and 6 cm respectively and conductivities are K and 3K. Temperature of their outer faces are 20°C and -5°C respectively. Evaluate their common temperature instead state.
Solution:
Given, d₁ = 3cm, d₂= 6cm, K₁ = K,K₂ = 3K, θ₁ = 20°C,θ₂ = -5°C
From formula, θ = \(\frac{K_{1} d_{2} \theta_{1}+K_{2} d_{1} \theta_{2}}{K_{2} d_{1}+K_{1} d_{2}} \)
∴ θ = \(\frac{K \times 6 \times 20+3 K \times 3 \times(-5)}{3 K \times 3+K \times 6} \)
= \(\frac{120 K-45 K}{15 K} \) = \(\frac{75 K}{15 K}\) = 5°C

Question 16.
Two vessels of same shape, length, breadth and height are made of to different metals. They are filled with same amount of ice at 0°C. They talc 25 and 20 minutes respectively for melting the ice completely. Compare the thermal conductivities of two metals.
Solution:
Given:t₁ =25 minutes, t₂ = 20 minutes
Since, equal quantity of ice is filled in two vcssels, hence same quantity of heat is required for melting the ice.
θ = \(\frac{K_{1} A\left(\theta_{1}-\theta_{2}\right) t_{1}}{d}\) = \(\frac{K_{2} A\left(\theta_{1}-\theta_{2}\right) t_{2}}{d} \)
A θ₁ – θ₂ and d are some for both.
∴ K₁t₁ = K₂t₂
or
\(\frac{K_{1}}{K_{2}} \) = \(\frac{t_{2}}{t_{1}}\)
= \(\frac{20}{25} \) = \(\frac{4}{5}\)
∴ K₁ :K₂ = 4:5

Question 17.
The surface area of a box is 5000 cm2 and its thickness is 6 mm. It is filled with the ice at 0°C and kept in a room of temperature 20°C. If 1 kg ice ¡s melted in 20 min. then calculate the conductivity of box. (Latent heat of ice = 80 cal/gm)
Solution:
Given m = 1kg = 1000 gm, L = 80 cal/gm, d = 6 mm = 0.6 cm,
A= 5000 cm2, θ1 – θ2=20 – 0=20°C t=20min = 20 × 60s

Thermal Properties of Matter  Class 11 Important Questions Objective Type

1. Multiple- choice questions:

Question 1.
Magnitude of absolute zero in Fahrenheit scale is :
(a) 273° F
(b) 459° F
(c) 182° F
(d) 0° F.
Answer:
(b) 459° F

Question 2.
Two system whose temperature is T_A and T_B are in thermal equilibrium, the relation between T_A and T_A is :
(a) T_A > T_B
(b)T_A < T_B
(C) T_A =T_B
(d) None of these.
Answer:
(C) T_A =T_B

Question 3.
A hole is done in a rectangular copper plate, when the plate is heated, then the size of hole will be :
(a) Not change
(b) Increase
(c) Decrease
(d) Nothing can be said.
Answer:
(b) Increase

Question 4.
There is a spherical cavity inside a solid sphere of metal, when the sphere is heated the volume of spherical cavity will:
(a) Increase
(b) Decrease
(c) No change
(d) Shape will change.
Answer:
(a) Increase

Question 5.
If it is coefficient of superficial expansion of a metal then its coefficient of volume expansion will be :
(a) \(\frac{\beta}{2}\)
(b) 3β
(c) \(\frac{3 \beta}{2}\)
(d) \(\frac{2 \beta}{3} \)
Answer:
(c) \(\frac{3 \beta}{2}\)

Question 6.
If the temperature of a diseased person is 40°C, then his temperature in Farenheit will be:
(a) 72°F
(b) 96°F
(c) 100°F
(d) 104°F.
Answer:
(d) 104°F.

Question 7.
Temperature of sun is measured by:
(a) Gas thermometer
(b) Pyrometer
(c) Platinium thermometer
(d) Constant pressure thermometer.
Answer:
(b) Pyrometer

Question 8.
For whom medium is not required for transmission of heat:
(a) Conduction
(b) Convection
(c) Radiation
(d) All.
Answer:
(c) Radiation

Question 9.
Coefficient of thermal conductivity for ideal conductor is:
(a) 0
(b) 1
(c) -1
(d) ∞ .
Answer:
(d) ∞

Question 10.
Unit of coefficient of thermal conductivity is :
(a) joule metre/ second ×°C
(b) joule × metre × second / °C
(c) joule / metre × second × °C
(d) joule × second / metre × °C.
Answer:
(c) joule / metre × second × °C

Question 11.
In steady state, heat stored in metai does not depends on :
(a) Temperature gradient
(b) Area of cross-section
(c) Length
(d) Thermal capacity.
Answer:
(d) Thermal capacity.

Question 12.
An equivalent rod is made by joining two rod of same length. If their coefficient of thermal conductivity are K₁ and K₂, then thermal conductivity of equivalent rod is:
(a) K₁+K₂
(b) K₁– K₂
(c) 2K₁ K₂/ (k₁+ K₂)
(d) (K₁ – K₂)/(k₁+ K₂).
Answer:
(c) 2K₁ K₂/ (k₁+K₂)

Question 13.
Monsoon is due to:
(a) Conduction
(b) Convection
(c) Radiation
(d) All the above.
Answer:
(b) Convection

Question 14.
PolIshed body are for thermal radiation:
(a) Good absorber and bad reflector
(b) Good absorber and good reflector
(c) Bad absorber and good reflector
(d) Bad absorber and bad reflector.
Answer:
(c) Bad absorber and good reflector

Question 15.
Whose specific heat is more :
(a) Iron
(b) Water
(c) Copper
(d) Glass.
Answer:
(b) Water

2. Fill in the blanks:

1. Coefficient of linear expansion of solid depend upon ……………. .
Answer:
Nature of matter

2. SI Unit of temperature gradient is……………. .
Answer:
joule / metre

3. Density of water is maximum at …………… temperature.
Answer:
4°C

4. α: β: γ = ……………
Answer:
1:2 : 3

5. The point where all three phases in solid, liquid and gas co-exist is called …………… .
Answer:
Triple point

6. Wavelength of light radiation is …………… than heat radiation.
Answer:
Less

7. Conversion of matter from solid-state to vapour state directly is called ……………
Answer:
Sub-limation.

3. Match the following:

Answer:
1.  (c) Only for solid
2. (e) Solid, liquid and gas.
3. (d) Raise of temperature by 1°C
4. (b) Rate of cooling ∝ difference temperature
5. (a) kilogram.

4. state true or False:

1. Volume of water at 4°C is maximum.
Answer:
False

2. At triple point of water, temperature is 273.16K.
Answer:
True

3. On increasing temperature the volume of mercury increases uniformly.
Answer:
True

4. Gases has two type of specific heat.
Answer:
True

5. Cp is always less than Cv.
Answer:
False

6. For monoatomic gas the value of Y is 1.67.
Answer:
True

7. In convection molecules changes their position for transfer of heat.
Answer:
True

8. Absorbtive power of ideal black body is one.
Answer:
True.

Students get through the MP Board Class 11th Physics Important Questions Chapter 6 Work, Energy and Power which are most likely to be asked in the exam.

MP Board Class 11th Physics Important Questions Chapter 6 Work, Energy and Power

Work, Energy and Power Class 11 Important Questions Very Short Answer Type

Question 1.
What is work? Is it scalar or vector quantity?
Answer:
Work is said to be done by a force when the force produces a displacement of the body on which it acts in any direction except perpendicular to the direction of force, i.e.,
W = Fs cosθ
It is a scalar quantity.

Question 2.
Write the unit and dimensional formula of work.
Answer:
Unit: SI unit of work is joule.
Dimensional formula : Work = Force × Displacement
= [MLT^-2][L]
= [ML²T^-2].

Question 3.
Define 1 joule and 1 erg.
Answer:
1 joule : When a force of 1 N is applied on a body and is displaced 1 m in the direction of force, then 1 joule work is done.
1 erg : When a force of 1 dyne is applied on a body and is displaced 1 cm in the direction of force, then 1 erg work is done.

Question 4.
What is zero work? Give one example.
Answer:
If force applied on an object is perpendicular to its displacement then the work is zero.
∴ W = Fs cos90° = 0
Example: A coolie carrying suitcase on his head. The work done against gravity is zero.

Question 5.
When does the work done become maximum by a force?
Answer:
When the direction of force and displacement are same.

Question 6.
Can a satellite do any work against gravitational force, while revolving around the earth? Why?
Answer:
No, because the direction of motion of satellite is always perpendicular to the gravitational force.

Question 7.
What is energy? Give examples. Write its unit and dimensional formula.
Answer:
The efficiency of work done of a body is called energy.
Examples: Mechanical energy, heat energy, light energy etc.
SI unit: joule.
Dimensional formula: [ML²T^-2].

Question 8.
Write the statement of laws of conservation of energy.
Answer:
According to this law energy is neither created nor destroyed but transformed from one form to another.

Question 9.
A man walk 10 metre with a luggage in his head. How much work will be done by him against gravitational force.
Answer:
Zero, because displacement of the man is perpendicular to the direction of gravitational force.

Question 10.
What do you mean by variable force? Explain with example.
Answer:
When the magnitude and direction of a force are changing with time then such force is called variable force.
Example :
The gravitational force acting between two bodies depends upon the distance between the bodies, hence it is variable force.

Question 11.
What do you mean by mechanical energy? What are the types of it?
Answer:
The energy due to mechanical work is called mechanical energy. The mechanical energy are of two types :

• Kinetic energy,
• Potential energy.

Question 12.
If velocity of an object is doubled then what will be the increase in its kinetic energy?
Answer:
Kinetic energy will become four time as
K = \(\frac{1}{2}\) mν²

Question 13.
Write the formula for gravitational potential energy at height ‘h’.
Answer:
Gravitational potential energy at height ‘h’ of a body of mass ‘m’ is mgh.

Question 14.
When a bullet strike a target and get stopped, then in which form its kinetic energy get changes?
Answer:
Kinetic energy of the bullet get changes into sound energy and heat energy.

Question 15.
What is power? Write its S.I. unit.
Answer:
Rate of doing work is defined as power. Its S.I. unit is watt. It is a scalar quantity.
power = \(\frac{\text { Work }}{\text { Time }}\)

Question 16.
Define watt.
Answer:
If the rate of work done is 1 joule per second, then the power is called watt.

Question 17.
Write the magnitude of one horse power.
Answer:
1 H.P. = 746 watt.

Question 18.
What do you mean by collision? Which quantities are transferred during the collision?
Answer:
The interaction between two bodies due to impact, is called collision. During the collision momentum and kinetic energy are transferred.

Question 19.
Do the bodies touch each other compulsorily in collision? Give example.
Answer:
Not necessary. When alpha particles (₂He⁴), which are positively charged particles are made to strike the nucleus, then they get deviated from their path without contact with each other.

Question 20.
How many types of collisions are there? Describe them.
Answer:
Collisions are of two types :
(i) Elastic collision: The collisions in which the momentum and kinetic energy of the system conserved are called elastic collision.
Example : The collisions between the molecules and atoms. In daily life collision of glass or ivory balls are considered as elastic collision.

(ii) Inelastic collision : The collisions in which the momentum of the system is conserved but not kinetic energy.
Example : Spreading of mud and stick to wall. A bullet embedded with a wooden block.

Question 21.
Write the magnitude of one kilowatt and one megawatt.
Answer:
1 kilowatt = 1000 watt = 10³ watt.
1 megawatt = 10⁶ watt.

Question 22.
How kinetic energy is measured?
Answer:
Kinetic energy is measured by formula \(\frac{1}{2}\) mν²

Question 23.
What do you mean by variable force? write the formula for work done by variable force.
Answer:
When the magnitude and direction of a force are changing with time, then such force is called variable force.
Work done by variable force W = F. d cosθ.

Question 24.
Power is a scalar quantity or vector quantity. Write its dimensional formula.
Answer:
Power is a scalar quantity. Its dimensional formula is [ML² T^-3 ].

Question 25.
An object is displaced by applying force on it. When will be the work done maximum?
Answer:
When the displacement of an object is in direction of applied force, then work done is maximum.

Question 26.
In which direction force must be applied on a body so that work done is zero?
Answer:
When the force is applied perpendicular to the displacement work done is zero.
Since, W = F.d cosθ
If θ = 90°, cos 90° = 0.
∴ W = 0.

Question 27.
A car is moving on the plane road with uniform velocity, does the car work, why?
Answer:
Yes, the car does work against frictional force.

Question 28.
What type of energy is stored in the spring of a watch?
Answer:
Elastic potential energy.

Question 29.
What is spring constant? What is its S.I. unit?
Answer:
The spring constant of a spring is defined by the force, which changes its length by unity. Its S.I. unit is Nm^-1.

Question 30.
Write the formula for potential energy of a compressed spring.
Answer:
Elastic potential energy of a compressed spring is =\(\frac{1}{2}\) kx² where k is spring constant and x is displacement from normal state.

Question 31.
When a bulb glow, does electric energy get transformed into any energy?
Answer:
Yes, electrical energy get transformed into light energy and thermal energy.

Question 32.
Write the different forms of energy.
Answer:
Different form of energy are Mechanical energy, Thermal energy, Light energy, Magnetic energy, Electrical energy, Sound energy, Chemical energy, Nuclear energy etc.

Question 33.
What is kilowatt hour ? Write its magnitude.
Answer:
kilowatt hour is a unit of energy.
1 kilowatt hour = 1000 watt × 1 hour
= 1000 joule/second × (3600 second) = 3.6 × 10⁶ joule.

Question 34.
Write the relation between momentum (p) and kinetic energy (E). The K.E. of a body is increased 4 times. What will be the change in momentum?
Answer:
p = \(\sqrt{2 m E}\)
∴ p ∝√E
or
\(\frac{p_{1}}{p_{2}}=\sqrt{\frac{E_{1}}{E_{2}}}=\sqrt{\frac{E}{4 E}}=\frac{1}{2}\)
∴p₂ = 2p₁
Hence, momentum will be doubled.

Question 35.
if the momentum of a body becomes 4 times. How many time its K.E. will be ?
Answer:
We have, \(\frac{p_{1}}{p_{2}}=\sqrt{\frac{E_{1}}{E_{2}}}\)
or \(\frac{E_{1}}{E_{2}}=\frac{p_{1}^{2}}{p_{2}^{2}}=\left(\frac{p_{1}}{4 p_{1}}\right)^{2}=\frac{1}{16}\)
or E₂= 16E₁
Hence, K.E. will be 16 times.

Question 36.
The quantity whose unit is kWh?
Answer:
Energy.

Question 37.
What do you mean by mass-energy equivalence?
Answer:
It means that mass can be converted into energy and energy can be converted into mass.
i.e., E = mc²
Where, c is velocity of light.
This relation is called mass-energy equivalence equation.

Question 38.
What is the equivalent energy of 1 gm mass?
Answer:
1 gm = 9 × 10¹³ joule.

Question 39.
How much energy is produced by the absorption of an electron and a positron?
Answer:
1.02 MeV.

Question 40.
Which mechanical energy increases or decreases when a body is thrown vertically upwards?
Answer:
Potential energy and kinetic energy.

Question 41.
Two bodies of same mass are moving with a velocity of u₁ and u₂ collide with each other, what will be the velocity after collision?
Answer:
Let the velocity after collision be ν₁ and ν₂.
According to conservation of momentum,
mu₁ + mu₂ = mν₁ + mν₂
or
u₁+u₂ = ν₁+ν₂ …(1)
and for elastic collision ν₁ – ν₂ = – (u₁– u₂) …(2)
Solving eqns. (1) and (2), we get
ν₁ = u₂ and ν₂ = u₁.

Question 42.
Write the forms of energy we get from the following :

• Dynamo,
• Electric motor,
• Photoelectric cell,
• Loudspeaker,
• Glow of electric bulb,
• Free falling stone.

Answer:

• Dynamo : Mechanical energy into electrical energy.
• Electric motor: Electrical energy into mechanical energy.
• Photoelectric cell: Light energy into electrical energy.
• Loudspeaker: Electrical energy into sound energy.
• Glow of electric bulb : Electrical energy into light energy.
• Free falling stone : Potential energy into kinetic energy.

Question 43.
Does a satellite do any work against gravitation while it revolve in its orbit? Why?
Answer:
No, because motion of a satellite is always perpendicular to the direction of gravitation. ‘
Since, W = F.d cosθ
and θ= 90°,
∴ cosθ= cos 90° = 0
∴ W = 0.

Question 44.
When a body is thrown upward and when it fall downward, what type of mechanical energy is increased?
Answer:
When a body is thrown upward potential energy increases and when body fall downward its kinetic energy increases.

Question 45.
Force of repulsion acts between two similar charge, when they bought closer does their potential energy increases?
Answer:
Yes, there potential energy increases as work is done against force of repulsion.

Work, Energy and Power Class 11 Important Questions Short Answer Type

Question 1.
Explain geometrical meaning of scalar product of two vectors and write its three properties.
Answer:
Geometrical meaning of the scalar product of two vectors : If \(\vec{a} \text { and } \vec{b}\) are the two given vectors and θ is the angle between them, then
\(\vec{a} \cdot \vec{b}\) = abcosθ = a(bcosθ) -b(acosθ)

Now, b cos θ is the components or projection of the vector \(\vec{b}\) in the direction of the vector \(\vec{a}\) and a cos θ is the component of the vector \(\vec{a}\) in the direction of the vector \(\vec{b}\)

Thus, the scalar product of two vectors is equal to the multiplication of the magnitude of one vector and the magnitude of the component or projection of the other vector in the direction of the first vector.

Scalar product of two equal vectors :
\(\vec{a} \cdot \vec{b}\) = a.acos0° = a²
\(\vec{a} \cdot \vec{a}\) is also called \(\overrightarrow{a^{2}}\)

\(\vec{a} \cdot \vec{a}\) = \(\overrightarrow{a^{2}}\) = a ²
i.e., the dot product of a vector itself is equal to the square of its magnitude.

Properties of the scalar product of two vectors :
(i) The scalar product of two vectors is commutative i.e., the order of the vectors can be interchanged i.e.,
\(\vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}\)
Proof:
\(\vec{a} \cdot \vec{b}\) = ab cosθ
Now, \(\vec{b} \cdot \vec{a}\) = ba cosθ
⇒ ab cosθ = \(\vec{a} \cdot \vec{b}\)

(ii) The scalar product of vectors follows the distributive law in relation to the addition. Thus,
\(\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}\)

(iii) The scalar product of two vectors follows the multiplicative law in relation to scalar. Thus,
\((m \vec{a}) \cdot \vec{b}=m(\vec{a} \cdot \vec{b})=\vec{a} \cdot(m \vec{b})\)

Question 2.
Explain the geometrical meaning of cross product of two vectors and write its properties.
Answer:
Geometrical meaning of vector product of two

vectors : Let two vectors \(\vec{a} \text { and } \vec{b}\) are represented by the adjacent sides of parallelogram OACB, i.e.,
\(\overrightarrow{O A}=\vec{a} \text { and } \overrightarrow{O B}=\vec{b}\)
Now, draw perpendicular BD from point B, on OA.
Area of parallelogram OACB = OA × BD
= OA × OB sinθ [∵ sinθ = \(\frac{B D}{O B}\)
or
absinθ =\(|\vec{a} \times \vec{b}|\)
Hence, modulus of \(\vec{a} \times \vec{b}\) represents the area of parallelogram whose adjacent sides are \(\vec{a} \text { and } \vec{b}\).

Properties of vector or cross product of two vectors :
(i) The vector multiplication of two vectors is non-commutative i.e.,
\(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)
As
\(\vec{a} \times \vec{b}=a b \sin \theta \hat{n}\)
and \(\vec{b} \times \vec{a}=a b \sin \theta(-\hat{n})\)
\(=-a b \sin \theta \hat{n}\)
\(=-\vec{a} \times \vec{b}\)

(ii) The vector multiplication of two vectors does not follow the law of association Le.,
\(\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c}\)

(iii) The vector multiplication of two vectors follows the law of association in relation to scalar i.e.,
\(m \times(\vec{a} \times \vec{b})=(m \vec{a}) \times \vec{b}=\vec{a} \times(m \vec{b})\)

(iv) The vector multiplication of two vectors follows the law of distribution in relation to addition i.e.,
\(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\)

(v) If \(\vec{a} \text { and } \vec{b}\) are two non-zero vectors and are parallel also i.e., θ = 0 or π, then ,
sinθ =0
∴ \(\vec{a} \times \vec{b}=a b \sin \theta \hat{n}\)
\(=0 \times \hat{n}=\overrightarrow{0}\)
Therefore, the vector product of two parallel vectors is a zero or null vector.

Question 3.
Explain with example work done upon a body and work done by a body.
Answer:
When the displacement of the body is in the direction of external force applied the work is done upon the body and when the displacement is opposite to the direction of force then work is done by the body.
Example: When a stone is thrown vertically upward a force equivalent to its weight is applied upon it in the vertically upward direction, the work is done upon the stone while it fall downward work is done by the stone.

Question 4.
What do you mean by kinetic energy? Derive the expression for it.
Answer:
The efficiency of work done by the motion of a body is called its kinetic energy. Let m = Mass of a body at rest (i.e., u = 0)
F = Force applied on the body.
a = Acceleration produced in the body in the direction of applied force,
ν = Velocity acquired by the body in moving thrown a distance s.
From ν² -u² = 2as
ν²– 0 = 2as
or a = \(\frac{v^{2}}{2 s}\)
As F = m.a
∴ F = m\(\left(\frac{v^{2}}{2 s}\right)\)

Work done on the body,
W= Force × Displacement
or W = m\(\left(\frac{v^{2}}{2 s}\right)\) ×s
or W = \(\frac{m \cdot v^{2}}{2}\) = \(\frac{1}{2}\) mν²
This work done on the body is measure of kinetic energy (K.E.) of the body.
∴ K.E. = \(\frac{1}{2}\) mν²

Question 5.
What is work-energy theorem? Prove it.
Or
Write work-energy theorem and prove it for constant force.
Answer:
According to this theorem, the work done by a force is equal to the. change in kinetic energy of a body.
Let a force F is applied on a body of mass m so that its velocity becomes v.
If the body is displaced to distance d, then work done,
W = F.d
But, F = ma
∴ W = mad …(1)
Now, by the equation of motion ν² = u² + 2as, we get
2ad=ν² – u², ∴ (s =d)
or ad = \(\frac{1}{2}\)(ν² – u²)
Putting this value in eqn. (1), we get
W = m.\(\frac{1}{2}\)(ν² – u²)
or W = \(\frac{1}{2}\)mν² – \(\frac{1}{2}\)mu²
Work done = Change in K.E.

Question 6.
Momentum of a lighter body and heavier body is same, whose kinetic energy will be more?
Answer:
Let a body of mass m moves with a velocity of ν.
∴ Momentum of the body pm = m.ν …(1)
and Kinetic energy of the body K = \(\frac{1}{2}\)mν² …(2)
From eqn(1), ν = \(\frac{p}{m}\)
∴ K = \(\frac{1}{2}\)m.\(\frac{p^{2}}{m^{2}}\)
or
K = \(\frac{1}{2}\)\(\frac{p^{2}}{m}\)
or
P² = 2Km
or
p = \(\sqrt{2 m K}\) …(3)
Since, momentum of both bodies are same.
∴ From eqn. (3),
mK = Constant
or
K = \(\frac{\text { Constant }}{m}\)
K = \(\frac{1}{m}\)
∴ Kinetic energy of lighter body will be more.

Question 7.
What do you mean by variable force? Derive the expression for the work done by variable force.
Answer:
Variable force : When the magnitude and direction of a force are changing with time then such force is called variable force.
Example :
The gravitational force acting between two bodies depends upon the distance between the bodies, hence it is variable force.

Work done by variable force :
Case 1. Only magnitude of force changes not direction :

Let a variable force F(x) is applied along the direction of X-axis and body is displaced from x₁, to x_n in the direction of X-axis.
Graph (a) and (b) show the relation between force F (x) and displacement x.
Now the total displacement can be divided into so many small parts of Δx. As Δx is very small, the force during this displacement may be considered constant.
∴ Work done for the displacement Δx:
ΔW=F.Δx
Now, the work done for the displacement from x₁, to x_n will be the sum of work done of all displacements such as Δx.
∴ Total work done W = \(\sum_{x_{1}}^{x_{n}}\) F(x)Δx …(1)
If the displacement Δx is very small then the eqn. (1) can be written in the form of integration as
W = \(\int_{x_{1}}^{x_{n}}\)F{x)dx
or W = Area enclosed between the curve and displacement axis.

Case 2. When the magnitude and direction both are changing :
Let a particle is moving in irregular path and the coordinates of initial point are (x₁ , y₁, z₁,) and that of final point are (x₂ ,y₂ ,z₂).
∴ W\(\int_{x_{1}}^{x_{2}}\) F_xdx + \(\int_{y_{1}}^{y_{2}}\)F_ydy + \(\int_{z_{1}}^{z_{2}}\)F_zdz
Where \(\vec{F}=F_{x} \hat{i}+F_{y} \hat{j}+F_{z} \hat{k}\)

Question 8.
Kinetic energy of a lighter and heavier body are same, whose momentum will be more ?
Answer:
From the formula,
p = \(\sqrt{2 m K}\)
When the kinetic energy of both bodies are same, then
p ∝√ m
or
p²∝m
∴ \(\frac{p_{1}^{2}}{p_{2}^{2}}=\frac{m_{1}}{m_{2}}\)
Since, m₁>m₂
∴ p₁²> p₂²
∴ Momentum of heavier body will be more than lighter body.

Question 9.
Derive the expression for gravitational potential energy.
Answer:
Let, m = Mass of the body, g = Acceleration due to gravity on the surface of earth, h = Height through which the body is raised.
If we assume that height h is not too large and the value of g is practically constant, then the force applied just to overcome gravitational attraction is
F= mg …(1)

As the distance moved in the direction of the force applied, therefore F h Work done = Force x Distance
or
W = F × h
or
W= mg × h = mgh
Hence, gravitational potential energy = mgh.

Question 10.
Derive the formula for the work done in stretching a spring.
Or
What will be the work done in compressing a spring of force constant k by an amount x?
Answer:
For an ideal spring the force is proportional to the displacement. Hence, the graph between force and displacement is a straight line.
Consider a point A on the force-displacement curve.
∴ The corresponding force of point
A = F = kx
∴ Work done in stretching the spring,
W = Area of Δ ABO
or
W = \(\frac{1}{2}\) OB × AB
= \(\frac{1}{2}\) × kx
= \(\frac{1}{2}\)kx²
∴P.E. of stretched spring = \(\frac{1}{2}\)kx².

Question 11.
What is meant by collision ? Differentiate elastic and inelastic collision.
Answer:
Collision : The striking of one particle with another or an interaction between them is called collision.

For example: Hitting a ball by a bat, striking carrom-coin by a striker, etc.
Difference between Elastic collision and Inelastic collision:

Question 12.
If the momentum of a body is increased by four times, what will be in
crease in its kinetic energy?
Answer:
We know,

ie., kinetic energy will increase by 16 times.

Question 13.
If the kinetic energy of a body is increased by four times, then what will be
increase in Its momentum?
Answer:
We know,

i.e., momentum will increase by 2 times.

Question 14.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket
(b) Work done by gravitational force in the above case.
(c) Work done by friction on a body sliding down an inclined plane.
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Answer:
We know that the work done is given by W = F.s = Fs cos θ, where θ = smaller angle between force F and displacement s.
(a) To lift the bucket, force equal to the weight of the bucket has to be applied vertically upwards and the bucket moves along the same direction, thus 0 – 0, so W= FscosO = Fs i.e., positive.

(b) Here as the bucket moves in a direction opposite to the gravitational force which always acts vertically downwards.
∴ θ = 180°, so W= Fs cos 180° = Fs (- 1) = -Fs
i.e., W = negative.

(c) Work done by friction on a body sliding down an inclined plane is negative as friction opposes the relative motion,
Thus, θ = 180° ∴ W = Fs cos 180° = – Fs.

(d) As the body moves in the same direction in which the force is applied, so e = 0, thus W=Fs cos 0° = Fs i.e., it is positive.

(e) Work done is negative as the direction of the resistive force of air on the vibrating pendulum is opposite to the direction of displacement {i.e., motion) of the bob.

Question 15.
What do you mean by positive, negative and zero work? Give one example of each.
Answer:
Positive work:
When the angle between force and displacement is zero or acute angle then the work is positive.
Examples :

• When a bucket of .water from a well is pulled, the directions of force and displacement are same and hence positive work is done.
• A body falling freely under gravity. Then work done by gravity force is positive.

Negative work :
When the angle between force and displacement is 180° or obtuse angle then the work done is negative.
∴ Fs cos 180° = -Fs
Examples :

• When a bucket full of water is pulled from the well, the work done by the gravity force is negative, because the directions of force of gravity and displacement are opposite, (gravity acts downwards and displacement up).
• The work done by the friction force on an inclined plane is negative.

Zero work:
If the force applied is perpendicular to its displacement, then the work is
zero.
W = Fs cosθ = Fs cos 90° = 0
Examples :

• A coolie carrying suitcase on his head. The work done against gravity is zero.
• A body moving in circular motion. The centripetal force is perpendicular to its velocity. Hence, the work done by the centripetal force is zero.

Question 16.
A lighter body moving with velocity u. Collide with a heavy body at rest. What will be its velocity after collision?
Answer:
Let the velocity of lighter body (mass m) be ν₁ and the velocity of heavy body (mass M) be ν₂ after collision.
∴ From laws of conservation of momentum,
m u = m ν₁+ Mν₂ and For elastic collision
ν₁ – ν₂, = -(u – 0)
or
ν₂ – ν₁
On solving,
ν₁ = \(-\left(\frac{M-m}{M+m}\right)\) u
and ν₂ = \(\frac{2 m u}{(M+m)}\)
If m<< M, then
ν₁ = -u and ν₂ = 0.

Question 17.
(a) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

(b) In fig.-given below
(i) the man walks 2m carrying a mass of 15 kg on his hands. In fig.
(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Answer:
(a) When an artificial satellite comes closer to earth, its gravitational P.E. decreases (as its height from the earth surface decreases). By law of conservation of energy, the sum of kinetic and potential energy should remain constant. To keep the total energy constant, kinetic energy should increase and so, velocity of satellite should increase. But, total energy of satellite continuously decreases at a very small rate due to atmospheric resistance.

(b) In case (a) : No work is done against gravity as the displacement of 2m (horizontal)
and the weight of 15 kg (acting vertically downwards) are perpendicular to each other
(θ=90°),
W = Fs cos 90° = 0.
In case (b): Work is done against gravity
W = mgh ,
= 15 × 9.8 × 2=294 J.
Apart from the work to be done against. Friction while moving through a distance of 2m.
So, the work done in case (b) is more than that of case (a).

Work, Energy and Power Class 11 Important Questions Long Answer Type

Question 1.
Two bodies of masses m₁and m₂ are moving in straight line with velocities
u₁ and u₂ respectively, collide each other. After collision calculate the velocities of the bodies. What will happen ¡f:
(i) The masses of the bodies are equal.
(ii) One of the body is at rest.
(iii) Lighter body is at rest.
(iv) Heavier body is at rest.
Answer:
Consider two bodies A and B of masses m₁and m₂ moving with velocities u₁ and u₂ (u₁> u₂) in the same direction.
After collision the bodies move in the same direction with velocities ν₁ and ν₂ respectively.
The collision is perfectly elastic, hence the momentum and K.E. of the system will be
conserved.

As the momentum is conserved:
Momentum before collision = Momentum after collision
m₁u₁ + m₂u₂ = m₁ν₁ + m₂ν₂
or
m₁u₁ – m₁ν₁ = m₂ν₂ – m₂u₂
or
m₁(u₁ – ν₁)=m₂(ν₂ – u₂) …(1)

As the K.E. is conserved :
K.E. before collision K.E. after collision


or
u₁+ ν₁ = ν₂ + u₂
or
u₁ – u₂ = ν₂ – ν₁, =-(ν₁ – ν₂) … (3)
So, relative velocity before collision = – (Relative velocity after collision) i.e., The direction of relative velocity is reversed after collision.

Calculation of velocities after collision :
(i) Velocity of A: From eqn. (3), we have
ν₂ = u₁ – u₂ + ν₁
Putting the value of ν₂ in eqn. (1), we get

(ii) Velocity of B : From eqn. (3), we have
ν₁ = ν₂ + u₂ – u₁
Putting the value of ν₁ in eqn. (1), we get

Case (i) : When m₁ = m₂
Then, from eqn. (4), ν₁ = u₂
and from eqn. (5), ν₂ = u₁
Hence, when the mass of the bodies are equal then they interchange their velocities after collision.

Case (ii) : m₂ is rest i.e., u₂ = 0
Then, from eqn. (4), ν₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)u₁ …(6)
and From eqn. (5), ν₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\)u₁ …(7)
If m₁=m₂=m, then
From eqn. (6), we get ν₁ = 0
and From eqn. (7), ν₂ = u₁
Thus, when the masses are equal and one body is at rest, then after collision second body comes to rest and first body gains the velocity of second.