Category: Class 9

MP Board Class 9th Science Solutions Chapter 8 Motion

Motion Intext Questions

Motion Intext Questions Page No. 100

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes, if an object start from a point and finally, reaches the same point. Then, it has distance but displacement is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer:

Suppose farmer starts from A of the square field.
Now, he covered 10 m in 40 sec.
Now, 2 minutes 20 seconds =140 sec.
So, he cover in 140 seconds = \(\frac { 10 }{ 40 }\) × 140 m
= \(\frac { 140 }{ 4 }\)m
= 35 m.
Now, in square field = 35m
AB + BC + CD + DE (\(\frac { 1 }{ 2 }\)AD)
So, he will reach at E.
Then, his displacement is AE = 5 m.

Question 3.
Which of the following is true for displacement?

1. It cannot be zero.
2. Its magnitude is greater than the distance travelled by the object.

Answer:

1. False
2. False.

Motion Intext Questions Page No. 102

Question 1.
Distinguish between speed and velocity.
Answer:

Question 2.
Under what condition (s) is magnitude of average velocity of an object equal to its average speed?
Answer:
When a body is moving in a straight line in a particular direction.

Question 3.
What does the odometer of an automobile measure?
Answer:
It measures the distance travelled by the vehicle.

Question 4.
What does the path of an object look like when it is in uniform motion?
Answer:
Straight line.

Question 5.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3 × 10⁸ m/s.
Answer:
Time taken by signal to reach ground station = 5 minutes
= 5 × 60 sec.
= 300 seconds.
Speed of the signal = 3 × 10⁸ m/s.
∴ Distance between spaceship and ground station = speed × time
= 3 × 10⁸ × 300
= 9 × 10¹⁰ m.

Motion Intext Questions Page No. 103

Question 1.
When will you say a body is in

1. Uniform acceleration?
2. Non – uniform acceleration?

Answer:

1. A body is in uniform acceleration if its velocity changes uniformly with equal intervals of time. Example: Freely falling object.
2. A body is in non – uniform acceleration if its velocity changes non – uniformly with equal intervals of time. Example: A bus running in a city.

Question 2.
A bus decreases its speed from 80 km/h^-1 to 60 km/h^-1 in 5 s. Find the acceleration of the bus.
Answer:
Initial velocity of the bus, u = 80 km/h
= \(\frac { 80\times 100 }{ 60\times 60 } \) m/s
u = \(\frac { 800 }{ 36} \) m/s
Final velocity of the bus, v = 60 km/h
= \(\frac { 60\times 100 }{ 60\times 60 } \) m/s
v = \(\frac { 600 }{ 36 } \) m/s
Time taken by bus to change its speeds, t = 5 seconds
∴ Acceleration, a = \(\frac { v-u }{ t } \)
= \(\frac { 600 }{ 36 } \) – \(\frac { 800 }{ 36 } \) ÷ 5
= \(\frac { 600-800 }{ 36 } \) ÷ 5
= \(\frac { 200 }{ 36 } \) × \(\frac { 1 }{ 5 } \) m/s²
= \(\frac { -10 }{ 9 } \) m/s²
a = -1.11 m/s².

Question 3.
A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial velocity, u = 0
Final velocity, v = 40 km/h
= \(\frac { 40\times 1000 }{ 60\times 60 } \) m/s
= \(\frac { 400 }{ 36 } \) m/s
= \(\frac { 200 }{ 18 } \) m/s.
Time taken, t = 10 minutes
= 10 × 10 seconds = 100 seconds
a = \(\frac { v-u }{ t } \)
= \(\frac { 200 }{ 18 } \) – 0 ÷ 100
= \(\frac { 200 }{ 18 } \) ÷ 100
= \(\frac { 200 }{ 18 } \) × \(\frac { 1 }{ 100 } \) m/s²
= \(\frac { 1 }{ 9 } \) m/s²
a = 0.018 m/s².

Motion Intext Questions Page No. 107

Question 1.
What is the nature of the distance – time graphs for uniform and non – uniform motion of an object?
Answer:
For uniform motion, it is a straight line.

For non – uniform motion, it is a curved line.

Question 2.
What can you say about the motion of an object whose distance – time graph is a straight line parallel to the time axis?
Answer:
It shows the body at rest.

Question 3.
What can you say about the motion of an object if its speed – time graph is a straight line parallel to the time axis?
Answer:
It means body has constant speed or uniform motion with zero acceleration.

Question 4.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Distance travelled.

Motion Intext Questions Page No. 109 – 110

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
Answer:
Initial velocity, u = 0
Acceleration, a = 0.1 m/s²
Time, t = 2 min = 2 × 60 seconds
= 120 seconds
(a) Velocity, v = ?
We know, v = u + at
v = 0 + 0.1 × 120 m/s
= 12 m/s.

(b) Distance, s = ?
We Know, s = ut + \(\frac { 1 }{ 2 }\) at²
= 0 × 120 + \(\frac { 1 }{ 2 }\) × 0.1 × 120 × 120
= \(\frac { 1 }{ 2 }\) × \(\frac { 0.1 }{ 10 }\) × 120 × 120 m
= 720 m.

Question 2.
A train is travelling at a speed of 90 km/h^-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s². Find how far the train wilt go before it is brought to rest.
Answer:
Initial velocity, u = 90 km/h
= \(\frac { 90\times 5 }{ 18 } \) m/s
= 25 m/s
Final velocity, v = 0
Acceleration, a = -0.5 m/s².
Distance, s = ?
We know,
v² = u² + 2as
0² = (25)² + 2 × (-0.5) × s
0 = 625 – 1 × s
= 625 – 1 × s
= 625 – s
∴ s = 625 m.
∴ Distance travelled by the train = 625 m.

Question 3.
A trolley while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity in 3 s after the start?
Answer:
Acceleration, a = 2 cm/s²
Time, t = 3 sec.
Initial velocity, u = 0
Final velocity, v = ?
Now, v= u + at
v = 0 + 2 × 3
= 6 cm/s.

Question 4.
A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?
Answer:
Acceleration, a = 4 m/s²
Time, t = 10 sec,
Initial velocity, u = 0
Distance, s = ?
Now, s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 10 + \(\frac { 1 }{ 2 }\) × 4 × (10)²
s = \(\frac { 1 }{ 2 }\) × 4 × 100 m
s = 200 m.
So, the distance covered is 200 m.

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 m/s^-1. If the acceleration of the stone during its motion is 10 m/s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Initial velocity, u = 5 m/s
Final velocity, v = 0
Acceleration, a = -10 m/s² (∴ It is from opposite direction)
Height attained, s = ?
Time taken, t = ?
Now, V² = u² + 2 as
(0)² = (5)² + 2 × (-10) × s
0 = 25 – 20s
⇒ 20s = 25
∴ s = \(\frac { 25 }{ 20 }\) = \(\frac { 5 }{ 4 }\) = 1025m
Now, v = u + at
0 = 5 + (-10) × t
0 = 5 – 10 × t
-5 = -10 × t
∴ t = \(\frac { -5 }{ -10 }\) = \(\frac { 1 }{ 2 }\) = 0.5 sec
So, height attained by the stone is 1.25 m and time taken to reach there is 0.5 sec.

Motion NCERT Textbook Exercises

Question 1.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:

Diameter of the track = 200 m
So, radius = \(\frac { 200 }{ 2 }\) m
= 100m
Circumference = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 100 m
= \(\frac { 4400 }{ 7 }\) m.
Now, athlete covers 200 m in 40 s.
So, rounds covered in 40 seconds = 1
Rounds covered in 1 sec = \(\frac { 1 }{ 40 }\)
Rounds covered in 140 sec = \(\frac { 1 }{ 40 }\) × 140
= 3.5 rounds
So, distance covered = 3.5 × circumference
= \(\frac { 3.5 }{ 10 }\) × \(\frac { 4400 }{ 7 }\) m
= 2200 m.
And, displacement after 3.5 rounds
= Diameter of track
= 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:

(a) A to B,
Time taken, t = 2 min, 30 sec
= 150 sec.

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s

= \(\frac { 300 }{ 150 }\) m/s = 2 m/s.

(b) A to C,
Time taken = Time from A to B + Time from B to C
= 2 min 30 sec + 1 min
= 150 sec + 60 sec
= 210 sec.
Total distance covered (AB + BC) = (300 + 100) m
= 400 m
Displacement = AC = (300 – 100) m
= 200 m

= \(\frac { 400 }{ 210 }\) m/s = 1.9 m/s

= \(\frac { 200 }{ 210 }\) m/s = 0.95 m/s.

Question 3.
Abdul, while driving to school, computes the average speed for his trip to be 20 kmh^-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh^-1. What is the average speed for Abdul’s trip?
Answer:
Let the distance between Abdul’s home and school = x km.
Lets time taken from home to school = t₁ hr.
and time taken from school to home = t₂ hr.
Now,

∴ t₁ = \(\frac { x }{ 20 }\) hr.
t₂ = \(\frac { x }{ 30 }\) hr.
Now,

∴ Average Speed = \(\frac { 120 }{ 5 }\) 24 km/h.

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Initial velocity, u = 0
Acceleration, a = 3.0 m/s²
= 3 m/s²
Time, t = 8 sec
Now,
s = ut + \(\frac { 1 }{ 2 }\) at²
s = 0 × 8 + \(\frac { 1 }{ 2 }\) × 3 × (8)²
= 0 + \(\frac { 1 }{ 2 }\) × 3 × 64
s = 96 m.
∴ Distance Travelled by Boat is 96 m.

Question 5.
A driver of a car travelling at 52 kmh^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 34 kmh^-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Answer:
We have,
Speed of first driver = 52 km/h
= \(\frac { 52\times 1000 }{ 60\times 60 } \) m/s = 14.4 m/s
Speed of second driver = 3 km/h
= \(\frac { 3\times 1000 }{ 60\times 60 } \) m/s = 0.83 m/s
Now, distance covered by first driver = Area of ΔAOB
= \(\frac { 1 }{ 2 }\) × OA × OB
= \(\frac { 1 }{ 2 }\) × 14.4 × 5 = 36.1 m
Distance covered by second driver = Area of ΔMON
= \(\frac { 1 }{ 2 }\) × OM × ON
= \(\frac { 1 }{ 2 }\) × O.83 × 10 = 4.16 m
So, first driver covers more distance.

Question 6.
Figure below shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

1. Which of the three is travelling the fastest?
2. Are all three ever at the same point on the road?
3. How far has C travelled when B passes A?
4. How far has B travelled by the time it passes C?

Answer:

1. B is travelling the fastest.
2. No.
3. C was at around 9 km from origin when B passes A.
4. B travelled around 5.5 km when it passes C.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Distance, s = 20 m
Initial velocity, u = 0
Acceleration, a = 10 m/s²
Now, s = ut +\(\frac { 1 }{ 2 }\) at²
20 = 0 × t + \(\frac { 1 }{ 2 }\) × 10 × t²
20 = 5 × t²
∴ t² = \(\frac { 20 }{ 5 }\) = 4
∴ t = 2 sec.
Also, v = u + at
v = 0 + 10 × 2
y = 20 m/s.
So, it will strike the ground after 2 seconds with the velocity of 20 m/s.

Question 8.
The speed – time graph for a car is shown in figure below:

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a) The distance travelled by car in first 4 seconds
= Area of triangular like region OAB
= \(\frac { 1 }{ 2 }\) × AB × OB = \(\frac { 1 }{ 2 }\) × 6.9 × 4
= 12 m (approximately).

(b) Part of the graph after 5.5 seconds shows the uniform motion.

Question 9.
State which of the following situations are possible and give an example for each of these:

1. an object with a constant acceleration but with zero velocity.
2. an object moving in a certain direction with an acceleration in the perpendicular direction. Both are possible.

Answer:

1. Freely falling.
2. Motion of an object in a circular path.

Question 10.
An artificial, satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Radius of the orbit = 42250 km = 42250 × 1000 m.
Time taken to complete one revolution = 24 hours.
= 24 × 60 × 60 sec
Distance covered by satellite to complete 1 revolution = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 42250 × 1000 m

Motion Additional Questions

Motion Multiple Choice Questions

Question 1.
Deceleration of a body is expressed in _______ .
(a) m
(b) ms^-1
(c) ms^-2
(d) -ms^-2
Answer:
(c) ms^-2.

Question 2. A car goes from town A to another town B with a speed of 40 km/h and returns back to the town A with a speed of 60 km/h. The average speed of the car during the complete journey is _______ .
(a) 48 km/h
(b) 50 km/h
(c) Zero
(d) None of these.
Answer:
(a) 48 km/h

Question 3.
A ball is thrown vertically upwards. It rises to a height of 50 m and comes back to the thrower, _______ .
(a) The total distance covered by the ball is zero.
(b) The net displacement of the ball is zero.
(c) The displacement is 100 m.
(d) None of the above.
Answer:
(b) The net displacement of the ball is zero.

Question 4.
The SI unit of velocity is _______ .
(a) ms^-1
(b) ms^-2
(c) ms^-3
(d) Nm^-1.
Answer:
(a) ms^-1

Question 5.
In 12 minutes a car whose speed is 35 km/h travels a distance of _______ .
(a) 7 km
(b) 3.5 km
(c) 14 km
(d) 28 km.
Answer:
(a) 7 km

Question 6.
A 50 m long train passes over a 250 m long bridge at a velocity of 60 km/h. How long will it take to pass completely over the bridge?
(a) 18 s
(b) 20 s
(c) 24 s
(d) none of these.
Answer:
(a) 18 s

Question 7.
The initial velocity of a train which is stopped in 20 s by applying brakes (retardation due to brakes being 1.5 ms^-2) is _______ .
(a) 30 ms^-1
(b) 30 cm^-1
(c) 20 cm^-1
(d) 24 ms^-1.
Answer:
(a) 30 ms^-1

Question 8.
A wooden slab starting from rest, slides down a 10 m long inclined plane with an acceleration of 5 ms². What would be its speed at the bottom of the inclined plane?
(a) 10 ms^-1
(b) 12 ms^-1
(c) 10 cm s^-1
(d) 12 cm s^-1.
Answer:
(a) 10 ms^-1

Question 9.
m/s² is the SI unit of _______ .
(a) Distance
(b) Displacement
(c) Velocity
(d) Acceleration.
Answer:
(d) Acceleration.

Question 10.
A car increases its speed from 20 km/h to 50 km/h in 10 seconds. Its acceleration is _______ .
(a) 30 m/s²
(b) 3 m/s²
(c) 18 m/s²
(d) 0.83 m/s².
Answer:
(d) 0.83 m/s².

Motion Very Short Answer type Questions

Question 1.
Define uniform motion.
Answer:
When an object covers equal distances in equal intervals of time: it is said to be in uniform motion.

Question 2.
Define speed.
Answer:
It is defined as the distance travelled by an object in unit time. Its unit is m/s.

Question 3.
Define average speed.
Answer:
The total distance travelled by an object divided by the total time taken.

Question 4.
Define velocity.
Answer:
Velocity is the speed of an object moving in definite direction.

Question 5.
Define acceleration.
Answer:
Change in the velocity of an object per unit time.
a = \(\frac { v-u }{ t } \), S.I. unit is m/s².

Motion Short Answer type Questions

Question 1.
While plotting a distance – time graph, why do we plot time on x – axis.
Answer:
While plotting a distance – time graph, we should plot time on x – axis as, all independent quantities are plotted on x – axis.

Question 2.
What is odometer?
Answer:
Odometer is a device fitted in the automobiles to show the distance travelled.

Question 3.
Draw a distance – time graph that represents uniform speed.
Answer:

Question 4.
With the help of distance – time graph show that the object is stationary.
Answer:

Question 5.
What does the area under velocity – time graph represent?
Answer:
The area under the velocity – time graph represents the displacement.

Question 6.
A body travels a distance from A to B, its physical quantity is measured to be -15 m/s. Is it speed or velocity? Give reason for the answer.
Answer:
The physical quantity is -15 m/s. It represents velocity because sign of speed cannot be a negative. Negative sign indicates the opposite direction.

Question 7.
What type of velocity – time graph will you get for a particle moving with a constant acceleration?
Answer:
For a particle moving with a constant acceleration velocity – time graph will be a straight line inclined to x – axis.

Question 8.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

Question 9.
What is deceleration?
Answer:
When the speed of an object decreases, the object is said to be decelerating. In this case, the direction of acceleration is opposite to the velocity of the object.

Question 10.
What conclusion do you draw about acceleration of the particle in motion from given velocity – time graphs?

Answer:
(a) No acceleration.
(b) Uniform acceleration.
(c) Non – uniform acceleration.

Question 11.
With the help of graph, show the uniform acceleration and uniform retardation of a body.
Answer:

• OA – Shows uniform acceleration.
• CB – Shows uniform retardation.

Question 12.
Give difference between acceleration and velocity.
Answer:
table

Motion Long Answer Type Questions

Question 1.
How can we get speed from distance – time graph?
Answer:
Let us assume that an object moves with a uniform speed. The distance – time graph will be a straight line.
Let d₁ be the distance covered in time t₁.
Let d₂ be the distance covered in time t₂.
∴ d₂ – d₁ will be the distance covered in t₂ – t₂.

But, here \(\frac { BC }{ AC }\) = slope of the graph.
Conclusion:
To find the speed from distance – time graph, we can get its slope which is equal to the speed of the object.

Question 2.
Represent a graph that shows the following:
(a) Uniform speed
(b) Non – uniform speed
(c) Stationary object.
Answer:
Time is taken on x – axis as it is independent quantity and all dependent quantities are taken along y – axis.

Question 3.
(a) What is motion?
(b) State types of motion.
(c) Derive the unit for acceleration.
Answer:
(a) Motion: An object is said to be in motion when its position changes with time. We describe the location of an object by specifying a reference point. Motion is relative. The total path covered by an object is said to be the distance travelled by it.

(b) There are two types of motion: Uniform motion and non – uniform motion

• Uniform motion: When an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• Non – uniform motion: Motion where object cover unequal distances in equal intervals of time.

(c) Acceleration, ‘a’ is change in velocity per unit time.

Question 4.
(a) What is uniform circular motion?
(b) An athlete runs on a circular track, whose radius is 50 m with a constant speed. It takes 50 seconds to reach point B from starting point A. Find:

1. the distance covered.
2. the displacement.
3. the speed.

Answer:
(a) When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

(b) Radius = 50 m
Time = 50 s
The distance covered by an athlete A to B i.e., semicircle of the track
∴ Circumference = 2πr
circumference half of = πr.

Question 5.
Derive the equation for velocity – time relation (v = u + at) by graphical method.
Answer:

Let, the initial velocity of a body be, u = OA = CD.
The final velocity of a body be, v = OE = CB
Time, t = OC = AD
AD is parallel to OC
BC = BD + DC
= BD + OA
v = BD + u
∴ BD =v – u …….. (1)
In velocity – time graph, slope gives acceleration.
∴ a = \(\frac { BD }{ AD }\) = \(\frac { BD }{ OC }\) = OC = t
we get, a = \(\frac { BD }{ t }\)
∴ BD = at …….. (2).

Question 6.
A car accelerates uniformly from 20 km/h to 35 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Answer:
u =20 km/h = 5.5 m/s
v =35 km/h = 9.72 m/s
t = 5 s
a = \(\frac { v-u }{ t } \)
= \(\frac { 9.7-5.5 }{ 5 } \)
a = \(\frac { 4.2 }{ 5 } \)
= 0.84 m/s².

(ii) The distance covered by the car s = ut + \(\frac { 1 }{ 2 }\) at²
= 5.5 × 5 + \(\frac { 1 }{ 2 }\)(0.84)(5)²
= 27.5 + \(\frac { 1 }{ 2 }\) × 0.84 × 25
= 27.5 + 10.5
s = 38 m.

Question 7.
Draw graph to show the following:
(a) Uniform acceleration.
(b) Non – uniform acceleration.
(c) Uniform motion.
Answer:

Uniform acceleration – A body travels equal distances in equal intervals of time.

Non – uniform acceleration – A body travels unequal distances in equal intervals of time.

Uniform motion or zero acceleration A body moves with constant speed in same line.

Question 8.
How can you get the distance travelled by an object from its speed – time graph?
Answer:

Suppose an object is moving with a same speed v, the distance covered by this object when it was at (point A) t₁ to t₂ is:
AB = t₂ – t₁
AD = BC = v

= AD . AB
∴ s = Area of rectangle ABCD
∴ To find the distance travelled by a body from its speed – time graph, we need to find the area enclosed by the graph.

Motion Higher Order Thinking Skills (HOTS)

Question 1.
A blacksmith strikes a nail with a hammer of mass 500 g moving with a velocity of 20 ms^-1. The hammer comes to rest in 0.02 s after striking the nail. Calculate the force exerted by the nail on the hammer.
Answer:
m = 500 g = \(\frac { 1 }{ 2 }\) kg
u = 20 ms^-1 , v = 0
t = 0.02 s, F = ?

Therefore, force exerted by nail on the hammer = 500 N.

Motion Value Based Question

Question 1.
A jeep was moving at a speed of 120 km/h on an highway. All of a sudden tyre of the jeep got punctured and burst The driver of the jeep did not apply sudden brakes but he slowed down its speed and finally stopped it at the road side. All the passengers in the jeep were shocked but happy with the driver’s decison.

1. What would have happened if the driver applied sudden brakes?
2. On applying brakes in which direction would the passengers fall?
3. What value of driver is reflected in this act?

Answer:

1. If the driver would have applied sudden brakes, then all the passengers would get hurt and even other vehicles moving on the highway would have collided with this jeep from behind.
2. On applying brake, the passengers would fall in the forward direction.
3. The driver showed the presence of mind, concern and responsible behaviour and quick decision taking capability.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.
The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all Ijbe angles of the quadrilateral.

Solution:
∠A = 3x, ∠B = 5x, ∠C = 9x, ∠D = 13x
In quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
(∴ Sum of all the angles of ♢ is equal to 360°)

3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac{360^{\circ}}{30^{\circ}}\)
Let angle in ratio be x then angles are x= 12°
∠A = 3 x 12° = 36°
∠B = 5 x 12° = 60°
∠C = 9 x 12° = 108°
∠D = 13 x 12° = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given
ABCD is a parallelogram in which
AC =DB
To prove:
ABCD is a rectangle.
Proof
In ∆DAB and ∆CBA
DB = CA (given)
AB = BA (common)
AD = BC (∴ opposites sides of ∥^gm are equal)
∆DAB = ∆CBA (by SSS)
and so ∠DAB = ∠CBA
AD∥BC and AB is the transversal (by CPCT)
∴ ∠A + ∠B = 180° (CIA’s)
⇒ ∠A + ∠A = 180° (∴ ∠A = ∠B)
∴ ∠A = 90°
∠A = ∠C = 90°
and ∠B = ∠D = 90°
In ∥^gm ABCD, all the angles are right angles.
ABCD is a rectangle.

Question 3.
Show that if the diagonals ofa quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Given
OA = OC, OB = OD and ∠AOD = 90°
To prove:
ABCD is a rhombus.
Proof:
In ∆AOD and ∆COB

OA = OC (given)
OD = OB (given)
∠1 = ∠2 (V.O.A.’s)
∴ ∆AOD = ∆COB (by SAS)
and so AD = CB (by CPC 7)
∠3 = ∠4 (by CPCT)
∠3 and ∠4 are A.I.A.’s and are equal
∴ AD ∥ PC (proved)
AD = BC
∴ ABCD is a parallelogram
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠1 = ∠5 = 90°
(∴ ∠1 + ∠5 = 180° ⇒ 90° + ∠5 = 180° ∴ ∠5 = 90°)
∴ ∠AOD = ∠COD (by SAS)
and so AD = CD (by CPCT)
In ∥^gm APCD, all the sides are equal.
ABCD is rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given
ABCD is a square.
To prove:

1. AC = BD
2. OA = OC and OB = OD
3. ∠AOD = 90°

Proof:
In ∆DAB and ACBA,
DA = CB (given)
AB = BA (common)
∠A = ∠B (each 90°)
∴ ∆DAB = ∆CBA (by SAS)
and so BD =AC (by CPCT)

2. In ∆AOD and ∆COB, AD = CB(given)

∠4 = ∠5 (A.I.A.’s)
∠6 = ∠7 (A.I.A. ’s)
∆AOD = ∆COB (by ASA)
and so OA=OC , (byCPCT)
OD = OB (byCPCT)

3. In ∆AOD and ∆COD,
AO = CO (proved)
OD = OD (common)
AD = CD (given)
∆AOD = ∆COD (by SSS)
and so ∠1 = ∠3 (byCPCT)
∠1 + ∠3 = 180° (LPA’S)
⇒ ∠1 + ∠1 = 180° (∠1 = ∠3)
⇒ 2∠1 = 180°
∴ ∠1 = \(\frac{180^{\circ}}{2}\)

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:

Give
ABCD is in which
AC = BD
OA = OC
and OB = OD
∠AOD = 90°
To prove
ABCD is a square
Proof:
In ∆AOD and ∆COB,
OA = OC (given)
OD = OB (given)
∠7 = ∠8 (V.CXA.’s)
∴ ∆AOD = ∠COB
and so AD = BC
and ∠3 = ∠1 (byCPCT)
∠3 and ∠1 are A.I.A.’s and are equal
∴ AD ∥ BC
Similarly, AB ∥ CD
∴ ABCD is a parallelogram.
In ∆AOD and ∆COD,
OA = OC (given)
OD = OD (common)
∠7 = ∠9 (each 90°)
∆AOD = ∠COD (by SAS)
AD = CD (byCPCT)
In ∥^gmABCD, adjacent sidesAD = CD
∴ ABCD is a rhombus
In ∆DAB and ∆CBA,
DA = CB (proved)
AB = BA (common)
DB = CA (given)
∆DAB = ∆CBA (bySSS)
∠A = ∠B (by CPCT)
∠A + ∠B = 180° (CIA’s)
2∠A = 180°
∠A =90°
ABCD is a square.

Question 6.
Diagonals AC of a parallelogram ABCD bisects ∠A (see Fig). Show that

1. it bisects ∠C also,
2. ABCD is a rhombus.

Solution:
Given
ABCD is a parallelogram in which ∠1 = ∠2
To prove:

1. ∠3 = ∠4
2. ABCD is a rhombus.

Proof
1. ∠1 = ∠4 (A.I.A.’s) ….(i)
(∴ AD ∥ SC and AC is the transversal)
∠2 = ∠3 (A.I.A.’s) …(ii)
(∴ AB ∥ DC and AC is the transversal)
∠1 = ∠2 (given) …(iii)

From (i), (ii) and (iii), we get
∠4 = ∠3
(ii) From (ii) and (iii), we get,
∠1 = ∠3
In ∆ADC,
∠1 = ∠3
∴ AD = DC (In a A, sides opposites to equal angles are equal) and so ABCD is a rhombus
(∴ In a ∥^gm, if adjacent sides are equal then it is a rhombus)

Question 7.
ABCD is a rhombus, show that diagonal AC biusects ∠A as well as ∠C and diagonal BD biusects ∠B as well as ∠D.
Solution:
Given
ABCD is a rhombus.
To prove
∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8.
Proof:
In ∆ADC and ∆ABC.

AD = AB (Adj. sides of a rhombus)
DC = BC (Adj. sides of a rhombus)
AC = AC (common)
∴ ∆ADC = ∆ABC (by SSS)
so ∠1 = ∠2 (by CPCT)
and ∠3 = ∠4 (by CPCT)
∴ AC bisects ∠A and ∠C
Similarly, ∠5 = ∠6 and ∠7 = ∠8
BD bisects ∠B and ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that

1. ABCD is a square
2. diagonal BD bisects ∠B as well as ∠D.

Solution:
Given
ABCD is a rectangle.
∠1 = ∠2 and ∠3 = ∠4
To prove:

1. ABCD is a square
2. ∠5 = ∠6 and ∠7 = ∠8

Proof:
1. ∠A = ∠C
(∵ Rectangle is ∥^gm and in a ∥^gm opp. angles are equal.)

\(\frac{1}{2}\)∠A = \(\frac{1}{2}\)∠C
∠A = ∠C
∠2 = ∠4
In ∆ABC, ∠2 = ∠4
AB = BC
(In a A, sides opp. to equal angles are always equal)
ABCD is a rectangle in which adjacent sides are equal.
∴ ABCD is a square.

2. In ∆ABD,
AB = AD (∴ ABCD is a square)
∴ ∠5 = ∠7 (∴ In a A, angles opp. to equal sides are equal) ….(1)
AB ∥ DC and BD is the transversal
∴ ∠6 = ∠7 …(2)
AD ∥ BC and BD is the transversal.
∴ ∠5 = ∠8 …(3)
From (1) and (3), we get
∠7 = ∠8
From (1) and (2), we get
∠5 = ∠6
Diagonal BD bisects ∠B as well as ∠D.

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP A
5. APCQ is a parallelogram

Given
ABCD is a parallelogram.
∴ AD = BC, AB= DC and DP = BQ
To prove:

1. ∆APD = ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP
5. APCQ is a parallelogram

Proof:
In ∆APD and ∆CQB
PD = QB (given)
AD = CB (given)
∠2 = ∠1 (AIA’s)
∆APD = ∆CQB (by SAS)
and so AP = CQ (by CPCT)

In ∆AQB and ∆CPD,
AB = CD (given)
∠3 = ∠4 (AIA’s)
BQ = DP (given)
∆AQB = ∆CPD (by SAS)
and so AQ = CP (by CPCT)
In quadrilaterals AQCP,
AP = CQ
AQ = CP
AQCP is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD (see Fig.). Show that

1. ∆APB = ∆CQD
2. AP = CQ

Solution:
Given
ABCD is a ∥^gm in which Ap ⊥ BD and CQ ⊥ BD.
To prove:

1. ∆APB = ∆CQD
2. AP = CQ

Proof:
In ∆APB and ∆CQD,
∠P = ∠Q (each 90°)
∠1 = ∠2 (AIA’s)
AB = CD (given)
∆APB ≅ ∆CQD (byAAS)
and so AP = CQ (by CPCT)

Question 11.
In ∆ABC and ∆DEF, AB = DE, AB ∥ DE, BC = EF and BC ∥ EE. Vertices A, B and C are joined to vertices, D,E and F respectively (see Fig.). Show that

1. quadrilateral ABED is a parallelogram
2. quadrilateral BEFC is a parallelogram
3. AD ∥ CF and AD = CF
4. quadrilateral ACFD is a parallelogram
5. AC = DF
6. ∆ABC = ∆DEF.

Solution:
Given
AB = DE and AB ∥ DE
BC = EF and BC ∥ EF
To prove

1. ABED is a ∥^gm
2. BEFC is a ∥^gm
3. AD ∥ CF and AD – CF
4. ACFD is a ∥^gm
5. AC =DF
6. ∆ABC ≅ ∆DEF

Proof:
1. AB = DE and AB ∥ DE (given)
ABED is ∥^gm
and so AD ∥ BE and AD = BE …(1)

2. BC = EF and BC ∥ FC (given)
BEFC is a ∥^gm
and so BE ∥ CF and BE = CF …..(2)

3. From (1) and (2), we get
AD ∥ CF and AD = CF

4. AD ∥ CF and AD = CF (proved)
ACFD is a ∥^gm

5. and so AC = DF
(In a parallelogram, opp. sides are equal)

6. In ∆ABC and ∆DEF,
AB = DE (given)
BC = EF (given)
AC = DF (proved)
∆ABC = ∆DEF (by SSS)

Question 12.
ABCD is a trapezium in which AB ∥ CD and AD = BC (see Fig.). Show that:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. Diagonal AC = diagonal BF)

[Hint: Extend AB and draw line through C parallel to DA intersecting AB produced at E.]
Solution:
Given
AB ∥ CD, AD = BC
To prove:

1. ∠A = ∠B
2. ∠C = ∠D
3. ∆ABC ≅ ∆BAD
4. diagonal AC = diagonal BD

Construction:
Draw a line CE ∥ DA which intersect AB produced at E.
Proof:
1. In quadrilateral ADCE,
AD ∥ EC (by const)
and AE ∥ DC ( AB ∥ DC)
∴ ADCE is a parallelogram
and so AD = EC (opp. sides of a ∥ are equal)…(i)
AD = BC (given) ….(ii)
From (i) and (ii), we get
BC = EC
In ∆BCE BC = EC (proved)
∠4 = ∠3
(∴ In a ∆, angles opp. to equal sides are equal)
∠2 + ∠3 = 180° (LPA’s) …(iii)
∠1 + ∠4 = 180° (CIA’s) …(iv)
From (iii) and (iv), we get
∠2 + ∠3 = ∠1 + ∠4
∠2 = ∠1 (∠3 = ∠4)
i.e., ∠A = ∠B

2. ∠3 = ∠5 (AIA’s) …(v)
∠6 = ∠4
(∴ ADCE is a ∥^gm and in a ∥^gm opp. angles are equal) …(vi)
∠4 = ∠3 (proved) …(vii)
From (v), (vi) and (vii), we get
∠5 = ∠6
i.e., ∠C = ∠D

3. In ∆ABC and ∆BAD,
AB = BA (common)
BC = AD (given)
∠2 = ∠1 (proved)
∆ABC ≅ ∆BAD (by SAS)
and so AC =BD (by CPCT)

4. diagonal AC = diagonal BD (proved)

Mid Point Theorem:
The line segment joining the mid-points of the sides of a triangle is parallel to the third side and equal to half of it.
Given.
ABC is a A in which D and E are the mid-points of sides AB and AC respectively.
To prove.
DE ∥ BC and DE = \(\frac{1}{2}\) BC
Construction:
Extend DE uptoFsuch that DE = EF. Join CF.
Proof:
In ∆AED and ∆CEE
AE = CE (E is the mid – point of AC)
∠AED = ∠CEF (VOA’s)
DE = FE (By constriction)
∆AED = ∆CEF , (By SAS)
and so ∠DAE = ∠FCE (By CPCT)
AD = CF (By CPCT)
∠DAE and ∠FCE are alternate interior angles and are equal.

AD ∥ FC
⇒ DB ∥ FC
Now, AD = DB and AD = FC
DB = FC
In BCFD, DB ∥ FC and DF = BC
BCFD is a ∥^gm
and so DF ∥ BC and DF = BC
⇒ DF ∥ BC and 2DF = BC
DE ∥ BC and DE = \(\frac{1}{2}\) BC

Converse of mid point theorem:
The line drawn through the mid – point of one side of a triangle and parallel to another side, bisects the third side.
Given
ABC is a A in which D is the mid-point of AB and DE ∥ BC.
To Prove:
E is the mid – point of AC.
Construction:
Mark a point F on AC and join DF.
Proof:
Let E be not the mid – point of AC. Let us assume that F be the mid – point of AC.
Then by mid-point theorem

DF ∥ BC
DE ∥ BC (Given)
From (i) and (ii), we get
DE ∥ DF
But lines DE and DF are intersecting lines, intersecting at D. This is a contradiction. So our supposition is wrong. Hence E is the mid – point of AC.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Given
ABC is a right angle A.
To prove:
AC > AB and AC > BC

Proof:
In ∆ABC
∠B = 90°
∴ ∠A + ∠C = 90° (by ASP)
and so ∠B > ∠A and ∠B > ∠C
AC > BC and AC > AB
(In a A, sides opposite to large angle are always longer).

Question 2.
In Fig. given below, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
Given
∠PRC < ∠QCB To prove AC > AB

Proof:
In ∆ ABC
Exterior angle is equal to sum of two opposite angles
∠PBC = ∠1 + ∠3 and
∠QCB = ∠1 + ∠2
∠QCB > ∠PBC (given)
⇒ ∠1 + ∠2 > ∠1 + ∠3
⇒ ∠2 > ∠3
∴ AC > AB
(∴ In a A, side opposite to larger angle is always longer).

Question 3.
In Fig. below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Given
∠B < ∠A i.e., ∠A > ∠B
∠C < ∠D i.e., ∠D> ∠C
To prove: AD > BC
i.e., BC > AD
Proof:
In ∆OCD
∠D > ∠C
OC > OD
(∴ In a ∆, sides opposite to larger angle are always longer) …(1)
In ∆OBA
∠A > AB
OB > OA
(∴ In a A, sides oppositedo larger angle are always longer) …(2)
Adding (1) and (2), we get
OC + OB > OD + OA
BC > AD

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. below). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Given
AB is the smallest siBe and CD is the longest side
To prove:

1. ∠A > ∠C and
2. ∠B > ∠D

Construction:
Join AC

1. Proof:
In, ∆ABC
BC > AB (∴ AB is the smallest side) ∠1 > ∠3
(∴ In a ∆ angle opposite to longer side is always larger) …..(1)
In ∆ACD
CD > AD (∴ CD is the largest side) ∠2 > ∠4
(∴ In a ∆ angles opposite to longer side are always larger) …(2)
Adding (1) and (2), we get
∠1 + ∠2 > ∠3 + ∠4
∠A > ∠C

2. To prove: AB > AD
Construction:
Join BD
Proof:
In ∆ABD
AD > AB (AB is the smallest side)
∠5 > ∠7 …(3)
In ∆BCD
CD > BC (CD is the longest side)

∠6 > ∠8 …(4)
Adding (3) and (4), we get
∠5 + ∠6 > ∠7 + ∠8
∠B > ∠D

Question 5.
In Fig. below, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Sol.
Given:
PR > PQ
∠1 = ∠2
To prove:
∠PSR > ∠PSQ
Proof:
In ∆PSQ
∠PSR = ∠1 + ∠Q (EAP)
In ∆PSR
∠PSQ = ∠2 + ∠R (EAP)
= ∠1 + ∠R (∠1 = ∠2)
In ∆PQR
PR > PQ (given)
∠Q > ∠R
(∴ In a ∆, angle opposite to longer side is always larger)
Adding ∠1 on both sides
∠Q + ∠l > ∠R + ∠1
∴ ∠PSR > ∠PSQ
(∴ ∠PSR = ∠1 + ∠Q and ∠PSQ = ∠1 + ∠R)

Question 6.
Show that of all the line segments drawn from a give point not on it, the perpendicular line segment is the shortest.
Solution:
Given
Let us consider the ∆PMN such that ∠M = 90°
Since, ∠M + ∠N + ∠P = 180° [Sum of angles of a triangle]

∠M = 90° [PM ⊥ l]
∠N < ∠M
PM < PN …..(1)
Similarly PM < PN₁ …..(2)
PM < PN₂ …..(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l.
Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. below). If AD is extended to intersect BC at P, show that

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. AP bisects ∠A as well as ∠D.
4. AP is the perpendicular bisector of BC.

Solution:
Given
AB = AC
DB = DC
To prove:

1. ∆ABD = ∆ACD
2. ∆ABP = ∆ACP
3. ∠1 = ∠2 and ∠5 = ∠6
4. ∠3 = ∠4 = 90° and BP = PC

Proof
1. In ∆ABD and ∆ACD
AB = AC (given)
BD = CD (given)
AD =AD (common)
∆ABD = ∆ACD (by SSS)
and so ∠1 = ∠2 (by CPCT)

2. In ∆ABP and ∆ACP
AB = AC (given)
∠1 = ∠2 (proved)
AP = AP (common)
∆ABP ≅ ∆ACP (by SAS)
and so ∠3 = ∠4
and BP = CP (by CPCT)

∠1 = ∠2 AP bisects ∠A

3. In ∆DBP and ∆DCP
BP = CP (proved)
∠3 = ∠4 (proved)
DP = DP (common)

∆DBP ≅ ∆DCP (by SAS)
and so ∠5 = ∠6 (by CPCT)
∴ AP bisects ∠D.

4. BP = CP (proved)
∠3 = ∠4 (proved)
∠3 + ∠4 = 180° (LPA’s)
∠3 + ∠3 = 180° (∠3 = ∠4)
2∠3 = 180°
∠3 = 90°
∠3 = ∠4 (each 90°)
and therefore, AP is the perpendicular bisector of BC

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

1. AD bisects BC
2. AD bisects ∠A.

Solution:
Given
AB = AC and AD ⊥ BC
To prove
∠1 = ∠2 and
BD = CD
Proof:
In ∆ABD and ∆ACD
AB = AC
AD = AD
∠3 = ∠4
∆ABD = ∆ADC
and so BD = CD and ∠1 = ∠2

Question 3.
Two sides AS and BC and median AM on one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR (see-Fig. below). Show that:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Solution:
Given
AB = PQ
BC = QR
AM = PN
To prove:

1. ∆ABM ≅ ∆PQN
2. ∆ABC ≅ ∆PQR

Proof:
BC = QR (given)
\(\frac{1}{2}\) BC = \(\frac{1}{2}\)QR
BM = QN

1. In ∆ABM and ∆PQN
AB = PQ (given)
BM = QN (proved)
AM = PN (given)
∆ABM ≅ ∆PQN (by SSS)
and so ∠ABC = ∠PQR (by CPCT)

2. In ∆ABC and ∆PQR
AB = PQ (given)
BC = QR (given)
∠B = ∠Q (proved)
∆ABC ≅ ∆PQR by (SAS)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that triangle ABC is isosceles.

Solution:
Given
∠E = ∠F
BE = CF
To prove
AB = AC
Proof:
In ∆FBC and ∆ECB
BE = CF (given)
∠F = ∠E (each 90°)
BC = CB (common)
∴ ∆FBC = ∆ECB (by RHS)
and so ∠B = ∠C (by CPCT)
In ∆ABC, ∠B = ∠C
AB = AC
(sides opposite to equal angles of a A are equal)
and so ABC is isosceles.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given
AB = AC
∠APB = ∠APC = 90°
To prove:
∠B = ∠C

Proof:
In ∆ABP and ∆ACP
AP = AP (common)
∠APB = ∠APC (each 90°)
AB = AC (given)
∴ ∆ABP = ∆ACP (by RHS)
and ∠B = ∠C (by CPCT)

Theorem 7.6.
If two angles of a triangle are equal, then the sides opposite to them are also equal.
Given
In ∆ABC, ∠C = ∠B

To prove: AB = AC
Construction:
Draw the bisector of ∠A and let it meet BC an D
Proof:
In ∆s ABD and ACD, we have ∠B = ∠C (Given)
∠BAD = ∠CAD (Construction)
AD = AD (Common)
∴ ∆ABD ≅ ∆ACD (AAS Cong. Criterion)
Hence, AB = AC (CPCT)

Theorem 7.7.
If two sides of a triangle are unequal, the longest side has greater angle opposite to it.
Given:
In ∆ABC; AC > AB
To prove: ∠ABC > ∠ACB.

Construction:
Mark point D on AC such that AB = AD. Join BD.
Proof:
In ∆ABD,
AB = AD
∴ ∠1 = ∠2 (Const. As opp. equal sides) ….(1)
But ∠2 is an exterior angle of ABCD.
∠2 > ∠ACB (Exterior Angle Theorem) …(2)
From (1) and (2), we have
∠1 > ∠ACB (Const.)
But ∠ABC > ∠1
∴ ∠ABC > ∠ACB

Theorem 7.8. (Converse of Theorem 7.7)
In a triangle the greater angle has the longer side opposite to it.
Given:
In ∆ABC, ∠ABC > ∠ACB
To prove: AC > AB

Proof:
For ∆ABC, there are only three possibilities of which exactly one must be true.

1. AC = AB
2. AC < AB (iii) AC > AB.

Case 1:
If AC = AB, then ∠ABC = ∠ACB, which is contrary to what is given.
AB ≠ AC

Case 2:
If AC < AB, the longer side has the greater angle opposite to it.
∴ ∠ACB > ∠ABC.
This is also contrary to what is given.

Case 3:
We are left with the only possibility, namely AC > AB which is true.
AC > AB.

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 9 Force and Laws of Motion

Force and Laws of Motion Intext Questions

Force and Laws of Motion Intext Questions Page No. 118

Question 1.
Which of the following has more inertia:

1. a rubber ball and a stone of the same size?
2. a bicycle and a train?
3. a five – rupees coin and a one – rupee coin?

Answer:
As we know, inertia is the calculated value for the mass of the body. It is proportional to mass of the body:

1. Inertia of the stone is greater than that of a rubber ball as mass of a stone is more than the mass of a rubber ball for the same size.
2. Inertia of the train is greater than that of the bicycle. As mass of a train is more than the mass of a bicycle.
3. Mass of a five rupee coin is more than that of a one – rupee coin. Hence, inertia of the five – rupee coin is greater than that of the one – rupee coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kick it towards a player of his own team”. Also, identify the agent supplying the force in each case.
Answer:
Four times:

• First, when a football player kicks to another player. Agent supplying the force: First case – First player. Second when that player kicks the football to the goalkeeper. Agent supplying the force. Second case – Second player.
• Third when the goalkeeper stops the football. Agent supplying the force: Third case – Goalkeeper.
• Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force: Fourth case – Goalkeeper.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When we shake any tree’s branches vigorously some leaves of that tree get detached because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
Due to inertia of motion, we fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest.

1. Case I: Since the driver applies brakes and bus comes to rest. But, the passenger tries to maintain its inertia of motion. As a result, a forward force is exerted on him.
2. Case II: The passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus.

Force and Laws of Motion Intext Questions Page No. 126

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
According to Newton’s third law of motion, a force is exerted by the Earth on the horse in the forward direction while horse pushes the ground in the backward direction. As a result, the cart moves forward.

Question 2.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
According to Newton’s third law of motion, a reaction force is exerted over fireman by the ejecting water in the backward direction when a fireman holds a hose, which is ejecting large amounts of water at a high velocity. As a result of the backward force, the stability of the fireman get affected. Hence, it is difficult for him to remain stable while holding the hose.

Question 3.
From a rifle of mass 4kg, a bullet of mass 50g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Answer:
Given,
Mass of the rifle, m₁ = 4kg
Mass of the bullet, m₂ = 50g = 0.05kg
Recoil velocity of the rifle = v₁
Bullet is fired with an initial velocity, v₂= 35 m/s
Condition:
Initially, the rifle is at rest.
Thus, its initial velocity, v = 0
Total initial momentum of the rifle and bullet system = (m₁+ m₂)v = 0
Total momentum of the rifle and bullet system after firing = m₁v₁ + m₂v₂
= 0.05 × 35 = 4v₁ + 1.75
As per law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
⇒ 4v₁ + 1.75 = 0
v₁ = –\(\frac { 1.75 }{ 4 }\) = -0.4375 m/s
So, the rifle recoils backwards with a velocity of 0.4375 m/s because value is negative.

Question 4.
Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 ms^-1 and 1  ms^-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms^-1. Determine the velocity of the second object.
Answer:
Given,
m₁ = 100g = 0.1kg
m₂ = 200g = 0.2kg
Velocity of m₁ before collision, v₁ = 2 m/s
Velocity of m₂ before collision, v₂ = 1 m/s
Velocity of m₁ after collision, v₃ = 1.67 m/s
Velocity of m₂ after collision = v₄
As per the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Hence,
m₁v₁ + m₂v₂ = m₁v₃+ m₂v₄
Putting values
2(0.1) + 1(0.2) = 1.67(0.1) + v₄(0.2)
0.4 = 0.167 + 0.2v₄
v₄ = 1.165 m/s
Velocity of the second object = 1.165 m/s.

Force and Laws of Motion NCERT Textbook Exercises

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non –  zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, an object may travel with a non – zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of air. The net force on the drop is zero.

Question 2.
When a carpet is beaten with a stick, dust comes out of it Explain.
Answer:
When we beat the carpet with a stick, it comes into motion. But the dust particles continue to be at rest due to inertia and get detached from the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
Due to sudden jerks or due to the bus taking sharp turn on the road, the luggage may fall down from the roof because of its tendency to continue moving in the original direction. To avoid this, the luggage is tied with a rope on the roof.

Question 4.
A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would come to rest.
Answer:
(c) there is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Answer:
Here, u = 0, s = 400m, t = 20 s, a = ?, F = ?.
m = 7 tonnes
= 7 × 1000kg
= 7000kg
⇒ s = ut + \(\frac { 1 }{ 2 }\)at²
400 = (0 + 20) + \(\frac { 1 }{ 2 }\)a(20)²
a = \(\frac { 400\times 2 }{ { 20 }^{ 2 } } \)
∴ a = 2 m/s²
Force,
F = ma
= 7000 × 2 = 14,000 N.

Question 6.
A stone of 1kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?
Answer:
m = 1kg, u = 20 m/s, s = 50m, v = 0, F = ? a = ?.
⇒ v² – u² = 2as
(0)² – (20)² = 2a (50)
∴ – 400 = 100a
= \(\frac { 400 }{ 100 }\) – 4 m/s²
Force of friction, F = m × a
= 1kg × -4 m/s² = -4 N

Question 7.
A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer:
m = 8,000 + 5 × 2,000 = 18,000kg
(a) The net accelerating force,
F = Engine force – friction force
= 40,000 – 5,000 = 35,000 N.

(b) The acceleration of the train,
a = \(\frac { F }{ m }\) = \(\frac { 35,000 }{ 18,000 }\) = \(\frac { 35 }{ 18 }\) = 1.94 ms^-2

(c) The force of wagon 1 on wagon 2
= The net accelerating force – mass of wagon × acceleration
= 35,000 – 2,000 × \(\frac { 35 }{ 18 }\)
= 35,000 – 3888.8 = 31,111.2 N.

Question 8.
An automobile vehicle has a mass of 1500kg. What must be the force between the vehicle and road if the vehicle is stopped with a negative acceleration of 1.7 ms^-2?
Answer:
Here, mass = 1500kg
a = -1.7 ms²
F = m × a
= 1500 × (-1.7)
= -2550 N
The force between the vehicle and the road is 2,550 is, m a direction opposite to the direction of the vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v? Choose correct option.
(a) (mv)²
(b) mv²
(c) \(\frac { 1 }{ 2 }\) × mv²
(d) mv.
Answer:
(d) mv.

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
The cabinet will move with constant velocity only when the net force on it is zero.
Force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.

Question 11.
Two objects, each of mass 1.5kg, ate moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms^-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Here, m₁ = m₂ = 1.5kg,
u₁ = 2.5 ms^-1, u₂ = – 2.5 ms^-1
Let v be the velocity of the combined object after the collision.
By conservation of momentum,
Total momenta after collision = Total momenta before collision
= (m₁ + m₂) v = m₁u₁ + m₂u₂
= (1.5 + 1.5) v = 1.5 × 2.5 + 1.5 × (-2.5)
= 3.0 v = 0
or
⇒ v = 0 ms^-1

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so, the truck does not move.

Question 13.
A hockey ball of mass 200g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Here, m = 200g = 0.2kg,
u = 10 ms^-1,
v = -5 ms^-1
change in momentum = m (v – u)
= 0.2 (- 5 – 10) = -3kg ms^-1.

Question 14.
A bullet of mass 10g travelling horizontally with a velocity of 150 ms^-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
Here m = 10g = 0.01kg,
u = 150 ms^-1,
v = 0, t = 0.03 s
a = \(\frac { v-u\quad }{ t } \) = \(\frac { 0-150\quad }{ 0.03 } \) = -5,000 ms^-1
The distance of penetration of the bullet into the block,
s = ut + \(\frac { 1 }{ 2 }\)at²
150 × 0.03 + \(\frac { 1 }{ 2 }\) × (-5,000) × (0.03)
= 4.5 – 2.25 = 2.25
The magnitude of the force exerted by the wooden block on the bullet
= ma = 0.01 × 5,000 = 50 N.

Question 15.
An object of mass 1kg travelling in a straight line with a velocity of 10 ms^-1 collides with, and sticks to, a stationary wooden block of mass 5kg. Then, they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Here, m₁ = 1kg, u₁ = 10 ms^-1, m₂ = 5kg, u₂ = 0
Let v be the velocity of the combined object after the collision.
Total momentum just before the impact
= m₁u₁ + m₂u₂  = 1 × 10 + 5 × 0 = 10kg ms^-1
Total momentum just after the impact = (m₁ + m₂)v = (1 + 5)v
= 6v kg ms^-1 by conservation of momentum,
6v = 10
= v = \(\frac { 10 }{ 6 }\) = \(\frac { 5 }{ 3 }\) ms^-1
∴ Total momentum just after the impact
= 6 × \(\frac { 5 }{ 3 }\) = 10 ms^-1

Question 16.
An object of mass 100kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Here, m = 100kg,
u = 5 ms^-1,
v = 8 ms^-1,
t = 6 s
Initial momentum, P₁ = mu = 100 × 5 = 500kg ms^-1
Final momentum, P₂ = mu = 100 × 8 = 800kg ms^-1
The magnitude of the force exerted on the object.
F = \(\frac { { P }_{ 2 }-{ P }_{ 1 } }{ t } \) = \(\frac { 800-500 }{ 6 }\) = \(\frac { 300 }{ 6 }\) = 50 N.

Question 17.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
Both, the motorcar and insect experience the equal force and hence, a same change in their momentum. So, we agree with Rahul. But due to smaller mass or inertia, the insect dies.

Question 18.
How much momentum will a dumb – bell of mass 10kg transfer to the floor if it falls from a height of 80cm? Take its downward acceleration to be 10 ms^-2.
Answer:
Here, m = 10kg, u = 0,
s = 80cm = 0.80m,
a = 10 m/s^-2
Let v be the velocity gained by the dumb-bell as it reaches the floor.
As v² – u² = 2as
v² – 0² = 2 × 10 × 0.80 = 16
or
v = 4 ms^-1
Momentum transferred by the dumb-bell to the floor
p = mv = 10 × 4 = 40kg ms^-1

Force and Laws of Motion Additional Questions

Force and Laws of Motion Multiple Choice Questions

Question 1.
Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing.
(b) Its velocity always changes.
(c) It always goes away from the earth.
(d) A force is always acting on it.
Answer:
(c) It always goes away from the earth.

Question 2.
According to the third law of motion, action and reaction ________ .
(a) Always act on the same body.
(b) Always act on different bodies in opposite directions.
(c) Have same magnitude and directions.
(d) Act on either body at normal to each other.
Answer:
(b) Always act on different bodies in opposite directions.

Question 3.
A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to ________ .
(a) Exert larger force on the ball.
(b) Reduce the force exerted by the ball on hands.
(c) Increase the rate of change of momentum.
(d) Decrease the rate of change of momentum.
Answer:
(b) Reduce the force exerted by the ball on hands.

Question 4.
The inertia of an object tends to cause the object ________ .
(a) To increase its speed.
(b) To decrease its speed.
(c) To resist any change in its state of motion.
(d) To decelerate due to friction.
Answer:
(c) To resist any change in its state of motion.

Question 5.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is ________ .
(a) Accelerated
(b) Uniform
(c) Retarded
(d) Along circular tracks.
Answer:
(a) Accelerated

Question 6.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is ________ .
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N.
Answer:
(b) 0 N

Question 7.
Rocket works on the principle of conservation of ________ .
(a) Mass
(b) Energy
(c) Momentum
(d) Velocity.
Answer:
(c) Momentum

Question 8.
A water tanker filled up to \(\frac { 2 }{ 3 }\) of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would ________ .
(a) Move backward
(b) Move forward
(c) Be unaffected
(d) Rise upwards.
Answer:
(b) Move forward

Question 9.
Which of the following represents example(s) of potential energy?
(a) A moving car
(b) A moving fan
(c) A book resting on the table
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c).

Question 10.
Unit of force is ________ .
(a) Ampere
(b) Volt
(c) Joule
(d) Hertz.
Answer:
(c) Joule

Question 11.
Product of mass and acceleration of a body is called ________ .
(a) Acceleration
(b) Work
(c) Power
(d) Energy.
Answer:
(b) Work

Question 12.
Which of the following is correct about energy?
(a) Energy is not required to do work.
(b) Work can be expressed as Force × Displacement.
(c) Unit of power is joule.
(d) Power is the amount of work done per unit time.
Answer:
(c) Unit of power is joule.

Question 13.
An object of mass 3kg is falling from the height of 1m. The kinetic energy of the body will be when it touches the ground ________ .
(a) 29.4 N
(b) 29.4 J
(c) 30 N
(d) 15 J
Answer:
(b) 29.4 J

Question 14.
Two objects with masses 1kg and 9kg, and equal momentum. Calculate the ratios of their kinetic energies ________ .
(a) 3 : 1
(b) 9 : 1
(c) 1 : 1
(d) 1 : 2
Answer:
(b) 9 : 1

Question 15.
Considering air resistance negligible, the sum of potential and kinetic energies of the free falling body would be ________ .
(a) zero
(b) infinite
(c) would decrease
(d) remains fixed.
Answer:
(d) remains fixed.

Force and Laws of Motion Very Short Answer Type Questions

Question 1.
What do we call to the product of mass and velocity of an object?
Answer:
Momentum.

Question 2.
Define inertia.
Answer:
The property by which an object tends to remain in the state of rest or of uniform motion unless acted upon by some force is called inertia.

Question 3.
Which property has S.I. unit kilogram metres per second i.e., 1kg m/s?
Answer:
Momentum.

Question 4.
Give an example of scalar quantity.
Answer:
Mass.

Question 5.
Give an example of vector quantity.
Answer:
Momentum.

Question 6.
Calculate the total momentum of the bullet and the gun before firing.
Answer:
For both, it would be zero because both of them are at rest.

Question 7.
Which force slows down a moving bicycle when we try to stop?
Answer:
The force of friction.

Question 8.
Which kind of force of gravity work when an object is under free fall?
Answer:
Unbalanced force.

Question 9.
Which property of an object resist a change in their state of rest or motion?
Answer:
Inertia.

Question 10.
Which law of Newton is also known as Galileo’s law of inertia?
Answer:
First law.

Question 11.
Is force a vector quantity?
Answer:
Yes.

Question 12.
Which force of motion opposes motion of an object?
Answer:
Force of friction.

Question 3.
When action and reaction forces act on two different bodies, what kind of magnitude they have?
Answer:
Action and reaction forces act on two different bodies but they are equal in magnitude.

Question 14.
When gun moves in the backward direction, which kind of velocity is generated?
Answer:
Recoil.

Question 15.
Which factor of body is dependent on its mass?
Answer:
Inertia of a body depends on its mass.

Force and Laws of Motion Short Answer Type Questions

Question 1.
There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia?
Answer:
Steel has highest inertia because it has greatest density and greatest mass, therefore, it has highest inertia.

Question 2.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer.
Answer:
If the breaks are applied suddenly then, the balls will start rolling in the direction in which the train was moving. Due to the application of the brakes, the train comes to rest but due to inertia the balls try to remain in motion, therefore, they begin to roll. Direction and speed of all balls will not be same because the masses of the balls are not the same, therefore, the inertial forces are not same on both the balls. Thus, the balls will move with different speeds.

Question 3.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Answer:
According to law of conservation of momentum or explanation by Newton’s laws of motion, light rifle will hurt the shoulder more.

Question 4.
A horse continues to apply a force in order to move a cart with a constant speed. Explain why?
Answer:
The force applied by the horse balances the force of friction

Question 5.
Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain.
Answer:
Law of conservation of momentum is applicable to isolated system (no external force is applied). In this case, the change in velocity is due to the gravitational force of earth.

Question 6.
In which of the following conditions work done will be equal to zero?
Answer:
In the absence of any one of the two conditions given below, work done will be equal to zero, that is work is not considered to be executed:

1. Force should act on the object.
2. Object must be displaced.

Question 7.
Define energy and explain its forms.
Answer:

1. Energy: Energy is the capacity of doing work. More the power, more will be energy and vice – versa. For example, a motorcycle has more energy than a bicycle.
2. Forms of energy: There are many forms of energy, such as kinetic energy, potential energy, mechanical energy, chemical energy, electrical energy etc.

Force and Laws of Motion Long Answer Type Questions

Question 1.
Give the formulation of work. In which conditions work can occur?
Answer:
Work = Force × Displacement
or W = F × s
where, W is work ‘F’ is force and ‘s’ is displacement.
If force, F = 0
Therefore, work done, W = 0, s = 0
If displacement, s = 0
Therefore, Work done, W = F × 0 = 0
It proves that, there are two conditions for work to occur or be done:

1. Force should act on the object.
2. Object must be displaced.

Question 2.
Give the conditions when work done become positive and negative.
Answer:
When force is applied in the direction of displacement, the work done is considered as positive.
i.e., W = F × s
When force is applied in opposite direction of displacement, the work done is considered as negative.
i.e., W = -F × s = -Fs.

Question 3.
Explain positive and negative work.
Answer:
1. Positive work:
If a force displaces the object in its direction, then the work done is positive.
Here,
W = Fd
Example:
Motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative work. If the force and the displacement are in opposite directions, then the work is said to be negative.
Here,
W = -Fd.
Example:
If a ball is thrown in upward direction but the force due to earth’s gravity is in the downward direction.

Question 4.
A cyclist moving along a circular path of radius 63m completes three rounds in 3minutes.
1. The total distance covered by him during this time.
2. Net displacement of cyclist.
3. The speed of the cyclist
Answer:
1. Total distance covered
s = 2πr × t
s = 2πr × 3
= 2 × \(\frac { 22 }{ 7 }\) × 63 × 3 = 1188m

2. Displacement = Zero

3. Speed = \(\frac { Distance }{ Time }\)
= \(\frac { 1188 }{ 180 }\)
= 6.6 m/s.

Force and Laws of Motion Higher Order Thinking Skills (HOTS)

Question 1.
As per Newton’s third law, every force is accompanied by equal and opposite force. How then can anything move?
Answer:
According to the Newton’s third law, action and reaction are two equal and opposite forces but they act on different bodies. This make the motion of a body possible.

Question 2.
The passengers travelling in a bus fall ahead when a speeding bus stops suddenly. Why?
Answer:
When the speeding bus stops suddenly lower part of the body, a long with the bus comes to rest while the upper part tends to remain in motion due to inertia of motion. That is why passengers fall ahead.

Question 3.
A player always runs some distance before taking a jump. Why?
Answer:
A player always runs for some distance before taking a jump because inertia of motion helps him to take a longer jump.

Force and Laws of Motion Value Based Question

Question 1.
Sushil saw his karate expert breaking a slate. He tried to break the slate but Sushil’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing such activity.

1. How can a karate expert break the slate without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Sushil’s friend is seen in the above case?

Answer:

1. A karate expert Sushil applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks.
2. To every action, there is an equal and opposite reaction, both act on different bodies.
3. Sushil’s friend showed the value of being responsible and caring friend.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

1. OB = OC
2. AO bisect ∠A.

Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:

1. OB = OC
2. ∠5 = ∠6

Proof:
In ∆ABC,
AB = AC
∠B = ∠C
\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:
Given
∠ABC = ∠ADC
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∠ADB = ∠ADC (given each 90°)
AD = AD (common)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

1. ∆ABE ≅ ∆ACF
2. AB = AC,

i. e., ABC is an isosceles triangle.

Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:

1. ∆ABE = ∆ACE
2. AB = AC

Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Construction: Join AD
Proof:
In ∆ABD and ∆ACD
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
AD = AB
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°

Proof:
AB = AC …..(1)
AB = AD …..(2)
From (1) and (2), we get
AC = AD
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
AC = AD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,

∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac{180^{\circ}}{3}\) = 60°
∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,

AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.

Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.

∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 7 Diversity in Living Organisms

Diversity in Living Organisms Intext Questions

Diversity in Living Organisms Intext Questions Page No. 80

Question 1.
Why do we classify organisms?
Answer:
Classification of organism make it easy to study the millions of organisms on this earth. Similarities among them is the basis to classify them into different classes. Classification makes study easier.

Question 2.
Give three examples of the range of variations that you see in life – forms around you.
Answer:
Variations observed in life are:

1. Size: Organisms vary greatly in size – from microscopic bacteria to elephants, whales and large trees.
2. Appearance: The colour of various animals is quite different. Number of pigments are found in plants. Their body – built also varies.
3. Life time: The life span of different organisms is varied.

Diversity in Living Organisms Intext Questions Page No. 82

Question 1.
Which do you think is a more basic characteristic for classifying organisms?
(a) the place where they live.
(b) the kind of cells they are made of. Why?
Answer:
(a) Different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

(b) The kind of cells they are. made of. Because placement of organism to other destination can create a easy confusion.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

Question 3.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals are very different from each other but main basis to differentiate is “Mode of nutrition’’. Plants are autotrophs. They can make their food own while animals are heterotrophs which are dependent on others for food. Locomotion, absence of chloroplasts etc. also make them different.

Diversity in Living Organisms Intext Questions Page No. 83

Question 1.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time. As per the body design, the primitive organisms which have simple structures are different from those so – called advanced organisms which have complex body structure and organization.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, they are developed from same ancestor once. They have relatively acquired their complexity recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Diversity in Living Organisms Intext Questions Page No. 85

Question 1.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The organisms belonging to kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic. This is the main criterion of their classification.

Question 2.
In which kingdom will you place an organism which is single – celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista.

Question 3.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:
In the hierarchy of classification, “species” will have the smallest number of organisms with a maximum of characteristics in common whereas “the kingdom” will have the largest number of organisms a Arthropoda.

Diversity in Living Organisms Intext Questions Page No. 88

Question 1.
Which division among plants has the simplest organisms?
Answer:
Division thallophyta.

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
1. Pteridophyta: They have inconspicuous or less differentiated reproductive organs. They produce naked embryos called spores.
Examples:

• Ferns
• marsilea
• equisetum, etc.

2. Phanerogams: They have well developed reproductive organs. They produce seeds.
Example:

• Pinus
• cycas
• fir etc.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
Gymnosperm:

1. They are non – flowering plants.
2. Naked seeds not enclosed inside fruits are produced.
3. Examples:
   • Pinus
   • Cedar
   • Fir
   • Cycas etc.

Angiosperm:

1. They are flowering plants.
2. Seeds are enclosed inside fruits.
3. Examples:
   • Coconut
   • Palm
   • Mango etc.

Diversity in Living Organisms Intext Questions Page No. 94

Question 1.
How do poriferan animals differ from coelenterate animals?
Answer:

Question 2.
How do annelid animals differ from arthropods?
Answer:

Question 3.
What are the differences between amphibians and reptiles?
Answer:

Question 4.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:
Most birds have feathers and they possess a beak.Mammals do not have feathers and the beak is also absent. Birds lay eggs. Hence, they are oviparous. Some mammals lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Diversity in Living Organisms NCERT Textbook Exercises

Question 1.
What are the advantages of classifying organisms?
Answer:
Advantages of classification:

1. Better categorization of living beings based on common characters.
2. Easier study for scientific research.
3. Better understanding of human’s relation and dependency on other organisms.
4. Helps in cross breeding and genetic engineering for commercial purposes.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
Gross character will form the basis of start of the hierarchy and fine character will form the basis of further steps of single hierarchy.
Examples:

• Presence of vertebral column in human beings can be taken under vertebrata.
• Presence of four limbs makes them members of Tetrapoda.
• Presence of mammary glands keeps them under mammalia.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
Basis of classification:

1. Number of cells: unicellular or multicellular.
2. Complexity of cell structure: Prokaryote and Eukaryote.
3. Presence or absence of cell wall.
4. Mode of nutrition.
5. Level of organization.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
Major divisions of Kingdom Plantae:

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
In plants, basic structure of their body is a major criteria based on which thallophytes are different from bryophytes. Apart from this, absence or presence of seeds is another important criteria. Gymnosperms and angiosperms are further segregated based on if seeds are covered or not. It is clear that it is the morphological character which makes the basis for classification of plants.

In animal, classification is based on more minute structural variations. So in place of morphology, cytology forms the basis. Animals are classified based on layers of cells, presence or absence of coelom. Further, higher hierarchy animals are classified based on the presence or absence of smaller features, like presence or absence of four legs.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
Vertebrata is divided into two super classes, viz. Pisces and Tetrapoda. Animals of pisces have streamlined body with fins and tails to assist in swimming. Animals of tetrapoda have four limbs for locomotion.
Tetrapoda is further classified into following classes:

1. Amphibia: Amphibians are adapted to live in water and on land. They can breathe oxygen through kin when under water.
2. Reptilia: These are crawling animals. Skin is hard to withstand extreme temperatures.
3. Aves: Forelimbs are modified into wings to assist in flying. Beaks are present. Body is covered with feathers.
4. Mammalia: Mammary glands are present to nurture young ones. Skin is covered with hair. Most of the animals are viviparous.

Diversity in Living Organisms Additional Questions

Diversity in Living Organisms Tissues Multiple Choice Questions

Question 1.
Find out incorrect sentence.
(a) Protista includes unicellular eukaryotic organisms.
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
(c) Both Monera and Protista may be autotrophic and heterotrophic.
(d) Monerans have well defined nucleus.
Answer:
(d) Monerans have well defined nucleus.

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms.
Answer:
(d) Gymnosperms.

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish.
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c) Mollusca
(d) Pisces.
Answer:
(b) Echinodermata

Question 6.
Which among the following animals have pores all over their body?
(a) Porifera
(b) Aves
(c) Mollusca
(d) Pisces.
Answer:
(a) Porifera

Question 7.
Which among the following have chi tin as cell wall?
(a) Sycon
(b) Yeast
(c) Jelly fish
(d) Euplectella.
Answer:
(c) Jelly fish

Question 8.
Which among the following is not a Monocotyledonous plant?
(a) Wheat
(b) Rice
(c) Maize
(d) Gram.
Answer:
(d) Gram.

Question 9.
Which among the following is not a dicotyledonous plant?
(a) Wheat
(b) Sunflower
(c) Mango
(d) Gram.
Answer:
(a) Wheat

Question 10.
An organism with a single cell is called _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(c) Unicellular

Question 11.
The amphibians of the plant is _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(b) Bryophyta

Question 12.
Plant bearing naked seeds are _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Gymnosperm.
Answer:
(d) Gymnosperm.

Diversity in Living Organisms Very Short Answer Type Questions

Question 1.
Name a saprophyte and also tell, why are they called so.
Answer:
Aspergillus: They are called so because they obtain their nutrition from dead and decaying matter.

Question 2.
Why are lichens called dual organisms?
Answer:
Lichens are permanent symbiotic association between algae and fungi. Therefore, they are called dual organisms.

Question 3.
State the phylum to which centipede and prawn belong.
Answer:
Arthropoda.

Question 4.
Name one reptile with four – chambered heart.
Answer:
Crocodile.

Question 5.
Identify kingdom in which organisms do not have well defined nucleus and do not show multicellular body designs.
Answer:
Monera.

Diversity in Living Organisms Short Answer Type Questions

Question 1.
Why do we differentiate organism, give two main basis?
Answer:
Due to variation in various characteristics, we differentiate organism. Two main basis are mode of nutrition and habitat.

Question 2.
Which kingdom generate food on earth and initiate food chain?
Answer:
Plantae.

Question 3.
Which kingdom do not have cell wall to their cell?
Answer:
Animalia.

Question 4.
What do you understand by biodiversity?
Answer:
Biodiversity: The variety of living beings found in a particular geographical area is called biodiversity of that area. Amazon rainforests is the largest biodiversity hotspot in the world.

Question 5.
Why classification is required?
Answer:
Classification is necessary for the study of living beings in easy way. Without proper classification, it would be impossible to study millions of organisms which exist on this earth.

Question 6.
What was the basis of classification of Ancient Greek philosopher Aristotle?
Answer:
Aristotle classified living beings on the basis of their habitat. He classified them into two groups, i.e. those living in water and those living on land.

Question 7.
How can we divide organism on the basis of mode of nutrition ?
Answer:
On this basis, organisms can be divided into two broad groups, i.e. autotrophs and heterotrophs.

Question 8.
Define Monocotyledonous plants. Give examples.
Answer:
Monocotyledonous: There is single seed leaf in a seed. A seed leaf is a baby plant.
Examples:

• wheat
• rice
• maize, etc.

Question 9.
Give example of Dicotyledonous plants.
Answer:
Dicotyledonous plants: Mustard, gram, mango etc.

Question 10.
Give one difference between prokaryotes and eukaryotes
Answer:

1. Prokaryotes: When nucleus is not organized, i.e., nuclear materials are not membrane bound; the organism is called prokaryote.
2. Eukaryotes: When nucleus is organized, i.e., nuclear materials are membrane bound; the organism is called eukaryote.

Question 11.
What is the difference between unicellular and multicellular organism?
Answer:

1. Unicellular organism: An organism with a single cell is called unicellular organism.
2. Multicellular organism: An organism with more than one cell is called multicellular organism.

Question 12.
Write short notes on the following:
(a) Thallophyta
(b) Bryophyta
Answer:
(a) Thallophyta: The plant body is thallus type. The plant body is not differentiated into root, stem and leaves. They are known as algae also.
Examples:

• Spirogyra
• chara
• volvox
• ulothtrix etc.

(b) Bryophyta: Plant body is differentiated into stem and leaf like structure. Vascular system is absent, which means there is no specialized tissue for transportation of water, minerals and food. Bryophytes are known as the amphibians of the plant kingdom, because they need water to complete a part of their life cycle.
Examples:

• Moss
• marchantia.

Question 13.
What are cryptogams and phanerogams?
Answer:
Plant body is differentiated into root, stem and leaf. Vascular system is present. They do not bear seeds and hence are called cryptogams. Plants of rest of the divisions bear seeds and hence are called phanerogams.
Examples:

• Marsilear
• ferns
• horse tails etc.

Question 14.
How gymnosperms are different from angiosperms?
Answer:

1. Gymnosperms: They bear seeds. Seeds are naked i.e., are not covered. The word ‘gymnos’ means naked and ‘sperma’ means seed.
2. Angiosperms: The seeds are covered. The word ‘angios’ means covered. There is great diversity in species of angiosperm.

Question 15.
What is porifera?
Answer:
Porifera: These animals have pores all over their body. The pores lead into the canal system. They are marine animals. Examples:

• Sycon
• Spongilla
• Euplectella, etc.

Question 16.
What is coelenterata?
Answer:
Coelenterata: The body is made up of a coelom (cavity) with a single opening. The body wall is made up of two layers of cells (diploblastic).
Examples:

• Hydra
• jelly fish
• sea anemone, etc.

Question 17.
What is Platyhelminthes?
Answer:
The body is flattened from top to bottom and hence the name platyhelminthes. These are commonly known as flatworms. The body wall is composed of three layers of cells (triploblastic).
Example:

• Planaria
• liver fluke
• tapeworm etc.

Question 18.
What is Nematohelminthes and Annelida?
Answer:
Nematohelminthes: Animals are cylindrical in shape and the body is bilaterally symmetric and there are three layers in the body wall.
Example:

• Roundworms
• pinworms
• filarial parasite (Wuchereria) etc.

Annelida: True body cavity is present in these animals. The body is divided into segments and hence the name annelida.
Example:

• Earthworm
• leech etc.

Question 19.
Explain the followings:
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Protochordata
(e) Chordata.
Answer:
(a) Arthropoda: Animals have jointed appendages which gives the name arthropoda. Exoskeleton is present which is made of chitin. This is the largest group of animals; in terms of number of species.
Examples:

• cockroach
• housefly
• spider
• prawn
• scorpion etc.

(b) Mollusca: The animal has soft body; which is enclosed in a hard shell. The shell is made of calcium carbonate.
Examples:

• Snail
• mussels
• octopus etc.

(c) Echinodermata: The body is covered with spines, which gives the name echinodermata. Body is radially symmetrical. The animals have well developed water canal system, which is used for locomotion.
Examples:

• Starfish
• sea urchins etc.

(d) Protochordata: Animals are bilaterally symmetrical, triploblastic and ceolomate. Notochord is present at least at some stages of life.
Examples:

• Balanoglossus
• herdmania
• amphioxus etc.

(e) Chordata: Animals have notochord, pharyngeal gill slits and post anal tail; for at least some stages of life. Phylum chordata is divided into many sub – phyla; out of which we shall focus on vertebrata.

Diversity in Living Organisms Long Answer Type Questions

Question 1.
What is the different levels of organizations in case multicellular organism?
Answer:
Level of organization: Even in case of multicellular organisms, there can be different levels of organization:
(a) Cellular level of organization: When a cell is responsible for all the life processes, it is called cellular level of organization.

(b) Tissue level of organization: When some cells group together to perform specific function, it is called tissue level of organization.

(c) Organ level of organization: When tissues group together to form some organs, it is called organ level of organization. Similarly organ system level of organization is seen in complex organisms.

Question 2.
“Classification of living organism is based on evolution.” Explain.
Answer:
It is a well – established fact that all the life forms have evolved . from a common ancestor. Scientists have proved that the life begun on the earth in the form of simple life forms. During the course of time, complex organism evolved from them. So, classification is also based on evolution. A simple organism is considered to be primitive while a complex organism is considered to be advanced.

Question 3.
Explain five kingdom classification by Robert Whittaker (1959).
Answer:
Five Kingdom Classification by Robert Whittaker (1959):
This is the most accepted system of classification. The five kingdoms and their key characteristics are given below:

1. Monera: These are prokaryotes; which means nuclear materials are not membrane bound in them. They may or may not have cell wall. They can be autotrophic or heterotrophic. All organisms of this kingdom are unicellular. Examples: bacteria, blue green algae (cyanobacteria) and mycoplasma.

2. Protista: These are eukaryotes and unicellular. Some organisms use cilia or flagella for locomotion. They can be autotrophic or heterotrophic. Examples: unicellular algae, diatoms and protozoans.

3. Fungi: These are heterotrohic and have cell wall. The cell wall is made of chitin. Most of the fungi are unicellular. Many of them have the capacity to become multicellular at certain stage in saprophytic. Some fungi live in symbiotic relationship with other organisms, while some are parasites as well.
Examples:

• Yeast
• penicillum
• aspergillus
• mucor etc.

4. Plantae: These are multicellular and autotrophs. The presence of chlorophyll is a distinct characteristic of plants, because of which they are capable of doing photosynthesis. Cell wall is present.

5. Animalia: These are multicellular and heterotophs. Cell wall is absent. They feed on decaying organic materials.

Diversity in Living Organisms Higher Order Thinking Skills (HOTS)

Question 1.
What are the differences between Platyhelminthes and Nematohelminthes?
Answer:

Question 2.
Differentiate between animals belonging to the Mammalia group and those in the Aves group.
Answer:
Differences between mammals and aves.

Diversity in Living Organisms Value Based Questions

Question 1.
Ashish, a IX class student, was studying chapter, ‘Diversity in Living Organisms’. He thought that all the fungi are harmful as these spoil food and cause various diseases. However, his elder sister Dimple told him that not all fungi are harmful as these are also used in making bread, vitamins and medicines.

1. Name any fungus which is the source of some medicine.
2. Name any fungus which is used in bread making.
3. What value are displayed by Ashish’s sister?

Answer:

1. Pencillium.
2. Yeast.
3. Dimple acted as elder sister and enhanced his younger brother’s scientific knowledge about fungi and their functions.

Question 2.
Coral is getting diminished in all the oceans due to global warming. People in Goa island protects their coral by not allowing people / tourist to take it away.

1. What is the phylum of coral?
2. What is coral made up of?
3. What value of people in Goa island is reflected here?

Answer:

1. Coelenterates is the phylum of coral.
2. It is made up of calcium carbonate.
3. They reflect the value of being responsible citizen, respecting environment and nature.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1

Question 1.
In quadrilateral ABCD, AC = AD and AB bisects ∠A (see below). Show that ∆ABC = ∆ABD. What can you say about BC and BD?

Solution:
Given AC = AD
AB is the bisector of ∠A
i. e., ∠1 = ∠2
To prove: ∆ABC = ∆ABD
Proof:
In ∆ABC and ∆ABD
∠1 = ∠2 (given)
AC = AD (given)
AB = AB (common)
∆ABC ≅ ∆ABD (by SAS)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see below). Prove that

1. ∆ABD = ∆BAC
2. BD = AC
3. ∠ABD = ∠BAC

Solution:
Given AD = BC
∠DAB = ∠CBA
To prove:

1. ∆ABD ≅ ∆BAC
2. BD = AC
3. ∠ABD = ∠BAC

Proof:
1. In ∆ABD and ∆BAC
AD = BC (Given)
∠DAB = ∠CBA (given)
AB = BA (common)
∆ABD = ∆BAC (by SAS)
and so 2. BD = AC (by CPCT)
and so 3. ∠ABD – ∠BAC (by CPCT)

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see below). Show that CD bisects AB.

Solution:
Given: ∠B = ∠A (each 90°)
AD = BC
To prove: OA = OB
Proof:
In ∆OBC and ∆OAD
∠B = ∠A (each 90°)
BC = AD (given)
∠BOC = ∠AOD (V.O.A’s)
∆OBC = ∆OAD (by SAS)
and so OA = OB (by CPCT)

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see below). Show that ∆ABC = ∆CDM.

Solution:
Given: l ∥ m and p ∥ q
To prove:
∆ABC = ∆CDA
Proof:
In quadrilateral ABCD
AB ∥ DC and BC ∥ AD
∴ ABCD is a parallelogram
and so AB = DC [Opposite sides]
BC = AD
In ∆ABC and ∆CDA
AB = CD (proved)
BC = DA (proved)
AC = CA (proved)
∆ABC = ∆CDA (by SSS)

Question 5.
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see below). Show that

1. ∆APB = ∆AQB
2. BP = BQ or B is equidistant from the arms of ∠A.

Solution:
Given
∠1 = ∠2
BQ ⊥ AC
and BP ⊥ AD
To prove:

1. ∆APB ≅ ∆AQB
2. BP = BQ

Proof:
In ∆APB and ∆AQB
∠2 = ∠1 (given)
∠P = ∠Q (each 90°)
AB = AB (common)
∆APB ≅ ∆AQB (byAAS)
BP = BQ (by CPCT)

Question 6.
In Fig. given below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
Given
AC = AE
AB = AD
∠BAD = ∠EAC i.e., ∠1 = ∠2
To prove: BC = DE
Proof:
In ∆ABC and ∆ADE
AB = AD (given)
AC = AE (given)
∠1 = ∠2 (given)
Adding ∠3 on both sides
∠1 + ∠3 = ∠2 + ∠3
∠BAC = ∠DAE
∆ABC = ∆ADE , (by SAS)
and so BC = DE (by CPCT)

Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see below). Show that

1. ∆DAP ≅ ∆EBP
2. AD = BE

Solution:
Given:
AP = BP
∠BAD = ∠ABE
∠EPA = ∠DPB
∠1 = ∠2
To prove:

1. ∆DAP ≅ ∆EBP
2. AD = BE

Proof:
∠1 = ∠2 (given)
Adding ∠3 on both sides
∠1 + ∠3 = ∠2 + ∠3
∠APD = ∠BPE
In ∆DAP and ∆EBP
∠APD = ∠BPE (proved)
AP = BP (given)
∠PAD = ∠PBE (given)
∆DAP ≅ ∆EBP (by ASA)
and so AD = BE (by CPCT)

Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see below). Show that:

1. ∆AMC = ∆BMD
2. ∠DBC is a right angle.
3. ∆DBC = ∆ACB
4. CM = \(\frac{1}{2}\) AB

Solution:
∠C = 90°
AM = BM
DM = CM
To prove:

1. ∆AMC ≅ ∆BMD
2. ∆DBC = 90°
3. ∆DBC ≅ ∆ACB
4. CM = \(\frac{1}{2}\) AB

Proof:
1. ∆AMC and ∆BMD
AM = BM (given)
MC = MD (given)
∠1 = ∠2 (V.O.A’s)
∆AMC = ∆BMD (by SAS)
and so AC = DB and ∠4 and ∠3 (by CPCT)

2. ∠4 = ∠3 (proved) [AIA’s]
∴ DB ∥ AC
AC ∥ BD and BC is the transversal
∠C + ∠B = 180° (C.I.A’s)
∠B = 180° – 90° = 90°

3. In DBC and ∆ACB
DB = AC (proved)
∠B = ∠C (each 90°)
BC = CB (common)
∆DBC = ∆ACB (by SAS)
and so DC = AB (by CPCT)

4. DC =AB (proved)
\(\frac{1}{2}\)DC = \(\frac{1}{2}\)AB
CM = \(\frac{1}{2}\)AB.

Theorem 7.3
AAS (Angle-Angle-Side) Congruence Theorem:
If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.
Given:
In ∆s ABC and DEF, we have

∠A = ∠D
∠B = ∠E
and BC = EF
To prove:
∆ABC = ∆DEF
Proof:
Since the sum of the angles of a triangle is 180°. We have
∠A + ∠B + ∠C = ∠D + ∠E + ∠F
Since ∠A = ∠D and ∠B – ∠E (Given)
∠C = ∠F …..(i)
Now, in ∆ABC and ∆DEF, we have
∠B = ∠E (Given)
∠C = ∠F [From, (i)]
and BC = EF (Given)
∴ ∆ABC ≅ ∆DEF [ASA Cong. Theorem]

MP Board Class 9th Maths Solutions

MP Board Class 9th Science Solutions Chapter 6 Tissues

Tissues Intext Questions

Tissues Intext Questions Page No. 69

Question 1.
What is a tissue?
Answer:
A group of cells that are similar in structure and / or work together to achieve a particular function form a tissue.

Question 2.
What is the utility of tissues in multi – cellular organisms?
Answer:
In multi – cellular organisms, different types of tissues perform different functions. Since a particular group of cells carry out only a particular function, they do it very efficiently. So, multi – cellular organisms possess a specific division of labour.

Tissues Intext Questions Page No. 74

Question 1.
Name types of simple tissues.
Answer:

1. Parenchyma
2. Collenchyma and
3. Sclerenchyma

Question 2.
Where is apical meristem found?
Answer:
At root tip and shoot tip.

Question 3.
Which tissue makes up the husk of coconut?
Answer:
Sclerenchymatous tissue.

Question 4.
What are the constituents of phloem?
Answer:
Phloem is made up of four types of elements: Sieve tubes, companion cells, phloem fibres and the phloem parenchyma.

Tissues Intext Questions Page No. 78

Question 1.
Name the tissue responsible for movement in our body.
Answer:
Muscular, tissue.

Question 2.
What does a neuron look like?
Answer:
A neuron is the unit of nervous tissue. It is elongated in shape and consists of three parts viz., dendrites, cell body and axon.

Question 3.
Give three features of cardiac muscles.
Answer:

1. Cardiac muscle cells are cylindrical, branched and uninucleate.
2. They show rhythmic contraction and relaxation throughout life.
3. They are involuntary muscles.

Question 4.
What are the functions of areolar tissue?
Answer:
Areolar tissue fills space inside the organs, supports internal organs and helps in repair of the tissues.

Tissues NCERT Textbook Exercises

Question 1.
Define the term ‘tissue’.
Answer:
The group of cells similar in structure that work together to achieve a particular function forms a tissue. This group of cells has a common origin.

Question 2.
How many types of elements together make up the xylem tissue? Name them.
Answer:
Xylem is formed of four types of elements.
They are:

1. tracheids
2. vessels
3. xylem parenchyma
4. xylem fibre.

Question 3.
How are simple tissues different from complex tissues in plants?
Answer:
Simple tissues are made of one type of cells which coordinate to perform a common function.
Complex tissues are made of more than one type of cells. All these coordinate to perform a common function.

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.
Answer:

Question 5.
What are the functions of the stomata?
Answer:
Functions of stomata:

1. Exchange of gases, particularly CO₂ and O₂ with atmosphere.
2. Loss of water in the form of vapours during transpiration.

Question 6.
Diagrammatically show the difference between the three types of muscle fibres.
Answer:

Question 7.
What is the specific function of the cardiac muscle?
Answer:
The specific function of cardiac muscle is to contract and relax rhythmically throughout life.

Question 8.
Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site / location in the body.
Answer:

Question 9.
Draw a labelled diagram of a neuron.
Answer:

Question 10.
Name the following:

1. Tissue that forms the inner lining of our mouth.
2. Tissue that connects muscle to bone in humans.
3. Tissue that transports food in plants.
4. Tissue that stores fat in our body.
5. Connective tissue with a fluid matrix.
6. Tissue present in the brain.

Answer:

1. Squamous epithelium.
2. Tendons.
3. Phloem.
4. Adipose tissue.
5. Blood.
6. Nervous tissue.

Question 11.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:

1. Skin: Striated squamous epithelium
2. Bark of tree: Cork (protective tissue)
3. Bone: Connective tissue
4. Lining of kidney tubule: Cuboidal epithelium tissue
5. Vascular bundle: Conducting tissue (xylem and phloem).

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
Parenchyma is found in cortex and pith of both root and stem. When it contains chlorophyll, it is called chlorenchyma, found in green leaves.

Question 13.
What is the role of epidermis in plants?
Answer:
Cells of epidermis form a continuous layer without intercellular spaces. It protects all the parts of the plant.

Question 14.
How does the cork act as a protective tissue?
Answer:
In older plants, the secondary meristem forms on its outerside several layered thick cork or the bark. Cork acts as a protective tissue because:

1. Its cells are dead and compactly arranged without intercellular spaces
2. Its cells also have deposition of suberin in their walls that makes them impervious to gases and water.

Thus, cork protects the underlying tissues from excessive loss of water, adverse external environment and mechanical injuries.

Question 15.
Complete the table:

Answer:

Tissues Additional Questions

Tissues Multiple Choice Questions

Question 1.
The spindle shaped cells, uninucleated and unbranched are present in muscular tissue of _________.
(a) Striated muscles
(b) Smooth muscles
(c) Cardiac muscles
(d) Both (a) and (b).
Answer:
(b) Smooth muscles

Question 2.
The size of the stem increases in the width due to _________.
(a) Apical meristem
(b) Intercalary meristem
(c) Primary meristem
(d) Lateral meristem.
Answer:
(d) Lateral meristem.

Question 3.
The cells of cork are dead and have a chemical in their walls that makes them impervious to gas and water. The chemical is _________.
(a) Lignin
(b) Suberin
(c) Cutin
(d) Wax.
Answer:
(b) Suberin

Question 4.
The tissue that helps in the movement of our body are _________.
(a) Muscular tissue
(b) Skeletal tissue
(c) Nervous tissue
(d) All of the above.
Answer:
(d) All of the above.

Question 5.
The connective tissue that connects muscle to bone is called _________.
(a) Ligament
(b) Tendon
(c) Cartilage
(d) Areolar.
Answer:
(b) Tendon

Question 6.
Cartilage and bone are types of _________.
(a) Muscular tissue
(b) Connective tissue
(c) Meristematic tissue
(d) Epithelial tissue.
Answer:
(b) Connective tissue

Question 7.
A tissue whose cells are capable of dividing and re – dividingis called _________.
(a) Complex tissue
(b) Connective tissue
(c) Permanent tissue
(d) Meristematic tissue.
Answer:
(d) Meristematic tissue.

Question 8.
The tissue that helps in the secretion and absorption and is found in the inner lining of the alimentary canal is?
(a) Ciliated epithelium
(b) Cuboidal epithelium
(c) Squamous epithelium
(d) Columnar epithelium.
Answer:
(d) Columnar epithelium.

Question 9.
The flexibility in plants is due to a tissue called?
(a) Chlorenchyma
(b) Parenchyma
(c) Sclerenchyma
(d) Collenchyma.
Answer:
(d) Collenchyma.

Question 10.
The tissue present in the lining of kidney tubules and ducts of salivary glands is?
(a) Squamous epithelium tissue
(b) Glandular epithelium tissue
(c) Cuboidal epithelium tissue
(d) Columnar epithelium tissue.
Answer:
(d) Columnar epithelium tissue.

Tissues Very Short Answer Type Questions

Question 1.
Name two main groups of plant tissue.
Answer:
Two groups are:

1. meristematic tissues and
2. permanent tissues.

Question 2.
Where will you look for meristematic cells in a plant body?
Answer:
The meristematic cells are present only at the growing regions such as shoot tips, root tips and at the base of the leaves or internodes.

Question 3.
What are simple permanent tissues?
Answer:
Simple permanent tissues consist of one type of cells which look like each other.

Question 4.
Name two types of simple permanent plant tissues.
Answer:
Simple permanent plant tissues are:

1. Parenchyma
2. Collenchyma and
3. Sclerenchyma.

Question 5.
Write four elements of xylem.
Answer:

1. Tracheids
2. Vessels
3. Xylem parenchyma
4. Xylem fibres.

Question 6.
What are the three types of muscle fibres (muscle cells)?
Answer:
Three types of muscle fibres are:

1. Striated muscle (skeletal muscle or voluntary muscle).
2. Unstriated muscle (smooth muscle or involuntary muscle).
3. Cardiac muscle.

Question 7.
What are the various functions of all types of epithelial tissues?
Answer:
Epithelial tissues help in

1. protection
2. absorption
3. excretion
4. exchange of respiratory gases
5. secretion.

Question 8.
What is basement membrane?
Answer:
It is a very thin non – cellular membrane on which cells of epithelial tissue rest. It also separates the epithelial tissue from the underlying tissues.

Question 9.
Of which type of muscles the stomach wall is made of? Whether they are voluntary or involuntary?
Answer:
Smooth or non-striated. They are involuntary muscles.

Question 10.
What is the function of connective tissue?
Answer:
Connective tissue connects different tissues and organs. It provides support to different parts of the body by forming packing around different organs of the body

Question 11.
What are the two main features of connective tissue?
Answer:

1. Cells are loosely spaced and are embedded in matrix.
2. Matrix may be jelly – like, fluid, dense or rigid.

Question 12.
Name one animal whose whole skeleton is made of cartilage.
Answer:
Shark fish.

Question 13.
List seven examples of connective tissue.
Answer:

1. Bone
2. Cartilage
3. Tendons
4. Ligaments
5. Areolar tissue
6. Adipose tissue and
7. Blood.

Question 14.
What are blood platelets?
Answer:
Blood platelets are minute (about 2 to 4μ in diameter), anucleated, disc like bodies. The main function of platelets is to help in clotting of blood.

Question 15.
Name the connective tissue that is found between skin and muscles.
Answer:
Areolar connective tissue fills the space between skin and muscles.

Question 16.
How would you recognize that the slide given to you is of striated muscle? Give only two important points.
Answer:

1. Striated muscles have light and dark bands (which gives it a striped look).
2. Striated muscle fibres are not branched like cardiac muscles.

Question 17.
From which part of a nerve cell, dendrites arise?
Answer:
The cell body.

Question 18.
Name the tissue present in the brain.
Answer:
Nervous tissue (formed of neuron cells).

Question 19.
Which cell is the longest in the body of an animal?
Answer:
Neuron (Nerve cell) which may be upto 1m long (in special cases).

Question 20.
Name a fluid, except blood, which circulates in the body.
Answer:
Lymph.

Question 21.
Why are striated muscles generally called skeletal muscles?
Answer:
In general, striated muscles are attached with the bones. They help in the movement of the bones. So, they are also called skeletal muscles.

Question 22.
Name one muscle which does not fatigue and works throughout life in normal conditions.
Answer:
Cardiac muscles.

Question 23.
How are ligaments different from tendons?
Answer:
Ligaments are elastic connective tissues which attach bones with each other to keep them in then placer Tendons arc less elastic connective tissues which attach, muscles to a bone.

Question 24.
What is the main function of meristematic tissue?
Answer:
The mam function of meristematic tissue is to form continuously new cells for increasing cell number, length and the girth of the plant.

Tissues Short Answer Type Questions

Question 1.
How do tissues formed in multicellular organisms?
Answer:
During development from zygote (formed by the union of egg and sperm), a large number of cells are formed. These cells undergo differentiation to produce various specialised tissues to perform different functions in the living body.

Question 2.
What are the characteristics of plant tissues?
Answer:
Plants are stationary or fixed, they do not need much energy. So, most of the tissues they have are supportive, which provide them with structural strength. Most of these tissues are dead; they do not contain living protoplasm.

Question 3.
What are structural and functional differences between plants and animals?
Answer:
As plants do not move, they do not need much energy. So, most of the tissues they have are dead and supportive. But animals need energy as they move around in search of food, mate and shelter. So, most of the animal tissues are living.

Question 4.
Why do meristematic cells lack vacuoles?
Answer:
Meristematic cells divide frequently to give rise to new cells. So, they need dense cytoplasm and soft cell wall. Vacuoles cause hindrance in cell division as they are full of cell sap and provide turgidity and rigidity to the cell.

Question 5.
What are characteristic structural features of meristematic cells ?
Answer:
Meristematic cells have:

1. Thin cell walls.
2. Abundant or dense cytoplasm and single large nucleus.
3. Spherical, oval, polygonal or rectangular shape.
4. No intercellular spaces between them.
5. Either no vacuoles at all or a few vacuoles.

Question 6.
List any four salient features of meristematic tissue.
Answer:

1. This tissue consists of actively dividing cells.
2. This tissue is present in growing regions of plants.
3. In this tissue, cells are packed closely without intercellular spaces.
4. Cells of this tissue have thin cell walls, dense cytoplasm and prominent nuclei.

Question 7.
What are permanent tissues?
Answer:
The cells of meristematic tissue lose the ability to divide and get differentiated into specialised cells. These differentiated cells form different types of tissues which are known as permanent tissues.
Some examples: permanent tissues are

• parenchyma
• sclerenchyma etc.

Question 8.
With the help of only schematic line diagram, show classification of main permanent tissues.
Answer:

Question 9.
Write the characteristics of collenchyma.
Answer:
The cells in this type of tissue are living, elongated and thickened with cellulose at the corners. There is very little intercellular space. This tissue provides flexibility and mechanical support to plants. It is found in hypodermis of stems and leaf stalks.

Question 10.
What are protective tissues?
Answer:
All parts of plants e.g., leaves, flowers, stems and roots are covered by a single outermost protective layer called epidermis. It consists of rectangular, closely fitted relatively flat cells which lack intercellular spaces. Usually, it is one cell thick and is covered with cutin. Epidermis protects internal tissues of the plant.

Question 11.
Does the outer most protective layer is throughout, continuous? If not, what interruptions are usually foundation leaves and herbaceous parts of the plant?
Answer:
Epidermis is the outermost layer of cells in leaves and green herbaceous stems. It is not a continuous layer as it contain numerous minute pores called stomata.

Question 12.
Write major functions of stomata present in the epidermis.
Answer:
Major functions of stomata:

1. Stomata are essential for exchange of gases – oxygen and carbon dioxide between the plant and the atmosphere.
2. Normally, plants remove excessive water in the form of vapour through stomata. This process is called transpiration.

Question 13.
What is the main function of vascular tissues in plants?
Answer:
Vascular tissues transport:

1. Water and dissolved minerals from toots to various parts of the plant (xylem).
2. Prepared food material from leaves to different plant parts (phloem).

Question 14.
What ire vascular bundles?
Answer:
In plants, complex tissues xylem and phloem altogether constitute a structure called vascular bundle. Their main function is transportation of water, salts and food materials within the plant body.

Question 15.
List four elements of phloem. Which one of them is most important and why?
Answer:
Phloem consists of following four elements:

1. Sieve tubes
2. Companion cells
3. Phloem parenchyma
4. Phloem fibres.

The most important element of phloem is sieve tubes because they translocate food materials from leaves to other parts of plant.

Question 16.
What is muscular tissue? What is their function?
Answer:
1. Muscular tissue: This is a specialised tissue which is composed of contractile, fibre – like cells. This tissue is responsible for movement in our body.

2. Function: The movement of the body or limbs is brought about by contraction and relaxation of contractile proteins present in the muscle cells.

Question 17.
In desert plants, epidermis is covered by waxy substances. Can you think why?
Answer:
Waxy substances which cover epidermis are waterproof and hence, they check loss of water during transpiration. Besides this, they also aid in protection against mechanical injury and invasion by parasitic fungi.

Question 18.
What are the four important types of tissues found in animals?
Answer:
The four animal tissues found in animals are:

1. Epithelial tissues.
2. Connective tissues.
3. Muscular tissues and.
4. Nervous tissue.

Question 19.
State one important function of each of the following:

1. Areolar tissue
2. Cuboidal epithelium.

Answer:

1. Areolar tissue: It fills the space inside the organs, supports internal organs and helps in repair of tissues.
2. Cuboidai epithelium: it forms the lining of kidney tubules and ducts of salivary glands where it provides mechanical support.

Question 20.
What are the two main components of blood? Why is blood considered a type of connective tissue?
Answer:
Blood has two main components:

1. Fluid (liquid) matrix called plasma.
2. Suspended red blood cells (RBCs), white blood cells (WBCs) and platelets.

Blood is considered as a connective tissue because:

1. It has the same Origin as the other connective tissues.
2. It flows to different parts of the body and thus, connects different parts of the body with one another to exchange materials and gases.

Question 21.
Why do blood and lymph called as connective tissues?
Answer:
The fluid tissue (blood and lymph) connects various parts (cells) of the body. It supplies nutritive materials, oxygen and hormones collected from different organs to the body tissues. It also collects waste materials from different body tissues and delivers to excretory organs.

Question 22.
Why is blood called as fluid connective tissue?
Answer:
Blood is considered as a connective tissue because it connects different parts of the body. It contains plasma as matrix in which blood cells remain suspended. Being matrix (plasma) in liquid state, blood is called fluid connective tissue.

Question 23.
Explain how the bark of a tree is formed. How does it act as a protective tissue?
Answer:
As trees grow old, a strip of secondary meristem replaces the epidermis of the stem. Cells on the outside are cut off from this layer. This forms the several layer thick cork or the bark of the tree. Bark is a mass of dead tissue lying in the peripheral region of the plant body as a hard dry covering. Its function is protection. It protects the inner tissues against the attack of fungi and insects, against loss of water by evaporation etc.

Question 24.
What is the difference between voluntary and involuntary muscles? Give example of each.
Answer:
Voluntary muscles are those which are directly under control of our will
e.g.

• skeletal muscle.

They work or contract on our command. Involuntary muscles are not under the command of our will. We cannot control contraction and relaxation of these muscles. They are of two types – smooth muscles and cardiac muscles. Smooth muscles are found in stomach, intestine, iris of eye, in ureters and in the bronchi of lungs. Another type of involuntary muscles that is, cardiac muscles are found in heart.

Tissues Long Answer type Questions

Question 1.
The following diagrams are of L.S. of shoot apex and L.S. of root apex. Observe the diagrams and answer the questions given below:

1. What type of cells shoot apex and root apex consists of?
2. What are the main characteristics of these cells?
   

Answer:

1. Meristematic cells.
   • (a) Cells have big nucleus and dense cytoplasm.
   • (b) These cells lack intercellular spaces and vacuoles.

Question 2.
How many types of meristems are present in plants, on the basis of position?
Answer:
On the basis of location of meristem, it is classified into three types:

1. Apical meristem: It is present at the tip of stem, root and their branches.

2. Intercalary meristem: It is found at the leaf base, above the nodes (i.e. at the base of internodes as in grasses) or below the nodes (i.e. at the upper most region of internode as in mint.)

3. Lateral meristem: Vascular cambium and cork cambium are the examples of lateral meristem. Vascular cambium is found in vascular bundles while cork cambium is found underneath the bark of trees. Both of these cause increase in the girth of plants.

Question 3.
Draw a labelled diagratn of shoot to show location of meristematic tissues in the plant body and write the functions of each of them.
Answer:
Functions:

1. Apical meristem: It increases the length of the stem and the root.
2. Lateral meristem (cambium): It increases the girth of the stems and roots.
3. Intercalary meristem: It is present at the base of the leaves or internode (on either side of the node).

It produces buds which develop in auxiliary branches or auxiliary floral branches.

Question 4.
Explain the structure of parenchyma. What are its major modifications?
Answer:
Parenchyma: It is the basic or fundamental tissue found in plants. Cells of this tissue are thin walled, circular or polygonal. They are living with a nucleus and a vacuole. Intercellular spaces are present between the cells of this tissue. Two modifications of parenchyma are chlorenchyma and aerenchyma.

Chlorenchyma: Sometimes cells of the parenchyma contain chlorophyll and perform photosynthesis. This kind of parenchyma is known as chlorenchyma.

Aerenchyma: In aquatic plants, parenchyma contains big air spaces in between them. Such a parenchyma tissue is known

(a) Parenchyma with intercellular spaces.
(b) Aerenchyma with big spaces

Question 5.
Write the main functions of parenchyma.
Answer:
Functions of the parenchyma:

1. They store food.
2. They provide temporary support by retaining the turgid condition of cells.
3. Chlorenchyma containing chloroplasts prepare food through photosynthesis.
4. In aquatic plants, aerenchyma gives buoyancy to plants and hence they can float easily.

Question 6.
Describe the structure of selerenchyma. Write its major functions.
Answer:
Selerenchyma: Selerenchyma is the chief mechanical tissue of plants. The cells of this tissue are usually long, narrow and pointed at both ends. Due to deposition of lignin, their walls are often very highly thickened. Hence, the lumen of cell cavity is nearly obliterated. They are usually provided with simple pits which may be oblique or straight.
The selerenchyma cells are dead cells and their main function is to provide strength and rigidity to the plant.

Question 7.
What is xylem? Explain its structure. Which one of its component is very important and why?
Answer:
Xylem is a complex plant tissue which transports water and dissolved minerals from roots to all other plant parts.

Structure: Xylem consists of four kinds of cells (also known as elements):
(i) Tracheids: A tracheid is an elongated, hollow cell with its both ends tapering. The walls of these cells are thickened by the deposition of lignin. At certain spots lignin is not present. These spots are termed as pits. The tracheids are dead cells.

(ii) Vessels: These are tube like structures formed by a number of cells placed end to end with their transverse walls dissolved. The side walls of these tubes also have deposition of lignin. The thickening of the walls show various kinds of patterns. They are also dead cells.

(iii) Xylem parenchyma: These are parenchymatous, thin walled, living cells. They help in lateral conduction of water and sap. They also store food.

(iv) Xylem fibres: They are lignified dead fibres which provide mechanical support to plant. The most important element of xylem is vessel because most of the water and minerals are carried upward through this component of xylem.

Question 8.
Describe the structure of phloem.
Answer:
Structure of phloem: The main conducting part of the phloem is sieve tube which is formed of elongated cylindrical cells arranged in vertical rows. The walls between the cells have many minute pores through which food material can pass from one cell to the next. The porous walls between the cells is termed as sieve plate. Each sieve tube is supported by a long parenchymatous cell called the companion cell which helps the sieve tubes in the conduction of food material.

The above figure shows that phloem also contains phloem fibres, which provide mechanical support. Another component of phloem is parenchyma cells which stores food.

Question 9.
What is epithelial tissue?
Answer:
Epithelial Tissue: Epithelial tissue forms covering of entire surface of the body and lines the internal organs. Because of this, epithelial tissue is also called protective tissue. It also forms a barrier to keep different body systems separate. In this tissue cells are closely associated and arranged on a very thin extracellular fibrous basement membrane. Epithelial tissue may be composed of one (simple epithelium) or more layers of cells (compound epithelium). The skin, lining of mouth and alimentary canal, lung alveoli etc. are made of epithelial tissues.

Question 10.
What are the four main functions of epithelial tissue?
Answer:
The four main functions of epithelial tissue are:

1. It forms the outer layer of skin and hence it protects the underlying cells from drying, injury, bacterial and chemical effects..
2. It forms lining of mouth, alimentary canal and other internal organs and thus protects these organs.
3. It helps in absorption of water and other nutrients in alimentary canal.
4. Some of them are greatly specialised and perform secretory function.

Question 11.
What are the categories of epithelial tissue depending upon the shape and function of epithelial cells?
Answer:
Depending upon the shape and function of the cells, epithelial tissue is classified as:

Question 12.
What are the various forms of epithelial tissues? Describe briefly.
Answer:
(i) Squamous epithelium: The cells of this tissue are thin and flat forming a delicate lining. It is found in the alveoli and blood vessels. Tongue, oesophagus and lining of mouth also made up of squamous epithelium.

(ii) Cuboidal epithelium: The cells of this tissue are cube like in appearance. It is found in the lining of kidney tubules and ducts of salivary glands etc.

(iii) Columnar epithelium: The cells of this tissue are more tall and wide, placed side by side. Their nuclei are situated near the bases. They may have finger like . projections – the microvilli on their free surfaces. This tissue usually lives in the internal surface of stomach and intestine.

(iv) Ciliated epithelium: The cells of this tissue are modifications of columnar epithelial cells. They have many small hair like projections called cilia on their free ends. This type of epithelium is found in trachea and oviduct.

(v) Striated squamous epithelium: Cells of this tissue are similar to squamous epithelial cells but in this tissue they are arranged in many layers to prevent wear and tear of parts. In skin, striated squamous epithelium is found.

(vi) Glandular epithelium: In this type of epithelium, cells secrete substances at the epithelial surface. Such type of epithelium is found in glands.

Question 13.
Explain the structure of three types of muscle fibres. Also write the locations where they are found in the body.
Answer:
The following are the three types of muscle cells:
(i) Unstriated muscles: (Also known as smooth, involuntary muscles). This type of muscular tissue consists of spindle – shaped, long uninucleate cells. These muscles are present in alimentary canal, blood vessels, iris of eye, in ureters and bronchi of lungs etc.

(ii) Striated muscles: Also known as voluntary muscles because of their function being in our control or will. This type of muscular cells are long multi – nucleated and, enclosed in a membrane known as sarcolemma. Each fibre has several longitudinal filaments embedded in cytoplasm. These filaments give these muscles striated appearance. These muscles are attached to the skeleton; so they are also called skeletal muscles.

(iii) Cardiac muscles: These muscles are found in heart. They are not under the control of the will. They contract rhythmically and involuntarily throughout life without the sign of fatigue. Structurally they show the characters of both unstriated and striated muscles. They are made up of branched fibres. These fibres are uninucleated and show alternate light and dark bands (striations).

Question 14.
What are three main categories of connective tissue?
Answer:
Categories of Connective Tissue:
(i) Connective tissue: There is a matrix in which generally two types of (white and yellow) fibres are present. In between these fibres some connective tissue cells are present.
Examples of this kind of connective tissues are

• areolar tissue and
• adipose tissue.

(ii) Skeletal tissue: This type of tissue forms the skeleton of an organism. It is of two types – cartilage and bone Cartilage has solid matrix, called chondrin, in which fibres and cells known as chondrocytes are present. Usually, cells are present in clusters of 2 – 3 cells in small spaces called lacunae. Cartilage is found in the regions of pinna, nose, trachea and larynx. In bones, matrix is formed of a protein called ossein impregnated with phosphate and carbonates of calcium and magnesium.

(iii) Fluid tissue: Blood and lymph are examples of fluid connective tissue. These are specialized connective tissues. It consists of liquid matrix with no fibres. In liquid matrix called plasma corpuscles remain suspended. Blood transports food material, gases and other substances to the various parts of the body.

Question 15.
What are fibrous connective tissue?
Answer:
Fibrous connective tissue. It is of two types:
(i) The white fibrous connective tissue: It consists of white, non-elastic, unbranched fibres which unite to form bundles called tendons. Tendons are strong, tough and smooth, rope like structures which serve to attach muscles with the bones.

(ii) The yellow fibrous connective tissue: It also consists of fibres which are fine thread like structures. These fibres are quite elastic. Like white elastic fibres, these fibres also form cords called ligaments. These ligaments connect two bones.

Question 16.
Describe the structure of cartilage and bone.
Answer:
1. Cartilage: It is a solid but semi – rigid and flexible connective tissue. It has large, bluntly angular cartilage cells called chondrocytes. They occur in clusters of 2 or 3 in small spaces (lacunae) scattered in the matrix. Cartilage smoothens bone surfaces at joints and is also present in the nose, ear, trachea and larynx.

2. Bone: Bone is a solid, rigid and strong connective tissue. Its matrix become hard due to the deposition of salts of calcium and phosphorus. Osteocytes or bone cells are present in irregular spaces – lacunae in the matrix, interconnected by fine canals called canaliculi. In this tissue, matrix deposits in concentric rings around narrow longitudinal cavities called Haversian canals. These canal carries blood vessels and nerves.

Question 17.
Explain the structure of a fluid connective tissue.
Answer:
Blood is a fluid connective tissue consists of:

1. Blood plasma: It is the fluid matrix which contains 85% to 90% water, 7% different types of proteins, 0.9% of salts, about 0.1% glucose and a very small amount of hormones, wastes etc. In the plasma, blood corpuscles (cells) are suspended.
2. Blood cells. Three kinds of blood cells are found suspended in the blood plasma.

These are:
(a) Red blood corpuscles (erythrocytes) or RBC
(b) White blood corpuscles (leucocytes) or WBC and
(c) Blood platelets.

(a) Red blood corpuscles (erythrocytes) or RBC:
The red blood corpuscles are biconcave, disc – like cells which are devoid of nucleus. They contain a substance called haemoglobin because of this they appear red in colour. The most important function of the RBC is the transport of oxygen and carbon dioxide.

(b) White blood corpuscles (leucocytes) or WBC: These cells are comparatively larger in size, colourless and irregular in appearance. They are devoid of haemoglobin. They protect our body from diseases by destroying germs.

(c) Blood platelets: These are small, 2 – 4μ in diameter. They are without nucleus. Their main function is to liberate some substances which helps in blood clotting.

Question 18.
What are areolar tissue and adipose tissue? Where are they located?
Answer:

1. Areolar tissue: It is a connective tissue which consists of matrix, several types of cells, collagen and elastin fibres.
2. Location: They are found between the skin and muscles, around blood vessels and nerves and in the bone marrow.
3. Functions: It fills space inside the organs, support internal organs and helps in repair of the tissues.
   
4. Adipose tissue: It consists of cells (adipocytes) which are filled with fat globules. These cells remain scattered in a matrix.
5. Location: It is found below the skin and between internal organs.
6. Function: It stores fats and act as insulator.
   

Question 19.
(i) What is a nervous tissue?
(ii) Draw a well labelled diagram of neuron. (label any 4 parts).
Answer:
(i) Nervous Tissue: Nervous tissue is made up of specialised cells called nerve cells or neurons. Each nerve cell is elongated in shape and consists of three parts: cell body, dendrites and axon.The cell body contains a distinct nucleus and has a special kind of granules, called Nissel’s granules in the cytoplasm. From the cell body a single long part called the axon and many branched parts called dendrites arise. In some nerve cells, the axon has a thick white covering called medullary sheath. The terminal end of axon is without sheath and branched. Each nerve cell receives message through the dendrites and sends them to the axon (see Fig. below).
(ii)

Tissues Higher Order Thinking Skills (HOTS)

Question 1.
Cutting of rose plant is done timely in gardens, but still it regain its length. Give reason.
Answer:
In rose plant, intercalary meristeim is present at the base of leaves or internodes. The cells of these tissues, therefore divide and increase the length of plant.

Question 2.
Why does water hyacinth floats on water?
Answer:
A type of parenchyma called aerenchyma is present in water hyacinth. This encloses a lot of air and makes the plant lighter than water. Therefore, it floats on water.

Tissues Value Based Question

Question 1.
A patient suffering from paralysis was unable to walk. His family members took the utmost care of him:

1. Name two tissues responsible for the movement of body?
2. What are the tissues present in brain and spine?
3. What value of the family members is seen in the above case?

Answer:

1. Two tissues responsible for movement of the body are nervous tissue and muscular tissue.
2. The tissues present in brain and spine are nervous tissues.
3. His family members showed the value of being kind, caring, dutiful and responsible.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In Fig. given below, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
Given
∠SPR = 135°
∠PQT = 110°
To find. ∠PRQ
Calculation:
∠Q = 110°
110° + ∠PQR = 180° (LPA’s)
∠1 = 70°
∠P = ∠1 + ∠PRQ
135° = 70° + ∠PRQ
135° – 70° = ∠PRQ
65° = ∠PRQ

Question 2.
In Fig. given below, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.

Solution:
Given
∠X = 62°, ∠XYZ = 54°
∠OYZ = \(\frac{1}{2}\)∠XYZ
∠OZY = \(\frac{1}{2}\)∠XZY
To find ∠OZY and ∠YOZ
Calculation:
In ∠XYZ
∠X + ∠Y + ∠Z = 180° (ASP)
62° + 54° + ∠Z = 180°
116° + ∠Z = 180°
∴ ∠Z = 64°
In ∠OYZ
∠OYZ + ∠OZY + ∠O = 180° (ASP)
⇒ \(\frac{1}{2}\)∠XYZ + \(\frac{1}{2}\)∠XZY + ∠O = 180°
32° + 27° + ∠O = 180°
59° + ∠O = 180°
∠O = 121°

Question 3.
In Fig. given below, if AS ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
Given
AS ∥ DE,
∠BAC = 35°
∠CDE = 53°
To find. ∠DCE
Calculation:
AB ∥ DE and AE is the transversal.
∠BAE = ∠AED = 35°
In ∆DEC
∠D + ∠E + ∠C = 180°
53° + 35° + ∠C = 180°
88° + ∠C = 180°
∠C = 180° – 88° = 92°

Question 4.
In Fig. given below, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
Given
∠PRT = 40°
∠RPT = 95°
∠TSQ = 75°
To find. ∠SQT
Calculation:
In ∆PRT
∠P + ∠R + ∠T= 180° (ASP)
95° + 40 ° + ∠T= 180°
∠T = 180° – 135° = 45°
∠T = ∠STQ = 45°
In ∆TSQ
∠STQ + ∠S + ∠Q = 180° (V.O.A’S)
45° + 75° + ∠Q = 180° (ASP)
∠Q = 180° – 120° = 60°

Question 5.
In Fig. given below, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:
Given
PQ ⊥ PS
PQ ∥ SR
∠SQR = 28°
∠QRT = 65°
To find, x and y
Calculation:
PQ ∥ SR and QR is the transversal
x + 28° = 65° (A.I.A’s)
x = 65° – 28° = 37°
In ∆PQS
x + y + 90° = 180° (ASP)
37° + y + 90° = 180°
∴ y = 180° – 127° = 53°

Question 6.
In Fig. given below, the side QR to ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Given
∠TQR = \(\frac{1}{2}\)∠PQR
∠TRS = \(\frac{1}{2}\)∠PRS
To prove. ∠QTR = \(\frac{1}{2}\)∠QPR
Proof:
In ∆PQR
∠PRS = ∠P + ∠Q (EAP)
In ∆TQR, ∠TRS = ∠T + ∠TQR (EAP)
\(\frac{1}{2}\)∠PRS = ∠T + \(\frac{1}{2}\)∠PQR
Multiplying both sides by 2
∠PRS = 2∠T+ ∠PQR
From (1) and (2) we get,
∠P + ∠Q = 2∠T + ∠PQR
∠P=2∠T
∠T= \(\frac{1}{2}\)∠P
∠QTR = \(\frac{1}{2}\)∠QPR
Proved.

MP Board Class 9th Maths Solutions