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Class VII Science T1 U2 Force and Motion Textbook Solutions
Unit 2 Force and Motion
Evaluation
I. Choose the best answer.
1. A particle is moving in a circular path of radius r. The displacement after half a circle
would be
a. Zero
b. R
c. 2 r
d. r / 2
Ans: c. 2 r
2. Which of the following figures represent uniform motion of a moving object correctly?
Ans:
3. Suppose a boy is enjoying a ride on a merry go round which is moving with a constant speed of 10 m/s. It implies that the boy is
a. at rest
b. moving with no acceleration
c. in accelerated motion
d. moving with uniform velocity
Ans: c. in accelerated motion
4. From the given v-t graph it can be inferred that an object is
a. in uniform motion
b. at rest
c. in non - uniform motion
d. moving with uniform accelerations
Ans: d. moving with uniform accelerations
5. How can we increase the stability of an object?
a. Lowering the centre of gravity
b. Raising the centre of gravity
c. Increasing the height of the object
d. Shortening the base of the object
Ans: a. Lowering the centre of gravity
II. Fill in the blanks.
1. The shortest distance between two places is ___________.
Ans: displacement
2. The rate of change of velocity is_________.
Ans: acceleration
3. If the velocity of an object increases with respect to time, then the object is said to be in__________ acceleration.
Ans: positive
4. The slope of the speed–time graph gives ___________.
Ans: acceleration
5. In ___________ equilibrium, the centre of gravity remains at the same height when it
is displaced.
Ans: neutral
III. Match the following.
IV. Analogy
1. Velocity : metre / second :: Acceleration : meter / second2.
2. Length of scale : metre :: Speed of aeroplane : knot. mile .
3. Displacement / Time : Velocity :: Speed / Time : Acceleration.
V. Answer very briefly.
1. Asher says all objects having uniform speed need not have uniform velocity. Give reason.
We can consider uniform circular motion as an example.
In this motion, speed will be at constant but as the direction changes continuously, its velocity can not be uniform.
Example.
A satellite moves at constant speed but its velocity constantly changes, as its direction is always changing.
2. Saphira moves at a constant speed in the same direction. Rephrase the same sentence in fewer words using concepts related to motion.
Saphira moves with uniform velocity.
3. Correct your friend who says that acceleration gives the idea of how fast the position changes.
Acceleration gives the idea of change in speed or velocity of an object over a period of time.
VI. Answer briefly.
1. Show the shape of the distance – time graph for the motion in the following cases.
a. A bus moving with a constant speed.
b. A car parked on a road side.
2. Distinguish between speed and velocity.
3. What do you mean by constant acceleration?
The velocity of an object changes(increase or decrease) by a constant magnitude over a fixed or same interval of time.
4. What is centre of gravity ?
The centre of gravity of an object is the point through which the entire weight of the object appears to act.
VII. Answer in detail.
1. Explain the types of stability with suitable examples.
Stability is a measure of the body’s ability to maintain its original position. Three types of stability are:
a. Stable equilibrium
b. Unstable equilibrium
c. Neutral equilibrium
Stable Equilibrium
In stable equilibrium, the frustum can be tilted through quite a big angle without toppling.
Its centre of gravity is raised when it is displaced.
The vertical line through its centre of gravity still falls within its base. So, it can return to its original position
Unstable Equilibrium
In this equilibrium, the frustum will topple with the slightest tilting. Its centre of gravity is lowered when it is displaced.
Here, the vertical line through its centre of gravity falls outside its base. So, it will not come back to its position.
Neutral Equilibrium
It causes frustum to topple.
The frustum will roll about but does not topple.
Its centre of gravity remains at the same height when it is displaced.
The body will stay at any position to which it has been displaced.
2. Write about the experiment to find the centre of gravity of the irregularly shaped
Plate.
Aim: To determine the centre of gravity of an irregular shaped object
Apparatus: Nail,Cardboard plate, Pendulum ball, string, Rod, Pencil.
Procedure:
Punch 3 small holes near the edge of the cardboard plate using a nail.
Ensure that the holes are big enough to rotate freely on the rod
Place the rod horizontally and make sure that it's stable.
Hang the cardboard plate on the rod at the location of the holes you made and tie a string attached to a pendulum ball on the rod.
When the string becomes stable, make a point behind the string.
Draw a line from the hole to the point you made.
Repeat this procedure with the other holes.
The point where the 3 lines intersect each other is the center of gravity of your irregularly shaped plate.
VIII. Numerical problems.
1. Geetha takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the distance between her house and the school.
Time taken to reach her house =15min
As the speed is given in seconds, convert 15 minutes to seconds.
1 min = 60 sec
15 min = 15 X 60 sec = 900sec
Distance = ?
Distance = Speed X Time
= 2 m/s X 900 sec
= 1800 m
Distance between Geetha’s house and her school = 1800m
2. A car starts from rest and it is travelling with a velocity of 20 m /s in 10 s. What is its acceleration?
Initial velocity = 0 m/s
Final velocity = 20 m/s
Time = 10s
Final velocity - Initial velocity
Acceleration =--------------------------------
Time
20 - 0
= --------
10
= 2 m/s2
Acceleration of the car = = 2 m/s2
3. A bus can accelerate with an acceleration of 1 m / s2. Find the minimum time for the
bus to attain the speed of 100 km / s from 50 km / s.
Acceleration of the bus (a) = 1 m/s2
Initial velocity (u) = 50 km/s
By converting km to m, it can be rewritten as 50 x 103m/s
Final velocity (v) = 100 km/s
By converting km to m, it can be rewritten as 100 x 103m/s
v – u
Time (t) = ------
t
100 x 103 – 50 x 103
= -----------------------------------------
1
(100 -50) x 103
= ---------------
1
Minimum time required to reach the speed of 100km = 50 x 103s
