Samacheer Kalvi 7th Science Term 1 Solutions Chapter 2 Pdf
Tamilnadu Board Samacheer Kalvi 7th Science Term 1 Solutions Chapter 2: Tamilnadu State Board Solution Class 7 Science Term 1 Chapter 2 – Force and Motion.
Samacheer Kalvi 7th Science Term 1 Solutions Chapter 2: Overview
Board |
Samacheer Kalvi |
Class |
7 |
Subject |
Science |
Chapter |
2 |
Chapter Name |
Force and Motion |
Samacheer Kalvi 7th Science Term 1 Solutions Chapter 2 Pdf
Force and Motion
Chapter 2
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
Answer : c. 2 r
Solution : The shortest distance between the two points is known as displacement. The shortest path is the diameter of circle when particle cover half a circle distance.
2.) Which of the following figures represent uniform motion of a moving object correctly ?
Answer :
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
Answer : b. moving with no acceleration
Solution : In uniform circular motion , object moves with constant speed i.e have zero acceleration as magnitude of velocity does not change .
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
Answer : a. in uniform motion
Solution : The uniform motion is defined as an object covers equal distance in equal interval of time.
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
Answer : a. Lowering the centre of gravity
Solution : Condition for Stability i]Lowering its centre of gravity ii]Increasing the area of its base iii] A heavy base lowers the centre of gravity iv] A broad base makes the object more stable.
II.) Fill in the blanks.
1.) The shortest distance between two places is _displacement ____.
2.) The rate of change of velocity is ___acceleration ___
3.) If the velocity of an object increases with respect to time, then the object is said to be in__ positive _ acceleration.
4.) The slope of the speed–time graph gives __acceleration ___.
5.) In _neutral ___ equilibrium, the centre of gravity remains at the same height when it is displaced.
III.) Match the following.
Displacement | Metre |
Light travelling through vaccum | Uniform velocity |
Speed of ship | Knot |
Centre of gravity of geometrical shaped objects | Geometric centre |
stability | Larger base area |
IV.) Analogy
1.) Velocity : metre / second :: Acceleration : __metre/second2____ .
2.) Length of scale : metre :: Speed of aeroplane : __metre/second____ .
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.
Answer : When an object undergoes in uniform circular motion its speed remains constant but its direction change at every point. Therefore, all objects having uniform speed need not have uniform velocity.
2.) Saphira moves at a constant speed in the same direction. Rephrase the same sentence in fewer words using concepts related to motion.
Answer : Saphira moves in uniform straight motion.
3.) Correct your friend who says that acceleration gives the idea of how fast the position changes.
Answer : Velocity gived the idea of how fast the position of an object changes whereas, acceleration shows change in the velocity of an object.
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.
Answer :
2.) Distinguish between speed and velocity.
Answer :
SPEED |
VELOCITY |
Speed is the rate of change
of distance. |
Velocity is the rate of change in displacement. |
Speed = Distance / Time | Velocity (v) = Displacement / Time |
SI unit of speed is metre/second (m/s). | SI unit of velocity is metre / second (m/s). |
3.) What do you mean by constant acceleration?
Answer : An object said to be under constant acceleration when the change (increase or decrease) in its velocity for every unit of time is the same.
4.) What is centre of gravity ?
Answer : 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.
Answer : The measure of the body’s ability to maintain its original position is called stability.
There are 3 types of stability :
I] Stable equilibrium: In stable equilibrium, the frustum can be tilted through quite a big angle without toppling . Eg : when we slightly tilt the bottle-shaped objects without toppling, we can see that the centre of gravity is raised when it is displaced. Observe that the vertical line through its centre of gravity still falls within its base, so that the object can return to its original position, as it has a large base.
II] unstable equilibrium : In this equilibrium, the frustum will topple with the slightest tilting. Its centre of gravity is lowered when it is displaced. Eg: The object will be unstable when it is tilted; the vertical line through its centre of gravity does not fall within its base. Due to the small base of an object, the vertical line through the centre of gravity does not fall within its base.
III] 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.
Answer : Apparatus: Irregularly shaped card, string, pendulum bob, stand
1.) Make three holes in the lamina.
2.) Suspend the lamina from the optical pin through one of the holes as shown in figure.
3.) Suspend the plumbline from the pin and mark the position of the plumbline on the lamina.
4.) Draw lines on the lamina representing the positions of the plumbline.
5.) Repeat the above steps for the other holes.
6.) Label the intersection of the three lines as X, the position of the centre of gravity of the lamina.
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.
Answer : Speed of the bicycle = 2m/s
Time taken = 15 minutes = 15 × 60 seconds = 900 seconds
Speed = (Distance travelled)/ (Time)
Distance = (Speed x Time)
= 2 x 900 =1800m
Also, 1km= 1000m
Therefore, 1800x (1/1000) = 1.8km.
2.) A car starts from rest and it is travelling with a velocity of 20 m /s in 10 s. What is its acceleration?
Answer : A car starts from rest therefore,
Initial velocity of the car (u) = 0 m/s
Final velocity of the car (v) = 20 m/s
time taken = 10s
Acceleration = change in velocity /time
= v—u /t
= 20 – 0/ 10 = 10 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.
Answer : Acceleration = 1m/s2
Initial speed (u) = 100km/s =100000m/s
[ 1km = 1000m ]
Final velocity (v) = 50km/s =50000m/s
Acceleration = change in velocity /time
Time = change in velocity / accleration
= 50000 – 100000/1
= 50000 s
IX.) Fill in the boxes.
SNo. | First Move | Second Move | Distance [m] |
Displacement |
1. | Move 4 metres east | Move 2 metres west | 6 | 2 m east |
2. | Move 4 metres north | Move 2 metres south | 6 | 2 m north |
3. | Move 2 metres east | Move 4 metres west | 6 | 2 m west |
4. | Move 5 metres east | Move 5 metres west | 10 | 0 same place |
5. | Move 5 metres south | Move 2 metres north | 7 | 3 m south |
6. | Move 10 metres west | Move 3 metres east | 13 | 7 m west |