1 Mark Questions
Question 1 : What is the weight of a body placed at the centre of the earth?
Answer : The weight of a body placed at the centre of the earth is zero, because the value of g at the centre of earth is zero.
Question 2 : State the SI unit of the momentum of the body.
Answer : The SI unit of momentum of the body is newton-second (N-s).
2 Marks Questions
Question 3 : A metallic ball is hanging by a string from a fixed support. Draw a neat labelled diagram showing the forces acting on the ball and the string.
Answer : where, T is tension in the string and w is weight of the ball, T=mg or T=w.
Question 4 : A force is applied on
i) a rigid body and
ii) a non-rigid body.
How does the effect of the force differ in the above two cases?
Answer : A force when applied on a rigid body does not change the inter-spacing between its constituent particles and therefore, it does not change the dimensions of the object, but causes motion in it. On the other hand, a force when applied on a non-rigid body, it changes the inter-spacing between its constituent particles and therefore, causes a change in its dimensions.
Question 5 : i) Where is the centre of gravity of a uniform ring situated?
ii) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’ State whether the statement is true or false.
Answer : i) The centre of gravity of a uniform ring is situated at the centre of ring.
ii) The statement is false, because it changes its position when the body is deformed.
Question 6 : One end of a spring is kept fixed while the other end is stretched by a force as shown in the diagram.
i) Copy the diagram and mark on it the direction of the restoring force.
ii) Name one instrument which works on the above principle.
Answer : i) i.e., F‘ = -Kx
[x = displacement, K= spring constant]
ii) Spring balance works on the above principle.
Question 7 : Give any two effects of a force on a non-rigid body.
Answer : The two main effects of a force on a non-rigid body are:
i) It can change the state of rest or of motion of the body.
ii) It can change the size or shape of the body.
Question 8 : A boy of mass 30 kg is sitting at a distance of 2 m from the middle of a see-saw. Where should a body of mass 40 kg sit so as to balance the see-saw?
Answer : Let the distance of boy of mass 40 kg from the mean position be x metre. Using principle of moments,
30 x 2 = 40 x x
x = 1.5 m
So, the boy of mass 40 kg should sit at a distance of 1.5 m from the middle of see-saw.
Question 9 : i) What is meant by the term moment of force?
ii) If the moment of force is assigned a negative sign then will the turning tendency of the force be clockwise or anti-clockwise?
Answer : i) The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force.
ii) If moment of the force is negative, then the turning tendency of the force is clockwise.
Question 10 : i) What are non-contact forces?
ii) How does the distance of separation between two bodies affect the magnitude of non-contact force between them?
Answer : i) The forces which are applied from a distance without actually touching the body are known as non-contact forces.
ii) When the distance of separation between the two bodies increases, the magnitude of non-contact force decreases.
Question 11 : i) Define 1 kgf
ii) How is it related to the SI unit of force?
Answer : i)1 kgf is the force with which the earth pulls an object of mass 1kg towards itself.
ii) It is related by, 1 kgf = mg = 9.8 N
(m = 1 kg, (g=9.80 m/s2)
Question 12 : A man can open a nut by applying a force of 150 N by using a level handle of length 0.4 m. What should be the length of the handle, if he is able to open it by applying a force of 60 N?
Answer : Given, Load L = 150 N
Load arm dL = 0.4 m, Effort arm dE = ?
Effort E = 60 N
By principle of moments,
L x dL = E x dE
Question 13 : Where does the position of centre of gravity lie for
i) a circular lamina?
ii) a triangular lamina?
Answer : i) The position of centre of gravity for a circular lamina lie at the centre of the circle.
ii) The position of centre of gravity for a triangular lamina is at the centroid or the point of intersection of the medians.
Question 14 : i) Define one newton.
ii) Write the relation between SI unit and CGS unit of force.
Answer : i) One newton is the amount of force required to produce an acceleration of 1 m/s2 in a body of mass 1 kg.
ii) The relation between SI unit and CGS unit of force is
1 N = 1kg x 1 m/s2 = 103 g x 102 cm/s2
(1 kg = 103 g, 1 m = 102 cm)
Therefore, 1 N = 105 dyne.
Question 15 : A body of mass 1.50 kg is dropped from the 2nd floor of a building which is at a height of 12 m. What is the force acting on it during its fall? (g=9.80 m/s2)
Answer : Given, Mass m = 1.5 kg, Height h = 12 m
Therefore, force acting on the body during its fall
= mg = 1.5 x 9.8 = 14.7 N
Question 16 : A uniform metre scale is kept in equilibrium when supported at the 60 cm mark and a mass M is suspended from the 90 cm mark as shown in the figure. State with reasons, whether the weight of the scale is greater than, less than or equal to the weight of mass M.
Answer : Given, mass = M, weight of M = WM
Therefore, Load arm dL = 90 – 60 = 30 m
Since the weight will act at centre of gravity of scale which is the mid-point of the scale is 50 cm.
Therefore, Effort arm dE = 60 – 50 = 10
Let the weight of scale be w.
By principle of moments,
L x dL = E x dE
WM x 30 = w x 10
w = 3WM
Since, weight of scale is three times that of mass M.
Therefore, weight of scale is greater than weight of M.
Question 17 : Mention any two differences between the mass and the weight of a body.
Answer : The two differences between mass and weight of the body are
| Mass | Weight |
| It is the quantity of matter contained in a body. | It is the force with whcih the centre of the earth attracts a body. |
| It remains constant. | It varies from place to place due to the value of acceleration due to gravity (g). |
Question 18 : The weights of two bodies are 2.0 N and 2.0 kgf respectively. What is the mass of each body? (g=9.80 m/s2)
Answer : Given, Weight of I body w1 = 2.0 N
Weight of II body
w2 = 2.0 kgf = 2 x 10 = 20 N
Mass of I body m1 = ?
Mass of II body m2 = ?
w1 = m1g
=> m1= w1/g = 2/10
= 0.2 kg
w2 = m2g
m2 = w2/g = 2/10
= 2 kg
Question 19 : What is the weight of a body of mass 12 kg? What is the force acting on it?
Answer : Given, Mass m = 12 kg
Weight w = ?
Weight of the body w = m x g
[g = acceleration due to gravity = 9.8 m/s2]
= 12 x 9.8 = 117.6 N
117.6 is the force acting on the body.
4 Marks Question
Question 20 : Two forces each of 5N act vertically upwards and downwards respectively. On the ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the mid-point.
Answer : Given, Forces F1 = F2 = 5 N
Distance d1 = d2 = 50 cm = 0.5 m
Resultant moment = Torque τ = ?
τ1 = F1 x d1 = 5 x 0.5 = 2.5 N-m
τ2 = F2 x d2 = 5 x 0.5 = 2.5 N-m
Therefore τ = τ1 + τ2 = 2.5 + 2.5 = 5 N-m