1 Mark Questions
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.
Answer : The SI unit of momentum of the body is newton-second (N-s).
2 Marks Questions
Answer : where, T is tension in the string and w is weight of the ball, T=mg or T=w.
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.
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.
Answer : i) i.e., F‘ = -Kx
[x = displacement, K= spring constant]
ii) Spring balance works on the above principle.
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.
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.
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.
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.
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)
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
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.
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.
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
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.
Answer : The two differences between mass and weight of the body are
|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).|
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
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
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