### 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/s^{2})

**Answer : **Given, Load L = 150 N

Load arm d_{L} = 0.4 m, Effort arm d_{E} = ?

Effort E = 60 N

By principle of moments,

L x d_{L} = E x d_{E}

**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/s^{2} 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/s^{2} = 10^{3} g x 10^{2} cm/s^{2}

(1 kg = 10^{3} g, 1 m = 10^{2} cm)

Therefore, 1 N = 10^{5} 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 = W_{M}

Therefore, Load arm d_{L} = 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 d_{E} = 60 – 50 = 10

Let the weight of scale be w.

By principle of moments,

L x d_{L} = E x d_{E}

W_{M }x 30 = *w* x 10

*w* = 3W_{M}

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

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). |

**Answer : **Given, Weight of I body *w _{1}* = 2.0 N

Weight of II body

*w _{2}* = 2.0 kgf = 2 x 10 = 20 N

Mass of I body *m _{1} = ?*

Mass of II body *m _{2} = ?*

*w _{1} = m_{1}g*

=> *m _{1}= w_{1}/g = 2/10*

= 0.2 kg

*w _{2} = m_{2}g*

*m _{2} = w_{2}/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/s^{2}]

= 12 x 9.8 = 117.6 N

117.6 is the force acting on the body.

### 4 Marks Question

**Answer : **Given, Forces *F _{1} = F_{2} =* 5 N

Distance *d _{1} = d_{2} *= 50 cm = 0.5 m

Resultant moment = Torque τ = *?*

*τ _{1} = F_{1} x d_{1} = 5 x 0.5 = 2.5* N-m

*τ _{2} = F_{2} x d_{2} = 5 x 0.5 = 2.5* N-m

*Therefore τ = τ _{1} + τ_{2} = 2.5 + 2.5 =* 5 N-m