A mass is performing a vertical circular motion (see figure.) If the average velocity of the particle is increased, then at which point the string will break?

1.	A	2.	B
3.	C	4.	D

The bob of a simple pendulum having length \(l,\) is displaced from the mean position to an angular position \(\theta\) with respect to vertical. If it is released, then the velocity of the bob at the lowest position will be:

1. \(\sqrt{2 g l \left(\right. 1 - \cos \theta \left.\right)}\)
2. \(\sqrt{2 g l \left(\right. 1 + \cos\theta)}\)
3. \(\sqrt{2 g l\cos\theta}\)
4. \(\sqrt{2 g l}\)

A ball is dropped from a height of \(5~\text {m}.\) If it rebounds up to a height of \(1.8~\text {m},\) then the ratio of velocities of the ball after and before the rebound will be:
1. \(\dfrac{3}{5}\)

2. \(\dfrac{2}{5}\)

3. \(\dfrac{1}{5}\)

4. \(\dfrac{4}{5}\)

A particle of mass m₁ is moving with a velocity v₁ and another particle of mass m₂ is moving with a velocity v₂. Both of them have the same momentum, but their kinetic energies are E₁ and E₂ respectively. If m₁ > m₂ then: 

1. $\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}$

2. ${\mathrm{E}}_1>{\mathrm{E}}_2$

3. ${\mathrm{E}}_1={\mathrm{E}}_2$

4. ${\mathrm{E}}_1<{\mathrm{E}}_2$

A mass of \(0.5~\text{kg}\) moving with a speed of \(1.5~\text{m/s}\) on a horizontal smooth surface, collides with a nearly weightless spring with force constant \(k=50~\text{N/m}.\) The maximum compression of the spring would be:
                   
1.  \(0.12~\text{m}\) 
2.  \(1.5~\text{m}\) 
3.  \(0.5~\text{m}\) 
4.  \(0.15~\text{m}\) 

If two springs, A and B $(K_A$ $=$ $2$ $K_B),$ are stretched by the same suspended weights, then the ratio of work done in stretching is equal to:
1.  1 : 2
2.  2 : 1
3.  1 : 1
4.  1 : 4

When a spring is subjected to 4 N force, its length is a metre and if 5 N is applied, its length is b metre. If 9 N is applied, its length will be:

1.  4b – 3a

2.  5b – a

3.  5b – 4a

4.  5b – 2a

If a stone is projected vertically upward from the ground at a speed of 10 m/s, then it's: (g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

1.  Potential energy will be maximum after 0.5 s

2.  Kinetic energy will be maximum again after 1 s

3.  Kinetic energy = potential energy at a height of 2.5 m from the ground

4.  Potential energy will be minimum after 1 s

The kinetic energy of a body is increased by 21%. The percentage increase in the magnitude of linear momentum of the body will be:

1.  10%

2.  20%

3.  Zero

4.  11.5%