The strings and pulleys shown in the figure are massless. The reading shown by the light spring balance \(S\) is:

        

1.	\(2.4\) kg	2.	\(5\) kg
3.	\(2.5\) kg	4.	\(3\) kg

What is the velocity of the block when the angle between the string and horizontal is \(30^\circ\) as shown in the diagram?

          
1. \(v_B=v_P\)
2. \(v_B=\frac{v_P}{\sqrt{3}}\)
3. \(v_B=2v_P\)
4. \(v_B=\frac{2v_P}{\sqrt{3}}\)

A bucket full of water tied with the help of a \(2\) m long string performs a vertical circular motion. The minimum angular velocity of the bucket at the uppermost point so that water will not fall will be:
1. \(2\sqrt{5}\) rad/s
2. \(\sqrt{5}\) rad/s
3. \(5\) rad/s
4. \(10\) rad/s

On the application of an impulsive force, a sphere of mass \(500\) grams starts moving with an acceleration of \(10\) m/s². The force acts on it for \(0.5\) s. The gain in the momentum of the sphere will be:
1. \(2.5\) kg-m/s
2. \(5\) kg-m/s
3. \(0.05\) kg-m/s
4. \(25\) kg-m/s

What is the acceleration of block \(A\), if the acceleration of \(B\) is \(4~\text{m/s}^2\) towards the right at the instant shown?

                          
1. \(2.5~\text{m/s}^2\)
2. \(4~\text{m/s}^2\)
3. \(5~\text{m/s}^2\)
4. zero

Two blocks of masses \(2\) kg and \(3\) kg placed on a horizontal surface are connected by a massless string. If \(3\) kg is pulled by \(10\) N as shown in the figure, then the force of friction acting on the \(2\) kg block will be: [Take \(g=10~\text{m/s}^2\)] $\mathrm{}$

Fundamentally, the normal force between two surfaces in contact is:

1. Electromagnetic

2. Gravitational

3. Weak nuclear force

4. Strong nuclear force

A particle of mass \(m\) is suspended from a ceiling through a massless string. The particle moves in a horizontal circle as shown in the given figure. The tension in the string is:
 

1. \(mg\)

2. \(2mg\)

3. \(3mg\)

4. \(4mg\)

Choose the incorrect alternative:

The kinetic energy 'K' of a particle moving in a circular path varies with the distance covered S as K = a${\mathrm{S}}^2$, where a is constant. The angle between the tangential force and the net force acting on the particle is: (R is the radius of the circular path)

1.  ${\tan }^{-1}\left(\frac{\mathrm{S}}{\mathrm{R}}\right)$

2.  ${\tan }^{-1}\left(\frac{\mathrm{R}}{\mathrm{S}}\right)$

3.  ${\tan }^{-1}\left(\frac{\mathrm{a}}{\mathrm{R}}\right)$

4.  ${\tan }^{-1}\left(\frac{\mathrm{R}}{\mathrm{a}}\right)$