A particle moving with velocity \(\vec{v}\) is acted by three forces shown by the vector triangle \(\mathrm{PQR}.\) The velocity of the particle will:

        

1.	change according to the smallest force \(\mathrm{\overrightarrow{Q R}}\)
2.	increase
3.	decrease
4.	remain constant

A mass m₁, placed on top of a trolley of mass m₃, is connected to another mass m₂ by means of string passing over a smooth pulley as shown in figure. The friction between surfaces is negligible. For m₁ ans m₂ not to move with respect to trolley, the horizontal force F to be applied on trolley is

1. F=m₃g

2. F=(m₁ + m₂)g

3. $\mathrm{F}=\left({\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3\right)\frac{{\mathrm{m}}_2\mathrm{g}}{{\mathrm{m}}_1}$

4. F= m₁g

A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string when the trolley rolls up a plane of inclination $\alpha$ with acceleration $a_0$ is 

1. $\theta ={\tan }^{-1} \alpha$

2. $\theta ={\tan }^{-1} \left(\frac{a_0}{g}\right)$

3. $\theta ={\tan }^{-1} \left(\frac{g}{a_0}\right)$

4. $\theta ={\tan }^{-1} \left(\frac{a_0+g \sin \alpha }{g \cos \alpha }\right)$

The pulleys and string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle \(\theta\) should be:

      

1. 0°
2. 30$°$
3. 45$°$
4. 60$°$

System shown in figure is in equilibrium and at rest. The spring and string are massless, now the stringis cut. The acceleration of mass 2m and m just after string is cut will be

1. g/2 upwards, g downwards

2. g upwards, g/2 downwards

3. g upwards, 2g downwards

4. 2g upwards, g downwards

In the system shown $m_1>m_2$. System is held at rest by thread BC. Just after lower thread is Burnt.

1. Acceleration of $m_2$ is upward

2. Magnitude of acceleration of both blocks will be $\left(\frac{m_1-m_2}{m_1+m_2}\right)g$

3. Acceleration of $m_1$ will be non-zero

4. Magnitude of acceleration of two blocks will be non-zero and unequal.

Two blocks of 4 kg and 6 kg are attached by springs, they are hanging in vertical position, lower spring breaks due to excessive force. Acceleration of 4 kg block just after breaking

1. 15 $m/s^2$

2. 25 $m/s^2$

3. 10 $m/s^2$

4. Zero

The tension in the spring is

1. Zero

2. 2.5 N

3. 5 N

4. 10 N

A block can slide on a smooth inclined plane of inclination $\theta$ kept on the floor of a lift. When the lift is descending with retardation a, the acceleration of the block relative to the incline is:

1.  (g + a) sin $\mathrm{\theta }$              
2.  (g – a)
3.  g sin $\mathrm{\theta }$
4.  (g – a) sin $\mathrm{\theta }$

A block of mass \(10~\text{kg}\) is in contact with the inner wall of a hollow cylindrical drum of radius \(1~\text{m}.\) The coefficient of friction between the block and the inner wall of the cylinder is \(0.1.\) The minimum angular velocity needed for the cylinder, which is vertical and rotating about its axis, will be: 
\(\left(g= 10~\text{m/s}^2\right )\)
1. \(10~\pi~\text{rad/s}\)
2. \(\sqrt{10}~\pi~\text{rad/s}\)
3. \(\frac{10}{2\pi}~\text{rad/s}\)
4. \(10~\text{rad/s}\)