The masses ${\mathrm{m}}_1<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{m}}_2<mspace\; width="1em"></mspace>\left(\mathrm{with}<mspace\; width="1em"></mspace>{\mathrm{m}}_1<mspace\; width="1em"></mspace><<mspace\; width="1em"></mspace>{\mathrm{m}}_2\right)$ are connected to the two ends of a compressed spring. Now, when the masses are released on a smooth surface, they will move away with

(1)  the same magnitude of the force.

(2)  the equal magnitude of linear momentum.

(3)  the greater kinetic energy of the first body than that of the second body.

(4)  All of these

Four masses are joined to light circular frames as shown in the figure. The radius of gyration of this system about an axis passing through the center of the circular frame and perpendicular to its plane would be:
(where '\(a\)' is the radius of the circle)
                        
1. \(\frac{a}{\sqrt{2}}\)
2. \(\frac{a}{{2}}\)
3. \(a\)
4. \(2a\)

A disc of mass \(M\) and radius \(R\) starts falling down as shown in the figure. The string unwinds without slipping on the disc. The instantaneous power developed by the tension is:

     
1. \((T\times R \omega)\)
2. \((T\times R \omega)/2\)
3. \(2(T\times R \omega)\)
4. zero

The rotational analouge of equation, \(F=\dfrac{mdv}{dt} \) is:

Internal forces cannot change:

(1)  the kinetic energy of a system.

(2)  the mechanical energy of a system.

(3)  the momentum of a system.

(4)  all of these

When a projectile of mass 2 kg is projected from the top of a tower of height 20 m with velocity 10 m/s in the horizontal direction, then angular momentum about the lowermost point of the tower when it touches the ground is:

(1) 600 kg ${\mathrm{m}}^2/\mathrm{s}$

(2) 400 kg ${\mathrm{m}}^2/\mathrm{s}$

(3) 800 kg ${\mathrm{m}}^2/\mathrm{s}$

(4) Zero

Two particles of mass \(5~\text{kg}\) and \(10~\text{kg}\) respectively are attached to the two ends of a rigid rod of length \(1~\text{m}\) with negligible mass. The centre of mass of the system from the \(5~\text{kg}\) particle is nearly at a distance of:
1. \(50~\text{cm}\)
2. \(67~\text{cm}\)
3. \(80~\text{cm}\)
4. \(33~\text{cm}\)

What would be the torque about the origin when a force \(3\hat{j}~\text N\) acts on a particle whose position vector is \(2\hat{k}~\text m?\)

A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle of inclination ${30}^{°}$. The centre of mass of the cylinder has a speed of 4 m/s. The distance travelled by the cylinder on the inclined surface will be $[take$ $g=$ $10$ $m/s^2]:$

A particle of mass \(5m\) at rest suddenly breaks on its own into three fragments. Two fragments of mass \(m\) each move along mutually perpendicular directions with speed \(v\) each. The energy released during the process is: