A circular ring of mass M and radius R is rotating about its axis with constant angular velocity ω. Two particles, each of mass m are attached gently to the opposite ends of the diameter of the ring. The angular velocity of the ring will now become:

1. \(\frac{m \omega}{M   +   2 m}\)

2. \(\frac{m \omega}{M   -   2 m}\)

3.  \(\frac{M \omega}{M   +   2 m}\)

4. \(\frac{M   +   2 m}{M \omega}\)

The moment of inertia of a uniform circular disc of radius 'R' and mass 'M' about an axis passing from the edge of the disc and normal to the disc is:

1. $\frac{1}{2}{\mathrm{MR}}^2$

2. $\frac{7}{2}{\mathrm{MR}}^2$

3. $\frac{3}{2}{\mathrm{MR}}^2$

4. ${\mathrm{MR}}^2$

For the diagram given below, a triangular lamina is shown. The correct relation between I₁, I₂ & I₃ is (I – moment of inertia)         
     

1. I₁ > I₂

2. I₂ > I₁

3. I₃ > I₁

4. I₃ > I₂

In the figure given below, O is the centre of an equilateral triangle ABC and $\overrightarrow{{\mathrm{F}}_1},$ $$ $\overrightarrow{{\mathrm{F}}_2},$ $$ $\overrightarrow{{\mathrm{F}}_3}$ are three forces acting along the sides AB, BC and AC. What should be the magnitude of $\overrightarrow{F_3}$ $$ so that total torque about O is zero?
        
1.  $|$ ${\overrightarrow{F}}_3|$ $=$ $\left|\overrightarrow{F_1}\right|$ $+$ $|\overrightarrow{F_2|}$

2. $|$ $\overrightarrow{F_3}|$ $=$ $\left|\overrightarrow{F_1}\right|$ $-$ $|\overrightarrow{F_2|}$

3. $|$ $\overrightarrow{F_3}|$ $=$ $\overrightarrow{F_1}$ $+$ $2\overrightarrow{F_2}$

4. Not possible

Two bodies have their moments of inertia I and 2I, respectively, about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio:

1. 1 : 2

2. $\sqrt{2}$ :1

3. 1: $\sqrt{2}$

4. 2 : 1

The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring will be:
1. \(2:1\)
2. $\sqrt{5}$ : $\sqrt{6}$
3. \(2:3\)
4. \(1:\) $\sqrt{2}$

A round disc of the moment inertia ${\mathrm{I}}_2$ about its axis perpendicular to its plane and passing through its centre is placed over another disc of moment of inertia ${\mathrm{I}}_1$ rotating with an angular velocity ω about the same axis. The final angular velocity of the combination of discs is:

1. ω

2. $\frac{{\mathrm{I}}_1\omega }{{\mathrm{I}}_1+{\mathrm{I}}_2}$

3. $\frac{({\mathrm{I}}_1+{\mathrm{I}}_2)\omega }{{\mathrm{I}}_1}$

4. $\frac{{\mathrm{I}}_2\omega }{{\mathrm{I}}_1+{\mathrm{I}}_2}$

Three particles, each of mass \(m\) gram, are situated at the vertices of an equilateral triangle ABC of side \(l\) cm (as shown in the figure). The moment of inertia of the system about a line AX, perpendicular to AB and in the plane of ABC, in gram-cm² units will be:

           
1. \(2ml^2\)
2. \(\frac{5}{4}ml^2\)
3. \(\frac{3}{2}ml^2\)
4. \(\frac{3}{4}ml^2\)

A wheel having a moment of inertia of \(2\) kg–m² about its vertical axis rotates at the rate of \(60\) rpm about the axis. The torque which can stop the wheel's rotation in one minute would be:
1. \(\frac{\pi }{12}\) N-m

2.  \(\frac{\pi }{15}\) N-m

3.  \(\frac{\pi }{18}\) N-m

4.  \(\frac{2\pi }{15}\) N-m

Consider a system of two particles having masses m₁ and m₂. If the particle of mass m₁ is pushed towards the mass centre of particles through a distance 'd', by what distance would the particle of mass m₂ move so as to keep the mass centre of particles at the original position ?

1. $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}\mathrm{d}$

2. d

3. $\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}$

4. $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2}\mathrm{d}$