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. \(\frac{M   +   2 m}{M \omega}\)

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

For a body, with angular velocity \( \vec{\omega }=\hat{i}-2\hat{j}+3\hat{k}\)  and radius vector \( \vec{r }=\hat{i}+\hat{j}++\hat{k},\)  its velocity will be:
1. \(-5\hat{i}+2\hat{j}+3\hat{k}\)
2. \(-5\hat{i}+2\hat{j}-3\hat{k}\)
3. \(-5\hat{i}-2\hat{j}+3\hat{k}\)
4. \(-5\hat{i}-2\hat{j}-3\hat{k}\)

When a stick is released (as shown in the figure below), its free end velocity when it strikes the ground is:

1. 4.2 m/s
2. 1.4 m/s
3. 2.8 m/s
4. $\sqrt{6}$ m/s

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

A circular disc is to be made by using iron and aluminium so that it acquires a maximum moment of inertia about its geometrical axis. It is possible with: 

A disc is rotating with angular speed \(\omega.\) If a child sits on it, what is conserved here?
1. Linear momentum
2. Angular momentum
3. Kinetic energy
4. Potential energy

If a rod of length 3 m with its mass acting per unit length, is directly proportional to distance x from one of its ends, then its centre of gravity from that end will be at: 

A thin circular ring \(\mathrm{M}\) and radius \(\mathrm{r}\) is rotating about its axis with a constant angular velocity ω. Four objects, each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

1. $\frac{\mathrm{M\omega }}{4\mathrm{m}}$

2. $\frac{\mathrm{M\omega }}{\mathrm{M}+4\mathrm{m}}$

3. $\frac{(\mathrm{M}+4\mathrm{m})\mathrm{\omega }}{\mathrm{M}}$

4. $\frac{(\mathrm{M}+4\mathrm{m})\mathrm{\omega }}{\mathrm{M}+4\mathrm{m}}$

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. \(1:\) $\sqrt{2}$