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 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: 

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. \(-5\hat{i}-2\hat{j}-3\hat{k}\)

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 wheel has an angular acceleration of 3.0 rad/s² and an initial angular speed of 2.00 rad/s. In a time of 2 s, it has rotated through an angle (in radian) of:
1. 6
2. 10
3. 12
4. 4

Two gear wheels that are meshed together have radii of 0.50 cm and 0.15 cm. The number of revolutions made by the smaller one when the larger one goes through 3 revolutions is:

1. 5 revolutions 

2. 20 revolutions 

3. 1 revolution

4. 10 revolutions

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

From a circular ring of mass \({M}\) and radius \(R\), an arc corresponding to a \(90^\circ\)$$sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \(K\) times \(MR^2\). The value of \(K\) will be:
1. \(\frac{1}{4}\)
2. \(\frac{1}{8}\)
3. \(\frac{3}{4}\)
4. \(\frac{7}{8}\)

A force \(\vec{F}=\hat{i}+2\hat{j}+3\hat{k}~\text{N}\) acts at a point \(\hat{4i}+3\hat{j}-\hat{k}~\text{m}\). Let the magnitude of the torque about the point \(\hat{i}+2\hat{j}+\hat{k}~\text{m}\) be \(\sqrt{x}~\text{N-m}\). The value of \(x\) is:

Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of the square where the axis of rotation is parallel to the plane of the square is :

1. $\frac{4}{5}{\mathrm{ma}}^2+2{\mathrm{mb}}^2$

2. $\frac{8}{5}m{a}^2+m{b}^2$

3. $\frac{8}{5}{\mathrm{ma}}^2+2{\mathrm{mb}}^2$

4. $\frac{4}{5}m{a}^2$