The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through:
 
1. B                   
2. C
3. D                   
4. A

Three masses are placed on the x-axis: \(300\) g at the origin, \(500\) g at \(x =40\) cm, and \(400\) g at \(x=70\) cm. The distance of the center of mass from the origin is:
1.     \(40\) cm
2.   \(45\) cm
3.     \(50\) cm
4. \(30\) cm

The instantaneous angular position of a point on

a rotating wheel is given by the equation

$\mathrm{\theta }\left(\mathrm{t}\right)=2{\mathrm{t}}^3-6{\mathrm{t}}^2$

The torque on the wheel becomes zero at

1. t=0.5 s

2. t=0.25 s

3. t=2 s

4. t=1 s

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its mid-point and perpendicular to its length is ${\mathrm{I}}_0$. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

1. ${\mathrm{I}}_0+{\mathrm{ML}}^2/4$

2. ${\mathrm{I}}_0+2{\mathrm{ML}}^2$

3. ${\mathrm{I}}_0+{\mathrm{ML}}^2$

4. ${\mathrm{I}}_0+{\mathrm{ML}}^2/2$

A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ${\omega }_i.$ Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotation disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ${\omega }_f.$ The energy lost by the initially rotating disc to friction is 

(1)$\frac{1}{2}\frac{I_b^2}{\left(I_t+I_b\right)}{\omega }_i^2$                                     

(2)$\frac{1}{2}\frac{I_t^2}{\left(I_t+I_b\right)}{\omega }_i^2$

(3)$\frac{1}{2}\frac{I_b-I_t}{\left(I_t+I_b\right)}{\omega }_i^2$                                     

(4)$\frac{1}{2}\frac{I_b{I}_t}{\left(I_t+I_b\right)}{\omega }_i^2$

Two particles which are initially at rest, move towards each other under the action of their mutual attraction.If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be 

1. 2v                                             2. 0

3. 1.5v                                          4. v

A man of 50 kg mass is standing in a gravity free space at a height of 10m above the floor. He throws a stone of 0.5 kg mass downwards with a speed $2<mspace\; width="1em"></mspace>m{s}^{-1}.$ When the stone reaches the floor, the distance of the man above the floor will be 

1. 9.9m                                 

2. 10.1m

3. 10m                                 

4. 20m                       

If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{\tau }}$ be the torque of this force about the origin, then 

1. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}\ne 0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$                  

2. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}>0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}<0$

3. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$

4. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}\ne 0$

Two bodies of mass 1kg and 3kg have position vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-3\hat{i}-2\hat{j}+\hat{k}$, respectively. The centre of mass of this system has a position vector -

1. $-2\hat{i}+2\hat{k}$

2. $-2\hat{i}-\hat{j}+\hat{k}$

3. $2\hat{i}-\hat{j}-2\hat{k}$

4. $-\hat{i}+\hat{j}+\hat{k}$

Four identical thin rods each of mass M and length t, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is 

1. $\frac{4}{3}M{t}^2$                                   

2. $\frac{2}{3}M{t}^2$

3. $\frac{13}{3}M{t}^2$                                 

4. $\frac{1}{3}M{t}^2$