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. $\frac{1}{3}M{t}^2$

The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is -

1. $\sqrt{3}:\sqrt{2}$                                         2. $1:\sqrt{2}$

3. $\sqrt{2}:1$                                            4. $\sqrt{2}:\sqrt{3}$

A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is ${90}^{°}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is 

1. $\frac{M{L}^2}{24}$

2. $\frac{M{L}^2}{12}$

3. $\frac{M{L}^2}{6}$

4. $\frac{\sqrt{2}M{L}^2}{24}$

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