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

From a circular disc of radius R and mass 9M, a small disc of mass M and radius $\frac{\mathrm{R}}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc about an axis perpendicular to the plane of the disc and passing through its centre is -

(a) $\frac{40}{9}M{R}^2$                                        (b) $M{R}^2$

(c) $4\mathit{ }M{R}^2$                                           (d) $\frac{4}{9}M{R}^2$

The solid cylinder and a hollow cylinder, both of the same mass and same external diameter are released from the same height at the same time on a inclined plane. Both roll down without slipping. Which one will reach the bottom first ?

(1) Both together only when angle of inclination of plane is $45°$

(2) Both together

(3) Hollow cylinder

(4) Solid cylinder 

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 

(a) $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}\ne 0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$                  

(b) $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}>0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}<0$

(c) $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$

(d) $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}\ne 0$

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) $\frac{1}{3}M{t}^2$

A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

1. Angular velocity 

2. Moment of inertia

3. Angular momentum 

4. Rotational kinetic energy

The bricks, each of length L and mass M, are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is :

(1) L/4

(2) L/2

(3) (3/2) L

(4) (11/12) L