Two persons of mass 55 kg and 65 kg respectively, are at the opposite ends of a boat.The length of the boat is 3.0m and weighs 100 kg.The 55 kg man walks up to the 65 kg man and sits with him.If the boat is in still water the centre of mass of the system shifts by 

1. 3.0m

2. 2.3m

3. zero

4. 0.75m

A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity 4 $m{s}^{-1}.$ It collides with a horizontal spring of force constant 200 $N{m}^{-1}$. The maximum compression produced in the spring will be

1. 0.5 m

2. 0.6 m

3. 0.7 m

4. 0.2 m

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:

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 $2 m{s}^{-1}.$ 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. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}\ne 0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$                  

2. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}>0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}<0$

3. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}=0$

4. $\overrightarrow{\mathrm{r}}.\overrightarrow{\mathrm{\tau }}=0 \mathrm{and} \overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{\tau }}\ne 0$