When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

1. a line perpendicular to the plane of rotation

2. the line making an angle of $45°$ to the plane of rotation

3. the radius

4. the tangent to the orbit

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

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

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                       

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 and passing through its centre is:

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

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

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) $-\hat{i}+\hat{j}+\hat{k}$