Angular momentum of a body is defined as the product of

1.  Mass and angular velocity

2.  Centripetal force and radius

3.  Linear velocity and angular velocity

4.  Moment of inertia and angular velocity

A rigid body rotates with an angular momentum L. If its rotational kinetic energy is made 4 times, its angular momentum will become

1.  4L

2.  16L

3.  $\sqrt{2}<mspace\; width="1em"></mspace>\mathrm{L}$

4.  2L

The speed of a uniform spherical shell after rolling down an inclined plane of vertical height h from rest is:

1.  \(\sqrt{\frac{10   g h}{7}}\)

2.    \(\sqrt{\frac{6   g h}{5}}\)

3.  \(\sqrt{\frac{4   g h}{5}}\)

4.  \(\sqrt{2   g h}\)

A rigid body rotates about a fixed axis with a variable angular velocity equal to \(\alpha -\beta t\), at the time \(t\), where \(\alpha , \beta\) are constants. The angle through which it rotates before it stops is:

Two masses \(m_1~\text{and}~m_2, ~m_1 >m_2\) are connected to the ends of massless rope and allowed to move as shown in the figure. The acceleration of the centre-of-mass assuming pulley is massless and frictionless, is:

1. \(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}} g\)

2. \(0\)

3. \(\left(\dfrac{m_1 -m_2}{m_1 + m_2 } \right)^2g \)

4. \(\left(\dfrac{m_1 +m_2}{m_1 - m_2 }\right)^2 g \)

Four particles of mass \(m_1 = 2m\), \(m_2=4m\), \(m_3 =m \), and \(m_4\) are placed at the four corners of a square. What should be the value of \(m_4\) so that the centre of mass of all the four particles is exactly at the centre of the square?

For the uniform T shaped structure, with mass \(3~ \text M\), the moment of inertia about an axis normal to the plane and passing through O would be

             

1. \({ 2 \over 3} \text {MI}^2\)
2. \(\text {MI}^2\)
3. \({ MI^2 \over 3}\)
4. None of these 

A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing thorugh A and Perpendicular to the plane of paper is

1. $\frac{2}{3}M{R}^2$

2. $M{R}^2$

3. 2$M{R}^2$

4. $\frac{3}{2}M{R}^2$