A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first \(2\) sec, it rotates through an angle \(\theta_1\). In the next \(2\) seconds, it rotates through an additional angle \(\theta_2\). The ratio of \(\frac{\theta_2}{\theta_1}\) is: 

1.	\(1\)	2.	\(2\)
3.	\(3\)	4.	\(5\)

Two rotating bodies A and B of masses m and 2m with moments of inertia $I_A$ and $I_B(I_B>I_A)$ have equal kinetic energy of rotation. If $L_A$ $$and $L_B$ be their angular momenta respectively, then:
1. $L_A=\frac{L_B}{2}$
2. $L_A=2L_B$
3. $L_B>L_A$
4. $L_A>L_B$

A force F=$a\hat{i}$$+3\hat{j}$$+6\hat{k}$ is acting at a point r=$2\hat{i}$-$6\hat{j}$-12$\hat{k}$. The value of α for which angular momentum is conserved about the origin is:
1. -1
2. 2
3. zero
4. 1

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:

Which of the following will not be affected if the radius of the sphere is increased while keeping mass constant?

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?

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:

The position of a particle is given by \(\vec r = \hat i+2\hat j-\hat k\) and momentum \(\vec P = (3 \hat i + 4\hat j - 2\hat k)\). The angular momentum is perpendicular to:

The centre of the mass of \(3\) particles, \(10\) kg, \(20\) kg, and \(30\) kg, is at \((0,0,0)\). Where should a particle with a mass of \(40\) kg be placed so that its combined centre of mass is \((3,3,3)\)?
1. \((0,0,0)\)
2. \((7.5, 7.5, 7.5)\)
3. \((1,2,3)\)
4. \((4,4,4)\)