1. \(\frac{5}{3}mL^2\)
2. \(4mL^2\)
3. \(\frac{1}{4}mL^2\)
4. \(\frac{2}{3}mL^2\)
1. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
2. | \(\vec{\tau}=(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}) \) N-m |
3. | \(\vec{\tau}=(17 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}})\) N-m |
4. | \(\vec{\tau}=(-41 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+16 \hat{\mathrm{k}})\) N-m |
In the three figures, each wire has a mass M, radius R and a uniform mass distribution. If they form part of a circle of radius R, then about an axis perpendicular to the plane and passing through the centre (shown by crosses), their moment of inertia is in the order:
1.
2.
3.
4.
1. | \(1\) rad/s | 2. | \(2\) rad/s |
3. | \(3\) rad/s | 4. | \(4\) rad/s |
The value of \(M\), as shown, for which the rod will be in equilibrium is:
1. | \(1\) kg | 2. | \(2\) kg |
3. | \(4\) kg | 4. | \(6\) kg |
Particles \(A\) and \(B\) are separated by \(10\) m, as shown in the figure. If \(A\) is at rest and \(B\) started moving with a speed of \(20\) m/s then the angular velocity of \(B\) with respect to \(A\) at that instant is:
1. | \(1\) rad s-1 | 2. | \(1.5\) rad s-1 |
3. | \(2\) rad s-1 | 4. | \(2.5\) rad s-1 |
A uniform cubical block of side L rests on a rough horizontal surface with coefficient of friction . A horizontal force F is applied on the block as shown. If there is sufficient friction between the block and the ground, then the torque due to normal reaction about its centre of mass is:
1.
2.
3.
4.
A bomb is projected from the ground at a horizontal range of \(R\). If the bomb explodes mid-air, then the range of its centre of mass is:
1. \(\frac{R}{2}\)
2. \(R\)
3. \(2R\)
4. \(\frac{2R}{3}\)
The law of conservation of angular momentum is valid when:
1. | The net force is zero and the net torque is non-zero | 2. | The net force is non-zero and the net torque is non zero |
3. | Net force may or may not be zero and net torque is zero | 4. | Both force and torque must be zero |
A man hangs from a rope attached to a hot-air balloon. The man's mass is greater than the mass of the balloon and its contents. The system is stationary in the air. If the man now climbs up to the balloon using the rope, the centre of mass of the "man plus balloon" system will:
1. | remain stationary |
2. | move up |
3. | move down |
4. | first moves up and then return to its initial position |