Which of the following is the value of the torque of force \(F\) about origin \(O:\)
1. \(\vec{\tau}=5(1-\sqrt{3}) \hat{k}\) N-m
2. \(\vec{\tau}=5(1-\sqrt{3}) \hat{j}\) N-m
3. \(\vec{\tau}=5(\sqrt{3}-1) \hat{i}\) N-m
4. \(\vec{\tau}=\sqrt{3} \hat{j}\) N-m
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 |