Q. No.What would be the torque about the origin when a force \(3\hat{j}\) N acts on a particle whose position vector is \(2\hat{k}\) m?
1. \(6\hat{j}\) N-m
2. \(-6\hat{i}\) N-m
3. \(6\hat{k}\) N-m
4. \(6\hat{i}\) N-m
1. \(6\hat{j}\) N-m
2. \(-6\hat{i}\) N-m
3. \(6\hat{k}\) N-m
4. \(6\hat{i}\) N-m
Two particles of mass \(5~\text{kg}\) and \(10~\text{kg}\) respectively are attached to the two ends of a rigid rod of length \(1~\text{m}\) with negligible mass. The centre of mass of the system from the \(5~\text{kg}\) particle is nearly at a distance of:
1. \(50~\text{cm}\)
2. \(67~\text{cm}\)
3. \(80~\text{cm}\)
4. \(33~\text{cm}\)
The angular speed of the wheel of a vehicle is increased from \(360~\text{rpm}\) to \(1200~\text{rpm}\) in \(14\) seconds. Its angular acceleration will be:
| 1. | \(2\pi ~\text{rad/s}^2\) | 2. | \(28\pi ~\text{rad/s}^2\) |
| 3. | \(120\pi ~\text{rad/s}^2\) | 4. | \(1 ~\text{rad/s}^2\) |
Three identical spheres, each of mass \(M\), are placed at the corners of a right-angle triangle with mutually perpendicular sides equal to \(2~\text{m}\) (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of the centre of mass.
1. \(2(\hat{i}+\hat{j})\)
2. \(\hat{i}+\hat{j}\)
3. \(\frac{2}{3}(\hat{i}+\hat{j})\)
4. \(\frac{4}{3}(\hat{i}+\hat{j})\)
A particle of mass \(5m\) at rest suddenly breaks on its own into three fragments. Two fragments of mass \(m\) each move along mutually perpendicular directions with speed \(v\) each. The energy released during the process is:
1. \(\frac{3}{5}mv^2\)
2. \(\frac{5}{3}mv^2\)
3. \(\frac{3}{2}mv^2\)
4. \(\frac{4}{3}mv^2\)
A particle starting from rest moves in a circle of radius \(r\). It attains a velocity of \(v_0~\text{m/s}\) on completion of \(n\) rounds. Its angular acceleration will be:
| 1. | \( \frac{v_0}{n} ~\text{rad} / \text{s}^2\) | 2. | \( \frac{v_0^2}{2 \pi {nr}^2}~ \text{rad} / \text{s}^2 \) |
| 3. | \( \frac{v_o^2}{4 \pi {n}{r}^2}~ \text{rad} / \text{s}^2 \) | 4. | \( \frac{v_0^2}{4 \pi {nr}} ~\text{rad} / \text{s}^2 \) |
An object flying in the air with velocity \((20 \hat{i}+25 \hat{j}-12 \hat{k})\) suddenly breaks into two pieces whose masses are in the ratio of \(1:5.\) The smaller mass flies off with a velocity \((100 \hat{i}+35 \hat{j}+8 \hat{k})\). The velocity of the larger piece will be:
1. \( 4 \hat{i}+23 \hat{j}-16 \hat{k}\)
2. \( -100 \hat{i}-35 \hat{j}-8 \hat{k} \)
3. \( 20 \hat{i}+15 \hat{j}-80 \hat{k} \)
4. \( -20 \hat{i}-15 \hat{j}-80 \hat{k}\)
From a circular ring of mass \({M}\) and radius \(R\), an arc corresponding to a \(90^\circ\)sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is \(K\) times \(MR^2\). The value of \(K\) will be:
1. \(\frac{1}{4}\)
2. \(\frac{1}{8}\)
3. \(\frac{3}{4}\)
4. \(\frac{7}{8}\)
A uniform rod of length \(200~ \text {cm}\) and mass \(500~ \text g\) is balanced on a wedge placed at \(40~ \text{cm}\) mark. A mass of \(2~\text{kg}\) is suspended from the rod at \(20~ \text{cm}\) and another unknown mass \( \text m\) is suspended from the rod at \(160~\text{cm}\) mark as shown in the figure. What would be the value of \( \mathrm{m}\) such that the rod is in equilibrium? (Take \(g=10~( \text {m/s}^2)\)
| 1. | \({1 \over 6}~ \text{kg}\) | 2. | \({1 \over 12}~ \text{kg}\) |
| 3. | \({1 \over 2}~ \text{kg}\) | 4. | \({1 \over 3}~ \text{kg}\) |
A disc of radius \(2~\text{m}\) and mass \(100~\text{kg}\) rolls on a horizontal floor. Its centre of mass has a speed of \(20~\text{cm/s}\). How much work is needed to stop it?
1. \(1~\text{J}\)
2. \(3~\text{J}\)
3. \(30~\text{J}\)
4. \(2~\text{J}\)
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