Q. No.The instantaneous angular position of a point on a rotating wheel is given by the equation,
\(\theta(t)=2t^{3}-6t^{2}\)
The torque on the wheel becomes zero at:
1. \(t=0.5\) s
2. \(t=0.25\) s
3. \(t=2\) s
4. \(t=1\) s
The moment of inertia of a thin uniform rod of mass \(M\) and length \(L\) about an axis passing through its mid-point and perpendicular to its length is \(I_0\). Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is:
1. \(I_0+\frac{ML^2}{4}\)
2. \(I_0+2ML^2\)
3. \(I_0+ML^2\)
4. \(I_0+\frac{ML^2}{2}\)
A circular disk of a moment of inertia \(\mathrm{I_t}\) is rotating in a horizontal plane, about its symmetric axis, with a constant angular speed \(\omega_i.\) Another disk of a moment of inertia \(\mathrm{I_b}\) is dropped coaxially onto the rotating disk. Initially, the second disk has zero angular speed. Eventually, both the disks rotate with a constant angular speed \(\omega_f.\) The energy lost by the initially rotating disc due to friction is:
1. \( \frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2\)
2. \( \frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2\)
3. \( \frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}-\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2 \)
4. \( \frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}}}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2 \)
Two particles that are initially at rest, move towards each other under the action of their mutual attraction. If their speeds are \(v\) and \(2v\) at any instant, then the speed of the centre of mass of the system will be:
1. \(2v\)
2. \(0\)
3. \(1.5v\)
4. \(v\)
A man of \(50\) kg mass is standing in a gravity-free space at a height of \(10\) m above the floor. He throws a stone of \(0.5\) kg mass downwards with a speed of \(2\) ms-1. When the stone reaches the floor, the distance of the man above the floor will be:
1. \(9.9\) m
2. \(10.1\) m
3. \(10\) m
4. \(20\) m
If \(\vec F\) is the force acting on a particle having position vector \(\vec r\) and \(\vec \tau\) be the torque of this force about the origin, then:
| 1. | \(\vec r\cdot\vec \tau\neq0\text{ and }\vec F\cdot\vec \tau=0\) |
| 2. | \(\vec r\cdot\vec \tau>0\text{ and }\vec F\cdot\vec \tau<0\) |
| 3. | \(\vec r\cdot\vec \tau=0\text{ and }\vec F\cdot\vec \tau=0\) |
| 4. | \(\vec r\cdot\vec \tau=0\text{ and }\vec F\cdot\vec \tau\neq0\) |
A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity ω. If two objects each of mass m are attached gently to the opposite ends of the diameter of the ring, the ring will then rotate with an angular velocity:
| 1. | \(\frac{\omega(M-2 m)}{M+2 m} \) | 2. | \(\frac{\omega M}{M+2 m} \) |
| 3. | \(\frac{\omega(M+2 m)}{M} \) | 4. | \(\frac{\omega M}{M+m}\) |
Two bodies of mass \(1\) kg and \(3\) kg have position vectors \(\hat{i}+2\hat{j}+\hat{k}\) and \(-3\hat{i}-2\hat{j}+\hat{k}\) respectively. The centre of mass of this system has a position vector:
1. \(-2\hat{i}+2\hat{k}\)
2. \(-2\hat{i}-\hat{j}+\hat{k}\)
3. \(2\hat{i}-\hat{j}-2\hat{k}\)
4. \(-\hat{i}+\hat{j}+\hat{k}\)
2. \(\frac{2}{3}Ml^2\)
3. \(\frac{13}{3}Ml^2\)
4. \(\frac{1}{3}Ml^2\)
The ratio of the radii of gyration of a circular disc to that of a circular ring, each of the same mass and radius, around their respective axes is:
| 1. | \(\sqrt{3}:\sqrt{2}\) | 2. | \(1:\sqrt{2}\) |
| 3. | \(\sqrt{2}:1\) | 4. | \(\sqrt{2}:\sqrt{3}\) |
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