Q. No.A rod \(\mathrm{PQ}\) of mass \(M\) and length \(L\) is hinged at end \(\mathrm{P}\). The rod is kept horizontal by a massless string tied to point \(\mathrm{Q}\) as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:
1. \(\frac{g}{L}\)
2. \(\frac{2g}{L}\)
3. \(\frac{2g}{3L}\)
4. \(\frac{3g}{2L}\)
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along:
| 1. | a line perpendicular to the plane of rotation |
| 2. | the line making an angle of \(45^{\circ}\) to the plane of rotation |
| 3. | the radius |
| 4. | the tangent to the orbit |
Two persons of masses \(55\) kg and \(65\) kg respectively, are at the opposite ends of a boat. The length of the boat is \(3.0\) m and weighs \(100\) kg. The \(55\) kg man walks up to the \(65\) kg man and sits with him. If the boat is in still water, the centre of mass of the system shifts by:
1. \(3.0\) m
2. \(2.3\) m
3. zero
4. \(0.75\) m
\(\mathrm{ABC}\) is an equilateral triangle with \(O\) as its centre. \(F_1\), \(F_2\), and \(F_3\) represent three forces acting along the sides \(\mathrm{AB},\) \(\mathrm{BC}\) and \(\mathrm{AC}\) respectively. If the total torque about \(O\) is zero, then the magnitude of \(F_3\) is:
1. \(F_1+F_2\)
2. \(F_1-F_2\)
3. \(\frac{F_1+F_2}{2}\)
4. \(2F_1+F_2\)
\(\theta(t)=2t^{3}-6t^{2}\)
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. | \(\vec r\cdot\vec \tau=0\text{ and }\vec F\cdot\vec \tau\neq0\) |
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