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Q. No.In case of vertical circular motion of a particle by a thread of length \(r\) if tension in the thread is zero at an angle \(30^\circ\) shown in figure, the velocity at the bottom point \((A)\) of the circular path is: \((g =\) gravitational acceleration\()\)
1. \(\sqrt{5gr}\)
2. \(\sqrt{\dfrac{7}{2} {gr}}\)
3. \(\sqrt{4gr}\)
4.\(\sqrt{\dfrac{5}{2}{gr}}\)
Subtopic: Conservation of Mechanical Energy |
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Level 2: 60%+
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A small both \(A\) of mass \(m\) is attached to a massless rigid rod of length \(1~\text{m}\) pivoted at point \(P\) and kept at an angle of \(60^{\circ}\) with vertical as shown in figure. At distance of \(1~\text{m}\) below point \(P\), an identical bob \(B\) is kept at rest on a smooth horizontal surface that extends to a circular track of radius \(R\) as shown in figure. If bob \(B\) just manages to complete the circular path of radius \(R\) upto a point \(Q\) after being hit elastically by bob \(A\), then radius \(R\) is: (in m)
1. \(\dfrac{3}{5}\)
2. \(\dfrac{1}{5}\)
3. \(\dfrac{2+\sqrt{3}}{5}\)
4. \(\dfrac{2-\sqrt{3}}{5}\)
1. \(\dfrac{3}{5}\)
2. \(\dfrac{1}{5}\)
3. \(\dfrac{2+\sqrt{3}}{5}\)
4. \(\dfrac{2-\sqrt{3}}{5}\)
Subtopic: Conservation of Mechanical Energy |
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Level 2: 60%+
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Two blocks with masses \(100~\text{g}\) and \(200~\text{g}\) are attached to the ends of springs \(A\) and \(B\) as shown in figure. The energy stored in \(A\) is \(E\). The energy stored in \(B\), when spring constants \(k_A, k_B~\text{of}~A~\text{and}~B,\), respectively satisfy the relation \(4k_A = 3k_B\), is:
1. \(4E\)
2. \(2E\)
3. \(3E\)
4. \(\dfrac{4}{3} {E}\)
1. \(4E\)
2. \(2E\)
3. \(3E\)
4. \(\dfrac{4}{3} {E}\)
Subtopic: Elastic Potential Energy |
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Level 2: 60%+
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Given below are two statements:
In the light of the above statements, choose the correct answer from the options given below:
| Statement I: | An object moves from position \(r_1\) to position \(r_2\) under a conservative force field \(\vec{F}\). The work done by the force is \(W=-\int_{r_1}^{r_2} \vec{F} \cdot d\overrightarrow{r}\). |
| Statement II: | Any object moving from one location to another location can follow infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force. |
| 1. | Both Statement I and Statement II are True |
| 2. | Statement I is False but Statement II is True |
| 3. | Statement I is True but Statement II is False |
| 4. | Both Statement I and Statement II are False |
Subtopic: Concept of Work |
Level 3: 35%-60%
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In a perfectly inelastic collision, two spheres made of the same material with masses \(15~\text{kg}\) and \(25~\text{kg}\), moving in opposite directions with speeds of \(10~\text{m/s}\) and \(30~\text{m/s}\), respectively, strike each other and stick together. The rise in temperature (in \(^{\circ}\text{C}\)), if all the heat produced during the collision is retained by these spheres, is:
(specific heat of sphere material \(31\) cal/kg.oC and \(1\) cal = \(4.2\) J)
1. \(1.75\)
2. \(1.44\)
3. \(1.15\)
4. \(1.95\)
(specific heat of sphere material \(31\) cal/kg.oC and \(1\) cal = \(4.2\) J)
1. \(1.75\)
2. \(1.44\)
3. \(1.15\)
4. \(1.95\)
Subtopic: Collisions |
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Level 2: 60%+
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A body of mass \(2~\text{kg}\) is moving along \(x\text-\)direction such that its displacement as function of time is given by \(x(t)=\alpha t^2+\beta t+\gamma m,\) where \(\alpha=1 ~\text{m/s}^2\), \(\beta=1~\text{m/s}\) and \(\gamma=1~\text{m}.\) The work done on the body during the time interval \(t= 2~\text{s and}~3~\text{s},\) is: (in J)
1. \(49\)
2. \(42\)
3. \(24\)
4. \(12\)
1. \(49\)
2. \(42\)
3. \(24\)
4. \(12\)
Subtopic: Work done by constant force |
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Level 1: 80%+
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Three masses \(200~\text{kg}\), \(300~\text{kg}\) and \(400~\text{kg}\) are placed at the vertices of an equilateral triangle with sides \(20~\text{m}\). They are rearranged on the vertices of a bigger triangle of side \(25~\text{m}\) and with the same centre. The work done in this process: (in J)
(Gravitational constant \(G=6.7 \times 10^{-11} ~\text{N m}^2/ \text{kg}^2\))
1. \(9.86 \times 10^{-6} \)
2. \( 2.85 \times 10^{-7} \)
3. \(1.74 \times 10^{-7}\)
4. \(4.77 \times 10^{-7}\)
(Gravitational constant \(G=6.7 \times 10^{-11} ~\text{N m}^2/ \text{kg}^2\))
1. \(9.86 \times 10^{-6} \)
2. \( 2.85 \times 10^{-7} \)
3. \(1.74 \times 10^{-7}\)
4. \(4.77 \times 10^{-7}\)
Subtopic: Work done by constant force |
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Level 2: 60%+
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A body of mass \(1~\text{kg}\) moves along a straight line with a velocity \(v =2x^{2}.\) The work done by the body during displacement from \(x=0\) to \(5~\text{m}\) is: (in J)
1. \(0\)
2. \(250\)
3. \(1250\)
4. \(1000\)
1. \(0\)
2. \(250\)
3. \(1250\)
4. \(1000\)
Subtopic: Work Energy Theorem |
82%
Level 1: 80%+
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A mass of \(1~\text{kg}\) is kept on a inclined plane with \(30^\circ\) inclination with respect to horizontal plane and it is at rest initially. Then the whole assembly is moved up with constant velocity of \(4~\text{m/s}\). The work done by the frictional force in time \(2~\text{s}\) is: (in J) (Take \(g=10~\text{m/s}^2\))
1. \(20\)
2. \(25\)
3. \(30\)
4. \(10\)
1. \(20\)
2. \(25\)
3. \(30\)
4. \(10\)
Subtopic: Work done by constant force |
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Level 3: 35%-60%
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A particle of charge \(q\) and mass \(m\) is projected from origin with an initial velocity \(\vec{v}=\left(\dfrac{v_0}{\sqrt{2}} \hat{x}+\dfrac{v_0}{\sqrt{2}} \hat{y}\right)\). There exists a uniform magnetic field \(\vec{B}=B_0 \hat{z}\) and a space varying electric field \(\vec{E}=E_{{0}} {e}^{-\lambda x} \hat{x}\) within the region \(0 \leqslant x \leqslant L .\) After travelling a distance such that \(x\text-\)coordinate has changed from \(x=0\) to \(x=L,\) the changed in the kinetic energy is:
1. \(\dfrac{q E_0}{\lambda}\left[1-e^{-\lambda L}\right]\)
2. \(\left(\dfrac{v_0 q B_0}{2 \lambda}\right)\left[2-e^{-2 \lambda L}\right]\)
3. \(\dfrac{q E_0}{\lambda}\left[1+e^{-\lambda L}\right]\)
4. \(q\left(\dfrac{E_0+v_0 B_0}{\lambda}\right)\left[1-e^{-\lambda L / 2}\right]\)
1. \(\dfrac{q E_0}{\lambda}\left[1-e^{-\lambda L}\right]\)
2. \(\left(\dfrac{v_0 q B_0}{2 \lambda}\right)\left[2-e^{-2 \lambda L}\right]\)
3. \(\dfrac{q E_0}{\lambda}\left[1+e^{-\lambda L}\right]\)
4. \(q\left(\dfrac{E_0+v_0 B_0}{\lambda}\right)\left[1-e^{-\lambda L / 2}\right]\)
Subtopic: Work Energy Theorem |
Level 3: 35%-60%
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