Q. No.A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(\mathrm{AB}= D.\) The height \({h}\) is equal to:
1. \({3\over2}D\)
2. \(D\)
3. \({7\over4}D\)
4. \({5\over4}D\)
1. \({3\over2}D\)
2. \(D\)
3. \({7\over4}D\)
4. \({5\over4}D\)
2. \(\left(2 t^3+3 t^3\right) ~\text W\)
3. \(\left(2 t^3+3 t^5\right) ~\text W\)
4. \(\left(2 t^3+3 t^4\right) ~\text W\)
A ball is thrown vertically downward from a height of \(20~\text m\) with an initial velocity \(v_0.\) It collides with the ground, loses \(50\%\) of its energy in a collision, and rebounds to the same height. The initial velocity \(v_0\) is:
(Take, \(g=10~\text{ms}^{-2}\))
1. \(14~\text{ms}^{-1}\)
2. \(20~\text{ms}^{-1}\)
3. \(28~\text{ms}^{-1}\)
4. \(10~\text{ms}^{-1}\)
Two similar springs \(P\) and \(Q\) have spring constants \(k_P\) and \(k_Q\), such that \(k_P>k_Q\). They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and \(W_Q\) are related as, in case (a) and case (b), respectively:
| 1. | \(W_P=W_Q;~W_P>W_Q\) |
| 2. | \(W_P=W_Q;~W_P=W_Q\) |
| 3. | \(W_P>W_Q;~W_P<W_Q\) |
| 4. | \(W_P<W_Q;~W_P<W_Q\) |
1. \(475\) J
2. \(450\) J
3. \(275\) J
4. \(250\) J
| 1. | \( \sqrt{\frac{m k}{2}} t^{-1 / 2} \) | 2. | \( \sqrt{m k} t^{-1 / 2} \) |
| 3. | \( \sqrt{2 m k} t^{-1 / 2} \) | 4. | \( \frac{1}{2} \sqrt{m k} t^{-1 / 2}\) |
Two particles of masses \(m_1\) and \(m_2\) move with initial velocities \(u_1\) and \(u_2\) respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy \(E\). If the final velocities of particles are \(v_1\) and \(v_2\), then we must have:
| 1. | \(m_1^2u_1+m_2^2u_2-E = m_1^2v_1+m_2^2v_2\) |
| 2. | \(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2= \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\) |
| 3. | \(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2-E= \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\) |
| 4. | \(\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2+E = \frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2\) |
A uniform force of \((3 \hat{i} + \hat{j})\) newton acts on a particle of mass \(2~\text{kg}.\) Hence the particle is displaced from the position \((2 \hat{i} + \hat{k})\) metre to the position \((4 \hat{i} + 3 \hat{j} - \hat{k})\) metre. The work done by the force on the particle is:
1. \(6~\text{J}\)
2. \(13~\text{J}\)
3. \(15~\text{J}\)
4. \(9~\text{J}\)
| 1. | \(\dfrac{B}{A}\) | 2. | \(\dfrac{B}{2A}\) |
| 3. | \(\dfrac{2A}{B}\) | 4. | \(\dfrac{A}{B}\) |
| 1. | same as that of \(B.\) |
| 2. | opposite to that of \(B.\) |
| 3. | \(\theta = \text{tan}^{-1}\left(\frac{1}{2} \right)\) to the positive \(x\)-axis. |
| 4. | \(\theta = \text{tan}^{-1}\left(\frac{-1}{2} \right )\) to the positive \(x\)-axis. |
© 2025 GoodEd Technologies Pvt. Ltd.