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. \(\left(2 t^3+3 t^3\right) \) W
3. \(\left(2 t^3+3 t^5\right)\) W
4. \(\left(2 t^3+3 t^4\right) \) W
What is the minimum velocity with which a body of mass \(m\) must enter a vertical loop of radius \(R\) so that it can complete the loop?
1. \(\sqrt{2 g R}\)
2. \(\sqrt{3 g R}\)
3. \(\sqrt{5 g R}\)
4. \(\sqrt{ g R}\)
A ball is thrown vertically downward from a height of \(20\) 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, \(g=10~\mathrm{ms^{-2}}\))
1. \(14\) ms-1
2. \(20\) ms-1
3. \(28\) ms-1
4. \(10\) 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. | \(W_P<W_Q;~W_P<W_Q\) |
2. \(450\) J
3. \(275\) J
4. \(250\) J
A particle of mass \(m\) is driven by a machine that delivers a constant power of \(k\) watts. If the particle starts from rest, the force on the particle at time \(t\) is:
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. | \(\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2+E = \frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2\) |
On a frictionless surface, a block of mass \(M\) moving at speed \(v\) collides elastically with another block of the same mass \(M\) which is initially at rest. After the collision, the first block moves at an angle \(\theta\) to its initial direction and has a speed \(\frac{v}{3}\). The second block’s speed after the collision will be:
1. \(\frac{2\sqrt{2}}{3}v\)
2. \(\frac{3}{4}v\)
3. \(\frac{3}{\sqrt{2}}v\)
4. \(\frac{\sqrt{3}}{2}v\)
A body of mass (\(4m\)) is lying in the x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (\(m\)) move perpendicular to each other with equal speeds (\(u\)). The total kinetic energy generated due to explosion is:
| 1. | \(mu^2\) | 2. | \(1.5~mu^2\) |
| 3. | \(2~mu^2\) | 4. | \(3~mu^2\) |
© 2024 GoodEd Technologies Pvt. Ltd.