Consider a drop of rainwater having a mass of \(1~\text{gm}\) falling from a height of \(1~\text{km}\). It hits the ground with a speed of \(50~\text{m/s}\). Take \(g\) as constant with a value \(10~\text{m/s}^2.\) The work done by the
(i) gravitational force and the
(ii) resistive force of air is:
1. | \((\text{i})~1.25~\text{J};\) \((\text{ii})~-8.25~\text{J}\) |
2. | \((\text{i})~100~\text{J};\) \((\text{ii})~8.75~\text{J}\) |
3. | \((\text{i})~10~\text{J};\) \((\text{ii})~-8.75~\text{J}\) |
4. | \((\text{i})~-10~\text{J};\) \((\text{ii})~-8.75~\text{J}\) |
A particle moves from a point \(\left(\right. - 2 \hat{i} + 5 \hat{j} \left.\right)\) to \(\left(\right. 4 \hat{j} + 3 \hat{k} \left.\right)\) when a force of \(\left(\right. 4 \hat{i} + 3 \hat{j} \left.\right)\) \(\text{N}\) is applied. How much work has been done by the force?
1. | \(8\) J | 2. | \(11\) J |
3. | \(5\) J | 4. | \(2\) J |
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. \({5\over4}D\)
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. \(\sqrt{ g R}\)
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. | \(W_P<W_Q;~W_P<W_Q\) |
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. | \(\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2+E = \frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2\) |
A uniform force of \((3 \hat{i} + \hat{j})\) newton acts on a particle of mass \(2\) kg. Hence the particle is displaced from position \((2 \hat{i} + \hat{k})\) meter to position \((4 \hat{i} + 3 \hat{j} - \hat{k})\) meter. The work done by the force on the particle is:
1. | \(6\) J | 2. | \(13\) J |
3. | \(15\) J | 4. | \(9\) J |
The potential energy of a system increases if work is done:
1. | by the system against a conservative force |
2. | by the system against a non-conservative force |
3. | upon the system by a conservative force |
4. | upon the system by a non-conservative force |