A simple pendulum consisting of a bob of mass \(m\), and a string of length \(L\) is given a horizontal speed \(u\), at its lowest point as shown in the figure. As a result, it rises to \(B\), where it just comes to rest momentarily with \(OB\) horizontal. During the motion \(AB\text:\)

| 1. | Work done by the string is zero |
| 2. | Work done by gravity is \(-mgL\) |
| 3. | Change in K.E. of the bob is \(-\dfrac{1}{2}mu^2\) |
| 4. | All the above are true |
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. \(\sqrt{2 gR}\)
2. \(\sqrt{3 g R}\)
3. \(\sqrt{5 g R}\)
4. \(\sqrt{g R}\)
A body is falling freely under the action of gravity alone in a vacuum. Which of the following quantities remain constant during the fall?
| 1. | kinetic energy |
| 2. | potential energy |
| 3. | total mechanical energy |
| 4. | total linear momentum |
| 1. | \(2\sqrt{10}\) ms–1 | 2. | \(2\sqrt{5}\) ms–1 |
| 3. | \(4\sqrt{10}\) ms–1 | 4. | \(4\sqrt{5}\) ms–1 |
When a projectile is projected under a uniform gravitational field (air resistance is negligible):
| 1. | Its kinetic energy is conserved. |
| 2. | Its potential energy is conserved. |
| 3. | The sum of potential and kinetic energies is conserved. |
| 4. | Energy conservation is not valid for projectile motion. |

| Assertion (A): | According to the law of conservation of mechanical energy change in potential energy is equal and opposite to the change in kinetic energy. |
| Reason (R): | Mechanical energy is not a conserved quantity. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |