Q. No.A \(200~\text g\) ball is thrown upward by applying constant upward force \((F)\) over a distance of \(50~\text{cm}\) and then releasing it; the ball ascends to a height of \(10~\text m\) after it is released. The work done by the applied force \(F\) is: (take \(g=10~\text{m/s}^2\))
1. \(20~\text J\)
2. \(19~\text J\)
3. \(21~\text J\)
4. \(42~\text J\)
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| 1. | \(1~\text{m/s}\) | 2. | \(2~\text{m/s}\) |
| 3. | \(\sqrt2~\text{m/s}\) | 4. | \({\Large\frac{1}{\sqrt2}}~\text{m/s}\) |
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| 1. | \(v\propto t\) | 2. | \(v\propto \sqrt t\) |
| 3. | \(v\propto {\Large\frac{1}{\sqrt t}}\) | 4. | \(v\propto t^{3/2}\) |
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The relative velocity between the blocks, after the collision, has the magnitude:
1. \(2~\text{m/s}\)
2. \(1~\text{m/s}\)
3. \({\Large\frac12}~\text{m/s}\)
4. \({\Large\frac14}~\text{m/s}\)
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The momentum of the system, as a result of the collision:
1. increases
2. decreases
3. remains constant
4. decreases suddenly and then increases
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The velocity of the first block, after the collision, is:
| 1. | \({\Large\frac12}~\text{m/s}\) | 2. | \(1~\text{m/s}\) |
| 3. | \(2~\text{m/s}\) | 4. | \(0~\text{m/s}\) |
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| Assertion (A): | In a swinging simple pendulum, the power delivered by gravity is maximum when the pendulum is at the lowest point. |
| Reason (R): | The speed of the pendulum bob is highest when it reaches the lowest point and the power is given by \((\vec F.\vec v).\) |
| 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. |
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1. \(W_N=0,W_g>0\)
2. \(W_N>0,W_g=0\)
3. \(W_N=W_g=0\)
4. \(W_N+W_g=0\)
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1. \(E_1=E_2\)
2. \(E_1=2E_2\)
3. \(2E_1=E_2\)
4. \(E_1=4E_2\)
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1. \({\Large\frac12}\)
2. \({\Large\frac{1}{\sqrt2}}\)
3. \({\Large\frac14}\)
4. \(\sqrt2-1\)
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