Q. No.A particle is projected with a speed \(u\) so that it has the maximum horizontal range under gravity. The speed is increased to \(v\) (without changing the direction of its projection), so that, after projection, it passes above its previous point of impact but at a height which is equal to its previous maximum height. Then,
1. \(2u=v\)
2. \(3u=2v\)
3. \(2u = \sqrt 3 v\)
4. \(3u =\sqrt 2 v\)
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| Assertion (A): | If two particles move with uniform accelerations in different directions, then their relative velocity changes in direction. |
| Reason (R): | Since the acceleration are in different directions, there is a relative acceleration and hence the relative velocity changes. |
| 1. | (A) is True but (R) is False. |
| 2. | (A) is False but (R) is True. |
| 3. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 4. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
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| 1. | \(a<\dfrac{2 g}{5}\) |
| 2. | \(\dfrac{2 g}{5}< a< \dfrac{3 g}{5}\) |
| 3. | \(\dfrac {3g} {5} <a<g\) |
| 4. | \(a = g \) |
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| 1. | \(v_A~\text{cos}A=v_B~\text{cos}B\) |
| 2. | \(v_A~\text{sin}A=v_B~\text{sin}B\) |
| 3. | \(\dfrac{v_A}{\text{sin}A}=\dfrac{v_B}{\text{sin}B}\) |
| 4. | \(v_A~\text{tan}A=v_B~\text{tan}B\) |
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| 1. | \(ut\) | 2. | \(2ut\) |
| 3. | \(ut+\dfrac{1}{2}gt^2\) | 4. | \(2ut+gt^2\) |
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| 1. | \(2.4~\text{km/h}\) | 2. | \(4.8~\text{km/h}\) |
| 3. | \(2.4\sqrt2~\text{km/h}\) | 4. | \(\dfrac{2.4}{\sqrt2}~\text{km/h}\) |
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| 1. | \(2\) m/s | 2. | \(4\) m/s |
| 3. | \(2\sqrt3 \) m/s | 4. | \(4\sqrt3 \) m/s |
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| 1. | \(100\) m/s towards east and \(20\) m/s towards north. |
| 2. | \(100\) m/s towards east and \(20\) m/s towards south. |
| 3. | \(100\) m/s towards west and \(20\) m/s towards north. |
| 4. | \(100\) m/s towards west and \(20\) m/s towards south. |
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Consider the two situations shown in the figures. In figure (A), the particle \(P\) is given a velocity \(u\) up a smooth horizontal incline and it reaches a maximum vertical height \(h_A\). In figure (B), the same particle \(P\) is projected with a velocity \(u\) at an angle \(\theta\) (parallel to the previous incline) and reaches a maximum height \(h_B\).Then,
| 1. | \(h_A=h_B~\text{sin}\theta\) |
| 2. | \(h_A~\text{sin}\theta=h_B\) |
| 3. | \(h_A~\text{sin}^2\theta=h_B\) |
| 4. | \(\dfrac{h_A}{\text{sin}^2\theta}=h_B\) |
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It can be concluded that:
| 1. | \(v_1>v_2\) |
| 2. | \(v_1<v_2\) |
| 3. | \(v_1=v_2\) |
| 4. | Any of the above can be true depending on the angle of projection |
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