Q. No.A particle is released from the top of a smooth hemisphere of radius \(R,\) and it slides down along its surface. After it slides down a height \(\frac R5,\) its acceleration will be \(a,\) where:
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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| 1. | \(u_1=\dfrac{\sqrt3}{2}u_2\) | 2. | \(u_1=\dfrac{1}{2}u_2\) |
| 3. | \(u_1=\dfrac{1}{\sqrt2}u_2\) | 4. | \(u_1=\dfrac{1}{\sqrt3}u_2\) |
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1. \(v=4u\)
2. \(v=2u\)
3. \(v=u\)
4. \(v<u\)
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