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Q. No.Given below are two statements:
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).
Subtopic: Â Relative Motion |
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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. | \(a = g \) |
Subtopic: Â Circular Motion |
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Two particles \(A\), \(B\) are projected simultaneously from the base of a triangle \(ABC\). Particle \(A\) is projected from vertex \(A\) along \(AC,\) and particle \(B\) is projected from vertex \(B\) along \(BC\). Their respective velocities are \(v_A\) & \(v_B\) and they move with uniform velocities. For the particles to collide:
| 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. | \(v_A~\text{tan}A=v_B~\text{tan}B\) |
Subtopic: Â Relative Motion |
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Particles are simultaneously projected in all possible directions from a point in space, located in a uniform gravitational field. The initial speed of the particle is \(u.\) The maximum separation between any two particles, after a time \(t,\) is:
| 1. | \(ut\) | 2. | \(2ut\) |
| 3. | \(ut+\dfrac{1}{2}gt^2\) | 4. | \(2ut+gt^2\) |
Subtopic: Â Projectile Motion |
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A boat is rowed across a \(400~\text m\) wide river so that it can reach the opposite bank in a minimum of \(10\) minutes. No matter which direction the boat is rowed in, it cannot reach a point exactly opposite on the other bank, unless it is rowed at a slightly higher speed. The speed of flow of the river is:
| 1. | \(2.4~\text{km/h}\) | 2. | \(4.8~\text{km/h}\) |
| 3. | \(2.4\sqrt2~\text{km/h}\) | 4. | \(\dfrac{2.4}{\sqrt2}~\text{km/h}\) |
Subtopic: Â Relative Motion |
 59%
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Raindrops fall from the sky making an angle of \(30^\circ\) with the vertical. If a man runs at \(2\) m/s, he finds that the drops fall vertically. If he were to run in the opposite direction with the same speed\(,\) the raindrops will fall with a vertical speed of:
| 1. | \(2\) m/s | 2. | \(4\) m/s |
| 3. | \(2\sqrt3 \) m/s | 4. | \(4\sqrt3 \) m/s |
Subtopic: Â Relative Motion |
 63%
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Wind blows towards north with a speed of \(20\) m/s. A plane flies in air with a velocity relative to ground of \(100\) m/s, towards east. Then, the velocity of the plane relative to air is:
| 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. |
Subtopic: Â Relative Motion |
 69%
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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. | \(\dfrac{h_A}{\text{sin}^2\theta}=h_B\) |
Subtopic: Â Projectile Motion |
 53%
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The average velocity of a projectile from the point of projection to impact is \(v_1\) while the average velocity from projection to maximum height\((H)\) is \(v_2\).
It can be concluded that:
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 |
Subtopic: Â Projectile Motion |
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A man, swimming with a speed \(u_1,\) can cross a river fastest in a time, \(T\). His friend, who swims with a speed \(u_2,\) reaches the opposite bank in the same time when he swims at an angle of \(30^{\circ}\) with the bank. Then:
| 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. | \(u_1=\dfrac{1}{\sqrt3}u_2\) |
Subtopic: Â Relative Motion |
 58%
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