1. | \(3000\) m | 2. | \(2800\) m |
3. | \(2000\) m | 4. | \(1000\) m |
1. | \(4\sqrt2~\text{ms}^{-1},45^\circ\) | 2. | \(4\sqrt2~\text{ms}^{-1},60^\circ\) |
3. | \(3\sqrt2~\text{ms}^{-1},30^\circ\) | 4. | \(3\sqrt2~\text{ms}^{-1},45^\circ\) |
1. | \(20\) | 2. | \(10\sqrt3\) |
3. | zero | 4. | \(10\) |
1. | \(\overrightarrow v\) is a constant; \(\overrightarrow a\) is not a constant. |
2. | \(\overrightarrow v\) is not a constant; \(\overrightarrow a\) is not a constant. |
3. | \(\overrightarrow v\) is a constant; \(\overrightarrow a\) is a constant. |
4. | \(\overrightarrow v\) is not a constant; \(\overrightarrow a\) is a constant. |
1. | \(10\) ms–1 | 2. | zero |
3. | \(5\sqrt3\) ms–1 | 4. | \(5\) ms–1 |
Rain is falling vertically downward with a speed of \(35~\text{m/s}\). Wind starts blowing after some time with a speed of \(12~\text{m/s}\) in East to West direction. The direction in which a boy standing at the place should hold his umbrella is:
1. | \(\text{tan}^{-1}\Big(\frac{12}{37}\Big)\) with respect to rain |
2. | \(\text{tan}^{-1}\Big(\frac{12}{37}\Big)\) with respect to wind |
3. | \(\text{tan}^{-1}\Big(\frac{12}{35}\Big)\) with respect to rain |
4. | \(\text{tan}^{-1}\Big(\frac{12}{35}\Big)\) with respect to wind |
A car starts from rest and accelerates at \(5~\text{m/s}^{2}\). At \(t=4~\text{s}\), a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at \(t=6~\text{s}\)? (Take \(g=10~\text{m/s}^2)\)
1. \(20\sqrt{2}~\text{m/s}, 0~\text{m/s}^2\)
2. \(20\sqrt{2}~\text{m/s}, 10~\text{m/s}^2\)
3. \(20~\text{m/s}, 5~\text{m/s}^2\)
4. \(20~\text{m/s}, 0~\text{m/s}^2\)
A particle moving in a circle of radius \(R\) with a uniform speed takes a time \(T\) to complete one revolution. If this particle were projected with the same speed at an angle \(\theta\) to the horizontal, the maximum height attained by it equals \(4R.\) The angle of projection, \(\theta\) is then given by:
1. \( \theta=\sin ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1/2}\)
2. \(\theta=\sin ^{-1}\left(\frac{2 {gT}^2}{\pi^2 {R}}\right)^{1 / 2}\)
3. \(\theta=\cos ^{-1}\left(\frac{{gT}^2}{\pi^2 {R}}\right)^{1 / 2}\)
4. \(\theta=\cos ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1 / 2}\)
The speed of a swimmer in still water is \(20\) m/s. The speed of river water is \(10\) m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes with respect to the north is given by:
1. | \(45^{\circ}\) west of north | 2. | \(30^{\circ}\) west of north |
3. | \(0^{\circ}\) west of north | 4. | \(60^{\circ}\) west of north |