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Q. No.Rain is falling vertically downward with a speed of \(35~\text{m/s}.\) The wind starts blowing after some time with a speed of \(12~\text{m/s}\) in the east to the 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 |
The speed of a swimmer in still water is \(20~\text{m/s}.\) The speed of river water is \(10~\text{m/s}\) 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 |
Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being \(100~\text{m}\) apart and being of the same height of \(200~\text{m}\)) with the same velocity of \(25~\text{m/s}.\) When and where will the two bullets collide?
\((g = 10~\text{m/s}^2)\)
| 1. | After \(2~\text{s}\) at a height of \(180~\text{m}\) |
| 2. | After \(2~\text{s}\) at a height of \(20~\text{m}\) |
| 3. | After \(4~\text{s}\) at a height of \(120~\text{m}\) |
| 4. | They will not collide. |
A ship \(A\) is moving westward with a speed of \(10~\text{kmph}\) and a ship \(B,\) \(100 ~\text{km}\) south of \(A,\) is moving northward with a speed of \(10~\text{kmph}.\) The time after which the distance between them becomes the shortest is:
1. \(0~\text{h}\)
2. \(5~\text{h}\)
3. \(5\sqrt{2}~\text{h}\)
4. \(10\sqrt{2}~\text{h}\)
Two particles \({A}\) and \({B}\), move with constant velocities \(\vec{v}_1\) and \(\vec{v}_2\) respectively. At the initial moment, their position vectors are \(\vec{r}_1\) and \(\vec r_2\) respectively. The conditions for particles \({A}\) and \({B}\) for their collision will be:
| 1. | \(\dfrac{\vec{r}_1-\vec{r}_2}{\left|\vec{r}_1-\vec{r}_2\right|}=\dfrac{\vec{v}_2-\vec{v}_1}{\left|\vec{v}_2-\vec{v}_1\right|}\) |
| 2. | \(\vec{r}_1 \cdot \vec{v}_1=\vec{r}_2 \cdot \vec{v}_2\) |
| 3. | \(\vec{r}_1 \times \vec{v}_1=\vec{r}_2 \times \vec{v}_2\) |
| 4. | \(\vec{r}_1-\vec{r}_2=\vec{v}_1-\vec{v}_2\) |
Two boys are standing at the ends \(A\) and \(B\) of the ground where \(AB =a.\) The boy at \(B\) starts running in a direction perpendicular to \(AB\) with velocity \(v_1.\) The boy at \(A\) starts running simultaneously with velocity \(v\) and catches the other boy in a time \(t,\) where \(t\) is:
| 1. | \(\frac{a}{\sqrt{v^2+v^2_1}}\) | 2. | \(\frac{a}{\sqrt{v^2-v^2_1}}\) |
| 3. | \(\frac{a}{v-v_1}\) | 4. | \(\frac{a}{v+v_1}\) |
The width of the river is \(1\) km. The velocity of the boat is \(5\) km/hr. The boat covered the width of the river with the shortest possible path in \(15\) min. Then the velocity of the river stream is:
1. \(3\) km/hr
2. \(4\) km/hr
3. \(\sqrt{29}\) km/hr
4. \(\sqrt{41}\) km/hr
Two particles are separated by a horizontal distance \(x\) as shown in the figure. They are projected at the same time as shown in the figure with different initial speeds. The time after which the horizontal distance between them becomes zero will be:
| 1. | \(\dfrac{x}{u}\) | 2. | \(\dfrac{u}{2 x}\) |
| 3. | \(\dfrac{2 u}{x}\) | 4. | None of the above |
The speed of a boat is \(5\) km/hr in still water. It crosses a river of width \(1\) km along the shortest possible path in \(15\) minutes. The velocity of the river water is:
1. \(3\) km/hr
2. \(4\) km/hr
3. \(5\) km/hr
4. \(2\) km/hr
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Q. No.
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