Two particles A and B are moving in a uniform circular motion in concentric circles of radii $r_A and r_B$ with speeds $v_A and v_B$ respectively. Their time periods of rotation are the same. The ratio of angular speed of A  to that of B will be:

1. 1: 1

2. $r_A : r_B$

3. $v_A : v_B$

4. $r_B : r_A$

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 north is given by:

1. 45$°$ west of north

2. 30$°$ west of north

3. 0$°$ west of north

4. 60$°$ west of north

A particle starting from rest moves in a circle of radius 'r'. It attains a velocity of $v_0$ m/s on completion of n rounds. Its angular acceleration will be:

1. $\frac{v_0}{n} rad/s^2$

2. $\frac{v_0^2}{2{\mathrm{\pi nr}}^2} rad/s^2$

3. $\frac{v_o^2}{4{\mathrm{\pi nr}}^2} rad/s^2$

4. $\frac{v_0^2}{4\mathrm{\pi nr}} rad/s^2$

Two bullets are fired horizontally and simultaneously towards each other from rooftops of two buildings (building being 100 m apart and being of the same height of 200 m) with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/$s^2$)?

1. after 2s at a height of 180 m

2. after 2s at a height of 20 m

3. after 4s at a height of 120 m

4. they will not collide.

A car starts from rest and accelerates at $5 \mathrm{m}/{\mathrm{s}}^2$. At t = 4s, 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 s?

(Take g = 10 m/s²)

1. $20\sqrt{2} \mathrm{m}/\mathrm{s}, \mathrm{0 }\mathrm{m}/{\mathrm{s}}^2$

2. $20\sqrt{2} \mathrm{m}/\mathrm{s}, 10 \mathrm{m}/{\mathrm{s}}^2$

3. $20 \mathrm{m}/\mathrm{s}, 5 \mathrm{m}/{\mathrm{s}}^2$

4. $20 \mathrm{m}/\mathrm{s}, 0$

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 $\text{'}\mathrm{\theta }\text{'}$ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\mathrm{\theta }$, is then given by:

1. $\mathrm{\theta }={\sin }^{-1}{\left(\frac{{\mathrm{\pi }}^2\mathrm{R}}{{\mathrm{gT}}^2}\right)}^{1/2}$

2. $\mathrm{\theta }={\sin }^{-1}{\left(\frac{2{\mathrm{gT}}^2}{{\mathrm{\pi }}^2\mathrm{R}}\right)}^{1/2}$

3. $\mathrm{\theta }={\cos }^{-1}{\left(\frac{{\mathrm{gT}}^2}{{\mathrm{\pi }}^2\mathrm{R}}\right)}^{1/2}$

4. $\mathrm{\theta }={\cos }^{-1}{\left(\frac{{\mathrm{\pi }}^2\mathrm{R}}{{\mathrm{gT}}^2}\right)}^{1/2}$

Rain is falling vertically downward with a speed of 35 m/s. Wind starts blowing after some time with a speed of 12 m/s in East to West direction. The direction in which a boy standing at the place should hold his umbrella is:
1. \(tan^{-1}\Big(\frac{12}{37}\Big)\)with respect to rain
2. \(tan^{-1}\Big(\frac{12}{37}\Big)\) with respect to wind
3. \(tan^{-1}\Big(\frac{12}{35}\Big)\) with respect to rain
4. \(tan^{-1}\Big(\frac{12}{35}\Big)\) with respect to wind