Drops of water fall from the roof of a building 20 m high at regular intervals

of time. The first drop reaching the ground at the same instant fifth drop

starts its fall. What are the distance  of second and third drops from roof? (g =

10 $m/s^2$)

1. 5.0 m and 1.25 m

2. 5.0 m and 2.25 m

3. 11.25 m and 5.0 m

4. 11.25 and 1.25 m

A particle is dropped vertically from rest from a height. The time taken by it to

fall through successive distances of 1 m will the be

A ball rolls off the top of stairway with a horizontal velocity of magnitude 1.8 m/s. The steps are 0.20 m high and 0.20 m wide. Which step will the ball hit first ?

1. First

2. Second

3. Third

4. Fourth

A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t, the distance travelled by the car is:

1. $\frac{1}{2}\left(\frac{\alpha \beta }{\alpha +\beta }\right)t^2$

2. $\frac{1}{2}\left(\frac{\alpha +\beta }{\alpha \beta }\right)t^2$

3. $\frac{1}{2}\left(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right)t^2$

4. $\frac{1}{2}\left(\frac{{\alpha }^2-{\beta }^2}{\alpha \beta }\right)t^2$

A train accelerates from rest at a constant rate $\mathrm{\alpha }$ for distance ${\mathrm{x}}_1$ and time ti. After that, It retards to rest at a constant rate $\mathrm{\beta }$ for distance ${\mathrm{x}}_2$ and time ${\mathrm{t}}_2$. Then it is found that

1. $\frac{{\mathrm{x}}_1}{{\mathrm{x}}_2} = \frac{\mathrm{\alpha }}{\mathrm{\beta }} = \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$

2. $\frac{{\mathrm{x}}_1}{{\mathrm{x}}_2} = \frac{\mathrm{\beta }}{\mathrm{\alpha }} = \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$

3. $\frac{{\mathrm{x}}_1}{{\mathrm{x}}_2} = \frac{\mathrm{\beta }}{\mathrm{\alpha }} = \frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$

4. $\frac{{\mathrm{x}}_1}{{\mathrm{x}}_2} = \frac{\mathrm{\alpha }}{\mathrm{\beta }} = \frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}$

A man in a lift ascending with an acceleration throws a ball vertically upwards with a velocity u and catches it after time ${\mathrm{t}}_1$. Afterwards, when the lift is descending with the same acceleration, the man again throws the ball vertically upwards and catches it after time ${\mathrm{t}}_2$. The velocity of the projection of the ball:

$1. \frac{{\mathrm{gt}}_1{\mathrm{t}}_2}{\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)}$
$2. \frac{{\mathrm{gt}}_1{\mathrm{t}}_2}{\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)}$
$3. \frac{{\mathrm{gt}}_1{\mathrm{t}}_2}{2\left({\mathrm{t}}_1-{\mathrm{t}}_2\right)}$
$4. \frac{{\mathrm{gt}}_1{\mathrm{t}}_2}{2\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)}$

A vertical circular disc has different grooves along various chords as shown in fig. The particles are released from upper end O. The times taken by the particles to reach the ends of the grooves OA, OB and OC respectively are in the ratio

1. 1:1:1

2. 1:2:$\sqrt{3}$

3. 1: $\sqrt{3}$:2

4. 1: $\sqrt{3}$: $\sqrt{2}$

A ball is dropped vertically from a height above the ground. It hits the ground and bounced up vertically to a height d/2. neglecting subsequent motion and the air resistance, its velocity v with height h above the ground can be represented as 

1.  

2.  

3.  

4.  

In the given figure, \(a=15\) m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius \(R=2.5\) m at a given instant of time. The speed of the particle is:

              

1. \(4.5\) m/s
2. \(5.0\) m/s
3. \(5.7\) m/s
4. \(6.2\) m/s

A particle moves in the x-y plane according to rule $\mathrm{x}=\mathrm{asin\omega t}$ and $\mathrm{y}=\mathrm{acos\omega t}$. The particle follows: