Question

A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of the ball is

• A. 1 : 2 : 3
• B. 1 : 4 : 9
• C. 1 : 3 : 5
• D. 1 : 5 : 3

Answer 1

Answer: (C)

Work done $=mgh$
Since,work done is proportional to distance fall $(h)$.
We know
$S=ut+\frac{1}{2}g{t}^2;$
here $u=0,g=10m{s}^{-2}$
$S=5t^2$
if $t=1s;S=5.1^2;S=5$
if $t=2s;S=5.2^2-5.1^2;S=20-5;S=15$
if $t=3s;S=5.3^2-5.2^2;S=45-15;S=25$
Hence, the ratio is $1:3:5$