Q. A ball is released from the top of a tower. The ratio of work done by force of gravity in 1$^{st}$ second,2$^{nd}$ second and 3$^{rd}$ second of the motion of ball is

A) 1 : 2 : 3
B) 1 : 4 : 16
C) 1 : 3 : 5
D) 1 : 9 : 25
Explanation

(c) Initial velocity of ball is zero $ie\ u = 0$
$\therefore$ Displacement of ball in $t^{th}$ second
$s = gt - \frac{1}{2}g = g\left( t - \frac{1}{2} \right)$
$s \propto \left( t - \frac{1}{2} \right)$
or $s_{1}:s_{2}:s_{3} = \left( 1 - \frac{1}{2} \right):\left( 2 - \frac{1}{2} \right):\left( 3 - \frac{1}{2} \right)$
$= 1:3:5$
Now, $W = mgs$
$W \propto s$
$\therefore W_{1}:W_{2}:W_{3} = 1:3:5$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.