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. 1:02:03
• B. 1:04:09
• C. 1:03:05
• D. 1:05:03

Answer 1

Answer: (C)

1:03:05

Answer 2

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