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Q.
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
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 $1: 3: 5$