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
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a
1 : 2 : 3
b
1 : 4 : 9
c
1 : 3 : 5
d
1 : 5 : 3
answer is C.
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Detailed Solution
When the balt is released from the top of tower n then ratio of distance covered by the balt in first, second and third secondh1:hn:hm=1:3:5 [Because hn∝(2n−1)∴ Ratio of work donemghI : mghII : mghII : mghIII= 1 : 3 : 5