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Question

Let A, B, C, D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the time interval of descent through AB, BC and CD are in the ratio:

A
1:32:3+2
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B
1:21:32
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C
1:21:3
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D
1:2:31
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Solution

The correct option is B 1:21:32
We have a standard expression for time of descent when a body is released from rest. t=2hg ..........(i)
Let AB=hAC=2h and AD=3h
Time for descent up to B from A: tAB=2hg
Time for descent up to C from A: tAC=4hg=2hg×2
Time for descent up to D from A: tAD=6hg=2hg×3
Time taken along BC: tBC=tACtAB=2hg×(21)
Time taken along CD: tCD=tADtAC=2hg×(32)
Hence the ratio tAB:tBC:tCD=1:21:32
Note: Equation (i) can be derived from the laws of motion with uniform acceleration
Consider s=ut+12at2.
When a body is released from rest from a height h and acceleration due to gravity is g, we have s=h;a=g; t being the time of flight which gives h=12gt2

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