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Question

# A particle is released from rest from a tower of height 3h.The ratio of times to fall equal height h,i.e., t1: t2 :t3 is

A
3:2: 1
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B
3 : 2 : 1
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C
9 :4 : 1
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D
1 : (2 - 1) :(321)
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Solution

## The correct option is D 1 : (√2 - 1) :(√3−√2−1)Time in falling first height h will be t1=T1=2√2hgTime in falling first height 2h will be T2=2√2(2h)gso the time in falling second h will be t2=T2−T1=2√hg(2√4−2√2)Time in falling in height 3h will be T3=2√2(3h)gso the time in falling third h will be t3=T3−(T1+T2)=2√hg(2√6−2√4−2√2)the ratio will be t1:t2:t3=2√2:2√2(2√2−1):2√2(2√3−2√2−1)2√2 is common in all so can be cancelled so the answer is option D.

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