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Water drops fall at regular intervals fr...

Water drops fall at regular intervals from a roof. At an instant when a drop is about to leave the roof, the separations between 3 successive drops below the roof are in the ratio

A

1:2:3

B

1:4:9

C

1:3:5

D

1:5:13

Text Solution

Verified by Experts

The correct Answer is:
B
Suppose t be the time interval over which successive drops are released from roof
`h= ut+(1)/(2)gt^(2)=(1)/(2) gt^(2)`
`therefore (u=0)`
Thus , we find that `halphat^(2)`
For the first drop `t_(1) = t`
For the second drop `t_(2)` = 2t
For the third drop `t_(3) = 3t`
Hence the corresponding height are
`h_(1)alphat_(1)^(2). h_(2)aplhat_(2)^(2), h_(3)alphat_(3)^(2)`
`therfore (h_(1)/(h_(2)= ((t_(1)/(t_(2))^(2)= (1)/(4)` and `(t_(1)/(t_(3))^(1)=(1)/(9)` and `(h_(2)/(h_(3)) = (t_(2))/(t_(3))^(2) = (4)(/(9)`
Thus it is clear that , `H_(1): H_(2), : H_(3)= 1: 4: 9` ie
Correct option is (2)
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