A particle is dropped from rest vertically from a height The time taken by it to fall through successive distances of 1 m cach will then be :

Options

A. All equal being equal to `sqrt(2/g)`sec.
B. In the ratio of square roots of integers 1,2,3,4
C. In the ratio of the difference in the square roots of integers i.e. `(sqrt(1) - sqrt(0)),(sqrt(2) - sqrt(1)) ,(sqrt(3)-sqrt(2))`
D. In the ratio of `(1)/(sqrt(1)):(1)/(sqrt(2)):(1)/(sqrt(3))`

Answer

c

Explanation

Here let `t_1,t_2,t_3` be the time in which the body covers 1m , 2m 3m …. Starting from rest . Then
`1=(1)/(2)g.t_(1)^(2)` or `t_1sqrt((2)/(g))`
`:.` Interval to cover first meter distance
`=(sqrt((2)/(g)-sqrt(0)))=sqrt((2)/(g))(sqrt(1)-sqrt(0))`
Similarly intervel to cover third meter
`=sqrt((2)/(g))(sqrt(3)-sqrt(2))`
Ration of intervals `=(sqrt(1)-sqrt(0)):(sqrt(2)-sqrt(1)):(sqrt(3)-sqrt(2))`