A particle is dropped from rest from a large height Assume `g` to be constant throughout the motion. The time taken by it to fall through successive distance of `1 m` each will be :

Options

A. all equal, being equal to `sqrt(2//g)` second
B. in the ratio of the square roots of the integers `1,2, 3,`…..
C. in the ratio of the difference in the square roots of the integers, i.e.,
`sqrt(1), (sqrt(2) - sqrt(1)), (sqrt(3) - sqrt(2)), (sqrt(4) - sqrt(3))`, ….
D. in the ratio of the reciprocals of the square roots of the integers, i.e., `(1)/(sqrt(1)),(1)/(sqrt(2)),(1)/(sqrt(3))`,…..

Answer

C

Explanation

Time taken to cover first `n m` is given by
`n = (1)/(2) g t_n^2` or `t_n = sqrt((2 n)/(g))`
Total taken to cover first `(n + 1) m` si given by
`t_(n + 1) = sqrt((2(n + 1))/(g))`
So time taken to cover `( n + 1)^(th) m` is given by
`t_(n+1) - t_n =sqrt(2(n + 1))/(g) -sqrt((2n)/(g))= sqrt((2)/(g))[sqrt(n + 1) -sqrt(n)]`
This gives the required ratio as :
`sqrt(1),(sqrt(2) -sqrt(1)), (sqrt(3) - sqrt(2))`, ...etc.
(starting from n = 0).