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 roots of the integers `1, 2, 3,.............
C. In the ratio of the disfference in the square roots of the integers, i.e., sqrt1, (sqrt2, -sqrt1),(sqrt3-sqrt2), (sqrt4-sqrt3)`......
D. In the ratio of the rectiprocals of the square roots of the integers, i.e., (1)/(sqrt1), (1)/(sqrt2), (1)/(sqrt(3)`,........

Answer

C

Explanation

Time taken to cover first `n` meter is givenby
`n=(1)/(2) g t_(n)^(2)` or `t_(n) =sqrt((2n)/(g))`
Time taken to cover (n+1 )th `m` is givenby
`t _(n +t)= 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)-sqrtn]`
This gives the required ratio as
`sqrt1, sqrt2 -sqrt1), sqrt3- sqrt2),...,etc.., ("starting from n=0")`.