A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then 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 reciprocal of the square roots of the integers i.e. `1/(sqrt(1)) ,1/(sqrt2), 1/(sqrt3), 1/(sqrt4)`.....

Answer

C

Explanation

`h = ut + 1/2 g t^2 implies 1 = 0 xx t_1 + 1/2 g t_1^2 implies t_1 = sqrt(2//g)`
Velocity after travelling Im distance
`v^2 = u^2 + 2gh implies v^2 = (0)^2 + 2g xx 1 implies v = sqrt(2g)`
For second 1 meter distance
`1 = sqrt(2g) xx t_2 + 1/2 g t_2^2 implies g t_2^2 + 2sqrt(2 g t_2) - 2 = 0`
`t_2 =(-2sqrt(2 g) pm sqrt(8g + 8g))/(2g) = (-sqrt(2) pm 2)/(sqrtg)`
Taking +ve sign `t_2 = (2 - sqrt(2))//sqrt(g)`
`therefore (t_1)/(t_2) = (sqrt(2//g))/((2 - sqrt(2))//sqrt(g)) = 1/(sqrt(2) - 1)` and so on.