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 of square roots of the integers `1,2,3,4`
C. in the ration of the differences in the square roots of integers i.e. ` (sqrt 1 - sqrt 0 ), ( sqrt 2)- sqrt 1), (sqrt 3 - sqrt 2 ), (sqrt 4- sqrt 3 ),…`
D. in the ration ` 1/( sqrt 1): 1/(sqrt2) : 1/(sqrt3) : 1/ (sqrt 4)`

Answer

C

Explanation

Using the relation , ` S= ut + 1.2 at^@`, time to fakk
` dustabce (S) whenc ` u = 0, a = g ` will be ` t= sqrt ( 2S)/g`
Let ` T_1, t_2 , t_3 ,…….be the tiem taken to fall ` 1 m, 2 m, 3 m……..respectively Then
` t_1 = sqrt ( 2xx1)/g , t_2 = sqrt (2 xx2)/g , t_3 = sqrt (2 xx3)/g`
so the time taken to fall ` 1 m = t_1 -0`
` = sqrt 2 / g (sqrt 1-0) , time taken to fall ` 2ad metre
` `= t_1 - t_2 = sqrt2 // g 9sqrt 2 - sqrt 1), time taken to fall ` 3rd metre ` = (t_3-t_2) = sqrt 2//g (sqrt3 - sqrt2)` and so one .