A student measures that distance traversed in free fall of a body, initially at rest in given time. He uses this data to estimated `g`, the acceleration due to gravity. If the maximum percentage error in measurement of the distance and the time are `e_(1)` and `e_(2)`, respectively, the percentage error in the estimation of `g` is

Options

A. `e_2 - e_1`
B. `e_1+ 2e_2`
C. `e_1 + e_2`
D. `e_1 - 2e_2`

Answer

(d)

Explanation

Given `(Delta s)/(s) = e_1 and (Delta t)/(t) =e_2`
When a body falls vertically for time t seconds, The vertical distance covered is, `s = (1)/(2) gt^2`
Differentating it, we have
`Delta s = (1)/(2) [Delta gxxt^2 + g 2t Delta t]`
`(Delta s)/(s) = (1)/(2) [(Delta gxxt^2)/((1)/(2)gt^2) + (2g t Delta t)/((1)/(2)gt^2)]`
`= (1)/(2) [(2 Delta g)/(g) + (4 Delta t)/(t)] = (Delta g)/(g) +(2 Delta t)/(t)`
`:. (Delta g)/(g) = (Delta s)/(s) = e_1 - 2e_2`