While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value and a of time period. His percentage error in the measurement of g by the relation `g = 4pi^(2)(l//T^(2))` will be

Options

A. `2%`
B. `4%`
C. `7%`
D. `10%`

Answer

C

Explanation

`T= 2pi sqrt((l)/(g)) rArr g = (4pi^(2) l)/( T^(2))`
`(Delta g)/(g) xx 100 = (Delta l)/(l ) xx 100 + (2 Delta T)/(T) xx 100`
`" "= 1 % + 2(3 % ) = 7%`