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 of time period . His percentage error in the measurement of `g` by the relation ` g = 4 pi^(2) ( l // T^(2))` will be

Options

A. 0.02
B. 0.04
C. 0.07
D. 0.1

Answer

C

Explanation

`T=2pisqrt((l)/(g))rArrg=(4pi^(2)l)/(T^(2))`
`rArr(Deltag)/(g)=(Deltal)/(l)+2(DeltaT)/(T)`