While measuring 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 the time period. His percentage error in the measurment of the value of g 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)`