While measuring acceleration due to gravity by simpe 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 measurement of the value of g will be -

Options

A. `5%`
B. `7%`
C. `4%`
D. `2%`

Answer

B

Explanation

`T=2pil^(1//2) g^(-1//2)`
`g=4pi^2T^(-2) l % (Deltag)/g=(2DeltaT)/T+(Deltal)/l`
`%Deltag=2% DeltaT+%l`
`%Deltag`=6 +1 = 7%