Question

While measuring the acceleration due to gravity by a simple pendulum, a student makes an error of 1% in the measurement of length and an error of 2% in the measurement of time. If he uses the formula for g as `g=4pi^(2)((L)/(T^(2)))`, then the percentage error in the measurement of g will be
A. `3%`
B. `4%`
C. `5%`
D. `6%`

Answer 1

Correct Answer - c
`g=4pi^(2)((L)/(T^(2)))`
`therefore (Deltag)/(g)=(DeltaL)/(L)+2(DeltaT)/(T)=(1)/(100)+(2xx2)/(100)=5%`