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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%`

Text Solution

Verified by Experts

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