If the length and tiome period of an oscillating pendulum have errors of 1% and 3% respectgively then the error in measurement of acceleratinon due to gravity is

Options

A. `4%`
B. `5%`
C. `6%`
D. `7%`

Answer

D

Explanation

`"Time period T"=2pi sqrt((l)/(g)), T^(2)=4pi^(2)(l)/(g)`.
`g=4pi^(2)(l)/(T^(2))`
`therefore (Deltag)/(g)=(Deltal)/(l)+2(DeltaT)/(T)`
`" Given "(Deltal)/(l)=1%`
`(DeltaT)/(T)=3%`
`=1%+2(3%)=7%`