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The time period of oscillation of a simp...

The time period of oscillation of a simple pendulum is given by `T=2pisqrt(l//g)`
The length of the pendulum is measured as `1=10+-0.1` cm and the time period as `T=0.5+-0.02s`. Determine percentage error in te value of g.

Text Solution

Verified by Experts

The correct Answer is:
`9%`
`T=2pi sqrt(l/g)`
`T^(2)=4pi^(2) l/g`
`:. g=4pi^(2) l/T^(2)`
`(Deltag)/g=(Deltal)/l+(2 Delta T)/T`
Hence, `((Deltag)/g)%=[0.1/10xx100+2xx0.02/0.5xx100]%`
`=(1+8)%=9%`
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