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In a simple pendulum experiment for dete...

In a simple pendulum experiment for determination of acceleration due to gravity `(g)`, time taken for `20` oscillations is measure by using a watch of `1` second least count. The means value of time taken comes out to be 30s. The length pf pendulum is measured by using a meter scale of least count 1mm and the value obtined is `55.0 cm` The percentage error in the determination of `g` is close to :

A

` 0.7 % `

B

` 6.8 % `

C

`3.5 % `

D

` 0.2 % `

Text Solution

Verified by Experts

` T = 2pi sqrt (( l ) /( g)) rArr g = ( 4pi ^ 2 l ) /( T ^ 2) `
`(Delta g ) /(g) xx 100 = ( Delta l) /(l ) xx 100 + 2 ( Delta T ) /(T) xx 100 `
` = (1)/(550) xx 100 + 2 xx (1 ) /(30) xx 100 ~~ 6.8 % `
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