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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 oscillation is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of `g` is close to :

A

`0.2% `

B

`0.7%`

C

`6.8% `

D

`3.5%`

Text Solution

Verified by Experts

The correct Answer is:
C
`g = (4pi^2 l)/(T^2)`
`(Deltag)/g =(Deltal)/l+2(DeltaT)/T`
`=(0.1)/55+2(1/30)`
`(Deltag)/g xx100 =10/55 +20/3 =6.8%`
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