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The length of a simple pendulum is about...

The length of a simple pendulum is about 100 cm known to an accuray of 1 mm. its period of oscillation oscillations using a clock of 0.1 resolution. What is the accuray in the determined value of g?

A

`0.2%`

B

`0.5%`

C

`0.1%`

D

`2%`

Text Solution

Verified by Experts

The correct Answer is:
A
Time period of a simple pendulum
`T=2pisqrt(l//g)`
`g=(4pi^(2)L^(2))/(T^(2))` . . (i)
Differentiating Eq (i) we have
`(Deltag)/(g)=(DeltaL)/(L)+(2DeltaT)/(T)` . . (ii)
Given `L=100cm` T=2s
`DeltaT=(0.1)/(100)=0.001`s
`DeltaT=(0.1)/(100)=0.001s`
`DeltaT=(0.1)/(100)=0.001s`
`DeltaL=1mm=0.1cm`
substituting the value in Eq. (ii) we have
`|(Deltag)/(g)|_(max)=(DeltaL)/(L)+(2DeltaT)/(T)`
`=(0.1)/(100)+2xx(0.001)/(2)`
Thus, maximum percentage error
`|(Deltag)/(g)|_(max)xx100=((0.1)/(100)xx100)+((2xx0.001)/(2)xx100)`
`=0.1%+0.1%`
`=0.2%`
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