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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is T =`2pisqrt(L/g)`. Measuted value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

A

`3%`

B

`1%`

C

`5%`

D

`2%`

Text Solution

Verified by Experts

The correct Answer is:
A
`g=4pi^(2).(l)/(T^(2))`
`implies(Deltag)/(g)xx100=(Deltal)/(l)xx100+2(DeltaT)/(T)xx100`
`(Deltal)/(l)xx100+2.(Deltat)/(t)xx100`
`(0.1)/(20.0)xx100+2xx(1)/(90)xx100`
`(100)/(200)+(200)/(90)=(1)/(2)+(20)/(9)=3%`
So correct choice is `(a)`.
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