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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)).` Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is

A

`2%`

B

`3%`

C

`1%`

D

`5%`

Text Solution

Verified by Experts

The correct Answer is:
B
g =`4 pi^(2)(L)/(T^(2))`
or, Ing = In `(4pi^(2))` + In L -2 In T
Differentiating we get
`(dg)/(g)=(dL)/(L) +2 ((-dT)/(T))`
= `(1 "mm")/(20.0"cm")+2xx(1s)/(90s)` [To determine the maximum error, -dT has been taken as 1s]
`=(1)/(200)+(2)/(90)= 0.0272 = 2.72% ~~ 3%`
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