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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is given by `T=2pisqrt((l)/(g))` where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

A

`0.1%`

B

`1%`

C

`0.2%`

D

`0.8%`

Text Solution

Verified by Experts

The correct Answer is:
C
`T=2pisqrt(l//g)impliesT^(2)=4pi^(2)l//gimpliesg=(4pi^2l)/T^2`
Here % error in `l=(1"mm")/(100 cm) xx100 =(0.1)/(100) xx100 =0.1`
ad % error in T `=(0.1)/(2xx100) xx100=0.05%`
`:.%` error in g = % error l + 2 (%error in T )
` = 0.1 +2xx0.05 =0.2%`
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