An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:

Options

A. `1.7%`
B. `2.7%`
C. `4.4%`
D. `2.27%`

Answer

B

Explanation

According to the question.
`t-(90pm1) or, (Deltat)/(t)=(1)/(90)`
`l-(20 pm0.1)or, (Deltal)/(l)=(0.1)/(20)`
As we know,
`t=2pi sqrt((l)/(g)) rArr g=(4pi^(2)l)/(t^(2))`
`"or, "(Deltag)/(g)=pm ((Deltal)/(l)+2(Deltat)/(t))=((0.1)/(20)+2xx(1)/(90))=0.027`
`therefore" "(Deltag)/(g)%=2.7%`