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

Time period of simple pendulum on earth having effective length is l
`T^(2)=(4pi^(2)l)/(g) implies g= 4pi^(2)(l)/(T^2)`
`(triangleg)/(g)xx100=((trianglel)/(l)xx100)+2((triangleT)/(T)xx100)`
`=((0.1)/(20)xx100)+2((0.01)/(0.9)xx100)`
`=0.5+2xx(10)/(9)=0.5+2.2=2.7%`