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

Options

A. `0.1%`
B. `1%`
C. `0.2%`
D. 0.8%`

Answer

C

Explanation

`T = 2 pi sqrt(1//g)rArr T^(2) = 4 pi^(2) 1//g rArr g = (4 pi^(2)l)/(T^(2))`
`Here % error in l = (1 mm)/(100cm) xx 100 = (0.1)/(100) xx 100 = 0.1%`
and `% "error in" T = (0.1)/(2 xx 100) xx 100 = 0.05%`
`:. % "error in" g = % "error in" l + 2(% "error in" T)`.