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

(3) `T = 2 pi sqrt((l)/(g)) implies g = (4 pi^(2) l)/(T^(2))`
`(Delta g)/(g) = (Delta l )/(l) + 2 (Delta T)/(T)`
`(Delta g)/(g) xx 100 = (Delta l)/(l) xx 100 + 2 (Delta T)/(T) xx 100`
`(0.1)/(100) xx 100 + 2 (0.1)/(200) xx 100`
`= 0.1 + 0.1 = 0.2%`