The period of oscillation of a simple pendulum is `T=2pisqrt((L)/(g))`. Measured value of L is 10 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of 1 s resoltuion. What is the accuracy in the determination of g?

Options

A. 1
B. 2
C. 5
D. `0.25`

Answer

C

Explanation

Given`T= 2pi sqrt((l)/(g))`
`or T^(2)=4pi^(2) (l)/(g)`
`therefore 2 (Delta T)/(T)= (Delta l)/(l) + (Delta g)/(g)`…(i)
Now, l=50cm, `Delta l=2mm= 0.2cm`
`(Delta g)/(g)= 1.1% = (1.1)/(100)`
Put these value of Eq. (i), then we get
`(Delta T)/(T) = (1)/(2) [(0.2)/(50) + (1.1)/(100)]`
`=7.5 xx 10^(-3) s or 7.5ms`
`because` In 100s, resolution of clock is 7.5ms
`therefore` In 60s resolution of clock is `(7.5xx 60)/(100)~~5ms`
Hence, option (c ) is correct