Question

The length of a simple pendulum is about 100 cm known to have an accuracy of 1mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1s resolution. What is the accuracy in the determined value of g?
A. 0.002
B. 0.005
C. 0.001
D. 0.02

Answer 1

Correct Answer - A
`T = t/n` and `DeltaT = (Deltat)/(n)`
` :. (DeltaT)/(T) xx 100 = (Deltat)/(t) xx 100`
`= (0.1)/(2 xx 100) xx 100 = 0.05%`
`(Deltal)/(l) xx 100 = (0.1cm)/(100cm) xx 100 `
`=0.1%`
Now, `T = (2pi)sqrt((i)/(g))`
or `g= (4pi^2l)/(T^2) prop (l)/(T^(2))`
`:. %` error in `g =` (`%` error in `l`) `+2`(`%` error in `T`)
`= 0.1% + 2(0.05%) = 0.2%`.