The period of oscillation of a simple pendulum is `T = 2pisqrt(L//g)`. Measured value of L is `20.0 cm` known to `1mm` accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

Answer

B

Explanation

`g = 4 pie^2L//T^2`
Here, `T= (t)/(n)` and `(DeltaT) = (Deltat)/(n)`. Therefore, `(DeltaT)/(t) = (Deltat)/(t)`.
The errors in both L and t are the least count errors.
Therefore,
` (Delta g)/g = (DeltaL)/L+2(DeltaT)/T `
`= (Delta L)/L + 2(Delta t)/t `
` = (0.1)/(20.0) + 2((1)/(90)) = 0.027 `
Thus, the percentage error in g is `
`( Delta g)/gxx100 = 2.7%`