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?

Options

A. `2%`
B. `3%`
C. `4%`
D. `5%`

Answer

D

Explanation

Here `T= 2pisqrt((L)/g)`
Squaring both sides, we get `T^(2)= 4pi(L)/(g)` or `g= 4pi^(2)(L)/(T^(2))` ….(i)
Take log and differentiate both sides of equation (i), we get
`(Deltag)/(g)= (DeltaL)/(L)- 2((DeltaT)/(T))`
For maximum relative error, the individual errors should be added
`:. (Deltag)/(g)= (DeltaL)/(L)+ 2((DeltaT)/(T))` Here `T= (t)/(n), DeltaT= (Deltat)/(t):. ((DeltaT)/(T))= (Deltat)/(t)`
As errors in both L and t are the least count errors.
`:. (Deltag)/(g)= (0.1)/(10)+ 2((1)/(50))= 0.01+ 0.04= 0.05`