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

Options

A. `4%`
B. `3%`
C. `2%`
D. `1%`

Answer

B

Explanation

`g=(4pi^(2)L)/(T^(2))` Here, `T=(t)/(m) and DeltaT=(Deltat)/(n), (DeltaT)/(T)=(Deltat)/(t)=(1)/(90)`
Therefore, `((Deltag)/(g))=((DeltaL)/(L))+2((DeltaT)/(T))=(0.1)/(20.0)+2((1)/(90))=0.027`
Thus, the percentage error in g is `3%`