The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity g from the above measurement.

Answer

Length of simple pendulum (l) = 20 cm
absolute error in length `(Deltal)` = 2 mm = 0.2 cm
Time taken for 50 oscillation (t) = 40 s
error in time `DeltaT = 1` s

Explanation

Time period for one oscillation (T)
`= t/n = (40)/(50) s`
`:. DeltaT = (Deltat)/(n)`
So, `(DeltaT)/(T) = (Deltat//n)/(t//n) = (Deltat)/(t)`
`Deltal, DeltaT` are least count errors `(Deltag)(g) = (Deltal)/(l) + 2 (DeltaT)/(T) = (0.2)/(20) +2 ((1)/(40)) = (1.2)/(20) = 0.06`
Hence, the percentage error in g is
`((Deltag)/(g)) xx 100 = ((Deltal)/(l) +2 (DeltaT)/(T)) xx 100% = 0.06 xx 100% = 6%`