It is known that for a simple pendulum, time period is given by T=2πLg If the measured value of L is 10 cm known to 1 mm accuracy and also time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of 1 s resolution, then find the error in the determination of g.

Here, 2πLgsquaring both sides, we get,T2=4π2Lg or g=4π2LT2The relative error in g is,Δgg=ΔLL+2ΔTTHere,T=tn and ΔT=Δtn∴ΔTT=ΔttThe errors in both L and t are the least count errors.∴Δgg=0.110+2150=0.01+0.04 = 0.05The percentage error in g isΔgg×100=ΔL¯L×100+2ΔT¯T × 100=ΔLL+2ΔTT×100=0.05×100=5%