The periodic time of simple pendulum is T=2pi sqrt((l)/(g)). The length (l) of the pendulum is about 100 cm measured with 1mm accuracy. The periodic time is about 2s. When 100 oscillations are measured by a stop watch having the least count 0.1 second. Calcaulte the percentage error in measurement of g.

The periodic time of simple pendulum is T=2pi sqrt((l)/(g)). The lengt

1.	The periodic time of simple pendulum is T=2pi sqrt((l)/(g)). The length (l) of the pendulum is about 100 cm measured with 1mm accuracy. The periodic time is about 2s. When 100 oscillations are measured by a stop watch having the least count 0.1 second. Calcaulte the percentage error in measurement of g.
Answer» `0.1%` 0.01 `0.2%` `0.8%` Solution :`T=nT=100xx2=200s and Deltat=0.1s` `T=2pisqrt((l)/(g))"":.T=(4pi^(2)l)/(g)` `:.g=(4pi^(2)l)/(T^(2))` `4pi^(2)`= CONSTANT =dimensionless `:.(Deltag)/(g)=(Deltal)/(l)+(2DeltaT)/(T)` `=(Deltal)/(l)xx100+2(Deltat)/(t)xx100` `=[(0.1)/(100)xx100]+[2xx(0.1)/(2xx100)xx100]` `=0.1+0.1=0.2%`