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.

Options

A. `0.1%`
B. 0.01
C. `0.2%`
D. `0.8%`

Answer

B

Explanation

`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%`