The time period of a simple pendulum is given by `T = 2 pi sqrt(L//g)`, where `L` is length and `g` acceleration due to gravity. Measured value of length is `10 cm` known to `1 mm` accuracy and time for 50 oscillations of the pendulum is 80 s using a wrist watch of 1 s resloution. What is the accuracy in the determination of `g`?

Options

A. 0.02
B. 0.03
C. 0.04
D. 0.05

Answer

D

Explanation

Here, `T=2pisqrt((L)/(g))`
Squaring both sides, we get, `T^(2)=(4pi^(2)L)/(g) ` or `g=(4pi^(2)L)/(T^(2))`
The relative error in g is, `(Deltag)/(g)=(DeltaL)/(L)+2(DeltaT)/(T)`
Here, `T=(t)/(n) and DeltaT=(Deltat)/(n) therefore(DeltaT)/(T)=(Deltat)/(t)`
The errors in both L and t are the least count errors
`therefore(Deltag)/(g)=(0.1)/(10)+2((1)/(50))=0.01+0.04=0.05`
The percentage error in g is
`(Deltag)/(g)xx100=(DeltaL)/(L)xx100+2((DeltaT)/(T))xx100`
`=[(DeltaL)/(L)+2((DeltaT)/(T))]xx100=0.05xx100=5%`