The time `T` of oscillation of as simple pendulum of length `l` is given by `T=2pi sqrt(l/g)`. The percentage error in `T` corresponding to an error of 2% in the value of `l` is

Options

A. 0.02
B. 0.01
C. 0.03
D. 0.012

Answer

B

Explanation

Given `T=2pi sqrt(l/g)`……………….i
On taking log both sides of Eq. (i) we get
`logT=log2+logpi+1/2log l -1/2 log g`……………..ii
On differentiating both sides of Eq. (ii) w.r.t `l` we get
`1/T(dT)/(dl)=1/(2l)implies1/T . (dT)/(dl) Deltal=1/(2l)Deltal`
`implies1/T Delta T=1/(2l). Deltal [ :' DeltaT=(dT)/(dl). Deltal]`
`implies((DeltaT)/Txx100)=1/2xx2=1`
`:.` Percentage error in `T=1%`