Question

If the time period of a simple pendulum is T = 2π√l/g , then the fractional error in acceleration due to gravity is

(A) \(\frac{4\pi^2Δl}{ΔT^2}\) 

(B) \(\frac{Δl}{l}-2\frac{ΔT}{T}\)

(C) \(\frac{Δl}{l}+2\frac{ΔT}{T}\)

(D) None of these

Answer 1

Answer is  (C) \(\frac{Δl}{l}+2\frac{ΔT}{T}\)

We have;