Question

The period of oscillation of a simple pendulum is T = `2π sqrt"L"/"g"`. The measured value of L is 20.0 cm known to have 1 mm accuracy and the time for 100 oscillations of the pendulum is found to be 90 s using a wristwatch of ls resolution. The accuracy in the determination of g is:

Options

• 1%

• 5%

• 2%

• 3%

Answer 1

3%

Explanation:

As, g = `(4^2 "l")/"T"^2`

So, `(Δ"g")/"g" xx 100 = (Δ"l")/"l" xx 100 + 2 (Δ"T")/"T" xx 100`

= `0.1/20 xx 10 + 2 xx 1/90 xx 100 = 2.72`

= 3%