The period of oscillation of a simple pendulum is T = 2π (l)/(g) . Measured value of l is 20 . 0cm , known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wristwatch of 1s

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Correct answer: C. 3 %

∴ T = 2π (l)/(g) ⇒ g = 4 π ^ 2 l T^ 2 ∴ Error in~ g can be calculated as ( )/(g) = ( )/(l) + (2Δ T)/(T) .~ ∴ Total time for~ n oscillation is t = nT where T = time for oscillation. ⇒ ( )/(t) = (Δ T)/(T) ⇒ ( )/(g) = ( )/(l) + (2 )/(t) Given that = 1mm = 10 ^ - 3 m, l = 20 × 10 ^ - 2 m = 1s, t = 90s .~ %erroring is ( )/(g) × 100 = (( )/(l) + (2 )/(t)) × 100 = ( 10 ^ - 3 20 × 10 ^ - 2 + (2 × 1)/(90)) × 100= (1)/(2) + (20)/(9)= 0.5 + 2.22= 2.72% ≅ 3%

The period of oscillation of a simple pendulum is T = 2π (l)/(g) ....

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