A student uses a simple pendulum of exactly 1 m length to determine g , the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations.

Question

A student uses a simple pendulum of exactly 1 m length to determine g , the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statement(s) is/are true?

TestHub · Accepted Answer

T = 20 40 s ​ = 2 s . Further, t = n T = 20 T or Δ t = 20Δ T ∴ t Δ t ​ = T Δ T ​ or Δ T = t T ​ ⋅ Δ t = ( 40 2 ​ ) ( 1 ) = 0.05 s Further, T = 2 π g l ​ ​ or T ∝ g − 1/2 ∴ T Δ T ​ × 100 = − 2 1 ​ × g Δ g ​ × 100 or % error in determination of g is g Δ g ​ × 100 ​ = − 200 × T Δ T ​ = − 2 200 × 0.05 ​ = − 5% ​ ∴ correct options are (a) and (c).

Physics - Error Question with Solution | TestHub

Options:(select one or more)

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