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

Engineering

Physics

Errors

Question

Percentage error in the determination of g is 2.5%

Percentage error in the determination of g is 5%

Error $∆$ T in measuring T, the time period, is 0.05 seconds

Error $∆$ T in measuring T, the time period, is 1 second

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$T = 2 \pm \frac{1}{20} = 2 \pm 0 .05 \sec$

$g = \frac{4 π^{2} ℓ}{T^{2}}$

$\frac{Δg}{g} = \frac{Δℓ}{ℓ} + \frac{2 ΔT}{T} \Rightarrow \frac{Δg}{g} = \frac{0 .05 \times 2}{2} \Rightarrow % = 5 %$