The length of a simple pendulum is about $100 cm$ known to have an accuracy of $1 mm$ . Its period of oscillation is $2 s$ determined by measuring the time for 100 oscillations using a clock of $0.1 s$ resolution. What is the accuracy in the determined value of g?

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0.2%

0.1%

0.5%

answer is A.

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Detailed Solution

$T = \frac{t}{n} a n d Δ T = \frac{Δ t}{n} \frac{Δ T}{T} \times 100 = \frac{Δ t}{t} \times 100 = \frac{0.1}{2 \times 100} \times 100 = 0.05 % % \frac{Δ l}{l} \times 100 = \frac{0.1 cm}{100 cm} \times 100 = 0.1 % T = 2 π \sqrt{\frac{l}{g}} \Rightarrow g = \frac{4 π^{2} l}{T^{2}} \propto \frac{I}{T^{2}} % error in g = (% error in l) + 2 (% error in T) = 0.1 % + 2 (0.05 %) = 0.2 %$