The percentage errors in the measurements of the length of a simple pendulum and its time period are 3% and 5% respectively. The maximum error in the value of the acceleration due to gravity obtained from these measurements is

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13%

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15%

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Solution

The correct option is D 13%

$T = 2 π \sqrt{\frac{l}{g}} o r, g = 4 π^{2} \frac{l}{T^{2}} or, log (g) = log (4 π^{2}) + log (l) - 2 log (T)$
The maximum error in g, $\frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T} = 3 % + (2 \times 5 %) = 13 %$

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The correct option is D 13% T=2π√lg or, g=4π2lT2or, log(g)=log(4π2) + log (l) −2log (T) The maximum error in g, Δgg=Δll+2ΔTT=3% +(2×5%)=13%

The correct option is D 13%

$T = 2 π \sqrt{\frac{l}{g}} o r, g = 4 π^{2} \frac{l}{T^{2}} or, log (g) = log (4 π^{2}) + log (l) - 2 log (T)$
The maximum error in g, $\frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T} = 3 % + (2 \times 5 %) = 13 %$