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A student measures the value of g with the help of a simple pendulum using the formula
$g = \frac{4 π^{2} L}{T^{2}}$
The errors in the measurements of L and T are $∆ L$ and $∆$ T respectively. In which of the following cases is the error in the value of g the minimum?

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$∆ L = 0.5 c m, ∆ T = 0.5 s$

$∆ L = 0.2 c m, ∆ T = 0.2 s$

$∆ L = 0.1 c m, ∆ T = 1.0 s$

$∆ L = 0.1 c m, ∆ T = 0.1 s$

answer is D.

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Detailed Solution

The proportionate error in the measurement of g is

$\frac{∆ g}{g} = \frac{∆ L}{L} + 2 \frac{∆ T}{T}$

Hence $∆$ g will be minimum if $∆$ L and $∆$ T are minimum.

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