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A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_1$ and $e_2$ respectively, the percentage error in the estimation of g is
A
$e_2 - e_1$
B
$e_1 + 2e_2$
C
$e_1 + e_2$
D
$e_1 - 2e_2$
Detailed Solution
For free fall from rest: $h = \frac{1}{2}gt^2$, so $g = \frac{2h}{t^2}$
Taking logarithms: $\ln g = \ln2 + \ln h - 2\ln t$
Differentiating, and remembering that maximum errors always add: $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$
$\left(\frac{\Delta g}{g}\times100\right)_{max} = \frac{\Delta h}{h}\times100 + 2\times\frac{\Delta t}{t}\times100$
Percentage error in g = $e_1 + 2e_2$
Taking logarithms: $\ln g = \ln2 + \ln h - 2\ln t$
Differentiating, and remembering that maximum errors always add: $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$
$\left(\frac{\Delta g}{g}\times100\right)_{max} = \frac{\Delta h}{h}\times100 + 2\times\frac{\Delta t}{t}\times100$
Percentage error in g = $e_1 + 2e_2$