Question

A student measures the distance covered and time taken by a body under free fall which was initially at rest. He uses this data for estimating $g$ , 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, then the maximum possible percentage error in the estimation of $g$ is

• A. $e_{2}-e_{1}$
• B. $e_{1}+ \, 2e_{2}$
• C. $e_{1}+e_{2}$
• D. $e_{1}-2e_{2}$

Answer 1

Answer: (B)

From the relation
$ h = ut +\frac{1}{2} gt ^{2} $
$h =\frac{1}{2} gt ^{2} \Rightarrow g =\frac{2 h }{ t ^{2}} (\because$ body initially at rest $)$
Taking natural logarithms on both sides, we get
$\ln (g)=\ln (h)-2 \ln (t)+\ln (2)$
Differentiating, $\frac{\Delta g }{ g }=\frac{\Delta h }{ h }-2 \frac{\Delta t }{ t }$
For maximum permissible errors, we add both the errors. So
$ \left(\frac{\Delta g }{ g } \times 100\right)_{\max }=\left(\frac{\Delta h }{ h } \times 100\right)+2 \times\left(\frac{\Delta t }{ t } \times 100\right) $
$\frac{\Delta h }{ h } \times 100= e _{1} $ and $ \frac{\Delta t }{ t } \times 100= e _{2}$
Therefore, $\left.\left(\frac{\Delta g }{ g } \times 100\right)\right)_{\max }= e _{1}+2 e _{2}$