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
$h=ut+\frac{1}{2}g{t}^2$
$h=\frac{1}{2}g{t}^2\Rightarrow g=\frac{2h}{t^2}(\because$ 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, ${\left(\frac{\Delta g}{g}\times 100\right))}_{\max }=e_1+2e_2$