Question

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. $e_1-2e_2$

Answer 1

The correct option is B $e_1+2e_2$

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 logarithm on both sides, we get:
$lng=lnh-2lnt$
Differentiating, $\frac{\Delta g}{g}=\frac{\Delta h}{h}-2\frac{\Delta t}{t}$

For maximum permissible error:
$(\frac{\Delta g}{g}\times 100)=(\frac{\Delta h}{h}\times 100)+2\times (\frac{\Delta t}{t}\times 100)$

According to problem:
$\frac{\Delta h}{h}\times 100=e_1$ and $\frac{\Delta t}{t}\times 100=e_2$

Therefore, $(\frac{\Delta g}{g}\times 100)=e_1+2e_2$