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

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 logarithm on both sides, we get
In $g=$ ln $h-2$ ln $t$
Differentiating, $\frac{\Delta g}{g}=\frac{\Delta h}{h}-2\frac{\Delta t}{t}$
For maximum permissible error.
or $(\frac{\Delta g}{g}\times 100{)}_{\max }=(\frac{\Delta h}{h}\times 100)+2\times (\frac{\Delta t}{t}\times 100)$
According to problem
$\frac{\Delta h}{h}\times 100=e_1$ and $\frac{\Delta t}{t}\times 100$
$=e_2$
Therefore,$(\frac{\Delta g}{g}\times 100{)}_{\max }$
$=e_1+2e_2$