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
$h=ut+\frac{1}{2}gt^2$
$h=\frac{1}{2}gt^2 \Rightarrow g=\frac{2h}{t^2}$ ($\because$ 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 $\big(\frac{\Delta g}{g}\times100 \big)_{\max}=\big(\frac{\Delta h}{h}\times100 \big)+2\times \big(\frac{\Delta t}{t}\times100 \big)$
According to problem
$\frac{\Delta h}{h}\times100 =e_1$ and $\frac{\Delta t}{t}\times100$
$ =e_2$
Therefore,$\big(\frac{\Delta g}{g}\times100 \big)_{\max}$
$=e_1+2e_2$