Question

The acceleration due to gravity $g$ is determined by dropping an object through a distance of exactly $10m$ . The time is to be measured so that the result is to be good to $0.1$ . If the absolute error is $n\times {{10}^{-4}}s$ , find $n$ . (Take $g=10m/{{s}^{2}}$ in the calculation)
A.$7$
B.$3$
C.$14$
D.$6$

Answer 1

As the object is dropped from rest, so we are to consider covered distance $y=-\dfrac{1}{2}g{{t}^{2}}$ . From this relation, calculate the time required to fall. Then, taking the derivative on $g$ , calculate the percentage error, and thereafter, find the absolute error in $t$ . Thus, we can find the value of $n$.

Complete step by step answer:
As the object is dropped from rest, so we have, $y=-\dfrac{1}{2}g{{t}^{2}}$ ,
Where $g$ = acceleration due to gravity,
And, $t$ = total time taken to fall.
But, in the problem, it is given that $y=-10m$ .
So, we have, $t=\sqrt{-\dfrac{2y}{g}}$
Or, $t=\sqrt{-2\left( \dfrac{-10}{10} \right)}$
Or, $t=\sqrt{2}$
Therefore, the time taken to fall, is $t=1.43s$ .
Now, taking partial derivative on $y=-\dfrac{1}{2}g{{t}^{2}}$ , we get,
$\delta y=-\dfrac{1}{2}\delta g{{t}^{2}}-gt\delta t$ .
But, as $y$ is measured exactly (i.e., without error), so, we can take, $\delta y=0$ .
Then, we have $\delta t=-\dfrac{\delta gt}{2g}$ .
So, the percentage error in $g$ which can be tolerated is two times that in $t$ .
Therefore, the percentage error in measuring the time $t$ is $=\dfrac{\delta t}{t}$ .
As it is given, that the time is to be measured so that the result is to be good to $0.1$ , so, we have,
\dfrac{\delta t}{t}=\dfrac{0.1%}{2}
$=0.05$
Therefore, the absolute error $=(0.05$
$=0.7ms$
$=7\times {{10}^{-4}}s$
As, in the problem, it is given that the absolute error is $n\times {{10}^{-4}}s$ ,
So, the value of $n$ is $7$ .
Therefore, the correct option is (A) $7$ .

Note:
The absolute error of a measurement depicts the actual degree of error. It is actually the magnitude of the difference between the measured value and the actual value of a certain quantity. It is generally denoted by the symbol $\left| \Delta a \right|$ (or mod value of delta a) since, even if the value of $\Delta a$ is negative, the mod value is always positive.