Question

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\delta T=0.01$ seconds and he measures the depth of the well to be $L=20$ meters. Take the acceleration due to gravity $g=10m{s}^{-2}$ and the velocity of sound is $300m{s}^{-1}$. Then the fractional error in the measurement, $\delta L/L$, is closest to :

• A. $2\%$
• B. $5\%$
• C. $1\%$
• D. $3\%$

Answer 1

The correct option is C $1\%$

Here $T=\sqrt{\frac{2L}{g}}+\frac{L}{V}$

A B

$\Delta T=\Delta A+\Delta B$

$\Delta T=\sqrt{\frac{2}{g}}\times \frac{\Delta L}{2\sqrt{L}}+\frac{\Delta L}{V}$

$\Delta T=\sqrt{\frac{1}{2gL}}\times \Delta L+\frac{\Delta L}{V}$

$0.01=\frac{1}{\sqrt{2\times 10\times 20}}\times \Delta L+\frac{\Delta L}{300}$

$0.01=\Delta L[\frac{1}{20}+\frac{1}{300}]$

On calculating we get

$\Delta L=0.1875$

$\frac{\Delta L}{L}\times 100=\frac{0.1875}{20}\times 100=1$%