A person measures the depth of a well by measuring the time interval between dropping a stone and receiving…

Total time taken,

$T= \sqrt{\frac{2L}{g}}+\frac{L}{c}$, $c$ is the sound speed in air.

Now, for an error $\delta L$ in L,

We have an error $\delta T$ in T

So, $T+\delta T= \sqrt{\frac{2\left(L+\delta L\right)}{g}}+\frac{\left(L+\delta L\right)}{c}$

$= \sqrt{\frac{2L}{g} \left(1+\frac{\delta L}{L}\right)}+\frac{L}{c} \left(1+\frac{\delta L}{L}\right)$

Since, $\frac{\delta T}{T}$ is very small, hence

$\frac{\delta L}{L}$  is also small, so taking binomial approximation

$T+\delta T= \sqrt{\frac{2L}{g}} \left(1+\frac{1}{2}\frac{\delta L}{L}\right)+\frac{L}{c} \left(1+\frac{\delta L}{L}\right)$

$\Rightarrow T+\delta T=\left(\sqrt{\frac{2L}{g}}\right)+ \sqrt{\frac{2L}{g}} \left(\frac{1}{2}\frac{\delta L}{L}\right)+\left(\frac{L}{c}\right)+\frac{L}{c} \left(\frac{\delta L}{L}\right)$

$\Rightarrow \delta T=\sqrt{\frac{2L}{g}} \left(\frac{1}{2}\frac{\delta L}{L}\right)+\frac{L}{c} \left(\frac{\delta L}{L}\right)$

$\Rightarrow \delta T=\left[\sqrt{\frac{L}{2g}}+\left(\frac{L}{c}\right)\right]\frac{\delta L}{L}$

$\Rightarrow 0.01=\left[\sqrt{\frac{20}{2\times 10}}+\left(\frac{20}{300}\right)\right]\frac{\delta L}{L}$

$$\begin{gathered} \Rightarrow 0.01=\left[1+\frac{1}{15}\right]\frac{\delta L}{L} \\ \Rightarrow \frac{\delta L}{L}=\frac{0.15}{16} \end{gathered}$$

$\% \mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}=\left(\frac{\delta L}{L}\right)\times 100\%$

$=\frac{15}{16}\%$

$\approx 1\%$