A person measures the depth of a well by measuring the time...

Total time taken,

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

Now, for an error $\deltaL$ in L,

We have an error $\deltaT$ in T

So, $T + \deltaT = \sqrt{\dfrac{2\left(L + \deltaL\right)}{g}} + \dfrac{\left(L + \deltaL\right)}{c}$

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

Since, $\dfrac{\deltaT}{T}$ is very small, hence

$\dfrac{\deltaL}{L}$ is also small, so taking binomial approximation

$T + \deltaT = \sqrt{\dfrac{2L}{g}}\left(1 + \dfrac{1}{2}\dfrac{\deltaL}{L}\right) + \dfrac{L}{c}\left(1 + \dfrac{\deltaL}{L}\right)$

$⇒ T + \deltaT = \left(\sqrt{\dfrac{2L}{g}}\right) + \sqrt{\dfrac{2L}{g}}\left(\dfrac{1}{2}\dfrac{\deltaL}{L}\right) + \left(\dfrac{L}{c}\right) + \dfrac{L}{c}\left(\dfrac{\deltaL}{L}\right)$

$⇒ \deltaT = \sqrt{\dfrac{2L}{g}}\left(\dfrac{1}{2}\dfrac{\deltaL}{L}\right) + \dfrac{L}{c}\left(\dfrac{\deltaL}{L}\right)$

$⇒ \deltaT = \left[\sqrt{\dfrac{L}{2g}} + \left(\dfrac{L}{c}\right)\right]\dfrac{\deltaL}{L}$

$⇒ 0 . 01 = \left[\sqrt{\dfrac{20}{2 \times 10}} + \left(\dfrac{20}{300}\right)\right]\dfrac{\deltaL}{L}$

$⇒ 0 . 01 = \left[1 + \dfrac{1}{15}\right]\dfrac{\deltaL}{L} ⇒ \dfrac{\deltaL}{L} = \dfrac{0 . 15}{16}$

$$

$= \dfrac{15}{16}$

$\approx 1$