1. Set up the time equation

If the depth of the well is $L$, then the total measured time $T$ consists of:

• time taken by the stone to fall: $t_1=\sqrt{\frac{2L}{g}}$
• time taken by sound to travel upward: $t_2=\frac{L}{v}$

So, $T=\sqrt{\frac{2L}{g}}+\frac{L}{v}$

Here, $g=10\,\text{m s}^{-2}, \qquad v=300\,\text{m s}^{-1}, \qquad L=20\,\text{m}$

2. Differentiate to relate error in $T$ and error in $L$

For small errors, $\delta T = \frac{dT}{dL}\,\delta L$

Now, $\frac{dT}{dL}=\frac{d}{dL}\left(\sqrt{\frac{2L}{g}}+\frac{L}{v}\right)$

First term: $\frac{d}{dL}\sqrt{\frac{2L}{g}}=\frac{1}{\sqrt{2gL}}$

Second term: $\frac{d}{dL}\left(\frac{L}{v}\right)=\frac{1}{v}$

Hence, $\frac{dT}{dL}=\frac{1}{\sqrt{2gL}}+\frac{1}{v}$

Therefore, $\delta L = \frac{\delta T}{\dfrac{1}{\sqrt{2gL}}+\dfrac{1}{v}}$

3. Substitute numerical values

Compute: $\sqrt{2gL}=\sqrt{2\times 10\times 20}=\sqrt{400}=20$

So, $\frac{1}{\sqrt{2gL}}=\frac{1}{20}=0.05$

Also, $\frac{1}{v}=\frac{1}{300}\approx 0.00333$

Thus, $\frac{dT}{dL}=0.05+0.00333=0.05333$

Now, $\delta L=\frac{0.01}{0.05333}\approx 0.1875\,\text{m}$

4. Find fractional error

$\frac{\delta L}{L}=\frac{0.1875}{20}=0.009375$

In percentage, $\frac{\delta L}{L}\times 100 \approx 0.9375\%$

This is closest to: $1\%$

5. Evaluate options

• A: $0.2\%$ — too small
• B: $1\%$ — correct
• C: $3\%$ — too large
• D: $5\%$ — too large

So the correct option is B.