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 $$g = 10\,m{s^{ - 2}}$$   and the velocity of sound is $$300\,m{s^{ - 1}}.$$   Then the fractional error in the measurement, $$\frac{{\delta L}}{L},$$  is closest to-

Solution :
$$\eqalign{ & T = \sqrt {\frac{{2L}}{g}} + \frac{L}{v} \cr & {\text{with error limits}} \cr & T + \delta T = \sqrt {\frac{{2\left( {L + \delta L} \right)}}{g}} + \frac{{L + \delta L}}{v} \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}\left( {1 + \frac{{\delta L}}{L}} \right)} + \frac{L}{v}\left( {1 + \frac{{\delta L}}{L}} \right) \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}} \times \left( {1 + \frac{{\delta L}}{{2L}}} \right) + \frac{L}{v}\left( {1 + \frac{{\delta L}}{L}} \right) \cr & \therefore T + \delta T = \sqrt {\frac{{2L}}{g}} + \sqrt {\frac{{2L}}{g}} \frac{{\delta L}}{{2L}} + \frac{L}{v} + \frac{L}{v}\frac{{\delta L}}{L} \cr & T + \delta T = T + \sqrt {\frac{{2L}}{g}} \frac{{\delta L}}{{2L}} + \frac{L}{v}\frac{{\delta L}}{L} \cr & \delta T = \frac{{\delta L}}{L}\left[ {\frac{1}{2}\sqrt {\frac{{2L}}{g}} + \frac{L}{v}} \right] \cr & {\text{Substituting }}\delta T = 0.015,\,L = 20\,m,\, \cr & g = 10\,m{s^{ - 2}},v = 300\,m{s^{ - 1}} \cr & {\text{We get}} \cr & \frac{{\delta L}}{L} = \frac{{15}}{{1600}} \cr & \therefore \frac{{\delta L}}{L} \times 100 = \frac{{15}}{{1600}} \times 100 = \frac{{15}}{{16}}\% \approx 1\% \cr} $$