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Units and Measurements

Section: Physics

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Question : 18 of 34

Marks: +1, -0

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\,\text{m/s}^2

and the velocity of sound is

300\,\text{m/s}

Then the fractional error in the measurement,

\frac{\delta L}{L}

, is closet to :

[JEE Adv 2017 P2]

0.2%
5%
1%
3%

Solution:

Time taken for the stone to reach the bottom of the well.

t_1 = \frac{\sqrt{2L}}{g}

time taken by the sound to reach from the bottom of the well to the observer

t_2 = \frac{L}{V_s}

Total time

T = t_1 + t_2

\Rightarrow T = \sqrt{\frac{2L}{g}} + \frac{L}{V_s}

\frac{dT}{dL} = -\frac{1}{2}\sqrt{\frac{2}{g}} \times \frac{1}{\sqrt{L}} + \frac{1}{V_s} =

\frac{1}{2} \times \sqrt{\frac{2}{10 \times 20}} + \frac{1}{300} = \frac{1}{20} + \frac{1}{300}

\frac{0.01}{dL} = \frac{15+1}{300} \Rightarrow dl

= \frac{0.01 \times 300}{16} = \frac{3}{16}

Percentage error,

\frac{dl}{l} \times 100 = \frac{3}{16} \times \frac{1}{20} \times 100 \approx 1\%

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