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 seconds and he measures the depth of the well to be mete

Question

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 $δ T = 0.01$ seconds and he measures the depth of the well to be $L = 20$ meters. Take the acceleration due to gravity $g = 10 {ms}^{- 2}$ and the velocity of sound is $300 {ms}^{- 1}$ . Then the fractional error in the measurement, $δ L / L$ , is closest to [2017]

Smart Achievers · Accepted Answer

(2)

$Depth of the well = L = 20 m$

$∴$ $g = 10 {ms}^{- 2}$

$t_{1} = \sqrt{\frac{2 L}{g}}$ $(using, L = \frac{1}{2} g t^{2})$

$Time taken by impact sound to reach the person,$

$t_{2} = \frac{L}{v}$

$Total time taken in the process is given by,$

$T = t_{1} + t_{2} = \sqrt{\frac{2 L}{g}} + \frac{L}{v}$ $∴$ $δ T = \sqrt{\frac{2}{g}} \cdot \frac{δ L}{2 \sqrt{L}} + \frac{δ L}{v}$

$On substituting the given values we get,$

$δ T = \frac{16}{300} δ L$

$∴$ $Fractional error, \frac{δ L}{L} \times 100 = \frac{300}{16} \cdot \frac{δ T}{L} \times 100$ $= \frac{300}{16} \times \frac{0.01}{20} \times 100 = 1 %$

Solution

1 0.2 %

2 1 %

3 3 %

4 5 %

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