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 \(g=10 \mathrm{~ms}^{-2}\) and the velocity of sound is \(300 \mathrm{~ms}^{-1}\). Then the fractional error in the measurement, \(\delta L / L\), is closest to

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Units & Measurement

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

(A)0.2%
(B)1%✓ Correct
(C)3%
(D)5%

Answer: (B)

Solution

Total time

t = \frac{2 L}{g} + \frac{L}{c}

. Differentiating:

\frac{d t}{d L} = \frac{1}{2 g L} + \frac{1}{c}

. For

L = 20

g = 10

m/s²,

c = 300

m/s:

\frac{d t}{d L} = \frac{1}{400} + \frac{1}{300} = \frac{1}{20} + \frac{1}{300} = \frac{16}{300}

. Therefore

\frac{d L}{L} = \frac{d t}{( d t / d L ) \cdot L} = \frac{0.01}{( 16/300 ) \cdot 20} = \frac{0.01 \cdot 300}{320} = \frac{3}{320} \approx 0.94% \approx 1%

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option

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JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Units & Measurement

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