A person measures the depth of a well by measuring the time inter… - Vidyadip

MCQ

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

A
$0.2 \%$
✓
$1 \%$
C
$3 \%$
D
$5 \%$

✓

Answer

Correct option: B.

$1 \%$

b
$\mathrm{t}=\sqrt{\frac{2 \mathrm{~L}}{\mathrm{~g}}}+\frac{\mathrm{L}}{\mathrm{C}}$

$\frac{\mathrm{dt}}{\mathrm{dL}}=\sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \times \frac{1}{2 \sqrt{\mathrm{L}}}+\frac{1}{\mathrm{C}}$

$\mathrm{dL}=\frac{\mathrm{dt}}{\frac{1}{\sqrt{2 \mathrm{gL}}}+\frac{1}{\mathrm{C}}}$

$\Rightarrow \frac{\mathrm{dL}}{\mathrm{L}} \times 100=\left(\frac{\mathrm{dt}}{\frac{1}{\sqrt{2 \mathrm{gL}}}+\frac{1}{\mathrm{C}}}\right) \frac{1}{\mathrm{~L}} \times 100$

$=\frac{15}{16 \%} \approx 1 \%$

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