JEE Challenger
More from Units and Measurements
PhyMedAdv 2017 • Paper 2

Fractional Error in Well Depth Measurement using Sound Arrival

An individual estimates the depth of a well by measuring the time elapsed between releasing a stone and hearing the sound of its splash at the bottom. The uncertainty in the time measurement is δT=0.01 s\delta T = 0.01\text{ s}, while the measured depth of the well is L=20 mL = 20\text{ m}. Taking acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2} and the speed of sound as 300 m s−1300\text{ m s}^{-1}, the fractional error in the depth measurement, δL/L\delta L/L, is closest to:

Options

A

0.2%0.2\%

B

1%1\%

Correct
C

3%3\%

D

5%5\%

Topics & Concepts

Units and MeasurementsKinematicsSound Waves

Step-by-Step Solution

The total time TT elapsed between dropping the stone and receiving the sound of the impact at the top is the sum of the time taken for the stone to fall through depth LL (t1t_1) and the time taken for the sound to travel back up through depth LL (t2t_2).

  1. Time taken by the stone to fall: t1=2Lgt_1 = \sqrt{\frac{2L}{g}}

  2. Time taken by sound to travel up: t2=Lvt_2 = \frac{L}{v}

Thus, the total time TT is given by: T=t1+t2=2Lg+LvT = t_1 + t_2 = \sqrt{\frac{2L}{g}} + \frac{L}{v}

Differentiating TT with respect to LL to find the relationship between the uncertainties δT\delta T and δL\delta L: δT=(dTdL)δL=(12gL+1v)δL\delta T = \left( \frac{\mathrm{d}T}{\mathrm{d}L} \right) \delta L = \left( \frac{1}{\sqrt{2gL}} + \frac{1}{v} \right) \delta L

Substitute the given values into the expression:

  • L=20 mL = 20\text{ m}
  • g=10 m s−2g = 10\text{ m s}^{-2}
  • v=300 m s−1v = 300\text{ m s}^{-1}
  • δT=0.01 s\delta T = 0.01\text{ s}

First, compute 2gL\sqrt{2gL}: 2gL=2×10×20=400=20 m s−1\sqrt{2gL} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\text{ m s}^{-1}

Now, calculate dTdL\frac{\mathrm{d}T}{\mathrm{d}L}: dTdL=120+1300=15+1300=16300=475 s m−1\frac{\mathrm{d}T}{\mathrm{d}L} = \frac{1}{20} + \frac{1}{300} = \frac{15 + 1}{300} = \frac{16}{300} = \frac{4}{75}\text{ s m}^{-1}

Using δT=0.01 s=1100 s\delta T = 0.01\text{ s} = \frac{1}{100}\text{ s}: 1100=475δL\frac{1}{100} = \frac{4}{75} \delta L

δL=75400=316 m\delta L = \frac{75}{400} = \frac{3}{16}\text{ m}

Now, the fractional error δLL\frac{\delta L}{L} is: δLL=3/1620=3320\frac{\delta L}{L} = \frac{3/16}{20} = \frac{3}{320}

Expressed as a percentage: δLL×100%=300320%=1516%=0.9375%\frac{\delta L}{L} \times 100\% = \frac{300}{320}\% = \frac{15}{16}\% = 0.9375\%

This value is closest to 1%1\%.

Correct Answer: B (1%1\%)

PreviousNext