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 δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 m s⁻² and the velocity of sound is 300 m s⁻¹ . Then the fractional error in the measurement, δL / L, is closest to
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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 δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 m s⁻² and the velocity of sound is 300 m s⁻¹ . Then the fractional error in the measurement, δL / L, is closest to

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G1 Physics · Accepted Answer

Measured time is T=√(2L/g)+L/vₛ. Differentiate with respect to L: dT/dL = 1/√(2gL)+1/vₛ. At L=20 m, g=10 m s⁻² and vₛ=300 m s⁻¹, dT/dL≈0.0533 s m⁻¹. Thus ΔL≈ΔT/(dT/dL)=0.01/0.0533≈0.188 m, so ΔL/L≈0.94%, closest to 1%. Correct answer: D.