A person measures the depth of a well by measuring the time interval between dropping a stone
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 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10ms–2 and the velocity of sound is 300ms–1. Then the fractional error in the measurement, δL/L, is closest to
(A) 0.2%
(B) 1%
(C) 3%
(D) 5%
1 Answers
Here T = \(\sqrt{\frac{2L}{g}}\)....(A) + \(\frac{L}{V}\).......(B)
ΔT = ΔA + ΔB
ΔT = \(\sqrt{\frac{2}{g}}\) x \(\frac{\Delta L}{2\sqrt L}\) + \(\Delta L\over V\)
ΔT = \(\sqrt{\frac{1}{2gL}}\) x ΔL + \(\Delta L\over V\)
0.01 = \(\frac{1}{\sqrt{2\times10\times20}}\) x ΔL + \(\Delta L\over 300\)
0.01 = ΔL\([\frac{1}{20}+\frac{1}{300}]\)
On calculating we get
ΔL = 0.1875
\(\Delta L\over L\) x 100 = \(\frac{0.1875}{20}\) x 100 = 1%