devarshi-dt-logo

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 seconds and he measures the depth of the well to be L=20 meters. Take the acceleration due to gravity g=10 m/s² and the velocity of sound is 300 m/s. Then the fractional error in the measurement, ΔL/L, is closest to:

2

5

1

3

Solution:

Here T = √(2L/g) + L/V
ΔT = ΔA + ΔB
ΔT = √(2/g) x ΔL / 2√L + ΔL/V
ΔT = (1/√(2gL)) x ΔL + ΔL/V
0.01 = (1/√(2 x 10 x 20)) x ΔL + ΔL/300
0.01 = ΔL[(1/20) + (1/300)]
On calculating we get ΔL = 0.1875
(ΔL/L) x 100 = 0.1875/20 x 100 = 1