80. 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 ST = 0.01 seconds and he measures
the depth of the well to be L=20 meters. Take the
acceleration due to gravity g = 10ms and the velocity of
sound is 300ms. Then the fractional error in the
measurement is closest to
Answers
Given: 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 acceleration due to gravity g = 10 ms and the velocity of sound is 300 ms.
To find: The fractional error in the measurement ΔL/L ?
Solution:
- Now the time will be:
- T = time taken by stone to reach bottom of well from top + time taken by sound to reach top of the well from bottom
time taken by stone to reach bottom of well from top = T = √2L/g .....A
time taken by sound to reach top of the well from bottom = L/V .......B
- So, T = √2L/g + L/V
T = A + B
ΔT = ΔA + ΔB
ΔT = ( √2/g x ΔL/2√L ) + ΔL/V
ΔT = √1/2gL x ΔL + ΔL/V
0.01 = ( 1 / √2 x 20 x 10 ) x ΔL + ΔL / 300
0.01 = ΔL { 1/20 + 1/300 }
0.01 = ΔL { 320/6000 }
0.01 = ΔL { 0.0533 }
ΔL = 0.01 / 0.0533
ΔL = 0.1875
- Now,
ΔL/L x 100
= 0.1875 / 20 x 100
= 0.9375%
Answer:
The fractional error in the measurement ΔL/L is 0.9375%.
Fractional error = 1%
Explanation:
Given: Error in time = δT = 0.01
L = 20m
g =10ms−2
Velocity of sound V = 300ms−1
Find: Fractional error in measurement
Solution:
T = √2L/g + L/V
ΔT = ΔA + ΔB
= √2/g * ΔL/ 2√L + ΔL/V
Substituting the values, we get:
ΔT = 1/√(2gL) * ΔL + ΔL/V
ΔT = 1/√(2*10*20) * ΔL + ΔL/300
0.01 = ΔL [ (1/20) + (1/300)
ΔL = 0.1875
ΔL / L * 100 = 0.1875 / 20 * 100 = 1%