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 delta t equals to 0.01 seconds and images the depth of the well to be equals to 20 m take the acceleration due to gravity g equals to 10 m per second square and the velocity of sound is 300 m per second then the fraction of error in the measurement delta l/l explanation with reason

Answer 1

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 m/s.

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%.