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 `eltaT=0.01` seconds and he measures the depth of the well to be `L=20` meters. Take the acceleration due to gravity `g=10 ms^(-2)` and the velocity of sound is `300 ms^(-1).` Then the fractional error in the measurement, `deltaL//L,`is closest to

Options

A. `0.2%`
B. `3%`
C. `5%`
D. `1%`

Answer

D

Explanation

Let L be the depth of well. Let `t_(1)` is the time taken by the stone to touch the surface of water then we can write the following :
`L= 1"/"2 g t^(2) implies t_(1) = sqrt((2L)/(g))=sqrt((2L)/(10))=sqrt((L)/(5))`
Now sound takes time `t_(2)` to reach the top and the same can be written as : `t_(2)= L"/"300` because souond moves with a speed `300 m"/"s`.
Total time taken by the sound to reach the person can be written as `t = t_(1)+t_(2)` :
`t= sqrt((L)/(5))+(L)/(300)`
On differentiating above equation we get the following :
`dt = (1)/(sqrt(5))(1)/(2)L^(-1/2) dL+((1)/(300)dL)`
`implies triangle t= (1)/(sqrt(5))(1)/(2)(triangleL)/(sqrt(L))+(triangleL)/(300)`
We can substitute L= 20 m and `trianglet= 0.01s` in above expression to get the following :
`implies 0.01 = triangleL((1)/(sqrt(20))(1)/(sqrt(20))+(1)/(300))`
`implies 0.01 = triangleL((1)/(20)+(1)/(300))= triangleL(15+1)/(300)`
`triangle L= (0.01xx300)/(16)=(3)/(16)`
`(triangleL)/(L)xx100=(3)/(16)xx(1)/(20)xx100= (15)/(16) approx 1%`