Discussion Forum

1)

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 $\delta T= 0.01s$  and he measures the depth of the well to be  L=20 m. 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

$\frac{\delta L}{L}$is closet to

Answer:

Option A

Explanation:

$t=\sqrt{\frac{L}{5}}+\frac{L}{300}$

$dt=\frac{1}{\sqrt{5}}\frac{1}{2}L^{-\frac{1}{2}}dL+(\frac{1}{300}dL)$

$dt=\frac{1}{2\sqrt{5}}\frac{1}{20}dL+\frac{dL}{300}=0.01$

  $dL(\frac{1}{20}+\frac{1}{300})=0.01$

       $dL(\frac{15}{300})=0.01$

       $dL=\frac{3}{16}$

$\frac{dL}{L}\times100=\frac{3}{16}\times\frac{1}{20}\times 100=\frac{15}{16}=1$%