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 $\delta T=0.01$ seconds and he measures the depth of the well to be $L=20$ meters. Take the acceleration due to gravity $g=10m{s}^{-2}$ and the velocity of sound $300m{s}^{-1}$. Then the fractional error in he measurement $\frac{\delta L}{L}$, is closest to

• A. $0.2\%$
• B. $3\%$
• C. $5\%$
• D. $0.1\%$

Answer 1

The correct option is D $0.1\%$

$\begin{array}{cc}S& =ut+\frac{1}{2}a{t}^2\\ \Rightarrow L& =0+\frac{1}{2}\times g\times t^2\\ \Rightarrow & L=\frac{1}{2}g{t}^2\\ \Rightarrow & t=\sqrt{2gL}⟶\left(1\right)\\ t\to & \text{time of falling of}\\ & \text{stone}\end{array}$

$\begin{array}{c}\text{also,}\\ \text{let time t' for sound}\\ \text{to travel}L\text{distance up}\\ t^{\prime }=\frac{L}{v_{\text{sound}}}⟶\left(ii\right)\\ \text{Time interval in dropping}\\ \text{and recieving sound}\\ T=t+t^{\prime }⟶\left(iii\right)\end{array}$

$\begin{array}{c}\text{Let,}\\ L=20\pm \Delta Lm\\ \text{From eq.}\text{(i)}\\ t=\sqrt{2\times 10\times 20}\\ t=20s\\ \text{From eqn}\left(ii\right)\\ t^{\prime }=\frac{20}{300}s\\ \text{From eqn (iii)}\\ T=20+\frac{20}{300}=20.067s\end{array}$

$\begin{array}{c}\text{also,}\\ T=t+t^{\prime }\\ \text{differentiating}\\ \Rightarrow dT=dt+d{t}^{\prime }⟶\text{(iv)}\\ \text{From}\left(1\right)\\ t=\sqrt{2gL}\Rightarrow t^2=2gL\\ \text{differentiating}\\ 2tdt=2gdL\Rightarrow dt=\frac{g}{t}\cdot dL⟶\left(v\right)\end{array}$

$\begin{array}{c}\text{From (ii)}\\ t^{\prime }=\frac{1}{V_{\text{sound}}}\\ \text{differentiating}\\ d{t}^{\prime }=dL/v_{\text{sound}}-\left(vi\right)\end{array}$

$\begin{array}{c}\text{Pulting (v) and (vi)}\\ \text{in (iv)}\\ dT=\frac{g}{t}dL+\frac{dL}{v_{\text{sound}}}-\left(vii\right)\\ dT=dL(\frac{g}{t}+\frac{1}{v_{\text{sound}}})\\ dT=0.01\text{s ............. given}\end{array}$

$\begin{array}{c}\Rightarrow 0.01=dL(\frac{10}{20}+\frac{1}{300})\\ \Rightarrow 0.01=dL(0.5+\frac{1}{300})\\ \Rightarrow 0.01=dL\times 0.503\\ \Rightarrow dL=\frac{0.01}{0.503}\\ \Rightarrow dL=0.02m\\ \Rightarrow \frac{dL}{L}=\frac{0.02}{20}\end{array}$

$\text{\% error}=0.1\%\text{.}$