Question

A student performs an experiment to determine the Young's modulus of a wire, exactly $2$m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8$mm with an uncertainly of $\pm 0.05$mm at a load of exactly $1.0$ kg. The student also measures the diameter of the wire to be $0.4$mm with an uncertainly of $\pm 0.01$mm. Take $g=9.8m/s^2$(exact). The Young's modulus obtained from the reading is?

• A. $(2.0\pm 0.3)\times {10}^{11}N/m^2$
• B. $(2.0\pm 0.2)\times {10}^{11}N/m^2$
• C. $(2.0\pm 0.1)\times {10}^{11}N/m^2$
• D. $(2.0\pm 0.05)\times {10}^{11}N/m^2$

Answer 1

The correct option is B $(2.0\pm 0.2)\times {10}^{11}N/m^2$

$\begin{array}{c}\text{Given, length of wire}=2m\\ \text{extension in wire}=0.8\pm 0.05mm\\ \text{mass}=1kg\\ \text{diameter of wire}=0.4\pm 0.01mm\end{array}$

$\begin{array}{cc}\frac{F}{A}=& y\left(\frac{\Delta L}{L}\right)\\ \text{Here}F=mg& \\ A& =\frac{\pi d^2}{4}\end{array}$\

$y=\frac{\left(9180\right)\left(2\right)\left(4\right)}{\left(3114\right){(4\times {10}^{-4})}^2(8\times {10}^{-4})}=2\times {10}^{11}N/m^2$

$\begin{array}{c}y=\frac{FL}{A\left(\Delta L\right)}\\ \text{Taking log on both sides :}\\ \log y=\log F+\log L-\log A-\log \left(\Delta L\right)\end{array}$

$\begin{array}{c}\text{Here}F\text{and}L\text{are fixed, so.}\\ \Delta F=0,\Delta L=0\end{array}$

$\frac{\Delta y}{y}=\frac{\Delta A}{A}+\frac{\Delta \left(\Delta L\right)}{\Delta L}$

$\begin{array}{c}A=\frac{\pi d^2}{4}\\ dA=\frac{\pi d}{2}\left(\Delta d\right)\\ \frac{dA}{A}=\frac{2\Delta d}{d}\end{array}$

$\begin{array}{cc}\frac{\Delta y}{y}& =\frac{2\left(0.01\right)}{0.40}+\frac{0.05}{0.80}\\ & =0.05+0.06\\ & =0.11co\\ \therefore \Delta y& =0.22\times {10}^{11}N/m^2\end{array}$

$\begin{array}{c}\therefore \text{The expression of}\\ \text{young's medulus is}\end{array}:(2+0.2)\times {10}^{11}N/m^2$