Question

A student performs an experiment to determine the Young's modulus of a wire exactly $2m$ long by Searle's method. In a particular reading, he measures the extension in the length of the wire to be $0.8mm$ with an uncertainty of $\pm 0.05mm$ at a load of exactly $1.0kg$. He also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm 0.01mm$. The Young's modulus obtained from the reading is:

(Given Young's Modulus of a wire is given by $Y=\frac{Fl}{Ae}$, where $F$ is the force applied on the wire, $l$ is the length of the wire, $A$ is the area of cross section of the wire and $e$ is the elongation produced in the wire due to the load. Take $g=9.8m/s^2$ )

• A. $(2\pm 0.3)\times {10}^{11}N/m^2$
• B. $(2\pm 0.2)\times {10}^{11}N/m^2$
• C. $(2\pm 0.1)\times {10}^{11}N/m^2$
• D. $(2\pm 0.5)\times {10}^{11}N/m^2$

Answer 1

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

$Y=\frac{4FL}{{\pi D}^{2}e}$

Taking log both the sides $\Rightarrow logY=log4FL-log\pi D^{2}e$

Now partially differentiating,

$\frac{\Delta Y}{Y}=-(\frac{2\Delta D}{D}+\frac{\Delta e}{e})$

$\frac{\Delta Y}{Y}=-(\frac{2\times 0.01}{0.4}+\frac{0.05}{0.8})$

$\frac{\Delta Y}{Y}=-0.1125$

Also, $Y=\frac{FL}{Ae}=\frac{9.8\times 2}{\pi \times {0.2}^2\times 0.8}=194.96\times {10}^9\approx 2\times {10}^{11}$

$\Rightarrow \Delta Y=-Y\times (\frac{2\Delta r}{r}+\frac{\Delta e}{e})$

$\Rightarrow \Delta Y=0.225\times {10}^{11}$

$\Rightarrow Y=(2\pm 0.2)\times {10}^{11}$