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, the student measures the extension in the length of the wire to be $0.8mm$ with an uncertainty of $0.05mm$ at a load of exactly $1.0kg$. The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $0.01mm$. 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 $(2.0\pm 0.2)\times {10}^{11}N{m}^{-2}$

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

Taking long both the sides

$logY=log4Fl-log\pi D^{2}e$

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

$\Delta Y=-Y\times (\frac{2\Delta r}{r}+\frac{\Delta e}{e})\Delta Y=0.225\times {10}^{11}Y=(2\pm 0.2)\times {10}^{11}N{m}^{-2}$