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 $\pm 0.05mm$ at a load of exactly $1.0kg$. The student also measures the diameter of the wire to be $0.4mm$ with an uncertainty of $\pm 0.01mm.$ Take $g=9.8m/s^2$ (exact). The Young's modulus obtained from the reading is close to

• A. $(2.0\pm ;0.3)10^{11}N/m^2$
• 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

Answer: (B)

$Y=\frac{FL}{Al}=\frac{4FL}{\pi d^{2}l}$
$=\frac{(4)(1.0\times 9.8)(2)}{\pi (0.4\times 10^{-3}{)}^2(0.8\times 10^{-3})}$
$=2.0\times 10^{11}N/m^2$
Further $\frac{\Delta ;Y}{Y}=2\left(\frac{\Delta ;d}{d}\right)+\left(\frac{\Delta ;l}{l}\right)$
$\therefore \Delta ;Y=\left\{2\left(\frac{\Delta ;d}{d}\right)+\left(\frac{\Delta ;l}{l}\right)\right\}Y$
$=\left\{2\times \frac{0.01}{0.4}+\frac{0.05}{0.8}\right\}\times 2.0\times 10^{11}$
$=0.225\times 10^{11}N/m^2$
$=0.2\times 10^{11}N/m^2$ (By rounding off )
or $(Y+\Delta ;Y)=(2+0.2)\times 10^{11}N/m^2$