Question

In an experiment of determine the Young's modulus of wire of a length exactly $1m$, the extension in the length of the wire is measured as $0.4mm$ with an uncertainty of $\pm 0.02mm$ when a load of $1kg$ is applied. The diameter of the wire is measured as $0.4mm$ with an uncertainty of $\pm 0.01mm$. The error in the measurement of Young's modulus $(\Delta Y)$ is found to be $X\times 10^{10}N{m}^{-2}$ . The value of $x$ is _______
[Take $g=10m/s^2$ ]

Answer 1

Answer: 2

$L=1m$
$\Delta L=0.4\times 10^{-3}m$
$m=1kg$
$d=0.4\times 10^{-3}m$
$\frac{F}{A}=Y\frac{\Delta L}{L}$
$Y=\frac{FL}{A\Delta L}=\frac{(mg)\cdot (1)}{\left(\frac{\pi d^2}{4}\right)0.4\times 10^{-3}}$
$\Rightarrow \frac{10\times 4}{\pi {\left(0.4\times 10^{-3}\right)}^2\times 0.4\times 10^{-3}}$
$Y=\frac{40}{\pi {\left(0.4\times 10^{-3}\right)}^3}$
$Y=\frac{40\times 7}{22\times 64\times 10^{-3}\times 10^{-9}}$
$Y=0.199\times 10^{-12}N/m^2$
$\frac{\Delta Y}{Y}=\frac{\Delta F}{F}+\frac{\Delta L}{L}+\frac{\Delta A}{A}+\frac{\Delta (\Delta L)}{(\Delta L)}$
$=\frac{0.02}{0.4}+2\frac{\Delta d}{d}=\frac{0.2}{4}+2\times \frac{0.01}{0.4}$
$=\frac{0.1}{2}+\frac{0.1}{2}=0.1$
$\Rightarrow \Delta Y=0.1\times Y$
$=0.199\times 10^{11}=1.99\times 10^{10}$