Question

In Searle's experiment to find Young's modulus, the diameter of wire is measured as $D=0.05cm$, the length of wire is $L=125cm,$ and when a weight, $m=20kg$ is the put, extension in wire was found to be $0.100cm$. Find the maximum permissible error in Young's modulus (Y).

Answer 1

$\frac{mg}{\pi d^2/4}=Y\left(\frac{x}{l}\right)\Rightarrow Y=\frac{mgl}{\left(\pi /4\right)d^{2}x}$
${\left(\frac{dY}{Y}\right)}_{max}=\frac{△m}{m}+\frac{△l}{l}+2\frac{△d}{d}+\frac{△x}{x}$
$m=20.0kg\Rightarrow △m=0.1kg$
$l=125m\Rightarrow △l=1cm$
$d=0.050cm\Rightarrow △d=0.001cm$
$x=0.100cm\Rightarrow △x=0.001cm$
${\left(\frac{dY}{Y}\right)}_{max}=(\frac{0.1kg}{20.0kg}+\frac{1cm}{125cm}+\frac{0.001cm}{0.05cm}+\frac{0.001cm}{0.100cm})\times 100\%=4.3\%$