Question

In Searle's experiment to find Young's modulus the diameter of wire is measured as $d=0.05cm$, length of wire is $l=125cm$ and when a weight,$m=20.0kg$ is put, extension in wire was found to be $0.100cm$. Find the maximum permissible error in Young's modulus $\left(Y\right)$. Use:$Y=\frac{mgl}{\left(\pi /4\right)d^{2}x}$.

• A. $6.3\%$
• B. $5.3\%$
• C. $2.3\%$
• D. $1\%$

Answer 1

The correct option is A $6.3\%$

$Y=\frac{mgl}{\left(\pi /4\right)d^{2}x}$ .....$\left(1\right)$
${\left[\frac{dY}{Y}\right]}_{max}$ $=\frac{\Delta m}{m}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}+\frac{\Delta x}{x}$
$m=20.0kg\Rightarrow \Delta m=0.1kg$
$l=125cm\Rightarrow \Delta l=1cm$
$d=0.050cm\Rightarrow \Delta d=0.001cm$
$x=0.100cm\Rightarrow \Delta x=0.001cm$
${\left[\frac{dY}{Y}\right]}_{max}=[\frac{0.1kg}{20.0kg}+\frac{1cm}{125cm}+2\times \frac{0.001cm}{0.05cm}+\frac{0.001cm}{0.100cm}]\times 100\%=6.3\%$