Question

In Searle's exp 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 the wire was found to be $0.100cm.$ Find a maximum permissible error in young's modulus (Y).

• A. $43$ %
• B. $6.3$%
• C. $0.63$%
• D. $630$%

Answer 1

The correct option is B $6.3$%

Given: $D=0.05cm$ ; $\Delta D=0.01cm$

$L=125.0cm$; $\Delta L=0.1cm$

Weight , $F=20\times 10=200N$

$\delta L=0.100cm$; $\Delta \left(\delta L\right)=0.001cm$

So, Young Modulus , $Y=\frac{FL}{\pi (\frac{D}{2}{)}^2\delta L}=\frac{200\times 1.25}{3.14\times (\frac{0.05\times {10}^{-2}}{2}{)}^2\times 0.1\times {10}^{-2}}=1.273\times {10}^{12}N{m}^{-2}$

By error method

$\frac{\Delta Y}{Y}=\frac{\Delta L}{L}+2\frac{\Delta D}{D}+\frac{\Delta \left(\delta L\right)}{\delta L}$

By Substituting the values

$\therefore \frac{\Delta Y}{Y}=0.063$

$\therefore$ Maximum permissible error in young's modulus $=\frac{\Delta Y}{Y}\times 100\text{\%}=6.3\text{\%}$