Question

If 'E' is assumed to be correct, an error of 1% is done in determination of G. If the correct value of Poisson's ratio $\left(\mu \right)$ is 0.25, the error involved in calculating ${}^{\prime }{\mu }^{\prime }$ is

• A. 1%
• B. 5%
• C. 8%
• D. 1.2%

Answer 1

The correct option is B 5%

$E=2G(1+\mu )$

Take partial differentiation,

$\delta E=2\delta G(1+\mu )+2G(0+\delta \mu )$

$0=2\delta G(1+\mu )+2G\delta \mu$

($\because \delta E=0,Since{}^{\prime }{E}^{\prime }doesnotvary$)

$2\delta G(1+\mu )=(-)2G\delta \mu$

$\delta \mu =(-)\frac{\delta G}{G}(1+\mu )$

$\therefore \%error=\frac{\delta \mu }{\mu }\times 100$

$=(-)\frac{\delta G}{G}\frac{(1+\mu )}{\mu }\times 100$

$\frac{1}{100}\times \left(\frac{1+0.25}{0.25}\right)\times 100=5\%$