A) 7.96%
B) 4.56%
C) 6.50%
D) 8.42%
Correct Answer: C
Solution :
\[Y=\frac{4MgL}{\pi {{D}^{2}}l}\] so maximum permissible error in Y =\[\frac{\Delta Y}{Y}\times 100=\left( \frac{\Delta M}{M}+\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta D}{D}+\frac{\Delta l}{l} \right)\times 100\] \[=\left( \frac{1}{300}+\frac{1}{981}+\frac{1}{2820}+2\times \frac{1}{41}+\frac{1}{87} \right)\times 100\] \[=0.065\times 100=6.5%\]
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