Question

A uniform wire of radius $r=0.5\pm 0.005cm$ length $l=5\pm 0.05cm$. The maximum percentage error in its volume is

• A. $30\%$
• B. $3\%$
• C. $2\%$
• D. $1.5\%$

Answer 1

The correct option is B $3\%$

The volume of the wire is $V=\pi r^{2}l$
Thus, the relative error, $\frac{\Delta V}{V}=\pm (2\frac{\Delta r}{r}+\frac{\Delta l}{l})$
The maximum % error in V $=\frac{\Delta V}{V}\times 100=(2\frac{\Delta r}{r}+\frac{\Delta l}{l})\times 100=(2\frac{0.005}{0.5}+\frac{0.05}{5})100=3\%$