Question

A wire has a mass $0.3\pm 0.003g$, radius $0.5\pm 0.005mm$ and length $6\pm 0.06cm$. The maximum percentage error in

• A. 2
• B. 4
• C. 1
• D. 3

Answer 1

The correct option is B 4

Given:
$m\pm \Delta m=0.3\pm 0.003g\frac{\Delta m}{m}=0.01r\pm \Delta r=0.5\pm 0.005mm\frac{\Delta r}{r}=0.01l\pm \Delta l=6\pm 0.06cm\frac{\Delta l}{l}=0.01$
As we know that,
$\rho =\frac{M}{V}=\frac{M}{\text{Volume of cylinder}}\text{volume of cylinder}=\pi r^{2}L$
Therfore,
$\rho =\frac{M}{\pi r^{2}L}\cdots \left(i\right)\frac{\Delta \rho }{\rho }\times 100=[\frac{\Delta M}{M}+2\frac{\Delta R}{R}+\frac{\Delta L}{L}]\times 100\cdots \left(ii\right)$
By putting the values in equation $\left(ii\right)$ we get,
$\frac{\Delta \rho }{\rho }\times 100=[0.01+2\times 0.01+0.01]\times 100\frac{\Delta \rho }{\rho }\times 100=4\%\frac{\Delta \rho }{\rho }\times 100=4\%$