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A cylindrical wire of mass \left ( 0.4\pm 0.01 \right )g has length \left ( 8 \pm 0.04\right ) cm and radius \left ( 6\pm 0.03 \right ) mm. The maximum error
in its density will be:

Option: 1

4%


Option: 2

1%


Option: 3

3.5%


Option: 4

5%


Answers (1)

best_answer

Cylindrical wire m = \left ( 0.4\pm 0.01 \right ) g
\begin{aligned} & \ell=(8 \pm 0.04) \mathrm{cm} \\ & \mathrm{r}=(6 \pm 0.03) \mathrm{mm} \end{aligned}
Density \rho=\frac{m}{\pi r^2 \ell} \Rightarrow \rho r^2 \ell m^{-1}=\frac{1}{\pi}=\text { const. }
Differentiating after taking log on both side
\begin{aligned} & \frac{\mathrm{d} \rho}{\rho}+\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{d} \ell}{\ell}-\frac{\mathrm{dm}}{\mathrm{m}}=0 \\ & \frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{m}}{\mathrm{m}}-\frac{\Delta \ell}{\ell}-\frac{2 \Delta \mathrm{r}}{\mathrm{r}} \\ & \left(\frac{\Delta \rho}{\rho}\right)_{\max }=\left[\frac{0.01}{0.4}+\frac{0.04}{8}+2\left(\frac{0.03}{6}\right)\right] \\ & \left(\frac{\Delta \rho}{\rho}\right)_{\max }=0.04 \end{aligned}
Percentage error = 0.04\times 100= 4%

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