Question

A student determines the density ρ of steel by taking measurements from a steel wire

Answer 1

Answer:

The percentage uncertainty in his calculated value of density is $\pm 0.713\,\%$.

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

$\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l$ (1)

Where:

$\frac{\partial \rho}{\partial m}$ - Partial derivative of the density with respect to mass, measured in $\frac{1}{mm^{3}}$.

$\frac{\partial \rho}{\partial d}$ - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

$\frac{\partial \rho}{\partial l}$ - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

$\Delta m$ - Mass uncertainty, measured in grams.

$\Delta d$ - Diameter uncertainty, measured in milimeters.

$\Delta l$ - Length uncertainty, measured in milimeters.

$\Delta \rho$ - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

$\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}$ (2)

$\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}$ (3)

$\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}$ (4)

And we expand (1) as follows:

$\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}$

$\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)$ (5)

If we know that $d = 2\,mm$, $l = 25\,mm$, $m = 6.2\,g$, $\Delta m = \pm 0.1\,g$, $\Delta d = \pm 0.01\,mm$ and $\Delta l = \pm 0.1\,mm$, then the absolute uncertainty is:

$\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right]$

$\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}$

And the expected density is:

$\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}$ (6)

$\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}$

$\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}$

The percentage uncertainty in his calculated value of density is:

$\%e = \frac{\Delta \rho}{\rho}\times 100\,\%$ (7)

If we know that $\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}$ and $\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}$, then the percentage uncertainty is:

$\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%$

$\%e = \pm 0.713\,\%$

The percentage uncertainty in his calculated value of density is $\pm 0.713\,\%$.