Respuesta :
Answer:
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
Explanation:
We can estimate the absolute uncertainty by the definition of total differential. That is:
[tex]\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[/tex] (1)
Where:
[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].
[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.
[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.
[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.
[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.
[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.
[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.
Partial derivatives are, respectively:
[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)
[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)
[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)
And we expand (1) as follows:
[tex]\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}}[/tex]
[tex]\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)[/tex] (5)
If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:
[tex]\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][/tex]
[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]
And the expected density is:
[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)
[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]
[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]
The percentage uncertainty in his calculated value of density is:
[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)
If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:
[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]
[tex]\%e = \pm 0.713\,\%[/tex]
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
