Question 124: Units , Dimensions and Error Analysis

Question

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

Accepted Answer

The density of a cylindrical wire is given by the formula:
$\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$
where $m$ is the mass, $r$ is the radius, and $l$ is the length.
The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density, this is given by:
$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$
Given that $\Delta m = 0.01$ g, $m = 0.4$ g, $\Delta r = 0.03$ mm, $r = 6$ mm, $\Delta l = 0.04$ cm, and $l = 8$ cm, we can substitute these values into the formula to find the relative error in the density:
$\frac{\Delta \rho}{\rho} = \frac{0.01}{0.4} + 2\frac{0.03}{6} + \frac{0.04}{8} = 0.025 + 0.01 + 0.005 = 0.04$
So the relative error in the density is 0.04, or 4%.

Answer

1%

Answer

5%

Answer

4%

Answer

3.5%

Units , Dimensions and Error Analysis

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

Choose the correct answer: