Question 63: Units , Dimensions and Error Analysis

Question

The maximum percentage error in the measurment of density of a wire is 
[Given, mass of wire $=(0.60 \pm 0.003) \mathrm{g}$
radius of wire $=(0.50 \pm 0.01) \mathrm{cm}$
length of wire $=(10.00 \pm 0.05) \mathrm{cm}]$

Accepted Answer

To determine the maximum percentage error in the density of the wire, follow these steps:
The density of the wire is given by the formula for a cylinder:
$$ho = \frac{m}{\pi r^2 l}$$
When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:
$$\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$
Here:
$$\frac{\Delta m}{m}$$ is the relative error in the mass.
$$2\frac{\Delta r}{r}$$ is due to the radius being squared.
$$\frac{\Delta l}{l}$$ is the relative error in the length.
Now plug in the given values:
Mass: $$m = 0.60 \, 	ext{g}$$ with an error $$\Delta m = 0.003 \, 	ext{g}$$
$$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$$ or 0.5%.
Radius: $$r = 0.50 \, 	ext{cm}$$ with an error $$\Delta r = 0.01 \, 	ext{cm}$$
$$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$$ or 2%. Since the radius is squared in the formula, multiply by 2:
$$2\frac{\Delta r}{r} = 2 	imes 0.02 = 0.04$$ or 4%.
Length: $$l = 10.00 \, 	ext{cm}$$ with an error $$\Delta l = 0.05 \, 	ext{cm}$$
$$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$$ or 0.5%.
Sum these contributions to find the overall maximum relative error:
$$\frac{\Delta ho}{ho} = 0.005 + 0.04 + 0.005 = 0.05$$
Converting this relative error into a percentage gives:
$$0.05 	imes 100\% = 5\%$$
Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.

Answer

7

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8

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5

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4

Units , Dimensions and Error Analysis

The maximum percentage error in the measurment of density of a wire is [Given, mass of wire =(0.60 \pm 0.003) \mathrm{g} radius of wire =(0.50 \pm 0.01) \mathrm{cm} length of wire =(10.00 \pm 0.05) \mathrm{cm}]

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