For a particular wire of $ mass = (0.6 \pm 0.003) $ gm, $ radius = (0.50 \pm 0.01) $ cm, and $ length = (10.00 \pm 0.05) $ cm, the maximum percentage error in the measurement of its density is:

Question

A physical quantity $ Q $ is found to depend on quantities $ a $, $ b $, $ c $ by the relation\[Q = \frac{a^4 b^3}{c^2}.\]The percentage errors in $ a $, $ b $, and $ c $ are 3%, 4%, and 5% respectively. Then, the percentage error in $ Q $ is:

Answer 1

Step 1: Write density in terms of measurements.
A wire is a cylinder, so its volume is $V = \pi r^2 l$ and density is $\rho = \dfrac{m}{\pi r^2 l}$.
Step 2: Recall the error rule.
For a quantity made of products and powers, the maximum relative errors simply add, with each power as a multiplier: \[ \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \] The factor $2$ on $r$ comes from $r$ being squared.
Step 3: Mass error.
$\dfrac{\Delta m}{m} = \dfrac{0.003}{0.6} = 0.005$.
Step 4: Radius error (doubled).
$2\dfrac{\Delta r}{r} = 2\times\dfrac{0.01}{0.50} = 2\times 0.02 = 0.04$.
Step 5: Length error.
$\dfrac{\Delta l}{l} = \dfrac{0.05}{10.00} = 0.005$.
Step 6: Add and convert to percent.
Total relative error $= 0.005 + 0.04 + 0.005 = 0.05$, which is $0.05\times 100\% = 5\%$.
\[ \boxed{5\%} \]

For a particular wire of \( mass = (0.6 \pm 0.003) \) gm, \( radius = (0.50 \pm 0.01) \) cm, and \( length = (10.00 \pm 0.05) \) cm, the maximum percentage error in the measurement of its density is:

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