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EEJ MAIN Physics QUESTION #999
Question 1
The maximum percentage error in the measurement of density of a wire is: [mass \(= (0.60\pm0.003)\ \text{g}\), radius \(= (0.50\pm0.01)\ \text{cm}\), length \(= (10.00\pm0.05)\ \text{cm}\)]
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Correct Answer Explanation
Density \(\rho = \frac{m}{\pi r^2 l}\). \(\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} = \frac{0.003}{0.60}+2\times\frac{0.01}{0.50}+\frac{0.05}{10.00} = 0.005+0.04+0.005 = 0.05 = 5\%\).