Question

# A wire has a mass (0.3 $$\pm$$ 0.003)g, radius (0.5 $$\pm$$ 0.005)mm and length (6 $$\pm$$ 0.06)cm. The maximum percentage error in the measurement of its density is -

A
1
B
2
C
3
D
4

Solution

## The correct option is D 4The volume of the wire is given by, $$V = \pi r^2 L$$Thus density of the wire, $$\rho = \dfrac{m}{\pi r^2 L}$$Taking log and then differentiate:              $$\implies \dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{\Delta m}{m} \times 100 + 2 \dfrac{\Delta r}{r} \times 100 + \dfrac{\Delta L}{L} \times 100$$$$\therefore$$  $$\dfrac{\Delta \rho}{\rho} \times 100 = \dfrac{0.003}{0.3} \times 100 + 2 \dfrac{0.005}{0.5} \times 100 + \dfrac{0.06}{6} \times 100 = 4$$%.$$\implies$$ Percentage error in density is $$4$$%.Physics

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