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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
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B
2
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C
3
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D
4
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Solution

The correct option is D 4
The 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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