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If resistance $R_{1}$ in resistance box is $300 Ω$ , then the balanced length is found to be $75.0 cm$ from end $A$ . The diameter of unknown wire is $1 mm$ and length of the unknown wire is $31.4 cm$ . Find the specific resistance of the unknown wire.

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$4.5 \times 10^{- 4} Ω \cdot m$

$1.5 \times 10^{- 4} Ω \cdot m$

$3.5 \times 10^{- 4} Ω \cdot m$

$2.5 \times 10^{- 4} Ω \cdot m$

answer is D.

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Detailed Solution

$\frac{R}{X} = \frac{l}{100 - l}$

$X = (\frac{100 - l}{l}) R = (\frac{100 - 75}{75}) (300) = 100 Ω$

When,

$X = \frac{ρL}{A} = \frac{ρL}{({πd}^{2} / 4)}$

$ρ = \frac{{πd}^{2} X}{4 L}$

$= \frac{(3.14) {(10^{- 3})}^{2} (100)}{(4) (0.314)}$

$= 2.5 \times 10^{- 4} Ω \cdot m$

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