A wire of length 10 cm and radius 7 × 10 - 4 m connected across the right gap of a meter bridge. When a resistance of 4 . 5 Ω is connected on the left gap by using a resistance box, the balance length

Question

A wire of length 10 cm and radius 7 × 10 - 4 m connected across the right gap of a meter bridge. When a resistance of 4 . 5 Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R × 10 - 7 Ω m , then value of R is :

TestHub · Accepted Answer

From the null point condition of a Meter Bridge, it follows that,

$\frac{4.5 Ω}{60} = \frac{X}{100 - 60} \Rightarrow X = \frac{40}{60} \times 4.5 Ω = 3 Ω$

Also,

$\begin{array}{rcl} X & = & \frac{ρ l}{A} \\ = & \frac{ρ l}{π r^{2}} . . . (1) \end{array}$

From equation (1), it follows that

$3 = ρ \times \frac{0.1}{π \times {(\sqrt{7} \times 10^{- 4})}^{2}} \Rightarrow ρ = \frac{3 \times π \times {(\sqrt{7} \times 10^{- 4})}^{2}}{0.1} \approx 66 \times 10^{- 7} Ω m$

Therefore, $R = 66$ .

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