A wire of length 10 c m and radius 7 × 1 0 − 4 m connected across the right gap of a meter bridge. When a resistance of 4.5Ω is connected on the left gap by using a resistance box, the balance lengt

Question

A wire of length 10 c m and radius 7 ​ × 1 0 − 4 m connected across the right gap of a meter bridge. When a resistance of 4.5Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 c m from the left end. If the resistivity of the wire is R × 1 0 − 7 Ω m , then value of R is :

TestHub · Accepted Answer

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

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

$X = \frac{ρl}{A} = \frac{ρl}{π r ^{2}} ... (1)$ From equation (1), it follows that

$3 = ρ \times \frac{0.1}{π \times ( 7 \times 1 0 ^{- 4} ) ^{2}} \Rightarrow ρ = \frac{3 \times π \times ( 7 \times 1 0 ^{- 4} ) ^{2}}{0.1} \approx 66 \times 1 0^{- 7} Ω m$

Therefore, $R = 66$ .

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