Question

A drawn wire of resistance 5 Ω is further drawn so that its diameter becomes one-fifth of the original. What is its resistance with volume remaining the same?
1. 25 Ω
2. 125 Ω
3. 625 Ω
4. 3125 Ω

Answer 1

Correct Answer - Option 4 : 3125 Ω

Concept:

The resistance of a wire is given by \(R = \frac{{\rho l}}{A}\)

Where ρ is the resistivity of the material

l is length

A is the cross-sectional area

\(A = \frac{{\pi {d^2}}}{4}\) for a given diameter of d.

Calculation:

Given that, the resistance of a wire (R₁) = 5 Ω

\(\Rightarrow {R_1} = \frac{{4\rho {l_1}}}{{\pi d_1^2}}\)

Now it further drawn so that its diameter becomes one-fifth of the original.

\(\Rightarrow {d_2} = \frac{1}{5}{d_1}\)

As the volume remains the same, we have

\({A_1}{l_1} = {A_2}{l_2}\)

\(\Rightarrow \frac{{\pi d_1^2{l_1}}}{4} = \frac{{\pi d_2^2{l_2}}}{4}\)

\(\Rightarrow \frac{{{l_1}}}{{{l_2}}} = \frac{{d_2^2}}{{d_1^2}} = \frac{1}{{25}}\)

\({R_2} = \frac{{4\rho {l_2}}}{{\pi d_2^2}} = \frac{{4\rho \left( {25{l_1}} \right)}}{{\pi \left( {\frac{{d_1^2}}{{25}}} \right)}} = 625 \times \frac{{4\rho {l_1}}}{{\pi d_1^2}} = 625{R_1}\)

New resistance (R₂) = 625R₁ = 625 × 5 = 3125 Ω

Alternate method:

\(R = \frac{{\rho l}}{A}\)

Multiply and divide by area,

\(R = \rho \frac{{lA}}{{{A^2}}} = \rho \frac{V}{{{A^2}}}\)

Where V is volume.

Here, ρ and V are constant.

\(\Rightarrow R \propto \frac{1}{{{A^2}}} \propto \frac{1}{{{d^4}}}\)

\(\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4}\) 

⇒ R₂ = 625R₁ = 625 × 5 = 3125 Ω