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Question

Physics Question on Current electricity

One kgkg of copper is drawn into a wire of 11 mm diameter and a wire of 22 mm diameter. The resistance of the two wires will be in the ratio

A

2:12 : 1

B

1:21 : 2

C

16:116 : 1

D

4:14 : 1

Answer

16:116 : 1

Explanation

Solution

Mass of 1st wire =(πr121)σ=\left(\pi r^{2}_{1} \ell_{1} \right) \sigma
Mass of Il nd wire =(πr121)σ=\left(\pi r^{2}_{1} \ell_{1} \right) \sigma
\therefore (πr121)σ=(πr222)σ\left(\pi r^{2}_{1} \ell_{1}\right) \sigma =\left(\pi r^{2}_{2} \ell_{2}\right)\sigma
\Rightarrow 12=(r2r1)2\frac{\ell_{1}}{\ell_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}
\therefore R1R2=ρ1A1ρ2A2\frac{R_{1}}{R_{2}}=\frac{\rho \frac{^{\ell_{1}}}{A_{1}}}{\rho \frac{\ell_{2}}{A_{2}}} =12×A2A1=\frac{\ell_{1}}{\ell_{2}} \times\frac{A_{2}}{A_{1}}