JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Current Electricity

A wire of resistance

160Ω

is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

Answer: (A)

Solution

When the wire is redrawn, its volume remains constant:

A_{1} L_{1} = A_{2} L_{2}

. Given

L_{2} = \frac{L _{1}}{4}

, we have

A_{1} L_{1} = A_{2} \frac{L _{1}}{4}

, so

A_{2} = 4 A_{1}

. Resistance is

R = \frac{ρ L}{A}

. The ratio of resistances is

\frac{R _{2}}{R _{1}} = \frac{L _{2}}{L _{1}} \cdot \frac{A _{1}}{A _{2}} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}

. Therefore,

R_{2} = \frac{R _{1}}{16} = \frac{160}{16} = 10Ω

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Single correct option