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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2023
  • Hard
  • Numerical

A wire of resistance \(160\,\Omega\) is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be \(......\Omega\)

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Answer & Solution

Correct Answer

(A) \(10\)

Step-by-step Solution

Detailed explanation

Volume \(=\) Constant

\(A _1 L _1= A _2 L _2\)

\(A _1 L = A _2 \frac{ L }{4}\)

\(4 A _1= A _2\)

\(R _1=\frac{\rho L_1}{ A _1} \quad R _2=\frac{\rho L_2}{ A _2}\)

\(\frac{ R _2}{ R _1}=\frac{ L _2 A _1}{ A _2 L _1}=\frac{ L }{4} \frac{ A _1}{4 A _1 L }\)

\(R _2=\frac{1}{16} R _1=10\,\Omega\)

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