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JEE Mains · Physics · STD 12 - 3. current electricity

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 \(......\Omega\)

  1. A \(10\)
  2. B \(640\)
  3. C \(40\)
  4. D \(16\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(10\)

Step-by-step Solution

Detailed explanation

Volume \(=\) Constant \(A _1 L _1= A _2 L _2\) \(A _1 L = A _2 \frac{ L }{4}\) \(4 A _1= A _2\) \(R _1=\frac{\rho L_1}{ A _1} \quad R _2=\frac{\rho L_2}{ A _2}\) \(\frac{ R _2}{ R _1}=\frac{ L _2 A _1}{ A _2 L _1}=\frac{ L }{4} \frac{ A _1}{4 A _1 L }\)…
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