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 \(......\Omega\)
Pick an option to check your answer.
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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