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 $......\Omega$
JEE MAIN 2023, Diffcult
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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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