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$A$ metal wire of circular cross-section has a n'sistance $R_{1}$. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now beromes $R_{2},$ then $R_{2}: R_{1}$ is
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$4: 1$
As we know that, $R_{1}=\rho \frac{1}{a}=\rho \frac{l^{2}}{V}$
where, $1=$ length of wire $a=$ area of cross-section of the wire and $V=$ volume of the wire $R_{1} \propto l^{2}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{l_{1}}{I_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2} \Rightarrow R_{2}: R_{1}=4: 1$
where, $1=$ length of wire $a=$ area of cross-section of the wire and $V=$ volume of the wire $R_{1} \propto l^{2}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{l_{1}}{I_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2} \Rightarrow R_{2}: R_{1}=4: 1$
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