A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and density is assumed to remain same. If the resistance of the wire now becomes $R_{2}$ , then $R_{2} : R_{1}$ is:

1 : 1

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1 : 2

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4 : 1

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1 : 4

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Solution

The correct option is B 4 : 1
As we know that, $R_{1} = ρ \frac{l}{a} = ρ \frac{l^{2}}{V}$ ,
where $l =$ length of wire, $a =$ area of cross-section of the wire and $V =$ volume of the wire.
So, $R_{1} \propto l^{2}$
$\Rightarrow \frac{R_{1}}{R_{2}} = {(\frac{l_{1}}{l_{2}})}^{2} = {(\frac{1}{2})}^{2} \Rightarrow R_{2} : R_{1} = 4 : 1$

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A metal wire of circular cross-section has a resistance R1. The wire is now stretched without breaking so that its length is doubled and density is assumed to remain same. If the resistance of the wire now becomes R2, then R2:R1 is:

The correct option is B 4 : 1As we know that, R1=ρla=ρl2V,where l= length of wire, a= area of cross-section of the wire and V= volume of the wire.So, R1∝l2⇒R1R2=(l1l2)2=(12)2⇒R2:R1=4:1

A metal wire of circular cross-section has a resistance $R_{1}$ . The wire is now stretched without breaking so that its length is doubled and density is assumed to remain same. If the resistance of the wire now becomes $R_{2}$ , then $R_{2} : R_{1}$ is: