Current Electricity

JEE Main level, single correct, medium

66 of 380

Question 66 of 380

A wire of resistance R1{R}_{1}R1R_{1} is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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Answer: (A) 9:19:1

As per the given question, the change in the length is ∆l=2l∆l=2l.
Resistance of a wire is given by, R=ρlAR=\dfrac{\rho l}{A}
Therefore, R1=ρl1A1{R}_{1}=\rho \dfrac{{l}_{1}}{{A}_{1}}
When the length is starched by 2l12{l}_{1}, final length will be l2=l1+2l1=3l1{l}_{2}={l}_{1}+2{l}_{1}=3{l}_{1}. As the volume remains constant l2A2=l1A1⇒A2=A13{l}_{2}{A}_{2}={l}_{1}{A}_{1}\Rightarrow {A}_{2}=\dfrac{{A}_{1}}{3}. Therefore,
R2=ρ(3l1A13)=9ρl1A1{R}_{2}=\rho \left(\dfrac{3{l}_{1}}{\dfrac{{A}_{1}}{3}}\right)=9\rho \dfrac{{l}_{1}}{{A}_{1}}
∴R2R1=9∴\dfrac{{R}_{2}}{{R}_{1}}=9

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