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JEE Main 2023 · Physics · Current Electricity · medium

A uniform wire of resistance R is uniformly stretched to 1.2 times its original length (its volume stays constant). It is then cut into three equal…

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A uniform wire of resistance RR is uniformly stretched to 1.21.2 times its original length (its volume stays constant). It is then cut into three equal pieces and the pieces are joined in parallel. If the same potential difference is applied across this combination as across the original wire, the ratio of the power dissipated in the combination to the original power is
  1. A. 254\dfrac{25}{4}
  2. B. 94\dfrac{9}{4}
  3. C. 259\dfrac{25}{9}
  4. D. 1009\dfrac{100}{9}
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Answer: (A) 254\dfrac{25}{4}

Stretching by 1.21.2 at constant volume reduces the area by the factor 1/1.21/1.2, so R′=R×1.2×1.2=1.44RR'=R\times1.2\times1.2=1.44R. Each of the three pieces has resistance 1.44R3=0.48R\dfrac{1.44R}{3}=0.48R; in parallel they give 0.48R3=0.16R=4R25\dfrac{0.48R}{3}=0.16R=\dfrac{4R}{25}. At the same applied voltage P=V2/ReqP=V^2/R_{eq}, so P′P=R0.16R=254\dfrac{P'}{P}=\dfrac{R}{0.16R}=\dfrac{25}{4}. Taking R′=1.2RR'=1.2R (forgetting the change of area) would give R0.133R=7.5\dfrac{R}{0.133R}=7.5.
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