JEE Main 2018 · Physics · Calorimetry and Heat Transfer · easy

A uniform wire of resistance 8 Ω is connected across an ideal 16 V source. The wire is then drawn (without loss of material) to 1.5 times its…

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A uniform wire of resistance

8\ \Omega

is connected across an ideal

16\ \mathrm{V}

source. The wire is then drawn (without loss of material) to

1.5

times its original length and connected across the same source. The power dissipated in the wire becomes nearly

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Answer: (B) $14.2\ \mathrm{W}$

Volume is conserved, so area falls by the factor

1/1.5

while length rises by

1.5

. Resistance

R=\rho L/A

therefore scales as

1.5^2=2.25

, giving

R'=2.25\times8=18\ \Omega

. With the voltage fixed,

P'=\dfrac{V^2}{R'}=\dfrac{256}{18}=14.2\ \mathrm{W}

. The initial power was

32\ \mathrm{W}

, and

32/2.25=14.2\ \mathrm{W}

agrees. Forgetting the change in area gives

R'=12\ \Omega

and

21.3\ \mathrm{W}

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake