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JEE Main 2018 (15 Apr, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2018 (15 Apr, Shift 2)

Q45JEE Main 2018 · 15 APR · SHIFT 2Medium

Electrical Power & Heating Effect

A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:

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Answer: (d)

$R = \dfrac{\rho L}{\pi r^2}$ : halving $L$ and doubling $r$ gives $R' = \dfrac{R/2}{4} = R/8$ .
At constant $V$ , $P = V^2/R \Rightarrow P' = 8P$ .
So the answer is (d) increased 8 times.

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