Current Electricity

JEE Main level, single correct, medium

259 of 380

Question 259 of 380

A constant voltage 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: (A) Increased 8 times

Rate of heat i.e., Power developed in the
 wire =P=V2R\text { wire }=P=\frac{V^2}{R}
Resistance of the wire of length, LL
R1=ρLA=ρLπr2R_1=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}
∴\therefore Power, P1=V2R1P_1=\frac{V^2}{R_1}
Resistance of the wire when length is halved i.e., L/2L / 2
R2=ρL2π(2r)2=ρLπ8r2=R18R_2=\frac{\rho \frac{L}{2}}{\pi(2 r)^2}=\frac{\rho L}{\pi 8 r^2}=\frac{R_1}{8}
∴\therefore Power, P2=VR18=8VR1P_2=\frac{V}{\frac{R_1}{8}}=\frac{8 V}{R_1}
or, P2=8P1\mathrm{P}_2=8 \mathrm{P}_1 i.e., power increased 8 times of previous or original wire.

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