Question:
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 :
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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Rate of heat = \(V^2/R\) with \(R=\frac{\rho L}{\pi r^2}\); find how R changes.
Updated On: Nov 11, 2026
- Doubled
- Halved
- Unchanged
- Increased 8 times
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The Correct Option is D
Solution and Explanation
Power \(P=\frac{V^2}{R}\) with V constant, and \(R=\frac{\rho L}{\pi r^2}\).
New length L/2 and new radius 2r: \(R_2=\frac{\rho (L/2)}{\pi(2r)^2}=\frac{R}{8}\).
So \(P_2=\frac{V^2}{R/8}=8P\).
The rate of heat developed increases 8 times.
Answer: (4) Increased 8 times
New length L/2 and new radius 2r: \(R_2=\frac{\rho (L/2)}{\pi(2r)^2}=\frac{R}{8}\).
So \(P_2=\frac{V^2}{R/8}=8P\).
The rate of heat developed increases 8 times.
Answer: (4) Increased 8 times
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