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JEE Mains · Physics · STD 12 - 3. current electricity
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
- A Increased \(8\) times
- B Doubled
- C Halved
- D Unchanged
Answer & Solution
Correct Answer
(A) Increased \(8\) times
Step-by-step Solution
Detailed explanation
Rate of heat i.e., Power developed in the wire \(=P=\frac{V^{2}}{R}\) Resistance of the wire of length, \(L\) \(R_{1}=\frac{\rho L}{A}=\frac{\rho L}{\pi r^{2}}\) \(\therefore\) Power, \(P_{1}=\frac{V^{2}}{R_{1}}\) Resistance of the wire when length is halved i.e., \(L / 2\)…
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