A constant voltage is applied between the two ends of a uniform metallic wire. If both the length and radius of wire are doubled, then :

The heat developed in the wire will be doubled

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The electric field in the wire will be doubled

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The heat developed will remain the same

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The electric field in the wire will be halved

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Solution

The correct options are
A The heat developed in the wire will be doubled
C The electric field in the wire will be halved
Since radius is doubled,area becomes four times.
hence,
$L_{f} = 2 L_{i}$

$A_{f} = 4 A_{i}$

Hence, $R = \frac{ρ L}{A} ⟹ R_{f} = \frac{R_{i}}{2}$

By Joule's law,for constant voltage,

$P \propto \frac{1}{R} ⟹ P_{f} = 2 P_{i}$

By $E = \frac{V}{L}$ ,

we get $E_{f} = \frac{E_{i}}{2}$

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