Two metallic wire of same material and same length have different diameters when connected in series across a battery, the heat produced in thinner wire is $Q_{1}$ and that in thicker one is $Q_{2}$ . the correct relation is

Q_{1} = Q_{2}

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Q_{1} < Q_{2}

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Q_{1} > Q_{2}

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will depends on the emf of the battery.

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Solution

The correct option is C $Q_{1} > Q_{2}$
Heat produced in metallic wire is

$Q = I^{2} R$

$Q = (\frac{J}{A})^{2} R$

where variables have their usual meanings

$\Rightarrow Q \propto \frac{1}{A} \Rightarrow Q \propto \frac{1}{d}$

Hence, the heat produced in thinner wire is $Q_{1}$ and that in thicker one

is $Q_{2}$ . the correct relation is $Q_{1} > Q_{2}$ .

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