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(C) Let the voltage of the $AC$ mains be $V$. The resistance of a bulb is given by $R_b = \frac{V_{rated}^2}{P_{rated}}$.For a $50 \,W$ bulb,$R_1 = \f

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Increase

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(C) Let the voltage of the $AC$ mains be $V$. The resistance of a bulb is given by $R_b = \frac{V_{rated}^2}{P_{rated}}$.For a $50 \,W$ bulb,$R_1 = \frac{V_{rated}^2}{50}$.For a $100 \,W$ bulb,$R_2 = \frac{V_{rated}^2}{100} = \frac{R_1}{2}$.Let the resistance of the heater coil be $R_h$. The power output of the heater is $P = I^2 R_h = \left( \frac{V}{R_b + R_h} \right)^2 R_h$.In the first case,$P_1 = \left( \frac{V}{R_1 + R_h} \right)^2 R_h$.In the second case,$P_2 = \left( \frac{V}{R_2 + R_h} \right)^2 R_h = \left( \frac{V}{R_1/2 + R_h} \right)^2 R_h$.Since $R_1/2  P_1$.Therefore,the heater output will increase.

$A$ $50 \,W$ bulb connected in series with a heater coil is put to an $AC$ mains. Now the bulb is replaced by a $100 \,W$ bulb. The heater output will ...........

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