Two bulbs are rated $60 W, 220 V$ and $240 W, 220 V$ respectively. The ratio of their ratio of their resistances is

1 : 1

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1 : 4

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4 : 1

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2 : 1

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Solution

The correct option is C $4 : 1$
Power, $P = V I$ and $V = I R$ .... ( ohm's law)
$P = \frac{V^{2}}{R} ⟹ P_{1} = \frac{V_{1}^{2}}{R_{1}} & P_{2} = \frac{V_{2}^{2}}{R_{2}}$
Now, $\frac{P_{1}}{P_{2}} = \frac{V_{1}^{2}}{V_{2}^{2}} \times \frac{R_{2}}{R_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{V_{1}^{2}}{V_{2}^{2}} \times \frac{P_{2}}{P_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{220^{2}}{220^{2}} \times \frac{240}{60} = \frac{4}{1}$
$R_{1} : R_{2} = 4 : 1$

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