Explanation

(c) Resistance of a bulb $= \ \frac{(Rated\ voltage)^{2}}{(Rated\ power)}$
$R_{B_{1}} = \frac{(220)^{2}}{40}\Omega\ and\ R_{B_{2}} = \frac{(220)^{2}}{60}\Omega$
When the bulbs are connected in series,
$R_{S} = R_{B_{1}} + R_{B_{2}}$
$= \frac{(220)^{2}}{40} + \frac{(220)^{2}}{60} = (220)^{2}\left\lbrack \frac{1}{40} + \frac{1}{60} \right\rbrack$
$= (220)^{2}\left\lbrack \frac{60 + 40}{60 \times 40} \right\rbrack = (220)^{2}\left( \frac{100}{2400} \right) = \frac{(220)^{2}}{24}$
$\therefore P_{1} = \frac{V_{S}^{2}}{R_{s}} = (220)^{2} \times \frac{24}{(220)^{2}} = 24W$
When the bulbs are connected in parallel
$\frac{1}{R_{p}} = \frac{1}{R_{B_{1}}} + \frac{1}{R_{B_{2}}} \Rightarrow \frac{1}{R_{P}} = \frac{40}{(220)^{2}} + \frac{60}{(220)^{2}}$
$\frac{1}{R_{p}} = \frac{100}{(200)^{2}}\ or\ R_{p} = \frac{(220)^{2}}{100}$
$\therefore P_{2} = \frac{V_{S}^{2}}{R_{p}} = (220)^{2} \times \frac{100}{(200)^{2}} = 100\ W$
$\therefore\frac{P_{1}}{P_{2}} = \frac{24\ W}{100\ W} = 0.24$