QJEEM24Phe7a8: JEE Main 2024 Physics - Class 12 JEEM24Phe7a8 Solution: An electric bulb rated is connected across a supply. The power…

To find the power dissipation of the bulb when it's connected to a $100 \mathrm{V}$ supply instead of its rated $200 \mathrm{V}$ supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by $P = \frac{V^2}{R}$ . The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.

First, let's find the resistance of the bulb based on its rated conditions:

$P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P}$

Substituting the rated values, we get:

$R = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega$

Now, using this resistance, we can find the power dissipation when the bulb is connected to a $100 \mathrm{V}$ supply:

$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W$

This means the power dissipation of the bulb when connected to a $100 \mathrm{V}$ supply is $12.5 \, W$ .

Therefore, the correct answer is Option C: 12.5 W.