1. Use the bulb's rated values to find its resistance

For a bulb rated $220\,\text{V}$ and $1000\,\text{W}$,

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

So,

$R = \frac{V^2}{P} = \frac{(220)^2}{1000} = \frac{48400}{1000} = 48.4\,\Omega$

2. Now connect it to $110\,\text{V}$ supply

Assuming the resistance remains the same, the new power is

$P' = \frac{V'^2}{R} = \frac{(110)^2}{48.4}$

$P' = \frac{12100}{48.4} = 250\,\text{W}$

3. Check via ratio method

Since resistance is constant,

$P \propto V^2$

Therefore,

$\frac{P'}{P} = \left(\frac{110}{220}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

So,

$P' = 1000 \times \frac{1}{4} = 250\,\text{W}$

4. Evaluate options

• A: $750$ watt — incorrect
• B: $500$ watt — incorrect
• C: $250$ watt — correct
• D: $1000$ watt — incorrect

Hence, the correct answer is Option C.