An electric bulb is rated $$220$$ volt $$- 100$$ watt. The power consumed by it when operated on $$110$$ volt will be

The formula relating power ($$P$$), voltage ($$V$$), and resistance ($$R$$) is:

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

Rearranging to solve for resistance:

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

Given the rated values: 

Rated voltage, $$V_1 = 220 \text{ V}$$ 

Rated power, $$P_1 = 100 \text{ W}$$

Substitute these values to find the resistance ($$R$$):

$$ R = \frac{(220)^2}{100} $$

$$ R = \frac{48400}{100} $$

$$ R = 484\ \Omega $$

Now, we calculate the power consumed when the bulb is operated on the new voltage. 

New operating voltage, $$V_2 = 110 \text{ V}$$

Using the power formula again with the new voltage and the calculated constant resistance:

$$ P_2 = \frac{V_2^2}{R} $$

$$ P_2 = \frac{(110)^2}{484} $$

$$ P_2 = \frac{12100}{484} $$

$$ P_2 = 25 \text{ W} $$