The power consumed by a bulb is given by \(𝑃=\frac {𝑉^2}{R}\). Rearranging this formula to find resistance, we get \(𝑅 = \frac {𝑉^2}{P}\). Given a supply voltage \(V = 220 \ V\) and power \(P = 100\ W\), the resistance is calculated as \(𝑅 = \frac {(220)^2}{100} = 484\ Ω\). Assuming the bulb's resistance remains constant, if the supply voltage is reduced to \(110 V\), the energy consumed by it can be found using the power expression \(𝑃 = \frac {𝑉^2}{R}\). Substituting the new voltage and the calculated resistance, \(P= \frac {(110)^2}{484} = \frac {12100}{484} = 25 \ 𝑊\). Therefore, the correct option is (D): \(25\ W\).