If voltage across a bulb rated 220 volt-100 watt drops by `2.5 %` of its value, the percentage of the rated value by which the power would decrease is

Options

A. `10%`
B. `20%`
C. `2.5%`
D. `5%`

Explanation

To solve the problem, we need to determine how much the power of the bulb decreases when the voltage drops by 2.5% of its rated value. We can use the relationship between power, voltage, and resistance to find the solution.

### Step-by-Step Solution:

1. **Understand the rated values:**
- The rated voltage (V_rated) of the bulb is 220 volts.
- The rated power (P_rated) of the bulb is 100 watts.

2. **Calculate the resistance of the bulb:**
- The power formula is given by:
\[ P = \frac{V^2}{R} \]
- Rearranging the formula to find resistance (R):
\[ R = \frac{V^2}{P} \]
- Substituting the rated values:
\[ R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

3. **Calculate the new voltage after the drop:**
- The voltage drop is 2.5% of the rated voltage:
\[ \Delta V = 2.5\% \text{ of } 220 = \frac{2.5}{100} \times 220 = 5.5 \, \text{volts} \]
- Therefore, the new voltage (V_new):
\[ V_{new} = V_{rated} - \Delta V = 220 - 5.5 = 214.5 \, \text{volts} \]

4. **Calculate the new power with the reduced voltage:**
- Using the power formula again:
\[ P_{new} = \frac{V_{new}^2}{R} \]
- Substituting the new voltage and the resistance:
\[ P_{new} = \frac{(214.5)^2}{484} \]
- Calculating \( (214.5)^2 \):
\[ (214.5)^2 = 46006.25 \]
- Now substituting back:
\[ P_{new} = \frac{46006.25}{484} \approx 95.0 \, \text{watts} \]

5. **Calculate the percentage decrease in power:**
- The decrease in power (\(\Delta P\)):
\[ \Delta P = P_{rated} - P_{new} = 100 - 95 = 5 \, \text{watts} \]
- The percentage decrease in power:
\[ \text{Percentage decrease} = \frac{\Delta P}{P_{rated}} \times 100 = \frac{5}{100} \times 100 = 5\% \]

### Final Answer:
The percentage of the rated value by which the power would decrease is **5%**.