Two bulbs connected across a 110 V supply have resistances in the ratio 5:9. The ratio of brightness of the light from them is:

Options

A. `9:5`
B. `5:9`
C. `1:1`
D. `1:2`

Explanation

To solve the problem of finding the ratio of brightness of two bulbs connected across a 110 V supply with resistances in the ratio of 5:9, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between power and brightness**:
The brightness of a bulb is directly proportional to its power. Therefore, we need to find the power of each bulb to determine their brightness.

2. **Use the formula for power**:
The power (P) of a bulb connected to a voltage (V) is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( R \) is the resistance of the bulb.

3. **Define the resistances**:
Let the resistances of the two bulbs be \( R_1 \) and \( R_2 \). According to the problem, the ratio of their resistances is:
\[ \frac{R_1}{R_2} = \frac{5}{9} \]
This implies:
\[ R_1 = 5k \quad \text{and} \quad R_2 = 9k \]
for some constant \( k \).

4. **Calculate the power for each bulb**:
Using the power formula for each bulb:
- For bulb 1:
\[ P_1 = \frac{V^2}{R_1} = \frac{110^2}{R_1} = \frac{110^2}{5k} \]
- For bulb 2:
\[ P_2 = \frac{V^2}{R_2} = \frac{110^2}{R_2} = \frac{110^2}{9k} \]

5. **Find the ratio of powers**:
To find the ratio of the powers \( P_1 \) and \( P_2 \):
\[ \frac{P_1}{P_2} = \frac{\frac{110^2}{5k}}{\frac{110^2}{9k}} = \frac{9k}{5k} = \frac{9}{5} \]

6. **Conclusion**:
The ratio of brightness of the two bulbs, which is the same as the ratio of their powers, is:
\[ \frac{P_1}{P_2} = \frac{9}{5} \]

### Final Answer:
The ratio of brightness of the two bulbs is \( 9:5 \).

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