A boy has two spare light bulbs in his drawer. One is market 220V and 100W and the other is market 240V and 60W. He tries to decide which of the following assertions are correct?

Options

A. The 60W light bulb has more resistance and therefore burns less brightly
B. The 60W light bulb has less resistance and therefore burns less brightly
C. The 100W light bulb has more resistance and therefore burns more brightly
D. The 100W light bulb has less resistance and therefore burns more brightly

Explanation

To solve the problem, we need to analyze the two light bulbs based on their voltage and power ratings. We will calculate their resistances and determine which bulb is brighter based on the resistance values.

### Step 1: Identify the specifications of the bulbs
- **Bulb 1**: 220V, 100W
- **Bulb 2**: 240V, 60W

### Step 2: Calculate the resistance of each bulb
The resistance \( R \) of a bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage and \( P \) is the power.

#### For Bulb 1:
\[ R_1 = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

#### For Bulb 2:
\[ R_2 = \frac{(240)^2}{60} = \frac{57600}{60} = 960 \, \Omega \]

### Step 3: Compare the resistances
From our calculations:
- \( R_1 = 484 \, \Omega \)
- \( R_2 = 960 \, \Omega \)

Since \( R_2 > R_1 \), we conclude that the resistance of Bulb 2 is greater than that of Bulb 1.

### Step 4: Determine the brightness of the bulbs
The brightness of a bulb is related to the power it dissipates when connected to a circuit, which can be expressed as:
\[ P = I^2 R \]
where \( I \) is the current flowing through the bulb. If we assume the current remains constant, the bulb with the higher resistance will dissipate more power and thus be brighter.

Since \( R_2 > R_1 \), Bulb 2 (240V, 60W) will be brighter than Bulb 1 (220V, 100W) when both are connected to the same voltage source.

### Conclusion
- **Assertion**: Bulb 2 (240V, 60W) will be brighter than Bulb 1 (220V, 100W) due to its higher resistance.