An electric bulb marked `220 V, 100 W` will get fused if it is made to consume `150 W` or more. What voltage fluctuation will the bulb withstand?

Explanation

To solve the problem, we need to determine the voltage fluctuation that the electric bulb can withstand before it fuses. The bulb is rated at 220 V and 100 W, and it will fuse if it consumes 150 W or more.

### Step-by-Step Solution:

1. **Calculate the Resistance of the Bulb**:
The resistance \( R \) of the bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage rating of the bulb (220 V) and \( P \) is the power rating (100 W).

Substituting the values:
\[ R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

2. **Determine the Maximum Power Before Fusing**:
The bulb will fuse if it consumes 150 W or more. We can use the same formula for resistance to find the voltage at which the bulb consumes 150 W.

3. **Calculate the Voltage for 150 W**:
Rearranging the formula \( P = \frac{V^2}{R} \) gives us:
\[ V^2 = R \times P \]
Therefore, we can find \( V \) by taking the square root:
\[ V = \sqrt{R \times P} \]
Substituting \( R = 484 \, \Omega \) and \( P = 150 \, W \):
\[ V = \sqrt{484 \times 150} \]

Calculating:
\[ V = \sqrt{72600} \approx 270 \, V \]

4. **Conclusion**:
The bulb can withstand a voltage fluctuation up to approximately 270 V before it fuses.

### Final Answer:
The voltage fluctuation that the bulb can withstand is approximately **270 V**.