The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W is supplied, 60% of the extra power is converted into light and the remaining int heat. The bulb consumes 100 W at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from `220 V to 200 V`.

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Calculate the resistance of the bulb
The power consumed by the bulb at 220 V is given as 100 W. We can use the formula for power:

\[ P = \frac{V^2}{R} \]

Rearranging this formula to find the resistance \( R \):

\[ R = \frac{V^2}{P} \]

Substituting the values \( V = 220 \, V \) and \( P = 100 \, W \):

\[ R = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 2: Calculate the excess power at 220 V
The bulb requires 40 W to start glowing. The excess power when the bulb is consuming 100 W is:

\[ \text{Excess Power} = 100 \, W - 40 \, W = 60 \, W \]

### Step 3: Calculate the power converted into light at 220 V
According to the problem, 60% of the excess power is converted into light. Thus, the power converted into light is:

\[ \text{Power converted to light} = 0.6 \times \text{Excess Power} = 0.6 \times 60 \, W = 36 \, W \]

### Step 4: Calculate the power consumed at 200 V
Now, we need to find the power consumed when the voltage is reduced to 200 V. Using the same power formula:

\[ P = \frac{V^2}{R} \]

Substituting \( V = 200 \, V \) and \( R = 484 \, \Omega \):

\[ P = \frac{200^2}{484} = \frac{40000}{484} \approx 82.64 \, W \]

### Step 5: Calculate the excess power at 200 V
Now, we find the excess power at 200 V:

\[ \text{Excess Power} = 82.64 \, W - 40 \, W = 42.64 \, W \]

### Step 6: Calculate the power converted into light at 200 V
Again, using the 60% conversion rate for excess power:

\[ \text{Power converted to light} = 0.6 \times 42.64 \, W = 25.584 \, W \]

### Step 7: Calculate the change in power converted to light
Now we find the change in power converted to light between the two scenarios:

\[ \Delta P = 36 \, W - 25.584 \, W = 10.416 \, W \]

### Step 8: Calculate the percentage drop in light intensity
Finally, we calculate the percentage drop in light intensity:

\[ \text{Percentage drop} = \left( \frac{\Delta P}{\text{Power at 220 V}} \right) \times 100 = \left( \frac{10.416}{36} \right) \times 100 \approx 29.0\% \]

### Final Answer
The percentage drop in the light intensity at a point when the supply voltage changes from 220 V to 200 V is approximately **29%**.

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