A `500W ` heating unit is designed to operate from a `115 "volt"` line. If the line voltage drop to `110 "Volt"` , the percentage drop in heat output will be

Options

A. `10.20%`
B. `8.1%`
C. `8.6%`
D. `7.6%`

Explanation

To solve the problem step by step, we need to determine the percentage drop in heat output when the voltage drops from 115 volts to 110 volts.

### Step 1: Understand the relationship between power and voltage
The power (P) of a heating unit is related to the voltage (V) by the formula:
\[ P \propto V^2 \]
This means that power is directly proportional to the square of the voltage.

### Step 2: Write the power output at rated voltage
The rated power output at 115 volts is given as:
\[ P_{\text{rated}} = 500 \, \text{W} \]

### Step 3: Calculate the power output at the reduced voltage
Using the proportionality relationship, we can express the power output at the reduced voltage (110 volts) as:
\[ \frac{P}{P_{\text{rated}}} = \left( \frac{V}{V_{\text{rated}}} \right)^2 \]
Substituting the values:
\[ \frac{P}{500} = \left( \frac{110}{115} \right)^2 \]

### Step 4: Calculate the ratio
Now, calculate the ratio:
\[ \frac{110}{115} = 0.9565 \]
Now square this value:
\[ \left( 0.9565 \right)^2 \approx 0.9158 \]

### Step 5: Calculate the power output at 110 volts
Now, substituting back to find P:
\[ \frac{P}{500} \approx 0.9158 \]
Thus,
\[ P \approx 500 \times 0.9158 \approx 457.9 \, \text{W} \]

### Step 6: Calculate the percentage drop in heat output
Now, we can find the percentage drop in power output:
\[ \text{Percentage drop} = \left( \frac{P_{\text{rated}} - P}{P_{\text{rated}}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage drop} = \left( \frac{500 - 457.9}{500} \right) \times 100 \]
\[ \text{Percentage drop} = \left( \frac{42.1}{500} \right) \times 100 \approx 8.42\% \]

### Final Answer
The percentage drop in heat output is approximately **8.42%**.

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