Question

If the length of a filament of a heater is reduced by 10% , the power of the heater will:

• A. Increase by about 9%
• B. Decrease by about 10%
• C. Increase by 11%
• D. Decrease by 19%

Answer 1

The Correct Option is C

To determine how reducing the length of a filament by 10% affects the power of a heater, we need to consider how the resistance of the filament changes and how that impacts the power output.
Step 1: Understanding Resistance and Power Relationship
The resistance (\(R\)) of a filament is given by:
\[ R = \rho \frac{L}{A} \]
where:
- \(\rho\) is the resistivity of the material,
- \(L\) is the length of the filament,
- \(A\) is the cross-sectional area.
If the length (\(L\)) is reduced by 10%, the new length \(L'\) is:
\[ L' = 0.9L \]
Step 2: Effect on Resistance
The new resistance \(R'\) with the reduced length can be written as:
\[ R' = \rho \frac{L'}{A} = \rho \frac{0.9L}{A} = 0.9 \left( \rho \frac{L}{A} \right) = 0.9R \]
Step 3: Power Relationship
The power (\(P\)) of the heater is related to the voltage (\(V\)) and resistance (\(R\)) by:
\[ P = \frac{V^2}{R} \]
With the new resistance \(R'\), the new power \(P'\) is:
\[ P' = \frac{V^2}{R'} = \frac{V^2}{0.9R} \]
Step 4: Comparing Powers
The ratio of the new power to the original power is:
\[ \frac{P'}{P} = \frac{V^2 / 0.9R}{V^2 / R} = \frac{1}{0.9} = \frac{10}{9} \approx 1.11 \]
 Conclusion:
If the length of the filament of a heater is reduced by 10%, the power of the heater will increase by approximately 11%.
The correct answer is option  (C): Increase by 11%