One filament takes 10min to heat a kettle and another takes 15 min. If connected in parallel they combindly take………min to heat the same kettle:

Options

A. 6
B. 12.5
C. 25
D. 7.5

Explanation

To solve the problem of how long it takes for two filaments connected in parallel to heat a kettle, we can follow these steps:

### Step 1: Understand the given information
We have two filaments:
- Filament 1 takes \( T_1 = 10 \) minutes to heat the kettle.
- Filament 2 takes \( T_2 = 15 \) minutes to heat the kettle.

### Step 2: Relate time to resistance
The heat produced by each filament can be related to their resistances and time taken. The heat produced by a filament can be expressed as:
\[ H = \frac{V^2}{R} \times T \]
Where:
- \( H \) is the heat produced,
- \( V \) is the voltage,
- \( R \) is the resistance,
- \( T \) is the time.

From this, we can express the resistance of each filament:
- For Filament 1:
\[ R_1 = \frac{V^2}{H} \times T_1 \]
- For Filament 2:
\[ R_2 = \frac{V^2}{H} \times T_2 \]

### Step 3: Calculate effective resistance in parallel
When the two filaments are connected in parallel, the effective resistance \( R_{\text{effective}} \) can be calculated using:
\[ \frac{1}{R_{\text{effective}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

### Step 4: Substitute the expressions for resistances
Substituting the expressions for \( R_1 \) and \( R_2 \):
\[ \frac{1}{R_{\text{effective}}} = \frac{H}{V^2 \times T_1} + \frac{H}{V^2 \times T_2} \]

### Step 5: Simplify the equation
Factoring out \( \frac{H}{V^2} \):
\[ \frac{1}{R_{\text{effective}}} = \frac{H}{V^2} \left( \frac{1}{T_1} + \frac{1}{T_2} \right) \]

### Step 6: Relate effective resistance to time
The heat produced by the effective resistance can also be expressed as:
\[ H = \frac{V^2}{R_{\text{effective}}} \times T_{\text{effective}} \]

### Step 7: Set the equations equal
Equating the two expressions for heat:
\[ \frac{H}{V^2} \times T_{\text{effective}} = \frac{H}{V^2} \left( \frac{1}{T_1} + \frac{1}{T_2} \right) \]

### Step 8: Cancel out common terms
Since \( \frac{H}{V^2} \) is common on both sides, we can cancel it:
\[ T_{\text{effective}} = \frac{1}{\left( \frac{1}{T_1} + \frac{1}{T_2} \right)} \]

### Step 9: Substitute the values of \( T_1 \) and \( T_2 \)
Substituting \( T_1 = 10 \) minutes and \( T_2 = 15 \) minutes:
\[ T_{\text{effective}} = \frac{1}{\left( \frac{1}{10} + \frac{1}{15} \right)} \]

### Step 10: Calculate the effective time
Calculating the right-hand side:
\[ \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \]
Thus,
\[ T_{\text{effective}} = \frac{1}{\frac{1}{6}} = 6 \text{ minutes} \]

### Final Answer
The combined time taken by the two filaments connected in parallel to heat the kettle is **6 minutes**.

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