An electric kettle has tow heating coils. When one of the coils connected to an AC source, the water in the kettle boils in `10` min. when the other coil is used the water boils in `40` min. if both the coils are connected in parallel, the time taken by the same quantity of water of boil will be

Options

A. 25 min
B. 15 min
C. 8 min
D. 4 min

Answer

To solve the problem step by step, we will analyze the heating coils and their effects on the boiling time of water in the kettle.

### Step 1: Determine the Power of Each Coil
We know that the power \( P \) of a heating coil is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage and \( R \) is the resistance of the coil.

Let:
- \( R_1 \) be the resistance of the first coil (which boils water in 10 minutes).
- \( R_2 \) be the resistance of the second coil (which boils water in 40 minutes).

### Step 2: Calculate the Heat Produced by Each Coil
The heat produced by each coil can be expressed as:
\[ Q = P \cdot t \]
where \( t \) is the time in seconds.

For coil 1 (boils water in 10 minutes):
\[ Q_1 = P_1 \cdot 600 \quad \text{(since 10 min = 600 s)} \]
For coil 2 (boils water in 40 minutes):
\[ Q_2 = P_2 \cdot 2400 \quad \text{(since 40 min = 2400 s)} \]

### Step 3: Relate the Heat Produced to Resistance
From the power formula, we can express the heat produced in terms of resistance:
\[ Q_1 = \frac{V^2}{R_1} \cdot 600 \]
\[ Q_2 = \frac{V^2}{R_2} \cdot 2400 \]
Since both coils boil the same quantity of water, we can equate \( Q_1 \) and \( Q_2 \):
\[ \frac{V^2}{R_1} \cdot 600 = \frac{V^2}{R_2} \cdot 2400 \]

### Step 4: Simplify the Equation
Dividing both sides by \( V^2 \) (assuming \( V \neq 0 \)):
\[ \frac{600}{R_1} = \frac{2400}{R_2} \]
Cross-multiplying gives:
\[ 600 R_2 = 2400 R_1 \]
Simplifying this, we find:
\[ R_2 = 4 R_1 \]

### Step 5: Calculate the Equivalent Resistance When Both Coils are Connected in Parallel
When both coils are connected in parallel, the equivalent resistance \( R_{eq} \) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting \( R_2 = 4 R_1 \):
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{4 R_1} = \frac{4 + 1}{4 R_1} = \frac{5}{4 R_1} \]
Thus,
\[ R_{eq} = \frac{4 R_1}{5} \]

### Step 6: Calculate the Time Taken to Boil Water with Both Coils
The power when both coils are used is:
\[ P_{eq} = \frac{V^2}{R_{eq}} = \frac{V^2}{\frac{4 R_1}{5}} = \frac{5 V^2}{4 R_1} \]
The heat required to boil the water remains the same, so we can use the formula:
\[ Q = P \cdot t \]
Let \( t_{eq} \) be the time taken to boil the water with both coils:
\[ Q = P_{eq} \cdot t_{eq} \]
Setting this equal to the heat produced by the first coil:
\[ \frac{V^2}{R_1} \cdot 600 = \frac{5 V^2}{4 R_1} \cdot t_{eq} \]
Cancelling \( V^2 \) and \( R_1 \) from both sides:
\[ 600 = \frac{5}{4} t_{eq} \]
Multiplying both sides by \( \frac{4}{5} \):
\[ t_{eq} = \frac{600 \cdot 4}{5} = 480 \text{ seconds} = 8 \text{ minutes} \]

### Final Answer
The time taken by the same quantity of water to boil when both coils are connected in parallel is **8 minutes**.

---