It takes 12 minutes to boil 1 litre of water in an electric kettle. Due to some defect it becomes necessary to remove 20% turns of heating coil of the kettle. After repair, how much time will it take to boil 1 litre of water?

Options

A. 9.6 minute
B. 14.4 minute
C. 16.8 minute
D. 16.8 minute

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We know that it takes 12 minutes to boil 1 litre of water in the electric kettle with a heating coil of resistance \( R \).

### Step 2: Determine the effect of removing 20% of the turns
If we remove 20% of the turns from the heating coil, the remaining turns will be 80% of the original. The resistance of the heating coil is directly proportional to the number of turns. Therefore, if 20% of the turns are removed, the new resistance \( R_2 \) will be:
\[ R_2 = 0.8R \]

### Step 3: Apply the heat supply principle
The heat supplied to boil the water is given by the formula:
\[ H = \frac{V^2}{R} \cdot t \]
where \( H \) is the heat supplied, \( V \) is the voltage, \( R \) is the resistance, and \( t \) is the time.

Since the voltage \( V \) remains constant, we can set up the equation for the heat supplied in both scenarios:
\[ \frac{V^2}{R} \cdot T_1 = \frac{V^2}{R_2} \cdot T_2 \]
where \( T_1 \) is the initial time (12 minutes) and \( T_2 \) is the new time we want to find.

### Step 4: Substitute known values
Substituting \( R_2 = 0.8R \) into the equation gives:
\[ \frac{V^2}{R} \cdot 12 = \frac{V^2}{0.8R} \cdot T_2 \]

### Step 5: Cancel out common terms
We can cancel \( V^2 \) and \( R \) from both sides:
\[ 12 = \frac{T_2}{0.8} \]

### Step 6: Solve for \( T_2 \)
Now, we can solve for \( T_2 \):
\[ T_2 = 12 \times 0.8 = 9.6 \text{ minutes} \]

### Conclusion
After removing 20% of the turns from the heating coil, it will take approximately **9.6 minutes** to boil 1 litre of water.

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