It takes `16 min` to boil some water in an electric kettle. Due to some defect it becomes necessary to remove `10 %` turns of the heating coil of the kettle . After repairs , how much time will it take to boil the same mass of water ?

Options

A. `17.7` min
B. `14.4` min
C. `20.9` min
D. `13.9` min

Answer

To solve the problem, we need to analyze how the removal of 10% of the turns from the heating coil affects the resistance of the kettle and subsequently the time taken to boil the same mass of water.

### Step-by-Step Solution:

1. **Understanding the Resistance of the Heating Coil**:
The resistance \( R \) of the heating coil can be expressed as:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. The length of the wire is related to the number of turns \( n \) in the coil.

2. **Effect of Removing Turns**:
When 10% of the turns are removed, the remaining number of turns is:
\[ n' = n - 0.1n = 0.9n \]
The new resistance \( R' \) after removing 10% of the turns can be expressed as:
\[ R' = \frac{\rho L'}{A} \]
where \( L' \) is the new length of the wire corresponding to the new number of turns. Since the length is proportional to the number of turns, we can say:
\[ L' = 2\pi R \cdot n' = 2\pi R \cdot 0.9n = 0.9 \cdot L \]
Thus, the new resistance becomes:
\[ R' = \frac{\rho (0.9L)}{A} = 0.9R \]

3. **Power and Heat Transfer**:
The power \( P \) supplied by the kettle can be expressed as:
\[ P = \frac{V^2}{R} \]
The heat \( Q \) supplied to the water over time \( T \) is given by:
\[ Q = P \cdot T = \frac{V^2}{R} \cdot T \]
Since the heat required to boil the water remains constant, we can set up the equation for the initial and final scenarios:
\[ Q = \frac{V^2}{R} \cdot T = \frac{V^2}{R'} \cdot T' \]

4. **Relating Times with Resistance**:
Rearranging the above equation gives:
\[ \frac{V^2}{R} \cdot T = \frac{V^2}{R'} \cdot T' \]
This simplifies to:
\[ \frac{T}{T'} = \frac{R'}{R} \]
Substituting \( R' = 0.9R \):
\[ \frac{T}{T'} = \frac{0.9R}{R} = 0.9 \]
Therefore:
\[ T' = \frac{T}{0.9} \]

5. **Calculating the New Time**:
Given that the original time \( T = 16 \) minutes, we can calculate \( T' \):
\[ T' = \frac{16}{0.9} = \frac{16 \times 10}{9} = \frac{160}{9} \approx 17.78 \text{ minutes} \]

### Final Answer:
It will take approximately **17.78 minutes** to boil the same mass of water after removing 10% of the turns from the heating coil.