Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to `2//3` of its initial value, then the same amount of water will boil with the same supply voltage in

Options

A. 15 minutes
B. 12 minutes
C. 10 minutes
D. 8 minutes

Answer

To solve the problem step by step, we will analyze the relationship between the resistance of the heating wire, the time taken to boil water, and the length of the wire.

### Step 1: Understand the initial conditions
The water boils in 15 minutes with the initial resistance \( R \) of the heating wire. We know that the heat required to boil the water can be expressed using the formula:
\[ H = I^2 R t \]
where \( H \) is the heat required, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time in seconds.

### Step 2: Convert time to seconds
Since the time is given in minutes, we convert it to seconds:
\[ t = 15 \text{ minutes} = 15 \times 60 = 900 \text{ seconds} \]

### Step 3: Express heat in terms of voltage
Using Ohm's law, we can express the current \( I \) in terms of voltage \( V \) and resistance \( R \):
\[ I = \frac{V}{R} \]
Substituting this into the heat equation gives:
\[ H = \left(\frac{V}{R}\right)^2 R t = \frac{V^2}{R} t \]

### Step 4: Set up the equation for the new resistance
When the length of the heating wire is decreased to \( \frac{2}{3} \) of its initial value, the new resistance \( R' \) can be expressed as:
\[ R' = \frac{2}{3} R \]
This is because resistance is directly proportional to the length of the wire.

### Step 5: Set up the equation for the new time
The heat required to boil the water remains the same, so we can set up the equation for the new time \( t' \):
\[ H = \frac{V^2}{R'} t' \]
Since \( H \) is the same in both cases, we can equate the two expressions for \( H \):
\[ \frac{V^2}{R} t = \frac{V^2}{R'} t' \]

### Step 6: Cancel out \( V^2 \) and rearrange
Cancelling \( V^2 \) from both sides gives:
\[ \frac{t}{R} = \frac{t'}{R'} \]
Rearranging this gives:
\[ t' = t \cdot \frac{R'}{R} \]

### Step 7: Substitute the values
Now substituting \( R' = \frac{2}{3} R \) and \( t = 900 \text{ seconds} \):
\[ t' = 900 \cdot \frac{\frac{2}{3} R}{R} = 900 \cdot \frac{2}{3} = 600 \text{ seconds} \]

### Step 8: Convert back to minutes
Finally, convert \( t' \) back to minutes:
\[ t' = \frac{600}{60} = 10 \text{ minutes} \]

### Conclusion
Thus, the time taken to boil the same amount of water with the reduced length of the heating wire is **10 minutes**.