A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:

Options

A. Increased 8 times
B. Doubled
C. Halved
D. Unchanged

Explanation

To solve the problem step by step, we need to analyze how the rate of heat developed in the wire changes when the length is halved and the radius is doubled. The rate of heat developed in the wire can be expressed in terms of power (P), which is given by the formula:

\[ P = \frac{V^2}{R} \]

where \( V \) is the voltage applied across the wire and \( R \) is the resistance of the wire.

### Step 1: Understand the relationship between resistance, length, and area
The resistance \( R \) of a wire is given by the formula:

\[ R = \frac{\rho L}{A} \]

where:
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

For a circular wire, the area \( A \) can be expressed as:

\[ A = \pi r^2 \]

where \( r \) is the radius of the wire.

### Step 2: Calculate the initial resistance \( R_1 \)
Let the initial length of the wire be \( L \) and the initial radius be \( r \). The initial resistance \( R_1 \) can be expressed as:

\[ R_1 = \frac{\rho L}{\pi r^2} \]

### Step 3: Calculate the new resistance \( R_2 \)
Now, if the length is halved, the new length \( L' \) becomes:

\[ L' = \frac{L}{2} \]

If the radius is doubled, the new radius \( r' \) becomes:

\[ r' = 2r \]

The new cross-sectional area \( A' \) becomes:

\[ A' = \pi (r')^2 = \pi (2r)^2 = 4\pi r^2 \]

Now, substituting these values into the resistance formula gives us the new resistance \( R_2 \):

\[ R_2 = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{2}\right)}{4\pi r^2} = \frac{\rho L}{8\pi r^2} \]

### Step 4: Relate the new resistance to the initial resistance
We can express \( R_2 \) in terms of \( R_1 \):

\[ R_2 = \frac{R_1}{8} \]

### Step 5: Calculate the new power \( P_2 \)
Now we can calculate the new power \( P_2 \) using the new resistance \( R_2 \):

\[ P_2 = \frac{V^2}{R_2} = \frac{V^2}{\frac{R_1}{8}} = 8 \cdot \frac{V^2}{R_1} = 8P_1 \]

### Conclusion
Thus, the rate of heat developed in the wire increases by 8 times when the length is halved and the radius is doubled.

### Final Answer
The rate of heat developed in the wire will increase by 8 times.

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