A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

Options

A. both the lengths and radius of the wire are halved
B. both the length and radius of the wire are doubled
C. the radius of the wire is doubled
D. the length of the wire is doubled

Explanation

To solve the problem, we need to analyze how the heat developed in a metallic wire changes when certain physical parameters are altered. The heat developed in a wire due to the passage of electric current can be calculated using the formula:

\[ H = I^2 R t \]

Where:
- \( H \) is the heat developed,
- \( I \) is the current,
- \( R \) is the resistance of the wire,
- \( t \) is the time for which the current flows.

### Step 1: Express Current in terms of Voltage and Resistance
Using Ohm's law, we know that:

\[ I = \frac{V}{R} \]

Where \( V \) is the voltage applied across the wire.

### Step 2: Substitute Current in the Heat Equation
Substituting \( I \) in the heat equation:

\[ H = \left(\frac{V}{R}\right)^2 R t \]

This simplifies to:

\[ H = \frac{V^2}{R} t \]

### Step 3: Express Resistance in terms of Material Properties
The resistance \( R \) of the wire can be expressed as:

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

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

The cross-sectional area \( A \) can be expressed in terms of the radius \( r \) of the wire:

\[ A = \pi r^2 \]

### Step 4: Substitute Resistance in the Heat Equation
Substituting \( R \) in the heat equation:

\[ H = \frac{V^2 A}{\rho l} t \]

Substituting \( A \):

\[ H = \frac{V^2 \pi r^2}{\rho l} t \]

### Step 5: Analyze the Conditions for Doubling Heat
We want to find conditions under which the heat \( H \) is doubled, i.e., \( H' = 2H \).

#### Option A: Both Length and Radius are Doubled
If both length and radius are doubled:

- New length \( l' = 2l \)
- New radius \( r' = 2r \)

The new area \( A' \) becomes:

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

Now substituting these into the heat equation:

\[ H' = \frac{V^2 (4\pi r^2)}{\rho (2l)} t = \frac{2V^2 \pi r^2}{\rho l} t = 2H \]

So, in this case, the heat is doubled.

#### Option B: Both Length and Radius are Halved
If both length and radius are halved:

- New length \( l' = \frac{l}{2} \)
- New radius \( r' = \frac{r}{2} \)

The new area \( A' \) becomes:

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

Now substituting these into the heat equation:

\[ H' = \frac{V^2 \left(\frac{\pi r^2}{4}\right)}{\rho \left(\frac{l}{2}\right)} t = \frac{2V^2 \pi r^2}{4\rho l} t = \frac{1}{2}H \]

In this case, the heat is halved.

### Conclusion
The heat developed is doubled when both the length and radius of the wire are doubled. Thus, the correct answer is:

**Both length and radius of the wire are doubled.**

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