Wires `A` and `B` are made from the same material. `A` has twice the diameter and three times the length of `B`. If the elastic limits are not reached, when each is stretched by the same tension, the ratio of energy stored in `A` to that in `B` is

Options

A. `2:3`
B. `3:4`
C. `3:2`
D. `6:1`

Explanation

To find the ratio of energy stored in wires A and B when stretched by the same tension, we can follow these steps:

### Step 1: Understand the relationship between energy stored, tension, and elongation
The energy stored in a wire when it is stretched can be expressed using the formula:
\[ U = \frac{1}{2} F x \]
where \( U \) is the potential energy (energy stored), \( F \) is the force (tension), and \( x \) is the elongation of the wire.

### Step 2: Relate elongation to the properties of the wires
Using Young's modulus, we know that:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} \]
Rearranging gives:
\[ x = \frac{F L}{A Y} \]
where:
- \( F \) is the force (tension),
- \( A \) is the cross-sectional area,
- \( L \) is the original length,
- \( Y \) is Young's modulus (constant for both wires since they are made of the same material).

### Step 3: Calculate the cross-sectional area of each wire
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the wire.

For wire A:
- Diameter of A = 2 * Diameter of B
- Therefore, radius of A \( r_A = 2r_B \)
- Area of A \( A_A = \pi (2r_B)^2 = 4\pi r_B^2 = 4A_B \)

### Step 4: Set up the elongation for both wires
For wire A:
\[ x_A = \frac{F L_A}{A_A Y} \]
For wire B:
\[ x_B = \frac{F L_B}{A_B Y} \]

### Step 5: Substitute the lengths and areas
Given:
- Length of A \( L_A = 3L_B \)
- Area of A \( A_A = 4A_B \)

Substituting these into the elongation equations:
\[ x_A = \frac{F (3L_B)}{(4A_B) Y} = \frac{3F L_B}{4A_B Y} \]
\[ x_B = \frac{F L_B}{A_B Y} \]

### Step 6: Calculate the ratio of elongations
Now, we can find the ratio of elongations:
\[ \frac{x_A}{x_B} = \frac{\frac{3F L_B}{4A_B Y}}{\frac{F L_B}{A_B Y}} = \frac{3}{4} \]

### Step 7: Calculate the ratio of energy stored
Using the energy formula:
\[ U_A = \frac{1}{2} F x_A \quad \text{and} \quad U_B = \frac{1}{2} F x_B \]
The ratio of energies stored in A and B is:
\[ \frac{U_A}{U_B} = \frac{x_A}{x_B} = \frac{3}{4} \]

### Final Answer
Thus, the ratio of energy stored in wire A to that in wire B is:
\[ \frac{U_A}{U_B} = \frac{3}{4} \]