Final answer:
The ratio of energy stored in wire A to that in wire B, which have different dimensions but are made from the same material and stretched by the same tension, is 3:4. This is calculated using the elastic potential energy formula adapted for the varying dimensions of the wires. The correct option answer in the final answer is option B.
Step-by-step explanation:
The question deals with the concept of elastic potential energy stored in stretched wires, which relates to Hooke's Law and the physical properties of materials within the subject of physics. Given that wire A has twice the diameter and three times the length of wire B, and both are stretched by the same tension within their elastic limits, we can calculate the ratio of energy stored in wire A to that in wire B.
For a cylindrical wire, the elastic potential energy (U) stored when it is stretched by a force is proportional to the square of the force (F) applied and the length (L) of the wire, and inversely proportional to the cross-sectional area (A) of the wire. The formula for the potential energy (U) stored in a wire is given by U = (1/2) * (F^2 * L) / (A * Y), where Y is the Young's modulus of the material (which is the same for both wires).
For wire A, let's denote the diameter as 2d and the length as 3l. The cross-sectional area A would then be π * (2d/2)^2 = 4 * (π * d^2 / 4), which simplifies to π * d^2. The potential energy stored in wire A, U_A, is proportional to (F^2 * 3l) / (π * d^2).
For wire B, with a diameter of d and length of l, the cross-sectional area B is π * (d/2)^2 = π * d^2 / 4. The potential energy stored in wire B, U_B, is proportional to (F^2 * l) / (π * d^2 / 4), which simplifies to (4 * F^2 * l) / (π * d^2).
When we take the ratio of the energies stored in wires A and B, the forces and material properties cancel out, leaving us with the ratio U_A : U_B = 3l / d^2 : 4l / d^2 = 3 : 4. Therefore, the energy stored in wire A to that in wire B is in the ratio of 3:4, which corresponds to option B.