Final answer:
Wire D, with the smallest cross-sectional area and a reasonable length, will have the maximum increase in length when the same tension is applied, due to the direct proportionality to length and inverse proportionality to cross-sectional area in the formula for Young's modulus.
Step-by-step explanation:
To determine which wire will experience the maximum increase in length when the same tension is applied, we need to understand the concept of Young's modulus, tension, and how these are related to the dimensions of the wire. Young's modulus (Y) is a measure of the stiffness of a material and is defined as the ratio of the stress (force per unit area) to the strain (relative change in length).
Considering the formula for Young's Modulus (Y) given by:
Y = (FL) / (AΔL)
Where:
- F is the applied force (tension)
- L is the original length of the wire
- A is the cross-sectional area
- ΔL is the change in length (elongation)
We can see that for a given material (constant Y) and applied force (F), the change in length (ΔL) is directly proportional to the
original length (L)
and inversely proportional to the
cross-sectional area (A)
.
Since all the wires are made of the same material (and thus have the same Young's modulus) and the same tension is applied, the wire with the longest original length and smallest cross-sectional area will have the maximum increase in length. Cross-sectional area is given by πr^2, where r is the radius of the wire.
Let's calculate the cross-sectional area for each option:
- A: π(0.5mm)^2
- B: π(1mm)^2
- C: π(1.5mm)^2
- D: π(0.25mm)^2
Comparing the given options, we see that Wire D has both the smallest cross-sectional area and a reasonable length. Therefore, Wire D will experience the maximum increase in length when the same tension is applied.