The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied

Correct Answer: Length 50 cm, diameter 0.5 mm

To determine which wire will undergo the maximum increase in length under the same tension, we can use the relationship defined by Young's modulus of elasticity.

Young's modulus (Y) is given by the formula:
$Y=\frac{FL}{A\Delta L}$
where:
- $F$ is the tension (applied force)
- $L$ is the initial length of the wire
- $A$ is the cross-sectional area of the wire
- $\Delta L$ is the increase in length

Rearranging the formula to solve for the increase in length ($\Delta L$):
$\Delta L=\frac{FL}{AY}$

Since the wires are made of the same material, the Young's modulus (Y) is the same for all of them. Since the same tension (F) is applied, the force F is also constant.
Thus, the increase in length is directly proportional to the length and inversely proportional to the cross-sectional area:
$\Delta L\propto \frac{L}{A}$

The cross-sectional area of a wire with diameter d is $A=\frac{\pi d^2}{4}$. Therefore, the area is proportional to the square of the diameter:
$A\propto d^2$

Substituting this back, we get:
$\Delta L\propto \frac{L}{d^2}$

Now, we calculate the ratio $\frac{L}{d^2}$ for each wire to find which one has the maximum value:

1. For wire 1: Length $L=100\text{cm}$, diameter $d=1\text{mm}$
$\frac{L}{d^2}=\frac{100}{1^2}=100$

2. For wire 2: Length $L=200\text{cm}$, diameter $d=2\text{mm}$
$\frac{L}{d^2}=\frac{200}{2^2}=\frac{200}{4}=50$

3. For wire 3: Length $L=300\text{cm}$, diameter $d=3\text{mm}$
$\frac{L}{d^2}=\frac{300}{3^2}=\frac{300}{9}\approx 33.3$

4. For wire 4: Length $L=50\text{cm}$, diameter $d=0.5\text{mm}$
$\frac{L}{d^2}=\frac{50}{{0.5}^2}=\frac{50}{0.25}=200$

Comparing these values, the ratio $\frac{L}{d^2}$ is maximum for the fourth wire (value of 200). Therefore, the wire with length 50 cm and diameter 0.5 mm will undergo the maximum increase in length.