Both wires are made up of same material so they will have same density(d) and Young's modulus(Y)

Now,
Y=F×l/∆l×A=F×l²÷[∆l(m/d)]
∆l is directly proportional to l²/m

So, putting the values their ratio is 1:2

Explanation:

Given Data:
- Length of first wire (l₁) = 3m
- Mass of first wire (m₁) = 18gm
- Length of second wire (l₂) = 4m
- Mass of second wire (m₂) = 16gm

Key Concepts:
- The elongation of a wire is directly proportional to the force applied and inversely proportional to the cross-sectional area and modulus of elasticity of the material.
- The formula for elongation is:
\[ \frac{\text{Elongation of wire}}{\text{Original length of wire}} = \frac{F}{A \cdot Y} \]
where F is the force applied, A is the cross-sectional area, and Y is the Young's modulus of elasticity.

Calculations:
- Let's assume the force applied to both wires is the same, denoted as F.
- The elongation of the first wire (Δl₁) is given by:
\[ \frac{\Delta l₁}{l₁} = \frac{F}{A₁ \cdot Y} \]
- The elongation of the second wire (Δl₂) is given by:
\[ \frac{\Delta l₂}{l₂} = \frac{F}{A₂ \cdot Y} \]

Ratio of Elongations:
- The ratio of elongations of the two wires can be found by dividing the two equations above:
\[ \frac{\Delta l₁/\text{l₁}}{\Delta l₂/\text{l₂}} = \frac{F/(A₁ \cdot Y)}{F/(A₂ \cdot Y)} \]
\[ \frac{\Delta l₁/\text{l₁}}{\Delta l₂/\text{l₂}} = \frac{A₂}{A₁} \]

Given Data:
- Mass of a wire is proportional to its cross-sectional area, so we have:
\[ \frac{m₁}{m₂} = \frac{A₁}{A₂} \]
\[ \frac{18}{16} = \frac{A₁}{A₂} \]
\[ A₁ = \frac{9}{8}A₂ \]

Final Calculation:
- Substituting the value of A₁ in the ratio of elongations equation:
\[ \frac{\Delta l₁/\text{l₁}}{\Delta l₂/\text{l₂}} = \frac{9}{8} \]
\[ \frac{\Delta l₁}{3} : \frac{\Delta l₂}{4} = \frac{9}{8} \]
\[ \frac{\Delta l₁}{3} = \frac{9}{17} \]
\[ \frac{\Delta l₁}{\Delta l₂} = \frac{3}{4} \]
Hence, the ratio of elongation of the two wires is 1:2.