Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force FA and FB respectively they get equal increase in their lengths. Then the ratio FA/FB should be

The correct answer is 8:1 (or Option 8:1).

Let us break down the solution step-by-step using the concepts of elasticity and Young's modulus.

1. Understanding the Relationship (Young's Modulus Formula):
Young's modulus ($Y$) of a material is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{F\cdot L}{A\cdot \Delta L}$
where:
- $F$ is the stretching force applied,
- $L$ is the original length of the wire,
- $A$ is the cross-sectional area of the wire,
- $\Delta L$ is the increase in length (elongation).

Since the cross-sectional area of a wire with diameter $d$ is given by $A=\frac{\pi }{4}{d}^2$, we can rewrite the formula for Young's modulus as:
$Y=\frac{4FL}{\pi d^2\Delta L}$

2. Expressing Force:
Rearranging the formula to express the stretching force $F$, we get:
$F=\frac{Y\cdot \pi d^2\cdot \Delta L}{4L}$

3. Given Information:
- Both wires A and B are made of the same material, which means they have the same Young's modulus:
$Y_A=Y_B=Y$
- They experience the same increase in length:
$\Delta L_A=\Delta L_B$
- The ratio of their lengths is:
$\frac{L_A}{L_B}=\frac{1}{2}$
- The ratio of their diameters is:
$\frac{d_A}{d_B}=\frac{2}{1}$

4. Finding the Ratio of Forces:
Since $Y$, $\pi$, and $\Delta L$ are constants for both wires, the force is proportional to $\frac{d^2}{L}$:
$F\propto \frac{d^2}{L}$

Thus, the ratio of the forces is:
$\frac{F_A}{F_B}={\left(\frac{d_A}{d_B}\right)}^2\cdot \left(\frac{L_B}{L_A}\right)$

Substitute the given ratios into this equation:
$\frac{F_A}{F_B}={\left(\frac{2}{1}\right)}^2\cdot \left(\frac{2}{1}\right)$
$\frac{F_A}{F_B}=4\cdot 2=8$

Therefore, the ratio of the stretching forces $F_A/F_B$ is 8:1.