Question

The ratio of the lengths of two wires A and B of same material is 1: 2 and the ratio of their diameter is 2:1. They are stretched by the same force, then the ratio of increase in length will be

• A. 1 : 2
• B. 1 : 4
• C. 1 : 8
• D. 1 : 16

Answer 1

Answer: (C)

1 : 8

Answer 2

The extension of a wire is given by the formula:

$$\Delta L = \frac{F L}{A Y}$$

where $F$ is the applied force, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.

Since the wires are of the same material and are stretched by the same force, the extension $\Delta L$ is proportional to:

$$\Delta L \propto \frac{L}{A}$$

Let the length of wire A be $l$ and wire B be $2l$.

Given the ratio of their diameters is $2:1$, assume:

• Diameter of A, $d_A = 2k$
• Diameter of B, $d_B = k$

The cross-sectional areas (for circular cross-section) are:

$$A_A = \frac{\pi (2k)^2}{4} = \pi k^2 \quad \text{(proportional to } 4k^2\text{)}$$ $$A_B = \frac{\pi k^2}{4} \quad \text{(proportional to } k^2\text{)}$$

Now calculate the proportional factors (ignoring common constants like $\pi/4$):

• For wire A:

$$\frac{L_A}{A_A} \propto \frac{l}{4k^2}$$

• For wire B:

$$\frac{L_B}{A_B} \propto \frac{2l}{k^2}$$

Therefore, the ratio of increase in length (extension) of wire A to wire B is:

$$\frac{(\Delta L)_A}{(\Delta L)_B} = \frac{\frac{l}{4k^2}}{\frac{2l}{k^2}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$