Question

Two wires of same material have lengths in the ratio $1:2$ and diameters in the ratio $2:1$. If they are stretched by the same load force, the ratio of their extensions ($\Delta l_1 : \Delta l_2$) is:

• A. 1 : 4
• B. 1 : 8
• C. 1 : 2
• D. 4 : 1 Correct Answer: (B) 1 : 8

Hint:

To solve scaling questions instantly on physics exams, write out a clean proportional statement: $\mathbf{\Delta l \propto \frac{l}{d^2}}$. Halving the length drops the extension by a factor of $2$, and doubling the diameter drops it by another factor of $2^2 = 4$. Combining these independent drops together gives a total decrease of $2 \times 4 = \mathbf{8}$, directly pointing you to $1:8$!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When a solid metal wire is subjected to a longitudinal stretching force, it undergoes mechanical deformation, expanding its total length by a small amount called extension ($\Delta l$). According to Hooke's Law within the elastic limit, the ratio of tensile stress to tensile strain remains a constant value characteristic of the material itself. This material constant is known as Young's Modulus ($Y$). Because both wires in this problem are composed of the exact same material, their Young's Modulus values are perfectly identical ($Y_1 = Y_2 = Y$).

Step 2: Key Formula or Approach:
The definition of Young's Modulus is given by: $$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta l}{l}\right)} = \frac{F \cdot l}{A \cdot \Delta l} $$ Rearranging this formula to solve explicitly for the extension ($\Delta l$) yields: $$ \Delta l = \frac{F \cdot l}{A \cdot Y} $$ Since wires possess a circular cross-section, the cross-sectional area $A$ can be written in terms of its diameter $d$ as $A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$. Substituting this back into our extension relation gives: $$ \Delta l = \frac{4 F \cdot l}{\pi d^2 \cdot Y} $$ The problem states that both wires are stretched by the same load force ($F_1 = F_2 = F$). Since $F$, $Y$, and $\pi$ are constants across both scenarios, the extension is directly proportional to length and inversely proportional to the square of the diameter: $$ \Delta l \propto \frac{l}{d^2} $$

Step 3: Detailed Explanation:
Let's express the ratio of the extensions of the two wires using our proportionality relation: $$ \frac{\Delta l_1}{\Delta l_2} = \left(\frac{l_1}{l_2}\right) \times \left(\frac{d_2}{d_1}\right)^2 $$ From the text, we extract the following given ratios: - Length ratio: $\frac{l_1}{l_2} = \frac{1}{2}$ - Diameter ratio: $\frac{d_1}{d_2} = \frac{2}{1} \implies \frac{d_2}{d_1} = \frac{1}{2}$ Let's plug these numerical fractions directly into our ratio equation: $$ \frac{\Delta l_1}{\Delta l_2} = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right)^2 $$ $$ \frac{\Delta l_1}{\Delta l_2} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} $$ Thus, the calculated ratio of their extensions ($\Delta l_1 : \Delta l_2$) is $1 : 8$, which matches option (B).

Step 4: Final Answer:
The ratio of their extensions ($\Delta l_1 : \Delta l_2$) is $1 : 8$.