Two wires A and B are made of same material having ratio of lengths LA/LB = 1/3 and their diameters ratio dA/dB = 2. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q32 – JEE-MAIN 2025

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Home Physics Mechanical Properties of SolidsQuestion #296

Two wires A and B are made of same material having ratio of lengths

\frac{L_{A}}{L_{B}} = \frac{1}{3}

and their diameters ratio

\frac{d_{A}}{d_{B}} = 2

. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

Clue 1 of 3: Hooke's Law and Elasticity

Understand the relationship between stress, strain, Young's modulus, force, length, area, and elongation for a wire.

🥷

Ninja Strategy•Combined Proportionality

Recognize that elongation is directly proportional to length and inversely proportional to the square of the diameter, then correctly combine these ratios to quickly find the answer.

🧠 Full Solution Path

Step 1: Identify the relevant formula for elongation✦ Active

The elongation $Δ L$ of a wire under a force $F$ is derived from Young's modulus formula $Y = \frac{Stress}{Strain} = \frac{F / A}{Δ L / L}$ . Rearranging for $Δ L$ , we get $Δ L = \frac{F L}{A Y}$ . The cross-sectional area $A$ of a wire is $A = \frac{π d^{2}}{4}$ , where $d$ is the diameter. Substituting $A$ into the elongation formula:

Δ L = \frac{F L}{(\frac{π d^{2}}{4}) Y} = \frac{4 F L}{π d^{2} Y}

Step 2: Set up the ratio of elongations○ Expand

Since both wires are made of the same material ( $Y_{A} = Y_{B} = Y$ ) and stretched by the same force ( $F_{A} = F_{B} = F$ ), the constants $4, F, π, Y$ cancel out when taking the ratio of elongations for wire A and wire B:

\frac{Δ L_{A}}{Δ L_{B}} = \frac{\frac{4 F L_{A}}{π d_{A}^{2} Y}}{\frac{4 F L_{B}}{π d_{B}^{2} Y}} = \frac{L_{A}}{L_{B}} \cdot \frac{d_{B}^{2}}{d_{A}^{2}} = \frac{L_{A}}{L_{B}} \cdot {(\frac{d_{B}}{d_{A}})}^{2}

💡 Teacher's Secret Hint

Ensure to correctly handle the inverse square relationship with diameter.

Step 3: Substitute the given ratios and calculate○ Expand

We are given the ratio of lengths $\frac{L_{A}}{L_{B}} = \frac{1}{3}$ and the ratio of diameters $\frac{d_{A}}{d_{B}} = 2$ . From the diameter ratio, we can find $\frac{d_{B}}{d_{A}} = \frac{1}{2}$ . Substituting these values into the ratio of elongations:

\frac{Δ L_{A}}{Δ L_{B}} = \frac{1}{3} \cdot {(\frac{1}{2})}^{2} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}

Thus, the ratio of their respective elongations is $1 : 12$ .

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 32 – JEE-MAIN 2025

Clue 1 of 3: Hooke's Law and Elasticity

🧠 Full Solution Path