The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11. When the joined wire is kept under certain tension the elongations in the wires A and B are equal. If the length of wire A is 2.2 m, then the length of wire B is _______ m.

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q34 – JEE-MAIN 2026

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Home Physics Mechanical Properties of SolidsQuestion #292

The two wires

A

and

B

of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire

A

and wire

B

20 / 11

. When the joined wire is kept under certain tension the elongations in the wires

A

and

B

are equal. If the length of wire

A

2.2 m

, then the length of wire

B

is _______ m.

Clue 1 of 3: Understand the setup

The wires are joined together, implying they experience the same tension, and the problem states they have equal cross-sections and equal elongations.

🥷

Ninja Strategy•Relative Length Estimation

By understanding the inverse relationship between length and Young's modulus when elongation, force, and area are constant, one can deduce that $L_{B}$ must be less than $L_{A}$ because $Y_{B} < Y_{A}$ (since $Y_{B} / Y_{A} = 11 / 20$ ). This eliminates options greater than $L_{A} = 2.2 m$ .

🧠 Full Solution Path

Step 1: Relate elongation to Young's Modulus✦ Active

The elongation $Δ L$ of a wire under tension $F$ , with length $L$ , cross-sectional area $A$ , and Young's modulus $Y$ is given by:

Δ L = \frac{F L}{A Y}

Step 2: Apply conditions for both wires○ Expand

Given that the wires have equal cross-section ( $A_{A} = A_{B} = A$ ), are under the same tension ( $F_{A} = F_{B} = F$ ), and have equal elongations ( $Δ L_{A} = Δ L_{B}$ ), we can equate their elongation expressions:

\frac{F L_{A}}{A Y_{A}} = \frac{F L_{B}}{A Y_{B}}

This simplifies to:

\frac{L_{A}}{Y_{A}} = \frac{L_{B}}{Y_{B}}

💡 Teacher's Secret Hint

Remember that when wires are joined in series and under tension, the tension is the same throughout.

Step 3: Calculate the length of wire B○ Expand

Rearranging the simplified equation, we get $L_{B} = L_{A} (\frac{Y_{B}}{Y_{A}})$ . We are given $L_{A} = 2.2 m$ and the ratio $\frac{Y_{A}}{Y_{B}} = \frac{20}{11}$ , which implies $\frac{Y_{B}}{Y_{A}} = \frac{11}{20}$ . Substituting these values:

L_{B} = 2.2 \times \frac{11}{20} = \frac{24.2}{20} = 1.21 m

💡 Teacher's Secret Hint

Pay attention to the ratio given ( $Y_{A} / Y_{B}$ ) and ensure you use its reciprocal ( $Y_{B} / Y_{A}$ ) when calculating $L_{B}$ .

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

Physics Question 34 – JEE-MAIN 2026

Clue 1 of 3: Understand the setup

🧠 Full Solution Path