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MCQEasyJEE 2023Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2023 Question with Solution

A 100 m100 \, \text{m} long wire having cross-sectional area 6.25×10−4 m26.25 \times 10^{-4} \, \text{m}^2 and Young's modulus is 1010 N m−210^{10} \, \text{N m}^{-2} is subjected to a load of 250 N250 \, \text{N}, then the elongation in the wire will be:

  • A

    6.25×10−3 m6.25 \times 10^{-3} \, \text{m}

  • B

    4×10−4 m4 \times 10^{-4} \, \text{m}

  • C

    6.25×10−6 m6.25 \times 10^{-6} \, \text{m}

  • D

    4×10−3 m4 \times 10^{-3} \, \text{m}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Force on the wire is F=250 NF = 250 \, \text{N}, length is l=100 ml = 100 \, \text{m}, cross-sectional area is A=6.25×10−4 m2A = 6.25 \times 10^{-4} \, \text{m}^2, and Young's modulus is Y=1010 N m−2Y = 10^{10} \, \text{N m}^{-2}.

Find: The elongation δ\delta of the wire.

Use the relation

δ=FlAY\delta = \frac{Fl}{AY}

Substituting the given values,

δ=250×1006.25×10−4×1010\delta = \frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}}

Simplifying,

δ=4×10−3 m\delta = 4 \times 10^{-3} \, \text{m}

Therefore, the elongation in the wire is 4×10−3 m4 \times 10^{-3} \, \text{m}. The solution states the correct option is B, but this value matches option D in the listed options.

Common mistakes

  • Using an incorrect formula for strain or elongation. Young's modulus relates stress to strain, so the correct expression is δ=FlAY\delta = \frac{Fl}{AY}. Do not omit the length or area term.

  • Mishandling powers of 1010 while substituting 6.25×10−46.25 \times 10^{-4} and 101010^{10}. This leads to the wrong order of magnitude. Simplify the exponent terms carefully before evaluating.

  • Choosing the option label from the solution without checking the computed value. Here the working gives 4×10−3 m4 \times 10^{-3} \, \text{m}, which matches option D, not option B.

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