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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Physics
Chapter
Mechanical Properties of Solids

BITSAT 2023 · Paper 1 · Physics · Mechanical Properties of Solids

A 100 m long wire having cross-sectional area 6.25 × 10^ -4 m^2 and...

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

Options

  1. A6.25×10−3m6.25 \times 10^{-3} \mathrm{ m}
  2. B4×10−4m4 \times 10^{-4} \mathrm{ m}
  3. C6.25×10−6m6.25 \times 10^{-6} \mathrm{ m}
  4. D4×10−3m4 \times 10^{-3} \mathrm{ m}

Correct answer

D. 4×10−3m4 \times 10^{-3} \mathrm{ m}

Step-by-step solution

Δℓ=FℓYA=250×1001010×6.25×10−4=40×10−4m\Delta \ell=\frac{F \ell}{Y A}=\frac{250 \times 100}{10^{10} \times 6.25 \times 10^{-4}}=40 \times 10^{-4} \mathrm{ m}
=4×10−3m=4 \times 10^{-3} \mathrm{ m}
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