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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2023 · 24 January · Morning Shift · Physics · Mechanical Properties of Solids

A 100m long wire having cross-sectional area 6 . 25 × 10 ^ - 4 m ^...

A 100m100m long wire having cross-sectional area 6.25×10−4m26 . 25 \times {10}^{- 4}{m}^{2} and Young's modulus is 1010Nm−2{10}^{10}N{m}^{- 2} is subjected to a load of 250N250N, then the elongation in the wire will be :

Options

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

Correct answer

D. 4×10−3m4 \times {10}^{- 3}m

Step-by-step solution

As we know, elongation in wire is given by,

∆l=FlAY∆ l = \dfrac{Fl}{AY}

⇒∆l=250×1006.25×10−4×1010⇒ ∆ l = \dfrac{250 \times 100}{6 . 25 \times {10}^{- 4} \times {10}^{10}}

⇒∆l=4×10−3m⇒ ∆ l = 4 \times {10}^{- 3}m

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