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JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)

JEE Main 2023 Physics question on Properties of Matter

A 100 m long wire having cross-sectional area 6.25×10−46.25 \times 10^{-4} m2^2 and Young's modulus is 101010^{10} Nm−2^{-2} is subjected to a load of 250 N, then the elongation in the wire will be:
  1. A.
    6.25×10−66.25 \times 10^{-6} m
  2. B.
    6.25×10−36.25 \times 10^{-3} m
  3. C.
    4×10−34 \times 10^{-3} m
  4. D.
    4×10−44 \times 10^{-4} m
Show answer and solution

Correct option: C

Solution

Using the formula ΔL = FL/(AY):

ΔL = (250 × 100) / (6.25 × 10⁻⁴ × 10¹⁰)

ΔL = 25000 / (6.25 × 10⁶) = 4 × 10⁻³ m

The elongation is 4 × 10⁻³ m, which matches option 3.

[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]

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