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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

An iron rod of length 2 ~m and cross-sectional area of 50 ~mm^ 2...

An iron rod of length 2 m2 \mathrm{~m} and cross-sectional area of 50 mm250 \mathrm{~mm}^{2} stretched by 0.5 mm0.5 \mathrm{~mm}, when a mass of 250 kg250 \mathrm{~kg} is hung from its lower end. Young's modulus of iron rod is

Options

  1. A19.6×1020 N/\mathrmm219.6 \times 10^{20} \mathrm{~N} / \mathrmm^{2}
  2. B19.6×1018 N/\mathrmm219.6 \times 10^{18} \mathrm{~N} / \mathrmm^{2}
  3. C19.6×1010 N/\mathrmm219.6 \times 10^{10} \mathrm{~N} / \mathrmm^{2}
  4. D19.6×1015 N/\mathrmm219.6 \times 10^{15} \mathrm{~N} / \mathrmm^{2}

Correct answer

C. 19.6×1010 N/\mathrmm219.6 \times 10^{10} \mathrm{~N} / \mathrmm^{2}

Step-by-step solution

Y=F/AΔℓ/ℓ=250×9.850×10−60.5×10−32=250×9.850×10−6×20.5×10−3⇒19.6×1010 N/\mathrmm2Y=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrmm^{2}
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