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NEET 2024
Exam
NEET
Year
2024
Subject
Physics
Chapter
Mechanical Properties of Solids

NEET 2024 · Physics · Mechanical Properties of Solids

A metallic bar of Young's modulus, 0.5 × 10^ 11 ~N ~m^ -2 and...

A metallic bar of Young's modulus, 0.5×1011 N m−20.5 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} and coefficient of linear thermal expansion 10−5∘C−110^{-5}{ }^{\circ} \mathrm{C}^{-1}, length 1 m and area of cross-section 10−3 m210^{-3} \mathrm{~m}^2 is heated from 0∘C0^{\circ} \mathrm{C} to 100∘C100^{\circ} \mathrm{C} without expansion or bending. The compressive force developed in it is :

Options

  1. A50×103 N50 \times 10^3 \mathrm{~N}
  2. B100×103 N100 \times 10^3 \mathrm{~N}
  3. C2×103 N2 \times 10^3 \mathrm{~N}
  4. D5×103 N5 \times 10^3 \mathrm{~N}

Correct answer

A. 50×103 N50 \times 10^3 \mathrm{~N}

Step-by-step solution

 Thermal strain = Longitudinal strain =αΔT⇒ Longitudinal strain, δ=10−5×102=10−3⇒ Compressive stress =δ× Young’s Modulus =10−3×0.5×1011=0.5×108⇒ Compressive force =0.5×108×10−3=0.5×105=5×104×1010=50×103N\begin{aligned} & \text{ Thermal strain }=\text{ Longitudinal strain }=\alpha \Delta T \\ & \Rightarrow \quad \text{ Longitudinal strain, } \delta=10^{-5} \times 10^2=10^{-3} \\ & \Rightarrow \quad \text{ Compressive stress }=\delta \times \text{ Young's Modulus } \\ & =10^{-3} \times 0.5 \times 10^{11} \\ & =0.5 \times 10^8 \\ & \Rightarrow \quad \text{ Compressive force }=0.5 \times 10^8 \times 10^{-3}=0.5 \times 10^5 \\ & \quad=5 \times 10^4 \times \frac{10}{10} \\ & =50 \times 10^3 \mathrm{ N}\end{aligned}
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