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NEET PHYSICSThermal Properties of MatterMedium

Question

A metallic bar of Young's modulus 0.5×1011 N m20.5 \times 10^{11}\text{ N m}^{-2}, coefficient of linear thermal expansion 105 C110^{-5}\ ^\circ\text{C}^{-1}, length 1 m1\text{ m} and cross-sectional area 103 m210^{-3}\text{ m}^2 is heated from 0C0^\circ\text{C} to 100C100^\circ\text{C} without expansion or bending. The compressive force developed in the metallic bar is:

A

50×103 N50 \times 10^3\text{ N}

B

100×103 N100 \times 10^3\text{ N}

C

2×103 N2 \times 10^3\text{ N}

D

5×103 N5 \times 10^3\text{ N}

Step-by-Step Solution

When the metallic bar is heated and its expansion is prevented, thermal stress is developed in it. The thermal strain is given by ϵ=αΔT\epsilon = \alpha \Delta T. According to Hooke's law, Young's modulus Y=StressStrain=F/AαΔTY = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\alpha \Delta T}. Therefore, the compressive force developed is F=YAαΔTF = Y A \alpha \Delta T. Given values: Y=0.5×1011 N m2Y = 0.5 \times 10^{11}\text{ N m}^{-2} A=103 m2A = 10^{-3}\text{ m}^2 α=105 C1\alpha = 10^{-5}\ ^\circ\text{C}^{-1} ΔT=100C0C=100C\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C} Substituting the values into the formula: F=(0.5×1011)×(103)×(105)×(100)F = (0.5 \times 10^{11}) \times (10^{-3}) \times (10^{-5}) \times (100) F=0.5×10(1135+2)F = 0.5 \times 10^{(11 - 3 - 5 + 2)} F=0.5×105 NF = 0.5 \times 10^5\text{ N} F=50×103 NF = 50 \times 10^3\text{ N}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermal Properties of Mattermetallicyoungsmoduluscoefficientlinear

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