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JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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A steel bar 0.5 cm square [cross-sectional area] and 25 cm long at 30∘C30^\circ C fits into the space between two fixed supports. If the bar is now heated to 230∘C230^\circ C, the force it exerts against the supports is [αsteel=11×10−6/∘C  and  Ysteel=12×1011Nm2]\left[ {{\alpha _{\mathrm{steel}}} = 11 \times {{10}^{ - 6}}/^\circ C\,\,\mathrm{and}\,\,{Y_{\mathrm{steel}}} = 12 \times {{10}^{11}}\frac{N}{{{m^2}}}} \right]
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Correct option: B
22000 N
Option B (22000 N) is correct because the force exerted against the supports is calculated using the formula F=YAαΔtF = YA\alpha \Delta t, where Y is Young's modulus, A is the cross-sectional area, α\alpha is the coefficient of linear expansion, and Δt\Delta t is the change in temperature.

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