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JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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A brass rod and lead rod each 80 cm80 \mathrm{~cm} long at 0∘C0^{\circ} \mathrm{C} are clamped together at one end with their free ends coinciding. If the system is placed in a steam bath then the separation of free ends of the rods is:

(αBrass=18×10−6/∘C and αLead=28×10−6/∘C)\left(\alpha_{\text {Brass}}=18 \times 10^{-6} /{ }^{\circ} \mathrm{C} \text { and } \alpha_{\text {Lead}}=28 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)
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Correct option: B
0.8 mm0.8 \mathrm{~mm}
Option B (0.8 mm) is correct because it is the separation of the free ends of the rods calculated using the difference in their thermal expansion.

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