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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    A thin steel ring of inner diameter 40 cm and cross- sectional area \[1 m{{m}^{2}}\], is heated until it easily slides on a rigid cylinder of diameter 40.05 cm. [For steel,\[\alpha =1{{0}^{-5}}/{}^\circ C, Y= 200 GPa\]] When the ring cools down, the tension in the ring will be:               

    A) 1000 N

    B) 500 N

    C) 250 N

    D) 100 N

    Correct Answer: C

    Solution :

    [c] \[\frac{\Delta \ell }{\ell }=\frac{\pi \times 0.5}{\pi \times 40}=\frac{1}{800}\]  \[T=Y\frac{\Delta \ell }{\ell }\times A\] \[=200\times {{10}^{9}}\times \frac{1}{800}\times 1\times {{10}^{-6}}=250N\]


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