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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Topic Test - Thermometry, Calorimetry & Thermal Expansion

  • question_answer
    When a building is constructed at \[-10{}^\circ C\], a steel beam (cross-sectional area \[45\text{ }c{{m}^{2}}\]) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force in the beam when the temperature is \[25{}^\circ C\]? For this kind of    steel, \[\alpha =1.1\times {{10}^{-5}}\,{{\,}^{o}}{{C}^{-1}}\] and \[Y=2.0\times {{10}^{11}}N/{{m}^{2}}\].

    A) \[2.5\times {{10}^{3}}N\]                    

    B) \[3.5\times {{10}^{5}}N\]

    C) \[2.5\times {{10}^{5}}N\]                    

    D) \[3.5\times {{10}^{3}}N\]

    Correct Answer: B

    Solution :

    \[\frac{\Delta L}{{{L}_{0}}}=\alpha \,\Delta T=(1.1\times {{10}^{-5}}{{\,}^{o}}{{C}^{-1}})\,({{35}^{o}}C)\]
    \[[\because \,\,\Delta L={{L}_{0}}\,\alpha \,\Delta T]\] where, \[\Delta L=\]Change in length and \[{{L}_{0}}=\] original length \[=3.85\times {{10}^{-4}}\]
    So, \[F=YA\frac{\Delta L}{{{L}_{0}}}\]
    \[\left[ \because \,\,\text{Young }\!\!'\!\!\text{ s}\,\text{modulus},\,Y=\frac{F/A}{\Delta L/{{L}_{0}}} \right]\]
    \[=(2.0\times {{10}^{11}}\,N/{{m}^{2}})\] \[(45\times {{10}^{-4}}{{m}^{2}})\,(3.85\times {{10}^{-4}})\]
    \[=3.5\times {{10}^{5}}N\]


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