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

  • question_answer
    A pendulum clock loses 12 s a day if the temperature is \[40{}^\circ C\]and gains 4 s a day if the temperature is\[20{}^\circ C\]. The temperature at which the clock will show correct time, and the co- efficient of linear expansion \[(\alpha )\] of the metal of the pendulum shaft are respectively:

    A) \[30{}^\circ C;\,\alpha =1.85\times {{10}^{-3}}/{}^\circ C\]  

    B) \[55{}^\circ C;\,\alpha =1.85\times {{10}^{-2}}/{}^\circ C\]

    C) \[25{}^\circ C;\,\alpha =1.85\times {{10}^{-5}}/{}^\circ C\]

    D) \[60{}^\circ C;\,\alpha =1.85\times {{10}^{-4}}/{}^\circ C\]

    Correct Answer: C

    Solution :

    Time lost / gained per day \[=\frac{1}{2}\alpha \Delta \theta \times 86400\] second \[12=-\alpha \left( 40-8 \right)\times 86400\]           ....(i) \[4=-\alpha \left( \theta -20 \right)\times 86400\]         ....(ii) On dividing we get, \[3=\frac{40-\theta }{\theta -20}\] \[36-60=40-\theta \] \[4\theta =100\Rightarrow \theta =25{}^\circ C\]


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