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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Critical Thinking

  • question_answer
    A piece of metal weight 46 gm in air, when it is immersed in the liquid of specific gravity 1.24 at 27ºC it weighs 30 gm. When the temperature of liquid is raised to 42ºC the metal piece weight 30.5 gm, specific gravity of the liquid at 42ºC is 1.20, then the linear expansion of the metal will be [BHU 1995]

    A)            3.316 × 10?5/ºC                    

    B)            2.316 × 10?5/ºC

    C)                       4.316 × 10?5/ºC                    

    D)            None of these

    Correct Answer: B

    Solution :

                       Loss of weight at 27ºC is = 46 ? 30 = 16 = V1 × 1.24 rl × g ?(i) Loss of weight at 42ºC is = 46 ? 30.5 = 15.5 = V2 × 1.2 r l × g    ?(ii) Now dividing (i) by (ii), we get \[\frac{16}{15.5}\] = \[\frac{{{V}_{1}}}{{{V}_{2}}}\times \frac{1.24}{1.2}\] But \[\frac{{{V}_{2}}}{{{V}_{1}}}\] = 1 + 3a (t2 ? t1) = \[\frac{15.5\times 1.24}{16\times 1.2}\] = 1.001042 Þ 3a (42º ?  27º) = 0.001042 Þ a = 2.316 × 10?5/ºC.


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