NEET Physics Thermometry, Calorimetry & Thermal Expansion NEET PYQ-Thermometry, Calorimetry and Thermal Expansion

  • question_answer
    The coefficients of linear expansions of brass and steel are \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\] respectively. When we take a brass rod of length \[{{l}_{1}}\] and a steel rod of length \[{{l}_{2}}\] at \[{{0}^{o}}C,\] then the difference in their lengths \[({{l}_{2}}-{{l}_{1}})\] will remain the same at all temperatures if:               [AIPMT 1999]

    A) \[{{\alpha }_{1}}{{l}_{1}}={{\alpha }_{2}}{{l}_{2}}\]

    B)       \[{{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}\]

    C)  \[\alpha _{1}^{2}{{l}_{2}}=\alpha _{2}^{2}{{l}_{1}}\]

    D)                   \[{{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}\]

    Correct Answer: A

    Solution :

    Linear expansion coefficient
    \[=\frac{\text{change in length}}{\text{original length }\times \text{ rise temperature}}\]
    or         \[\alpha =\frac{\Delta l}{lt}\]
    or         \[\Delta l=l\,\alpha \,t\]
    For brass rod, \[\Delta {{l}_{1}}={{l}_{1}}{{\alpha }_{1}}t\]
    For steel rod, \[\Delta {{l}_{2}}={{l}_{2}}{{\alpha }_{2}}t\]
    Since, \[{{l}_{2}}-{{l}_{1}}=\] constant (give)
    So,       \[\Delta {{l}_{2}}-\Delta {{l}_{1}}=0\]
    or         \[\Delta {{l}_{2}}=\Delta {{l}_{1}}\]
    \[\therefore \]      \[{{l}_{2}}{{\alpha }_{2}}t={{l}_{1}}{{\alpha }_{1}}t\]
    As \[t\ne 0,\] hence
    \[{{l}_{2}}{{\alpha }_{2}}={{l}_{1}}{{\alpha }_{1}}\]

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