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Haryana PMT Haryana PMT Solved Paper-2010

  • question_answer
    A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 J-kg \[^{-10}{{C}^{-1}},g=10m{{s}^{-2}}\]

    A)  \[{{0.01}^{0}}C\]

    B)         \[{{0.1}^{0}}C\]

    C)  \[{{1}^{0}}C\]                                   

    D)  \[{{1.1}^{0}}C\]

    Correct Answer: B

    Solution :

                     According to energy conservation, change in potential energy of the ball, appears in the form of heat which raises the temperature of the ball. ie,           \[mg({{h}_{1}}-{{h}_{2}})=mc.\,\,\Delta \theta \] \[\Rightarrow \]               \[\Delta \theta =\frac{g({{h}_{1}}-{{h}_{2}})}{c}\]                 \[=\frac{10(10-5.4)}{460}={{0.1}^{o}}C\]


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