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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    A bullet of mass \[10\times {{10}^{-3}}kg\]moving with a speed of \[20\text{ }m{{s}^{-}}1\]hits an ice block (0°C) of 990 g kept at rest on a frictionless floor and gets embedded in it. If ice takes 50% of KE lost by the system, the amount of ice melted (in grams) approximately is: (J = 4.2 J/cal) (Latent heat of ice = 80 cal/g).

    A)  6                                            

    B)  3

    C) \[6\times {{10}^{-3}}\]                                  

    D)  \[3\times {{10}^{-3}}\]

    Correct Answer: D

    Solution :

                     Kinetic energy of bullet \[=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}\times 10\times {{10}^{-3}}\times 20\times 20\] \[=2J=\frac{2}{4.2}\,cal\] Heat gained by ice to melt \[=\frac{2}{4.2}\times \frac{50}{100}\]                 or            \[mL=\frac{1}{4.2}\]                 or            \[m=\frac{1}{4.2\times 80}g\]                                 \[=0.002976\,g\] \[=0.002976\,g\]


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