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EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    A lead bullet, of initial temperature 27°C and speed v km/h penetrates into a solid object and melts. If 50% of the kinetic energy is used to heat it, the value of v in km/h is (for lead melting point = 600 K, latent heat of fusion \[=2.5\times 104\text{ }J\text{ }k{{g}^{-1}};\] specific heat\[=125J\text{ }k{{g}^{-1}}{{K}^{-1}}\])

    A)  3600                                     

    B)  1800

    C)  1200                                     

    D)  1000

    Correct Answer: B

    Solution :

                     Heat energy required to just melt the bullet \[Q={{Q}_{1}}+{{Q}_{2}}\] Here, \[{{Q}_{1}}=ms\Delta \theta \]                 \[=m\times 125\times (327-27)\]                 \[=3.75\times {{10}^{4}}m\] \[{{Q}_{2}}=mL=m\times 2.5\times {{10}^{4}}\] \[=2.5\times {{10}^{4}}m\]                 \[\therefore \]  \[Q=6.25\times {{10}^{4}}\,m\] If v be the speed of bullet, then 50% of\[\frac{1}{2}m{{v}^{2}}\] should be equal to Q. Thus, \[0.5\times \frac{1}{2}m{{v}^{2}}=6.25\times {{10}^{4}}m\] \[\Rightarrow \]               \[v=\sqrt{\frac{6.25\times {{10}^{4}}\times 2}{0.5}}\] \[=500\,m/s=\,1800\,km/h\]


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