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EAMCET Medical EAMCET Medical Solved Paper-2010

  • question_answer
    A lead bullet of unknown mass is fired with a speed of 180 ms-1 into a tree in which it stops. Assuming that in this process two-third of heat produced goes into the bullet and one- third into wood. The temperature of the bullet rises by (Specific heat of lead\[=\text{ }0.120\text{ }J{{g}^{-1}}{{C}^{-1}}\])

    A)                 \[140{}^\circ C\]                              

    B)                 \[106{}^\circ C\]

    C)                 \[90{}^\circ C\]                                 

    D)                 \[100{}^\circ C\]

    Correct Answer: C

    Solution :

                     Since specific heat of lead is given in joule. Specific heat of lead \[=0.120\,J/g{{\,}^{o}}C=120\,J/kg\]                 The two-third of head produced goes into the bullet                 So,          \[m\times s\times \Delta \theta =\frac{2}{3}\times \frac{1}{2}m{{v}^{2}}\]                 \[\Delta \theta =\frac{{{v}^{2}}}{3\times s}\]                 \[=\frac{180\times 180}{3\times 120}=90{{\,}^{o}}C\]


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