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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    A lead bullet of unknown mass is fired-with a speed of \[180\,\,m{{s}^{-1}}\] 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

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

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

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

    D)  \[{{100}^{o}}C\]

    Correct Answer: C

    Solution :

    Specific heat of lead\[=0.120\,\,J/{{g}^{o}}C\]                 \[=120\,\,J/kg\] The two-third of heat 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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