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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A lead bullet strikes against a steel plate with a velocity\[200\text{ }m{{s}^{-1}}\]. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, then the rise in temperature of the bullet is (Specific heat capacity of lead\[=125\text{ }J\text{ }k{{g}^{-1}}{{K}^{-1}}\])

    A)  \[{{80}^{o}}C\]                

    B)         \[{{60}^{o}}C\]                

    C)  \[{{160}^{o}}C\]              

    D)         \[{{40}^{o}}C\]

    E)  \[{{120}^{o}}C\]

    Correct Answer: A

    Solution :

    Hence \[=\frac{1}{2}\times \frac{1}{2}m{{v}^{2}}=\frac{1}{4}m\times 4\times {{10}^{4}}\]                                                 \[=125\times \Delta T\times m\] \[\Delta T=4\times \frac{{{10}^{4}}}{500}={{80}^{o}}C\]


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