BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    A rubber cord catapult has cross-sectional area \[25\,m{{m}^{2}}\] and initial length of rubber cord is 10 cm. It is stretched to 5 cm and then released to project a missile of mass 5 g. Taking \[{{Y}_{rubber}}\,=5\times {{10}^{8}}N{{m}^{-2}},\] velocity of projected missile is

    A)  \[20\,m{{s}^{-1}}\] 

    B)                                         \[100\,m{{s}^{-1}}\]

    C)  \[250\,m{{s}^{-1}}\]    

    D)         \[200\,m{{s}^{-1}}\]

    Correct Answer: C

    Solution :

                    Potential energy stored in the rubber cord catapult will be converted into kinetic energy  of mass. \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\frac{YA{{l}^{2}}}{L}\]                 \[\Rightarrow \]               \[v=\sqrt{\frac{YA{{l}^{2}}}{mL}}\]                                 \[=\sqrt{\frac{5\times {{10}^{8}}\times 25\times {{10}^{-6}}\times {{(5\times {{10}^{-2}})}^{2}}}{5\times {{10}^{-3}}\times 10\times {{10}^{-2}}}}\] \[=250\,m{{s}^{-1}}\]


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