BCECE Medical BCECE Medical Solved Papers-2011

  • question_answer
    A rubber cord catapult has cross-sectional area 25 mm2 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\[{{Y}_{ruber}}=5\times {{10}^{8}}N{{m}^{-2}}\], 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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