BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    The escape velocity of a projectile on the earth's surface in\[11.2\text{ }km{{s}^{-1}}\]. A body is projected out with thrice this speed. The speed of the body far away form the earth will be:

    A)  \[22.4\text{ }km{{s}^{-1}}\]       

    B)  \[31.7\text{ }km\text{ }{{s}^{-1}}\]

    C)  \[33.6km{{s}^{-1}}\]     

    D)  none of these

    Correct Answer: B

    Solution :

                     Key Idea: Conservation of energy holds in the universe. By law of conservation of energy, energy at surface of earth = energy at infinity i.e.,        \[{{(U+K)}_{surface}}={{(U+K)}_{\inf inity}}\] or            \[-\frac{GMm}{R}+\frac{1}{2}m{{(3{{v}_{e}})}^{2}}=0+\frac{1}{2}m{{v}^{2}}\] or            \[-\frac{GM}{R}+\frac{9v_{e}^{2}}{2}=\frac{1}{2}{{v}^{2}}\] But escape velocity, \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\] \[\therefore \]  \[-\frac{v_{e}^{2}}{2}+\frac{9v_{e}^{2}}{2}=\frac{1}{2}{{v}^{2}}\] Or           \[{{v}^{2}}=8v_{e}^{2}\] Or           \[v=2\sqrt{2}\times 11.2=31.7\,km{{s}^{-1}}\]


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