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J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    The escape velocity of a body from the earth is \[{{v}_{e}}\]. If the radius of earth contracts to \[\frac{1}{4}th\] of its value, keeping the mass of the earth constant, the escape velocity will be

    A)  doubled      

    B)  halved

    C)  tripled        

    D)  unaltered

    Correct Answer: A

    Solution :

    Escape velocity \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\] If   \[R'=\frac{R}{4}\] \[v{{'}_{e}}=2\sqrt{\frac{2GM}{R}}\] Since, G and M are constant, hence \[v{{'}_{e}}=2{{v}_{e}}\]


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