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WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the Earth is (Here R is the radius of the earth)

    A)  \[\left( \frac{n}{n+1} \right)mgR\]                         

    B)  \[\left( \frac{n}{n-1} \right)mgR\]

    C)  \[nmgR\]                           

    D)  \[\frac{mgR}{n}\]

    Correct Answer: A

    Solution :

                     The acceleration due to gravity varies with height as \[g=\frac{g}{\left( 1+\frac{h}{R} \right)}\] Hence, \[h=nR\Rightarrow g=\frac{g}{\left( 1+\frac{nR}{R} \right)}=\frac{g}{(1+n)}\] The change in potential energy is equal to \[\Delta U=mgh=\frac{mg(nR)}{(1+n)}=\left( \frac{n}{n+1} \right)mgR\]


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