BHU PMT BHU PMT (Screening) Solved Paper-2007

  • question_answer
    The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of jump by the same person on the planet B?

    A)  6 m                                       

    B)  2/3 m

    C)   2/9 m                                 

    D)  18 m

    Correct Answer: D

    Solution :

                     It is given that, acceleration due to gravity on plane A is 9 times the acceleration due to gravity on planet B ie, \[{{g}_{A}}=9{{g}_{B}}\]                                                ?.. (i) From third equation of motion \[{{v}^{2}}=2gh\] At planet A, \[{{h}_{A}}=\frac{{{v}^{2}}}{2{{g}_{A}}}\]                  ...(ii) At planet B, \[{{h}_{B}}=\frac{{{v}^{2}}}{2{{g}_{B}}}\]                        ...(iii) Dividing Eq. (ii) by Eq. (iii), we have                 \[\frac{{{h}_{A}}}{{{h}_{B}}}=\frac{{{g}_{B}}}{{{g}_{A}}}\] From Eq. (i),   \[{{g}_{A}}=9{{g}_{B}}\] \[\therefore \]  \[\frac{{{h}_{A}}}{{{h}_{B}}}=\frac{{{g}_{B}}}{9{{g}_{B}}}=\frac{1}{9}\] or            \[{{h}_{B}}=9{{h}_{A}}=9\times 2=18\,m\]           \[(\because {{h}_{A}}=2m)\]


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