Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    Two bodies of masses 10 kg and 100 kg are separated by a distance of 2m. The gravitational potential at the mid-point on the line joining the two is

    A)  \[7.3\times {{10}^{-7}}\text{J/kg}\]  

    B)  \[7.3\times {{10}^{-8}}\text{J/kg}\]

    C)  \[7.3\times {{10}^{-9}}\text{J/kg}\]  

    D)  \[7.3\times {{10}^{-6}}\text{J/kg}\]

    Correct Answer: C

    Solution :

     Gravitational potential \[|V|=\frac{Gm}{r}\] So, gravitational potential at mid-point is \[V={{V}_{1}}+{{V}_{2}}\] \[=\frac{G{{m}_{1}}}{{{r}_{1}}}+\frac{G{{m}_{2}}}{{{r}_{2}}}\] \[=\frac{6.67\times {{10}^{-11}}\times 10}{1}+\frac{6.67\times {{10}^{-11}}\times 100}{1}\] \[=6.67\times {{10}^{-11}}\times 110\] \[=7.3\times {{10}^{-9}}J/kg\]


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