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9th Class Science Gravitation and Floatation Question Bank Gravitation IIT JEE Objective Problems

  • question_answer
    The mass of the Earth is 80 times the mass of a planet and diameter is twice that of the planet. Then the acceleration due to gravity on the planet's surface is

    A)  \[0.\text{49m}/{{\text{s}}^{\text{2}}}\]       

    B)  \[\text{9}.\text{8m}/{{\text{s}}^{\text{2}}}\]

    C)  \[\text{1}.\text{6 m}/{{\text{s}}^{\text{2}}}\]                  

    D)  \[0.\text{8m}/{{\text{s}}^{\text{2}}}\]

    Correct Answer: A

    Solution :

     
    Earth Planet
    \[{{M}_{1}}=80M\] \[{{M}_{2}}=M\]
    \[{{D}_{1}}=2D\]\[\Rightarrow \] \[{{R}_{1}}=2R\] \[{{D}_{2}}=D\]\[\Rightarrow \] \[{{R}_{2}}=R\]
    \[{{g}_{1}}=g=9.8m/{{s}^{2}}\] \[{{g}_{2}}=?\]
    \[g=\frac{GM}{{{R}^{2}}}\] Apply the formula to both cases. \[9.8=\frac{G\times 80\,M}{{{(2R)}^{2}}}\]                          \[{{g}_{2}}=\frac{G\times M}{{{R}^{2}}}\] \[\Rightarrow \]\[9.8=\frac{G\times 80M}{4{{R}^{2}}}\]                ??? (1) \[\Rightarrow \]\[{{g}_{2}}=\frac{GM}{{{R}^{2}}}\]                          ??? (2) Dividing (2) by (1), we get \[\frac{{{g}_{2}}}{9.8}=\frac{GM}{{{R}^{2}}}\times \frac{4{{R}^{2}}}{G\times 80\,M}\] \[\Rightarrow \frac{{{g}_{2}}}{9.8}=\frac{1}{20}\Rightarrow {{g}_{2}}=\frac{9.8}{20}\Rightarrow {{g}_{2}}=0.49\,m/{{s}^{2}}\]     


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