J & K CET Engineering J and K - CET Engineering Solved Paper-2015

  • question_answer
    Radius of Earth is \[6400\text{ }km\]and that of Mars is \[3200\text{ }km\]. Mass of Mars is \[0.1\] that of Earth's mass. Then, the acceleration due to gravity on Mars is nearly

    A) \[1m/{{s}^{2}}\]          

    B) \[2.5\text{ }m/{{s}^{2}}\]

    C) \[4m/{{s}^{2}}\]          

    D) \[5\,m/{{s}^{2}}\]

    Correct Answer: C

    Solution :

    Acceleration due to gravity of a planet is \[g=\frac{GM}{{{R}^{2}}}\] where, M = mass of planet R = radius of planet \[{{g}_{m}}={{g}_{Mas}}=\frac{GMm}{R_{m}^{2}}\] \[\Rightarrow \] \[{{g}_{e}}={{g}_{Earth}}=\frac{GMe}{R_{e}^{2}}\] \[\frac{{{g}_{m}}}{{{g}_{e}}}=\left( \frac{Mm}{Me} \right){{\left( \frac{{{R}_{e}}}{{{R}_{m}}} \right)}^{2}}=(0.1){{\left( \frac{6400}{3200} \right)}^{2}}\] \[=(0.1)\,\,{{(2)}^{2}}=4\times 10.1=0.4\] \[\Rightarrow \] \[{{g}_{m}}=(0.4)\,(10)=4\,\,m/{{s}^{2}}\]


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