J & K CET Medical J & K - CET Medical Solved Paper-2002

  • question_answer
    A man weighs 60 kg at earth surface. At what height above the earths surface weight becomes 30 kg? (Given, radius of earth is 6400 km.)

    A)  2624 km                             

    B)  3000 km

    C)  2020 km                             

    D)  none of these

    Correct Answer: A

    Solution :

                    From Newtons law of gravitation \[F=\frac{G{{M}_{e}}m}{R_{e}^{2}}\]                     ???.. (i) where G is gravitational constant,\[{{M}_{e}}\]the mass of earth,\[{{R}_{e}}\]the radius and m the mass of man. Also, F = mg = weight                      ...(ii) From Eqs. (i) and (ii), we get                 \[W=\frac{G{{M}_{e}}m}{R_{e}^{2}}\] \[\therefore \]  \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{R_{2}^{2}}{R_{1}^{2}}=\frac{{{({{R}_{e}}+h)}^{2}}}{R_{e}^{2}}\] Given,  \[{{W}_{1}}=60kg,{{W}_{2}}=30kg\] \[\therefore \]  \[\frac{60}{30}={{\left( \frac{{{R}_{e}}+h}{{{R}_{e}}} \right)}^{2}}\] Taking square root                 \[\sqrt{2}=\frac{{{R}_{e}}+h}{{{R}_{e}}}\] \[\Rightarrow \]               \[\sqrt{2}{{R}_{e}}={{R}_{e}}+h\] \[\Rightarrow \]               \[h=\sqrt{2}{{R}_{e}}-{{R}_{e}}\]                 \[h=1.41{{R}_{e}}-{{R}_{e}}\]                 \[h=0.41\,{{R}_{e}}\] Given,      \[{{R}_{e}}=6400\text{ }km\] \[\therefore \] \[h=0.41\times 6400=2624\text{ }km\]


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