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KVPY Sample Paper KVPY Stream-SX Model Paper-8

  • question_answer
    A bullet is fired vertically upwards with velocity \[v\]. From the surface of a spherical planet. When it Reaches its maximum height, its acceleration due to the planet's gravity is \[\frac{1}{4}th\] of its value of the Surface of the planet. If the escape velocity from The planet is \[{{V}_{esc}}=V\sqrt{N},\] then the value of N is (Ignore energy loss due to atmosphere)

    A) \[1/2\]                           

    B) \[2\]

    C) \[3\]                             

    D) \[1/3\]

    Correct Answer: B

    Solution :

    Let \[h\] be the height to which the bullet rises then, g? \[=g{{\left( 1+\frac{h}{R} \right)}^{-2}}\Rightarrow \frac{g}{4}=g{{\left( 1+\frac{h}{R} \right)}^{-2}}\]
    \[\Rightarrow h=R\]
    We know that \[{{v}_{e}}=\sqrt{\frac{2GM}{R}}\]\[=v\sqrt{N}\left( \operatorname{given} \right)..(i)\]
    Now applying conservation of energy for the throw
    Loss of kinetic energy =Gain in gravitational potential energy
    \[\therefore \frac{1}{2}m{{v}^{2}}=-\frac{GMm}{2R}-\left( -\frac{GMm}{R} \right)\]\[\therefore v=\sqrt{\frac{GM}{R}}(ii)\]
    Comparing \[\left( i \right)\text{ }\And \text{ }\left( ii \right)\text{ }N=2\]


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