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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