A) \[\frac{\pi {{v}^{2}}}{g}\]
B) \[\frac{{{v}^{2}}}{{{g}^{2}}}\]
C) \[\frac{\pi {{v}^{4}}}{{{g}^{2}}}\]
D) \[\frac{{{v}^{4}}}{{{g}^{2}}}\]
Correct Answer: C
Solution :
For maximum area \[\text{ }\!\![\!\!\text{ }\theta \text{=4}{{\text{5}}^{o}}\text{ }\!\!]\!\!\text{ }\] Range \[{{R}_{\max }}=\frac{{{v}^{2}}}{g}\] \[\because \]Area in which bullet will spread \[=\pi {{r}^{2}}\] \[\therefore \]Maximum area \[\pi R_{\max }^{2}=\pi \left( \frac{{{v}^{2}}}{g} \right)=\frac{\pi {{v}^{4}}}{{{g}^{2}}}\]
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