A) \[\pi \frac{{{v}^{2}}}{g}\]
B) \[\pi \frac{{{v}^{4}}}{{{g}^{2}}}\]
C) \[{{\pi }^{2}}\frac{{{v}^{4}}}{{{g}^{2}}}\]
D) \[{{\pi }^{2}}\frac{{{v}^{2}}}{{{g}^{2}}}\]
Correct Answer: B
Solution :
[b] Area in which bullet will spread = \[\pi {{r}^{2}}\] For maximum area, \[r={{R}_{\max }}=\frac{{{v}^{2}}}{g}[\text{when }\theta \,\,=45{}^\circ ]\] Maximum area \[\pi \ R_{\max }^{2}=\pi {{\left( \frac{{{v}^{2}}}{g} \right)}^{2}}=\frac{\pi {{v}^{4}}}{{{g}^{2}}}\]
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