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