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 \[[\theta ={{45}^{o}}]\] Range \[{{R}_{\max }}=\frac{{{v}^{2}}}{g}\] \[\because \] Area in which bullet will spread \[=\pi {{r}^{2}}\] \[\therefore \] Maximum area \[\pi R_{\max }^{2}=\pi {{\left( \frac{{{v}^{2}}}{g} \right)}^{2}}=\frac{\pi {{v}^{4}}}{{{g}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
