A) 1000 m
B) 2250 m
C) 500 m
D) 4500 m
Correct Answer: B
Solution :
in vertical direction aeroplane is in equilibrium |
\[\therefore F\sin 45{}^\circ =mg\] |
\[F=\frac{mg}{\sin 45{}^\circ }\] |
Lift force \[F=\frac{mg}{\sin 45{}^\circ }\] |
The centripetal force is provided by the component of lift force |
\[\Rightarrow F\cos 45{}^\circ =\frac{m{{v}^{2}}}{r}\]\[\Rightarrow \frac{mg}{\sin 45{}^\circ }\cos 45{}^\circ =\frac{m{{v}^{2}}}{r}\] |
\[r=\frac{m{{v}^{2}}}{mg}=\frac{{{v}^{2}}}{g}=\frac{{{\left( 540\times \frac{5}{18} \right)}^{2}}}{10}=2250\,\,m\] |
You need to login to perform this action.
You will be redirected in
3 sec