question_answer

An aeroplane is flying in a horizontal circle at a speed of 540 km/h. Banked for this turn, the wings of the plane are tilted at an angle \[45{}^\circ \] from the horizontal. Assume that a lift force acting perpendicular to the wings holds the aircraft in the sky. The radius of the circle in which the plane is flying is \[(Take\,\,\,g=10m/{{s}^{2}})\]

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

\[\Rightarrow F\cos 45{}^\circ =\frac{m{{v}^{2}}}{r}\]\[\Rightarrow \frac{mg}{\sin 45{}^\circ }\cos 45{}^\circ =\frac{m{{v}^{2}}}{r}\]

\[r=\frac{m{{v}^{2}}}{mg}=\frac{{{v}^{2}}}{g}=\frac{{{\left( 540\times \frac{5}{18} \right)}^{2}}}{10}=2250\,\,m\]