• # 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

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