Q125 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
An aircraft executes a horizontal loop of radius 9 km at a constant speed of $540 \mathrm{kmh}^{-1}$. The wings of the aircraft are banked at an angle of (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
- A$\operatorname{cosec}^{-1}(4)$
- Correct answerB$\cot ^{-1}(4)$
- C$\tan ^{-1}(4)$
- D$\sec ^{-1}(4)$
Solution
For circular motion of aircraft,
$\mathrm{r}=9 \mathrm{~km}, \mathrm{v}=540 \mathrm{~km} \mathrm{~h}^{-1}=540 \times \frac{5}{18}=150 \mathrm{~ms}^{-1}$
Banked angle, $\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{150 \times 150}{9 \times 10^3 \times 10}=\frac{1}{4}$ $\therefore \quad \theta=\cot ^{-1}(4)$
Banked angle, $\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{150 \times 150}{9 \times 10^3 \times 10}=\frac{1}{4}$ $\therefore \quad \theta=\cot ^{-1}(4)$
Hence, the correct answer is option: 'B'.