TS EAMCET · Physics · Laws of Motion
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 \(\operatorname{cosec}^{-1}(4)\)
- B \(\cot ^{-1}(4)\)
- C \(\tan ^{-1}(4)\)
- D \(\sec ^{-1}(4)\)
Answer & Solution
Correct Answer
(B) \(\cot ^{-1}(4)\)
Step-by-step Solution
Detailed explanation
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, \(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{rg}}=\frac{150 \times 150}{9 \times 10^3 \times 10}=\frac{1}{4}\)…
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