KCET · Physics · Ray Optics
An aeroplane executes a horizontal loop at a speed of \( 720 \mathrm{kmph} \) with its wings banked at \( 45^{\circ} \).
What is the radius of the loop? Take \( \mathrm{g}=10 \mathrm{~ms}^{-2} \).
- A 4
- B \( 4.5 \mathrm{~km} \)
- C \( 7.2 \mathrm{~km} \)
- D \( 2 \mathrm{~km} \)
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
Given, speed of aeroplane \(=720 \mathrm{kmph}=200 \mathrm{~ms}^{-1}\)
angle of wings \(=45^{\circ} ; g=10 \mathrm{~ms}^{-2}\)
Now we know that
\(\tan \theta=\frac{v^{2}}{r g} \Rightarrow r=\frac{v^{2}}{\tan \theta \times g}\)
\(\Rightarrow r=\frac{(200)^{2}}{\tan 45^{\circ} \times 10}=\frac{(200)^{2}}{10}=400 \mathrm{~m}=4 \mathrm{~km}\)
Thus, radius of the loop is \(4 \mathrm{~km}\)
angle of wings \(=45^{\circ} ; g=10 \mathrm{~ms}^{-2}\)
Now we know that
\(\tan \theta=\frac{v^{2}}{r g} \Rightarrow r=\frac{v^{2}}{\tan \theta \times g}\)
\(\Rightarrow r=\frac{(200)^{2}}{\tan 45^{\circ} \times 10}=\frac{(200)^{2}}{10}=400 \mathrm{~m}=4 \mathrm{~km}\)
Thus, radius of the loop is \(4 \mathrm{~km}\)
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