Question

An aircraft executes a horizontal loop with a speed of $150\,m{s^{ - 1}}$ with its wings banked at an angle of ${12^o}$. The radius of the loop is (\[g = 10\,m{s^{ - 2}},tan{12^o} = 0.21\])
A. $10.7\,km$
B. $9.6\,km$
C. $7.4\,km$
D. $5.8\,km$

Answer 1

Hint:In order to solve this question, we will use the general relation between banking angle, velocity and radius of the circular path for a body to turn safely and using this we will solve for the radius of the loop and then we will determine the correct option.

Formula used:
If $\theta $ is the angle required for a body to turn safely in the circular path of radius r and if v is the velocity of the body then the relation between banking angle, radius and velocity is,
$\tan \theta = \dfrac{{{v^2}}}{{rg}}$
where g is the acceleration due to gravity and tan is the tangent of the angle $\theta $

Complete step by step solution:
According to the question, we have given that the velocity of the aircraft is $v = 150\,m{s^{ - 1}}$ and it has a banking angle of $\theta = {12^o}$ and let r is the radius of the loop for the aircraft and we have given that \[g = 10\,m{s^{ - 2}},tan{12^o} = 0.21\].

So, using the relation $\tan \theta = \dfrac{{{v^2}}}{{rg}}$ and solving for r we get,
$\tan {12^o} = \dfrac{{{{(150)}^2}}}{{r(10)}} \\
\Rightarrow r = 10.7 \times {10^3}m \\
\therefore r = 10.7\,km $
So, the radius of the circular loop for the aircraft is $10.7\,km$.

Hence, the correct answer is option A.

Note: It should be remembered that, the basic unit of conversion used are $1\,km = 1000\,m$ and here acceleration due to gravity is given to us $g = 10\,m{s^{ - 2}}$ but when the value is not given in such a problem always used the accepted value of acceleration due to gravity which is $g = 9.8\,m{s^{ - 2}}$.