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Question

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A. $\sqrt{3gr}$

B. $\sqrt{2gr}$

C. $\sqrt{gr}$

D. $\sqrt{gr/2}$

Answer

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Here in the above diagram we see, the fighter plane B is at the top of the circle that is at point A.

Now, the arrow indicates the direction of the fighter plane, and there are two forces working on the airplane when it is at the top position, which is at point A. These two forces are counterbalancing each other and stop the fighter plane from falling down.

And this diagram is the free body diagram of that plane in its highest position.

Now, according to the problem

we know that, for a moving object in a circle of radius ‘r’,

At the highest point, the weight of the fighter plane will act downwards and so in order to keep the plane in motion and in the trajectory, another force acting upward should be balancing it. Let us take this force as $F_g$ which is equal to

$F_g$ = $\dfrac{m{{v}^{2}}}{r}$ ,

Now, according to the problem

mg $\le $$\dfrac{m{{v}^{2}}}{r}$,

Here the m cancels out, and

g $\le $$\dfrac{{{v}^{2}}}{r}$,

gr $\le $${{v}^{2}}$,

$\sqrt{gr}\le v$ ,

We had to find the minimum velocity at point A, hence

$v = \sqrt{gr}$

Try to figure out the forces acting on the body properly as it is the most important step of the answer, $F_g$ that is the force counterfeiting the force of gravitation is the greater force because without that force the fighter plane will simply fall down, when we got the answer it was in the relation of greater than equals to but we need it in equals to form so we are considering it as equals to because in the question it was written to get the minimum velocity.