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**Hint:**The force acting on a particle undergoing a circular motion is called centripetal force. This centripetal force directed inwards is what keeps the particle moving in its circular path without getting thrown off/losing the circular path.

**Formula Used:**

1) The centripetal force acting on a particle is given by, $F = \dfrac{{m{v^2}}}{r}$ where $m$ is the mass of the particle, $v$ is its linear velocity and $r$ is the radius of the circular path along which the particle moves.

2) The relation between the linear velocity $v$ and angular velocity $\omega $ of a particle undergoing circular motion is given by, $v = r\omega $ ; $r$ is the radius of the circular path along which the particle moves.

**Complete step by step answer:**

Step 1: Write down the parameters involved in the problem at hand.

The mass of the particle performing circular motion is represented by $m$ .

It is given that a force represented by $F$ acts on the particle and is directed inwards.

Also, it is given that the radius of the circular path along which the particle moves is represented by $r$ .

Step 2: Express the relation of the force acting on the particle.

The force acting on the particle is the centripetal force and it is given by, $F = \dfrac{{m{v^2}}}{r}$ ------ (1)

where $m$ is the mass of the particle, $v$ is its linear velocity and $r$ is the radius of the circular path along which the particle moves.

The linear velocity $v$ of the particle can be expressed in terms of its angular velocity $\omega $ as $v = r\omega $ .

When the above relation for the linear velocity is substituted in equation (1) we get, $F = \dfrac{{m{{\left( {r\omega } \right)}^2}}}{r} = mr{\omega ^2}$ ------- (2)

Now, the angular velocity $\omega $ can be expressed in terms of the period of the motion $T$ as $\omega = \dfrac{{2\pi }}{T}$ .

Substituting the expression $\omega = \dfrac{{2\pi }}{T}$ in equation (2) we get, $F = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2}$ ------ (3)

Step 3: Find the square root of the force $F$ using equation (3).

Equation (3) gives us $F = mr{\left( {\dfrac{{2\pi }}{T}} \right)^2}$

Taking the square root of equation (3) on both sides we get, $\sqrt F = \dfrac{{2\pi }}{T}\sqrt {mr} $

$\therefore$ The square root of the force acting on the particle is $\dfrac{{2\pi }}{T}\sqrt {mr} $ . Hence, the correct option is (A)

**Note:**Here, the linear velocity of the particle at every point along the circular path will be directed along the tangent of the circle at each point. The centripetal force can also be described as a force that constantly pulls the particle inwards. The apparent force that is directed away from the centre is called the centrifugal force. Also one should note that the square root of the centrifugal force is what is asked and not only the centrifugal force.

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