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**Hint:**We draw the free body diagram of the following system to solve this question. We find the sum of forces acting upwards and downwards and equate them since there is horizontal force. Both the horizontal components of the two sides of the string get cancelled. So, the weight acting downwards is equal to the sum of upward forces.

The angle between the string and the x-axis should be zero for the body to be horizontal. Substituting this condition in the final equation we get the answer.

**Complete step by step solution:**The free body diagram of the system is

The weight is at the centre of the string so the tension is distributed equally on both sides of the string. The tension force on both the sides of the string is resolved into

$T = Tcos\theta + T\sin \theta $

As shown in the diagram

The horizontal components are equal and opposite in direction so they get cancelled.

$Tcos\theta - Tcos\theta = 0$

Since the body is at equilibrium the sum of upwards forces is equal to downwards force (weight of the body)

Forces acting upwards are the vertical components of the tension force

$T\sin \theta + T\sin \theta = 2T\sin \theta $

Weight of the body is $\omega \ $

Upward force is equal to downwards force

$\omega \ = 2T\sin \theta $

For the rope to be perfectly horizontal $\theta $ should be $0^\circ $

Substituting $\theta = 0^\circ $

$ \omega \ = 2T\sin \theta $

$ \omega \ = 2T\sin 0^\circ $

$ T = \dfrac{{\omega \ }}{{2\sin 0^\circ }} = \dfrac{{\omega \ }}{0} $

Anything by zero is infinite

Hence the force applied on each end should be infinitely large.

**Option (D) infinitely large is the correct answer.**

**Note:**The angle $\theta $ is also equal to the angle between the tension force and the Horizontal which is the reason we can resolve the tension force with angle $\theta $

Students might go wrong by taking tensions in the two sides of the string with different values. The tension remains the same on both the sides because the weight is being suspended from the center.

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