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(D) Let the tension in the rope be $T$ and the angle the rope makes with the horizontal be $	heta$. For the weight $m = 15\, kg$ to be in equilibrium

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infinitely large

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(D) Let the tension in the rope be $T$ and the angle the rope makes with the horizontal be $	heta$. For the weight $m = 15\, kg$ to be in equilibrium at the midpoint,the vertical components of the tension must balance the weight: $2T \sin 	heta = mg$ Solving for tension $T$: $T = \frac{mg}{2 \sin 	heta}$ To straighten the rope completely,the rope must be horizontal,which means the angle $	heta$ must be $0^\circ$. As $	heta 	o 0^\circ$,$\sin 	heta 	o 0$. Therefore,$T = \frac{mg}{2 	imes 0} = \infty$. Thus,an infinitely large tension is required to make the rope perfectly horizontal.

Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. $A$ $15\, kg$ weight is attached to the rope at the mid-point,which now no longer remains horizontal. The minimum tension required to completely straighten the rope is

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