Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m/s²)

The correct answer is D, infinitely large. A perfectly horizontal rope has no upward component of tension, so no finite tension can hold up the 200 N weight with the rope fully straight.

Let the two halves of the rope make an angle θ with the horizontal. For the mass to be in equilibrium, the upward components of the two tensions must balance its weight: 2T sin θ = mg = 20 × 10 = 200 N. So T = 100 ÷ sin θ. As the rope is pulled straighter, θ gets smaller and sin θ approaches zero, so T grows without limit. At θ = 0, the tension would have to be infinite. In practice the rope always sags a little, however hard it is pulled.

Option A is wrong because 200 N is only the weight of the mass. Option B is wrong because 20 N is the mass value, not a force at all here. Option C is wrong because 10 N is just g written as a force.

Exam tip: 2T sin θ = mg; as θ approaches zero, T becomes infinite, so a loaded rope can never be made perfectly straight.