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Sahay Sir > Question Answers > JEE Mains Physics 2002-2019 Solved Video Solutions > Chapter 3 - Laws of Motion > A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45^∘ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g=10ms^−2)

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45^∘ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g=10ms^−2)

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point the magnitude of the force applied is (g=10ms^−2) the rope deviated at an angle of 45^∘ at the roof point. If the suspended mass is at equilibrium