• # question_answer 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 $45{}^\circ$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $\left( g=10\text{ }m{{s}^{-}}^{2} \right)$ [JEE Main 09-Jan-2019 Evening] A) 100 N B) 70 N C) 140 N D) 200 N

 [a] $\frac{f}{\sin \,(180{}^\circ -45{}^\circ )}\,\,=\,\,\frac{Mg}{\sin \,(90{}^\circ +45{}^\circ )}$ $\frac{f}{\sin \,45{}^\circ }\,\,=\,\,\frac{Mg}{\cos \,45{}^\circ }$ $\Rightarrow \,\,\,f=Mg=10g=100\,N$