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 Online Paper (Held On 09-Jan-2019 Evening]

A) 100 N          

B) 70 N

C) 140 N               

D)               200 N

Correct Answer: A

Solution :

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