question_answer

A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of \[45{}^\circ \] with the vertical. Then F equals: (Take \[g=10\text{ }m{{s}^{2}}\] and the rope to be massless) [JEE MAIN Held on 07-01-2020 Evening]

A) 75 N    

B) 70 N

C) 90 N    

D) 100 N

Correct Answer: D

Solution :

[d] \[{{T}_{2}}\text{ }cos\,45{}^\circ =100\text{ }N\]                       ...(i)

\[{{T}_{2}}\text{ }sin\,45{}^\circ =F\]                          ...(ii)

\[\Rightarrow \,\,F=100\text{ }N\]