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QJEEM20Ph245c: JEE Main 2020 Physics - Class 11 JEEM20Ph245c Solution: A mass of 10 kg is suspended by a rope of length 4 m, from the…
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 45o with the vertical. Then F equal : (Take g = 10 ms–2 and the rope to be massless)
Multiple Choice Options
- A 100 N
- B 75 N
- C 90 N
- D 70 N
Step-by-Step Verified Solution
For equilibrium,
T sin 45o = F ....(1)
and T cos 45o = 10g ....(2)
Performing (1)/(2)
we get F = 10g = 100 N