JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 11)
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)
100 N
75 N
90 N
70 N
Explanation
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For equilibrium,
T sin 45o = F ....(1)
and T cos 45o = 10g ....(2)
Performing (1)/(2)
we get F = 10g = 100 N
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