A 1 kg mass is suspended from the ceiling by a rope of length 4 m . A horizontal force ' F ' is applied at the mid point of the rope so that the rope makes an angle of 45∘ with respect to the vertical axis as shown in figure. The magnitude of F is :
(A)210N
(B)1 N
(C)10×21N
(D)10 N✓ Correct
Answer: (D)
Solution
At the midpoint of the rope, the horizontal force F and the two tension segments (each making 45∘ with the vertical) balance; the tension in the lower segment supports the hanging mass, so T2=mg=10 N, and by symmetry of the 45∘ angle,
F=T2tan45∘=10 N
JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option