T1sin45∘=FT1cos45∘=T2=1×g∴tan45∘=gF∴F=10 N
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…
JEE Main 2024 — Physics Mechanics
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 : (Assume that the system is in equilibrium and g=10 m/s2 ) 
Held on 9 Apr 2024 · Verified 6 Jul 2026.
Options
- A
10 N
- B
210 N
- C
1 N
- D
10×21N
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