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) (a) 1 N (b) 10 N (c) (10)/(√(2)) N (d) (1)/(10 × √(2)) N

Answer: (b). At the midpoint: T₁cos 45° = mg and T₁sin 45° = F. F = mgtan 45° = 1 × 10 × 1 F = 10 N So the answer is (b) 10 N.

A 1 kg mass is suspended from the ceiling by a rope of length 4 m. A… | JEE Main 2024 (9 Apr, Shift 2)

Q9JEE Main 2024 · 9 APR · SHIFT 2Medium

Equilibrium, Third Law & Frames

A $1\text{ kg}$ mass is suspended from the ceiling by a rope of length $4\text{ m}$ . A horizontal force ' $F$ ' is applied at the mid point of the rope so that the rope makes an angle of $45^\circ$ with respect to the vertical axis as shown in figure. The magnitude of $F$ is:

(Assume that the system is in equilibrium and $g = 10\text{ m/s}^2$ )

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Answer: (b)

At the midpoint: $T_1\cos 45^\circ = mg$ and $T_1\sin 45^\circ = F$ .
$F = mg\tan 45^\circ = 1 \times 10 \times 1$
$F = 10\text{ N}$
So the answer is (b) 10 N.

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