A body of mass m is suspended by two strings making angles θ₁ and θ₂ with the horizontal ceiling with tensions T₁ and T₂ simultaneously. T₁ and T₂ are related by T₁ = √(3)T₂, the angles θ₁ and θ₂ are (a) θ₁ = 30°, θ₂ = 60° with T₂ = 3mg/4 (b) θ₁ = 45°, θ₂ = 45° with T₂ = 3mg/4 (c) θ₁ = 30°, θ₂ = 60° with T₂ = 4mg/5 (d) θ₁ = 60°, θ₂ = 30° with T₂ = mg/2

Answer: (d). Horizontal: T₁cos θ₁ = T₂cos θ₂ ⇒ √(3)cos θ₁ = cos θ₂; this holds for θ₁ = 60°, θ₂ = 30°. Vertical: T₁sin 60° + T₂sin 30° = mg √(3)T₂·√(3)/2 + T₂/2 = 2T₂ = mg ⇒ T₂ = mg/2 So the answer is (d).

A body of mass m is suspended by two strings making angles θ₁ and θ₂… | JEE Main 2025 (4 Apr, Shift 1)

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Equilibrium, Third Law & Frames

A body of mass $m$ is suspended by two strings making angles $\theta_1$ and $\theta_2$ with the horizontal ceiling with tensions $T_1$ and $T_2$ simultaneously. $T_1$ and $T_2$ are related by $T_1 = \sqrt{3}T_2$ , the angles $\theta_1$ and $\theta_2$ are

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Answer: (d)

Horizontal: $T_1\cos\theta_1 = T_2\cos\theta_2 \Rightarrow \sqrt{3}\cos\theta_1 = \cos\theta_2$ ; this holds for $\theta_1 = 60^\circ, \theta_2 = 30^\circ$ .
Vertical: $T_1\sin 60^\circ + T_2\sin 30^\circ = mg$
$\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2} + \frac{T_2}{2} = 2T_2 = mg \Rightarrow T_2 = \frac{mg}{2}$
So the answer is (d).

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