A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions T₁ and T₂, respectively, are (in N): (Take acceleration due to gravity 10 m/s²) (a) 5√(3), 5 (b) 5√(3), 5√(3) (c) 5,5 (d) 5, 5√(3)

Answer: (a). Horizontal: T₁cos 60° = T₂cos 30° ⇒ T₁ = √(3)T₂. Vertical: T₁sin 60° + T₂sin 30° = mg = 10 √(3)T₂·√(3)/2 + T₂/2 = 2T₂ = 10 ⇒ T₂ = 5 N T₁ = 5√(3) N So the answer is (a) 5√(3), 5.

A body of mass 1 kg is suspended with the help of two strings making… | JEE Main 2025 (2 Apr, Shift 2)

Q3JEE Main 2025 · 2 APR · SHIFT 2Easy

Equilibrium, Third Law & Frames

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitudes of tensions $T_1$ and $T_2$ , respectively, are (in N):

(Take acceleration due to gravity $10\text{ m/s}^2$ )

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Answer: (a)

Horizontal: $T_1\cos 60^\circ = T_2\cos 30^\circ \Rightarrow T_1 = \sqrt{3}T_2$ .
Vertical: $T_1\sin 60^\circ + T_2\sin 30^\circ = mg = 10$
$\sqrt{3}T_2\cdot\frac{\sqrt{3}}{2} + \frac{T_2}{2} = 2T_2 = 10 \Rightarrow T_2 = 5\text{ N}$
$T_1 = 5\sqrt{3}\text{ N}$
So the answer is (a) $5\sqrt{3}, 5$ .

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