JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Laws of Motion

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions

T_{1}

and

T_{2}

, respectively, are (in N) :

(A) $5, 53$
(B) $53, 5$ ✓ Correct
(C) $53, 53$
(D)5,5

Answer: (B)

Solution

The block stays in equilibrium under its weight

m g

and the two string tensions, with

T_{1}

6 0^{\circ}

to the horizontal and

T_{2}

3 0^{\circ}

. Balancing horizontal components,

T_{1} cos 6 0^{\circ} = T_{2} cos 3 0^{\circ}

, which gives

T_{1} = 3 T_{2}

. Balancing vertical components,

T_{1} sin 6 0^{\circ} + T_{2} sin 3 0^{\circ} = m g

. Substituting,

3 T_{2} \cdot \frac{3}{2} + T_{2} \cdot \frac{1}{2} = \frac{3 T _{2}}{2} + \frac{T _{2}}{2} = 2 T_{2} = m g

. With

m g = 1 \times 10 = 10

T_{2} = 5

N and

T_{1} = 53

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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