JEE Main 2025 Apr 4 Shift 1, Physics Q12: Newton's Laws Of Motion

JEE Main2025Apr 4, Shift 1Physics

A body of mass $m$ is suspended by two strings making angles $θ_{1}$ and $θ_{2}$ with the horizontal ceiling with tensions $T_{1}$ and $T_{2}$ simultaneously. $T_{1}$ and $T_{2}$ are related by $T_{1} = 3 T_{2}$ . The angles $θ_{1}$ and $θ_{2}$ are

A
$θ_{1} = 3 0^{\circ}, θ_{2} = 6 0^{\circ}$ with $T_{2} = \frac{3 m g}{4}$
B
$θ_{1} = 6 0^{\circ}, θ_{2} = 3 0^{\circ}$ with $T_{2} = \frac{m g}{2}$
C
$θ_{1} = 4 5^{\circ}, θ_{2} = 4 5^{\circ}$ with $T_{2} = \frac{3 m g}{4}$
D
$θ_{1} = 3 0^{\circ}, θ_{2} = 6 0^{\circ}$ with $T_{2} = \frac{4 m g}{5}$

Solution

Equilibrium conditions: horizontal — $T_{1} cos θ_{1} = T_{2} cos θ_{2}$ ; vertical — $T_{1} sin θ_{1} + T_{2} sin θ_{2} = m g$ .

With $T_{1} = 3 T_{2}$ , the horizontal equation gives $3 cos θ_{1} = cos θ_{2}$ . This is satisfied by $θ_{1} = 6 0^{\circ}$ , $θ_{2} = 3 0^{\circ}$ (then $3 \cdot \frac{1}{2} = \frac{3}{2}$ ✓).

Vertical: $3 T_{2} sin 6 0^{\circ} + T_{2} sin 3 0^{\circ} = m g \Rightarrow T_{2} (\frac{3}{2} + \frac{1}{2}) = m g \Rightarrow T_{2} = \frac{m g}{2}$ .

$∴$ Option (2): $θ_{1} = 6 0^{\circ}$ , $θ_{2} = 3 0^{\circ}$ , $T_{2} = m g /2$ .

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JEE Main 2025 Apr 4 Shift 1, Physics Q12: Newton's Laws Of Motion

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