A body of mass $m$ is suspended by two strings making angles $\theta_{1}$ and $\theta_{2}$ with the horizontal ceiling with tensions $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ simultaneously. $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are related by $T_{1}=\sqrt{3} T_{2}$. the angles $\theta_{1}$ and $\theta_{2}$ are

(B) $\theta_{1}=60^{\circ} \theta_{2}=30^{\circ}$ with $\mathrm{T}_{2}=\frac{\mathrm{mg}}{2}$

(A) $\theta_{1}=30^{\circ} \theta_{2}=60^{\circ}$ with $\mathrm{T}_{2}=\frac{3 \mathrm{mg}}{4}$

(C) $\theta_{1}=45^{\circ} \theta_{2}=45^{\circ}$ with $\mathrm{T}_{2}=\frac{3 \mathrm{mg}}{4}$

(D) $\theta_{1}=30^{\circ} \theta_{2}=60^{\circ}$ with $\mathrm{T}_{2}=\frac{4 \mathrm{mg}}{5}$

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Laws of Motion

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 $T_{2} = \frac{3 mg}{4}$
(B) $θ_{1} = 6 0^{\circ} θ_{2} = 3 0^{\circ}$ with $T_{2} = \frac{mg}{2}$ ✓ Correct
(C) $θ_{1} = 4 5^{\circ} θ_{2} = 4 5^{\circ}$ with $T_{2} = \frac{3 mg}{4}$
(D) $θ_{1} = 3 0^{\circ} θ_{2} = 6 0^{\circ}$ with $T_{2} = \frac{4 mg}{5}$

Answer: (B)

Resolving the two string tensions for the suspended body, horizontal and vertical balance give

T_{1} cos θ_{1} = T_{2} cos θ_{2}, T_{1} sin θ_{1} + T_{2} sin θ_{2} = m g .

With

T_{1} = 3 T_{2}

, the horizontal equation becomes

3 cos θ_{1} = cos θ_{2}

, satisfied by

θ_{1} = 6 0^{\circ}, θ_{2} = 3 0^{\circ}

, since

3 cos 6 0^{\circ} = \frac{3}{2} = cos 3 0^{\circ}

. Substituting in the vertical equation,

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

T_{2} = m g /2

. The strings make angles

θ_{1} = 6 0^{\circ}

and

θ_{2} = 3 0^{\circ}

with

T_{2} = m g /2

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option