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 $\mathrm{T}_{1}=\sqrt{3} \mathrm{~T}_{2}$. the angles $\theta_{1}$ and $\theta_{2}$ are
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 $\mathrm{T}_{1}=\sqrt{3} \mathrm{~T}_{2}$. the angles $\theta_{1}$ and $\theta_{2}$ are
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- $\theta_{1}=30^{\circ} \theta_{2}=60^{\circ}$ with $\mathrm{T}_{2}=\frac{3 \mathrm{mg}}{4}$
- $\theta_{1}=60^{\circ} \theta_{2}=30^{\circ}$ with $\mathrm{T}_{2}=\frac{\mathrm{mg}}{2}$
- $\theta_{1}=45^{\circ} \theta_{2}=45^{\circ}$ with $\mathrm{T}_{2}=\frac{3 \mathrm{mg}}{4}$
- $\theta_{1}=30^{\circ} \theta_{2}=60^{\circ}$ with $\mathrm{T}_{2}=\frac{4 \mathrm{mg}}{5}$
The Correct Option is B
Approach Solution - 1
2. Substitute $\mathrm{T}_{1}$: \[ \sqrt{3} \mathrm{~T}_{2} \sin \theta_{1} + \mathrm{T}_{2} \sin \theta_{2} = \mathrm{mg} \] \[ \mathrm{T}_{2} (\sqrt{3} \sin \theta_{1} + \sin \theta_{2}) = \mathrm{mg} \]
3. For $\theta_{1} = 60^{\circ}$ and $\theta_{2} = 30^{\circ}$: \[ \mathrm{T}_{2} (\sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}) = \mathrm{mg} \] \[ \mathrm{T}_{2} = \frac{\mathrm{mg}}{2} \] Therefore, the correct answer is (2) $\theta_{1}=60^{\circ} \theta_{2}=30^{\circ}$ with $\mathrm{T}_{2}=\frac{\mathrm{mg}}{2}$.
Approach Solution -2
A body of mass \( m \) is suspended by two strings making angles \( \theta_1 \) and \( \theta_2 \) with the horizontal ceiling. The tensions in the strings are \( T_1 \) and \( T_2 \), related by \( T_1 = \sqrt{3} T_2 \). We are to determine the angles \( \theta_1 \) and \( \theta_2 \), and the value of \( T_2 \).
Concept Used:
For a body in equilibrium under the action of two tensions and its weight:
- Horizontally: \( T_1 \cos\theta_1 = T_2 \cos\theta_2 \)
- Vertically: \( T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg \)
Given \( T_1 = \sqrt{3} T_2 \), we can use these equations to find \( \theta_1 \) and \( \theta_2 \).
Step-by-Step Solution:
Step 1: Write the horizontal equilibrium equation.
\[ T_1 \cos\theta_1 = T_2 \cos\theta_2 \] \[ \sqrt{3} T_2 \cos\theta_1 = T_2 \cos\theta_2 \] \[ \cos\theta_2 = \sqrt{3} \cos\theta_1 \]
Step 2: The above equation suggests \( \theta_1 > \theta_2 \), because \( \cos\theta_2 > \cos\theta_1 \) implies \( \theta_2 < \theta_1 \).
Step 3: Try possible standard angles that satisfy \( \cos\theta_2 = \sqrt{3} \cos\theta_1 \).
If we assume \( \theta_1 = 60^\circ \) and \( \theta_2 = 30^\circ \):
\[ \cos 30^\circ = \sqrt{3} \times \cos 60^\circ \] \[ \frac{\sqrt{3}}{2} = \sqrt{3} \times \frac{1}{2} \]
This is true. Hence, \( \theta_1 = 60^\circ \) and \( \theta_2 = 30^\circ \) satisfy the horizontal equilibrium.
Step 4: Use the vertical equilibrium equation to find \( T_2 \):
\[ T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg \] \[ \sqrt{3} T_2 \sin 60^\circ + T_2 \sin 30^\circ = mg \] \[ \sqrt{3} T_2 \left( \frac{\sqrt{3}}{2} \right) + T_2 \left( \frac{1}{2} \right) = mg \] \[ \left( \frac{3}{2} + \frac{1}{2} \right) T_2 = mg \] \[ 2 T_2 = mg \] \[ T_2 = \frac{mg}{2} \]
Final Computation & Result:
The angles and tensions are:
\[ \boxed{\theta_1 = 60^\circ, \; \theta_2 = 30^\circ, \; T_2 = \frac{mg}{2}} \]
Final Answer: \( \theta_1 = 60^\circ, \; \theta_2 = 30^\circ, \; T_2 = \frac{mg}{2} \)
Correct Option: (2)
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