Question:
A traffic light of mass $10\sqrt{3}$ kg is suspended by two cables making $30^{\circ}$ with the vertical. The tension in each cable is ________.
A traffic light of mass $10\sqrt{3}$ kg is suspended by two cables making $30^{\circ}$ with the vertical. The tension in each cable is ________.
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For symmetrical suspension, $2T\cos\theta = \text{Weight}$.
Updated On: Jun 26, 2026
- 10 N
- 9.8 N
- 98 N
- 19.6 N
- 20 N
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The Correct Option is C
Solution and Explanation
Step 1: Concept
At equilibrium, the vertical components of tension balance the weight ($2T \cos\theta = mg$).
Step 2: Meaning
$\theta = 30^{\circ}$, $m = 10\sqrt{3}$ kg, and $g \approx 9.8~ms^{-2}$.
Step 3: Analysis
$2T \cos 30^{\circ} = 10\sqrt{3} \times 9.8$. $2T (\frac{\sqrt{3}}{2}) = 10\sqrt{3} \times 9.8 \implies T\sqrt{3} = 98\sqrt{3}$.
Step 4: Conclusion
$T = 98 \text{ N}$. Final Answer: (C)
At equilibrium, the vertical components of tension balance the weight ($2T \cos\theta = mg$).
Step 2: Meaning
$\theta = 30^{\circ}$, $m = 10\sqrt{3}$ kg, and $g \approx 9.8~ms^{-2}$.
Step 3: Analysis
$2T \cos 30^{\circ} = 10\sqrt{3} \times 9.8$. $2T (\frac{\sqrt{3}}{2}) = 10\sqrt{3} \times 9.8 \implies T\sqrt{3} = 98\sqrt{3}$.
Step 4: Conclusion
$T = 98 \text{ N}$. Final Answer: (C)
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