AP EAMCET · PHYSICS · Laws of Motion
If the tension in the horizontal wire shown in the figure is 30 N, then the weight W and tension in the wire OA are respectively

- A \(30 \sqrt{3} \mathrm{~N}, 30 \mathrm{~N}\)
- B \(30 \sqrt{3} \mathrm{~N}, 60 \mathrm{~N}\)
- C \(60 \sqrt{3} \mathrm{~N}, 30 \mathrm{~N}\)
- D \(60 \sqrt{3} \mathrm{~N}, 60 \mathrm{~N}\)
Answer & Solution
Correct Answer
(B) \(30 \sqrt{3} \mathrm{~N}, 60 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(T_{OA} \cos 60^\circ = 30 \mathrm{~N}\) \(T_{OA} = \frac{30}{\cos 60^\circ} = \frac{30}{1/2} = 60 \mathrm{~N}\) \(W = T_{OA} \sin 60^\circ\) \(W = 60 \times \frac{\sqrt{3}}{2} = 30 \sqrt{3} \mathrm{~N}\) Weight W = \(30 \sqrt{3} \mathrm{~N}\), Tension in wire OA =…
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