JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A flexible chain of mass \(m\) hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is \(30^{\circ}\). Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is _________ .
- A \(\frac{\sqrt{3}}{2} m g\)
- B \(\frac{1}{2} m g\)
- C mg
- D \(\sqrt{3}mg\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{2} m g\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Tsin} 30^{\circ}=\frac{ m }{2} g\) \(T \cos 30^{\circ}= T _0\) \(\tan 30^{\circ}=\frac{ mg }{2 T_0}\) \(T _0=\frac{\sqrt{3}}{2} mg\)
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