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Sol. . At equilibrium ⇒ T = mg When it moves to an angle θ, and released, the tension the T’ at lowest point is ⇒ T’ = mg + mv2/r The change in tension is due to centrifugal force ∆T = mv2/r ….(1) ⇒ Again, by work energy principle, ⇒ 1/2mv2 – 0 = mgr(1 - cosθ) ⇒ v2 = 2gr (1- cosθ) So, ∆T = m[2gr(1 – cos θ)]/r = 2mg(1 – cos θ) ⇒ F = ∆T ⇒ F = YA ∆L/L = 2mg – 2mg cos θ ⇒ 2mg cos θ = 2mg - YA ∆L/L = cos θ = 1 - YA ∆L/L(2mg)
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