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JEE Main 2020 · Physics · Kinematics and Laws of Motion · medium

A bob of mass m hangs from a light inextensible string of length L. It is given a horizontal speed √(5gL) at the lowest point. The tension in the…

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A bob of mass mm hangs from a light inextensible string of length LL. It is given a horizontal speed 5gL\sqrt{5gL} at the lowest point. The tension in the string at the instant the string makes an angle of 60∘60^\circ with the downward vertical is
  1. A. 4 mg4\,mg
  2. B. 3.5 mg3.5\,mg
  3. C. 4.5 mg4.5\,mg
  4. D. 5 mg5\,mg
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Answer: (C) 4.5 mg4.5\,mg

Energy conservation up to height L(1−cos⁡60∘)=L/2L(1-\cos 60^\circ) = L/2: v2=5gL−2g(L/2)=4gLv^2 = 5gL - 2g(L/2) = 4gL. Radial equation towards the centre (the weight component mgcos⁡60∘mg\cos 60^\circ points away from the centre): T−mgcos⁡60∘=mv2/LT - mg\cos 60^\circ = mv^2/L. So T=4mg+0.5 mg=4.5 mgT = 4mg + 0.5\,mg = 4.5\,mg. Using T−mgT - mg instead of T−mgcos⁡θT - mg\cos\theta (a frequent mistake) would give 5mg5mg, and subtracting the component gives 3.5mg3.5mg.
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Original question written in the JEE Main 2020 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake