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NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice

A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle α . A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in the figure. If the particle doesn't undergo a complete vertical circle, then the maximum angle θ < π 2 through which the string swings after hitting the pin is

Options

  1. Acos - 1 ⁡ L cos ⁡ α + l L + l
  2. Bcos - 1 ⁡ L cos ⁡ α - l L - l
  3. Ccos - 1 &#8289; L cos &#8289; &#945; + l &#160; L - l
  4. Dcos - 1 &#8289; L cos &#8289; &#945; - l &#160; L + l

Correct answer

B. cos - 1 ⁡ L cos ⁡ α - l L - l

Step-by-step solution

At the bottom most point, square of speed of bob, v 2 = 2 g L 1 – c o s α It will rise further to a height, h = v 2 2 g = L 1 - cos ⁡ α or L - l 1 - cos ⁡ θ = L 1 - cos ⁡ α ∴ θ = cos - 1 ⁡ L cos ⁡ α - l L - l

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