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Question

A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle $$\theta$$ of thread deflection in the extreme position will be:


A
tan1(2)
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B
tan1(2)
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C
tan1(12)
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D
2tan1(12)
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Solution

The correct option is C $$2\tan ^{ -1 }{ \left( \cfrac { 1 }{ 2 } \right) } $$
At highest point velocity is zero $$a_N =0$$
                                                      $$a_T =gsin\theta$$

At lowest point $$a_N = \dfrac{v^2}{l}$$
                          $$a_T =0$$ 

Applying law of conventional energy
       $$mgl(1-cos\theta) = \dfrac{1}{2}mv^2$$
$$\dfrac{v^2}{l} = 2g(1-cos\theta)$$
$$gsin\theta = 2g(1-cos\theta)$$
$$2sin\dfrac{\theta}{2}cos\dfrac{\theta}{2} = 2 \times 2 {sin}^2\dfrac{\theta}{2}$$
$$\theta = 2{tan}^{-1}\dfrac{1}{2}$$

Physics

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