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Question

The bob of a simple pendulum is released when its string makes an angle $$\theta $$ with vertical. If the tension in the string is equal to thrice the weight of bob when the bob is at lowest point, the value of $$\theta $$ is:


A
30°
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B
45°
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C
60°
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D
90°
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Solution

The correct option is D $$90°$$
Apply WET between A and B,
$$mg L( 1 - cos\theta)  =  \dfrac{1}{2} mv^{2}$$
$$v  =  \sqrt{2gL(1 - cos \theta) }$$
At pt. A,
$$T -  mg  =  \dfrac{mv^{2}}{L} $$
 Also, $$T  =  3\ mg$$
$$3\ mg - mg   =    \dfrac{mv^{2}}{L} $$
$$    \sqrt{2gL}     =  v  =  \sqrt{2gL \big(1 - cos \theta\big) } $$
$$  1  =  1 - cos \theta$$
$$cos\theta  =  0$$
$$\theta  =  90^{0}$$

Physics

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