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Q.

A small sphere of mass m suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released, then 

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a

total acceleration of sphere as a function of θ is g 1+3cos2θ

b

thread tension as a function of e is T =3mg cos θ

c

the angle θ between the thread and the vertical at the moment when the total acceleration vector of the sphere is directed horizontally is cos-1 1/ 3

d

the thread tension at the moment when the vertical component of the sphere's velocity is maximum will be mg

answer is A, B, C.

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Detailed Solution

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Between A and B

 ,MgLcosθ= 12 mvB2 vB2=2gLcosθ ar=vB2L=2gcosθ  & at=gsinθ  a=at2+ar2=g1+3cos2θ  

Now, at BTB- mg cosθ =mvB2L

put vB  TB= 3mg cosθ

When total acceleration vector directed horizontally

tan(90°-θ)=atar=12tanθ

On solving  θ=cos-113 

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