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$V_{b}=7gr $

Work Energy Theorem applied between top and bottom.

$mg(2R)=21 m(V_{b})_{2}−21 m(V_{T})_{2}$

$2mgr=27 mgr−21 m(V_{T})_{2}$

$V_{T}=3gr $

At top point:

$T=rmV_{T} −mg$

$=mr3gr −mg$

$=2mg$

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