Let a small body of mass 'm' attached to a string and revolved in a vertical CIRCLE of radius 'r'. We know that weight of the body always acts vertically downward and tension in the string towards the centre of the circular path.Let v 2 is the speed of the body and T 2 is the tension in the string at the lowest point B. So at the lowest point T 2 = rmrm+mg ..(1)

+mg ..(1)TOTAL energy at bottom

+mg Total energy at bottom=PE+KE=0+ 21 MV 22= 21 mv 22 ...(2)

+mg ..Let v 1 is the speed and T 1 is the tension in the string at highest point A.

+mg ..(1)Total energy at bottom=PE+KE=0+ 21 mv 22= 21 mv 22 ...(2)Let v 1 is the speed and T 1 is the tension in the string at highest point A.So, T 1 = rmv 12−mg ....(3)Total energy at A=PE+KE =2mgr+ 21 mmv1

....(4)From equations (1) and (3)T 2 −T 1 = rmv 22+mg−( rmv 12 -mg) = rr (v 22−v12 )+2mg

2 )+2mg ...(5)By law of conservation of energy

2 )+2mg ..

12=4rg ...(6)Putting this value in EQUATION (5)TT −T 1 = rm

−T 1 = rm (4rg)+2mg

−T 1 = rm (4rg)+2mg =4mg+2mgT -T 1 =6mg

Hope this will help you...

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