53. a) Let the velocity at B be v₂

1/2 mv₁² = 1/2 mv₂² + mgL

implies 1/2 m (10 gL) = 1/2 mv₂² + mgL

V₂²=8gL

So, the tension in the string at horizontal position T=MV²/R = 8mgL/L = 8mg   (B) Let the velocity at C be V₃   1/2 mv₁² = 1/2 mv₃² + mg(2L)  implies  1/2m(10*g*L) = 1/2 mv₃² + mg(2L)  implies V₃² = 6mgL .So, the tension in the string is given by T_c = mv²/L - mg = 6gLm/L * mg = 5mg  (C) Let the velocity at point D be v₄ 1/2 mv₁² = 1/2 mv₄² + mgh  1/2 * m * (10 gL) = 1.2 mv₄² + mgL (1 + cos 60°) impliesV₄² = 7gL  So, the tension in the string is T = (mv²/L) – mg cos 60°  implies  m*7gL/L– l – 0.5 mg   implies 7 mg – 0.5 mg = 6.5 mg.