Suppose, the ball goes up with acceleration w_1 and the rod comes down with the acceleration w_2 .As the length of the thread is constant, 2w_1=w_2 (1)From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives, T-mg=mw_1 (2)and Mg-T^\\prime=Mw_2 (3)but T=2T (because pulley is massless) (4)From equations (1), (2), (3) and (4) \\displaystyle w_1=\\frac{(2M-m)g}{m+4M}=\\frac{(2-\\eta)g}{\\eta+4} (in upward direction)and \\displaystyle w_2=\\frac{2(2-\\eta)g}{\\eta+4} (downwards)From kinematical equation in projection form, we get \\displaystyle l=\\frac{1}{2}(w_1+w_2)t^2 as, w_1 and w_2 are in the opposite direction.Putting the values of w_1 and w_2 , the sought time becomes \\displaystyle t=\\sqrt{\\frac{2l(\\eta+4)}{3(2-\\eta)g}}=1.4\\:s \\implies 5t=7