(1) 4L/9
A1 = Area of cross–section of cylinder πR2
A2 = Area of cylinder in air = (R/2)2 πR2 /4
A3 = Area of bottom of cylinder inside liquid = A1 - A2 = 3/4 πR2
Let P0 be the atmospheric pressure. The forces acting on cylinder as shown in Fig. Obvisouly