Question

The solid steel ball shown in the figure is submerged between the interface of two liquids with sg of 1.2 and 1.5. The ball has a diameter of 215 mm and specific gravity of 7.4 and is attached by a rope. Note: The interface of the two liquids coincides the center of the sphere.

Determine the buoyant force acting on the ball in N.

Answer 1

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Given, Let, liquid 6 has specific gravity, = 1,2 = SG, loisu liquid @ has specific O.SV gravity, SG₂ = 1.5 diamety of ball, d=215mm radius, & 107.5mm 001075 m specific gravity of ball, Saba 7.4 Weight of ball, Wball - Phall x Uball og P ball - Sabx Pw - 7.48 103 3 Wball = 7.4 x 10°* 9.81 * ( * 3.14% CO-104533) Whall - 377.56N o Volume of sphery Tr3 Now, ti 2 JE u bao buoyant force acting on ball = BFT - BF, + BF2 i.e. BF, = buoyant force acting on ball due to liquido BF2 of liquid BF - weight, displaced by ball. = P. vig Bri => ( sa,x 10°)x (0.5x * x** C0.1095)) * 9.81 c) [1,2 X 107 x 0.5 x 4 x 3 + x Co. 10s)* 19. 8)

BF = 30.62 N BF₂ = CSC2 x 103) & Co.sxy 4 x am (0.1075)3) * 981 = (1.5x 103 x 0.5 x 놓 3.14x Co.10753 9.81 BF₂ = 38.27N Total buoyant force, BFI - BF, + BF2 68.89 N Answer.