A 60 cm tall bucket is filled half way with water and then topped off with oil h

Question

A 60 cm tall bucket is filled half way with water and then topped off with oil having a density of 0.76 kg per liter. A ball has a mass of 420 grams and a radius of 6 cm. It is tied to a string, which is tied to the bottom of the bucket. The ball naturally wants to float to the top of the bucket, but the string holds it submerged exactly halfway below the oil-water interface, as shown. (a) Find the gauge pressure at the top of the ball. (b) Find the gauge pressure at the bottom of the ball. (c) Find the tension in the cable holding the ball submerged exactly at the center of the oil-water border.

Explanation / Answer

let

m = 420 g = 0.42 kg

rho_oil = 760 kg/m^3

rho_water = 1000 kg/m^3

r = 6 cm = 0.06 m

Volume of the sphere, V = (4/3)*pi*r^3

= (4/3)*pi*0.06^3

= 9.05*10^-4 m^3

a) gauge pressure at the top of the ball = rho_oil*g*h

= rho_oil*g*(H/2 - r)

= 760*9.8*(0.6/2 - 0.06)

= 1788 pa

b) gauge pressure at the bottom of the ball = rho_oil*g*H/2 + rho_water*g*r

= 760*9.8*0.6/2 + 1000*9.8*0.06

= 2822 pa

c) net force acting on the ball = 0

B - m*g - T = 0 (here buoynat force)

T = B - m*g

= weight of the fluid dispalced - m*g

= rho_oil*(V/2)*g + rho_water*(V/2)*g - m*g

= 760*(9.05*10^-4/2)*9.8 + 1000*(9.05*10^-4/2)*9.8 - 0.42*9.8

= 3.7 N

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