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An air bubble of radius 1 cm in water experiences an upward acceleration of 9.8 cm/s². The density of water is 1 g/cm³ and water exerts negligible drag on the bubble. Taking g = 980 cm/s², what is the mass of the bubble?
- 3.15 g
- 4.51 g
- 4.15 g
- 1.52 g
Correct answer: 4.15 g
Solution
Volume V = (4/3)*pi*r³ = (4/3)*pi*(1)³ = 4.1888 cm³. Buoyancy = rho_w*V*g = 1*4.1888*980. Newton: rho_w*V*g - m*g = m*a, so m = rho_w*V*g/(g+a) = (4.1888*980)/(980+9.8) = 4105/989.8 ~ 4.15 g.
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