Question

Bowling balls are roughly the same size but come in a variety of weights. Given that the official radius a bowling ball should have is roughly 0.111 m, calculate the weight of the heaviest bowling ball that will float in a fluid of density 1.100×103 kg/m3.

I used m = ρV = (1000 kg/m³) x [4/3 π (0.111)³m³] = 5.7287 Kg
I then multiplied by g to get the weight in N, which gave me 56.199 N

This was incorrect. Can anyone tell me why?

Answer 1

V = 5.7287 * 10*-3 meters^3
mass of fluid of that volume = 1.1 *10^3 * 5.7287*10^-3
= 6.3016 kg
the mass of the ball must be the mass of the fluid displaced by its volume to
be neutrally buoyant. You used pure water. This stuff is 10% heavier.