Question

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calculate the heaviest bowling ball that will float in a fluid of density 1.100.

Answer 1

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = $\frac{4}{3}\pi r^3$

The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

Weight will be $6.1328\times 9.81=60.16284\ N$

Answer 2

The weight of the bowling ball that will float in the fluid is 0.06 N.

Weight of the ball that will float in the fluid

The weight of the ball that will float in the given fluid is calculated as follows;

W = mg

where;

• m is mass of the object
• g is acceleration due to gravity

W = (ρV)g

where;

• ρ is density of the fluid = 1.100 kg/m³
• V is volume of the ball

Volume of the spherical ball

V = ⁴/₃πr³

V = ⁴/₃ x π x (0.11)³

V = 5.575  x 10⁻³ m³

W = (1.1 x 5.575  x 10⁻³) x (9.8)

W = 0.06 N

Thus, the weight of the bowling ball that will float in the fluid is 0.06 N.

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