A ping-pong ball has an average density of 0.140 g/cm3 and a diameter of 3.80 cm. What force would be required to keep the ball completely submerged under water?

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To tackle this problem, start by identifying the key forces on the ping-pong ball when it is fully submerged in water.
Option A: 0.92N. This value is much larger than the buoyant force expected for a small ball; it would imply a downward force that far exceeds the upward buoyancy, which is not consistent with the ball’s small size and density.
Option B: 0.66N. While closer, this still overestimates the needed force; the calculation must account for both the buoyant force and the ball’s own weight, and then take their difference if you’re applying a downward force to maintain submersion.
Option C: 0.54N. This is also too high; it indicates an incorrect balancing of forces, likely by miscomputing either the buoyant force or the weight component.
Option D: 0.24N. This matches the correct balance: the buoyant force equals the weight of water displaced, which for a sphere of diameter 3.80 cm gives a volume, and thus a buoyant force around 0.28 N; the ball’s weight is smaller, about 0.039 N, so the needed downward force is B − W ≈ 0.281 N − 0.039 N ≈ 0.242 N, i.e., roughly 0.24 N.
Option E: 1.11N. This is far too large for a ping-pong ball; it would require an enormous downward force, inconsistent with the physics of such a small object in water.

Now, let’s walk through the relevant calculations to see how the 0.24 N value arises. The buoyant force equals the weight of the displaced water, so use Archimedes’ principle: F_b = ρ_water × V_ball × g. The ball is a sphere with diameter 3.80 cm, so radius r = 1.90 cm = 0.019 m. The volume is V = (4/3)πr^3, which evaluates to about 28.75 cm^3 or 2.875×10^−5 m^3. The density of water is about 1.00 g/cm^3, i.e., 1000 kg/m^3, so the mass of displaced water is roughly 0.02875 kg, giving F_b ≈ 0.02875 kg × 9.8 m/s^2 ≈ 0.281 N.
The ball’s own weight is W = m_ball × g, where m_ball = ρ_ball × V_ball. With ρ_ball = 0.140 g/cm^3 (140 kg/m^3) and V ≈ 2.875×10^−5 m^3, m_ball ≈ 4.03×10^−3 kg, so W ≈ 0.0395 N.
To keep the ball submerged and stationary, the downward external force F must balance the excess buoyant force over weight: F = F_b − W ≈ 0.281 N − 0.0395 N ≈ 0.2415 N, rounding to about 0.24 N.

In summary, the option that aligns with this calculation is the one giving a downward force around 0.24 N; the other values do not reflect the correct difference between buoyancy and weight for this ball.

2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 2205 Final Exam

A ping-pong ball has an average density of 0.140 g/cm3 and a diameter of 3.80 cm. What force would be required to keep the ball completely submerged under water?

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