A Ping-Pong ball has a diameter of 4.78 cm
and average density of 0.0843 g/cm3
.
What force would be required to hold it
completely submerged under water? The acceleration of gravity is 9.8 m/s
2
.
Answer in units of N.


To find the force required to hold the Ping-Pong ball completely submerged under water, we can use the principle of buoyancy. The buoyant force exerted on the ball by the water is equal to the weight of the water displaced by the ball. The force required to hold the ball submerged is the difference between the buoyant force and the weight of the ball.

Here are the steps:

### 1. Volume of the Ping-Pong ball
The ball is a sphere, and the formula for the volume \( V \) of a sphere is:

\[
V = \frac{4}{3} \pi r^3
\]

Where \( r \) is the radius. Since the diameter is 4.78 cm, the radius is:

\[
r = \frac{4.78}{2} = 2.39 \, \text{cm}
\]

Now, calculate the volume:

\[
V = \frac{4}{3} \pi (2.39)^3 = \frac{4}{3} \pi (13.637) \approx 57.24 \, \text{cm}^3
\]

### 2. Buoyant force (Archimedes' Principle)
The buoyant force \( F_b \) is the weight of the water displaced by the ball, which is:

\[
F_b = \rho_{\text{water}} \cdot V \cdot g
\]

Where:
- \( \rho_{\text{water}} \) is the density of water (\( \approx 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \)),
- \( V \) is the volume of the ball in \( \text{m}^3 \),
- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).

First, convert the volume of the ball from \( \text{cm}^3 \) to \( \text{m}^3 \):

\[
V = 57.24 \, \text{cm}^3 = 57.24 \times 10^{-6} \, \text{m}^3 = 5.724 \times 10^{-5} \, \text{m}^3
\]

Now calculate the buoyant force:

\[
F_b = 1000 \, \text{kg/m}^3 \times 5.724 \times 10^{-5} \, \text{m}^3 \times 9.8 \, \text{m/s}^2
\]
\[
F_b = 0.561 \, \text{N}
\]

### 3. Weight of the Ping-Pong ball
The weight \( W \) of the ball is:

\[
W = \rho_{\text{ball}} \cdot V \cdot g
\]

Where \( \rho_{\text{ball}} \) is the density of the ball (\( 0.0843 \, \text{g/cm}^3 = 84.3 \, \text{kg/m}^3 \)).

Now, calculate the weight:

\[
W = 84.3 \, \text{kg/m}^3 \times 5.724 \times 10^{-5} \, \text{m}^3 \times 9.8 \, \text{m/s}^2
\]
\[
W = 0.0473 \, \text{N}
\]

### 4. Force required to hold the ball submerged
The net force required is the difference between the buoyant force and the weight of the ball:

\[
F_{\text{required}} = F_b - W = 0.561 \, \text{N} - 0.0473 \, \text{N}
\]
\[
F_{\text{required}} \approx 0.514 \, \text{N}
\]

Thus, the force required to hold the Ping-Pong ball completely submerged under water is approximately **0.514 N**.

---

Would you like further details on any step or have any questions?

### 5 Relative Questions:
1. How would the force change if the ball's density increased?
2. What if the ball was submerged in a different liquid like oil?
3. How does the buoyant force depend on the shape of the object?
4. How can Archimedes' principle be applied to floating objects?
5. What would happen if the ball were submerged on a planet with different gravity?

### Tip:
Always ensure that units are consistent (e.g., converting cm³ to m³) when calculating physical quantities, especially for buoyancy problems.

A Ping-Pong ball has a diameter of 4.78 cm and average density of 0.0843 g/cm³. What force would be required to hold it completely submerged under water? The acceleration of gravity is 9.8 m/s². Answer in units of N.

Learn how to calculate the force required to hold a Ping-Pong ball submerged under water using buoyancy principles. The problem involves the volume of a sphere, buoyant force, and weight calculations, suitable for students in Grades 10-12. Given data includes a diameter of 4.78 cm, density of 0.0843 g/cm³, and gravity of 9.8 m/s².

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