Water Level Increase Calculation

This solution explains how to find the approximate increase in water level when a solid spherical ball is immersed into a cylindrical container partially filled with water.

Cylinder and Sphere Dimensions

• Cylindrical container radius: $r_c = 20$ cm
• Initial water height: $h_i = 25$ cm
• Solid spherical ball radius: $r_s = 7$ cm

Volume Displacement Principle

When an object is submerged in a fluid, it pushes aside (displaces) a volume of fluid equal to the object's own volume. This displaced fluid causes the fluid level to rise. To find the increase in water level, we need to equate the volume of the submerged object (the sphere) to the volume of the water that rises in the container (which takes the shape of a cylinder).

Steps to Calculate Water Level Rise

1. Calculate the volume of the spherical ball. The formula for the volume of a sphere is $V_s = \frac{4}{3}\pi r_s^3$.
2. Determine the volume of the displaced water. This volume is equal to the volume of the sphere since the ball is fully immersed.
3. Relate displaced volume to water level rise. The displaced water occupies a cylindrical shape within the container. If the increase in water level is $\Delta h$, the volume of this rise is $V_{disp} = \pi r_c^2 \Delta h$.
4. Equate volumes and solve for $\Delta h$. Since $V_{disp} = V_s$, we have $\pi r_c^2 \Delta h = \frac{4}{3}\pi r_s^3$. Solving for $\Delta h$ gives $\Delta h = \frac{4 r_s^3}{3 r_c^2}$.

Applying the Formula with Values

Now, let's substitute the given values into the formula for $\Delta h$:

• $r_s = 7$ cm
• $r_c = 20$ cm

The formula becomes:

$\Delta h = \frac{4 \times (7 \text{ cm})^3}{3 \times (20 \text{ cm})^2}$

First, calculate the cubes and squares:

• $(7 \text{ cm})^3 = 343 \text{ cm}^3$
• $(20 \text{ cm})^2 = 400 \text{ cm}^2$

Now substitute these back into the equation:

$\Delta h = \frac{4 \times 343 \text{ cm}^3}{3 \times 400 \text{ cm}^2}$

$\Delta h = \frac{1372 \text{ cm}^3}{1200 \text{ cm}^2}$

Finally, perform the division:

$\Delta h \approx 1.1433$ cm

Approximate Water Level Rise

The approximate increase in the water level in the cylindrical container after the solid spherical ball is fully immersed is approximately $1.14$ cm.