Calculating the Rise in Water Level When a Metal Sphere is Immersed in a Cylinder

The scenario presented involves a metal sphere and a right cylindrical container. The question at hand is: how much will the water level rise if a metal sphere with a radius of 4 cm is fully submerged in a cylindrical container with a base radius of 6 cm?

Understanding the Problem Statement

The problem statement is as follows: A metal sphere of radius 4 cm is immersed in water inside a right cylindrical container. If the base radius of the container is 6 cm, what is the rise in water level?

Volume of the Metal Sphere

The volume of a sphere is given by the formula:

V_s  frac{4}{3}pi r^3

Given the radius of the sphere ( r 4 ) cm, we substitute this into the formula:

V_s  frac{4}{3}pi (4)^3  frac{4}{3}pi 64  frac{256}{3}pi approx 268.08 text{ cm}^3

Volume of Water Displaced

When the sphere is fully immersed, it displaces an equal volume of water equal to its volume. This displaced volume will increase the water level in the cylindrical container. The volume of a cylinder is given by:

V_c  pi R^2h

Where ( R 6 ) cm (the radius of the base of the cylinder) and ( h ) cm is the rise in water level that we need to determine.

The displaced volume of water corresponds to the volume of the sphere:

pi R^2h  frac{4}{3}pi r^3

Solving for the Rise in Water Level

We need to solve for ( h ) in the equation:

pi (6)^2h  frac{4}{3}pi (4)^3

Canceling out the common factors ( pi ) and simplifying, we get:

(6)^2h  frac{4}{3}(4)^3

Calculating the right-hand side:

(6)^2h  frac{4}{3} cdot 64  frac{256}{3}

Now, solve for ( h ):

h  frac{256/3}{(6)^2}  frac{256}{3 cdot 36}  frac{256}{108}  frac{64}{27} approx 2.37 text{ cm}

Conclusion

The rise in water level when the metal sphere with a radius of 4 cm is fully submerged in a cylindrical container with a base radius of 6 cm is approximately 2.37 cm.

Final Equation

The final equation for the rise in water level ( h ) is:

h  frac{4r^3}{3R^2}

Where ( r 4 ) cm and ( R 6 ) cm.

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