A right circular cylinder is partially filled with water. Two solid iron spherical balls are completely immersed so that the height of the water in the cylinder rises by 4 cm. If the radius of one ball is half of the radius of the other and the diameter of the cylinder is 18 cm, what are the radii (in cm) of the two spherical balls?

Introduction / Context:
This is a classic volume displacement question. When solid spheres are immersed in water inside a cylindrical container, the water level rises by an amount directly related to the total volume of the immersed spheres. Using the volume formulas for spheres and the cylinder, we can determine the radii of the two spherical balls.

Given Data / Assumptions:
• Radius of one sphere is half of the radius of the other.
• Diameter of the cylinder is 18 cm, so its radius Rc = 9 cm.
• Rise in water level inside the cylinder is Δh = 4 cm.
• Water rise is caused only by the volume of the two spheres, with no overflow.
• All solids are perfectly rigid and water does not compress.

Concept / Approach:
The additional volume of water occupied equals the combined volume of the two spheres. For the cylinder, additional volume is base area × rise in height. For spheres, the volume formula is V = (4/3)πr3. Let radii be r and 2r. Sum the two spherical volumes and equate this to the cylindrical volume to solve for r.

Step-by-Step Solution:
Step 1: Volume increase in the cylinder = πRc2Δh = π × 92 × 4. Step 2: Compute Rc2 = 92 = 81, so cylinder volume increase = π × 81 × 4 = 324π cubic centimetres. Step 3: Let small sphere radius be r. Then larger sphere radius is 2r. Step 4: Volume of small sphere = (4/3)πr3. Step 5: Volume of large sphere = (4/3)π(2r)3 = (4/3)π × 8r3. Step 6: Total spherical volume = (4/3)πr3 + (4/3)π × 8r3 = (4/3)π × 9r3 = 12πr3. Step 7: Equate volumes: 12πr3 = 324π, so r3 = 324 / 12 = 27. Step 8: Cube root of 27 is 3, so r = 3 cm and the larger radius is 2r = 6 cm.

Verification / Alternative check:
Compute the spherical volumes numerically. Vsmall = (4/3)π × 27 = 36π. Vlarge = (4/3)π × 216 = 288π. Sum = 324π, which matches the cylinder volume rise 324π. This confirms that the radii 3 cm and 6 cm are correct.

Why Other Options Are Wrong:
• 6 cm and 12 cm give too large a combined volume, causing a greater rise than 4 cm.
• 4 cm and 8 cm or 2 cm and 4 cm give total volumes that do not equal 324π and therefore do not match the observed water rise.

Common Pitfalls:
Students sometimes apply the radius ratio but forget that volume depends on the cube of the radius. Others may miscalculate the cylinder volume by using diameter instead of radius. Being careful with radius values and cube operations is crucial.

Final Answer:
The radii of the spherical balls are 3 cm and 6 cm.