Sphere immersed in cylindrical vessel — compute water rise: A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is completely immersed. By how much (in cm) does the water level rise?

Introduction / Context:Again use volume displacement: the water level rise times the cross-sectional area of the cylinder equals the volume of the immersed sphere.

Given Data / Assumptions:

• Cylinder radius R = 4 cm
• Sphere radius r = 3 cm

Concept / Approach:πR^2 * h_rise = (4/3)πr^3 ⇒ h_rise = (4/3) r^3 / R^2.

Step-by-Step Solution:h_rise = [(4/3)*27] / 16 = 36/16 = 9/4 cm

Verification / Alternative check:Numerically 9/4 = 2.25 cm; units are consistent.

Why Other Options Are Wrong:4/9 cm and 2/3 cm invert or mis-square radii; 4/5 cm is unrelated.

Common Pitfalls:Dropping the π factors prematurely without keeping track (they cancel, but after setup).

Final Answer:9/4 cm