A cylindrical tank with a radius of 5 feet and a height of 10 feet is filled with water. If a spherical ball of radius 2 feet is submerged, how much does the water level rise?

Tue, 05 May 2026 07:30:49 +0000

**How Does Water Level Rise When a Sphere Is Submerged in a Large Cylindrical Tank?**
*A cylindrical tank with a radius of 5 feet and a height of 10 feet is filled with water. If a spherical ball of radius 2 feet is submerged, how much does the water level rise? This question blends practical engineering insight with everyday curiosity—and lately, it’s been gaining attention across the U.S. as more people explore how fluid dynamics impact storage systems, design efficiency, and everyday problem-solving.* ### Why This Question Matters Now
In a time when water storage and space optimization shape everything from backyard rain systems to industrial warehouses, understanding how objects affect water levels is more relevant than ever. The tank dimensions mentioned—5 feet in radius and 10 feet tall—represent common capacities found in municipal infrastructure, agricultural tanks, and even large-scale consumer products. People naturally ask: how much water shifts when a foreign object displaces space inside? This isn’t just theoretical—it’s essential knowledge for engineers, designers, and consumers alike. ### The Science Behind the Rise
When the spherical ball of radius 2 feet is fully submerged, it displaces a volume of water equal to its own total volume. Using the formula for the volume of a sphere—V = (4/3)πr³—we calculate the displaced water:
V = (4/3) × π × (2)³ = (4/3) × π × 8 ≈ 33.51 cubic feet. Since the tank’s cross-sectional area is π × r² = π × 25 ≈ 78.54 square feet, the rise in water level is simply volume divided by area:
Rise = Displaced Volume ÷ Base Area ≈ 33.51 ÷ 78.54 ≈ 0.426 feet.