Let rAl represent the density of aluminum and rFe that S of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equal-arm balance.
Answer
🔍 Radius of an Aluminum Sphere That Balances an Iron Sphere
🧠 Concept Overview
To determine the radius of a solid aluminum sphere that balances a solid iron sphere on an equal-arm balance, we use the principle that both sides of the balance must have equal mass.
⚖️ Equal Mass Condition:
Mass of aluminum sphere = Mass of iron sphere
Mass of aluminum sphere = Mass of iron sphere
📐 Step-by-Step Derivation
The mass of a sphere is given by the product of its density and volume.
✅ Volume of a Sphere:
V = (4/3) × π × r³✅ Let:
ρAl= Density of AluminumρFe= Density of IronrAl= Radius of Aluminum sphere (unknown)rFe= Radius of Iron sphere (given)
🧾 Set Up the Equation:
Using Mass = Density × Volume,
ρAl × (4/3)πrAl³ = ρFe × (4/3)πrFe³Simplify both sides (cancel out common terms):
ρAl × rAl³ = ρFe × rFe³📌 Solving for rAl:
rAl³ = (ρFe / ρAl) × rFe³ rAl = [ (ρFe / ρAl) × rFe³ ]1/3
✅ Final Answer:
rAl = rFe × (ρFe / ρAl)1/3
🧪 Example Calculation
Suppose:
- ρFe = 7.87 g/cm³
- ρAl = 2.70 g/cm³
- rFe = 6 cm
Plug into the formula:
rAl = 6 × (7.87 / 2.70)1/3Calculate cube root:
rAl ≈ 6 × (2.915)1/3 ≈ 6 × 1.42 ≈ 8.52 cm
🔢 Therefore, the aluminum sphere must have a radius of approximately 8.52 cm to balance the iron sphere of 6 cm.
You can replace the densities and radius values with any valid numbers for different materials or sizes.