A solid metallic sphere of radius 10 cm is melted and recast into 125 identical spheres. What is the ratio of the surface area of the original sphere to the total surface area of 6 smaller spheres so formed?

49:108

25:6

109:84

25:96

APEDIA

NTPC Graduate Tier 1

2025 • 05 Jun, 2025 • Shift-3

A solid metallic sphere of radius 10 cm is melted and recast into 125 identical spheres. What is the ratio of the surface area of the original sphere to the total surface area of 6 smaller spheres so formed?

Correct Answer

25:6

[Volume Preservation Law]: During melting and recasting, the total physical volume remains strictly constant. The original sphere's volume (R=10) equates to the......

💡 Analysis & Explanation

[Volume Preservation Law] ▼

During melting and recasting, the total physical volume remains strictly constant. The original sphere's volume (R=10) equates to the combined volume of 125 tiny spheres (radius 'r').

[Radius Deduction Phase] ▼

Equation: (4/3)π(10)³ = 125 × (4/3)π(r)³. Canceling (4/3)π from both sides simplifies this to 1000 = 125r³. Dividing yields r³ = 8, which means the new radius 'r' is exactly 2 cm.

[Surface Area Calculation] ▼

The formula for a sphere's surface area is 4πr².
Original Area = 4 × π × (10)² = 400π.
Surface Area of ONE small sphere = 4 × π × (2)² = 16π.
The area of SIX such smaller spheres combined = 6 × 16π = 96π.

[Ratio Simplification] ▼

The requested ratio is 400π to 96π. Dropping Pi leaves 400:96. Dividing both numbers by their highest common factor (16) reduces the ratio neatly.

Conclusion ▼

The final simplified mathematical ratio is precisely 25:6.

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A solid metallic sphere of radius 10 cm is melted and recast into 125 identical spheres. What is the ratio of the surface area of the original sphere to the total surface area of 6 smaller spheres so formed?