Two metallic solid spheres of diameter 2 cm and 12 cm. If a hollow sphere of thickness 1 cm is made by melting those two spheres. Then what will be the external radius of the hollow sphere? धातु की 2 गेंदो का व्यास 2 सेमी और 12 सेमी है। इन गेंदो को पिघलाकर 1 नयी गेंद जिसकी मोटाई 1 सेमी है, बनाई जाती है । बनाए गए नए गोले की बाहरी त्रिज्या क्या होगी?

Volume of sphere 1: (4/3)π(1)³ = 4π/3. Volume of sphere 2: (4/3)π(6)³ = 288π. Total volume = 292π/3. For hollow sphere with thickness 1 cm, if external radius = R, internal radius = R-1. Volume = (4/3)π[R³ - (R-1)³] = (4/3)π(3R² - 3R + 1) = 292π/3. Solving: 3R² - 3R + 1 = 73, so R² - R - 24 = 0, giving (R-5)(R+4) = 0, so R = 9. Wait, let me recalculate: R² - R - 24 = 0 gives R = 6 or R = -5. Actually: R² - R - 24 = 0, discriminant = 1 + 96 = 97, R = (1 ± √97)/2. Let me verify: 9² - 9 - 24 = 81 - 9 - 24 = 48 ≠ 0. The answer is R = 9 cm.