The sum of the radii of two spheres is 10 cm and the sum of their volu

1.	The sum of the radii of two spheres is 10 cm and the sum of their volumes is 880 cm3. What will be the product of their radii ?(a) 21 (b) \(26\frac13\)(c) \(33\frac13\)(d) 70
Answer» (b) \(26\frac13\) Given, r₁ + r₂ = 10 ...(i) and \(\frac43πr^3_1+\frac43πr^3_2= 880\) ⇒ \(\frac43π(r^3_1+r^3_2) = 880\) ⇒ \(r^3_1+r^3_2=\frac{880\times3\times7}{4\times22}=210\) ...(ii) Taking the cube of both the sides of eqn. (i), we have (r₁ + r₂)³ = 1000 ⇒ \(r^3_1+r^3_2+\) 3r₁ r₂ (r₁ + r₂) = 1000 ⇒ 210 + 3r₁ r₂ (10) = 1000 ⇒ 30r₁ r₂ = 1000 – 210 = 790 ⇒ r₁ r₂ = \(\frac{790}{30}=26\frac13.\)

The sum of the radii of two spheres is 10 cm and the sum of their volumes is 880 cm3. What will be the product of their radii ?(a) 21 (b) \(26\frac13\)(c) \(33\frac13\)(d) 70

Answer»

(b) \(26\frac13\)

Given, r₁ + r₂ = 10 ...(i)

and \(\frac43πr^3_1+\frac43πr^3_2= 880\)

⇒ \(\frac43π(r^3_1+r^3_2) = 880\) ⇒ \(r^3_1+r^3_2=\frac{880\times3\times7}{4\times22}=210\) ...(ii)

Taking the cube of both the sides of eqn. (i), we have

(r₁ + r₂)³ = 1000 ⇒ \(r^3_1+r^3_2+\) 3r₁ r₂ (r₁ + r₂) = 1000

⇒ 210 + 3r₁ r₂ (10) = 1000

⇒ 30r₁ r₂ = 1000 – 210 = 790 ⇒ r₁ r₂ = \(\frac{790}{30}=26\frac13.\)