Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r′ of the new sphere,  (ii) ratio of S and S′

Solution:

Since 27 solid iron spheres are melted to form a single solid sphere, the volume of the newly formed sphere will be equal to the volume of 27 solid iron spheres together.

The surface area of a sphere = 4πr²

The volume of a sphere = 4/3πr³

Therefore, The volume of 27 solid spheres with radius r = 27 × [4/3πr³] = 36πr³ ---------------- (1)

Also, the volume of the new sphere with radius r' = 4/3πr'³----------------- (2)

(i) Volume of the new sphere = Volume of 27 solid spheres

(4/3) πr'³ = 36πr³ [From equation (1) and (2)]

⇒ r'³ = 36πr³ × 3/4π

⇒ r'³ = 27r³

⇒ r' = ∛27r³

r' = 3r

Radius of the new sphere, r' = 3r

(ii) Ratio of S and S′

Now, surface area of each iron sphere, S = 4πr²

Surface area of the new sphere, S' = 4πr'² = 4π(3r)² = 36πr²

The ratio of the S and S’ = 4πr²/36πr² = 1/9

Hence, the radius of new sphere is 3r and the ratio of S and S’ is 1:9.

☛ Check: NCERT Solutions for Class 9 Maths Chapter 13

Video Solution:

Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the i) radius r′ of the new sphere, ii) ratio of S and S′

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Question 9

Summary:

It is given that twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. We have found that the radius r of the new sphere is 3r and the ratio of S and S’ is 1:9

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