Answer :

Sum of the volumes of 3 gives spheres.

\( = \ \frac{4}{3} \pi ((r_1)^3 \ + \ (r_2)^3 \ + \ (r_3)^3) \)

\(= \ \frac{4}{3} \pi (6^3 \ + \ 8^3 \ + \ 10^3) \)

\(= \ \frac{4}{3} \pi (216 \ + \ 512 \ + \ 1000) \)

\(= \ \frac{4}{3} \pi (1728) \) cm^{3}

Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.

\( \Rightarrow \ \frac{4}{3} \pi R^3 \ = \ \frac{4}{3} \pi (1728) \)

\(\Rightarrow \ R^3 \ = \ 1728 \ => \ R \ = \ 12 \) cm

\(\therefore \) The radius of the resulting sphere is 12 cm.

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- NCERT solutions for class 10 maths chapter 1 Real Numbers
- NCERT solutions for class 10 maths chapter 2 Polynomials
- NCERT solutions for class 10 maths chapter 3 Pair of linear equations in two variables
- NCERT solutions for class 10 maths chapter 4 Quadratic Equations
- NCERT solutions for class 10 maths chapter 5 Arithmetic Progressions
- NCERT solutions for class 10 maths chapter 6 Triangles
- NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
- NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
- NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
- NCERT solutions for class 10 maths chapter 10 Circles
- NCERT solutions for class 10 maths chapter 11 Constructions
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- NCERT solutions for class 10 maths chapter 14 Statistics
- NCERT solutions for class 10 maths chapter 15 Probability

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- NCERT solutions for class 10 science chapter 2 Acids, Bases and Salts
- NCERT solutions for class 10 science chapter 3 Metals and Non Metals
- NCERT solutions for class 10 science chapter 4 Carbon and its Compounds
- NCERT solutions for class 10 science chapter 5 Periodic Classification of Elements
- NCERT solutions for class 10 science chapter 6 Life Processes
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