Kavita Kaushik
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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:

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NCERT Solutions for Class 9 Maths Chapter 13
Important NCERT Questions
Surface Areas and Volumes
NCERT Books for Session 2022-2023
CBSE Board, UP Board and Other state Boards
EXERCISE 13.8
Page No:236
Questions No:9

(i) radius r ′ of the new sphere,

(ii) ratio of S and S′.

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2 Answers

  1. (i) Given that the radius of solid sphere is r and the radius of new sphere is r’.
    Volume of solid sphere = 4/3 πr³
    Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³
    According to question :
    Volume of new sphere = volume of 27 solid spheres
    ⇒ 4/3π (r’)³ = 27 × 4/3πr³
    ⇒ (r’)³ = 27 × r³
    ⇒ r’ = 3 × r
    Hence, the radius of new sphere is 3r.

    (ii) Surface area of new sphere S’ = 4πr²
    Surface area of new sphere S’ = 4π(r’)² = 4π(3r)² = 36πr²
    Therefore,
    s/s’ = 4πr²/36πr² = 1/9
    Hence, the ratio of S and S’ is 1:9.

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  2. Get Hindi Medium and English Medium NCERT Solution for Class 9 Maths to download.
    Please follow the link to visit website for first and second term exams solutions.
    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/chapter-13/

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