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Ex 13.8, 9 - Twenty seven solid iron spheres, each of radius - Volume Of Sphere

  1. Chapter 13 Class 9 Surface Areas and Volumes
  2. Serial order wise
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Ex 13.8, 9 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 radius r ′ of the new sphere, Whenever one shape is melted to another shape, volume remains constant Volume of 27 sphere of radius r = Volume of big sphere of radius r’ 27 × (4/3 πr3) = 4/3 π(r’)3 27r3 = r’3 r’3 = 27r3 r’3 = (3r)3 r ‘ = 3r. Ex 13.8, 9 (ii) ratio of S and S′. 𝑆/𝑆′ = (𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝑤𝑖𝑡ℎ 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟)/(𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒 𝑤𝑖𝑡ℎ 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟^′ ) = 4𝜋𝑟2/(4𝜋(𝑟^′ )^2 ) = 𝑟2/( (𝑟^′ )^2 ) = 𝑟2/(3𝑟)2 = 𝑟2/9𝑟2 = 1/9 Therefore, ratio of S & S’ is 1 : 9

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