# Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 10, 2019 by Teachoo

Last updated at Dec. 10, 2019 by Teachoo

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Ex 9.4, 19 The volume of spherical balloon being inflated changes at a constant rate. if initially its radius is 3 units and after 3 seconds it is 6 units . Find the radius of balloon after ๐ก seconds . Let volume of the spherical balloon = V V = 4/3 ๐๐^3 Since volume changes at ๐ constant rate, โด ๐๐/๐๐ก=๐ ๐/๐๐ก (4/3 " " ๐๐^3 ) = k 4/3 ๐ (๐๐^3)/๐๐ก = k 4/3 ๐ 3r2 ๐๐/๐๐ก = k 4๐r2 ๐๐/๐๐ก = k 4๐r2 ๐๐ = k dt Integrating both sides 4๐โซ1โใ"r2 " ๐๐" = k " ใ โซ1โ๐๐ก (4๐๐^3)/3 = kt + C At T = 0, r = 3 units ใ4๐(3)ใ^3/3 = k(0) + C ใ"4" ๐(3)ใ^2 = C "4" ๐(9) = C 36๐ = C C = 36ฯ โฆ(1) Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36ฯ in equation (1) (4๐๐^3)/3 = kt + C ใ4๐(6)ใ^3/3 = 3k + 36๐ (4๐(216))/3 = 3k + 36๐ 288๐ = "3k + 36" ๐ 252๐ = "3k " 84๐ = "k" k = 84ฯ Putting value of k & C in equation (1) (4๐๐^3)/3 = kt + C ใ4๐๐ใ^3/3 = 84๐๐ก + 36๐ ใ"4" ๐๐ใ^3 = 3[84๐๐ก+36๐] ใ"4" ๐๐ใ^3 = 252๐t + 108๐ ๐^3 = (252๐๐ก+ 108๐" " )/4๐ ๐^3 = 63t + 27 r = ("63t + 27" )^(1/3) โด Radius of the balloon after t seconds is ("63t + 27" )^(๐/๐) units

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Chapter 9 Class 12 Differential Equations (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.