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

Last updated at Dec. 10, 2019 by Teachoo

Ex 9.4

Ex 9.4, 1
Important

Ex 9.4, 2

Ex 9.4, 3

Ex 9.4, 4 Important

Ex 9.4, 5

Ex 9.4, 6

Ex 9.4, 7 Important

Ex 9.4, 8

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Ex 9.4, 11 Important

Ex 9.4, 12

Ex 9.4, 13

Ex 9.4, 14

Ex 9.4, 15 Important

Ex 9.4, 16

Ex 9.4, 17 Important

Ex 9.4, 18

Ex 9.4, 19 Important You are here

Ex 9.4, 20 Important

Ex 9.4, 21

Ex 9.4, 22 Important

Ex 9.4, 23 (MCQ)

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Dec. 10, 2019 by Teachoo

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 𝑡 seconds . Let volume of the spherical balloon = V V = 4/3 𝜋𝑟^3 Since volume changes at 𝑎 constant rate, ∴ 𝑑𝑉/𝑑𝑡=𝑘 𝑑/𝑑𝑡 (4/3 " " 𝜋𝑟^3 ) = k 4/3 𝜋 (𝑑𝑟^3)/𝑑𝑡 = k 4/3 𝜋 3r2 𝑑𝑟/𝑑𝑡 = k 4𝜋r2 𝑑𝑟/𝑑𝑡 = k 4𝜋r2 𝑑𝑟 = k dt Integrating both sides 4𝜋∫1▒〖"r2 " 𝑑𝑟" = k " 〗 ∫1▒𝑑𝑡 (4𝜋𝑟^3)/3 = kt + C At T = 0, r = 3 units 〖4𝜋(3)〗^3/3 = k(0) + C 〖"4" 𝜋(3)〗^2 = C "4" 𝜋(9) = C 36𝜋 = C C = 36π …(1) Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36π in equation (1) (4𝜋𝑟^3)/3 = kt + C 〖4𝜋(6)〗^3/3 = 3k + 36𝜋 (4𝜋(216))/3 = 3k + 36𝜋 288𝜋 = "3k + 36" 𝜋 252𝜋 = "3k " 84𝜋 = "k" k = 84π Putting value of k & C in equation (1) (4𝜋𝑟^3)/3 = kt + C 〖4𝜋𝑟〗^3/3 = 84𝜋𝑡 + 36𝜋 〖"4" 𝜋𝑟〗^3 = 3[84𝜋𝑡+36𝜋] 〖"4" 𝜋𝑟〗^3 = 252𝜋t + 108𝜋 𝑟^3 = (252𝜋𝑡+ 108𝜋" " )/4𝜋 𝑟^3 = 63t + 27 r = ("63t + 27" )^(1/3) ∴ Radius of the balloon after t seconds is ("63t + 27" )^(𝟏/𝟑) units