# Ex 9.3, 19 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.3

Ex 9.3, 1
Important

Ex 9.3, 2

Ex 9.3, 3

Ex 9.3, 4 Important

Ex 9.3, 5

Ex 9.3, 6

Ex 9.3, 7 Important

Ex 9.3, 8

Ex 9.3, 9 Important

Ex 9.3, 10 Important

Ex 9.3, 11 Important

Ex 9.3, 12

Ex 9.3, 13

Ex 9.3, 14

Ex 9.3, 15 Important

Ex 9.3, 16

Ex 9.3, 17 Important

Ex 9.3, 18

Ex 9.3, 19 Important You are here

Ex 9.3, 20 Important

Ex 9.3, 21

Ex 9.3, 22 Important

Ex 9.3, 23 (MCQ)

Last updated at April 16, 2024 by Teachoo

Ex 9.3, 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 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