# Ex 9.4, 19 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 10, 2019 by Teachoo

Last updated at Dec. 10, 2019 by Teachoo

Transcript

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

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

Ex 9.4, 8

Ex 9.4, 9 Important

Ex 9.4, 10 Important

Ex 9.4, 11

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

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

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.