Ex 9.4, 19 - Volume of spherical balloon being inflated changes - Variable separation - Statement given

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  1. Chapter 9 Class 12 Differential Equations
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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 š‘” seconds . Let volume of the spherical balloon = V V = 4ļ·®3ļ·Æ šœ‹ š‘Ÿļ·®3ļ·Æ Since volume changes at š‘Ž constant rate, āˆ“ š‘‘š‘‰ļ·®š‘‘š‘”ļ·Æ=š‘˜ š‘‘ļ·®š‘‘š‘”ļ·Æ 4ļ·®3ļ·Æ šœ‹ š‘Ÿļ·®3ļ·Æļ·Æ = k 4ļ·®3ļ·Æ šœ‹ š‘‘ š‘Ÿļ·®3ļ·Æļ·®š‘‘š‘”ļ·Æ = k 4ļ·®3ļ·Æ šœ‹ 3r2 š‘‘š‘Ÿļ·®š‘‘š‘”ļ·Æ = k 4šœ‹r2 š‘‘š‘Ÿļ·®š‘‘š‘”ļ·Æ = k 4šœ‹r2 š‘‘š‘Ÿ = k dt Integrating both sides 4šœ‹ ļ·®ļ·®r2 š‘‘š‘Ÿ = k ļ·Æ ļ·®ļ·®š‘‘š‘”ļ·Æ 4šœ‹ š‘Ÿļ·®3ļ·Æļ·®3ļ·Æ = kt + C At T = 0, r = 3 units 4šœ‹(3)ļ·®3ļ·Æļ·®3ļ·Æ = k(0) + C 4šœ‹(3)ļ·®2ļ·Æ = C 4šœ‹(9) = C 36šœ‹ = C C = 36Ļ€ Also, At T = 3, r = 6 units Putting t = 3, r = 6 and C = 36Ļ€ in equation (1) 4šœ‹(6)ļ·®3ļ·Æļ·®3ļ·Æ = 3k + 36šœ‹ 4šœ‹(216)ļ·®3ļ·Æ = 3k + 36šœ‹ 288šœ‹ = 3k + 36šœ‹ 252šœ‹ = 3k 84šœ‹ = k k = 84Ļ€ Putting value of k & C in equation (1) 4šœ‹š‘Ÿļ·®3ļ·Æļ·®3ļ·Æ = 84šœ‹š‘” + 36šœ‹ 4šœ‹š‘Ÿļ·®3ļ·Æ = 3 84šœ‹š‘”+36šœ‹ļ·Æ 4šœ‹š‘Ÿļ·®3ļ·Æ = 252šœ‹t + 108šœ‹ š‘Ÿļ·®3ļ·Æ = 252šœ‹š‘”+ 108šœ‹ ļ·®4šœ‹ļ·Æ š‘Ÿļ·®3ļ·Æ = 63t + 27 r = 63t + 27ļ·Æļ·® 1ļ·®3ļ·Æļ·Æ āˆ“ Radius of the balloon after 3 seconds is 63t + 27ļ·Æļ·® šŸļ·®šŸ‘ļ·Æļ·Æ units

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.