The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds,, it becomes 7 units, then its radius…

Question

The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds,, it becomes 7 units, then its radius after 9 seconds is :

Accepted Answer

Correct answer: A. (a) Let r be the radius of spherical balloon S = Surface area S=4 π r^2 d Sd t=8 π r× d rd t=k(constant) 4 π r^2=k t+C(C is constant of integration) For t = 0,r = 3 ⇒36π = C For t =5,r = 7 ⇒K = 32π 4πr^2=32πt + 36π r^2=8t + 9 for t = 9 r^2 = 81 r = 9

$The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds,, it becomes 7 units, then its radius after 9 seconds is :$

Solution

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