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Q.

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 :

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a

11

b

12

c

10

d

9

answer is A.

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Detailed Solution

Let r be the radius of spherical balloon
S = Surface area
S = 4πr²
$\frac{dS}{dt} = 8 πr \times \frac{dr}{dt} = k (constant)$
$4 {πr}^{2} = kt + C$ (C is constant of integration)
For t = 0 , r = 3 ⇒ 36π = C
For t = 5 , r = 7 ⇒K = 32π
$\begin{array}{l} 4 {πr}^{2} = 32 πt + 36 π \\ r^{2} = 8 t + 9 \\ for t = 9 \\ r^{2} = 81 \\ r = 9 \end{array}$

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