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 a

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

TestHub · Accepted Answer

Let surface area of the spherical balloon $A = 4 π r^{2}$

$\frac{d A}{d t} = 8 π r \frac{d r}{d t} = k$ (let) $. . . (1)$

On integrating on both sides w.r.t $t$ , we get

$4 π r^{2} = k t + C$ .

Given that, at $t = 0, r = 3$ .

$\Rightarrow 36 π = C$

Also given that, at $t = 5, r = 7$

$\Rightarrow 4 π \times 49 = 5 k + 36 π$

$\Rightarrow 5 k = 4 π (49 - 9)$ $\Rightarrow 5 k = 4 π \times 40$

$\Rightarrow k = 32 π$

On substituting $k$ value in equation $(1)$ we get, $4 π r^{2} = 32 π t + 36 π$

$\Rightarrow r^{2} = 8 t + 9$

Given $t = 9$ .

$\Rightarrow r^{2} = 81 \Rightarrow r = 9$ .

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