The volume of spherical balloon is increasing at the rate of $4 \text{ cm}^3/ \text{sec}$. The rate of increase of its surface area, when the radius is 3cm will be :-

The volume of a spherical balloon is increasing at $4 \text{ cm}^3/\text{sec}$.

Given: $\dfrac{dV}{dt} = 4 \text{ cm}^3/\text{sec}$ and $r = 3 \text{ cm}$

For a sphere:

Volume: $V = \dfrac{4}{3}\pi r^3$

Surface Area: $A = 4\pi r^2$

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Differentiating the volume with respect to time:

$\dfrac{dV}{dt} = \dfrac{4}{3}\pi \times 3r^2 \times \dfrac{dr}{dt}$

$\dfrac{dV}{dt} = 4\pi r^2 \times \dfrac{dr}{dt}$

Substituting $\dfrac{dV}{dt} = 4$ and $r = 3$:

$4 = 4\pi (3)^2 \times \dfrac{dr}{dt}$

$4 = 4\pi (9) \times \dfrac{dr}{dt}$

$4 = 36\pi \times \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{4}{36\pi}$

$\dfrac{dr}{dt} = \dfrac{1}{9\pi} \text{ cm/sec}$

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Differentiating the surface area with respect to time:

$\dfrac{dA}{dt} = 4\pi \times 2r \times \dfrac{dr}{dt}$

$\dfrac{dA}{dt} = 8\pi r \times \dfrac{dr}{dt}$

Substituting $r = 3$ and $\dfrac{dr}{dt} = \dfrac{1}{9\pi}$:

$\dfrac{dA}{dt} = 8\pi \times 3 \times \dfrac{1}{9\pi}$

$\dfrac{dA}{dt} = \dfrac{24\pi}{9\pi}$

$\dfrac{dA}{dt} = \dfrac{24}{9}$

$\dfrac{dA}{dt} = \dfrac{8}{3} \text{ cm}^2/\text{sec}$

Therefore, the rate of increase of surface area is $\dfrac{8}{3}$ cm²/sec or approximately $2.67$ cm²/sec.