The volume of a spherical balloon is increasing at the rate of 6 cm^3…

The correct answer is Option (3) → 6 \(\mathrm{cm}^{2}\)/sec

For a sphere of radius r:

    \( \displaystyle V = \frac{4}{3}\pi r^{3},\quad S = 4\pi r^{2} \)

Differentiating V with respect to time (rate of change):

    \( \displaystyle \frac{dV}{dt} = 4\pi r^{2}\,\frac{dr}{dt} \)

At r = 2 with \(\frac{dV}{dt} = 6\):

    \( \displaystyle 6 = 16\pi\,\frac{dr}{dt} \)

    \( \displaystyle \frac{dr}{dt} = \frac{3}{8\pi} \)

Differentiating S:

    \( \displaystyle \frac{dS}{dt} = 8\pi r\,\frac{dr}{dt} \)

    \( \displaystyle = 8\pi \times 2 \times \frac{3}{8\pi} = 6 \)

So the surface area increases at 6 \(\mathrm{cm}^{2}\)/sec.

Shortcut: Dividing the two rates, \(\frac{dS}{dt} = \frac{2}{r}\,\frac{dV}{dt} = \frac{2}{2} \times 6 = 6\).