The volume of spherical balloon is increasing at the rate of 4 cm^3/s…

The correct answer is Option (2) → \(2.66\, cm^2/sec\)

Both rates are linked through the radius r, using the chain rule.

    \( \displaystyle V = \frac{4}{3}\pi r^3 \)

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

At r = 3, with \(\frac{dV}{dt} = 4\):

    \( \displaystyle 4 = 36\pi\,\frac{dr}{dt} \)

    \( \displaystyle \frac{dr}{dt} = \frac{1}{9\pi} \)

For the surface area:

    \(S = 4\pi r^2\)

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

    \( \displaystyle \frac{dS}{dt} = 8\pi \times 3 \times \frac{1}{9\pi} = \frac{8}{3} \)

So the surface area increases at \(\frac{8}{3} = 2.66\ldots\) cm²/sec.

Shortcut: Dividing the two rates removes \(\frac{dr}{dt}\):

    \( \displaystyle \frac{dS}{dV} = \frac{8\pi r}{4\pi r^2} = \frac{2}{r} \)

    \( \displaystyle \frac{dS}{dt} = \frac{2}{3} \times 4 = \frac{8}{3} \)