\[\begin{array}{1 1}(1)1&(2)\frac{1}{2\pi}\\(3)4\pi&(4)\frac{4\pi}{3}\end{array}\]

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Let S be the surface area, V be the volume, r be the radius of a sphere at time t

Given $\large\frac{dv}{dt} =\frac{dr}{dt}$

$V= \large\frac{4}{3}$$\pi r^3$

Surface area $S= 4 \pi r^2$

$V= \large\frac{4}{3} $$\pi r^3$

$\large\frac{dV}{dt} =\frac{4}{3} $$3 \pi ^2 \large\frac{dr}{dt}$

$\large\frac{dV}{dt}$$ =4 \pi ^2 \large\frac{dV}{dt}$

$1=S\qquad => S=1$

Surface area= 1

Hence 1 is the correct answer.

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