The radius of a sphere is increasing at the rate of 0.5 cm/minute. The rate of change of the surface area of (in cm2/ minute) the sphere when the radius is 20 cm is

$20 \pi$ $40 \pi$ $160 \pi$ $80 \pi$

$80 \pi$

Surface area = $4 \pi r^2 = S$ $\Rightarrow \frac{d s}{d t}=8 \pi r \frac{d r}{d t}$ $\frac{dr}{dt} = 0.5 \frac{cm}{min}$ so  at r = 20 $\frac{ds}{dt} = 8 \pi × 20 × 0.5 = 80 \pi cm^2/min$