If the surface area of a cube is increasing at a rate of $3.6 \mathrm{~cm}^{2} / \mathrm{sec}$, retaining its shape; then the rate of change of its volume (in $\mathrm{cm}^{3} / \mathrm{sec}$ ), when the length of a side of the cube is $10 \mathrm{~cm}$, is:
Correct Option: 1
$\frac{\mathrm{d}}{\mathrm{dt}}\left(6 \mathrm{a}^{2}\right)=3.6 \Rightarrow 12 \mathrm{a} \frac{\mathrm{da}}{\mathrm{dt}}=3.6$
$\mathrm{a} \frac{\mathrm{da}}{\mathrm{dt}}=0.3$
$\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}^{3}\right)=3 \mathrm{a}\left(\mathrm{a} \frac{\mathrm{da}}{\mathrm{dt}}\right)$
$=3 \times 10 \times 0.3=9$
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