# If the surface area of a cube is increasing at a rate of $3.6\; cm^{2}/sec,$ retaining its shape ; then the rate of change of its volume (in $cm^{2}/sec$) , when the length of a side of the cube is $10\; cm$, is : Option: 1 Option: 2 Option: 3 Option: 4

$\\ \frac{d}{d t}\left(6 a^{2}\right)=3.6 \Rightarrow 12 a \frac{d a}{d t}=3.6 \\ \text { a } \frac{d a}{d t}=0.3 \\ \frac{d v}{d t}=\frac{d}{d t}\left(a^{3}\right)=3 a\left(a \frac{d a}{d t}\right) \\ =3 \times 10 \times 0.3=9$

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