Question:
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: , 4
Solution:
Let the side of cube be $a$.
$S=6 a^{2} \Rightarrow \frac{d S}{d t}=12 a \cdot \frac{d a}{d t} \Rightarrow 3.6=12 a \cdot \frac{d a}{d t}$
$\Rightarrow 12(10) \frac{d a}{d t}=3.6 \Rightarrow \frac{d a}{d t}=0.03$
$V=a^{3} \Rightarrow \frac{d V}{d t}=3 a^{2} \cdot \frac{d a}{d t}=3(10)^{2} \cdot\left(\frac{3}{100}\right)=9$