If the surface area of a cube is increasing at a rate

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 :

  1. (1) 18

  2. (2) 10

  3. (3) 20

  4. (4) 9


Correct Option: , 4

Solution:

Let the side of cube be $a$.

$S=6 a^{2} \Rightarrow \frac{d S}{d t}=12 a \cdot \frac{d a}{d t} \Rightarrow 3.6=12 a \cdot \frac{d a}{d t}$

$\Rightarrow 12(10) \frac{d a}{d t}=3.6 \Rightarrow \frac{d a}{d t}=0.03$

$V=a^{3} \Rightarrow \frac{d V}{d t}=3 a^{2} \cdot \frac{d a}{d t}=3(10)^{2} \cdot\left(\frac{3}{100}\right)=9$

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