\[(A)\; 0.06 x^3 \: m^3 \quad (B)\; 0.6 x^3 \: m^3 \quad (C)\; 0.09 x^3 \: m^3 \quad (D)\; 0.9 x^3 \: m^3\]

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- Let $y=f(x)$
- $\Delta y=f(x+\Delta x)-f(x)$
- $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
- Volume of the cube=$x^3m^3$

Step 1:

Side of a cube =$x$

Increase in the side=3%

Volume of the cube=$x^3m^3$

$\Delta x$=3% of $x=0.03x$

$\Delta V$ be the change in volume.

Step 2:

$\Delta V=\large\frac{dV}{dx}$$\times \Delta x$----(1)

$V=x^3$

$\large\frac{dV}{dx}=$$3x^2$[Differentiating with respect to x]

Step 3:

Substitute the value of $\large\frac{dV}{dx}$ in equation (1) we get,

$\Delta V=3x^2\times 0.03x$

$\quad\;\; =0.09x^3m^3$

$\Delta V=0.09x^3m^3$

Part (C) is the correct answer.

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