Question

If an edge of a cube measure $2$ m with a possible error of $0.5$ cm. Find the corresponding error in the calculated volume of the cube.

• A. $0.6m^3$
• B. $0.06m^3$
• C. $0.006m^3$
• D. $0.0006m^3$

Answer 1

Answer: (B)

$0.06m^3$

Let $x$ be the length of an edge of a cube and $V$ be the volume of that cube.
Thus $V=x^3$
On differentiating w.r.t. $x$, we get
$\frac{dV}{dx}=3x^2$
Let $\delta V$ be error in $V$ and corresponding error $\delta x$ in $x$.
$\therefore \delta V=\frac{dV}{dx}\delta x=3x^2\delta x$
Given that, $x=2$ m and $\delta x=0.5$ cm $=\frac{0.5}{100}$ m
$\therefore \delta V=3(2{)}^2\left(\frac{0.5}{100}\right)$
$=\frac{12\times 0.5}{100}=\frac{6}{100}=0.06c{m}^3$