Question

If there is an error of $k\%$ in measuring the edge of a cube, then the percent error in estimating its volume is

• A. $k$
• B. $3k$
• C. $\frac{k}{3}$
• D. None of these

Answer 1

Answer: (B)

$3k$

Answer 2

We know that, the volume V of a cube of side x is given by
$V = x^3$
On differenting w. r. t, we get
$\Rightarrow \frac{dv}{dx} = 3 x^2$
Let the change in $x$ be $\Delta x = K \%$ of $x = \frac{kx}{100}$
Now, the change in volume.
$\Delta V = \left( \frac{dV}{dx}\right) \Delta x = 3x^2 ( \Delta x)$
$= 3x^{2} \left(\frac{kx}{100}\right) = \frac{3x^{2} . k}{100}$
$\therefore$ Approximate change in volume
$=\frac{3kx^{3}}{100} = \frac{3k}{100} . x^{3}$
$= 3k \%$ of original volume