The percentage error in calculating the volume of 1 cubical box if an error of 1% is made in measuring the length of edges of the cube, is

Solution:

Let $x$ be the length of an edge of the cube and $y$ be its volume. Then, $y=x^3$.

Let $∆x$ be the error in x and $∆y$ be the corresponding error in $y.$

Then,

$\frac{\mathrm{\Delta }x}{x}\times 100=1$                                   [Given]

$\Rightarrow \frac{dx}{x}\times 100=1$                             $[\because dx\cong \mathrm{\Delta }x]\dots \text{ (i) }$

We have to find $\frac{\mathrm{\Delta }y}{y}\times 100$.

Now, $y=x^3\Rightarrow \frac{dy}{dx}=3x^2$

$\therefore dy=\frac{dy}{dx}dx$

$\begin{array}{l}\Rightarrow dy=3x^2\cdot dx\Rightarrow \frac{dy}{y}=\frac{3x^2}{y}dx\Rightarrow \frac{dy}{y}=\frac{3x^2}{x^3}dx\left[\because y=x^3\right]\\ \Rightarrow \frac{dy}{y}=3\frac{dx}{x}\Rightarrow \frac{dy}{y}\times 100=3\left(\frac{dx}{x}\times 100\right)=3[U\sin g(i\left)\right]\\ \Rightarrow \frac{\mathrm{\Delta }y}{y}\times 100=3 [\because dy\cong \mathrm{\Delta }y]\\ \end{array}$

So, there is 3% error in calculating the volume of the cube.