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If there is an error of $\mathrm{k} \%$ in measuring the edge of a cube, then the percent error in estimating its volume is
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Verified Answer
The correct answer is:
$3 \mathrm{k}$
Volume V of a cube of side $x$ is given by
$\begin{aligned}
\mathrm{V} &=\mathrm{x}^{3} \\
\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}} &=3 \mathrm{x}^{2}
\end{aligned}$
Let the change in $x$ be $\Delta x=K \%$ of
$x=\frac{k x}{100}$
Now, the change in volume,
$\begin{array}{c}
\Delta \mathrm{V}=\left(\frac{\mathrm{dV}}{\mathrm{dx}}\right) \Delta \mathrm{x}=3 \mathrm{x}^{3}(\Delta \mathrm{x}) \\
=\quad 3 \mathrm{x}^{2}\left(\frac{\mathrm{kx}}{100}\right)=\frac{3 \mathrm{x}^{3} \cdot \mathrm{k}}{100}
\end{array}$
$\therefore \quad$ Approximate change in volume
$=\frac{3 \mathrm{kx}^{3}}{100}=\frac{3 \mathrm{k}}{100} \cdot \mathrm{x}^{3}$
$=3 \mathrm{~K} \%$ of original volume
$\begin{aligned}
\mathrm{V} &=\mathrm{x}^{3} \\
\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}} &=3 \mathrm{x}^{2}
\end{aligned}$
Let the change in $x$ be $\Delta x=K \%$ of
$x=\frac{k x}{100}$
Now, the change in volume,
$\begin{array}{c}
\Delta \mathrm{V}=\left(\frac{\mathrm{dV}}{\mathrm{dx}}\right) \Delta \mathrm{x}=3 \mathrm{x}^{3}(\Delta \mathrm{x}) \\
=\quad 3 \mathrm{x}^{2}\left(\frac{\mathrm{kx}}{100}\right)=\frac{3 \mathrm{x}^{3} \cdot \mathrm{k}}{100}
\end{array}$
$\therefore \quad$ Approximate change in volume
$=\frac{3 \mathrm{kx}^{3}}{100}=\frac{3 \mathrm{k}}{100} \cdot \mathrm{x}^{3}$
$=3 \mathrm{~K} \%$ of original volume
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