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If there is an error of $\mathrm{m} \%$ in measuring the edge of cube, then the percent error in estimating its surface area is
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Verified Answer
The correct answer is:
$2 \mathrm{~m}$
Surface area $\mathrm{A}$ of a cube of side $\mathrm{x}$ is given by $A=6 x^{2}$,
On differentiating w.r.t. $x$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=12 \mathrm{x}$
Let the change in $x$ be $\Delta x=m \%$ of $x$
$=\frac{\mathrm{mx}}{100}$
Change in surface area,
$\begin{array}{l}
\Delta \mathrm{A}=\left(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}\right) \Delta \mathrm{x}=(12 \mathrm{x}) \Delta \mathrm{x} \\
=12 \mathrm{x}\left(\frac{\mathrm{mx}}{100}\right)=\frac{12 \mathrm{x}^{2} \mathrm{~m}}{100}
\end{array}$
$\therefore \quad$ The approximate change in surface area
$\begin{array}{l}
=\frac{2 \mathrm{~m}}{100} \times 6 \mathrm{x}^{2} \\
=2 \mathrm{~m} \% \text { of original surface area }
\end{array}$
On differentiating w.r.t. $x$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=12 \mathrm{x}$
Let the change in $x$ be $\Delta x=m \%$ of $x$
$=\frac{\mathrm{mx}}{100}$
Change in surface area,
$\begin{array}{l}
\Delta \mathrm{A}=\left(\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}\right) \Delta \mathrm{x}=(12 \mathrm{x}) \Delta \mathrm{x} \\
=12 \mathrm{x}\left(\frac{\mathrm{mx}}{100}\right)=\frac{12 \mathrm{x}^{2} \mathrm{~m}}{100}
\end{array}$
$\therefore \quad$ The approximate change in surface area
$\begin{array}{l}
=\frac{2 \mathrm{~m}}{100} \times 6 \mathrm{x}^{2} \\
=2 \mathrm{~m} \% \text { of original surface area }
\end{array}$
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