KCET · Maths · Indefinite Integration
If the side of a cube is increased by \(5 \%\), then the surface area of a cube is increased by
- A \(10 \%\)
- B \(60 \%\)
- C \(6 \%\)
- D \(20 \%\)
Answer & Solution
Correct Answer
(A) \(10 \%\)
Step-by-step Solution
Detailed explanation
We know that the surface area of cube is given by \(S=6 x^{2}\)
\(\begin{aligned}
\therefore \quad \frac{d S}{d x} &=12 x^{2} \\
\Delta x &=5 \% \text { of } x=0 \cdot 05 x
\end{aligned}\)
The surface area of a cube is increased by
\(\begin{aligned}
=\frac{\Delta S}{S} \times 100 \% & \\
=& \frac{\left(\frac{d S}{d x}\right) \Delta x}{S} \times 100 \% \\
=& \frac{12 x \times(0 \cdot 05 x)}{6 x^{2}} \times 100 \% \\
=& 0 \cdot 10 \times 100 \%=10 \%
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \frac{d S}{d x} &=12 x^{2} \\
\Delta x &=5 \% \text { of } x=0 \cdot 05 x
\end{aligned}\)
The surface area of a cube is increased by
\(\begin{aligned}
=\frac{\Delta S}{S} \times 100 \% & \\
=& \frac{\left(\frac{d S}{d x}\right) \Delta x}{S} \times 100 \% \\
=& \frac{12 x \times(0 \cdot 05 x)}{6 x^{2}} \times 100 \% \\
=& 0 \cdot 10 \times 100 \%=10 \%
\end{aligned}\)
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