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The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is
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Verified Answer
The correct answer is:
$\frac{3}{2} \alpha$
$\frac{3}{2} \alpha$
Relative error in Surface area, $\frac{\Delta \mathrm{s}}{\mathrm{s}}=2 \times \frac{\Delta \mathrm{r}}{\mathrm{r}}=\alpha$ and relative error in volume, $\frac{\Delta \mathrm{v}}{\mathrm{v}}=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}}$
$\therefore$ Relative error in volume w.r.t. relative error in area,
$$
\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{3}{2} \alpha
$$
$\therefore$ Relative error in volume w.r.t. relative error in area,
$$
\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{3}{2} \alpha
$$
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