The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error…

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
  1. $\frac{2}{3} \alpha$
  2. $\frac{2}{3} \alpha$
  3. $\frac{3}{2} \alpha$
  4. $\alpha$

Solution

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 $

Asked in: JEE Main 2018 (15 Apr Shift 1 Online)

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