The radius of a sphere is measured to be

Question:

The radius of a sphere is measured to be $(7.50 \pm 0.85) \mathrm{cm}$. Suppose the percentage error in its volume is $x$. The value of $x$, to the nearest $x$, is__________.

Solution:

$\because \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3}$

taking $\log \&$ then differentiate

$\frac{\mathrm{dV}}{\mathrm{V}}=3 \frac{\mathrm{dr}}{\mathrm{r}}$

$=\frac{3 \times 0.85}{7.5} \times 100 \%=34 \%$

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