A spherical body of radius r and density sigm | vedclass

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(B) The terminal velocity $v_0$ of a spherical body is given by Stokes' Law: $v_0 = \frac{2}{9} \frac{r^2 g}{\eta} (\sigma - \rho)$. Rearranging the f

Vedclass · Accepted Answer

$2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$

Vedclass · Answer

(B) The terminal velocity $v_0$ of a spherical body is given by Stokes' Law: $v_0 = \frac{2}{9} \frac{r^2 g}{\eta} (\sigma - \rho)$. Rearranging the formula to solve for viscosity $\eta$: $\eta = \frac{2}{9} \frac{r^2 g}{v_0} (\sigma - \rho)$. Since $\sigma, \rho$, and $g$ are constants with negligible errors, the relative error in $\eta$ is determined by the variables $r$ and $v_0$. Using the rules of error propagation for multiplication and division, the maximum relative error is: $\frac{\Delta \eta}{\eta} = 2 \frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$.

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