A spherical body of radius r and density sigma$\sigma$ falls freely through a viscous liquid having density rho$\rho$ and viscosity eta$\eta$ and attains a terminal velocity v_0$v_{0}$. Estimated maximum error in the quantity eta$\eta$ is : (Ignore errors associated with sigma$\sigma$, rho$\rho$ and g, gravitational acceleration)

### Related Formula v_0 = frac29fracr^2 geta (sigma - rho)$$v_0 = \frac{2}{9}\frac{r^2 g}{\eta} (\sigma - \rho)$$ ### Core Logic Rearranging the formula for viscosity eta$\eta$: eta = frac29 fracr^2 gv_0 (sigma - rho)$$\eta = \frac{2}{9} \frac{r^2 g}{v_0} (\sigma - \rho)$$ Since frac29$\frac{2}{9}$, g$g$, sigma$\sigma$, and rho$\rho$ are constants or have negligible error, the relative error depends only on r$r$ and v_0$v_0$. ### Step 1: Error Propagation Taking natural logarithm on both sides: ln(eta) = ln(textconstant) + 2ln(r) - ln(v_0)$$\ln(\eta) = \ln(\text{constant}) + 2\ln(r) - \ln(v_0)$$ Differentiating for maximum possible error (adding absolute errors): fracDelta etaeta = 2fracDelta rr + fracDelta v_0v_0$$\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$ ### Step 2: Final Conclusion The maximum fractional error is frac2Delta rr + fracDelta v_0v_0$\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$. ### Pattern Recognition For maximum fractional error of a derived quantity Z = X^a Y^b$Z = X^a Y^b$, the formula is Delta Z/Z = |a|Delta X/X + |b|Delta Y/Y$\Delta Z/Z = |a|\Delta X/X + |b|\Delta Y/Y$. Always sum the absolute relative errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Units and Measurements Class 11 Physics: Mechanical Properties of Fluids