JEE Main 2026 — Mechanical Properties of Fluids Question with Solution

The terminal velocity $v_0$ of a spherical body falling through a viscous liquid is given by Stokes' Law:
$v_0=\frac{2r^2(\sigma -\rho )g}{9\eta }$
Rearranging the formula to express viscosity $\eta$ in terms of other variables:
$\eta =\frac{2r^2(\sigma -\rho )g}{9v_0}$
Since we are told to ignore errors associated with $\sigma$, $\rho$, and $g$, these quantities along with the numerical factor $\frac{2}{9}$ are treated as constants.
The expression for $\eta$ can be written as $\eta =k\cdot \frac{r^2}{v_0}$, where $k=\frac{2(\sigma -\rho )g}{9}$ is a constant.
Taking the natural logarithm on both sides:
$\ln \eta =\ln k+2\ln r-\ln v_0$
Differentiating both sides to find the relative error:
$\frac{d\eta }{\eta }=0+2\frac{dr}{r}-\frac{d{v}_0}{v_0}$
For the estimated maximum fractional error, we take the absolute sum of the individual relative errors:
$\frac{\Delta \eta }{\eta }=2\frac{\Delta r}{r}+\frac{\Delta v_0}{v_0}$
This matches option (1).