Derivation of Viscosity Formula

Terminal velocity ($v_0$) is achieved when the net force on the sphere is zero. The forces involved are gravitational ($F_g$), buoyancy ($F_b$), and viscous drag ($F_d$).

• $F_g = \frac{4}{3}\pi r^3 \sigma g$
• $F_b = \frac{4}{3}\pi r^3 \rho g$
• $F_d = 6\pi \eta r v_0$

Setting $F_g - F_b = F_d$ gives:

Solving for the viscosity ($\eta$):

Error Propagation Analysis

We analyze the error in $\eta$ considering its dependence on radius $r$ and terminal velocity $v_0$. The other terms ($\sigma$, $\rho$, $g$) are assumed to have negligible errors.

Let $\eta$ be represented as:

where $K = \frac{2 (\sigma - \rho)g}{9}$ is treated as a constant.

The change in $\eta$, denoted $\Delta \eta$, can be approximated using partial derivatives:

Calculate the partial derivatives:

• $ \frac{\partial \eta}{\partial r} = \frac{\partial}{\partial r} \left( K \frac{r^2}{v_0} \right) = K \frac{2r}{v_0} $
• $ \frac{\partial \eta}{\partial v_0} = \frac{\partial}{\partial v_0} \left( K \frac{r^2}{v_0} \right) = - K \frac{r^2}{v_0^2} $

Substitute these into the error approximation:

Calculating Relative Error

The question asks for the error in $\eta$. The options provided are in terms of relative errors. We calculate the relative error $\frac{\Delta \eta}{\eta}$:

Simplifying the expression:

This result matches the expression in Option D.