Fluid Mechanics

JEE Main level, single correct, medium

164 of 192

Question 164 of 192

A spherical body of radius rr and density σ\sigma falls freely through a viscous liquid having density ρ\rho and viscosity η\eta and attains a terminal velocity v0v_{0}. Estimated maximum error in the quantity η\eta is : (Ignore errors associated with σ,ρ\sigma, \rho and gg, gravitational acceleration)

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Answer: (A) 2Δrr+Δvovo\frac{2 \Delta r}{r}+\frac{\Delta v_{\mathrm{o}}}{v_{\mathrm{o}}}

The terminal velocity v0v_0 of a spherical body falling through a viscous liquid is given by Stokes' Law:
v0=2r2(σ−ρ)g9ηv_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}
Rearranging the formula to express viscosity
η\eta in terms of other variables:
η=2r2(σ−ρ)g9v0\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}
Since we are told to ignore errors associated with
σ\sigma, ρ\rho, and gg, these quantities along with the numerical factor 29\frac{2}{9} are treated as constants.
The expression for
η\eta can be written as η=k⋅r2v0\eta = k \cdot \frac{r^2}{v_0}, where k=2(σ−ρ)g9k = \frac{2(\sigma - \rho)g}{9} is a constant.
Taking the natural logarithm on both sides:

ln⁡η=ln⁡k+2ln⁡r−ln⁡v0\ln \eta = \ln k + 2\ln r - \ln v_0
Differentiating both sides to find the relative error:

dηη=0+2drr−dv0v0\frac{d\eta}{\eta} = 0 + 2\frac{dr}{r} - \frac{dv_0}{v_0}
For the estimated maximum fractional error, we take the absolute sum of the individual relative errors:

Δηη=2Δrr+Δv0v0\frac{\Delta \eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}
This matches option (1).

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