GATE 2023 BT – Question 62
A dilatant fluid has consistency index K=0.415 (CGS) and flow index n=1.23. Its apparent viscosity ($\mathrm{g\ cm^{-1}s^{-1}}$) at shear rate $60\ \mathrm{s^{-1}}$, rounded to the nearest integer, is ___________.
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Correct answer: 1
Explanation
For a power-law (Ostwald-de Waele) fluid the shear stress is $\tau=K\dot\gamma^{\,n}$, so the **apparent viscosity** is
$$\mu_{app}=\frac{\tau}{\dot\gamma}=K\,\dot\gamma^{\,n-1}.$$
A dilatant (shear-thickening) fluid has $n>1$, and here $n=1.23$.
**Numbers.**
$$\mu_{app}=0.415\times(60)^{1.23-1}=0.415\times60^{0.23}=0.415\times2.564=1.064\ \text{g cm}^{-1}\text{s}^{-1}.$$
To the nearest integer, the apparent viscosity is **1**.