Question

The kinematic viscosity of glycerin and kerosene are 1.2 times and 0.95 times of that of water, respectively. Glycerin and kerosene flow through two identical porous media having same hydraulic gradient. Assuming Darcy’s law is valid for the porous media, the ratio of flow rate of kerosene to that of glycerin is:

• A. 1.052
• B. 1.140
• C. 0.792
• D. 1.263

Hint:

- Use dynamic viscosity in Darcy’s law.
- If only kinematic viscosity values are provided (without densities), take the ratio directly.

Answer 1

The Correct Option is D

Step 1: Recall Darcy’s law.
According to Darcy’s law: \[ Q \propto \frac{1}{\mu} \] where $Q$ is flow rate and $\mu$ is dynamic viscosity.
Step 2: Relation with kinematic viscosity.
Kinematic viscosity: $\nu = \frac{\mu}{\rho}$. Thus, $\mu = \nu . \rho$.
Step 3: Ratio of flow rates.
\[ \frac{Q_{kerosene}}{Q_{glycerin}} = \frac{\mu_{glycerin}}{\mu_{kerosene}} = \frac{\nu_{glycerin} . \rho_{glycerin}}{\nu_{kerosene} . \rho_{kerosene}} \] Given: - $\nu_{glycerin} = 1.2 \nu_{water}$ - $\nu_{kerosene} = 0.95 \nu_{water}$ Assume densities relative to water: - $\rho_{glycerin} \approx 1.26 \rho_{water}$ - $\rho_{kerosene} \approx 0.82 \rho_{water}$ So: \[ \frac{Q_{kerosene}}{Q_{glycerin}} = \frac{(1.2 . 1.26)}{(0.95 . 0.82)} = \frac{1.512}{0.779} \approx 1.94 \] But if only viscosity ratios are considered (ignoring density): \[ \frac{Q_{kerosene}}{Q_{glycerin}} = \frac{1.2}{0.95} \approx 1.263 \] Note: Since the question specifies “kinematic viscosity” and Darcy’s law depends on dynamic viscosity, the simplified ratio using given data (ignoring densities) gives the correct exam choice. Final Answer: \[ \boxed{\text{(D) 1.263}} \]