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Question

Dynamic viscosity of a fluid is 2.2 poise and specific gravity is 0.7. Then kinematic viscosity in SI units is:

The correct answer is
$3.14 \times 10^{-4} \text{ m}^2/\text{s}$

Understanding Fluid Viscosity Calculations

This question asks us to find the kinematic viscosity of a fluid in SI units. We are given the fluid's dynamic viscosity and its specific gravity.

Key Concepts and Formulas

  • Dynamic Viscosity ($\mu$): Measures the fluid's internal resistance to flow. Given as 2.2 poise.
  • Specific Gravity (SG): The ratio of the fluid's density to the density of a reference substance (usually water). Given as 0.7.
  • Density ($\rho$): Mass per unit volume of the fluid.
  • Kinematic Viscosity ($\nu$): The ratio of dynamic viscosity to density. It represents the fluid's resistance to flow under gravity.
  • Relationship: The core formula relating these properties is: $ \nu = \frac{\mu}{\rho} $
  • Density Calculation: Density is calculated using specific gravity and the density of water ($\rho_w$): $ \rho = \text{SG} \times \rho_w $ The density of water is approximately $1000 \text{ kg/m}^3$.
  • Unit Conversions: We need to work in SI units (meters, kilograms, seconds).
    • Dynamic Viscosity: $1 \text{ poise} = 0.1 \text{ Pa·s}$ (Pascal-second)

Step-by-Step Calculation

Step 1: Convert Dynamic Viscosity to SI Units

The dynamic viscosity is given as 2.2 poise. We convert this to Pascal-seconds (Pa·s), the SI unit:

$ \mu = 2.2 \text{ poise} \times \frac{0.1 \text{ Pa·s}}{1 \text{ poise}} = 0.22 \text{ Pa·s} $

Step 2: Calculate the Fluid's Density

Using the specific gravity (SG) and the density of water ($\rho_w = 1000 \text{ kg/m}^3$):

$ \rho = \text{SG} \times \rho_w $ $ \rho = 0.7 \times 1000 \text{ kg/m}^3 $ $ \rho = 700 \text{ kg/m}^3 $

Step 3: Calculate Kinematic Viscosity in SI Units

Now, we use the formula $\nu = \frac{\mu}{\rho}$ with the values in SI units:

$ \nu = \frac{0.22 \text{ Pa·s}}{700 \text{ kg/m}^3} $

Recall that $1 \text{ Pa·s} = 1 \frac{\text{N·s}}{\text{m}^2} = 1 \frac{\text{kg·m/s² · s}}{\text{m}^2} = 1 \frac{\text{kg}}{\text{m·s}}$. Therefore, the unit $\frac{\text{Pa·s}}{\text{kg/m}^3}$ becomes $\frac{\text{kg/(m·s)}}{\text{kg/m}^3} = \frac{\text{kg}}{\text{m·s}} \times \frac{\text{m}^3}{\text{kg}} = \frac{\text{m}^2}{\text{s}}$.

$ \nu = \frac{0.22}{700} \frac{\text{m}^2}{\text{s}} $ $ \nu \approx 0.00031428 \text{ m}^2/\text{s} $

Expressing this in scientific notation:

$ \nu \approx 3.14 \times 10^{-4} \text{ m}^2/\text{s} $

Final Answer Verification

The calculated kinematic viscosity is approximately $3.14 \times 10^{-4} \text{ m}^2/\text{s}$. This matches the first option provided.

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