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.
Important Questions from Mixed Topic (CUET PG)
- Who was the founder of Bolshevik Communist party?
- What is the key guide to statecraft in the realist tradition?
- Chronologically arrange the events in the Cold War period.
A. Berlin Wall is constructed
B. Communist China joins the UN
C. Soviet invasion of Czechoslovakia
D. Berlin Blockade
Choose the correct answer from the options given below: Morgenthau's principles of political realism are:
A. Politics is rooted in permanent and unchanging human nature which is basically self centred, self-regarding and self-interested
B. Politics is an autonomous sphere of action and cannot therefore be reduced to morals
C. International Politics is an arena of conflicting self-interests
D. The ethics of international relations is situational ethics which is very different from private morality
Choose the correct answer from the options given below:Who among the following political thinkers consider the anarchical self help system to be a compelling factor for States to maximise their relative power positions?
