Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

SI units of coefficient of viscosity are (a) \(\mathrm{kg} \mathrm{s}^{-1} \mathrm{~m}^{-2}\) (b) \(\mathrm{g} \mathrm{m}^{-1} \mathrm{~s}^{-1}\) (c) \(\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-1}\) (d) \(\mathrm{kg} \mathrm{cm}^{-1} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The SI unit of coefficient of viscosity is option (c) kg·m⁻¹·s⁻¹.

Step 1: Understanding Viscosity

Viscosity is a measure of a fluid's resistance to flow. The coefficient of viscosity is a physical quantity that describes the internal friction of a moving fluid. The SI unit of viscosity, specifically dynamic viscosity, is the pascal-second (Pa·s).

Step 2: Relation to Other Units

The pascal is the SI unit for pressure and is equivalent to one newton per square meter. Therefore, 1 Pa = 1 N/m². Because 1 N (newton) = 1 kg·m/s², we have 1 Pa = 1 kg/(m·s²). Thus, multiplying by 's' for the second in Pa·s, we find the unit for viscosity is 1 kg/(m·s).

Step 3: Analyzing the Options

We have identified the SI unit of viscosity as kg/(m·s) or kg·m⁻¹·s⁻¹. Now, let's match this with the given options: (a) kg·s⁻¹·m⁻² (b) g·m⁻¹·s⁻¹ (c) kg·m⁻¹·s⁻¹ (d) kg·cm⁻¹·s⁻¹ Option (c) matches exactly with kg·m⁻¹·s⁻¹.

Step 4: Concluding the Solution

Since option (c) accurately represents the SI unit for kinematic viscosity, select this option as the correct answer.

Key Concepts

SI Units

In the world of physics, SI units, or the International System of Units, are critical for ensuring standardization across all scientific disciplines. This uniformity allows scientists around the globe to communicate findings and validate results without ambiguity. The SI units are built on seven base units: meter (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, Kelvin (K) for temperature, mole (mol) for amount of substance, and candela (cd) for luminous intensity.

The importance of using SI units lies in their universal acceptance, allowing exact replication of experiments and reliable data comparison. In the context of viscosity, this turns the focus to the derived unit "pascal-second" (Pa·s), which incorporates the base units of mass, length, and time, providing a consistent measure to describe fluid dynamics.

Dynamic Viscosity

Dynamic viscosity, often simply called viscosity, measures a fluid's internal resistance to flow. Picture a fluid as a layer cake; each layer of the fluid slides over the others. The ease or difficulty with which these layers move past one another defines the fluid's viscosity. High viscosity fluids like honey flow more sluggishly compared to low viscosity ones like water.

The mathematical expression for dynamic viscosity is given by \( \eta = \frac{F}{A} \cdot \frac{L}{v} \), where \( F \) is the force applied, \( A \) is the area, \( L \) is the distance the layer is moved, and \( v \) is the velocity. The SI unit, pascal-second (Pa·s), is derived from this equation. It embodies the newton (force) per square meter (pressure) times second, linking the physical impact of a force to the resulting fluid flow rate.

Physics Problem Solving

Tackling physics problems, especially those involving units like viscosity, requires a systematic approach. Begin by identifying the relevant physical laws and the quantities involved. For instance, if faced with a problem related to viscosity, understand that you're dealing with the dynamic viscosity concept and you'll likely use its SI unit, pascal-second.

Next, consider dimensional analysis. This powerful tool checks the consistency of equations and conversions between units. By ensuring that each side of your equation has the same base dimensions, you eliminate potential errors.

• Identify what is being asked.
• Visualize or diagram the scenario to understand the physical interactions.
• Apply the physical laws relevant to dynamic viscosity.
• Check your units, ensuring they remain consistent throughout the process.
• Review and verify your final answer to see if it makes sense with the context of the problem.

Developing this structured method not only aids in solving specific problems but also builds foundational skills for more complex physics challenges.