Question

Test if the following equation are dimensionally correct \((\mathrm{S}=\) surface tension \(\rho=\) density \(\mathrm{P}=\) pressure \(\mathrm{v}=\) volume \(\mathrm{n}=\) coefficient of viscosity \(\mathrm{r}=\) radius \()\) (a) \(\mathrm{h}=[(2 \mathrm{~S} \cos \theta) /(\rho \mathrm{rg})]\) (b) \(\mathrm{v}=\sqrt{(p / \rho)}\) (c) \(\mathrm{v}=\left[\left(\pi \mathrm{pr}^{4} \mathrm{t}\right) /(8 \mathrm{n} \ell)\right]\) (d) all correct

Short answer

The short answer based on the step-by-step solution is: After performing dimensional analysis for each of the given equations (a), (b), and (c), we find that the dimensions match on both sides for all the equations, which implies that they are dimensionally correct. Therefore, the correct answer is (d) "all correct".

Step-by-step solution

Identify the dimensions of each physical quantity

We first need to determine the dimensions of each physical quantity in the given equations: - Surface tension (S): \( [M^1 L^0 T^{-2}] \) - Density (ρ): \( [M^1 L^{-3} T^0] \) - Pressure (P): \( [M^1 L^{-1} T^{-2}] \) - Volume (v): \( [M^0 L^3 T^0] \) - Coefficient of viscosity (n): \( [M^1 L^{-1} T^{-1}] \) - Radius (r): \( [M^0 L^1 T^0] \) - Height (h): \( [M^0 L^1 T^0] \) - Gravitational acceleration (g): \( [M^0 L^1 T^{-2}] \) - Angle (θ): Dimensionless - Time (t): \( [M^0 T^1 L^0] \) Now, we will apply dimensional analysis to each equation to verify their correctness. Equation (a): \(h = \frac{2 S \cos\theta}{\rho rg}\)

Check dimensions in Equation (a)

Now we check if the dimensions are equal on both sides of the equation (a): - Left side: Dimension of h = \( [M^0 L^1 T^0] \) - Right side: Dimension of S = \( [M^1 L^0 T^{-2}] \), Dimension of ρ = \( [M^1 L^{-3} T^0] \), Dimension of r = \( [M^0 L^1 T^0] \), Dimension of g = \( [M^0 L^1 T^{-2}] \) The right side of the equation has dimensions: \(\frac{[M^1 L^0 T^{-2}]}{[M^1 L^{-3} T^0] \cdot [M^0 L^1 T^0] \cdot [M^0 L^1 T^{-2}]} = [M^0 L^1 T^0]\) The dimensions match on both sides of Equation (a), so it is dimensionally correct. Equation (b): \(v=\sqrt{\frac{p}{\rho}}\)

Check dimensions in Equation (b)

Now we check if the dimensions are equal on both sides of the equation (b): - Left side: Dimension of v = \( [M^0 L^3 T^0] \) - Right side: Dimension of P = \( [M^1 L^{-1} T^{-2}] \), Dimension of ρ = \( [M^1 L^{-3} T^0] \) The right side of the equation has dimensions: \(\sqrt{\frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^{-3} T^0]}} = [M^0 L^3 T^0]\) The dimensions match on both sides of Equation (b), so it is dimensionally correct. Equation (c): \(v=\frac{\pi Pr^4 t}{8n\ell}\) Here, l represents length which has the dimensions \( [M^0 L^1 T^0] \).

Check dimensions in Equation (c)

Now we check if the dimensions are equal on both sides of the equation (c): - Left side: Dimension of v = \( [M^0 L^3 T^0] \) - Right side: Dimension of P = \( [M^1 L^{-1} T^{-2}] \), Dimension of r = \( [M^0 L^1 T^0] \), Dimension of t = \( [M^0 T^1 L^0] \), Dimension of n = \( [M^1 L^{-1} T^{-1}] \), Dimension of l = \( [M^0 L^1 T^0] \) The right side of the equation has dimensions: \(\frac{[M^1 L^{-1} T^{-2}] \cdot [M^0 L^1 T^0]^4 \cdot [M^0 T^1 L^0]}{[M^1 L^{-1} T^{-1}] \cdot [M^0 L^1 T^0]} = [M^0 L^3 T^0]\) The dimensions match on both sides of Equation (c), so it is dimensionally correct. Thus, all the given equations (a), (b), and (c) are dimensionally correct, making the answer (d) "all correct".

Key concepts

Surface Tension

Surface tension is a fascinating phenomenon that occurs at the surface of a liquid in contact with a gas, or between two immiscible liquids.
It is what allows certain insects to walk on water or causes a droplet of water to form a bead on a flat surface.
Surface tension is essentially the result of the cohesive forces between liquid molecules.
The Basic Idea
At the interface, molecules experience a net inward force because they are pulled more strongly by similar molecules in the liquid than by molecules in the surrounding air.
To depict this phenomenon mathematically, surface tension is represented as a force per unit length, with dimensional formula \([M^1 L^0 T^{-2}]\).
This is crucial for understanding how forces act along the liquid’s surface to maintain stability or shape. The higher the surface tension, the more energy required to increase the surface area of the liquid.

• Helps liquids resist external force
• Causes droplets to form spheres
• Measured in force per unit length

Density

Density is a fundamental concept in physics and chemistry that describes how much mass an object has in a given volume.
It is an intrinsic property, meaning it doesn’t change regardless of how much substance you have.
The formula to calculate density is simple: it’s mass divided by volume.
In formula terms, density is denoted by \( \rho \) and has the dimensional formula \([M^1 L^{-3} T^0]\).
Why Density Matters
The concept of density explains why some objects float while others sink.
It plays a crucial role in buoyancy, which determines whether an object will sink or float in a fluid.
For example, a balloon filled with helium floats because helium is less dense than the surrounding air. On the other hand, a rock sinks in water because it has a higher density than water.

• Mass per unit volume
• Critical for buoyancy calculations
• Varies with temperature and pressure

Pressure

Pressure is an important concept in fluid mechanics that refers to the force exerted per unit area.
It is what happens when a force is applied perpendicular to the surface of an object, spreading over that area.
Commonly seen in phenomena like atmospheric pressure or pressure in fluids, it is measured in units such as Pascals (Pa).
Pressure, represented as \( P \), possesses the dimensional formula \([M^1 L^{-1} T^{-2}]\).
Understanding Pressure
Think of pressure as the result of the collective force of fluid molecules colliding with surfaces.
It explains why deep-sea creatures need special adaptations to survive; pressure increases with depth as the weight of the water above rises.
This is crucial in many real-world applications: in hydraulic systems and weather forecasting, where understanding atmospheric pressure helps predict weather changes.

• Force exerted per area unit
• Higher at greater depths
• Important in weather and engineering

Viscosity

Viscosity is a measure of a fluid's resistance to deformation or its "thickness."
Imagine honey dripping slowly from a spoon; it’s more viscous than water, which flows freely.
This property is essential in understanding the behavior of different fluids in motion.
The coefficient of viscosity, often denoted by \( n \), carries the dimensional formula \([M^1 L^{-1} T^{-1}]\).
What Affects Viscosity?
The temperature of a fluid can significantly impact its viscosity.
As a general rule, increasing temperature lowers viscosity in liquids but increases it in gases.
This is why warm honey flows more easily than when it's cold, while air becomes less fluid at higher temperatures.
Viscosity is crucial in formulae that govern the flow of fluids, such as in pipelines or blood through veins.
It affects not only industrial processes but also biological systems, aiding in the design of devices and systems that can mimic natural fluid movements effectively.

• Resistance to flow
• Affected by temperature
• Crucial for fluid dynamics