Applications Of Dimensional Analysis

Dimensional analysis is a powerful mathematical tool that goes beyond the classroom, finding practical applications in various real-life scenarios. By examining the dimensions of physical quantities, we can simplify complex problems, verify equations, and even predict relationships between variables. For instance, engineers use dimensional analysis to scale models of structures, ensuring they behave similarly to full-sized versions. In medicine, it aids in converting units and calculating dosages accurately. Whether in designing safe bridges or administering the correct drug dosage, dimensional analysis serves as a bridge between theory and practical application, making abstract concepts tangible and relevant.

Applications of Dimensional Analysis

We can find the dimension of a physical constant by substituting the dimensions of physical quantities in the given equation

Gravitational Constant

The gravitational constant, denoted by GGG, is a fundamental physical constant that appears in Newton's law of universal gravitation. It quantifies the strength of the gravitational force between two masses. Discovered by Sir Isaac Newton, the gravitational constant is crucial for calculating the force that governs the motion of planets, stars, and galaxies.

$\begin{aligned} & F=G \frac{m_1 m_2}{r^2} \Rightarrow G=\frac{F r^2}{m_1 m_2} \\ & G=\frac{\left[M L T^{-2}\right]\left[L^2\right]}{[M][M]}=\left[M^{-1} L^3 T^{-2}\right] \\ & F \rightarrow \text { force of Gravitation } \\ & G \rightarrow \text { Universal Gravitational Constant } \\ & r \rightarrow \text { distance between two masses } \\ & m_1, m_2 \rightarrow \text { two masses }\end{aligned}$

Planck's Constant(h)

Planck's constant (h) is a fundamental constant in quantum mechanics that sets the scale for the quantization of energy, momentum, and angular momentum in the microscopic world. Discovered by Max Planck in 1900, it marked the beginning of quantum theory, revolutionizing our understanding of physics.

$E=h v \Rightarrow h=\frac{E}{v}$

Dimensional formula $M^1 L^2 T^{-1}$

SI unit- Joule-sec

Rydberg Constant (R)

The Rydberg constant (R) is a fundamental physical constant that appears in the formula used to describe the wavelengths of light emitted or absorbed by electrons in a hydrogen atom as they transition between energy levels. Named after the Swedish physicist Johannes Rydberg, this constant plays a crucial role in atomic physics and spectroscopy.
$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$
Dimension- $M^0 L^{-1} T^0$
Unit- $m^{-1}$

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Solved Examples Based on Applications of Dimensional Analysis

Example 1:Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

1)[ M ]=[ T⁻¹ C⁻² h ]

2)[ M ]=[ T⁻¹ C² h ]

3)[ M ]=[ T⁻¹ C⁻² h⁻¹ ]

4)[ M ]=[ T C⁻² h ]

Solution:

$\begin{aligned} & \text { Dimension of length }[L]=[C T] \\ & \text { Dimension of mass }=\frac{[\text { Angular Momentum }]}{[\text { Velocity }][\text { Length }]} \\ & =\frac{[h]}{[C][C T]}=\left[C^{-2} T^{-1} h\right]\end{aligned}$

Hence, the answer is option 1.

Example 2: Out of the following pairs which one does not have identical dimensions is

1) moment of inertia and moment of force

2)work and torque

3)angular momentum and Planck’s constant

4)impulse and momentum

Solution:

Dimension of Work, Potential Energy, Kinetic Energy, Torque is

and that of Momentum and Impulse -

and that of angular momentum & Plank's Constant (h) is -

Moment of inertia is defined as (I) = Mass (M) x radius²(r)

The dimensional formula is=ML²

Know that the moment of force, T = radius (r) x force (F)

The dimensional formula for the moment of a force is- ML²T^-2

Therefore the dimension of a moment of inertia and a moment of force does not have an identical dimension and torque is also called a moment of force.

Hence, the answer is the option (1).

Example 3: The speed of light (c), gravitational constant (G) and Planck's constant (h) are taken as fundamental units in a system. The dimensions of the time in the new system should be

1) $G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-5}{2}}$
2) $G^{\frac{-1}{2}} h^{\frac{1}{2}} c^{\frac{1}{2}}$
3) $G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-3}{2}}$
4) $G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{1}{2}}$

Solution:

The speed of light (c), gravitational constant (G) and Planck's constant (h) are taken as fundamental units in a system. The dimension of Time T is
$
\begin{aligned}
& T \propto c^a G^b h^d \\
& T=k c^a G^b h^d \\
& {[T]=[k][c]^a[G]^b[h]^d} \\
& {[T]=\left[L T^{-1}\right]^a\left[M^{-1} L^3 T^{-2}\right]^b\left[M L^2 T^{-1}\right]^d}
\end{aligned}
$
By comparing the power of $M, L \& T$.

$
\begin{aligned}
& -b+d=0 \Rightarrow b=d \\
& a+3 b+2 d=0 \quad-(2) \\
& -a-2 b-d=1
\end{aligned}
$
By solving (1), (2) and (3)

$
\begin{aligned}
& a=\frac{-5}{2} \quad b=d=\frac{1}{2} \\
& T=k G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-5}{2}}
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is :

1) $\frac{c h}{2 \pi \epsilon_0^2}$
2) $\frac{e^2}{2 \pi \epsilon_0 G m_e^2}\left(m_e=\right.$ mass of electron $)$
3) $\frac{\mu_0 \epsilon_0}{c^2} \frac{G}{h e^2}$
4) $\frac{2 \pi \sqrt{\mu_0 \epsilon_0}}{c e^2} \frac{h}{G}$

Solution:

Physical quantity, $\mathrm{nu}=$ constant

$
n_1 u_1=n_2 u_2=\text { constant }
$
The permittivity of free space

$
\epsilon_o=M^1 L^3 T^{-4} A^2
$

wherein its unit is

$
C^{-2} N^1 m^{-2}
$

Hence, the answer is the option (2).

Example 5: The dimension formula of the Rydberg constant (R) is

1) $\left[M L^2 T^{-2}\right]$
2) $\left[M L^{-1} T^0\right]$
3) $\left[M^0 L^{-1} T^0\right]$
4) $\left[M^1 L^{-1} T^1\right]$

Solution:

Rydberg constant (R)

$\begin{aligned} & \text { wherein unit - } m^{-1} \\ & \frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \\ & \text { Here } \lambda \rightarrow \text { wavelength, } n_1 \& n_2 \rightarrow \text { Quantum no. } \\ & \qquad=\left[\frac{v_\lambda}{\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]}\right] \\ & \text { So Dimension formula }[\mathrm{R}]= \\ & {[R]=\left[M^0 L^{-1} T^0\right]}\end{aligned}$

Hence, the answer is the option (3).

Summary

Dimensional analysis is a technique deployed in the simplification or solving of complex problems by checking the consistency of units and also obtaining the relationship between physical quantities. Some fields where the technique has been applied are engineering, physics, and chemistry in everyday activity. It does, for example, help in the design of systems where it checks the dimensional correctness of formulas and calculations. In fluid flow, it helps to understand the flow pattern using dimensionless numbers like the Reynolds number and, most critically, in scaling experiments from a model to REAL LIFE such that the scaling is ACCURATE. By breaking down a problem into its most fundamental elements, Dimensional Analysis provides a rapid route to the solution of most problems in a direct, straightforward manner.