JEE Main Cutoff for IIIT Srirangam 2024 - Check Here

Applications Of Dimensional Analysis - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Application of Dimensional analysis (I)- To find dimension of physical constant is considered one the most difficult concept.

  • Application of Dimensional analysis (II)- To convert a physical quantity from one system to other, Application of Dimensional analysis (V)- As a research tool to derive new relations is considered one of the most asked concept.

  • 70 Questions around this concept.

Solve by difficulty

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:

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

Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length?

The physical quantities not having same dimensions are :

Strain has same dimension as that of:

A physical quantity Q depends on the variables x, y \ and \ z as Q = kx^2 y^3/z^2, where  k is a constant of proportionality. If the dimensions of x, y \ and \ z are L, M and T, respectively, then the dimensions of k in terms of L, M and T are :

Concepts Covered - 5

Application of Dimensional analysis (I)- To find dimension of physical constant

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

  1. Gravitational constant

$
\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]
\end{aligned}
$

$F \rightarrow$ force of Gravitation
$G \rightarrow$ Universal Gravitational Constant
$r \rightarrow$ distance between two masses

$
m_1, m_2 \rightarrow \text { two masses }
$

2. Planck's Constant(h):-

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


Dimensional formula- $M^1 L^2 T^{-1}$
SI unit- Joule-sec
3. Rydberg constant ( R )

$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
$


Dimension- $M^0 L^{-1} T^0$

 

Application of Dimensional analysis (II)- To convert a physical quantity from one system to other

As we know, the measure of a physical quantity is constant, i.e., nu=constant.

$
n_1\left[u_1\right]=n_2\left[u_2\right]
$


If the dimension of a quantity in one system is $\left[M_1^a L_1^b T_1^c\right]$ and in another system, dimension is $\left[M_2^a L_2^b T_2^c\right]$, then

$
n_2=n_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c
$
 

Application of Dimensional analysis (III)- Check the dimensional correctness

It is based on the principle of homogeneity. According to this principle, both sides of an equation must be the same.

L.H.S = R.H.S

It also states that only those physical quantities can be added or subtracted which have the same dimensions.

If the dimension of each term on both sides is the same, then the equation is dimensionally correct, otherwise not.

A dimensionally correct equation may or may not be physically correct.

Application of Dimensional analysis (IV)- To find the unit of physical quantity in a given system"

Let physical quantity be a force
So $[\mathrm{F}]=M L T^{-2}$
If we replace $M, L, T$ in the dimensional formula of the physical quantity by fundamental units of the required system, we will get the unit of that physical quantity.

Now we want to find the unit of Force in SI system
Which is $k g m\left(\mathrm{sec}^{-2}\right)$ or Newton

Application of Dimensional analysis (V)- As a research tool to derive new relations

The method of dimensional analysis can be used to derive new relations.
For example, we can derive a relation for the Time period of a simple pendulum.

$
{ }^{1 \mathrm{f}} T=K m^a l^b g^c
$

where

$
T=\text { time period }
$

$l=$ length
$g=$ acceleration due to gravity
So
Equating exponents of similar quantities

$
a=0 b=1 / 2 c=-1 / 2
$


We get

$
\therefore T=2 \pi \sqrt{\frac{l}{g}}
$
 

Study it with Videos

Application of Dimensional analysis (I)- To find dimension of physical constant
Application of Dimensional analysis (II)- To convert a physical quantity from one system to other
Application of Dimensional analysis (III)- Check the dimensional correctness

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top