Applications Of Dimensional Analysis MCQ - Practice Questions with Answers

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

1. Gravitation constant

$
\begin{aligned}
& F=G \frac{m_1 m_2}{r^2}=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 $m_1, m_2 \rightarrow$ 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- $M^0 L^{-1} T^0$
Unit- $m^{-1}$

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, the dimension is $\left[M_2^a L_2^b T_2^c\right]_{\text {,then }}$

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

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.

Let physical quantity is a force

$\mathrm{So}[\mathrm{F}]=M L T^{-2}$
If we replace $\mathrm{M}, \mathrm{L}, \mathrm{T}$ 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 Sl system
Which is $K g m\left(\mathrm{sec}^{-2}\right)$ or Newton

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.

$$
\text { If } T=K m^a l^b g^0
$$

where

$
\begin{aligned}
& T=\text { time period } \\
& l=\text { length } \\
& g=\text { acceleration due to gravity }
\end{aligned}
$

So 

Equating exponents of similar quantities

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

We get 

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