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Applications Of Dimensional Analysis MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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.

  • 47 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

The value of plank's constant is :

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The dimension of Rydberg constant is -

The dimensional formula for gravitational constant G 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 dimesnions of (\mu_{o} \epsilon_{o} )^{\frac{-1}{2}}\;are

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The pair of quantities having same dimensions is :

The density of material in the CGS system of units is 4g/cm3. In a system of units in which the unit of length is 10 cm and the unit of mass is 100g, the value of the density of the material will be 

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The unit of work  is -

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. Gravitation constant

 F= G\frac{m_{1}m_{2}}{r^{2}}= G= \frac{Fr^{2}}{m_{1}m_{2}}

G= \frac{\left [ MLT^{-2} \right ]\left [ L^{2} \right ]}{\left [ M \right ]\left [ M \right ]}= \left [ M^{-1}L^{3}T^{-2} \right ]

F\rightarrow force \: of \: Gravitation

\dpi{100} G\rightarrow Universal \: \: Gravitational \: Constant

r\rightarrow distance\: between\: two\: masses

m_{1},m_{2}\rightarrow two\: masses

 

  1. Planck's Constant(h):-

E=h\upsilon \Rightarrow h=\frac{E}{\upsilon }

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

 SI unit- Joule-sec

 

  1. Rydberg constant (R)

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

Unit-  m^{-1}

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 and in another system, dimension is ,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 is a force

So [F]=

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 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.

If T=Km^{a}l^{b}g^{0}

where 

T= 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{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

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