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Density has dimensional formula: $[ML_{−3}]$

To convert density of the object in CGS system.

Density= $1000/10_{6}$

= $10_{−3}$ $g/cc$

Then dimensional formula of $A_{x}B_{y}$ is given by $[A_{x}B_{y}]=[M_{xam+ybm}L_{xal+ybl}T_{xat+ybt}]$

Example:

Angular momentum of a physical quantity is given by $I=mvr$. Its dimensional formula can be found as:

$[m]=[M_{1}L_{0}T_{0}]$

$[v]=[M_{0}L_{1}T_{−1}]$

$[r]=[M_{0}L_{1}T_{0}]$

Hence, $[I]=[mvr]=[M_{1}L_{2}T_{−1}]$

Example: Finding time-period of a simple pendulum ($T$) given it depends on length of the pendulum ($l$) and acceleration due to gravity ($g$).

Dimensional formulae of the quantities are:

$[T]=[M_{0}L_{0}T_{1}]$

$[l]=[M_{0}L_{1}T_{0}]$

$[g]=[M_{0}L_{1}T_{−2}]$

Let $T=kl_{A}g_{B}$ where $k,A,B$ are constants.

Then, $[T]=[k][l_{A}][g_{B}]$

$[M_{0}L_{0}T_{1}]=[M_{0}L_{A+B}T_{−2B}]$

Equating the powers on LHS and RHS,

$A+B=0$

$−2B=1$

Solving, $A=21 ,B=2−1 $

Hence, time-period is given by: $T=kl_{1/2}g_{−1/2}$

Note:

The established relation between the physical quantities is not unique and hence may or may not be absolutely correct.

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