Name: Conversion of Units Using Dimensional Analysis
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: How do you solve problems if the units of quantity are different? Is there any conversion method? Let us discuss the conversion using dimensional analysis in this video.

Question

(i) What are the applications of dimensional analysis ? The position $(x)$ of a particle depends on its velocity ( $v$ ) and time $(t)$ as given by the relation. $x = P v + \frac{Q}{P + t} .$ Find the dimensions of $P Q$ .

(ii) If kinetic energy $K$ depends on velocity $v,$ acceleration $a$ and density $(\rho)$ of an object, find the expression for $K .$

Accepted Answer

(i) Application of Dimensional Analysis

Checking the dimensional consistency of physical formulae and equation.

Each of the terms appearing in a mathematical relation should have the same dimensions. If it is the not so the relation is proved wrong. This is called the principle of dimensional homogeneity.

Deducing relation among various physical quantities.

Given equation $x = P v + \frac{Q}{P + t}$

Since $P$ is added to $t,$ therefore $[P] = [t]$

$\Rightarrow [P] = [T^{1}]$ ...(i)

Also, all the terms in an equation should have the same dimensions

$∴ [\frac{Q}{P + t}] = [x]$ $\Rightarrow \frac{[Q]}{[T]} = [L]$ $\Rightarrow [Q] = [L T]$ ...(ii)

From (i) and (ii)

$∴ [PQ] = [T^{1}] [LT] = [{LT}^{2}]$

(ii) Let the required expression be

$K \propto v^{x} ρ^{y} a^{z}$

$[K] = {[{LT}^{- 1}]}^{x} {[{ML}^{- 3}]}^{y} {[{LT}^{- 2}]}^{z}$

$[{ML}^{2} T^{- 2}] = [M^{y} L^{x - 3 y + z} T^{- x - 2 z}]$

Equating the dimensions of like quantities on both sides.

$\Rightarrow y = 1$
$\Rightarrow x - 3 y + z = 2$
$\Rightarrow x + z = 5$
$- x - 2 z = - 2$
$\Rightarrow x + 2 z = 2$

or $z = - 3$

and $x = 8$
$∴ K \propto v^{8} ρ a^{- 3}$

$K \propto \frac{v^{8} ρ}{a^{3}}$

Important Questions on Units and Measurements

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