How to derive the Formula by Dimensional Analysis method?
How to derive the Formula by Dimensional Analysis method?

Derive the Formula by Dimensional Analysis method



Let a physical quantity $x$ depends on the other quantity $P, Q$ and $R$.

Then $x \propto(\mathrm{P})^{\mathrm{a}}(\mathrm{Q})^{\mathrm{b}}(\mathrm{R})^{\mathrm{c}}$

$x=\mathrm{k}(\mathrm{P})^{a}(\mathrm{Q})^{b}(\mathrm{R})^{c}$

Now consider dimensional formula of each quantity in both side –

$a x_{1}+b x_{2}+c x_{3}=x$    …(2)

$a y_{1}+b y_{2}+c y_{3}=y$    …(3)

$a z_{1}+b z_{2}+c z_{3}=z$     …(4)

After solving equation $(2),(3)$ and $(4)$ value of $a, b$ and $c$ will be $m, n$ and o may be find out

Now substitute the values of $x, y$ and $z$ in equation (1)

Then obtained formula will be $-x=(P)^{m}(Q)^{n}(R)^{0}$

Ex. The time of oscillation (T) depends upon the density ‘d’ radius ‘r’ and surface Tension (s). Obtain the formula for $T$ by the dimensional method.

Sol. $\mathrm{T} \propto(\mathrm{d})^{\mathrm{a}}(\mathrm{r})^{\mathrm{b}}(\mathrm{s})^{\mathrm{c}} \quad \Rightarrow \quad \mathrm{T}=\mathrm{k}(\mathrm{d})^{\mathrm{a}}(\mathrm{r})^{\mathrm{b}}(\mathrm{s})^{\mathrm{c}}$    …(1)

Taking dimension of each quantity in both sides.

$M^{0} L^{0} T^{1}=\left[M^{1} L^{-3} T^{0}\right]^{a}\left[L^{1}\right]^{b} \quad\left[M^{1} L^{0} T^{-2}\right]^{c}$

$\Rightarrow \quad \mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}=\left[\mathrm{M}^{\mathrm{a}+\mathrm{o}}\right]\left[\mathrm{L}^{-3 \mathrm{a}+\mathrm{b}}\right]\left[\mathrm{T}^{-2 \mathrm{c}}\right]$

$\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=\left[\mathrm{M}^{\mathrm{a}}+\mathrm{c} \mathrm{L}^{-3 \mathrm{a}+\mathrm{b}} \mathrm{T}^{-2 \mathrm{c}}\right]$

comparing the dimensions of both sides.

$a+c=0$    …(2)

$-3 a+b=0$    …(3)

$-2 c=1 \quad$ or $\quad c=-1 / 2$      …(4)

Substituting value of $c$ in equation (3)

$a+(-1 / 2)=0 \Rightarrow a=1 / 2$

Now putting $a=1 / 2$ in equation $(3) \quad \Rightarrow-3\left(\frac{1}{2}\right)+b=0 \quad \Rightarrow b=3 / 2$

on substituting value $a, b$ and $c$ in equation $(1) T=k(d)^{1 / 2}(r)^{3 / 2}(s)^{-1 / 2} \Rightarrow T=\sqrt{\frac{r^{3} d}{s}}$

Ex. The kinetic energy of rotation $\mathrm{k}$ depends on the angular momentum $\mathrm{J}$ and moment of inertia I. Find the expression for Kinetic Energy.

Sol. Let $\quad K \propto J^{a} I^{b}$

the $\quad K=C \cdot J^{a} I^{b}$

Writing dimensions on both the sides, we get

$\left[\mathrm{M} L^{2} T^{-2}\right]=\left[\mathrm{M} L^{2} T^{-1}\right]^{\mathrm{a}} \cdot\left[\mathrm{M} L^{2}\right]^{\mathrm{b}}$

$\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{\mathrm{a}+\mathrm{b}} \mathrm{L}^{2 \mathrm{a}+2 \mathrm{~b}} \mathrm{~T}^{-\mathrm{a}}\right]$

Comparing powers of $T$, we get

$-a=-2 \quad$ or $\quad a=2$

Comparing powers of $M$, we get

$a+b=1 \quad$ or $\quad 2+b=1 \quad$ or $\quad b=-1$

Putting these values of ‘a’ and ‘b’ in eq. (i), we get

$\mathrm{K}=\frac{\mathrm{C} \cdot \mathrm{J}^{2}}{\mathrm{I}}$

The value of constant $C$ cannot be found.

Ex. If $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}^{3}}{3 \mathrm{Yq}}}$ then find the dimensions of q. Where $\mathrm{T}$ is the time period of bar of mass $\mathrm{M}$, length $\mathrm{L}$ and Young modulus $Y$.

Sol. $\quad T=2 \pi \sqrt{\frac{M L^{3}}{3 Y q}}$, writing dimensions of both the sides, we get

$[T]=\left[\frac{M L^{3}}{M L^{-1} T^{-2} q}\right]^{1 / 2} \quad$ or $\quad q=\left[L^{4}\right]$

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