Units and Measurements — Physics STD 11 Science — Question
Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements4 Marks
Question
State the uses of dimensional analysis.
✓
Answer
Uses of dimensional analysis: To check the correctness of a physical equation.
Correctness of a physical equation by dimensional analysis:
$1.$ A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
$2.$ For example, consider the equation of motion.
$v = u + at ……………. (1)$
$3.$Writing the dimensional formula of every term, we get
Dimensions of $\text{LH.S.} [v] [L^1M^0T^{-1}],$
Dimensions of $\text{R.H.S.}= [u] + [at]$
$= [L^1M^0T^{-1}] + [L^1M^0T^{-2}] [L^1M^0T^{-1}]$
$= [L^1M^0T^{-1}] + [L^1M^0T^{-1}]$
$\Rightarrow \text{[L.HS.] = [R.H.S.]}$
$4.$As dimensions of both side of equation is same, physical equation is dimensionally correct.
To derive the relationship between related physical quantities.
Expression for time period of a simple pendulum by dimensional analysis:
$1.$Time period $(T)$ of a simple pendulum depends upon length $(l)$ and acceleration due to gravity $(g)$ as follows:
$T ∝ l^a g^b$
i.e., $T = k l^a g^b ………… (1)$
where, $k =$ proportionality constant, which is dimensionless.
$2.$The dimensions of $T = [L^0M^0T^1)$
The dimensions of $l = [L^1M^0T^0]$
The dimensions of $g = [L^1M^0T^2]$
Taking dimensions on both sides of equation $(1),$
$[L^0M^0T^1] = [L^1M^0T^0]^a [L^1M^0T^{-2}]^b$
$[L^0M^0T^1] = [L^{a+b}M^0T^{-2b}]$
$3.$Equating corresponding power of $L, M$ and $T$ on both sides, we get
$a + b = 0 …………. (2)$
and $-2b = 1$
$\therefore b = -\frac{1}{2}$
$4.$Substituting $‘b \ ’$ in equation $(2),$ we get
$a = \frac{1}{2}$
$5.$Substituting values of $a$ and $b$ in equation$ (1),$
we have,
$T = k l^{\frac{1}{2}} g{ }^{-\frac{1}{2}}$
$\therefore T = k \frac{l^{\frac{1}{2}}}{g^{\frac{1}{2}}}= k \left(\frac{l}{g}\right)^{\frac{1}{2}}= k \sqrt{\frac{l}{g}}$
$6.$ Experimentally, it is found that $k = 2\pi$
$\therefore T =2 \pi \sqrt{\frac{l}{g}}$
This is the required expression for time period of a simple pendulum.
To find the conversion factor between the units of the same physical quantity in two different systems of units.
Conversion factor between units of same physical quantity:
$1.$ let $‘n \ ’$ be the conversion factor between the units of work.
$\therefore 1 J = n \ erg ………….. (1)$
$2.$ Dimensions of work in $\text{S.l.}$ system are $\left[ L _1^2 M _1^{\prime} T _1^{-2}\right]$ and in $\text{CGS}$ system are $\left[ L _2^2 M _2^1 T_2^{-2}\right]$
$3.$ From $(1),$
$1\left[L_1^2 M _1^1 T_1^{-2}\right]= n \left[ L _2^2 M _2^1 T_2^{-2}\right]$
$\therefore n =\left[\frac{ L _1^2 M _1^1 T_1^{-2}}{L_2^2 M _2^1 T_2^{-2}}\right]$
$ =\left[\frac{ L _1}{L_2}\right]^2\left[\frac{ M _1}{ M _2}\right]^1\left[\frac{ T _1}{T_2}\right]^{-2} .........(2)$
$4.$ By expressing $L , M$ and $T$ into its corresponding unit we have,
$n =\left[\frac{ m }{ \ cm }\right]^2\left[\frac{ \ kg }{ g }\right]^1\left[\frac{\text { second }}{\text { second }}\right]^{-2}........(3)$
$5.$ Since, $1 m=100 \ cm$ and $1 \ kg=1000 g$, we have,
$n =\left(\frac{100 \ cm}{ \ cm }\right)^2\left(\frac{1000 g}{ g }\right)(1)^{-2}$
$n= 10^4 \times 10^3 \times 1 = 10^7$
Hence, the conversion factor, $n = 10^7$
There fore, from equation $(1),$ we have,
$\therefore 1 J = 10^7 \ erg.$
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