The volume of a liquid flowing out per second of a pipe is given by $V=\frac{\pi}{8} \frac{p r^4}{\eta l}$
Dimension of $$ V=\frac{\text { Dimension of volume }}{\text { Dimension of time }}=\frac{\left[\mathrm{L}^3\right]}{[T]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right] $$
$(\because V$ is the volume of liquid flowing out per second $)$
Dimension of $p=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
Dimension of $\eta=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$
Dimension of $l=[\mathrm{L}]$
Dimension of $r=[\mathrm{L}]$
Dimensions of LHS, $[V]=\frac{\left[L^3\right]}{[T]}=\left[L^3 \mathrm{~T}^{-1}\right]$
Dimensions of RHS, $\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^4\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \times[\mathrm{L}]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right]$
As dimensions of LHS is equal to the dimensions of RHS. Therefore, equation is correct dimensionally.
A physical quantity $X$ is related to four measurable quantities $a$, $b$, $c$ and $d$ as follows $X = a^2 b^3 c^{5/2} d^{-2}$. The percentage error in the measurement of $a$, $b$, $c$ and $d$ are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity $X$? If the value of $X$ calculated on the basis of the above relation is 2.763, to what value should you round off the result?
Given, physical quantity is $X = a^2 b^3 c^{5/2} d^{-2}$.
Maximum percentage error in $X$ is
$$\frac{\Delta X}{X} \times 100 = \pm \left[ 2 \left(\frac{\Delta a}{a} \times 100 \right) + 3 \left(\frac{\Delta b}{b} \times 100 \right) + \frac{5}{2} \left(\frac{\Delta c}{c} \times 100 \right) + 2 \left(\frac{\Delta d}{d} \times 100 \right) \right]$$
$= \pm \left[2(1) + 3(2) + \frac{5}{2}(3) + 2(4) \right]$ %
$= \pm \left[2 + 6 + 7.5 + 8 \right] = \pm 23.5 \%$
Thus, percentage error in quantity $X$ = ± 23.5%.
Mean absolute error in $X$ = $\pm 0.235 \pm 0.24$ (rounding-off up to two significant digits).
The calculated value of $x$ should be rounded off to two significant digits.
$\therefore X$ = 2.8
In the expression $P=E l^2 m^{-5} G^{-2}$, $E$, $m$, $l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.
Given, expression is
$P=E L^2 m^{-5} G^{-2}$
where $E$ is energy
$$[E] = [ML^2 T^{-2}]$$
$m$ is mass
$$[m] = [M]$$
$L$ is angular momentum
$$[L] = [ML^2 T^{-1}]$$
$G$ is gravitational constant
$$[G] = [M^{-1}L^3 T^{-2}]$$
Substituting dimensions of each term in the given expression,
$\begin{aligned} & {[P]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^2 \times[\mathrm{M}]^{-5} \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-2}} \\\\ & =\left[\mathrm{M}^{1+2-5+2} \mathrm{~L}^{2+4-6} \mathrm{~T}^{-2-2+4}\right]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right] \end{aligned}$
Therefore, $P$ is a dimensionless quantity.
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.
We know that, dimensions of
$$ \text { (h) }=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] $$
Dimensions of
(c) $=\left[\mathrm{LT}^{-1}\right]$
Dimensions of gravitational constant $(G)=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$
(i) Let $$ m \propto c^x h^y G^z $$
$$ \Rightarrow m=k c^x h^y G^z $$ ....(i)
where, $k$ is a dimensionless constant of proportionality.
Substituting dimensions of each term in Eq. (i), we get
$$ \begin{aligned} {\left[\mathrm{ML}^0 \mathrm{~T}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$
Comparing powers of same terms on both sides, we get
$$ \begin{array}{r} y-z=1 \,\,\,\, ....(ii) \\ x+2 y+3 z=0 \,\,\,\, ....(iii) \\ -x-y-2 z=0 \,\,\,\, ....(iv) \end{array} $$
Adding Eqs. (ii), (iii) and (iv), we get
$$ 2 y=1 \Rightarrow y=\frac{1}{2} $$
Substituting value of $y$ in Eq. (ii), we get
$$ z=-\frac{1}{2} $$
From Eq. (iv)
$$ x=-y-2 z $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2} $$
Putting values of $x, y$ and $z$ in Eq. (i), we get
$$ \begin{aligned} & m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2} \\\\ & \Rightarrow \quad m=k \sqrt{\frac{c h}{G}} \\ & \end{aligned} $$
$$ \begin{aligned} & \text{Let} \quad L \propto C^x h^y G^z \\\\ & \Rightarrow L=k c^x h^y G^z \,\,\,\, ....(v) \end{aligned} $$
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (v), we get
$$ \begin{aligned} {\left[\mathrm{M}^0 \mathrm{LT}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$
On comparing powers of same terms, we get
$$ \begin{array}{r} y-z=0 \,\,\,\,....(vi)\\ x+2 y+3 z=1 \,\,\,\,....(vii)\\ -x-y-2 z=0 \,\,\,\,....(viii) \end{array} $$
Adding Eqs. (vi), (vii) and (viii), we get
$$ \begin{array}{r} 2 y=1 \\\Rightarrow y=\frac{1}{2} \end{array} $$
Substituting value of $y$ in Eq. (vi), we get
$$ z=\frac{1}{2} $$
From Eq. (viii),
$$ x=-y-2 z $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)=-\frac{3}{2} $$
Putting values of $x, y$ and $z$ in Eq. (v), we get
$$ \begin{aligned} & L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2} \\\\ & L=k \sqrt{\frac{h G}{c^3}} \end{aligned} $$
(iii) Let $T \propto C^x h^y G^z$
$$ \Rightarrow \quad T=k c^x h^y G^z $$ ....(ix)
where, $k$ is a dimensionless constant.
Substituting dimensions of each term in Eq. (ix), we get
$$ \begin{aligned} {\left[M^0 L^0 T\right.} & =\left[L T^{-1}\right]^x \times\left[M^2 T^{-1}\right]^y \times\left[M^{-1} L^3 T^{-2}\right]^z \\\\ & =\left[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}\right] \end{aligned} $$
On comparing powers of same terms, we get
$$ \begin{array}{r} y-z=0 \,\,\,\,....(x) \\ x+2 y+3 z=0 \,\,\,\,....(xi) \\ -x-y-2 z=1 \,\,\,\,....(xii) \end{array} $$
Adding Eqs. (x), (xi) and (xii), we get
$$ \begin{aligned} 2 y & =1 \\ \Rightarrow y & =\frac{1}{2} \end{aligned} $$
Substituting value of $y$ in Eq. (x), we get
$$ z=y=\frac{1}{2} $$
From Eq. (xii),
$$ x=-y-2 z-1 $$
Substituting values of $y$ and $z$, we get
$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2} $$
Putting values of $x, y$ and $z$ in Eq. (ix), we get
$$ \begin{aligned} T & =k c^{-5 / 2} h^{1 / 2} G^{1 / 2} \\\\ T & =k \sqrt{\frac{h G}{c^5}} \end{aligned} $$
An artificial satellite is revolving around a planet of mass $M$ and radius $R$, in a circular orbit of radius $r$. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis, that $T = \frac{k}{R}\sqrt{\frac{r^3}{g}}$, where $k$ is a dimensionless constant and $g$ is acceleration due to gravity.
By Kepler’s third law, $T^2 \propto r^3 \Rightarrow T \propto r^{3/2}$
We know that $T$ is a function of $R$ and $g$.
Let $T \propto r^{3/2}R^ag^b \Rightarrow T = kr^{3/2}R^ag^b$ .....(i)
where, $k$ is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq. (i), we get
$[M^0L^0 T] = k[L^3]^{1/2}[L]^{a}[LT^{-2}]^b = k[L^{a + b + 3/2}T^{-2b}]$
On comparing the powers of same terms, we get
$a + b + 3/2 = 0$ ...(ii)
$-2b = 1 \Rightarrow b = -1/2$ ...(iii)
From Eq. (ii), we get $a-1 / 2+3 / 2=0 \Rightarrow a=-1$
Substituting the values of $a$ and $b$ in Eq. (i), we get
$T = kr^{3/2}R^{-1}g^{-1/2}$
$ T = \frac{k}{R}\sqrt{\frac{r^3}{g}} $
Note: When we are applying formulae, we should be careful about $r$ (radius of orbit) and $R$ (radius of planet).