Question 35: Units , Dimensions and Error Analysis

Question

$ 	ext { Match List - I with List - II. } $
$
	ext { List - I }
$
$
	ext { List - II }
$
A.
Meter (L)
I.
$
\sqrt{\frac{h c}{G}}
$
B.
Second (S)
II.
$
\sqrt{\frac{G h}{c^5}}
$
C.
Kilogram (M)
III.
$
\sqrt{\frac{K^2 L^2 c^3}{G h}}
$
D.
Kelvin (K)
IV.
$
\sqrt{\frac{G h}{c^3}}
$
where h (Planck's constant), G (gravitational constant) and c (speed of light in vacuum) as fundamental units.
Choose the correct answer from the options given below :

Accepted Answer

The speed of light (c) is $\frac{	ext { distance }}{	ext { time }}$.
So, the dimensions of speed of light is $[\mathrm{c}]=\left[\mathrm{L} \mathrm{T}^{-1}ight]$
From $\mathrm{E}=\mathrm{h} 
u$, where E is energy $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]$ and $
u$ is frequency $\left[\mathrm{T}^{-1}ight]$.
So, the dimensions of Planck's constant is,
$ [\mathrm{h}]=\frac{[\mathrm{E}]}{[v]}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{\left[\mathrm{T}^{-1}ight]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight] $
From Newton's law of gravitation, $\mathrm{F}=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}$, where F is gravitational force $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}ight], \mathrm{G}$ is the universal gravitational constant, m is the mass $[\mathrm{m}]=\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0ight]$ and r is the distance $[\mathrm{r}]=\left[\mathrm{M}^0 \mathrm{LT}^0ight]$
So, the dimensions of G is,
$ [\mathrm{G}]=\frac{[\mathrm{F}]\left[\mathrm{r}^2ight]}{\left[\mathrm{m}^2ight]}=\frac{\left[\mathrm{MLT}^{-2}ight]\left[\mathrm{L}^2ight]}{\left[\mathrm{M}^2ight]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight] $
Let a physical quantity Q be proportional to $\mathrm{h}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{c}^{\mathrm{z}}$. Using principle of dimensional homogeneity
$ [\mathrm{Q}]=[\mathrm{h}]^{\mathrm{x}}[\mathrm{G}]^{\mathrm{y}}[\mathrm{c}]^{\mathrm{z}} $
$\Rightarrow $ $[Q]=\left[M L^2 T^{-1}ight]^x\left[M^{-1} L^3 T^{-2}ight]^y\left[L T^{-1}ight]^z$
$\Rightarrow $ $ [Q]=M^{x-y} L^{2 x+3 y+z} T^{-x-2 y-z} $
For Kilogram (mass), dimensions are $\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0ight]$.
On comparing the powers of dimensions,
$ x-y=1 \Rightarrow x=1+y \ldots 	ext { (i) } $
$ 2 x+3 y+z=0 \ldots (ii)$
$ -x-2 y-z=0 \Rightarrow z=-x-2 y $....(iii)
Substituting z into equation (i):
$ 2 x+3 y+(-x-2 y)=0 \Rightarrow x+y=0 \Rightarrow x=-y $
From equation (i),
$ -y=1+y \Rightarrow 2 y=-1 \Rightarrow y=-1 / 2 $
Then $\mathrm{x}=1 / 2$.
And $z=-\left(\frac{1}{2}ight)-2\left(-\frac{1}{2}ight)=-\frac{1}{2}+1=\frac{1}{2}$
$ 	ext { kilogram }=h^{1 / 2} G^{-1 / 2} c^{1 / 2}=\sqrt{\frac{h c}{G}} $
So, $\mathrm{M} \propto \sqrt{\frac{\mathrm{hc}}{\mathrm{G}}} \Rightarrow(\mathrm{C}) ightarrow(\mathrm{I})$
For meter (length), the dimensions are $\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0ight]=\left[\mathrm{M}^{\mathrm{x}-\mathrm{y}} \mathrm{L}^{2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}} \mathrm{T}^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}}ight]$
On comparing the powers of dimensions,
$ x-y=0 \Rightarrow x=y $....(i)
$ 2 x+3 y+z=1 \ldots $(ii)
$ -x-2 y-z=0 \Rightarrow z=-x-2 y $.....(iii)
Substituting z into equation (ii):
$ 2 x+3 y-x-2 y=1 \Rightarrow x+y=1 $
Since $\mathrm{x}=\mathrm{y}$, then $2 \mathrm{x}=1 \Rightarrow \mathrm{x}=\frac{1}{2}, \mathrm{y}=\frac{1}{2}$.
Then $\mathrm{z}=-\frac{1}{2}-2\left(\frac{1}{2}ight)=-\frac{3}{2}$
Thus, Meter (L) $=[\mathrm{h}]^{1 / 2}[\mathrm{G}]^{1 / 2}[\mathrm{c}]^{-3 / 2}$
So, $\mathrm{L} \propto \sqrt{\frac{\mathrm{Gh}}{\mathrm{c}^3}} \Rightarrow(\mathrm{~A}) ightarrow(\mathrm{IV})$
For Second (time), the dimensions are $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1ight]$.
On comparing the powers of dimensions,
$ x-y=0 \Rightarrow x=y $ ....(i)
$ \begin{gathered} 2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}=0 \Rightarrow 2 \mathrm{x}+3 \mathrm{x}+\mathrm{z}=0 \Rightarrow \mathrm{z}=-5 \mathrm{x} \ldots (ii)\ -\mathrm{x}-2 \mathrm{y}-\mathrm{z}=1 \Rightarrow-\mathrm{x}-2 \mathrm{x}-\mathrm{z}=1 \Rightarrow-3 \mathrm{x}-\mathrm{z}=1......(iii) \end{gathered} $
Substituting z :
$ -3 x-(-5 x)=1 \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}, y=\frac{1}{2} $
Then $\mathrm{z}=-5\left(\frac{1}{2}ight)=-\frac{5}{2}$.
Thus, Second $(S)=[h]^{1 / 2}[G]^{1 / 2}[c]^{-5 / 2}$
So, $\mathrm{S} \propto \sqrt{\frac{\mathrm{Gh}}{\mathrm{c}^5}} \Rightarrow(\mathrm{~B}) ightarrow(\mathrm{II})$
Let a physical quantity $	heta$ represent the Kelvin be proportional to $\mathrm{h}^{\mathrm{x}} \mathrm{G}^{\mathrm{y}} \mathrm{c}^{\mathrm{z}} \mathrm{K}^{\mathrm{w}}$. Using principle of dimensional homogeneity
$ [	heta]=[\mathrm{h}]^{\mathrm{x}}[\mathrm{G}]^{\mathrm{y}}[\mathrm{c}]^{\mathrm{z}}[\mathrm{~K}]^{\mathrm{w}} $
$\Rightarrow $ $[	heta]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight]^{\mathrm{x}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight]^{\mathrm{y}}\left[\mathrm{LT}^{-1}ight]^{\mathrm{z}}[\mathrm{K}]^{\mathrm{w}}$
$\Rightarrow $ $ [	heta]=M^{x-y} L^{2 x+3 y+z} T^{-x-2 y-z} K^w $
The dimensions of temperature is $[	heta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~K}^1ight]=\left[\mathrm{M}^{\mathrm{x}-\mathrm{y}} \mathrm{L}^{2 \mathrm{x}+3 \mathrm{y}+\mathrm{z}} \mathrm{T}^{-\mathrm{x}-2 \mathrm{y}-\mathrm{z}} \mathrm{K}^{\mathrm{w}}ight]$ On comparing the powers,
$ x-y=0 \Rightarrow x=y \ldots(i) $
$ 2 x+3 y+z=0 \Rightarrow 2 x+3 x+z=0 \Rightarrow z=-5 x \ldots (ii)$
$-x-2 y-z=0 \ldots$ (iii)
$ \mathrm{w}=1 \ldots 	ext { (iv) } $
On simplifying $\sqrt{\frac{\mathrm{K}^2 \mathrm{~L}^2 \mathrm{c}^3}{\mathrm{Gh}}}=\sqrt{\frac{\mathrm{K}^2 \mathrm{~L}^2\left[\mathrm{LT}^{-1}ight]^2}{\left(\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight)\left[\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight]ight]}}=\sqrt{\mathrm{K}^2}=\mathrm{K}$ which is kelvin. Thus, Kelvin $(\mathrm{K})=\sqrt{\frac{\mathrm{K}^2 \mathrm{~L}^2 \mathrm{c}^3}{\mathrm{Gh}}} \Rightarrow(\mathrm{D}) ightarrow(\mathrm{III})$
A - IV, B - II, C - I, D - III
Therefore, the correct option is (B).

Answer

A-II, B-IV, C-I, D-III

Answer

A-IV, B-II, C-I, D-III

Answer

A-IV, B-I, C-II, D-III

Answer

A-III, B-I, C-II, D-IV

Units , Dimensions and Error Analysis

Choose the correct answer: