Question

A physical quantity of the dimensions of length
ATEO
that can be formed out of c G and is lc is
velocity of ight, G is universal constant of
gravitation and e is charge) INEET-2017) with explain​

Answer 1

Answer:

Given:

Given quantities are c , G and c²/4πε .

To find:

Relationship between these quantities such that a quantity with dimension of length can be formed.

Concept :

First try to find out the dimensions of each of these quantities by relating them with Famous Equation :

$\boxed{ \huge{ \green{ \dfrac{1}{4\pi \epsilon} \dfrac{ {e}^{2} }{ {r}^{2} } = force}}}$

So Dimensions of the required Quantity will be :

$\boxed{ \large{ \green{ \dfrac{ {e}^{2} }{4\pi \epsilon} = force \times {r}^{2} }}}$

$\large{ \green{ = > \dfrac{ {e}^{2} }{4\pi \epsilon} = \bigg \{ML {T}^{ - 2} \times {L}^{2} \bigg \} }}$

$\large{ \green{ = > \dfrac{ {e}^{2} }{4\pi \epsilon} = \bigg \{M {L}^{3} {T}^{ - 2} \bigg \} }}$

Similarly :

$\boxed{ \huge{ \red{force = G \dfrac{ {m}^{2} }{ {r}^{2} } }}}$

$\large{ \red{ = > G = \dfrac{force \times {r}^{2} }{ {m}^{2} }}}$

$\large{ \red{ = > G = \bigg \{\dfrac{ML {T}^{-2} \times {L}^{2} }{ {M }^{2} } \bigg \}}}$

$\large{ \red{ = > G = \bigg \{ {M}^{ - 1} {L}^{3} {T}^{ - 2} \bigg \}}}$

Similarly :

$\large{ \blue{c = \bigg \{L {T}^{ - 1} \bigg \} }}$

If we see option 1 )

$\boxed{ \orange{\dfrac{1}{ {c}^{2} } \bigg( { \dfrac{G {e}^{2} }{4\pi \epsilon} \bigg) }^{ \frac{1}{2} } = L }}$

So option 1) is correct

Answer 2

Question:

A physical quantity of the dimensions of length that can be formed out of c, G and $\sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$ is.

[c is velocity of light, G is the universal constant of gravitation and e is charge]

$\sf a) \: c^2\Bigg[G \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\Bigg]^{1/2}$

$\sf b) \: \dfrac{1}{c^2}\Bigg[ \dfrac{ {e}^{2} }{G4\pi\varepsilon_{0}}\Bigg]^{1/2}$

$\sf c) \: \dfrac{1}{c}G \dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}$

$\sf d) \: \dfrac{1}{c^2}\Bigg[ G\dfrac{ {e}^{2} }{4\pi\varepsilon_{0}}\Bigg]^{1/2}$

Solution:

Let's suppose the physical quantity formed by c, G and $\sf \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$ is $[c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c$

According to the question, it is a physical quantity for length.

So,

$L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c$  

Use dimensional analysis,

Dimensions of the following are:

• L= [L]
• $c=[LT^-1]$
• $G=[M^{-1}L^3T{-2}]$
• $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]$

Why $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}} =[ML^3T^{-2}]?$

Use dimensional analysis for each quantity in $\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}$

Here we have

$e^2=[q] ^2=[AT]^2$

$\epsilon_0=[M^{-1}L^{-3}T^4A^2]$

Rest are Constants

Combining these two according to the formula we get,

$\dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}=\frac{[AT]^2}{[M^{-1}L^{-3}T^4A^2]}=\frac{1}{[M^{-1}L^{-3}T^2]} = [ML^3T^{-2}]$

Now, equating the dimensional formulas for

$L= [c]^a[G] ^b \bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^c \\\\ L=[LT^-1]^a[M^{-1}L^3T{-2}]^b[ML^3T^{-2}] ^c \\\\L=[L^aT^{-a}] [M^{-b} L^{3b}T^{-2b}][M^cL^{3c}T^{-2c}] \\\\ L=[L^{a+3b+3c}][M^{-b+c}][T^{-a-2b-2c}]$

Now,equating the exponent power we have,

$a+3b+3c=1 - - - - [i]\\\\ -b+c=0 - - - - [ii] \\\\ - a-2b-2c=0 - - - - [iii] \\\\ From [ii], we~have~b=c,~using~this~value~in~[iii] ~we, ~get \\\\ - a-2c-2c=0 \\\\ - a-4c=0 a=-4c \\\\ Using~a=-4c~and~b=c~in~[i]~we~have~\\\\ - 4c+3c+3c=1 \\\\ 2c=1 \\\\ c=\frac{1}{2} =b \\\\ We~ know~that~a=-4c ~so, ~a=-4 \times \frac{1}{2} ~, a=-2$  

The physical quantity so obtained is

$L= [c]^-2[G] ^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ => \frac{1}{c^2} G^{\frac{1}{2}}\bigg[ \dfrac{ {e}^{2} }{4\pi \varepsilon_{0}}\bigg]^{\frac{1}{2}} \\\\ =\frac{1}{c^2} \bigg[G \frac{e^{2} }{4\pi \varepsilon_{0}} \bigg]^{\frac{1}{2}}$