Question 209: Units , Dimensions and Error Analysis

Question

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :

Accepted Answer

Let t $ \propto $ G
x
h
y
c
z
$ 	herefore $   [t] = [G]
x
[h]
y
[c]
z
. . . . . (1)
We know,
F = ${{G{M^2}} \over {{R^2}}}$
$ \Rightarrow $  G = ${{F{R^2}} \over {{M^2}}}$
$ 	herefore $  [G] = ${{\left[ {ML{T^{ - 2}}} ight]\left[ {{L^2}} ight]} \over {\left[ {{M^2}} ight]}}$
[G] = $[{M^{ - 1}}{L^3}{T^{ - 2}}]$
Also,
E = hf
$ 	herefore $  [h] = ${{[E]} \over {[F]}}$
= ${{\left[ {M{L^2}{T^{ - 2}}} ight]} \over {[{T^{ - 1}}]}}$
= $\left[ {M{L^2}{T^{ - 1}}} ight]$
[C] = $\left[ {{M^o}L\,{T^{ - 1}}} ight]$
From equation (1) we get,
$\left[ {{M^o}\,{L^o}\,{T^1}} ight] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} ight]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} ight]^y}{\left[ {{M^o}L{T^{ - 1}}} ight]^z}$
$\left[ {{M^o}\,{L^o}\,{T^1}} ight] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} ight]$
By comparing the power of M, L, T
$-$ x + y = 0
$ \Rightarrow $  x = y
3x + 2y + z = 0
$ \Rightarrow $  5x + z = 0 . . . . .                  (2)
$-$ 2x $-$ y $-$ z = 1
$ \Rightarrow $  $-$ 3x $-$ z = 1  . . . . (3)
By solving (2) and (3), we get,
x = ${1 \over 2}$ = y  and z = $-$ ${5 \over 2}$
$ 	herefore $  t $ \propto $ $\sqrt {{{Gh} \over {{C^5}}}} $

Answer

$\sqrt {{{h{c^5}} \over G}} $

Answer

$\sqrt {{{{c^3}} \over {Gh}}} $

Answer

$\sqrt {{{Gh} \over {{c^5}}}} $

Answer

$\sqrt {{{Gh} \over {{c^3}}}} $

Units , Dimensions and Error Analysis

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :

Choose the correct answer: