Question 181: 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: