Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If G, c and h are the fundamental constant of physics, then the unit of time is expressed as:A) $\sqrt {\left( {\dfrac{{hc}}{G}} \right)}$B) $\sqrt {\left( {\dfrac{{hG}}{{{c^5}}}} \right)}$C) $\left( {\dfrac{{hc}}{G}} \right)$D) $hG{c^3}$

Last updated date: 13th Jun 2024
Total views: 53.1k
Views today: 0.53k
Verified
53.1k+ views
Hint: In this question, we will use the concept of the dimensional analysis. First we will write the given quantities in the dimensional units and then compare their fundamental units with the fundamental unit of the time to express them in the time unit.

Complete step by step solution:
In this question, we have given $G$ , $c$ and $h$ as the elemental constant of physics and we need to express them in the unit of the time.
Let time for the physical constant we expressed as,
$\Rightarrow t \propto {h^a}{c^b}{G^c}$
Now we will remove the directly proposal sign,
$\Rightarrow t = k{h^a}{c^b}{G^c}$
Where, $k$ is dimension constant in the equation, and $a,b,c$ are the exponent.
Now we are equating dimension both the side
$\Rightarrow \left[ {{M^0}{L^0}{T^1}} \right] = {\left[ {M{L^2}{T^{ - 1}}} \right]^a}{\left[ {L{T^{ - 1}}} \right]^b}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^c}$
Simplify the above dimensions and we get,
$\left[ {{M^\circ }{L^\circ }{T^1}} \right] = {\left[ M \right]^{a - c}}{\left[ L \right]^{2a + b + 3c}}\left[ {{T^{ - a - b - 2c}}} \right]$
Now we get the relations,
$\Rightarrow a - c = 0$
Then equal both the constant terms,
$\Rightarrow a = c$
Now we put the value of (a) in equation (2)
$\Rightarrow - a - b - 2c = 1$
Simplify the equation,
$\Rightarrow - c - b - 2c = 1$
$\Rightarrow - 3c - b = 1$.............(2)
Now, put the value of (b) in equation (2)
$\Rightarrow 2a + b + 3c = 0$
Simplify the equation and we get
$2c + b + 3c = 0$
Adding the constant terms then we get,
$\Rightarrow 5c = - b$
Then, we will get the value of c and a.
$- 3c - ( - 5c) = 1$
Simplify the equation,
$\Rightarrow - 3c + 5c = 1$…………… (3)
While adding the values we get,
$c = \dfrac{1}{2},$ and $a = \dfrac{1}{2}$
To find the value of (b), put the value of (c) in equation (3)
$\Rightarrow b = - 5c$
While substitute the values we get,
$\Rightarrow b = \dfrac{{ - 5}}{2}$
On comparing we will get the answer $\sqrt {\left( {\dfrac{{hG}}{{{c^5}}}} \right)}$.

So, the correct answer is option (B).

Note: As we know that the fundamental dimension is the form of the $MLT$ that shows the mass and length and the time of the dimensions. While comparing the values of the given data it shows the physical quantity of the units.