Question

Consider a new system of units in which c (speed of light in vacuum), h (Planck's constant), and G (gravitational constant) are taken as fundamental units. Which of the following would correctly represent mass in this new system?
 $ \left( A \right){\text{ }}\sqrt {\dfrac{{hc}}{G}} $
 $ \left( B \right){\text{ }}\sqrt {\dfrac{{Gc}}{h}} $
 $ \left( C \right){\text{ }}\sqrt {\dfrac{{hG}}{c}} $
 $ \left( D \right){\text{ }}\sqrt {hGc} $

Answer 1

Hint: For solving this question we will use and the concept of dimensional analysis. And for this, we will assume the mass is proportional to the $ {c^x}{G^y}{h^z} $ . And from here we will be able to calculate the mass and by substituting the dimension of mass in the equation formed will be the dimension of the new system.

Formula used
The dimension of the mass is given by, $ m = \left[ {{M^1}{L^0}{T^0}} \right] $
Here, $ m $ will be the mass
 $ M $ , will be the dimension of mass
 $ L $ , will be the dimension of length
 $ T $ , will be the dimension of time
The dimension of the speed of light is given by, $ c = \left[ {L{T^{ - 1}}} \right] $
Here, $ c $ will be the speed of light
The dimension of gravitational constant is given by, $ G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right] $
The dimension of Planck’s constant is given by, $ h = \left[ {M{L^2}{T^{ - 1}}} \right] $

Complete step by step answer
So here in this question we have the units given as the velocity of light, Planck’s constant, and gravitational constant. By using this we have the relation which is given by,
 $ \Rightarrow m \prec {c^x}{G^y}{h^z} $ , since we know that the mass is directly proportional to all of these units.
So by equating the above relation we will get the equation as
 $ \Rightarrow m = K{c^x}{G^y}{h^z} $
Here, the term $ K $ will be constant.
Since we know that the constant is a dimensionless quantity. Therefore, on equating the above equation with the dimensions, we will get the equation as
 $ \Rightarrow \left[ {{M^1}{L^0}{T^0}} \right] = {\left[ {L{T^{ - 1}}} \right]^x}{\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^y}{\left[ {M{L^2}{T^{ - 1}}} \right]^z} $
Now by using the power properties the above equation can be written as
 $ \Rightarrow \left[ {{M^1}{L^0}{T^0}} \right] = \left[ {{M^{ - y + z}}{L^{x + 3y + 2z}}{T^{ - x - 2y - z}}} \right] $
Now from here, we will compare the powers, so we will get the equations as
 $ \Rightarrow - y + z = 1 $ , and we will name it equation $ 1 $
Similarly, $ x + 3y + 2z = 0 $ , and we will name it equation $ 2 $ and
 $ \Rightarrow - x - 2y - z = 0 $ , and we will name it equation $ 3 $
Now on adding the equation $ 2\& 3 $ , and we will get the equation
Since, the same term will cancel out so,
 $ \Rightarrow y + z = 0 $ , and we will name it equation $ 4 $
Now on solving the equation $ 1\& 4 $ , we will get value as,
 $ \Rightarrow 2z = 1 $
And on solving it we get
 $ \Rightarrow z = \dfrac{1}{2} $
And putting this value in the equation $ 4 $
We get
 $ \Rightarrow y = \dfrac{{ - 1}}{2} $
Similarly for solving the value of $ x $ , we will substitute the value in the equation $ 3 $ so we get
 $ \Rightarrow - x - 2 \times \dfrac{{ - 1}}{2} - \dfrac{1}{2} = 0 $
And on solving it we get
 $ \Rightarrow - x + 1 - \dfrac{1}{2} = 0 $
And from here on solving it, the value for the $ x $ will be equal to
 $ \Rightarrow x = \dfrac{1}{2} $
Since we know $ m = K{c^x}{G^y}{h^z} $
And we also know mass is constant therefore its value will be equal to one.
Hence on substituting the values, we will get the equation as
 $ \Rightarrow m = 1 \cdot {c^{\dfrac{1}{2}}}{G^{ - \dfrac{1}{2}}}{h^{\dfrac{1}{2}}} $
And the above equation can also be written as
 $ \Rightarrow m = {\text{ }}\sqrt {\dfrac{{hc}}{G}} $
Therefore, the correct answer is an option $ \left( A \right) $ .

Note
As we can see from this question the solution part becomes very easy if we already had remembered the value of the different dimensions. Also while solving the last value $ x $ at that time we could have used any of the equations from them.