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Question: If dimensions of length are expressed as \({G^x}{C^y}{h^z}\) where \(G\), \(C\) and \(h\) are univer...

If dimensions of length are expressed as GxCyhz{G^x}{C^y}{h^z} where GG, CC and hh are universal gravitational constant, speed of light and Planck's constant respectively. Then
(A) x=yx = y
(B) x=zx = z
(C) y=−32y = - \dfrac{3}{2}
(D) z=12z = \dfrac{1}{2}

Explanation

Solution

Hint
Here, in this question as the dimensions of length is expressed as GxCyhz{G^x}{C^y}{h^z},as we know that the dimension of length is [M0L1T0]\left[ {{M^0}{L^1}{T^0}} \right], and gravitational constant is [M−1L3T−2]\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right], speed of light is [M0L1T−1]\left[ {{M^0}{L^1}{T^{ - 1}}} \right], Planck’s constant is [M1L2T−1]\left[ {{M^1}{L^2}{T^{ - 1}}} \right], on putting the values in given equation and then equating the equation, we get the desired result.

Complete step by step solution
Here, it is given that the length is expressed as L=GxCyhzL = {G^x}{C^y}{h^z} …………………….. (1)
Where, GG is the gravitational constant,
CC is the speed of light,
hh is the Planck’s constant.
As we know that, the dimension of length is L=[M0L1T0]L = \left[ {{M^0}{L^1}{T^0}} \right]
The dimension of gravitation constant is G=Fr2M1M2=[MLT−2][L2][M2]=[M−1L3T−2]G = \dfrac{{F{r^2}}}{{{M_1}{M_2}}} = \dfrac{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {{M^2}} \right]}} = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]
The dimension of speed of light is C=[M0L1T−1]C = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]
The dimension of Planck’s constant is h=energyCλ=[M1L2T−2][LT−1][L]=[M1L2T−2]h = \dfrac{{energy}}{{\dfrac{C}{\lambda }}} = \dfrac{{\left[ {{M^1}{L^2}{T^{ - 2}}} \right]}}{{\dfrac{{\left[ {L{T^{ - 1}}} \right]}}{{\left[ L \right]}}}} = \left[ {{M^1}{L^2}{T^{ - 2}}} \right]
Now substituting the values in given equation (1), we get
[M0L1T0]=[M−1L3T−2]x[LT−1]y[M1L2T−1]z\left[ {{M^0}{L^1}{T^0}} \right] = {\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^x}{\left[ {L{T^{ - 1}}} \right]^y}{\left[ {{M^1}{L^2}{T^{ - 1}}} \right]^z}
Now, equating the powers on both sides, we get
−x+z=0,3x+y+2z=1,−2x−y−z=0- x + z = 0,3x + y + 2z = 1, - 2x - y - z = 0
Now, on solving the above equations, we get
x=zx = z
Put this value in equation −2x−y−z=0 - 2x - y - z = 0 and 3x+y+2z=13x + y + 2z = 1, we get
−3z−y=0- 3z - y = 0 and 5x+y=15x + y = 1
Solving above equations, we get
y=−32y = - \dfrac{3}{2}
x=12x = \dfrac{1}{2}
z=12z = \dfrac{1}{2}
Hence, here options (B), (C), (D) are correct.

Note
in these types of questions we can find the dimensions by using the direct formula and substitute the dimensions in it. After solving we get the required dimensions and it must be noticed that the dimensions are always written in the square bracket.