Question

$\displaystyle \text{Give answer if you know only}\\\\\\\displaystyle \text{Planck's constant h, speed of light c and gravitational constant}\\\\\\\displaystyle \text{ G are used to form a unit of length L and a unit of mass M.}$

$\displaystyle \text{Then the correct option(s) is/are}\\\\\\\displaystyle a.\ M \propto\sqrt C \\\\\\\displaystyle b. \ M \propto\sqrt G\\\\\\\displaystyle c. \ M\propto\sqrt H\\\\\\\displaystyle b. \ L \propto\sqrt G\\\\\\\displaystyle \text{Kindly give good and content quality answer with great explain.}$

Answer 1

First we find mass is proportional to which powers of each h, c and G.

$\text{Let}\ \ M\propto\ h^x\ c^y\ G^z.$

Taking dimensions,

$[M]=[h]^x\ [c]^y\ [G]^z$

We have,

$[M]=M\\ \\ \[[h]=ML^2T^{-1}\\ \\ \[[c]=LT^{-1}\\ \\ \[[G]=M^{-1}L^3T^{-2}$

So,

$M=(ML^2T^{-1})^x\ (LT^{-1})^y\ (M^{-1}L^3T^{-2})^z\\ \\ M=M^x\ L^{2x}\ T^{-x}\ L^y\ T^{-y}\ M^{-z}\ L^{3z}\ T^{-2z}\\ \\ M^1\ L^0\ T^0=M^{x-z}\ L^{2x+y+3z}\ T^{-x-y-2z}$

Now, we're equating powers of similar quantities in both sides of this equation.

$x-z=1\quad\quad\longrightarrow\quad(1)\\ \\ -x-y-2z=0\ \implies\ x+y+2z=0\\ \\ 2x+y+3z=x+y+2z=0\ \implies\ x+z=0\quad\quad\longrightarrow\quad(2)\\ \\ (1)\ \&\ (2)\ \implies\ \underline{\underline{x=\dfrac{1}{2}\quad;\quad z=-\dfrac{1}{2}\quad;\quad\therefore\ y=\dfrac{1}{2}}}$

So,

$M\propto\ h^{\frac{1}{2}}\ c^{\frac{1}{2}}\ G^{-\frac{1}{2}}\ \ \implies\ \ M\propto\ \sqrt{\dfrac{hc}{G}}\\ \\ \\ \therefore\ M\propto\ \sqrt{h}\quad;\quad M\propto\ \sqrt{c}\quad;\quad M\propto\ \sqrt{\dfrac{1}{G}}$

So options (a) and (c) are correct but (b) is incorrect.

Now let's derive an equation for L with respect to h, c and G.

$\textsf{Let}\ \ L\propto\ h^x\ c^y\ G^z.$

On taking dimensions, finally we get,

$M^0\ L^1\ T^0=M^{x-z}\ L^{2x+y+3z}\ T^{-x-y-2z}$

So, equating powers of similar quantities,

$-x-y-2z=0\ \implies\ x+y+2z=0\\ \\ x-z=x+y+2z=0\ \implies\ y+3z=0\\ \\ (2x+y+3z)-(y+3z)=1\ \implies\ 2x=1\\ \\ \implies\ \underline{\underline{x=\dfrac{1}{2}\quad;\quad z=\dfrac{1}{2}\quad;\quad y=-\dfrac{3}{2}}}$

Now,

$L\propto\ h^{\frac{1}{2}}\ c^{-\frac{3}{2}}\ G^{\frac{1}{2}}\ \ \implies\ \ L\propto\ \sqrt{\dfrac{hG}{c^3}}\\ \\ \\ \therefore\ L\propto\ \sqrt h\quad;\quad L\propto\sqrt G\quad;\quad L\propto\sqrt{\dfrac{1}{c^3}}$

So the option (d) [the fourth option] is also correct.