Dimension of length using h,G,c

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Dimension of length using h,G,c

[h] = [F r]

##[G] =[ \frac { Fr^2}{m^2} ]

\\ [\frac { hG}c] = [L] ##

So, the answer is option (b).

Is this correct?

 

[Pushoam]

Mistake: [h ] ## \neq ## [Fr]

[h] = [mv r]

[L] = [ ## \sqrt{ \frac { hG}{c^3}}] ##

Is this correct?

 

[rude man]

Keep guessing & if we say 'yes' or 'no' you'll eventually hit it, won't you?
So rather than answer 'yes' or 'no' we prefer that you show how you arrived at your answer.
BTW why do you distinguish "r" from "L"? They're the same. For example, [G] =M^-1L³T^-2 etc. , don't need the "r". In SI (mks) there are only 4 dimensions, to wit, M,L,T and Q. (In cgs there are only the first three).

 

[haruspex]

Is this correct?

Yes, but as rude man says your working would be clearer if you were to first reduce all the parameters to the standard set M, L, T... and introduce unknowns for the exponents of tne parameters.

 

[Pushoam]

Keep guessing & if we say 'yes' or 'no' you'll eventually hit it, won't you?
So rather than answer 'yes' or 'no' we prefer that you show how you arrived at your answer.
BTW why do you distinguish "r" from "L"? They're the same. For example, [G] =M^-1L³T^-2 etc. , don't need the "r". In SI (mks) there are only 4 dimensions, to wit, M,L,T and Q. (In cgs there are only the first three).

I felt that I could solve the question without converting the dimensions into M,L,T. So, I went that way. I had the impression that that approach is faster. It may be that this impression is wrong.
I think you are suggesting me to do the following way:
I should write dimensions of G, h, c in M,L, T respectively.
And then I should solve the following dimensional equation.
[L] = ##[G]^p[h]^q [c]^r ##
I will get one equation for each dimension. This will give me the values of p,q,r. Thus I will reach the answer.