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Dimensional Analysis Using G h and c JEE Physics Question

Solve this JEE Physics question based on dimensional analysis and the principle of homogeneity of dimensions. Learn how to find unknown powers using

 

❓ Question

If

F=kGxhyczF=kG^x h^y c^z

where GG is the gravitational constant, hh is Planck's constant, cc is the speed of light and kk is a dimensionless constant, find the values of x,y,zx,y,z.

Dimensional Analysis Using G h and c JEE Physics Question


✍️ Solution

Given,

F=kGxhyczF=kG^x h^y c^z

Dimensional formula of gravitational constant:

[G]=[M−1L3T−2][G]=[M^{-1}L^3T^{-2}]

Dimensional formula of Planck's constant:

[h]=[ML2T−1][h]=[ML^2T^{-1}]

Dimensional formula of speed of light:

[c]=[LT−1][c]=[LT^{-1}]

Also,

[F]=[MLT−2][F]=[MLT^{-2}]

Therefore,

[MLT−2]=[M−1L3T−2]x[ML2T−1]y[LT−1]z[MLT^{-2}] = [M^{-1}L^3T^{-2}]^x [ML^2T^{-1}]^y [LT^{-1}]^z

Comparing powers of M,L,TM,L,T:

−x+y=1-x+y=1
3x+2y+z=13x+2y+z=1
−2x−y−z=−2-2x-y-z=-2

Solving these equations, we get

x=−12,y=12,z=−52x=-\frac12,\qquad y=\frac12,\qquad z=-\frac52

Dimensional Analysis Using G h and c JEE Physics Question
✅ Final Answer

x=−12,y=12,z=−52\boxed{x=-\frac12,\quad y=\frac12,\quad z=-\frac52}

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