The correct option is C √hGc3
Let l∝cxhyGz;l=kcxhyGz
where k is a dimensionless constant and x, y and z are the exponents.
Equating dimensions on both sides, we get
[M0LT0]=[LT−1]x[ML2T−1]y[M−1L3T−2]z
=[My−zLx+2y+3zT−x−y−2z]
Applying the principle of homogeneity of dimensions, we get
y−z=0 ....(i)
x+2y+3z=1 ....(ii)
−x−y−2z=0 ...(iii)
On solving Eqs. (i), (ii) and (iii), we get
x=−32,y=12z=12∴l=√hGc3