Planck’s constant \(h\), speed of light in vacuum \(c\), and Newton’s gravitational constant \(G\), are three fundamental constants. Which of the following combinations of these has the dimension of length?
[2016-II]
The Planck length formula is \(l_P = \sqrt{\frac{hG}{c^3}}\). Checking option c: \(\frac{\sqrt{hG}}{c^{3/2}} = h^{1/2}G^{1/2}c^{-3/2}\). \(h=[ML^2T^{-1}]\), \(G=[M^{-1}L^3T^{-2}]\), \(c=[LT^{-1}]\). Thus, \([M^{1/2}L^1T^{-1/2}] [M^{-1/2}L^{3/2}T^{-1}] [L^{-3/2}T^{3/2}] = [M^0L^{1+3/2-3/2}T^{-1/2-1+3/2}] = [L]\).
If energy \((E)\), velocity \((V)\), and time \((T)\), are chosen as the fundamental quantities, the dimensional formula of surface tension will be:
[2015]
Surface tension \(gamma\) has dimensions \(MT^{-2}\). Given fundamental quantities are energy \(E = [ML^2T^{-2}]\), velocity \(V = [LT^{-1}]\), and time \(T = [T]\). Let \(\gamma = E^x V^y T^z\).
If dimension of critical velocity of liquid flowing through a tube are expressed as \(v_c \propto \eta^x \rho^y r^z\) where \(\eta\) and \(\rho\) are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of \(x, y\) and \(z\) are given by:
[2015-Re]
Critical velocity \(v_c = [LT^{-1}]\). Coefficient of viscosity \(\eta = [ML^{-1}T^{-1}]\). Density \(\rho = [ML^{-3}]\). Radius \(r = [L]\). Assume \(v_c \propto \eta^x \rho^y r^z\). Equating dimensions: \([LT^{-1}] = ([ML^{-1}T^{-1}])^x ([ML^{-3}])^y ([L])^z = [M^{x+y} L^{-x-3y+z} T^{-x}]\). Comparing powers: \(x+y = 0\), \(-x-3y+z = 1\), \(-x = -1\). From \(-x = -1\), we get \(x=1\). From \(x+y = 0\), we get \(1+y = 0 ⇒ y=-1\). From \(-x-3y+z = 1\), we get \(-1-3(-1)+z = 1⇒ -1+3+z=1 ⇒ 2+z=1 ⇒ z=-1\). Therefore, \(x=1, y=-1, z=-1\).
A dimensional constant is a physical constant with dimensions. The Gravitational constant \(G\) is a dimensional constant with dimensions \(M^{-1}L^3T^{-2}\).