Question

If velocity of c, Plank’s constant h and gravitational constant G are taken as fundamental quantities then the dimensions of length will be

• A. $\sqrt{\frac{ch}{G}}$
• B. $\sqrt{\frac{hG}{c^{5}}}$
• C. $\sqrt{\frac{hG}{c^{3}}}$
• D. $\sqrt{\frac{hc^{3}}{G}}$

Answer 1

Answer: (C)

$\sqrt{\frac{hG}{c^{3}}}$

Answer 2

Let $l \propto c^{x}h^{y}G^{z}$

$l = kc^{x}h^{y}G^{z}$

Where k is a dimensionless constant and x, y and z are the exponents.

Equating dimension on both sides, we get$\lbrack M^{0}LT^{0}\rbrack = \lbrack LT^{- 1}\rbrack^{x}\lbrack ML^{2}T^{- 1}\rbrack^{y}\lbrack M^{- 1}L^{3}T^{- 2}\rbrack^{z} = \lbrack M^{y - z}L^{x + 2y + 3z}T^{- x - y - 2z}\rbrack$

Applying the principle of homogeneity of dimensions,

We get

y – z = 0 ….. (i)

x + 2y + 3y = 1 ….. (ii)

-x – y – 2z = 0 …… (iii)

On solving Eqs. (i), (ii) and (iii), we get

$x = - \frac{3}{2},y = \frac{1}{2},z = \frac{1}{2}\therefore l = \sqrt{\frac{hG}{c^{3}}}$