Ncert Exercises & Solutions

40. If the velocity of light c, Plancks constant h, and gravitational constant G are taken as fundamental quantities, then express mass, length, and time in terms of dimensions of these quantities.

Hint: Use the principle of the derivation of formula through dimensional analysis.

Step 1: Find the dimensions of individual quantities.

$\begin{array}{ll} We know that dimensions of & (h) = [{ML}^{2} T^{- 1}] \\ Dimensions of & (c) = [{LT}^{- 1}] \\ Dimensions of gravitational constant (G) & = [M^{- 1} L^{3} T^{- 2}] \end{array}$
$(i) Step 2 : Define the dependence of m with C, h and G .$
$\begin{array}{ll} Let & m \propto C^{x} h^{y} G^{z} \\ \Rightarrow & m = {kC}^{x} h^{y} G . . . (i) \end{array}$

where k is a dimensionless constant of proportionality.

Step 3: Put the dimensions of each quantity.
Substituting dimensions of each term in Eq (i), we get,

$\begin{array}{l} [{ML}^{0} T^{0}] = [{LT}^{- 1}]^{x} \times [{ML}^{2} T^{- 1}]^{γ} [M^{- 1} L^{3} T^{- 2}]^{2} \\ = [M^{y - z} L^{x + 2 y + 3 z} T^{- x - y - 2 z}] \end{array}$

Comparing powers of the same terms on both sides, we get,

y - z = 1 ...(I)
x + 2y + 3z = 0 ...(iii)
- x - y - 2z = 0 ...(iii)

Adding Eqs. (i). (i) and (v), we get,
$2 y = 1 \Rightarrow y = \frac{1}{2}$
Substituting the value of y in Eq. (i), we get,
$z = - \frac{1}{2}$
From Eq. (iv)
x = - y - 2z
Substituting values of y and z, we get
$x = - \frac{1}{2} - 2 (- \frac{1}{2}) = \frac{1}{2}$
Putting values of x, y, and z in Eq. (i), we get

$\begin{array}{l} m = {kc}^{1 / 2} h^{1 / 2} G^{- 1 / 2} \\ \Rightarrow m = k \sqrt{\frac{ch}{G}} \end{array}$

(ii) $Step 4 : Define the dependence of L with C, h and G .$

$\begin{array}{ll} Let & L \propto c^{x} h^{y} G^{z} \\ \Rightarrow & L = {kc}^{x} h^{y} G^{z} \end{array}$

where K is a dimensionless constant.

Step 3: Put the dimensions of each quantity.

Substituting dimensions of each term in Eq. (v), we get

$[M^{0} {LT}^{0}] = [{LT}^{- 1}]^{x} \times [{ML}^{2} T^{- 1}]^{γ} \times [M^{- 1} L^{3} T^{- 2}]^{z}$
$= [M^{y - z} ⌊^{x + 2 y + 3 z} T^{- x - y - 2 z}]$

On comparing powers of same terms, we get
y - z = 0 ...(vi)
x + 2y + 3z = 1 ...(vii)
- x - y - 2z = 0 ...(viii)

(iii) Let $T \propto c^{a} h^{b} G^{c}$

$\Rightarrow T = c^{a} h^{b} G^{c} . . . . . . . . . . . (ix)$

where, k is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$[M^{0} L^{0} T^{1}] = {[{LT}^{- 1}]}^{a} \times {[{ML}^{2} T^{- 1}]}^{b} \times {[M^{- 1} L^{3} T^{- 2}]}^{c}$
$= [M^{b - c} L^{a + 2 b + 3 c} T^{- a - b - 2 c}]$

On comparing powers of the same terms, we get

$b - c = 0 . . . (x)$
$a + 2 b + 3 c = 1 . . . (xi)$
$- a - b - 2 c = 1 . . . (xii)$

Adding Eqs. (x), (xi) and (xii), we get

$2 b = 1 \Rightarrow b = \frac{1}{2}$

Substituting the value of b in Eq. (x), we get

$c = b = \frac{1}{2}$

From Eq. (xii),

$a = - b - 2 c - 1$

Substituting values of b and c, we get

$a = - \frac{1}{2} - 2 (\frac{1}{2}) - 1 = - \frac{5}{2}$

Putting values of a, b and c in Eq. (ix), we get

$T = {kc}^{- 5 / 2} h^{1 / 2} G^{1 / 2} = k \sqrt{\frac{hG}{c^{5}}}$