To determine the relationships between the fundamental dimensions of mass M M M L L L h h h c c c G G G M \text{M} M L \text{L} L T \text{T} T
Speed of light (c c c
[ c ] = L T − 1 [c] = \text{L T}^{-1} [ c ] = L T − 1
Planck's constant (h h h
Since energy E = h ν ⟹ [ h ] = [ E ] [ ν ] = M L 2 T − 2 T − 1 E = h \nu \implies [h] = \frac{[E]}{[\nu]} = \frac{\text{M L}^2 \text{T}^{-2}}{\text{T}^{-1}} E = h ν ⟹ [ h ] = [ ν ] [ E ] = T − 1 M L 2 T − 2 [ h ] = M L 2 T − 1 [h] = \text{M L}^2 \text{T}^{-1} [ h ] = M L 2 T − 1
Newton's gravitational constant (G G G
Since force F = G m 1 m 2 r 2 ⟹ [ G ] = [ F ] [ r ] 2 [ m ] 2 = ( M L T − 2 ) L 2 M 2 F = \frac{G m_1 m_2}{r^2} \implies [G] = \frac{[F][r]^2}{[m]^2} = \frac{(\text{M L T}^{-2}) \text{L}^2}{\text{M}^2} F = r 2 G m 1 m 2 ⟹ [ G ] = [ m ] 2 [ F ] [ r ] 2 = M 2 ( M L T − 2 ) L 2 [ G ] = M − 1 L 3 T − 2 [G] = \text{M}^{-1} \text{L}^3 \text{T}^{-2} [ G ] = M − 1 L 3 T − 2
Step 1: Finding the expression for Length L L L
Let L = h a c b G c L = h^a c^b G^c L = h a c b G c
L = ( M L 2 T − 1 ) a ( L T − 1 ) b ( M − 1 L 3 T − 2 ) c \text{L} = (\text{M L}^2 \text{T}^{-1})^a (\text{L T}^{-1})^b (\text{M}^{-1} \text{L}^3 \text{T}^{-2})^c L = ( M L 2 T − 1 ) a ( L T − 1 ) b ( M − 1 L 3 T − 2 ) c L = M a − c L 2 a + b + 3 c T − a − b − 2 c \text{L} = \text{M}^{a-c} \text{L}^{2a+b+3c} \text{T}^{-a-b-2c} L = M a − c L 2 a + b + 3 c T − a − b − 2 c
Equating the exponents of M \text{M} M L \text{L} L T \text{T} T
For M \text{M} M a − c = 0 ⟹ a = c a - c = 0 \implies a = c a − c = 0 ⟹ a = c
For L \text{L} L 2 a + b + 3 c = 1 2a + b + 3c = 1 2 a + b + 3 c = 1
For T \text{T} T − a − b − 2 c = 0 ⟹ a + b + 2 c = 0 -a - b - 2c = 0 \implies a + b + 2c = 0 − a − b − 2 c = 0 ⟹ a + b + 2 c = 0
Substitute c = a c = a c = a L \text{L} L T \text{T} T 5 a + b = 1 ⟹ b = 1 − 5 a 5a + b = 1 \implies b = 1 - 5a 5 a + b = 1 ⟹ b = 1 − 5 a 3 a + b = 0 ⟹ b = − 3 a 3a + b = 0 \implies b = -3a 3 a + b = 0 ⟹ b = − 3 a
Equating the two expressions for b b b 1 − 5 a = − 3 a ⟹ 2 a = 1 ⟹ a = 1 2 1 - 5a = -3a \implies 2a = 1 \implies a = \frac{1}{2} 1 − 5 a = − 3 a ⟹ 2 a = 1 ⟹ a = 2 1
Thus:
c = 1 2 c = \frac{1}{2} c = 2 1 b = − 3 ( 1 2 ) = − 3 2 b = -3\left(\frac{1}{2}\right) = -\frac{3}{2} b = − 3 ( 2 1 ) = − 2 3
Therefore, the unit of length L L L L = h 1 / 2 c − 3 / 2 G 1 / 2 = h G c 3 L = h^{1/2} c^{-3/2} G^{1/2} = \sqrt{\frac{h G}{c^3}} L = h 1/2 c − 3/2 G 1/2 = c 3 h G
From this relation, we see that:
L ∝ h L \propto \sqrt{h} L ∝ h L ∝ G L \propto \sqrt{G} L ∝ G
This makes options (C) and (D) correct.
Step 2: Finding the expression for Mass M M M
Let M = h x c y G z M = h^x c^y G^z M = h x c y G z
M = ( M L 2 T − 1 ) x ( L T − 1 ) y ( M − 1 L 3 T − 2 ) z \text{M} = (\text{M L}^2 \text{T}^{-1})^x (\text{L T}^{-1})^y (\text{M}^{-1} \text{L}^3 \text{T}^{-2})^z M = ( M L 2 T − 1 ) x ( L T − 1 ) y ( M − 1 L 3 T − 2 ) z M = M x − z L 2 x + y + 3 z T − x − y − 2 z \text{M} = \text{M}^{x-z} \text{L}^{2x+y+3z} \text{T}^{-x-y-2z} M = M x − z L 2 x + y + 3 z T − x − y − 2 z
Equating the exponents of M \text{M} M L \text{L} L T \text{T} T
For M \text{M} M x − z = 1 ⟹ x = z + 1 x - z = 1 \implies x = z + 1 x − z = 1 ⟹ x = z + 1
For L \text{L} L 2 x + y + 3 z = 0 2x + y + 3z = 0 2 x + y + 3 z = 0
For T \text{T} T − x − y − 2 z = 0 ⟹ x + y + 2 z = 0 -x - y - 2z = 0 \implies x + y + 2z = 0 − x − y − 2 z = 0 ⟹ x + y + 2 z = 0
Substitute x = z + 1 x = z + 1 x = z + 1 L \text{L} L T \text{T} T 2 ( z + 1 ) + y + 3 z = 0 ⟹ 5 z + y + 2 = 0 ⟹ y = − 5 z − 2 2(z+1) + y + 3z = 0 \implies 5z + y + 2 = 0 \implies y = -5z - 2 2 ( z + 1 ) + y + 3 z = 0 ⟹ 5 z + y + 2 = 0 ⟹ y = − 5 z − 2 ( z + 1 ) + y + 2 z = 0 ⟹ 3 z + y + 1 = 0 ⟹ y = − 3 z − 1 (z+1) + y + 2z = 0 \implies 3z + y + 1 = 0 \implies y = -3z - 1 ( z + 1 ) + y + 2 z = 0 ⟹ 3 z + y + 1 = 0 ⟹ y = − 3 z − 1
Equating the two expressions for y y y − 5 z − 2 = − 3 z − 1 ⟹ 2 z = − 1 ⟹ z = − 1 2 -5z - 2 = -3z - 1 \implies 2z = -1 \implies z = -\frac{1}{2} − 5 z − 2 = − 3 z − 1 ⟹ 2 z = − 1 ⟹ z = − 2 1
Thus:
x = − 1 2 + 1 = 1 2 x = -\frac{1}{2} + 1 = \frac{1}{2} x = − 2 1 + 1 = 2 1 y = − 3 ( − 1 2 ) − 1 = 1 2 y = -3\left(-\frac{1}{2}\right) - 1 = \frac{1}{2} y = − 3 ( − 2 1 ) − 1 = 2 1
Therefore, the unit of mass M M M M = h 1 / 2 c 1 / 2 G − 1 / 2 = h c G M = h^{1/2} c^{1/2} G^{-1/2} = \sqrt{\frac{h c}{G}} M = h 1/2 c 1/2 G − 1/2 = G h c
From this relation, we see that:
M ∝ c M \propto \sqrt{c} M ∝ c M ∝ 1 G M \propto \frac{1}{\sqrt{G}} M ∝ G 1 M ∝̸ G M \not\propto \sqrt{G} M ∝ G
This makes option (A) correct and option (B) incorrect.
Conclusion
The correct options are A, C, and D .