Question

Planck's constant $h$, speed of light $c$ and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. The relation between $M$, $c$, $G$, $h$ are as under

• A. M $\propto$ $\sqrt{c}$
• B. M $\propto$ $\sqrt{G}$
• C. L $\propto$ $\sqrt{h}$
• D. L $\propto$ $\sqrt{G}$

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

This problem can be solved using dimensional analysis. We are given Planck's constant ($h$), the speed of light ($c$), and the gravitational constant ($G$). We need to find the relations for a unit of length ($L$) and a unit of mass ($M$) formed from these constants.

First, let's determine the dimensions of the given constants in terms of Mass (M), Length (L), and Time (T):

1. Planck's constant ($h$): Dimensions of energy $\times$ time. Energy has dimensions $[ML^2T^{-2}]$. So, $[h] = [ML^2T^{-1}]$.
2. Speed of light ($c$): $[c] = [LT^{-1}]$.
3. Gravitational constant ($G$): From $F = G\frac{m_1m_2}{r^2}$, we get $[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]$.

We assume that the unit of mass $M$ and the unit of length $L$ are formed by some combination of $h$, $c$, and $G$. These fundamental units formed from $h, c, G$ are known as Planck units.

Let's find the expression for Planck mass ($M_P$) and Planck length ($L_P$). Assume $M_P = k_1 h^x c^y G^z$, where $k_1$ is a dimensionless constant. The dimensions of $M_P$ are $[M]$. $[M] = ([ML^2T^{-1}])^x ([LT^{-1}])^y ([M^{-1}L^3T^{-2}])^z$ $[M] = [M^{x-z} L^{2x+y+3z} T^{-x-y-2z}]$

Equating the exponents on both sides: For M: $1 = x - z$ (1) For L: $0 = 2x + y + 3z$ (2) For T: $0 = -x - y - 2z$ (3)

Adding (2) and (3): $(2x+y+3z) + (-x-y-2z) = 0 \implies x+z = 0 \implies x = -z$. Substitute $x=-z$ into (1): $1 = -z - z \implies 1 = -2z \implies z = -1/2$. Then $x = -z = 1/2$. Substitute $x=1/2$ and $z=-1/2$ into (3): $0 = -(1/2) - y - 2(-1/2) \implies 0 = -1/2 - y + 1 \implies y = 1/2$.

So, the Planck mass is proportional to $h^{1/2} c^{1/2} G^{-1/2}$. $M_P \propto \sqrt{\frac{hc}{G}}$.

Now let's find the Planck length ($L_P$). Assume $L_P = k_2 h^a c^b G^d$. The dimensions of $L_P$ are $[L]$. $[L] = ([ML^2T^{-1}])^a ([LT^{-1}])^b ([M^{-1}L^3T^{-2}])^d$ $[L] = [M^{a-d} L^{2a+b+3d} T^{-a-b-2d}]$

Equating the exponents on both sides: For M: $0 = a - d \implies a = d$. For L: $1 = 2a + b + 3d$. For T: $0 = -a - b - 2d$.

From the T equation, $b = -a - 2d$. Substitute $a=d$: $b = -d - 2d = -3d$. Substitute $a=d$ and $b=-3d$ into the L equation: $1 = 2d + (-3d) + 3d \implies 1 = 2d \implies d = 1/2$. So, $a = d = 1/2$, and $b = -3d = -3/2$.

The Planck length is proportional to $h^{1/2} c^{-3/2} G^{1/2}$. $L_P \propto \sqrt{\frac{hG}{c^3}}$.

Now let's examine the given options: (A) $M \propto \sqrt{c}$ From $M_P \propto \sqrt{\frac{hc}{G}}$, we can write $M_P = K_M \sqrt{\frac{h}{G}} \sqrt{c}$. Thus, $M_P$ is proportional to $\sqrt{c}$. This statement is correct.

(B) $M \propto \sqrt{G}$ From $M_P \propto \sqrt{\frac{hc}{G}}$, we have $M_P \propto G^{-1/2}$, which means $M_P$ is inversely proportional to $\sqrt{G}$. This statement is incorrect.

(C) $L \propto \sqrt{h}$ From $L_P \propto \sqrt{\frac{hG}{c^3}}$, we can write $L_P = K_L \sqrt{\frac{G}{c^3}} \sqrt{h}$. Thus, $L_P$ is proportional to $\sqrt{h}$. This statement is correct.

(D) $L \propto \sqrt{G}$ From $L_P \propto \sqrt{\frac{hG}{c^3}}$, we can write $L_P = K_L \sqrt{\frac{h}{c^3}} \sqrt{G}$. Thus, $L_P$ is proportional to $\sqrt{G}$. This statement is correct.