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`. Then the correct option `(s)` is `(are)`

Options

A. `M prop sqrt(c )`
B. `M prop sqrt(G )`
C. `L prop sqrt(h )`
D. `L prop sqrt(G)`

Explanation

To solve the problem, we need to express the units of length \( L \) and mass \( M \) in terms of Planck's constant \( h \), the speed of light \( c \), and the gravitational constant \( G \). We will derive the relationships step by step.

### Step 1: Determine Dimensions of Constants
1. **Planck's Constant \( h \)**:
- From the relation \( E = h \nu \), where \( E \) is energy and \( \nu \) is frequency:
\[ h = \frac{E}{\nu} \]
- The dimension of energy \( E \) is \( [E] = ML^2T^{-2} \) and the dimension of frequency \( \nu \) is \( [\nu] = T^{-1} \).
- Therefore, the dimension of \( h \) is:
\[ [h] = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1} \]

2. **Speed of Light \( c \)**:
- The dimension of speed is:
\[ [c] = LT^{-1} \]

3. **Gravitational Constant \( G \)**:
- From the formula \( F = \frac{G m_1 m_2}{r^2} \):
\[ G = \frac{F r^2}{m_1 m_2} \]
- The dimension of force \( F \) is \( [F] = ML T^{-2} \) and the dimension of radius \( r \) is \( L \).
- Thus, the dimension of \( G \) is:
\[ [G] = \frac{MLT^{-2} \cdot L^2}{M^2} = M^{-1}L^3T^{-2} \]

### Step 2: Express Length \( L \) in Terms of \( h \), \( c \), and \( G \)
Assume:
\[ L = h^a c^b G^c \]
Substituting the dimensions:
\[ [L] = [h]^a [c]^b [G]^c \]
This gives:
\[ L = (ML^2T^{-1})^a (LT^{-1})^b (M^{-1}L^3T^{-2})^c \]
Expanding this:
\[ L = M^{a-c} L^{2a+b+3c} T^{-a-b-2c} \]

### Step 3: Equate Dimensions
For \( L \):
- The left-hand side has dimensions \( M^0 L^1 T^0 \).
- Therefore, we set up the equations:
1. \( a - c = 0 \) (for mass)
2. \( 2a + b + 3c = 1 \) (for length)
3. \( -a - b - 2c = 0 \) (for time)

### Step 4: Solve the Equations
From the first equation:
\[ a = c \]
Substituting \( c = a \) into the other two equations:
- From the second equation:
\[ 2a + b + 3a = 1 \implies 5a + b = 1 \implies b = 1 - 5a \]
- From the third equation:
\[ -a - (1 - 5a) - 2a = 0 \implies -a - 1 + 5a - 2a = 0 \implies 2a = 1 \implies a = \frac{1}{2} \]
Thus, \( c = \frac{1}{2} \) and substituting \( a \) back:
\[ b = 1 - 5 \cdot \frac{1}{2} = 1 - \frac{5}{2} = -\frac{3}{2} \]

### Step 5: Write the Dependence of Length
So, we have:
\[ L \propto h^{1/2} c^{-3/2} G^{1/2} \]
This means:
\[ L \propto \sqrt{h} \cdot \frac{1}{c^{3/2}} \cdot \sqrt{G} \]

### Step 6: Express Mass \( M \) in Terms of \( h \), \( c \), and \( G \)
Assume:
\[ M = h^a c^b G^c \]
Following similar steps as above, we will derive:
1. \( a - c = 1 \)
2. \( 2a + b + 3c = 0 \)
3. \( -a - b - 2c = 0 \)

Solving these will yield:
\[ a = \frac{1}{2}, b = \frac{1}{2}, c = -\frac{1}{2} \]
Thus:
\[ M \propto h^{1/2} c^{1/2} G^{-1/2} \]
This means:
\[ M \propto \sqrt{h} \cdot \sqrt{c} \cdot \frac{1}{\sqrt{G}} \]

### Conclusion
The correct options based on the derived relationships are:
- \( L \propto \sqrt{h} \) and \( L \propto \sqrt{G} \)
- \( M \propto \sqrt{h} \) and \( M \propto \sqrt{c} \)