If velocity of light, acceleration due to gravity and normal atmospheric pressure are taken as the fundamental units, what will be the units of mass, length and time, given velocity of light as `3xx10^(8)` `ms^(-1)` acceleration due to gravity `10ms^(-2)` and normal pressure `=10^(5)Nm^(-2)`

Explanation

To solve the problem, we need to express the units of mass, length, and time in terms of the fundamental units given: the velocity of light (c), acceleration due to gravity (g), and atmospheric pressure (P).

### Step 1: Define the given quantities
1. **Velocity of Light (c)**: \( c = 3 \times 10^8 \, \text{m/s} \)
2. **Acceleration due to Gravity (g)**: \( g = 10 \, \text{m/s}^2 \)
3. **Normal Atmospheric Pressure (P)**: \( P = 10^5 \, \text{N/m}^2 \)

### Step 2: Write the dimensional formulas
1. **Velocity (v)**:
\[ [v] = \frac{L}{T} \quad \text{(where L is length and T is time)} \]
Therefore, \( [v] = L T^{-1} \).

2. **Acceleration (a)**:
\[ [a] = \frac{L}{T^2} \]
Therefore, \( [g] = L T^{-2} \).

3. **Pressure (P)**:
Pressure is defined as force per unit area. The dimensional formula for force is \( [F] = M L T^{-2} \).
Thus,
\[ [P] = \frac{[F]}{[A]} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 3: Set up equations based on the given values
1. From the velocity of light:
\[ L T^{-1} = 3 \times 10^8 \quad \text{(Equation 1)} \]

2. From acceleration due to gravity:
\[ L T^{-2} = 10 \quad \text{(Equation 2)} \]

3. From atmospheric pressure:
\[ M L^{-1} T^{-2} = 10^5 \quad \text{(Equation 3)} \]

### Step 4: Solve for time (T)
From Equation 1:
\[ L = 3 \times 10^8 T \]

Substituting \( L \) into Equation 2:
\[ (3 \times 10^8 T) T^{-2} = 10 \]
\[ 3 \times 10^8 T^{-1} = 10 \]
\[ T = \frac{3 \times 10^8}{10} = 3 \times 10^7 \, \text{s} \]

### Step 5: Solve for length (L)
Substituting \( T \) back into the expression for \( L \):
\[ L = 3 \times 10^8 (3 \times 10^7) = 9 \times 10^{15} \, \text{m} \]

### Step 6: Solve for mass (M)
Now substituting \( L \) and \( T \) into Equation 3:
\[ M (9 \times 10^{15})^{-1} (3 \times 10^7)^{-2} = 10^5 \]
\[ M \cdot \frac{1}{9 \times 10^{15}} \cdot \frac{1}{9 \times 10^{14}} = 10^5 \]
\[ M \cdot \frac{1}{81 \times 10^{29}} = 10^5 \]
\[ M = 10^5 \cdot 81 \times 10^{29} = 81 \times 10^{34} \, \text{kg} \]

### Final Results
- **Mass (M)**: \( 81 \times 10^{34} \, \text{kg} \)
- **Length (L)**: \( 9 \times 10^{15} \, \text{m} \)
- **Time (T)**: \( 3 \times 10^{7} \, \text{s} \)