Taking foce, length and time to be the fundamental quantities find the dimension of
a. density, b. pressure,
c. momentum and d. energy

Explanation

To find the dimensions of density, pressure, momentum, and energy in terms of force (F), length (L), and time (T), we will follow these steps:

### Step 1: Finding the Dimension of Density
Density (ρ) is defined as mass (m) per unit volume (V):
\[ \rho = \frac{m}{V} \]
Volume (V) can be expressed in terms of length (L):
\[ V = L^3 \]
Thus, the dimension of density becomes:
\[ \rho = \frac{m}{L^3} \]

Next, we need to express mass (m) in terms of force (F), length (L), and time (T). From Newton's second law, we know:
\[ F = m \cdot a \]
Where acceleration (a) can be expressed as:
\[ a = \frac{dv}{dt} = \frac{L}{T^2} \]
Thus, we can rearrange to find mass:
\[ m = \frac{F}{a} = \frac{F \cdot T^2}{L} \]

Now substituting this expression for mass back into the density equation:
\[ \rho = \frac{F \cdot T^2 / L}{L^3} = \frac{F \cdot T^2}{L^4} \]

So, the dimension of density is:
\[ \text{Dimension of Density} = F L^{-4} T^2 \]

### Step 2: Finding the Dimension of Pressure
Pressure (P) is defined as force (F) per unit area (A):
\[ P = \frac{F}{A} \]
Area (A) can be expressed in terms of length (L):
\[ A = L^2 \]
Thus, the dimension of pressure becomes:
\[ P = \frac{F}{L^2} \]

So, the dimension of pressure is:
\[ \text{Dimension of Pressure} = F L^{-2} \]

### Step 3: Finding the Dimension of Momentum
Momentum (p) is defined as mass (m) multiplied by velocity (v):
\[ p = m \cdot v \]
We already have the dimension of mass (m) as:
\[ m = \frac{F \cdot T^2}{L} \]
And velocity (v) can be expressed as:
\[ v = \frac{L}{T} \]

Now substituting these dimensions into the momentum equation:
\[ p = \left(\frac{F \cdot T^2}{L}\right) \cdot \left(\frac{L}{T}\right) = \frac{F \cdot T^2 \cdot L}{L \cdot T} = F \cdot T \]

So, the dimension of momentum is:
\[ \text{Dimension of Momentum} = F L^{0} T^{1} \]

### Step 4: Finding the Dimension of Energy
Energy (E) can be defined in terms of mass (m) and velocity (v):
\[ E = \frac{1}{2} m v^2 \]
Using the dimensions we have:
\[ E = \frac{1}{2} \cdot m \cdot v^2 = \frac{1}{2} \cdot \left(\frac{F \cdot T^2}{L}\right) \cdot \left(\frac{L}{T}\right)^2 \]
\[ = \frac{1}{2} \cdot \left(\frac{F \cdot T^2}{L}\right) \cdot \left(\frac{L^2}{T^2}\right) = \frac{F \cdot T^2 \cdot L^2}{L \cdot T^2} = F \cdot L \]

So, the dimension of energy is:
\[ \text{Dimension of Energy} = F L^{1} T^{0} \]

### Summary of Results
- Dimension of Density: \( F L^{-4} T^{2} \)
- Dimension of Pressure: \( F L^{-2} \)
- Dimension of Momentum: \( F L^{0} T^{1} \)
- Dimension of Energy: \( F L^{1} T^{0} \)