A formula is given as `P=(b)/(a)sqrt(1+(k.theta.t^(3))/(m.a))`
where P = pressure, k = Boltzmann's constant,
`theta=` temperature, t= time, 'a' and 'b' are constants.
Dimensional formula of 'b' is same as

Options

A. Force
B. Linear momentum
C. Angular momentum
D. Torque

Explanation

To determine the dimensional formula of 'b' in the given formula \( P = \frac{b}{a} \sqrt{1 + \frac{k \theta t^3}{m a}} \), we will follow these steps:

### Step 1: Understand the components of the equation
The equation involves pressure \( P \), constants \( a \) and \( b \), Boltzmann's constant \( k \), temperature \( \theta \), time \( t \), and mass \( m \).

### Step 2: Identify the dimensions of pressure \( P \)
Pressure is defined as force per unit area. The dimensional formula for force is given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = M L T^{-2} \]
The dimensional formula for area is:
\[ \text{Area} = L^2 \]
Thus, the dimensional formula for pressure \( P \) is:
\[ [P] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 3: Analyze the term inside the square root
The term inside the square root is \( 1 + \frac{k \theta t^3}{m a} \). For this entire expression to be dimensionless, the dimensions of \( \frac{k \theta t^3}{m a} \) must also be dimensionless.

### Step 4: Determine the dimensions of \( k \), \( \theta \), \( t \), \( m \), and \( a \)
- The Boltzmann constant \( k \) has the dimensions of energy per temperature. Thus, its dimensions are:
\[ [k] = \frac{M L^2 T^{-2}}{\Theta} \quad (\text{where } \Theta \text{ is the dimension of temperature}) \]
- The temperature \( \theta \) has the dimension:
\[ [\theta] = \Theta \]
- The time \( t \) has the dimension:
\[ [t] = T \]
- The mass \( m \) has the dimension:
\[ [m] = M \]
- The constant \( a \) is yet to be determined.

### Step 5: Set up the equation for dimensional analysis
Now, we can set up the dimensional formula for the term \( \frac{k \theta t^3}{m a} \):
\[ \frac{k \theta t^3}{m a} = \frac{\left( \frac{M L^2 T^{-2}}{\Theta} \right) \cdot \Theta \cdot T^3}{M \cdot [a]} \]
This simplifies to:
\[ \frac{M L^2 T^{-2} \cdot T^3}{M \cdot [a]} = \frac{L^2 T}{[a]} \]
For this to be dimensionless, we must have:
\[ [a] = L^2 T \]

### Step 6: Find the dimensions of \( b \)
Now, substituting \( a \) back into the original equation:
\[ P = \frac{b}{L^2 T} \sqrt{1 + \text{dimensionless term}} \]
Thus, the dimensions of \( b \) must match those of pressure \( P \):
\[ [P] = \frac{b}{L^2 T} \]
Rearranging gives:
\[ b = P \cdot (L^2 T) = (M L^{-1} T^{-2}) \cdot (L^2 T) = M L^{1} T^{-1} \]

### Conclusion
The dimensional formula of \( b \) is:
\[ [b] = M L^{1} T^{-1} \]