If pressure can be expressed as `P=(b)/(a)sqrt(1+(kthetat^(3))/(ma))`
where k is the Boltzmann's constant, `theta` is the temperature, t is the time and a and b are constants, then dimensional formula of b is equal to the dimensional formula of

Options

A. linear momentum
B. force
C. angular momentum
D. torque

Answer

To find the dimensional formula of \( b \) from the given equation for pressure \( P \):

\[ P = \frac{b}{a} \sqrt{1 + \frac{k \theta t^3}{m a}} \]

where:
- \( k \) is the Boltzmann constant,
- \( \theta \) is the temperature,
- \( t \) is the time,
- \( a \) and \( b \) are constants.

### Step 1: Understand the Equation
The term inside the square root, \( 1 + \frac{k \theta t^3}{m a} \), must be dimensionless because it is added to 1. Therefore, the entire fraction \( \frac{k \theta t^3}{m a} \) must also be dimensionless.

### Step 2: Identify Dimensions of Each Quantity
1. **Boltzmann constant \( k \)** has the dimensions of energy per temperature:
\[ [k] = \frac{[M L^2 T^{-2}]}{[K]} = [M L^2 T^{-2} K^{-1}] \]

2. **Temperature \( \theta \)** has the dimension:
\[ [\theta] = [K] \]

3. **Time \( t \)** has the dimension:
\[ [t] = [T] \]

4. **Mass \( m \)** has the dimension:
\[ [m] = [M] \]

5. **Constant \( a \)** will have some dimension, which we will denote as \([A]\).

### Step 3: Combine Dimensions
Now, substituting these into the term \( \frac{k \theta t^3}{m a} \):
\[ \frac{k \theta t^3}{m a} = \frac{[M L^2 T^{-2} K^{-1}] [K] [T^3]}{[M][A]} \]
This simplifies to:
\[ \frac{[M L^2 T^{-2} K^{-1}] [K] [T^3]}{[M][A]} = \frac{[L^2 T^{1}]}{[A]} \]

### Step 4: Set the Fraction Dimensionless
For this fraction to be dimensionless:
\[ \frac{[L^2 T^{1}]}{[A]} = 1 \implies [A] = [L^2 T^{1}] \]

### Step 5: Relate Dimensions of \( P \), \( b \), and \( a \)
From the equation for pressure:
\[ P = \frac{b}{a} \sqrt{1 + \frac{k \theta t^3}{m a}} \]
Since \( \sqrt{1 + \text{(dimensionless)}} \) is dimensionless, we have:
\[ [P] = \frac{[b]}{[A]} \]
Thus,
\[ [P] = \frac{[b]}{[L^2 T^{1}]} \]

### Step 6: Find the Dimension of Pressure
Pressure \( P \) is defined as force per unit area:
\[ [P] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]

### Step 7: Solve for the Dimension of \( b \)
Now we can find the dimension of \( b \):
\[ [M L^{-1} T^{-2}] = \frac{[b]}{[L^2 T^{1}]} \]
Multiplying both sides by \([L^2 T^{1}]\):
\[ [b] = [M L^{-1} T^{-2}] \cdot [L^2 T^{1}] = [M L^{1} T^{-1}] \]

### Final Result
Thus, the dimensional formula of \( b \) is:
\[ [b] = [M L^{1} T^{-1}] \]