`(P+(a)/(y^(2)))(V-b)=RT` represents the equation of state of some gases.Where `P` is the pressure, `V` is the volume, `T` is the temperature and `a,b,R` are the constants.The physical quantity,which has dimensional formula as that of `(b^(2))/(a)`,will be

Options

A. Bulk modulus
B. Compressibility
C. Modulus of rigidity
D. Energy density

Answer

To solve the problem, we need to find a physical quantity that has the same dimensional formula as \( \frac{b^2}{a} \) given the equation of state \( (P + \frac{a}{V^2})(V - b) = RT \).

### Step 1: Identify the dimensions of the variables involved

1. **Pressure (P)**: The dimensional formula for pressure is given by:
\[ [P] = \frac{F}{A} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

2. **Volume (V)**: The dimensional formula for volume is:
\[ [V] = L^3 \]

3. **Temperature (T)**: The dimensional formula for temperature is:
\[ [T] = \Theta \]
(where \( \Theta \) is the dimension of temperature, which we will not need explicitly).

4. **Gas constant (R)**: The gas constant \( R \) has dimensions derived from the ideal gas law \( PV = nRT \):
\[ [R] = \frac{P \cdot V}{T} = \frac{ML^{-1}T^{-2} \cdot L^3}{\Theta} = ML^{2}T^{-2}\Theta^{-1} \]

### Step 2: Determine the dimensions of constants \( a \) and \( b \)

1. **Constant \( a \)**: In the equation, \( a \) is added to \( P \) divided by \( V^2 \). Thus, the dimensions of \( a \) can be derived from the term \( \frac{a}{V^2} \):
\[ \frac{a}{V^2} \text{ has dimensions of } P \implies [a] = [P] \cdot [V^2] = (ML^{-1}T^{-2}) \cdot (L^6) = ML^{5}T^{-2} \]

2. **Constant \( b \)**: The constant \( b \) is subtracted from \( V \), so it has the same dimensions as volume:
\[ [b] = [V] = L^3 \]

### Step 3: Calculate the dimensions of \( \frac{b^2}{a} \)

Now we can find the dimensions of \( \frac{b^2}{a} \):
\[ [b^2] = (L^3)^2 = L^6 \]
\[ [a] = ML^{5}T^{-2} \]
Thus,
\[ \frac{b^2}{a} = \frac{L^6}{ML^{5}T^{-2}} = \frac{L^{6-5}}{M T^{-2}} = \frac{L^1}{M T^{-2}} = ML^{-1}T^{2} \]

### Step 4: Identify the physical quantity with the same dimensions

The dimensional formula \( ML^{-1}T^{2} \) corresponds to the dimension of compressibility or specific volume. Therefore, the physical quantity that has the same dimensional formula as \( \frac{b^2}{a} \) is:

**Answer**: The physical quantity is compressibility.