In the equation `[X+(a)/(Y^2)] [Y-b]=RT,X ` is pressure,Y volume,R is universal gas constant and `T is temperature. The physical quantity equivalent to the ratio `(a)/(b)` is :

Options

A. Energy
B. Pressure gradient
C. Impulse
D. Coefficient of viscosity

Answer

To solve the problem, we need to analyze the given equation and determine the dimensions of the quantities involved. The equation provided is:

\[ \left[X + \frac{a}{Y^2}\right] (Y - b) = RT \]

where:
- \(X\) is pressure,
- \(Y\) is volume,
- \(R\) is the universal gas constant,
- \(T\) is temperature.

### Step 1: Identify the dimensions of pressure, volume, and temperature
- Pressure (\(X\)) has the dimension of force per unit area. The dimension of force is \(M L T^{-2}\) and area is \(L^2\). Therefore, the dimension of pressure is:
\[ [X] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

- Volume (\(Y\)) has the dimension of \(L^3\):
\[ [Y] = L^3 \]

- The universal gas constant \(R\) has dimensions of energy per temperature. Energy has the dimension of \(M L^2 T^{-2}\) and temperature has the dimension of \(\Theta\). Therefore:
\[ [R] = \frac{M L^2 T^{-2}}{\Theta} \]

### Step 2: Analyze the equation
From the equation, we can see that \(X + \frac{a}{Y^2}\) must have the same dimensions as \(RT\). Therefore, we need to find the dimensions of \(a\) and \(b\).

### Step 3: Determine the dimensions of \(a\)
The term \(\frac{a}{Y^2}\) must have the same dimensions as pressure \(X\):
\[ \left[\frac{a}{Y^2}\right] = [X] = M L^{-1} T^{-2} \]
Since \(Y^2 = (L^3)^2 = L^6\), we can write:
\[ \frac{[a]}{L^6} = M L^{-1} T^{-2} \]
Thus, the dimension of \(a\) is:
\[ [a] = M L^{-1} T^{-2} \cdot L^6 = M L^{5} T^{-2} \]

### Step 4: Determine the dimensions of \(b\)
The term \(Y - b\) must also have the same dimensions as \(Y\). Therefore, \(b\) must have the same dimensions as \(Y\):
\[ [b] = [Y] = L^3 \]

### Step 5: Find the ratio \(\frac{a}{b}\)
Now we can find the ratio \(\frac{a}{b}\):
\[ \frac{a}{b} = \frac{M L^{5} T^{-2}}{L^3} = M L^{2} T^{-2} \]

### Step 6: Identify the physical quantity equivalent to \(\frac{a}{b}\)
The dimensions \(M L^{2} T^{-2}\) correspond to energy. Therefore, the physical quantity equivalent to the ratio \(\frac{a}{b}\) is:

**Energy.**

### Final Answer:
The physical quantity equivalent to the ratio \(\frac{a}{b}\) is **Energy**.

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