In the equation `((1)/(pbeta))=(y)/(k_(B)T)`, where `p` is the pressure, `y` is the distance, `k_(B)` is Boltzmann constant and `T` is the tempreture. Dimensions of `beta` are

Options

A. `["M"^(-1)"L"^(1)"T"^(2)]`
B. `["M"^(0)"L"^(2)"T"^(0)]`
C. `["M"^(1)"L"^(-1)"T"^(-2)]`
D. `["M"^(0)"L"^(0)"T"^(0)]`

Explanation

To find the dimensions of β in the equation

\[ \frac{1}{p \beta} = \frac{y}{k_B T} \]

we will follow these steps:

### Step 1: Identify the dimensions of the quantities involved
- **Pressure (p)** has dimensions:
\[ [p] = M L^{-1} T^{-2} \]
- **Distance (y)** has dimensions:
\[ [y] = L \]
- **Boltzmann constant (k_B)** has dimensions:
\[ [k_B] = M L^2 T^{-3} \]
- **Temperature (T)** has dimensions:
\[ [T] = \Theta \quad \text{(where Θ represents the dimension of temperature)} \]

### Step 2: Rearranging the equation to isolate β
From the given equation, we can rearrange it to find β:

\[ \beta = \frac{y}{p \cdot k_B \cdot T} \]

### Step 3: Substitute the dimensions into the equation
Now we substitute the dimensions we identified:

\[ [\beta] = \frac{[y]}{[p] \cdot [k_B] \cdot [T]} \]

Substituting the dimensions:

\[ [\beta] = \frac{L}{(M L^{-1} T^{-2}) \cdot (M L^2 T^{-3}) \cdot \Theta} \]

### Step 4: Simplifying the dimensions
Now we simplify the expression:

\[ [\beta] = \frac{L}{M^2 L^{1} T^{-5} \Theta} \]

This can be rewritten as:

\[ [\beta] = \frac{L}{M^2 L^{1} T^{-5} \Theta} = \frac{L^1}{M^2 L^1 T^{-5} \Theta} = \frac{1}{M^2 T^{-5} \Theta} \]

### Step 5: Final simplification
Now, simplifying further, we get:

\[ [\beta] = M^{-2} L^0 T^5 \Theta^{-1} \]

Thus, the dimensions of β are:

\[ [\beta] = M^{-2} T^5 \Theta^{-1} \]

### Final Answer
The dimensions of β are \( M^{-2} T^5 \Theta^{-1} \).

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