Find the dimensions of `alpha//beta` in equation `P=(alpha-t^2)/(betax)` where P= pressure, x is distance and t is time.

Answer

To find the dimensions of \(\frac{\alpha}{\beta}\) in the equation \(P = \frac{\alpha - t^2}{\beta x}\), where \(P\) is pressure, \(x\) is distance, and \(t\) is time, we will follow these steps:

### Step 1: Identify the dimensions of pressure \(P\)
Pressure is defined as force per unit area. The dimensions of force can be expressed as:
\[ \text{Force} = \text{mass} \times \text{acceleration} \]
The dimension of mass is \(M\), and acceleration is given by \(\frac{\text{length}}{\text{time}^2}\), which has dimensions \(L T^{-2}\). Therefore, the dimensions of force are:
\[ [M][L T^{-2}] = M L T^{-2} \]
The area has dimensions \(L^2\). Thus, the dimensions of pressure \(P\) are:
\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 2: Analyze the equation \(P = \frac{\alpha - t^2}{\beta x}\)
From the equation, we can see that the dimensions of the numerator \((\alpha - t^2)\) must match the dimensions of the denominator \((\beta x)\) in order for the equation to be dimensionally consistent.

### Step 3: Determine the dimensions of \(t^2\)
The dimension of time \(t\) is \(T\), so:
\[ \text{Dimension of } t^2 = T^2 \]

### Step 4: Set up the dimensions for \(\alpha\)
Since \(\alpha - t^2\) must have the same dimensions as \(\beta x\), we can express:
\[ \text{Dimension of } \alpha = \text{Dimension of } t^2 = T^2 \]

### Step 5: Determine the dimensions of \(\beta x\)
The dimension of \(x\) (distance) is \(L\). Therefore, the dimensions of \(\beta x\) can be expressed as:
\[ \text{Dimension of } \beta x = \text{Dimension of } \beta \cdot L \]
Thus, we can write:
\[ \text{Dimension of } \beta = \frac{\text{Dimension of } P}{\text{Dimension of } x} = \frac{M L^{-1} T^{-2}}{L} = M L^{-2} T^{-2} \]

### Step 6: Find the dimensions of \(\frac{\alpha}{\beta}\)
Now we can find the dimensions of \(\frac{\alpha}{\beta}\):
\[ \frac{\alpha}{\beta} = \frac{T^2}{M L^{-2} T^{-2}} = \frac{T^2 \cdot L^2}{M} = \frac{L^2 T^4}{M} \]

### Final Answer
The dimensions of \(\frac{\alpha}{\beta}\) are:
\[ \frac{L^2 T^4}{M} \]

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