Which of the following physical quantities represent the dimensions of `(b)/(a)` in the relaction `P=(x^(2)-b)/(at )` , where p is power x is distance and t is time

Options

A. power
B. Energy
C. Torsional constant
D. Force

Answer

To solve the problem, we need to analyze the given equation \( P = \frac{x^2 - b}{at} \) and determine the dimensions of \( \frac{b}{a} \).

### Step 1: Identify the dimensions of power \( P \)
Power is defined as work done per unit time. The dimensional formula for work done is given by:
\[ \text{Work} = \text{Force} \times \text{Distance} = (m \cdot a) \cdot d = (m \cdot \frac{l}{t^2}) \cdot l = m l^2 t^{-2} \]
Thus, the dimensional formula for power \( P \) is:
\[ P = \frac{\text{Work}}{\text{Time}} = \frac{m l^2 t^{-2}}{t} = m l^2 t^{-3} \]

### Step 2: Identify the dimensions of \( x^2 \)
Since \( x \) represents distance, its dimensional formula is:
\[ [x] = l \]
Thus, the dimensional formula for \( x^2 \) is:
\[ [x^2] = l^2 \]

### Step 3: Set up the equation for dimensions
From the equation \( P = \frac{x^2 - b}{at} \), we can express the dimensions on the right-hand side:
\[ \text{Dimensions of } P = \text{Dimensions of } \frac{x^2 - b}{at} \]
This implies:
\[ m l^2 t^{-3} = \frac{[x^2] - [b]}{[a][t]} \]
Since \( [x^2] = l^2 \), we have:
\[ m l^2 t^{-3} = \frac{l^2 - [b]}{[a][t]} \]

### Step 4: Rearranging the equation
Multiplying both sides by \( [a][t] \):
\[ m l^2 t^{-3} \cdot [a][t] = l^2 - [b] \]
This leads to:
\[ m l^2 t^{-2} [a] = l^2 - [b] \]

### Step 5: Equating dimensions
Since \( l^2 \) and \( b \) must have the same dimensions (as they are being subtracted), we can write:
\[ [b] = l^2 \]
Now we need to find the dimensions of \( a \).

### Step 6: Find dimensions of \( a \)
Rearranging the equation:
\[ l^2 = m l^2 t^{-2} [a] \]
Dividing both sides by \( l^2 \):
\[ 1 = m t^{-2} [a] \]
Thus, we can express \( [a] \) as:
\[ [a] = m^{-1} t^2 \]

### Step 7: Find dimensions of \( \frac{b}{a} \)
Now we can find the dimensions of \( \frac{b}{a} \):
\[ \frac{b}{a} = \frac{l^2}{m^{-1} t^2} = l^2 \cdot m t^2 = m^1 l^2 t^2 \]

### Conclusion
The dimensions of \( \frac{b}{a} \) are \( m^1 l^2 t^2 \), which corresponds to the dimensions of a torsional constant.