A physical quantity `P=B^(2)l^(2)//m` where B is the magnetic induction, l is the length and m is the mass. The dimensions of P is

Options

A. `MLT^(-3)`
B. `ML^(2)T^(-4)I^(-2)`
C. `M^(2)L^(2)T^(-4)I`
D. `MLT^(-2)I^(-2)`

Answer

To find the dimensions of the physical quantity \( P = \frac{B^2 L^2}{m} \), we will follow these steps:

### Step 1: Identify the dimensions of each variable
- **Magnetic Induction (B)**: The dimension of magnetic induction \( B \) is given as \( [B] = M^1 T^{-2} I^{-1} \).
- **Length (L)**: The dimension of length \( L \) is \( [L] = L^1 \).
- **Mass (m)**: The dimension of mass \( m \) is \( [m] = M^1 \).

### Step 2: Substitute the dimensions into the formula for P
We can now substitute these dimensions into the formula for \( P \):
\[ P = \frac{B^2 L^2}{m} \]

### Step 3: Calculate \( B^2 \)
To find \( B^2 \):
\[ [B^2] = (M^1 T^{-2} I^{-1})^2 = M^{2} T^{-4} I^{-2} \]

### Step 4: Calculate \( L^2 \)
To find \( L^2 \):
\[ [L^2] = (L^1)^2 = L^{2} \]

### Step 5: Substitute the dimensions into P
Now substituting \( B^2 \) and \( L^2 \) into the equation for \( P \):
\[ [P] = \frac{M^{2} T^{-4} I^{-2} L^{2}}{M^1} \]

### Step 6: Simplify the expression
When we divide by \( M^1 \), we subtract the exponents:
\[ [P] = M^{2-1} T^{-4} I^{-2} L^{2} = M^{1} T^{-4} I^{-2} L^{2} \]

### Final Answer
Thus, the dimensions of the physical quantity \( P \) are:
\[ [P] = M^1 L^2 T^{-4} I^{-2} \]

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