The S.I unit of a physical quantity is `[J.m^(-2)]`. The dimensional formula for that quantity is

Options

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

Answer

To find the dimensional formula for the physical quantity whose SI unit is \( [J \cdot m^{-2}] \), we will follow these steps:

### Step 1: Understand the SI unit
The SI unit given is \( J \cdot m^{-2} \) (Joule per meter squared).

### Step 2: Break down the SI unit
Recall that:
- The unit of work (or energy) is Joule (J).
- The unit of length is meter (m).

Thus, we can express the unit as:
\[ [J \cdot m^{-2}] = [J] \cdot [m]^{-2} \]

### Step 3: Find the dimensional formula for Joule
The work done (W) is defined as:
\[ W = F \cdot d \]
where \( F \) is force and \( d \) is displacement.

### Step 4: Find the dimensional formula for force
Force (F) is given by:
\[ F = m \cdot a \]
where \( m \) is mass and \( a \) is acceleration. The dimensional formula for acceleration is:
\[ [a] = [L \cdot T^{-2}] \]
Thus, the dimensional formula for force becomes:
\[ [F] = [M] \cdot [L \cdot T^{-2}] = [M^1 L^1 T^{-2}] \]

### Step 5: Find the dimensional formula for work done
Now substituting the dimensional formula of force into the work done:
\[ [W] = [F] \cdot [d] = [M^1 L^1 T^{-2}] \cdot [L^1] = [M^1 L^2 T^{-2}] \]

### Step 6: Combine with the meter squared in the denominator
Now we need to combine this with \( m^{-2} \):
\[ [A] = [W] \cdot [L^{-2}] = [M^1 L^2 T^{-2}] \cdot [L^{-2}] = [M^1 L^{2-2} T^{-2}] = [M^1 L^0 T^{-2}] \]

### Step 7: Write the final dimensional formula
Thus, the dimensional formula for the physical quantity is:
\[ [M^1 L^0 T^{-2}] \]

### Step 8: Identify the correct option
From the options given:
- Option A: \( M^1 L^{-2} \)
- Option B: \( M^1 L^0 T^{-2} \) (Correct)
- Option C: \( M^1 L^2 T^{-1} \)
- Option D: \( M^1 L^{-1} T^{-2} \)

The correct answer is **Option B: \( M^1 L^0 T^{-2} \)**.

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