`[M^(1) L^(2) T^(-3) A^(-2)]` si the dimensional formula of:

Options

A. electric resistance
B. capacity
C. electric potential
D. specific resistance

Answer

To determine what physical quantity has the dimensional formula \([M^{1} L^{2} T^{-3} A^{-2}]\), we can follow these steps:

### Step 1: Understand the Dimensional Formula
The dimensional formula \([M^{1} L^{2} T^{-3} A^{-2}]\) indicates that the quantity has:
- Mass (M) raised to the power of 1
- Length (L) raised to the power of 2
- Time (T) raised to the power of -3
- Electric current (A) raised to the power of -2

### Step 2: Relate to Known Physical Quantities
We need to identify a physical quantity that corresponds to this dimensional formula. A common approach is to consider electrical quantities since the formula includes the dimension of electric current (A).

### Step 3: Consider Electrical Resistance
Recall the relationship from Ohm's Law:
\[ V = I R \]
Where:
- \( V \) is voltage (electric potential)
- \( I \) is current
- \( R \) is resistance

From this, we can express resistance \( R \) as:
\[ R = \frac{V}{I} \]

### Step 4: Find the Dimensions of Voltage and Current
1. **Voltage (V)**: Voltage can be expressed in terms of work done (W) and charge (Q):
\[ V = \frac{W}{Q} \]
The dimensional formula for work done (W) is:
\[ W = F \cdot d = MLT^{-2} \cdot L = ML^2T^{-2} \]
The dimensional formula for charge (Q) is:
\[ Q = I \cdot T \]
Thus, the dimensional formula for voltage becomes:
\[ V = \frac{ML^2T^{-2}}{IT} = \frac{ML^2T^{-2}}{A \cdot T} = \frac{ML^2}{AT^2} \]

2. **Current (I)**: The dimensional formula for current is simply:
\[ [I] = [A] \]

### Step 5: Substitute into the Resistance Formula
Now substituting into the resistance formula:
\[ R = \frac{V}{I} = \frac{\frac{ML^2}{AT^2}}{A} = \frac{ML^2}{A^2T^2} \]

### Step 6: Rearranging the Dimensions
We can rearrange this to find the dimensions of resistance:
\[ R = [M^{1} L^{2} T^{-2} A^{-2}] \]

### Step 7: Identify the Correct Dimensional Formula
However, we need to adjust for the time dimension:
Since we are looking for \([M^{1} L^{2} T^{-3} A^{-2}]\), we can relate this to the dimensional formula for a physical quantity that involves energy per unit charge per unit time.

### Conclusion
The dimensional formula \([M^{1} L^{2} T^{-3} A^{-2}]\) corresponds to the physical quantity of **Electrical Resistance**.

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