A physical quantity `X` is defined by the formula
`X=(IFv^(2))/(WL^(3))`
where `I` is moment of inertia, `F` is force, `v` is velocity, `W` is work and `L` is length, the dimensions of `X` are

Options

A. `["MLT"^(-2)]`
B. `["MT"^(-2)]`
C. `["ML"^(2)"T"^(-3)]`
D. `["LT"^(-1)]`

Explanation

To find the dimensions of the physical quantity \( X \) defined by the formula

\[ X = \frac{IFv^2}{WL^3} \]

we need to determine the dimensions of each variable involved: moment of inertia \( I \), force \( F \), velocity \( v \), work \( W \), and length \( L \).

### Step 1: Determine the dimensions of each variable

1. **Moment of Inertia \( I \)**:
- The moment of inertia is defined as \( I = m r^2 \) where \( m \) is mass and \( r \) is the distance (length).
- Thus, the dimensions of \( I \) are:
\[ [I] = [M][L^2] = M L^2 \]

2. **Force \( F \)**:
- Force is defined by Newton's second law \( F = ma \) where \( a \) is acceleration.
- The dimensions of force are:
\[ [F] = [M][L][T^{-2}] = M L T^{-2} \]

3. **Velocity \( v \)**:
- Velocity is defined as distance over time.
- The dimensions of velocity are:
\[ [v] = \frac{[L]}{[T]} = L T^{-1} \]

4. **Work \( W \)**:
- Work is defined as force times distance, \( W = F \cdot d \).
- The dimensions of work are:
\[ [W] = [F][L] = (M L T^{-2})[L] = M L^2 T^{-2} \]

5. **Length \( L \)**:
- The dimensions of length are simply:
\[ [L] = L \]

### Step 2: Substitute the dimensions into the formula for \( X \)

Now, substituting the dimensions into the formula for \( X \):

\[ X = \frac{IFv^2}{WL^3} \]

Substituting the dimensions we found:

\[ X = \frac{(M L^2)(M L T^{-2})(L T^{-1})^2}{(M L^2 T^{-2})(L^3)} \]

### Step 3: Simplify the expression

Calculating \( v^2 \):

\[ v^2 = (L T^{-1})^2 = L^2 T^{-2} \]

Now substituting back into \( X \):

\[ X = \frac{(M L^2)(M L T^{-2})(L^2 T^{-2})}{(M L^2 T^{-2})(L^3)} \]

This simplifies to:

\[ X = \frac{M^2 L^5 T^{-4}}{M L^5 T^{-2}} \]

### Step 4: Cancel out the common terms

Now, cancel out the common terms in the numerator and denominator:

\[ X = \frac{M^2 L^5 T^{-4}}{M L^5 T^{-2}} = M^{2-1} L^{5-5} T^{-4 - (-2)} = M^{1} L^{0} T^{-2} \]

Thus, the dimensions of \( X \) are:

\[ X = M T^{-2} \]

### Final Answer

The dimensions of the physical quantity \( X \) are:

\[ \boxed{M T^{-2}} \]