If Force=(x/density)+C is dimensionally correct, the dimension of x are -

Options

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

Answer

To solve the problem, we need to determine the dimensions of \( x \) in the equation given:

\[ \text{Force} = \frac{x}{\text{density}} + C \]

### Step-by-Step Solution:

1. **Understanding the Equation**:
- We know that for the equation to be dimensionally correct, all terms on the right side must have the same dimensions as the left side (Force).

2. **Identifying Dimensions of Force**:
- The dimension of force \( F \) is given by the formula:
\[ F = m \cdot a \]
where \( m \) is mass and \( a \) is acceleration.
- The dimension of mass \( m \) is \( [M] \) and the dimension of acceleration \( a \) is \( [L][T]^{-2} \).
- Therefore, the dimension of force \( F \) is:
\[ [F] = [M][L][T]^{-2} = M L T^{-2} \]

3. **Identifying Dimensions of Density**:
- Density \( \rho \) is defined as mass per unit volume:
\[ \rho = \frac{m}{V} \]
- The dimension of volume \( V \) is \( [L^3] \).
- Thus, the dimension of density \( \rho \) is:
\[ [\rho] = \frac{[M]}{[L^3]} = M L^{-3} \]

4. **Setting Up the Dimensional Equation**:
- From the equation \( F = \frac{x}{\rho} + C \), we can isolate \( \frac{x}{\rho} \):
\[ \frac{x}{\rho} \text{ must have the same dimension as } F \]
- Therefore, we can write:
\[ [F] = \left[\frac{x}{\rho}\right] \]

5. **Finding the Dimension of \( x \)**:
- Rearranging gives us:
\[ [x] = [F] \cdot [\rho] \]
- Substituting the dimensions we found:
\[ [x] = (M L T^{-2}) \cdot (M L^{-3}) = M^2 L^{-2} T^{-2} \]

6. **Conclusion**:
- The dimension of \( x \) is:
\[ [x] = M^2 L^{-2} T^{-2} \]

### Final Answer:
The dimensions of \( x \) are \( M^2 L^{-2} T^{-2} \).