An object is moving through the liquid. The viscous damping force acting on it is proportional to the velocity. Then dimensions of constant of proportionality are

Options

A. (a)`ML^(-1)T^(-1)`
B. (b)`MLT^(-1)`
C. (c)`M^(0)LT^(-1)`
D. (d)`ML^(0)T^(-1)`

Answer

To find the dimensions of the constant of proportionality for the viscous damping force that is proportional to the velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The problem states that the viscous damping force (F) acting on an object moving through a liquid is proportional to its velocity (V). This can be expressed mathematically as:
\[ F \propto V \]
To convert this proportionality into an equation, we introduce a constant of proportionality (k):
\[ F = kV \]

2. **Identify the Dimensions of Force and Velocity**:
- The dimension of force (F) is given by Newton's second law, which states that force is mass times acceleration:
\[ F = m \cdot a \]
The dimension of mass (m) is \( [M] \) and the dimension of acceleration (a) is \( [L][T^{-2}] \). Therefore, the dimension of force is:
\[ [F] = [M][L][T^{-2}] = [M L T^{-2}] \]
- The dimension of velocity (V) is defined as distance divided by time:
\[ [V] = \frac{[L]}{[T]} = [L][T^{-1}] \]

3. **Express the Dimensions of the Constant k**:
Rearranging the equation \( F = kV \) gives us:
\[ k = \frac{F}{V} \]
Now we substitute the dimensions of force and velocity into this equation:
\[ [k] = \frac{[F]}{[V]} = \frac{[M L T^{-2}]}{[L T^{-1}]} \]

4. **Simplify the Dimensions**:
When we divide the dimensions, we perform the following calculations:
\[ [k] = \frac{[M L T^{-2}]}{[L T^{-1}]} = [M L^{1-1} T^{-2 - (-1)}] = [M L^{0} T^{-1}] \]
Thus, we find that:
\[ [k] = [M][L^0][T^{-1}] = [M][T^{-1}] \]

5. **Conclusion**:
The dimensions of the constant of proportionality \( k \) are:
\[ [M^1 L^0 T^{-1}] \]
Therefore, the correct option is \( M^1 L^0 T^{-1} \).