Assume that the largest stone of mass 'm' that can be moved by a flowing river depends upon the velocity of flow v, the density d & the acceleration due to gravity g. I 'm' varies as the `K^(th)` power of the velocity of flow, then find the value of K.

Explanation

To solve the problem, we will use dimensional analysis to find the value of \( K \) in the relationship between the mass \( m \) of the stone that can be moved by the river and the velocity \( v \), density \( d \), and acceleration due to gravity \( g \).

### Step-by-Step Solution:

1. **Identify the relationship**: We know that the mass \( m \) of the stone depends on the velocity \( v \), density \( d \), and acceleration due to gravity \( g \). We can express this relationship as:
\[ m = k \cdot v^K \cdot d^b \cdot g^c \]
where \( k \) is a constant, and \( K \), \( b \), and \( c \) are the powers we need to determine.

2. **Write down the dimensions**:
- The dimension of mass \( m \) is \( [M] \).
- The dimension of velocity \( v \) is \( [L][T^{-1}] \).
- The dimension of density \( d \) is \( [M][L^{-3}] \).
- The dimension of acceleration due to gravity \( g \) is \( [L][T^{-2}] \).

3. **Express the dimensions**:
- For \( m \):
\[ [m] = [M] \]
- For \( v^K \):
\[ [v^K] = [L^K][T^{-K}] \]
- For \( d^b \):
\[ [d^b] = [M^b][L^{-3b}] \]
- For \( g^c \):
\[ [g^c] = [L^c][T^{-2c}] \]

4. **Combine the dimensions**:
Combining all the dimensions on the right-hand side, we have:
\[ [M] = [M^b][L^K][T^{-K}][M^0][L^{-3b}][L^c][T^{-2c}] \]
This simplifies to:
\[ [M] = [M^{b}][L^{K - 3b + c}][T^{-K - 2c}] \]

5. **Set up equations for each dimension**:
We can equate the powers of each dimension on both sides:
- For mass \( M \):
\[ b = 1 \quad \text{(Equation 1)} \]
- For length \( L \):
\[ K - 3b + c = 0 \quad \text{(Equation 2)} \]
- For time \( T \):
\[ -K - 2c = 0 \quad \text{(Equation 3)} \]

6. **Substitute \( b \) into the equations**:
From Equation 1, we have \( b = 1 \). Substitute this into Equation 2:
\[ K - 3(1) + c = 0 \implies K - 3 + c = 0 \implies K + c = 3 \quad \text{(Equation 4)} \]

7. **Solve for \( K \) and \( c \)**:
From Equation 3:
\[ -K - 2c = 0 \implies K = -2c \quad \text{(Equation 5)} \]
Substitute Equation 5 into Equation 4:
\[ -2c + c = 3 \implies -c = 3 \implies c = -3 \]
Now substitute \( c = -3 \) back into Equation 5:
\[ K = -2(-3) = 6 \]

8. **Final Result**:
Thus, the value of \( K \) is:
\[ K = 6 \]