A small spherical ball of radius r falls with velocity `upsilon` through a liquid having coeffiecinet of viscosity `eta.` find viscous darg F on the wall if it depends or `r, upsilon, eta. Take K = 6 pi`

Explanation

To find the viscous drag \( F \) on a small spherical ball falling through a liquid, we will use dimensional analysis. The viscous drag \( F \) depends on the radius \( r \) of the ball, the velocity \( \upsilon \), and the coefficient of viscosity \( \eta \). We are given that the constant \( K = 6\pi \).

### Step-by-Step Solution:

1. **Assume the form of the equation:**
\[ F \propto r^a \upsilon^b \eta^c \]
When the proportionality sign is replaced by an equality, we introduce a constant \( K \):
\[ F = K r^a \upsilon^b \eta^c \]

2. **Write the dimensions of each quantity:**
- Force \( F \): \([F] = MLT^{-2}\)
- Radius \( r \): \([r] = L\)
- Velocity \( \upsilon \): \([\upsilon] = LT^{-1}\)
- Viscosity \( \eta \): \([\eta] = ML^{-1}T^{-1}\)

3. **Express the dimensions of the right-hand side (RHS):**
\[ [r^a \upsilon^b \eta^c] = [L^a] [LT^{-1}]^b [ML^{-1}T^{-1}]^c \]
Simplify the dimensions:
\[ [r^a \upsilon^b \eta^c] = L^a \cdot L^b T^{-b} \cdot M^c L^{-c} T^{-c} \]
Combine the exponents:
\[ = M^c L^{a+b-c} T^{-b-c} \]

4. **Equate the dimensions of LHS and RHS:**
\[ MLT^{-2} = M^c L^{a+b-c} T^{-b-c} \]

5. **Compare the powers of \( M \), \( L \), and \( T \):**
- For \( M \):
\[ c = 1 \]
- For \( L \):
\[ a + b - c = 1 \]
- For \( T \):
\[ -b - c = -2 \]

6. **Solve the equations:**
- From \( c = 1 \):
\[ c = 1 \]
- Substitute \( c = 1 \) into \( a + b - c = 1 \):
\[ a + b - 1 = 1 \implies a + b = 2 \]
- Substitute \( c = 1 \) into \( -b - c = -2 \):
\[ -b - 1 = -2 \implies -b = -1 \implies b = 1 \]
- Substitute \( b = 1 \) into \( a + b = 2 \):
\[ a + 1 = 2 \implies a = 1 \]

7. **Substitute \( a \), \( b \), and \( c \) back into the equation:**
\[ F = K r^1 \upsilon^1 \eta^1 = K r \upsilon \eta \]

8. **Include the given constant \( K = 6\pi \):**
\[ F = 6\pi \eta r \upsilon \]

### Final Answer:
\[ F = 6\pi \eta r \upsilon \]