The viscous force on a spherical body, when it moves through a viscous liquid, depends on the radius of the body, the coefficient of viscosity of the liquid and the velocity of the body.Find an expression for the viscous force.

Answer

To find the expression for the viscous force \( F \) acting on a spherical body moving through a viscous liquid, we will use dimensional analysis. The viscous force depends on three variables: the radius of the body \( R \), the coefficient of viscosity \( \eta \), and the velocity of the body \( V \).

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( F \) be the viscous force.
- Let \( R \) be the radius of the spherical body.
- Let \( \eta \) be the coefficient of viscosity.
- Let \( V \) be the velocity of the body.

2. **Assign Dimensions**:
- The dimension of force \( F \) is given by:
\[ [F] = M^1 L^1 T^{-2} \]
- The dimension of radius \( R \) is:
\[ [R] = L^1 \]
- The dimension of velocity \( V \) is:
\[ [V] = L^1 T^{-1} \]
- The dimension of the coefficient of viscosity \( \eta \) is:
\[ [\eta] = M^1 L^{-1} T^{-1} \]

3. **Formulate the Relationship**:
- Assume the relationship can be expressed in the form:
\[ F = k \cdot R^x \cdot \eta^y \cdot V^z \]
- Here, \( k \) is a dimensionless constant, and \( x \), \( y \), and \( z \) are the powers we need to determine.

4. **Substitute Dimensions**:
- Substitute the dimensions into the equation:
\[ [F] = [R]^x \cdot [\eta]^y \cdot [V]^z \]
- This translates to:
\[ M^1 L^1 T^{-2} = (L^1)^x \cdot (M^1 L^{-1} T^{-1})^y \cdot (L^1 T^{-1})^z \]

5. **Expand and Simplify**:
- Expanding the right-hand side gives:
\[ M^1 L^1 T^{-2} = M^y \cdot L^{x - y + z} \cdot T^{-y - z} \]

6. **Equate Dimensions**:
- Now, we can equate the powers of \( M \), \( L \), and \( T \) from both sides:
- For \( M \):
\[ y = 1 \]
- For \( L \):
\[ x - y + z = 1 \]
- For \( T \):
\[ -y - z = -2 \]

7. **Solve the Equations**:
- From \( y = 1 \):
- Substitute \( y \) into the other equations:
- \( x - 1 + z = 1 \) → \( x + z = 2 \) (Equation 1)
- \( -1 - z = -2 \) → \( z = 1 \) (Equation 2)
- Substitute \( z = 1 \) into Equation 1:
- \( x + 1 = 2 \) → \( x = 1 \)

8. **Final Expression**:
- Now we have \( x = 1 \), \( y = 1 \), and \( z = 1 \). Thus, the expression for the viscous force is:
\[ F = k \cdot R^1 \cdot \eta^1 \cdot V^1 \]
- Therefore, the viscous force can be expressed as:
\[ F = \eta R V \]

### Final Answer:
\[ F = \eta R V \]