A copper ball of radius `r` travels with a uniform speed v in a viscous fluid if the ball is changed withh another ball of radius 2r then new uniform speed will be

Options

A. v
B. 2v
C. 4v
D. 8v

Answer

To solve the problem, we need to determine the new uniform speed of a copper ball when its radius is doubled while it travels through a viscous fluid.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity (uniform speed) of a sphere moving through a viscous fluid is reached when the net force acting on the sphere is zero. This means that the downward gravitational force is balanced by the upward buoyant force and the viscous drag force.

2. **Forces Acting on the Ball**:
- The weight of the ball (downward force) is given by \( mg \), where \( m \) is the mass of the ball.
- The buoyant force \( F_b \) acting upward can be expressed as \( F_b = V \cdot \rho_a \cdot g \), where \( V \) is the volume of the ball and \( \rho_a \) is the density of the fluid.
- The viscous drag force \( F_v \) acting upward is given by Stokes' law: \( F_v = 6 \pi \eta r v \), where \( \eta \) is the viscosity of the fluid, \( r \) is the radius of the ball, and \( v \) is the speed of the ball.

3. **Setting Up the Equation**:
At terminal velocity, the forces balance:
\[ mg = F_b + F_v \]
Substituting the expressions for these forces:
\[ \rho_c \cdot V \cdot g = \rho_a \cdot V \cdot g + 6 \pi \eta r v \]
Here, \( \rho_c \) is the density of the copper ball, and \( V = \frac{4}{3} \pi r^3 \).

4. **Rearranging the Equation**:
Rearranging gives:
\[ 6 \pi \eta r v = \rho_c \cdot V \cdot g - \rho_a \cdot V \cdot g \]
\[ 6 \pi \eta r v = Vg (\rho_c - \rho_a) \]
Substituting \( V = \frac{4}{3} \pi r^3 \):
\[ 6 \pi \eta r v = \left(\frac{4}{3} \pi r^3\right) g (\rho_c - \rho_a) \]

5. **Solving for Velocity**:
Simplifying further:
\[ v = \frac{2g r^2 (\rho_c - \rho_a)}{9 \eta} \]
This shows that the terminal velocity \( v \) is directly proportional to the square of the radius \( r^2 \).

6. **Finding New Speed for Radius \( 2r \)**:
If the radius is changed to \( 2r \):
\[ v' = \frac{2g (2r)^2 (\rho_c - \rho_a)}{9 \eta} \]
\[ v' = \frac{2g \cdot 4r^2 (\rho_c - \rho_a)}{9 \eta} \]
\[ v' = 4 \cdot \frac{2g r^2 (\rho_c - \rho_a)}{9 \eta} = 4v \]
Thus, the new uniform speed \( v' \) is \( 4v \).

### Final Answer:
The new uniform speed of the ball with radius \( 2r \) is \( 4v \).