A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 800 N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of `2 cm s^(-1)`. If a force of 2.4 kN is applied instead, the shaft would move with a speed of

Options

A. `2 cm s^(-1)`
B. `15 cm s^(-1)`
C. `6 cm s^(-1)`
D. none of these

Answer

To solve the problem, we will use the relationship between force and velocity for a Newtonian fluid. According to Newton's law of viscosity, the shear stress (τ) is proportional to the rate of shear strain (velocity gradient).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial Force (F₁) = 800 N
- Initial Velocity (V₁) = 2 cm/s
- New Force (F₂) = 2.4 kN = 2400 N

2. **Understand the Relationship:**
- For a Newtonian fluid, the relationship between force and velocity is linear. This means that:
\[ \frac{F_1}{F_2} = \frac{V_1}{V_2} \]
- Where:
- \(F_1\) is the initial force,
- \(F_2\) is the new force,
- \(V_1\) is the initial velocity,
- \(V_2\) is the new velocity we need to find.

3. **Substitute the Known Values:**
- Plugging in the values we have:
\[ \frac{800}{2400} = \frac{2}{V_2} \]

4. **Cross-Multiply to Solve for \(V_2\):**
- Cross-multiplying gives:
\[ 800 \cdot V_2 = 2400 \cdot 2 \]
- Simplifying the right side:
\[ 800 \cdot V_2 = 4800 \]

5. **Isolate \(V_2\):**
- Now, divide both sides by 800:
\[ V_2 = \frac{4800}{800} = 6 \text{ cm/s} \]

6. **Conclusion:**
- The new velocity \(V_2\) when a force of 2.4 kN is applied is **6 cm/s**.