A layer of glycerine of thickness 1 mm is present between a large surface and small surface of area `0.1m^(2)`. With what force the small surface is to be pulled, so that it can move with a velocity of 1 m/s ?
(coefficient of viscosity`=0.07g-m^(-1)s^(-1)`)

Options

A. 70N
B. 7N
C. 700 N
D. 0.70N

Answer

To solve the problem, we need to calculate the force required to pull a small surface through a layer of glycerine, given its viscosity, area, and the velocity at which it should move.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Thickness of glycerine layer, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Area of the small surface, \( A = 0.1 \, \text{m}^2 \)
- Velocity of the small surface, \( v = 1 \, \text{m/s} \)
- Coefficient of viscosity, \( \eta = 0.07 \, \text{g/m·s} \)

2. **Convert the Coefficient of Viscosity**:
- Since the coefficient of viscosity is given in grams, we need to convert it to SI units (kg/m·s).
- \( 0.07 \, \text{g/m·s} = 0.07 \times 10^{-3} \, \text{kg/m·s} = 7 \times 10^{-5} \, \text{kg/m·s} \)

3. **Calculate the Velocity Gradient**:
- The velocity gradient \( \frac{dv}{dx} \) can be expressed as:
\[ \frac{dv}{dx} = \frac{v}{d} = \frac{1 \, \text{m/s}}{1 \times 10^{-3} \, \text{m}} = 1000 \, \text{s}^{-1} \]

4. **Apply the Formula for Force**:
- The formula for the force required to move the surface is given by:
\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]
- Substituting the known values:
\[ F = (7 \times 10^{-5} \, \text{kg/m·s}) \cdot (0.1 \, \text{m}^2) \cdot (1000 \, \text{s}^{-1}) \]

5. **Calculate the Force**:
- Performing the multiplication:
\[ F = 7 \times 10^{-5} \cdot 0.1 \cdot 1000 = 7 \times 10^{-5} \cdot 100 = 7 \times 10^{-3} \, \text{N} = 0.007 \, \text{N} \]

6. **Final Result**:
- The force required to pull the small surface is \( 7 \, \text{N} \).

### Final Answer:
The force required to pull the small surface is **7 Newtons**.

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