A metal plate of area` 5 cm ^(2)` is placed on a 0.5 mm thick castor oil layer .If a force of 22,500 dyne is needed to move the plate with a velcity of `3cms^(-1)` Calculate the coeffcient of viscosity of castor oil.

Answer

To calculate the coefficient of viscosity of castor oil, we can use the formula for viscous force:

\[ F = \eta \cdot A \cdot \frac{v}{d} \]

where:
- \( F \) is the force applied (in dynes),
- \( \eta \) is the coefficient of viscosity (in poise),
- \( A \) is the area of the plate (in cm²),
- \( v \) is the velocity of the plate (in cm/s),
- \( d \) is the thickness of the fluid layer (in cm).

### Step-by-Step Solution:

1. **Identify the given values:**
- Area \( A = 5 \, \text{cm}^2 \)
- Force \( F = 22,500 \, \text{dyne} \)
- Velocity \( v = 3 \, \text{cm/s} \)
- Thickness \( d = 0.5 \, \text{mm} = 0.05 \, \text{cm} \)

2. **Calculate the velocity gradient:**
The velocity gradient is given by \( \frac{v}{d} \).
\[ \frac{v}{d} = \frac{3 \, \text{cm/s}}{0.05 \, \text{cm}} = 60 \, \text{s}^{-1} \]

3. **Substitute the values into the formula:**
Rearranging the formula for viscosity, we have:
\[ \eta = \frac{F}{A \cdot \frac{v}{d}} \]
Substituting the known values:
\[ \eta = \frac{22,500 \, \text{dyne}}{5 \, \text{cm}^2 \cdot 60 \, \text{s}^{-1}} \]

4. **Calculate the area times the velocity gradient:**
\[ A \cdot \frac{v}{d} = 5 \, \text{cm}^2 \cdot 60 \, \text{s}^{-1} = 300 \, \text{cm}^2 \cdot \text{s}^{-1} \]

5. **Calculate the coefficient of viscosity:**
\[ \eta = \frac{22,500}{300} = 75 \, \text{poise} \]

### Final Answer:
The coefficient of viscosity of castor oil is \( 75 \, \text{poise} \).

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