Viscosity of ethanol is 12.0 millipoise. Viscosity of ethanol in S.I system is equal to

Options

A. 1,2
B. `1.2 xx 10^(-3)`
C. `1.2 xx 10^(-2)`
D. `1.2 xx 10^(-2)`

Answer

To convert the viscosity of ethanol from millipoise to the SI unit (which is kg/m·s), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the given viscosity**:
The viscosity of ethanol is given as 12.0 millipoise.

2. **Convert millipoise to poise**:
Since 1 millipoise = \(10^{-3}\) poise, we can convert:
\[ 12.0 \text{ millipoise} = 12.0 \times 10^{-3} \text{ poise} \]

3. **Convert poise to SI units (kg/m·s)**:
We know that:
\[ 1 \text{ poise} = 1 \text{ g/(cm·s)} = 10^{-3} \text{ kg/(10^{-2} m)·s} = 10^{-3} \text{ kg/(0.01 m)·s} = 100 \text{ kg/m·s} \]
Therefore, we can express poise in SI units:
\[ 1 \text{ poise} = 0.1 \text{ kg/(m·s)} \]

4. **Calculate the viscosity in SI units**:
Now, substituting the value of poise into the SI unit conversion:
\[ 12.0 \times 10^{-3} \text{ poise} = 12.0 \times 10^{-3} \times 0.1 \text{ kg/(m·s)} \]
\[ = 1.2 \times 10^{-3} \text{ kg/(m·s)} \]

5. **Final answer**:
The viscosity of ethanol in SI units is:
\[ 1.2 \times 10^{-3} \text{ kg/(m·s)} \]