The surface tension of water at 20°C is 72.75 dyne `cm^(-1)` . Its value in SI system is

Options

A. `7.275 Nm^(-1)`
B. `0.7275 Nm^(-1)`
C. `0.07275 Nm^(-1)`
D. None of these

Answer

To convert the surface tension of water from dyne/cm to SI units (Newton/meter), we can follow these steps:

### Step 1: Understand the Conversion Factors
- 1 Newton (N) = \(10^5\) dyne
- 1 centimeter (cm) = \(10^{-2}\) meter (m)

### Step 2: Write Down the Given Value
The surface tension of water at 20°C is given as:
\[ \text{Surface Tension} = 72.75 \, \text{dyne/cm} \]

### Step 3: Convert Dyne to Newton
To convert dyne to Newton, we divide by \(10^5\):
\[ \text{Surface Tension in N/m} = \frac{72.75 \, \text{dyne/cm}}{10^5} \]

### Step 4: Convert cm to m
Since we are converting from dyne/cm to N/m, we also need to account for the conversion from cm to m. We know that:
\[ 1 \, \text{cm} = 10^{-2} \, \text{m} \]
So, we can express dyne/cm in terms of N/m:
\[ \text{Surface Tension in N/m} = \frac{72.75}{10^5} \times \frac{1}{10^{-2}} = \frac{72.75}{10^5} \times 10^2 \]

### Step 5: Simplify the Expression
Combining the powers of ten:
\[ \text{Surface Tension in N/m} = 72.75 \times 10^{-5 + 2} = 72.75 \times 10^{-3} \, \text{N/m} \]

### Step 6: Calculate the Final Value
Calculating the final value:
\[ 72.75 \times 10^{-3} = 0.07275 \, \text{N/m} \]

### Final Answer
The surface tension of water at 20°C in SI units is:
\[ \text{Surface Tension} = 0.07275 \, \text{N/m} \]

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