Surface tension of water is 73 dyne `cm^(-1)` at `20^(@)C`. If surface area is increased by `0.10" m"^(2)`, work done will be

Options

A. 73 ergs
B. 730 ergs
C. 7300 ergs
D. 73000 ergs

Explanation

To calculate the work done when the surface area of water is increased, we can use the formula:

\[ \text{Work Done} = \text{Surface Tension} \times \text{Change in Surface Area} \]

### Step 1: Convert Surface Tension to SI Units
The surface tension is given as 73 dyne/cm. We need to convert this to Newton/meter (N/m) for consistency in SI units.

1 dyne = \( 10^{-5} \) N, and 1 cm = \( 0.01 \) m.

So,

\[ 73 \, \text{dyne/cm} = 73 \times 10^{-5} \, \text{N/cm} = 73 \times 10^{-5} \times 100 \, \text{N/m} = 7.3 \, \text{N/m} \]

### Step 2: Identify the Change in Surface Area
The change in surface area is given as \( 0.10 \, \text{m}^2 \).

### Step 3: Calculate the Work Done
Now, we can substitute the values into the work done formula:

\[ \text{Work Done} = \text{Surface Tension} \times \text{Change in Surface Area} \]

\[ \text{Work Done} = 7.3 \, \text{N/m} \times 0.10 \, \text{m}^2 \]

\[ \text{Work Done} = 0.73 \, \text{J} \]

### Final Answer
The work done when the surface area is increased by \( 0.10 \, \text{m}^2 \) is \( 0.73 \, \text{J} \).

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