A beaker of radius 15 cm is filled with a liquid of surface tension 0.075 N/m. Force across an imaginary diameter on the surface of the liquid is

Options

A. 0.075 N
B. `1.5 xx 10^(-2) N`
C. `0.225 N`
D. `2.25 xx 10^(-2) N`

Explanation

To find the force across an imaginary diameter on the surface of the liquid in the beaker, we can use the formula relating surface tension to force. The surface tension (S) is defined as the force (F) per unit length (L).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the beaker (r) = 15 cm = 0.15 m
- Surface tension (S) = 0.075 N/m

2. **Calculate the Diameter**:
- The diameter (d) of the beaker is twice the radius.
\[ d = 2 \times r = 2 \times 0.15 \, \text{m} = 0.30 \, \text{m} \]

3. **Use the Formula for Force**:
- The force (F) across the imaginary diameter can be calculated using the formula:
\[ F = S \times L \]
where \( L \) is the length across which the surface tension acts. In this case, \( L \) is the diameter of the beaker.

4. **Substitute the Values**:
- Substitute the values of surface tension and diameter into the formula:
\[ F = 0.075 \, \text{N/m} \times 0.30 \, \text{m} \]

5. **Calculate the Force**:
\[ F = 0.075 \times 0.30 = 0.0225 \, \text{N} \]
- Converting this to scientific notation:
\[ F = 2.25 \times 10^{-2} \, \text{N} \]

6. **Final Answer**:
- The force across the imaginary diameter on the surface of the liquid is:
\[ F = 2.25 \times 10^{-2} \, \text{N} \]