The force required to lift a circular flat plate of radius 5 cm on the surface of water is: (Surface tension of water is 75 dyne/cm):-

Options

A. 30 dyne
B. 60 dyne
C. 750 dyne
D. 750 `pi` dyne

Answer

To find the force required to lift a circular flat plate of radius 5 cm from the surface of water, we will use the concept of surface tension. The force due to surface tension can be calculated using the formula:

\[ F = \gamma \times L \]

where:
- \( F \) is the force,
- \( \gamma \) is the surface tension of the liquid (in dyne/cm),
- \( L \) is the perimeter (circumference) of the circular plate (in cm).

### Step-by-step Solution:

1. **Calculate the Perimeter (Circumference) of the Plate:**
The formula for the circumference \( C \) of a circle is given by:
\[ C = 2 \pi r \]
where \( r \) is the radius of the circle.

Given that the radius \( r = 5 \) cm, we can substitute this value into the formula:
\[ C = 2 \pi \times 5 \text{ cm} = 10 \pi \text{ cm} \]

2. **Substitute the Values into the Force Formula:**
Now we can use the surface tension value and the circumference to find the force. The surface tension \( \gamma \) of water is given as 75 dyne/cm.

Using the formula for force:
\[ F = \gamma \times L \]
Substitute \( \gamma = 75 \) dyne/cm and \( L = 10 \pi \) cm:
\[ F = 75 \text{ dyne/cm} \times 10 \pi \text{ cm} \]

3. **Calculate the Force:**
Now we can calculate the force:
\[ F = 750 \pi \text{ dyne} \]

Using the approximate value of \( \pi \approx 3.14 \):
\[ F \approx 750 \times 3.14 \text{ dyne} \approx 2355 \text{ dyne} \]

### Final Answer:
The force required to lift the circular flat plate of radius 5 cm on the surface of water is approximately **2355 dyne**.