A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m^–1, then the excess force required to take it away from the surface is

The correct answer is 19.8 mN.

Understanding the Concept:
When a thin, flat circular disc is placed on the surface of water, the water wets its circular boundary. To lift the disc off the water surface, an excess force is required to overcome the force of surface tension. This surface tension force acts downwards along the entire perimeter (circumference) of the disc.

The formula for the excess force due to surface tension is given by:
$F=T\times L$
where:
• $T$ is the surface tension of the liquid.
• $L$ is the length of the line of contact. For a circular disc, the contact is along its circumference, so $L=2\pi R$, where $R$ is the radius of the disc.

Thus, the excess force formula becomes:
$F=T\times 2\pi R$

Step-by-Step Calculation:

1. Identify the given values and convert them to standard SI units:
• Radius of the disc, $R=4.5\text{cm}=4.5\times {10}^{-2}\text{m}$
• Surface tension of water, $T=0.07{\text{N m}}^{-1}$
• Acceleration due to gravity is not required here, and we use $\pi \approx 3.1416$.

2. Substitute the values into the force equation:
$F=0.07\times 2\times 3.1416\times 4.5\times {10}^{-2}$

3. Simplify the calculation:
$F=0.07\times 9\times 3.1416\times {10}^{-2}$
$F=0.63\times 3.1416\times {10}^{-2}$
$F\approx 1.9792\times {10}^{-2}\text{N}$

4. Convert the force to millinewtons (mN):
$F\approx 19.792\times {10}^{-3}\text{N}=19.792\text{mN}$
Rounding to three significant figures gives $19.8\text{mN}$.