Question

Water is kept in a beaker of radius 5.0 cm. Consider a diameter of the beaker on the surface of the water. Find the force by which the surface on one side of the diameter pulls the surface on the other side. Surface tension of water = 0.075 N m^-1.

• A. 0.75 N
• B. 7.5 × 10^-4 N
• C. 7.5 × 10^-3 N
• D. 7.5 × 10^-2 N

Answer 1

The correct option is C

7.5 × 10^-3 N

The length of the diameter is
$l=2r=10cm$
=0.1m
The surface tension is S =$\frac{F}{l}$.Thus,
F=Sl
=$\left(0.075N{m}^{-1}\right)\times \left(0.1m\right)=7.5\times {10}^{-3}N$