Chapter 10, Problem 62P

(a)

To determine

To Find:The formula for the surface tension.

Answer to Problem 62P

  $\gamma =\frac{F}{4\pi r}$

Explanation of Solution

Given info:

The force needed to lift the ring isF .

The radius of the circular ring from the surface of the liquid is r .

Calculation:

Surface tension is the force applied for unit length. Its general formula is:

  $\gamma =\frac{1}{2}\frac{F}{L}$

In the given case, the length would be equal to the circumference which is $2\pi r$ . Substitute the expression in the above formula;

  $\begin{array}{l}\gamma =\frac{F}{2\left(2\pi r\right)}\\ \gamma =\frac{F}{4\pi r}\end{array}$

Conclusion:

The surface tension formula is $\gamma =\frac{F}{4\pi r}$ .

(b)

To determine

To Calculate:The value of the surface tension.

Answer to Problem 62P

The value of surface tension is $23.885\times {10}^{-3}\text{N/m}$ .

Explanation of Solution

Given information:

Radius, $r=2.8\text{cm}=0.028\text{m}$

Force, $F=8.40\times {10}^{-3}\text{m}$

Formula The surface tension is given expressed as:

  $\gamma =\frac{F}{4\pi r}$

Here, $F$ is the force on the surface of the ring and $r$ is the radius of the ring.

Calculation:

Calculate the surface tension.

  $\begin{array}{c}\gamma =\frac{F}{4\pi r}\\ =\frac{8.40\times {10}^{-3}}{4\times 3.14\times 0.028\text{m}}\\ =23.885\times {10}^{-3}\text{N/m}\end{array}$

Conclusion:

The surface tension is $23.885\times {10}^{-3}\text{N/m}$ .