Question

Consider a circular ring of radius \(1.4\,\text{cm}\) lying on the surface of a liquid. If a vertical force of \(0.022\,\text{N}\) greater than the weight of the ring is required to lift this ring from the liquid surface, then the surface tension of the liquid is

• A. \(0.085\,\text{N m}^{-1}\)
• B. \(0.125\,\text{N m}^{-1}\)
• C. \(0.250\,\text{N m}^{-1}\)
• D. \(0.465\,\text{N m}^{-1}\)

Hint:

For a circular ring lifted from a liquid surface, surface tension acts along both the inner and outer circumferences. Hence, \[ F=4\pi rT. \]

Answer 1

The Correct Option is B

Step 1: Understand the extra force required.
The force greater than the weight of the ring is due to surface tension.
Given, \[ F=0.022\,\text{N} \] Radius of the ring is \[ r=1.4\,\text{cm} \] \[ r=1.4\times 10^{-2}\,\text{m} \]

Step 2: Use the surface tension force formula for a ring.
For a circular ring lifted from a liquid surface, the surface tension acts along both inner and outer circumferences.
Therefore, total length in contact is \[ L=2(2\pi r)=4\pi r \] Hence, \[ F=T(4\pi r) \] So, \[ T=\frac{F}{4\pi r} \]

Step 3: Substitute the given values.
\[ T=\frac{0.022}{4\pi(1.4\times 10^{-2})} \] Using \[ \pi=\frac{22}{7}, \] \[ T=\frac{0.022}{4\times \frac{22}{7}\times 1.4\times 10^{-2}} \] Since \[ 1.4=\frac{14}{10}, \] \[ 1.4\times 10^{-2}=0.014 \] Thus, \[ T=\frac{0.022}{4\times \frac{22}{7}\times 0.014} \] \[ T=\frac{0.022}{0.176} \] \[ T=0.125\,\text{N m}^{-1} \]

Step 4: Final conclusion.
Therefore, the surface tension of the liquid is \[ \boxed{0.125\,\text{N m}^{-1}} \]