Consider a circular ring of radius $1,4 \mathrm{~cm}$ lying on the surface of a liquid. If a vertical force of $0.022 \mathrm{~N}$ 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

  $0.085 \mathrm{Nm}^{-1}$
• Correct answer
  B

  $0.125 \mathrm{Nm}^{-1}$
• C

  $0.250 \mathrm{Nm}^{-1}$
• D

  $0.465 \mathrm{Nm}^{-1}$

Solution

The given situation is shown below.
Force required to lift the ring, $F=2 \pi(r+r) \times T+W$ $\Rightarrow \quad F=4 \pi r T+W$ Now given, $F=0.022+W$ and $\quad r=1.4 \mathrm{~cm}$ So, $0.022+W=4 \pi \times 1.4 \times T+W$ $0.022+4 \times \frac{22}{7} \times 1.4 \times T$ $\Rightarrow \quad \frac{0.022 \times 7}{4 \times 22 \times 1.4}=T$ or $T=125 \times 10^{-3}=0.125 \mathrm{~N} / \mathrm{m}$.

Hence, the correct answer is option: 'B'.