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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
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The correct answer is:
$0.125 \mathrm{Nm}^{-1}$
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 $T=125 \times 10^{-3}=0.125 \mathrm{~N} / \mathrm{m}$.
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 $T=125 \times 10^{-3}=0.125 \mathrm{~N} / \mathrm{m}$.
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