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A drop of some liquid of volume $0.04 \mathrm{cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 \mathrm{cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{-1}$ )
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Verified Answer
The correct answer is:
80
Let, thickness of layer be $x$
So, volume $V=$ Area $\times x$
$$
\begin{array}{ll}
V=A \times x & (\because)=2 r) \\
x=V / A
\end{array}
$$
$2 r=\frac{V}{A} \Rightarrow r=\frac{V}{2 A}$
(i)
and $\quad \Delta P=\frac{T}{r}$
We know that $F=\Delta P \times A$
$$
=\frac{T}{r} \times A
$$
$$
\begin{array}{l}
F=\frac{T}{\left(\frac{V}{2 A}\right)} \times A \\
T=\frac{F \times V}{2 A^{2}}
\end{array}
$$
[from
where $F=16 \times 10^{5}$ dyne $, V=0.04 \mathrm{cm}^{3}$
$$
\begin{aligned}
A=20 \mathrm{cm}^{2}=\frac{16 \times 10^{5} \times 0.04}{2 \times 20^{2}} \\
=\frac{8 \times 10^{5} \times 4}{20^{2} \times 100}=\frac{8 \times 10^{5} \times 4}{400 \times 100} \\
=8 \times 10^{5} \times 10^{-4} &=80 \mathrm{dyne} / \mathrm{cm} \\
&=80 \text { dyne } \mathrm{cm}^{-1}
\end{aligned}
$$
So, volume $V=$ Area $\times x$
$$
\begin{array}{ll}
V=A \times x & (\because)=2 r) \\
x=V / A
\end{array}
$$
$2 r=\frac{V}{A} \Rightarrow r=\frac{V}{2 A}$
(i)
and $\quad \Delta P=\frac{T}{r}$
We know that $F=\Delta P \times A$
$$
=\frac{T}{r} \times A
$$
$$
\begin{array}{l}
F=\frac{T}{\left(\frac{V}{2 A}\right)} \times A \\
T=\frac{F \times V}{2 A^{2}}
\end{array}
$$
[from
where $F=16 \times 10^{5}$ dyne $, V=0.04 \mathrm{cm}^{3}$
$$
\begin{aligned}
A=20 \mathrm{cm}^{2}=\frac{16 \times 10^{5} \times 0.04}{2 \times 20^{2}} \\
=\frac{8 \times 10^{5} \times 4}{20^{2} \times 100}=\frac{8 \times 10^{5} \times 4}{400 \times 100} \\
=8 \times 10^{5} \times 10^{-4} &=80 \mathrm{dyne} / \mathrm{cm} \\
&=80 \text { dyne } \mathrm{cm}^{-1}
\end{aligned}
$$
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