WBJEE · Physics · Mechanical Properties of Fluids
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 \(^{-1}\) )
- A 60
- B 70
- C 80
- D 90
Answer & Solution
Correct Answer
(C) 80
Step-by-step Solution
Detailed explanation
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 \(F=\Delta P \times A\)…
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