MHT CET · Physics · Mechanical Properties of Fluids
The excess pressure inside a soap bubble of radius \(2 \mathrm{~cm}\) is 50 dyne \(/ \mathrm{cm}^2\). The surface tension is
- A 25 dyne \(/ \mathrm{cm}\)
- B 60 dyne/ cm
- C 50 dyne \(/ \mathrm{cm}\)
- D 75 dyne/ cm
Answer & Solution
Correct Answer
(A) 25 dyne \(/ \mathrm{cm}\)
Step-by-step Solution
Detailed explanation
Given,
Excess pressure, \(\Delta P=50\) dyne \(/ \mathrm{cm}^2\)
Radius, \(r=2 \mathrm{~cm}\)
For a soap bubble liquid-air interface is crossed two times, so we have,
Excess pressure, \(\Delta P=\frac{4 T}{r}\), where \(\mathrm{T}\) is the surface tension.
\(\therefore T=\frac{\Delta P \times r}{4}=\frac{50 \mathrm{dyne} / \mathrm{cm}^2 \times 2 \mathrm{~cm}}{4}=25 \mathrm{dyne} / \mathrm{cm}\)
Excess pressure, \(\Delta P=50\) dyne \(/ \mathrm{cm}^2\)
Radius, \(r=2 \mathrm{~cm}\)
For a soap bubble liquid-air interface is crossed two times, so we have,
Excess pressure, \(\Delta P=\frac{4 T}{r}\), where \(\mathrm{T}\) is the surface tension.
\(\therefore T=\frac{\Delta P \times r}{4}=\frac{50 \mathrm{dyne} / \mathrm{cm}^2 \times 2 \mathrm{~cm}}{4}=25 \mathrm{dyne} / \mathrm{cm}\)
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