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Chapter 7: Problem 1002

A spherical drop of coater has radius \(1 \mathrm{~mm}\) if surface tension of contex is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) difference of pressures between inside and outside of the spherical drop is (A) \(35 \mathrm{~N} / \mathrm{m}^{2}\) (B) \(70 \mathrm{~N} / \mathrm{m}^{2}\) (C) \(140 \mathrm{~N} / \mathrm{m}^{2}\) (D) zero

Short Answer

Expert verified
The pressure difference between inside and outside of the spherical drop is \(140\) N/\(m^2\), which corresponds to option (C).

Step by step solution

01

Identify the given values

We are given the following values in the problem: - Radius of the spherical drop (r) = 1 mm = \(1 \times 10^{-3}\) m (converted to meters) - Surface tension of coater (T) = \(70 \times 10^{-3}\) N/m
02

Apply the formula for pressure difference

The formula for pressure difference (ΔP) between the inside and outside of a spherical drop due to surface tension is given by: ΔP = \(\frac{2T}{r}\) Now we can plug in the given values for surface tension and radius: ΔP = \(\frac{2 (70 \times 10^{-3})}{1 \times 10^{-3}}\)
03

Calculate the pressure difference

Simplify the expression: ΔP = \(\frac{140}{1}\) N/\(m^2\) ΔP = \(140\) N/\(m^2\) So, the pressure difference between inside and outside of the spherical drop is \(140\) N/\(m^2\), which corresponds to option (C).

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