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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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Most popular questions from this chapter

The force required to separate two glass plates of area $10^{-2} \mathrm{~m}^{2}\( with a film of water \)0.05 \mathrm{~mm}$ thick between them is (surface tension of water is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) ) (A) \(28 \mathrm{~N}\) (B) \(14 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(38 \mathrm{~N}\)

A big drop of radius \(R\) is formed by 1000 small droplets of coater then the radius of small drop is (A) \((\mathrm{R} / 2)\) (B) \((\mathrm{R} / 5)\) (C) \((\mathrm{R} / 6)\) (D) \((\mathrm{R} / 10)\)

The resistance of a resistance thermometer has values \(2.71\) and $3.70 \mathrm{ohm}\( at \)10^{\circ} \mathrm{C}\( and \)100^{\circ} \mathrm{C}$. The temperature at which the resistance is \(3.26 \mathrm{ohm}\) is (A) \(40^{\circ} \mathrm{C}\) (B) \(50^{\circ} \mathrm{C}\) (C) \(60^{\circ} \mathrm{C}\) (D) \(70^{\circ} \mathrm{C}\)

If the excess pressure inside a soap bubble is balanced by oil column of height \(2 \mathrm{~mm}\) then the surface tension of soap solution will be. \((\mathrm{r}=1 \mathrm{~cm}\) and density $\mathrm{d}=0.8 \mathrm{gm} / \mathrm{cc})$ (A) \(3.9 \mathrm{~N} / \mathrm{m}\) (B) \(3.9 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (C) \(3.9 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) (D) \(3.9 \times 10^{-1} \mathrm{~N} / \mathrm{m}\)

If a rubber ball is taken at the depth of \(200 \mathrm{~m}\) in a pool, Its volume decreases by \(0.1 \%\). If the density of the water is $1 \times 10^{3}\left(\mathrm{~kg} / \mathrm{m}^{3}\right) \& \mathrm{~g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)$. Then what will be the volume elasticity in ? (A) \(10^{8}\) (B) \(2 \times 10^{8}\) (C) \(10^{9}\) (D) \(2 \times 10^{9}\)
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