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Chapter 26: Problem 2807

To what height can mercury be filled in a vessel without any leakage if there is a pin hole of diameter \(0.1 \mathrm{~mm}\) at the bottom of the vessel. (Density of mercury \(=13.6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), surface tension of mercury $=550 \times 10 \mathrm{~h}^{-3} \mathrm{pNm}^{-1}\(, Angle of contact with the vessel for mercury \)=0^{\circ}$ ) (A) \(16.1 \mathrm{~cm}\) (B) \(18.5 \mathrm{~cm}\) (C) \(20 \mathrm{~cm}\) (D) \(12.5 \mathrm{~cm}\)

Short Answer

Expert verified
The maximum height of mercury that can be filled in the vessel without any leakage is approximately \(16.1 \mathrm{~cm}\) (Option A).

Step by step solution

01

Formula for height without leakage

We will use the Jurin's Law formula to find the maximum height of the liquid column that can be supported by the given surface tension without leakage: \[h = \frac{2S\cos{\theta}}{r\rho g}\] where \(h\) is the height of the liquid column, \(S\) is the surface tension of the liquid, \(\theta\) is the angle of contact between the liquid and the wall of the container, \(r\) is the radius of the hole or tube, \(\rho\) is the density of the liquid, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m~s}^{-2}\)).
02

Calculate the radius of the hole

The diameter of the hole is given as \(0.1 \mathrm{~mm}\). To find the radius, we will divide the diameter by 2: \[r = \frac{0.1 \mathrm{~mm}}{2} = 0.05 \mathrm{~mm} = 5 \times 10^{-5} \mathrm{~m}\]
03

Substitute known values into Jurin's Law formula

Now, we will substitute the values given in the problem into the Jurin's Law formula: \[h = \frac{2(550 \times 10^{-3} \mathrm{~N~m}^{-1})\cos{0^{\circ}}}{(5 \times 10^{-5} \mathrm{~m})(13.6 \times 10^{3} \mathrm{~kg~m}^{-3})(9.81 \mathrm{~m~s}^{-2})}\]
04

Solve for height

After substituting the known values into the formula, we simply need to solve for the height, \(h\): \[h \approx \frac{2(550 \times 10^{-3} \mathrm{~N~m}^{-1})(1)}{(5 \times 10^{-5} \mathrm{~m})(13.6 \times 10^{3} \mathrm{~kg~m}^{-3})(9.81 \mathrm{~m~s}^{-2})} = 0.16116 \mathrm{~m}\] Since we want the height in centimeters, we can convert meters to centimeters: \[h \approx 0.16116 \mathrm{~m} \times \frac{100 \mathrm{~cm}}{1\mathrm{~m}} = 16.116 \mathrm{~cm}\]
05

Choose the correct answer

Given the options in the question, we can see that our answer is very close to option (A) \(16.1~\mathrm{cm}\). Thus, the maximum height of mercury that can be filled in the vessel without any leakage is approximately \(16.1 \mathrm{~cm}\).

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Most popular questions from this chapter

A spherical soap bubble of radius \(2 \mathrm{~cm}\) attached to the outside of a spherical bubble of radius \(4 \mathrm{~cm}\). Then what is the radius of the common surface? (A) \(3.5 \mathrm{~cm}\) (B) \(4 \mathrm{~cm}\) (C) \(4.5 \mathrm{~cm}\) (D) \(3 \mathrm{~cm}\)

By how much depth will the surface of a liquid be depressed in a glass tube of radius \(0.2 \mathrm{~mm}\) if the angle of contact of the liquid is \(135^{\circ}\) and the surface tension is \(0.547 \mathrm{Nm}^{-1}\) ? (Density of the liquid is \(1.28 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}\) ) (A) \(3 \mathrm{~cm}\) (B) \(4 \mathrm{~cm}\) (C) \(5 \mathrm{~cm}\) (D) \(2 \mathrm{~cm}\)
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