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Chapter 7: Problem 1009

In a capillary tube water rises by \(1.2 \mathrm{~mm}\). The height of water that will rise in another capillary tube having half the radius of the first is (A) \(1.2 \mathrm{~mm}\) (B) \(2.4 \mathrm{~mm}\) (C) \(0.6 \mathrm{~mm}\) (D) \(0.4 \mathrm{~mm}\)

Short Answer

Expert verified
The height of water that will rise in the second capillary tube with half the radius of the first tube is 2.4 mm. Therefore, the correct answer is (B) \(2.4 \mathrm{~mm}\).

Step by step solution

01

Determine the formula for the height of water in the first tube

Using the equation for capillary action, the height of water in the capillary tube is: \(h_1 = \dfrac{2 \cdot S \cos \theta}{r_1 \cdot \rho g}\)
02

Find the formula for the height of water in the second tube

Since the second capillary tube has half the radius of the first one, we can write the radius as \(r_2 = \dfrac{r_1}{2}\). Now, using the equation for capillary action, the height of water in the second capillary tube is: \(h_2 = \dfrac{2 \cdot S \cos \theta}{r_2 \cdot \rho g}\)
03

Relate the heights of water in the two tubes

We can write the second equation in terms of the first equation: \(h_2 = \dfrac{2 \cdot S \cos \theta}{\dfrac{r_1}{2} \cdot \rho g}\) Now, multiplying both numerator and denominator by 2: \(h_2 = \dfrac{4 \cdot S \cos \theta}{r_1 \cdot \rho g}\) We can notice that all the physical property terms are the same for both tubes, i.e., \(S\), \(\theta\), \(\rho\), and \(g\). So, we can write the equation in terms of the heights: \(h_2 = 2h_1\)
04

Substitute the original height and find the new height

Now, we know the height of water in the first capillary tube is 1.2 mm. Substitute this value into the equation above: \(h_2 = 2(1.2)\) \(h_2 = 2.4\) So, the height of water that will rise in the second capillary tube with half the radius of the first tube is 2.4 mm. Therefore, the correct answer is (B) \(2.4 \mathrm{~mm}\).

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Most popular questions from this chapter

The ratio of two specific heats of gas $\mathrm{C}_{\mathrm{P}} / \mathrm{C}_{\mathrm{V}}\( for Argon is \)1.6\( and for hydrogen is \)1.4 .$ Adiabatic elasticity of Argon at pressure P is E. Adiabatic elasticity of hydrogen will also be equal to \(\mathrm{E}\) at the pressure. (A) \(\mathrm{P}\) (B) \((7 / 8) \mathrm{P}\) (C) \((8 / 7) \mathrm{P}\) (D) \(1.4 \mathrm{P}\)

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