Explain the phenomenon of capillarity. What factors determine the rise or fall of a liquid in a capillary tube?

4.22

Let us consider a vector $\stackrel{?}{P}$ . The equation can be written as

Px = Py = 1 $\left|\stackrel{?}{P}\right|$ = $\surd (P_x^2+P_y^2)$ $\left|\stackrel{?}{P}\right|$ = $\surd (1^2+1^2)\left|\stackrel{?}{P}\right|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\stackrel{?}{i}$ + $\stackrel{?}{j}$ = $\sqrt{2}$

Let $\theta$ be the angle made by vector $\stackrel{?}{P}$ , with the x axis as given in the above figure

$\tan ?\theta$ = $\left(Px/Py\right)\theta$ = ${\tan }^{-1}?(1/1)$ , $\theta$ = 45 $°$ with the x axis

Let $\stackrel{?}{Q}$ = $\stackrel{?}{i}$ - $\stackrel{?}{j}$

$\stackrel{?}{Q_x}\stackrel{?}{i}$ – $\stackrel{?}{Q_y}$ $\stackrel{?}{j}$ = ( $\stackrel{?}{i}$ – $\stackrel{?}{j})$

$\stackrel{?}{Q_x}$ = $\stackrel{?}{Q_y}$ = 1

$\left|\stackrel{?}{Q}\right|$ = $\sqrt{{Q_x}^2+{Q_y}^2}$ = $\surd 2$

Hence $\left|\stackrel{?}{Q}\right|$ = $\surd 2$ . Therefore the magnitude of ( $\stackrel{?}{i}$ + $\stackrel{?}{j})$ = $\surd 2$

Let $\theta$ be the angle made by vector $\stackrel{?}{Q}$ , with the x axis as given in the above figure

$\tan ?\theta$ = $\left(Qx/Qy\right)\theta$ = ${-\tan }^{-1}?(1/1)$ , $\theta$ = - 45 $°$ with the x axis

It is given that $\stackrel{?}{A}$ = 2 $\stackrel{?}{i}$ + 3 $\stackrel{?}{j}$ . We can write Ax $\stackrel{?}{i}$ + Ay $\stackrel{?}{j}$ = 2 $\stackrel{?}{i}$ + 3 $\stackrel{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$\left|\stackrel{?}{A}\mathrm{}\right|$ = $\sqrt{{(2}^2}+3^2$ ) = $\surd 13$

Let $\stackrel{?}{A_x}$ make an angle $\theta$ with the x axis, then

$\tan ?\theta$ = (Ax / Ay), $\theta$ = ${\tan }^{-1}(?3/2)$ = 56.31 $°$

So the angle between (2 $\stackrel{?}{i}$ + 3 $\stackrel{?}{j}$ ) and ( $\stackrel{?}{i}$ + $\stackrel{?}{j}$ )

$\theta \text{'}$ = 56.31 – 45 = 11.31 $°$

Component of vector $\stackrel{?}{A}$ , along the direction of $\stackrel{?}{P}$ , making an angle $\theta$

= ( A cos $\theta \text{'}$ ) $\stackrel{?}{P}$

= ( A cos 11.31) $\frac{(\mathrm{}\stackrel{?}{i}+\stackrel{?}{j}\mathrm{})}{\sqrt{2}}$

= $\sqrt{13}$ $\times \frac{0.9806}{\sqrt{2}}$ ( $\stackrel{?}{i}$ + $\stackrel{?}{j}$ )

= 2.5 $\times$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $\theta \text{'}$ be the angle between the vectors (2 $\stackrel{?}{i}$ + 3 $\stackrel{?}{j}$ ) and ( $\stackrel{?}{i}$ - $\stackrel{?}{j})$

$\theta "=45+56.31=101.31°$

Component of vector $\stackrel{?}{A}$ , along the direction of $\stackrel{?}{Q}$ , making an angle $\theta$

= (A cos $\theta "$ ) $\frac{\mathrm{}(\stackrel{?}{i}\mathrm{}-\mathrm{}\stackrel{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31°)\frac{\mathrm{}(\stackrel{?}{i}\mathrm{}-\mathrm{}\stackrel{?}{j})}{\sqrt{2}}$

= -0.5( $\stackrel{?}{i}$ - $\stackrel{?}{j})$

= $-\frac{1}{\sqrt{2}}$