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Water rises to a height of 3.2 cm in a glass capillary tube.

Find the height to which the same water rises in another capillary tube having half the area of cross-section.

Answer :

Answer:

The height to which the water rises in the second capillary tube is [tex]\( 3.2 \)[/tex] cm.

Explanation:

The height to which water rises in a capillary tube is given by the formula:

[tex]\[ h = \frac{{2T}}{{r\rho g}} \][/tex]

Where:

- [tex]\( h \)[/tex] is the height to which water rises,

- [tex]\( T \)[/tex] is the surface tension of water,

- [tex]\( r \)[/tex] is the radius of the capillary tube,

- [tex]\( \rho \)[/tex] is the density of water,

- [tex]\( g \)[/tex] is the acceleration due to gravity.

Let's denote the initial height as [tex]\( h_1 \)[/tex] and the radius of the first capillary tube as [tex]\( r_1 \)[/tex], and the height to which the water rises in the second capillary tube as [tex]\( h_2 \)[/tex], and the radius of the second capillary tube as [tex]\( r_2 \)[/tex].

Since the surface tension and the density of water remain constant, we can set up the equation:

[tex]\[ h_1 = \frac{{2T}}{{r_1 \rho g}} \][/tex]

For the second capillary tube, given that the area of cross-section is half, the radius [tex]\( r_2 \)[/tex] is related to the radius [tex]\( r_1 \)[/tex] by:

[tex]\[ r_2 = \sqrt{\frac{r_1^2}{2}} = \frac{r_1}{\sqrt{2}} \][/tex]

Now, we can use this relationship to find [tex]\( h_2 \)[/tex] :

[tex]\[ h_2 = \frac{{2T}}{{r_2 \rho g}} = \frac{{2T}}{{(r_1/\sqrt{2}) \rho g}} \][/tex]

Given that [tex]\( h_1 = 3.2 \)[/tex] cm, let's denote this value as [tex]\( h_1 = 3.2 \times 10^{-2} \)[/tex] m. We'll also assume the radius [tex]\( r_1 \)[/tex] is in meters.

[tex]\[ 3.2 \times 10^{-2} = \frac{{2T}}{{r_1 \rho g}} \][/tex]

[tex]\[ \frac{{2T}}{{r_1 \rho g}} = \frac{{2T}}{{(r_1/\sqrt{2}) \rho g}} \][/tex]

[tex]\[ \frac{{1}}{{r_1}} = \frac{{1}}{{r_1/\sqrt{2}}} \][/tex]

[tex]\[ r_1 = \frac{{r_1}}{{\sqrt{2}}} \][/tex]

[tex]\[ 1 = \frac{{1}}{{\sqrt{2}}} \][/tex]

Thus, [tex]\( h_2 = h_1 = 3.2 \times 10^{-2} \)[/tex] m.

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