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3 years ago

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Calculate (a) the maximum capillary rise of water between two vertical glassplates spaced 0.15 mm apart and (b) the minimum capi

llary drop if the sameplates, now spaced 0.20 mm apart, are in mercury. Assume room temperature.

1 answer:

miss Akunina [59]3 years ago

5 0

Answer:

The the maximum capillary rise of water between two vertical glass plates is 9.79 cm

The the maximum capillary rise of mercury between two vertical glass plates is -2.77 cm

Explanation:

Given that,

Distance = 0.15 mm

Suppose, The surface tension of water is 72 dynes/cm at 25°C for pure water and perfectly clean glass, the angle of contact is 0°.

(a). We need to calculate the maximum capillary rise of water between two vertical glass plates

Using formula of rise of liquid in capillary tube

$h=\dfrac{2S\cos\theta}{r\rho g}$

Put the value into the formula

$h=\dfrac{2\times72\cos0}{0.015\times1\times980}$

$h=9.79\ cm$

(b). If the same plates, now spaced 0.20 mm apart, are in mercury

The density of mercury is 13.6 g/cm³, the angle of contact of mercury with glass is 139°, and the surface tension of mercury is 490 dyne/cm and g = 980 cm/s².

We need to calculate the minimum capillary drop

$h=\dfrac{2\times490\cos(139)}{0.02\times13.6\times980}$

$h=-2.77\ cm$

Hence, The the maximum capillary rise of water between two vertical glass plates is 9.79 cm

The the maximum capillary rise of mercury between two vertical glass plates is -2.77 cm

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Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
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Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
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We need to do an analysis of horizontal and vertical forces acting on the object:

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Vertically the forces acting on the object:

Normal force $N$ (acting up from the object)

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so the sum of forces in the vertical axis "y" are:

$F_{y}=N-w\\F_{y}=N-mg$

from Newton's second Law we know that $F=ma$ , so:

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and since the object is not accelerating in the vertical direction (the movement is only horizontal) $a_{y}=0$ , and:

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and we are told that the crate moves at a steady speed, thus there is no acceleration: $a_{x}=0$

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