
Given that the surface tension of water is \[{\mathbf{75}}\;{\mathbf{dyne/cm}}\], its density \[{\mathbf{1}}\;{\mathbf{g/cc}}\] and angle of contact zero, the height to which water rises in a capillary tube of \[{\mathbf{1}}\;{\mathbf{mm}}\] diameters-
(Take\[{\mathbf{g = 10}}\;{\mathbf{m/}}{{\mathbf{s}}^{\mathbf{2}}}\])
A. \[{\mathbf{9}}{\mathbf{.0}}\;{\mathbf{cm}}\]
B. \[{\mathbf{7}}{\mathbf{.5}}\;{\mathbf{cm}}\]
C. \[{\mathbf{6}}{\mathbf{.0}}\;{\mathbf{cm}}\]
D. \[{\mathbf{3}}{\mathbf{.0}}\;{\mathbf{cm}}\]
Answer
570.9k+ views
Hint: In order to calculate the height, first convert all the units in a common form. Since all the options have units of cm, therefore the common unit will be cm. The height to which the water rises can be obtained by using the formula,
\[h = \dfrac{{4S\cos \theta }}{{\rho gd}}\].
Complete step by step answer:
Given,
Surface tension of water,\[S = 75\;{\text{dyne/cm}}\]
Density of water, \[\rho = \;1\;{\text{g/cc}}\]
Diameter of the capillary tube, \[d = 1\;{\text{mm}}\]
Acceleration due to gravity, \[g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}}\]
Convert the values of \[d\] and \[g\] in\[{\text{cm}}\].
Therefore,
\[d = 1\;{\text{mm}} = 0.1\;{\text{cm}}\]
\[g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}} = 1000\;{\text{cm/}}{{\text{s}}^2}\]
We can measure the height, over which water rises inside a capillary tube by using the formula,
\[h = \dfrac{{4S\cos \theta }}{{\rho gd}}\] …… (i)
Let us assume, the angle of contact, \[\theta = 0\]
Now substitute the values of \[s,\rho ,g,d\] in equation (i)
\[
h = \dfrac{{4 \times 75 \times \cos 0}}{{1 \times 1000 \times 0.1}} \\
= \dfrac{{300}}{{100}} \\
= 3\;{\text{cm}} \\
\]
Therefore, the required height to which the water rises is \[3\;{\text{cm}}\]
So, the correct answer is “Option D”.
Additional Information:
Surface tension: It is the inclination of the liquid surfaces to shrink to lowest possible surface area. Surface friction enables insects to float and slip on a water surface, which are typically denser than water.
The density of a material is its mass per unit volume. While the Latin letter D can also be used, the symbol most commonly used for density is the \[\rho \]. Density is calculated mathematically as mass divided by volume.
Note:
In this question we need to find the height of the water to which water reaches. Let us have a look at the options. All of them have units in cm. Do not just solve the problem. First, to obtain the result we convert the units of diameter and acceleration due to gravity into cm and then solve it.
\[h = \dfrac{{4S\cos \theta }}{{\rho gd}}\].
Complete step by step answer:
Given,
Surface tension of water,\[S = 75\;{\text{dyne/cm}}\]
Density of water, \[\rho = \;1\;{\text{g/cc}}\]
Diameter of the capillary tube, \[d = 1\;{\text{mm}}\]
Acceleration due to gravity, \[g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}}\]
Convert the values of \[d\] and \[g\] in\[{\text{cm}}\].
Therefore,
\[d = 1\;{\text{mm}} = 0.1\;{\text{cm}}\]
\[g = 10\;{\text{m/}}{{\text{s}}^{\text{2}}} = 1000\;{\text{cm/}}{{\text{s}}^2}\]
We can measure the height, over which water rises inside a capillary tube by using the formula,
\[h = \dfrac{{4S\cos \theta }}{{\rho gd}}\] …… (i)
Let us assume, the angle of contact, \[\theta = 0\]
Now substitute the values of \[s,\rho ,g,d\] in equation (i)
\[
h = \dfrac{{4 \times 75 \times \cos 0}}{{1 \times 1000 \times 0.1}} \\
= \dfrac{{300}}{{100}} \\
= 3\;{\text{cm}} \\
\]
Therefore, the required height to which the water rises is \[3\;{\text{cm}}\]
So, the correct answer is “Option D”.
Additional Information:
Surface tension: It is the inclination of the liquid surfaces to shrink to lowest possible surface area. Surface friction enables insects to float and slip on a water surface, which are typically denser than water.
The density of a material is its mass per unit volume. While the Latin letter D can also be used, the symbol most commonly used for density is the \[\rho \]. Density is calculated mathematically as mass divided by volume.
Note:
In this question we need to find the height of the water to which water reaches. Let us have a look at the options. All of them have units in cm. Do not just solve the problem. First, to obtain the result we convert the units of diameter and acceleration due to gravity into cm and then solve it.
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