
The sap in a tree rises in a system of capillaries of radius $2.5 \times {10^{ - 5}}\;{\text{m}}$. The surface tension of the sap is $7.28 \times {10^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}$ and angle of contact is $0^\circ $. The maximum height to which sap can rise in a tree through the capillary action is $\left( {{\rho _{sap}} = {{10}^3}\;{\text{kg}}{{\text{m}}^{ - 3}}} \right)$.
A) 0.21 m
B) 0.59 m
C) 0.87 m
D) 0.91 m
Answer
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Hint: The phenomenon of the capillarity is defined as the rise or fall in the level of the liquid surface in a tube compared to the normal level of the liquid around the tube. The diameter of the tube must be very less compared to the length of the tube for this effect. The rise or fall of the liquid varies with the density of the liquid, size of the tube, surface tension of the liquid and angle of contact.
Complete step by step answer:
Given: The radius of the tube is $r = 2.5 \times {10^{ - 5}}\;{\text{m}}$, surface tension of the sap is $\sigma = 7.28 \times {10^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}$, angle of contact is $\theta = 0^\circ $and density of the sap is ${\rho _{sap}} = {10^3}\;{\text{kg}}{{\text{m}}^{ - 3}}$.
The expression of maximum height raised by the sap in a tree due to capillary action is given as,
$h = \dfrac{{8\sigma \cos \theta }}{{{\rho _{sap}}gr}}......\left( 1 \right)$
Here, g is the gravitational acceleration and its value is $9.8\;{\text{m}}/{{\text{s}}^2}$.
Substitute $r = 2.5 \times {10^{ - 5}}\;{\text{m}}$, $\sigma = 7.28 \times {10^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}$, $\theta = 0^\circ $and ${\rho _{sap}} = {10^3}\;{\text{kg}}{{\text{m}}^{ - 3}}$in the expression (1) to find the maximum height raised by the sap in a tree.
$h = \dfrac{{4\left( {7.28 \times {{10}^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}} \right)\left( {\cos 0^\circ } \right)}}{{\left( {{{10}^3}\;{\text{kg}}{{\text{m}}^{ - 3}}} \right)\left( {9.8\;{\text{m}}/{{\text{s}}^2}} \right)\left( {2.5 \times {{10}^{ - 5}}\;{\text{m}}} \right)}}$
$h = 1.19\;{\text{m}}$
Thus, the maximum height raised by the sap in a tree due to capillary action is 1.19 m.
Additional Information: The surface tension is the effect that acts at the contact surface of two non-mixing fluids due to tension at the one fluid by another fluid. This effect varies with the temperature and impurity in the fluid.
Note: Be careful in substituting the values because sometimes the units of the variables are different, so change the units of all the variables in the same system. Check that radius or diameter which is given in the problem. The angle of contact should be in degree not in radians.
Complete step by step answer:
Given: The radius of the tube is $r = 2.5 \times {10^{ - 5}}\;{\text{m}}$, surface tension of the sap is $\sigma = 7.28 \times {10^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}$, angle of contact is $\theta = 0^\circ $and density of the sap is ${\rho _{sap}} = {10^3}\;{\text{kg}}{{\text{m}}^{ - 3}}$.
The expression of maximum height raised by the sap in a tree due to capillary action is given as,
$h = \dfrac{{8\sigma \cos \theta }}{{{\rho _{sap}}gr}}......\left( 1 \right)$
Here, g is the gravitational acceleration and its value is $9.8\;{\text{m}}/{{\text{s}}^2}$.
Substitute $r = 2.5 \times {10^{ - 5}}\;{\text{m}}$, $\sigma = 7.28 \times {10^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}$, $\theta = 0^\circ $and ${\rho _{sap}} = {10^3}\;{\text{kg}}{{\text{m}}^{ - 3}}$in the expression (1) to find the maximum height raised by the sap in a tree.
$h = \dfrac{{4\left( {7.28 \times {{10}^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}} \right)\left( {\cos 0^\circ } \right)}}{{\left( {{{10}^3}\;{\text{kg}}{{\text{m}}^{ - 3}}} \right)\left( {9.8\;{\text{m}}/{{\text{s}}^2}} \right)\left( {2.5 \times {{10}^{ - 5}}\;{\text{m}}} \right)}}$
$h = 1.19\;{\text{m}}$
Thus, the maximum height raised by the sap in a tree due to capillary action is 1.19 m.
Additional Information: The surface tension is the effect that acts at the contact surface of two non-mixing fluids due to tension at the one fluid by another fluid. This effect varies with the temperature and impurity in the fluid.
Note: Be careful in substituting the values because sometimes the units of the variables are different, so change the units of all the variables in the same system. Check that radius or diameter which is given in the problem. The angle of contact should be in degree not in radians.
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