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# 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 mB) 0.59 mC) 0.87 mD) 0.91 m

Last updated date: 16th Jul 2024
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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.

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 ${\rho _{sap}} = {10^3}\;{\text{kg}}{{\text{m}}^{ - 3}}$.
$h = \dfrac{{8\sigma \cos \theta }}{{{\rho _{sap}}gr}}......\left( 1 \right)$
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.
$h = \dfrac{{4\left( {7.28 \times {{10}^{ - 2}}\;{\text{N}}{{\text{m}}^{ - 1}}} \right)\left( {\cos 0^\circ } \right)}}{{\left( {{{10}^3}\;{\text{kg}}{{\text{m}}^{ - 3}}} \right)\left( {9.8\;{\text{m}}/{{\text{s}}^2}} \right)\left( {2.5 \times {{10}^{ - 5}}\;{\text{m}}} \right)}}$
$h = 1.19\;{\text{m}}$