Solveeit Logo
Login

Question

Question: The lower end of a capillary tube of diameter \(2mm\)is dipped \(8cm\)below the surface of water in ...

The lower end of a capillary tube of diameter 2mm2mmis dipped 8cm8cmbelow the surface of water in a beaker. The pressure required in the tube to blow a bubble at its end in water is:
(Twater=7.3×10−3Nm−1,ρwater=1000kgm−3,1atm=1.01×105Pa,g=9.8ms−2)\left( {{T_{water}} = 7.3 \times {{10}^{ - 3}}N{m^{ - 1}},{\rho _{water}} = 1000kg{m^{ - 3}},1atm = 1.01 \times {{10}^5}Pa,g = 9.8m{s^{ - 2}}} \right)
A) 2.19×105Pa2.19 \times {10^5}Pa
B) 1.02×105Pa1.02 \times {10^5}Pa
C) 4.9×105Pa4.9 \times {10^5}Pa
D) 7×105Pa7 \times {10^5}Pa

Explanation

Solution

In the capillary tube method, we immerse a capillary tube of radius rr vertically in a liquid to a depth h1{h_1}. The liquid under experiment will have a density ρ\rho .The meniscus will be forced down to the lower end of the capillary and is held there by a pressure. This pressure is measured using the capillary tube method. We use the equation for pressure below the meniscus (the concave surface of the liquid).

Formula used:
Pascal’s law,
P0=Pa+hρg{P_0} = {P_a} + h\rho g
Where, Po{P_o} stands for the pressure on the surface of the liquid, Pa{P_a} stands for the atmospheric pressure, hh stands for the height of the liquid, ρ\rho stands for the density of the liquid, and gg stands for the acceleration due to gravity.

Complete step by step solution:
The pressure just below the concave surface of the bubble will be less by 2Tr\dfrac{{2T}}{r} than the pressure just above the surface. Hence the bubble will have an excess pressure, 2Tr\dfrac{{2T}}{r} (where, TT stands for surface tension of the liquid and rr is the radius of the capillary tube)
Using Pascal’s law, the pressure on the surface of the liquid is given by,
P0=Pa+hρg{P_0} = {P_a} + h\rho g
The values are given as,
Pa=1.01×105Pa{P_a} = 1.01 \times {10^5}Pa
h=8×10−2mh = 8 \times {10^{ - 2}}m
ρ=1000kgm−3\rho = 1000kg{m^{ - 3}}
g=9.8ms−2g = 9.8m{s^{ - 2}}
Substituting these values in equation, we get
P0=1.01×105+8×10−2×1000×9.8{P_0} = 1.01 \times {10^5} + 8 \times {10^{ - 2}} \times 1000 \times 9.8
P0=1.01784×105Pa{P_0} = 1.01784 \times {10^5}Pa
Now the bubble will have an excess pressure 2Tr\dfrac{{2T}}{r}
Therefore, the pressure required to blow the bubble at its end in the water will be given by,
Pr=P0+2Tr{P_r} = {P_0} + \dfrac{{2T}}{r}
We have calculated the value of P0=1.01784×105Pa{P_0} = 1.01784 \times {10^5}Pa
The surface tension of water is given by Twater=7.3×10−3Nm−1{T_{water}} = 7.3 \times {10^{ - 3}}N{m^{ - 1}}
Radius of the tube is given by, r=2×10−32=1×10−3mr = \dfrac{{2 \times {{10}^{ - 3}}}}{2} = 1 \times {10^{ - 3}}m
Substituting the values in the equation for pressure, we get
Pr=1.01784×105+2×7.3×10−31×10−3Pa{P_r} = 1.01784 \times {10^5} + \dfrac{{2 \times 7.3 \times {{10}^{ - 3}}}}{{1 \times {{10}^{ - 3}}}}Pa Pr=1.01798×105{P_r} = 1.01798 \times {10^5}
This value can be rounded off into Pr=1.02×105Pa{P_r} = 1.02 \times {10^5}Pa

The answer is: option (B), 1.02×105Pa.1.02 \times {10^5}Pa.

Note: A capillary tube is a small tube of fixed length and very small diameter. Capillary action occurs when the adhesion between the walls of the tube dominate the cohesion between the molecules of the liquid. The capillary rise is proportional to the surface tension of the liquid and it is inversely proportional to the radius rr of the capillary tube and the acceleration due to gravity.