Question

A glass capillary tube of inner diameter $0.28\, mm$ is lowered vertically into water in a vessel. The pressure to be applied on the water in the capillary tube so that water level in the tube is same as that in the vessel is (surface tension of water $=0.07\, N / m$ and atmospheric pressure $\left.=10^{5} N / m ^{2}\right)$.

• A. $10^{3}$
• B. $99 \times 10^{3}$
• C. $100 \times 10^{3}$
• D. $101 \times 10^{3}$

Answer 1

Answer: (D)

Height of liquid in capillary $=\frac{2 T}{r \rho g}=h$
Pressure we need to apply $=\rho g h +P_{0}$
Substitute value of $h$
$P=\rho g \times \frac{2 T}{r \rho g}+P_{0}=\frac{2 T}{r}+P_{0}=\frac{4 T}{d}+P_{0}$
$\Rightarrow P=\frac{4 \times 0.07}{\left(0.28 \times 10^{-3}\right)}+P_{0}$
$=1000 Nm ^{-2}+10^{5} Nm ^{-2}$
$\Rightarrow P=\left(10^{3} +10^{5}\right) Nm ^{-2}$
$=101 \times 10^{3} Nm ^{-2}$
Where,
$T=$ Surface tension
$r=$ Radius of capillary
$\rho=$ Density of liquid
$P_{0}=$ Atmospheric pressure
Given.
$T=0.07\, N / m$
$d=0.28\, mm$