Question

Liquid reaches an equilibrium as shown in a capillary tube of internal radius $r=1\text{cm}$. If the surface tension of the liquid is $T=0.07\text{N/m}$. The angle of contact $\theta ={60}^{\circ }$ and density of liquid $\rho ={10}^3{\text{kg/m}}^3$, then the pressure difference between $P$ and $Q$ is


Take atmospheric pressure as ${10}^5{\text{N/m}}^2$.

• A. $0.7{\text{N/m}}^2$
• B. $3.5{\text{N/m}}^2$
• C. $7{\text{N/m}}^2$
• D. $0.2{\text{N/m}}^2$

Answer 1

The correct option is C $7{\text{N/m}}^2$

$P_A=$ Atmospheric Pressure $={10}^5{\text{N/m}}^2$
We know,
Pressure at $Q,P_Q=P_P-\frac{2T\cos \theta }{r}$

We know that,
$\Rightarrow P_Q=P_P-\frac{2\times 0.07\times \text{cos}{60}^{\circ }}{0.01}$
$\Rightarrow P_Q=P_P-7$
$\therefore P_P-P_Q=7{\text{N/m}}^2$