If a soap bubble expands, the pressure inside the bubble :
(2022)
The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{R}$. As the bubble expands, its radius $R$ increases. Therefore, the excess pressure (and total pressure inside) decreases.
A soap bubble, having radius of $1 \text{ mm}$, is blown from a detergent solution having a surface tension of $2.5 \times 10^{-2} \text{ N/m}$. The pressure inside the bubble equals at a point $Z_0$ below the free surface of water in a container. Taking $g = 10 \text{ m/s}^2$, density of water $= 10^3 \text{ kg/m}^3$, the value of $Z_0$ is :
(2019)
Pressure inside the bubble is $P = P_0 + \frac{4T}{r}$. Pressure at depth $Z_0$ is $P = P_0 + \rho g Z_0$. Equating them, $\rho g Z_0 = \frac{4T}{r}$. Solving gives $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 10^3 \times 10} = 10^{-2} \text{ m} = 1 \text{ cm}$.
A rectangular film of liquid is extended from $(4 \text{ cm} \times 2 \text{ cm})$ to $(5 \text{ cm} \times 4 \text{ cm})$. If the work done is $3 \times 10^{-4} \text{ J}$, the value of the surface tension of the liquid is:
(2016 – II)
Work done in stretching a liquid film is $W = T \times 2\Delta A$ (since it has two surfaces). The change in area is $\Delta A = (5 \times 4) - (4 \times 2) = 12 \text{ cm}^2 = 12 \times 10^{-4} \text{ m}^2$. Thus, $$T = \frac{3 \times 10^{-4}}{2 \times 12 \times 10^{-4}} = 0.125 \text{ Nm}^{-1}$$.
A liquid does not wet the solid surface if angle of contact is:
(2020-Covid)
For a liquid to not wet a solid surface, it must form an obtuse angle of contact. Therefore, the angle of contact must be greater than $90^\circ$.
A capillary tube of radius $r$ is immersed in water and water rises in it to a height $h$. The mass of the water in the capillary is $5\text{ g}$. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is:
(2020)
Mass of water risen in capillary $m = \pi r^2 h \rho$. Since $h \propto \frac{1}{r}$, we get $m \propto r$. Thus, $m_2 = m_1 \left(\frac{r_2}{r_1}\right) = 5 \times \left(\frac{2r}{r}\right) = 10\text{ g}$.
Three liquids of densities $\rho_1$, $\rho_2$ and $\rho_3$ (with $\rho_1 > \rho_2 > \rho_3$), having the same value of surface tension $T$, rise to the same height in three identical capillaries. The angles of contact $\theta_1$, $\theta_2$ and $\theta_3$ obey:
(2016 – II)
Capillary rise $h = \frac{2T \cos\theta}{r \rho g}$. Since $h, T, r, g$ are constant, $\cos\theta \propto \rho$. Since $\rho_1 > \rho_2 > \rho_3$, $\cos\theta_1 > \cos\theta_2 > \cos\theta_3$. For acute angles, this means $$0 \le \theta_1 < \theta_2 < \theta_3 < \frac{\pi}{2}$$.
Water rises to height ‘$h$’ in capillary tube. If the length of capillary tube above the surface of water is made less than ‘$h$’, then:
(2015 Re)
When a capillary tube is of insufficient length, the liquid rises to the top and changes its meniscus radius to maintain equilibrium ($hR = \text{constant}$). It does not overflow.
The wettability of a surface by a liquid depends primarily on:
(2013)
Wettability directly depends on the angle of contact. If the angle is acute, the liquid wets the solid; if it is obtuse, it does not wet the solid.
The velocity of a small ball of mass $M$ and density $d$, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $\frac{d}{2}$, then the viscous force acting on the ball will be:
(2021)
At constant terminal velocity, net force is zero. Viscous force $F_v = \text{Weight} - \text{Buoyant force}$. $$F_v = Vdg - V\left(\frac{d}{2}\right)g = \frac{Vdg}{2} = \frac{Mg}{2}$$.
A small sphere of radius ‘$r$’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to :
(2018)
Rate of heat production $P = F_v v_t$. We know $F_v = 6\pi\eta r v_t$ and $v_t \propto r^2$. Thus, $$P = (6\pi\eta r v_t) v_t \propto r (r^2)^2 \propto r^5$$.