Surface Tension, Excess Pressure and Capillarity — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Surface Tension, Excess Pressure and Capillarity MCQs with step-by-step solutions (7 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Surface Tension, Excess Pressure and Capillarity · medium · theory
A liquid does not wet the solid surface if angle of contact is
A. equal to $45^\circ$
B. equal to $60^\circ$
C. greater than $90^\circ$ ✓ Correct
D. zero
Solution: A liquid does not wet a solid surface when the angle of contact is obtuse, i.e. $\theta > 90^\circ$.
Q2 — Surface Tension, Excess Pressure and Capillarity · medium · numerical
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 tube is 5 g. Another capillary tube of radius $2r$ is immersed in water. The mass of water that will rise in this tube is
A. $5.0$ g
B. $10.0$ g
C. $20.0$ g
D. $2.5$ g ✓ Correct
Solution: Height of water in a capillary tube, $h=\dfrac{2S\cos\theta}{\rho g r}\Rightarrow h\propto\dfrac{1}{r}$, so $\dfrac{h_1}{h_2}=\dfrac{r_2}{r_1}=\dfrac{2r}{r}=2$. As mass $m=A\cdot h\cdot\rho$, $\dfrac{m_2}{m_1}=\dfrac{h_2}{h_1}=\dfrac{1}{2}$. Therefore $m_2=\dfrac{m_1}{2}=\dfrac{5}{2}=2.5$ g.
Q3 — Surface Tension, Excess Pressure and Capillarity · medium · numerical
A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of $2.5\times10^{-2}$ N/m. The pressure inside the bubble equals at a point $Z_0$ below the free surface of water in a container. Taking $g=10$ m/s$^2$, density of water $=10^3$ kg/m$^3$, the value of $Z_0$ is
A. $10$ cm
B. $1$ cm ✓ Correct
C. $0.5$ cm
D. $100$ cm
Solution: Excess pressure inside a soap bubble is $p=\dfrac{4T}{r}$, so the total pressure inside is $p_1=p_0+\dfrac{4T}{r}$. The pressure at depth $Z_0$ is $p_2=p_0+Z_0\rho g$. Equating gives $Z_0=\dfrac{4T}{r\rho g}=\dfrac{4\times2.5\times10^{-2}}{1\times10^{-3}\times10^3\times10}=10\times10^{-3}$ m $=1$ cm.
Q4 — Surface Tension, Excess Pressure and Capillarity · medium · numerical
A rectangular film of liquid is extended from ($4$ cm $\times$ 2 cm) to ($5$ cm $\times$ 4 cm). If the work done is $3\times10^{-4}$ J, the value of the surface tension of the liquid is
A. $0.250$ Nm$^{-1}$
B. $0.125$ Nm$^{-1}$ ✓ Correct
C. $0.2$ Nm$^{-1}$
D. $8.0$ Nm$^{-1}$
Solution: Increase in surface energy = increase in area $\times$ surface tension. As the film has two surfaces, $\Delta A=(5\times4-4\times2)\times2=24$ cm$^2=24\times10^{-4}$ m$^2$. From $W=T\cdot\Delta A$: $3\times10^{-4}=T\times24\times10^{-4}\Rightarrow T=\dfrac{1}{8}=0.125$ N/m.
Q5 — Surface Tension, Excess Pressure and Capillarity · medium · numerical
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
A. $\dfrac{\pi}{2}>\theta_1>\theta_2>\theta_3\geq0$
B. $0\leq\theta_1<\theta_2<\theta_3<\dfrac{\pi}{2}$ ✓ Correct
C. $\dfrac{\pi}{2}<\theta_1<\theta_2<\theta_3<\pi$
D. $\pi>\theta_1>\theta_2>\theta_3>\dfrac{\pi}{2}$
Solution: By the ascent formula for a capillary tube, $h=\dfrac{2T\cos\theta}{\rho g r}$. Since $h$, $T$ and $r$ are the same, $\dfrac{\cos\theta_1}{\rho_1}=\dfrac{\cos\theta_2}{\rho_2}=\dfrac{\cos\theta_3}{\rho_3}$, so $\cos\theta\propto\rho$. As $\rho_1>\rho_2>\rho_3$, $\cos\theta_1>\cos\theta_2>\cos\theta_3$, giving $0\leq\theta_1<\theta_2<\theta_3<\dfrac{\pi}{2}$.
Q6 — Surface Tension, Excess Pressure and Capillarity · medium · numerical
A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, then
A. energy $=4VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$ is released
B. energy $=3VT\left(\dfrac{1}{r}+\dfrac{1}{R}\right)$ is absorbed
C. energy $=3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$ is released ✓ Correct
D. energy is neither released nor absorbed
Solution: Energy released $=(A_f-A_i)T$, where the single big drop has area $A_i=4\pi R^2=\dfrac{3V}{R}$ and the small drops together have area $A_f=n\cdot4\pi r^2=\dfrac{3V}{r}$ (using $V=\dfrac{4}{3}\pi R^3$). Hence energy released $=3VT\left(\dfrac{1}{r}-\dfrac{1}{R}\right)$.
Q7 — Surface Tension, Excess Pressure and Capillarity · medium · theory
The wettability of a surface by a liquid depends primarily on
A. viscosity
B. surface tension
C. density
D. angle of contact between the surface and the liquid ✓ Correct
Solution: The wettability of a surface by a liquid depends primarily on the angle of contact between the surface and the liquid.