Surface Tension — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Surface Tension MCQs with step-by-step solutions (27 questions). Part of Mechanical Properties of Fluids. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Surface Tension · easy · theory
The SI unit of surface tension is:
A. $\text{N/m}^2$
B. $\text{N}\cdot\text{m}$
C. $\text{N}\cdot\text{m}^2$
D. $\text{N/m}$ ✓ Correct
Solution: Surface tension is the force acting per unit length of a line drawn on the liquid surface, so its unit is $\text{N/m}$ (equivalently $\text{J/m}^2$ as surface energy).
Q2 — Surface Tension · medium · theory
The dimensional formula of surface tension is:
A. $[M^1L^1T^{-2}]$
B. $[M^1L^0T^{-2}]$ ✓ Correct
C. $[M^1L^2T^{-2}]$
D. $[M^1L^{-1}T^{-2}]$
Solution: $T = \dfrac{\text{force}}{\text{length}} = \dfrac{[M^1L^1T^{-2}]}{[L]} = [M^1L^0T^{-2}]$.
Q3 — Surface Tension · easy · theory
The excess pressure inside a soap bubble of radius $R$ and surface tension $T$ is:
A. $\dfrac{8T}{R}$
B. $\dfrac{T}{R}$
C. $\dfrac{4T}{R}$ ✓ Correct
D. $\dfrac{2T}{R}$
Solution: A soap bubble has two liquid surfaces (inner and outer), so the excess pressure is twice that of a single surface: $\Delta P = \dfrac{4T}{R}$.
Q4 — Surface Tension · easy · theory
The excess pressure inside a spherical liquid drop of radius $R$ and surface tension $T$ is:
A. $\dfrac{2T}{R}$ ✓ Correct
B. $\dfrac{4T}{R}$
C. $\dfrac{T}{R}$
D. $\dfrac{T}{2R}$
Solution: A liquid drop has only one free surface, so $\Delta P = \dfrac{2T}{R}$ — half the value for a soap bubble of the same radius.
Q5 — Surface Tension · easy · numerical
The rise of a liquid in a capillary tube of radius $r$ is given by $h = \dfrac{2T\cos\theta}{r\rho g}$. If the radius of the tube is halved, the capillary rise becomes:
A. $\dfrac{h}{2}$
B. $2h$ ✓ Correct
C. $h$
D. $4h$
Solution: Since $h \propto \dfrac{1}{r}$, halving the radius doubles the height of rise.
Q6 — Surface Tension · hard · numerical
Water rises to a height $h$ in a vertical capillary tube. If the tube is now tilted so that it makes $60^\circ$ with the vertical, the length of the water column along the tube is:
A. $\dfrac{h}{2}$
B. $h\sqrt{3}$
C. $2h$ ✓ Correct
D. $h$
Solution: The vertical rise is still $h$, fixed by the pressure balance. The slanted length is $l = \dfrac{h}{\cos 60^\circ} = \dfrac{h}{0.5} = 2h$.
Q7 — Surface Tension · easy · theory
The angle of contact between pure water and clean glass is:
A. $180^\circ$
B. $90^\circ$
C. $135^\circ$
D. $0^\circ$ ✓ Correct
Solution: Water wets clean glass completely because adhesion to glass exceeds cohesion within the water, so the angle of contact is zero and the meniscus is concave.
Q8 — Surface Tension · medium · theory
Mercury is depressed in a glass capillary tube because the angle of contact between mercury and glass is:
A. Obtuse, so that $\cos\theta$ is negative ✓ Correct
B. Acute, so that $\cos\theta$ is positive
C. Exactly $0^\circ$
D. Exactly $90^\circ$
Solution: For mercury on glass the angle of contact is about $140^\circ$, making $\cos\theta$ negative in $h = \dfrac{2T\cos\theta}{r\rho g}$ and hence $h$ negative — a depression rather than a rise.
Q9 — Surface Tension · hard · numerical
The work done in blowing a soap bubble of surface tension $T$ from radius $R$ to radius $2R$ in air is:
A. $12\pi T R^2$
B. $8\pi T R^2$
C. $24\pi T R^2$ ✓ Correct
D. $6\pi T R^2$
Solution: A bubble has two surfaces. Initial area $= 2(4\pi R^2) = 8\pi R^2$; final area $= 2\left[4\pi(2R)^2\right] = 32\pi R^2$. So $W = T\,\Delta A = T(32 - 8)\pi R^2 = 24\pi T R^2$.
Q10 — Surface Tension · hard · numerical
Eight equal spherical droplets of water, each of radius $r$, coalesce into a single large drop. The surface energy of the system:
A. Increases by $50\%$
B. Decreases by $50\%$ ✓ Correct
C. Decreases by $25\%$
D. Remains unchanged
Solution: Volume conservation gives $R = 2r$. Initial energy $= 8(4\pi r^2 T) = 32\pi r^2 T$; final $= 4\pi(2r)^2 T = 16\pi r^2 T$. The energy halves, i.e. falls by $50\%$.
Q11 — Surface Tension · hard · numerical
Two soap bubbles of radii $3\text{ cm}$ and $4\text{ cm}$ coalesce isothermally inside a vacuum chamber. The radius of the resulting bubble is:
A. $3.5\text{ cm}$
B. $2.4\text{ cm}$
C. $7\text{ cm}$
D. $5\text{ cm}$ ✓ Correct
Solution: In vacuum the total surface area is conserved: $R^2 = r_1^2 + r_2^2 = 9 + 16 = 25$, so $R = 5\text{ cm}$.
Q12 — Surface Tension · medium · numerical
A needle of length $5\text{ cm}$ floats on the surface of water. The minimum downward force needed to pull it off the surface is ($T = 0.07\text{ N/m}$):
A. $7.0\text{ N}$
B. $1.4 \times 10^{-2}\text{ N}$
C. $7.0 \times 10^{-3}\text{ N}$ ✓ Correct
D. $3.5 \times 10^{-3}\text{ N}$
Solution: The surface contacts the needle along both its sides, so the effective length is $2l$. Hence $F = 2lT = 2(0.05)(0.07) = 7.0 \times 10^{-3}\text{ N}$.
Q13 — Surface Tension · hard · numerical
Two glass plates are separated by a thin water film of thickness $0.1\text{ mm}$ over an area of $10\text{ cm}^2$. The force required to pull them apart is ($T = 0.07\text{ N/m}$):
A. $2.8\text{ N}$
B. $1.4\text{ N}$ ✓ Correct
C. $0.7\text{ N}$
D. $14\text{ N}$
Solution: The curved film of radius $\dfrac{d}{2}$ produces an excess pressure $\Delta P = \dfrac{2T}{d}$. So $F = \Delta P \times A = \dfrac{2(0.07)(10 \times 10^{-4})}{0.1 \times 10^{-3}} = 1.4\text{ N}$.
Q14 — Surface Tension · medium · theory
The surface tension of a liquid as its temperature is raised:
A. First decreases and then increases
B. Decreases and becomes zero at the critical temperature ✓ Correct
C. Remains completely unchanged
D. Increases steadily without limit
Solution: Heating weakens the intermolecular cohesive forces responsible for surface tension. At the critical temperature the liquid and vapour become indistinguishable and the surface tension vanishes.
Q15 — Surface Tension · medium · theory
The pressure just below a concave meniscus of a liquid in a capillary tube is:
A. Greater than atmospheric pressure
B. Less than atmospheric pressure ✓ Correct
C. Zero
D. Equal to atmospheric pressure
Solution: Pressure is always higher on the concave side of a curved surface by $\dfrac{2T}{R}$. Here the air is on the concave side, so the liquid just beneath the meniscus is below atmospheric pressure — which is what draws the liquid up the tube.
Q16 — Surface Tension · easy · theory
Small liquid drops are spherical in shape because surface tension tends to:
A. Minimise the surface area for a given volume ✓ Correct
B. Keep the pressure inside equal to that outside
C. Maximise the surface area for a given volume
D. Minimise the volume for a given surface area
Solution: Surface energy is proportional to area, so a free drop settles into the shape of least area for its volume. For a fixed volume that shape is a sphere.
Q17 — Surface Tension · easy · theory
Adding detergent to water improves its cleaning action mainly because the detergent:
A. Increases the viscosity of the water
B. Increases the density of the water
C. Lowers the surface tension, so water spreads and wets the fabric better ✓ Correct
D. Raises the surface tension, so water forms tighter drops
Solution: A lower surface tension lets water penetrate the fine spaces in cloth and wet greasy surfaces instead of beading up on them.
Q18 — Surface Tension · medium · numerical
The excess pressure inside a soap bubble of radius $2\text{ cm}$ is ($T = 0.03\text{ N/m}$):
A. $1.5\text{ Pa}$
B. $12\text{ Pa}$
C. $6\text{ Pa}$ ✓ Correct
D. $3\text{ Pa}$
Solution: A soap bubble has two surfaces: $\Delta P = \dfrac{4T}{R} = \dfrac{4 \times 0.03}{0.02} = \dfrac{0.12}{0.02} = 6\text{ Pa}$.
Q19 — Surface Tension · medium · numerical
The excess pressure inside a spherical water drop of radius $1\text{ mm}$ is ($T = 0.07\text{ N/m}$):
A. $140\text{ Pa}$ ✓ Correct
B. $0.14\text{ Pa}$
C. $280\text{ Pa}$
D. $70\text{ Pa}$
Solution: A drop has a single surface: $\Delta P = \dfrac{2T}{R} = \dfrac{2 \times 0.07}{10^{-3}} = 140\text{ Pa}$.
Q20 — Surface Tension · hard · numerical
Water rises in a capillary tube of radius $0.2\text{ mm}$. Taking $T = 0.07\text{ N/m}$, $\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$ and zero angle of contact, the height of rise is:
A. $7\text{ cm}$ ✓ Correct
B. $3.5\text{ cm}$
C. $14\text{ cm}$
D. $0.7\text{ cm}$
Solution: $h = \dfrac{2T\cos\theta}{r\rho g} = \dfrac{2 \times 0.07}{(2 \times 10^{-4})(1000)(10)} = \dfrac{0.14}{2} = 0.07\text{ m} = 7\text{ cm}$.
Q21 — Surface Tension · hard · numerical
The work done in blowing a soap bubble of radius $5\text{ cm}$ is ($T = 0.03\text{ N/m}$):
A. $9.42 \times 10^{-4}\text{ J}$
B. $0.03\text{ J}$
C. $1.885 \times 10^{-3}\text{ J}$ ✓ Correct
D. $3.77 \times 10^{-3}\text{ J}$
Solution: A bubble has two surfaces, so $\Delta A = 2(4\pi R^2) = 8\pi(0.05)^2 = 0.0628\text{ m}^2$. Hence $W = T\Delta A = 0.03 \times 0.0628 \approx 1.885 \times 10^{-3}\text{ J}$.
Q22 — Surface Tension · medium · numerical
The excess pressure inside a soap bubble of radius $1\text{ cm}$ is ($T = 0.025\text{ N/m}$):
A. $2.5\text{ Pa}$
B. $10\text{ Pa}$ ✓ Correct
C. $20\text{ Pa}$
D. $5\text{ Pa}$
Solution: A soap bubble has two surfaces: $\Delta P = \dfrac{4T}{R} = \dfrac{4 \times 0.025}{0.01} = 10\text{ Pa}$.
Q23 — Surface Tension · medium · numerical
The excess pressure inside a liquid drop of radius $2\text{ mm}$ is ($T = 0.06\text{ N/m}$):
A. $120\text{ Pa}$
B. $30\text{ Pa}$
C. $15\text{ Pa}$
D. $60\text{ Pa}$ ✓ Correct
Solution: A drop has a single surface: $\Delta P = \dfrac{2T}{R} = \dfrac{2 \times 0.06}{2 \times 10^{-3}} = 60\text{ Pa}$.
Q24 — Surface Tension · hard · numerical
The excess pressure inside an air bubble of radius $0.5\text{ mm}$ formed inside a liquid of surface tension $0.075\text{ N/m}$ is:
A. $300\text{ Pa}$ ✓ Correct
B. $75\text{ Pa}$
C. $600\text{ Pa}$
D. $150\text{ Pa}$
Solution: An air bubble inside a liquid has one surface: $\Delta P = \dfrac{2T}{R} = \dfrac{2 \times 0.075}{5 \times 10^{-4}} = 300\text{ Pa}$.
Q25 — Surface Tension · hard · numerical
Water rises in a capillary tube of radius $0.1\text{ mm}$. Taking $T = 0.06\text{ N/m}$, $\rho = 1000\text{ kg/m}^3$, $g = 10\text{ m/s}^2$ and zero angle of contact, the rise is:
A. $12\text{ cm}$ ✓ Correct
B. $1.2\text{ cm}$
C. $6\text{ cm}$
D. $24\text{ cm}$
Solution: $h = \dfrac{2T}{r\rho g} = \dfrac{2 \times 0.06}{(10^{-4})(1000)(10)} = \dfrac{0.12}{1} = 0.12\text{ m} = 12\text{ cm}$.
Q26 — Surface Tension · hard · numerical
One thousand identical spherical droplets coalesce into a single large drop. The surface energy of the system:
A. Decreases by $90\%$ ✓ Correct
B. Increases by $90\%$
C. Remains unchanged
D. Decreases by $50\%$
Solution: Volume conservation gives $R = 1000^{1/3}r = 10r$. Initial energy $\propto 1000(4\pi r^2)$ and final $\propto 4\pi(10r)^2 = 100(4\pi r^2)$, so the energy falls to one-tenth — a $90\%$ decrease.
Q27 — Surface Tension · medium · numerical
A needle of length $10\text{ cm}$ rests on the surface of water of surface tension $0.072\text{ N/m}$. The minimum force needed to pull it off the surface is:
A. $1.44 \times 10^{-2}\text{ N}$ ✓ Correct
B. $0.72\text{ N}$
C. $7.2 \times 10^{-3}\text{ N}$
D. $1.44 \times 10^{-3}\text{ N}$
Solution: The surface pulls along both sides of the needle, so $F = 2lT = 2(0.1)(0.072) = 1.44 \times 10^{-2}\text{ N}$.