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601. Capillary action is the phenomenon in which:
ⓐ. Liquids always rise in narrow tubes regardless of liquid type
ⓑ. Liquids rise or fall in a narrow tube due to surface tension and adhesion
ⓒ. Liquids rise only due to density difference
ⓓ. Liquids move upward only due to buoyant force
Correct Answer: Liquids rise or fall in a narrow tube due to surface tension and adhesion
Explanation: Capillary action occurs because of the balance between cohesive and adhesive forces along with surface tension, leading to liquid rise (concave meniscus) or depression (convex meniscus).
602. Which equation gives the height of rise in a capillary tube?
ⓐ. $h = \frac{2T}{\rho g r}$
ⓑ. $h = \frac{2T \cos \theta}{\rho g r}$
ⓒ. $h = \frac{T \cos \theta}{\rho g}$
ⓓ. $h = \frac{2\rho g r}{T}$
Correct Answer: $h = \frac{2T \cos \theta}{\rho g r}$
Explanation: Height of capillary rise $h$ depends on surface tension $T$, liquid density $\rho$, gravity $g$, radius $r$, and contact angle $\theta$.
603. Capillary rise of a liquid is inversely proportional to:
ⓐ. Surface tension
ⓑ. Density of the liquid
ⓒ. Radius of the capillary tube
ⓓ. Acceleration due to gravity
Correct Answer: Radius of the capillary tube
Explanation: Smaller the radius, higher the rise, since $h \propto 1/r$.
604. For water in a clean glass tube, the meniscus is concave because:
ⓐ. Adhesion > cohesion
ⓑ. Cohesion > adhesion
ⓒ. Buoyant forces dominate
ⓓ. Gravitational force dominates
Correct Answer: Adhesion > cohesion
Explanation: Adhesive forces between water and glass molecules exceed water’s cohesive forces, creating a concave meniscus and capillary rise.
605. For mercury in glass, the meniscus is convex and mercury is depressed in the capillary tube because:
ⓐ. Adhesion > cohesion
ⓑ. Cohesion > adhesion
ⓒ. Gravity dominates
ⓓ. Density is very low
Correct Answer: Cohesion > adhesion
Explanation: Mercury’s cohesive forces exceed adhesive forces with glass, giving a convex meniscus and capillary depression.
606. If the radius of a capillary tube is halved, the height of capillary rise:
ⓐ. Doubles
ⓑ. Halves
ⓒ. Becomes four times
ⓓ. Remains same
Correct Answer: Doubles
Explanation: Since $h \propto 1/r$, halving radius doubles the rise.
607. Which condition leads to maximum capillary rise?
ⓐ. $\theta = 0^\circ$ (perfect wetting)
ⓑ. $\theta = 90^\circ$
ⓒ. $\theta = 135^\circ$
ⓓ. $\theta = 180^\circ$
Correct Answer: $\theta = 0^\circ$ (perfect wetting)
Explanation: At $\theta = 0^\circ$, $\cos \theta = 1$, so rise is maximum.
608. Which of the following liquids will rise highest in the same capillary tube?
ⓐ. Water ($T = 0.072 \, N/m, \rho = 1000 \, kg/m^3$)
ⓑ. Kerosene ($T = 0.025 \, N/m, \rho = 800 \, kg/m^3$)
ⓒ. Glycerin ($T = 0.063 \, N/m, \rho = 1260 \, kg/m^3$)
ⓓ. Mercury ($T = 0.48 \, N/m, \rho = 13600 \, kg/m^3$)
Correct Answer: Water
Explanation: Using $h = \frac{2T \cos \theta}{\rho g r}$, water has high surface tension, moderate density, and zero contact angle, producing maximum rise.
609. Which of the following is **not** an application of capillary rise?
ⓐ. Movement of water in soil
ⓑ. Ink rising in pens
ⓒ. Flow of blood in arteries
ⓓ. Rise of oil in lamp wicks
Correct Answer: Flow of blood in arteries
Explanation: Blood flow in arteries is due to pumping action of the heart, not capillary rise.
610. Why do paper towels absorb water?
ⓐ. Paper is porous and capillary action draws water into fibers
ⓑ. Adhesive forces are absent in paper
ⓒ. Cohesive forces are zero in water
ⓓ. Density difference causes flow
Correct Answer: Paper is porous and capillary action draws water into fibers
Explanation: Paper has tiny pores acting as capillaries. Adhesion and surface tension pull water upward into the fibers, enabling absorption.
611. The height of liquid rise in a capillary is directly proportional to:
ⓐ. Tube radius
ⓑ. Density of the liquid
ⓒ. Surface tension of the liquid
ⓓ. Gravitational force
Correct Answer: Surface tension of the liquid
Explanation: From $h = \tfrac{2T \cos \theta}{\rho g r}$, higher surface tension increases capillary rise for a given liquid and tube.
612. The height of capillary rise is inversely proportional to:
ⓐ. Liquid surface tension
ⓑ. Tube radius
ⓒ. Adhesion between liquid and solid
ⓓ. Contact angle
Correct Answer: Tube radius
Explanation: Narrower tubes give greater rise. Since $h \propto 1/r$, halving radius doubles the capillary rise.
613. If the radius of a capillary tube is doubled, the capillary rise will:
ⓐ. Double
ⓑ. Remain unchanged
ⓒ. Become half
ⓓ. Become four times
Correct Answer: Become half
Explanation: $h \propto 1/r$. Doubling $r$ reduces rise to half.
614. Which of the following decreases the height of capillary rise?
ⓐ. Increasing surface tension
ⓑ. Decreasing tube radius
ⓒ. Increasing density of liquid
ⓓ. Decreasing contact angle
Correct Answer: Increasing density of liquid
Explanation: From the formula, $h \propto 1/\rho$. Higher density reduces the height of rise.
615. For maximum capillary rise, the contact angle should be:
ⓐ. $0^\circ$
ⓑ. $45^\circ$
ⓒ. $90^\circ$
ⓓ. $180^\circ$
Correct Answer: $0^\circ$
Explanation: At $\theta = 0^\circ$, $\cos \theta = 1$, giving maximum height of rise.
616. Which of the following correctly describes the effect of gravity on capillary rise?
ⓐ. Increase in $g$ decreases rise
ⓑ. Increase in $g$ increases rise
ⓒ. Capillary rise is independent of $g$
ⓓ. Capillary rise first increases then decreases
Correct Answer: Increase in $g$ decreases rise
Explanation: From $h = \tfrac{2T \cos \theta}{\rho g r}$, rise is inversely proportional to gravitational acceleration.
617. If surface tension of water decreases due to soap solution, then capillary rise:
ⓐ. Increases
ⓑ. Decreases
ⓒ. Becomes infinite
ⓓ. Remains unchanged
Correct Answer: Decreases
Explanation: Since $h \propto T$, reducing $T$ lowers the height of capillary rise.
618. Which factor is responsible for mercury showing a depression in a capillary tube?
ⓐ. High density of mercury
ⓑ. High surface tension only
ⓒ. Obtuse contact angle ($\theta > 90^\circ$)
ⓓ. Low viscosity
Correct Answer: Obtuse contact angle ($\theta > 90^\circ$)
Explanation: With $\theta = 135^\circ$, $\cos \theta$ becomes negative, producing depression instead of rise.
619. In soil, capillary rise is more effective when:
ⓐ. Soil particles are large
ⓑ. Soil pores are wide
ⓒ. Soil pores are narrow
ⓓ. Soil is dry only
Correct Answer: Soil pores are narrow
Explanation: Narrow pores act like fine capillary tubes, allowing water to rise higher, supplying moisture to plants.
620. If the radius of a capillary tube is reduced to one-fourth, the height of capillary rise will become:
ⓐ. 2 times
ⓑ. 4 times
ⓒ. 8 times
ⓓ. 16 times
Correct Answer: 4 times
Explanation: Since $h \propto 1/r$, reducing radius to one-fourth increases rise by a factor of 4.
621. The upward movement of water in soil pores is mainly due to:
ⓐ. Buoyancy
ⓑ. Capillary action
ⓒ. Diffusion
ⓓ. Osmosis
Correct Answer: Capillary action
Explanation: Water rises through tiny pores between soil particles by capillary action, making moisture available to plant roots.
622. In plant physiology, the rise of water in xylem vessels of plants is partly due to:
ⓐ. Osmotic pressure only
ⓑ. Capillary action and adhesion to vessel walls
ⓒ. Gravity
ⓓ. Buoyant force
Correct Answer: Capillary action and adhesion to vessel walls
Explanation: Water wets xylem walls ($\theta < 90^\circ$) and rises by capillarity, aiding upward transport of water.
623. Which soil type will retain maximum water by capillary action?
ⓐ. Sandy soil
ⓑ. Clay soil
ⓒ. Loamy soil
ⓓ. Gravel
Correct Answer: Clay soil
Explanation: Clay has very fine particles and narrow pores, leading to maximum capillary rise and water retention.
624. Why do sandy soils drain water quickly compared to clay soils?
ⓐ. Larger pores cause weaker capillary rise
ⓑ. Lower density of sand
ⓒ. Higher surface tension of water in sand
ⓓ. Higher viscosity of water in sand
Correct Answer: Larger pores cause weaker capillary rise
Explanation: Capillary rise is inversely proportional to pore radius. Sandy soil has large pores, hence weaker water retention.
625. In very tall trees, capillary rise alone cannot account for full transport of water. What additional mechanism helps?
ⓐ. Osmosis in leaves
ⓑ. Cohesion-tension theory (transpiration pull)
ⓒ. Buoyancy of sap
ⓓ. Root pressure only
Correct Answer: Cohesion-tension theory (transpiration pull)
Explanation: In tall trees, capillary action helps, but the main force is transpiration pull combined with cohesion of water molecules.
626. Which natural phenomenon ensures seeds in soil can absorb moisture even without rainfall?
ⓐ. Diffusion of water vapor
ⓑ. Capillary rise of groundwater through soil pores
ⓒ. Buoyancy of soil particles
ⓓ. Condensation
Correct Answer: Capillary rise of groundwater through soil pores
Explanation: Capillary action in fine soil pores pulls up underground water, keeping seeds moist for germination.
627. Why do overwatered plants often die despite soil saturation?
ⓐ. Capillary action stops completely
ⓑ. Oxygen supply in soil pores is blocked
ⓒ. Cohesion of water molecules is destroyed
ⓓ. Surface tension becomes zero
Correct Answer: Oxygen supply in soil pores is blocked
Explanation: Excess water fills soil pores, preventing capillary spaces from holding air, suffocating roots.
628. In dry climates, plants with deep roots survive because:
ⓐ. Root hairs generate viscosity
ⓑ. Capillary action allows water movement from deep moist soil layers
ⓒ. Roots create buoyant force
ⓓ. Contact angle becomes obtuse
Correct Answer: Capillary action allows water movement from deep moist soil layers
Explanation: Even in dry surface soils, water rises in capillary pores from deeper moist layers, enabling survival of deep-rooted plants.
629. Which of the following agricultural practices improves water retention in soil through capillary action?
ⓐ. Ploughing to break soil particles
ⓑ. Mixing clay or organic matter
ⓒ. Increasing sand content
ⓓ. Using non-porous fertilizers
Correct Answer: Mixing clay or organic matter
Explanation: Clay and organic matter reduce pore size, enhancing capillary rise and water retention in soil.
630. Capillary action in soils is most important for:
ⓐ. Long-distance water transport in rivers
ⓑ. Storage of groundwater
ⓒ. Supply of moisture to plant roots
ⓓ. Condensation of dew
Correct Answer: Supply of moisture to plant roots
Explanation: Capillary rise in small soil pores provides water continuously to plant roots, especially when rainfall is scarce.
631. The primary role of detergents in reducing surface tension is:
ⓐ. Increasing adhesion between liquid and solid
ⓑ. Breaking cohesive forces between water molecules
ⓒ. Increasing density of water
ⓓ. Increasing viscosity of water
Correct Answer: Breaking cohesive forces between water molecules
Explanation: Detergent molecules insert themselves between water molecules, weakening hydrogen bonding and lowering surface tension.
632. When detergent is added to water, the contact angle on glass:
ⓐ. Increases
ⓑ. Decreases
ⓒ. Becomes 90°
ⓓ. Becomes 180°
Correct Answer: Decreases
Explanation: Reduced surface tension enhances adhesion between water and glass, lowering contact angle and improving wetting.
633. Why do detergents make clothes easier to wash?
ⓐ. They increase viscosity of water
ⓑ. They reduce surface tension, allowing water to spread into fabric pores
ⓒ. They increase buoyant force
ⓓ. They decrease density of water
Correct Answer: They reduce surface tension, allowing water to spread into fabric pores
Explanation: Lowered surface tension helps water penetrate and emulsify dirt and grease particles in clothes.
634. Which part of a detergent molecule lowers water’s surface tension?
ⓐ. Hydrophilic head only
ⓑ. Hydrophobic tail only
ⓒ. Both head and tail at interface
ⓓ. None of the above
Correct Answer: Both head and tail at interface
Explanation: Hydrophobic tails align away from water and hydrophilic heads stay in water, disrupting cohesive forces and reducing surface tension.
635. When detergent concentration increases beyond a certain point, surface tension:
ⓐ. Keeps decreasing indefinitely
ⓑ. Increases again
ⓒ. Reaches a constant minimum value
ⓓ. Becomes zero
Correct Answer: Reaches a constant minimum value
Explanation: Beyond the critical micelle concentration (CMC), additional detergent forms micelles and no longer reduces surface tension.
636. Which phenomenon demonstrates detergent’s role in reducing surface tension?
ⓐ. Floating of a needle on pure water
ⓑ. Needle sinking after adding detergent
ⓒ. Mercury depression in glass
ⓓ. Raindrop formation in clouds
Correct Answer: Needle sinking after adding detergent
Explanation: Detergent reduces water’s surface tension so that the surface can no longer support a light needle.
637. In soap bubbles, detergents help by:
ⓐ. Increasing density of liquid film
ⓑ. Decreasing surface tension to stabilize thin films
ⓒ. Increasing viscosity of film only
ⓓ. Neutralizing buoyant force
Correct Answer: Decreasing surface tension to stabilize thin films
Explanation: Detergents lower surface tension, allowing the bubble film to stretch without breaking.
638. Why do detergents spread rapidly over water surface?
ⓐ. High viscosity of detergent molecules
ⓑ. Low surface tension of detergent solution compared to water
ⓒ. Strong hydrogen bonding with water
ⓓ. High density of detergent
Correct Answer: Low surface tension of detergent solution compared to water
Explanation: Detergent molecules reduce local surface tension, causing spreading across the surface.
639. Which of the following is an industrial application of detergents reducing surface tension?
ⓐ. Fuel injection in engines
ⓑ. Inkjet printing
ⓒ. Paper coating
ⓓ. All of the above
Correct Answer: All of the above
Explanation: In all these applications, reduced surface tension ensures better spreading, coating, and atomization.
640. Why does dishwashing become more effective with hot water and detergent?
ⓐ. Both reduce surface tension significantly
ⓑ. Both increase viscosity
ⓒ. Both increase buoyancy
ⓓ. Both increase cohesion
Correct Answer: Both reduce surface tension significantly
Explanation: Detergents reduce surface tension, while hot water further lowers it, allowing grease to be emulsified and washed away more easily.
641. The cleaning action of detergents is mainly due to:
ⓐ. High density of detergent solution
ⓑ. Formation of micelles that trap grease and oil
ⓒ. Increase in viscosity of water
ⓓ. Evaporation of dirt particles
Correct Answer: Formation of micelles that trap grease and oil
Explanation: Detergent molecules form micelles where hydrophobic tails surround grease/oil and hydrophilic heads face water, emulsifying dirt for removal.
642. The hydrophilic part of a detergent molecule is:
ⓐ. Long hydrocarbon chain
ⓑ. Polar ionic head group
ⓒ. Non-polar tail
ⓓ. Entire molecule
Correct Answer: Polar ionic head group
Explanation: The hydrophilic head is water-loving and interacts with polar water molecules, making detergent soluble in water.
643. The hydrophobic part of a detergent molecule is:
ⓐ. Polar ionic head
ⓑ. Hydrocarbon tail
ⓒ. Oxygen atoms
ⓓ. Carboxylate group
Correct Answer: Hydrocarbon tail
Explanation: The long hydrocarbon chain repels water but interacts with non-polar grease/oil molecules, helping in cleaning.
644. Micelles form in detergent solutions only when:
ⓐ. Temperature is maximum
ⓑ. Concentration exceeds the critical micelle concentration (CMC)
ⓒ. Pressure is high
ⓓ. Surface tension becomes zero
Correct Answer: Concentration exceeds the critical micelle concentration (CMC)
Explanation: Micelle formation begins when detergent concentration crosses the CMC, allowing effective cleaning action.
645. In detergent cleaning, grease is removed because:
ⓐ. Grease is converted into gas
ⓑ. Grease dissolves in water directly
ⓒ. Grease is trapped inside hydrophobic cores of micelles
ⓓ. Grease evaporates due to heat
Correct Answer: Grease is trapped inside hydrophobic cores of micelles
Explanation: Hydrophobic detergent tails surround grease, while hydrophilic heads face water, suspending grease in solution.
646. Which of the following is a correct description of detergent action?
ⓐ. Detergents increase adhesion between dirt and cloth
ⓑ. Detergents lower water’s surface tension and emulsify oils
ⓒ. Detergents increase water viscosity for cleaning
ⓓ. Detergents dissolve dirt without chemical interaction
Correct Answer: Detergents lower water’s surface tension and emulsify oils
Explanation: Reduced surface tension improves wetting, and micelle formation emulsifies oils for easy removal.
647. Why are detergents more effective in hard water compared to soaps?
ⓐ. They form insoluble scum with calcium and magnesium ions
ⓑ. They do not form insoluble scum with calcium and magnesium ions
ⓒ. They increase water density
ⓓ. They increase water viscosity
Correct Answer: They do not form insoluble scum with calcium and magnesium ions
Explanation: Detergents contain sulfonate groups that remain soluble in hard water, unlike soaps that form insoluble salts (scum).
648. In micelle formation, the orientation of detergent molecules is such that:
ⓐ. Hydrophilic heads face water, hydrophobic tails face inward
ⓑ. Hydrophobic tails face water, hydrophilic heads face inward
ⓒ. Both heads and tails face water
ⓓ. Both heads and tails face inward
Correct Answer: Hydrophilic heads face water, hydrophobic tails face inward
Explanation: This arrangement stabilizes grease/oil in micelle cores, making them dispersible in water.
649. The minimum concentration of detergent required to form micelles is called:
ⓐ. Solubility limit
ⓑ. Critical micelle concentration (CMC)
ⓒ. Adsorption coefficient
ⓓ. Surface tension constant
Correct Answer: Critical micelle concentration (CMC)
Explanation: CMC is the threshold concentration above which detergent molecules aggregate into micelles.
650. Why do detergents work better in hot water?
ⓐ. Hot water increases density
ⓑ. Hot water reduces surface tension and enhances micelle formation
ⓒ. Hot water decreases adhesion of detergents
ⓓ. Hot water breaks micelles easily
Correct Answer: Hot water reduces surface tension and enhances micelle formation
Explanation: Heat lowers surface tension and accelerates detergent action, making micelles form more efficiently and improving cleaning.
651. Water flows through a horizontal pipe of cross-sectional areas $A_1 = 10^{-2} \, m^2$ and $A_2 = 5 \times 10^{-3} \, m^2$. The pressure difference between these sections is $4 \times 10^3 \, Pa$. Find the flow rate of water. ($\rho = 1000 \, kg/m^3$)
ⓐ. $0.02 \, m^3/s$
ⓑ. $0.025 \, m^3/s$
ⓒ. $0.030 \, m^3/s$
ⓓ. $0.035 \, m^3/s$
Correct Answer: $0.025 \, m^3/s$
Explanation: Using continuity equation: $A_1 v_1 = A_2 v_2$. Bernoulli: $P_1 – P_2 = \tfrac{1}{2}\rho(v_2^2 – v_1^2)$. Solve simultaneously: flow rate $Q = A_1 v_1 = 0.025 \, m^3/s$.
652. A spherical raindrop of radius $2 \, mm$ falls at terminal velocity. Viscosity of air = $1.8 \times 10^{-5} \, Ns/m^2$, density of water $= 1000 \, kg/m^3$, air density negligible. Find terminal velocity. (g = 9.8)
ⓐ. $0.40 \, m/s$
ⓑ. $0.50 \, m/s$
ⓒ. $0.60 \, m/s$
ⓓ. $0.70 \, m/s$
Correct Answer: $0.60 \, m/s$
Explanation: Using Stokes’ law: $v_t = \frac{2 r^2 g (\rho – \sigma)}{9\eta}$. Substituting values gives $\approx 0.6 \, m/s$.
653. A U-tube manometer contains mercury ($\rho = 13.6 \times 10^3 \, kg/m^3$) and is connected to a pipeline carrying water. The difference in mercury levels is $20 \, cm$. Find gauge pressure in the pipe. (g = 9.8)
ⓐ. $2.67 \times 10^4 \, Pa$
ⓑ. $2.76 \times 10^4 \, Pa$
ⓒ. $2.87 \times 10^4 \, Pa$
ⓓ. $2.96 \times 10^4 \, Pa$
Correct Answer: $2.76 \times 10^4 \, Pa$
Explanation: Pressure difference = $\Delta P = \rho g h = 13.6 \times 10^3 \times 9.8 \times 0.20 = 2.76 \times 10^4 \, Pa$.
654. A liquid of density $800 \, kg/m^3$ rises to $6 \, cm$ in a capillary tube of radius $0.5 \, mm$. Find the surface tension. ($\theta = 0^\circ, g = 9.8$)
ⓐ. $0.020 N/m$
ⓑ. $0.024 N/m$
ⓒ. $0.026 N/m$
ⓓ. $0.030 N/m$
Correct Answer: $0.024 N/m$
Explanation: Using $T = \frac{h \rho g r}{2 \cos \theta}$. Substituting: $T = (0.06 \times 800 \times 9.8 \times 0.0005)/2 = 0.024 N/m$.
655. A pipe of diameter $20 \, cm$ is connected to a pipe of diameter $5 \, cm$. The velocity of water in the large pipe is $2 \, m/s$. Find the velocity in the narrow pipe.
ⓐ. $24 m/s$
ⓑ. $30 m/s$
ⓒ. $32 m/s$
ⓓ. $40 m/s$
Correct Answer: $40 m/s$
Explanation: $A_1 v_1 = A_2 v_2$. $ \pi (0.1^2) \times 2 = \pi (0.025^2) \times v_2$. $v_2 = 40 \, m/s$.
656. A soap bubble of radius $2 \, cm$ has surface tension $0.04 \, N/m$. Find excess pressure inside the bubble.
ⓐ. 4 Pa
ⓑ. 6 Pa
ⓒ. 8 Pa
ⓓ. 10 Pa
Correct Answer: 8 Pa
Explanation: For bubble, $\Delta P = 4T/r = 4 \times 0.04 / 0.02 = 8 Pa$.
657. A tank is filled with water up to 5 m. Calculate the pressure at the bottom. ($\rho = 1000, g = 9.8$)
ⓐ. $4.9 \times 10^4 \, Pa$
ⓑ. $5.0 \times 10^4 \, Pa$
ⓒ. $5.1 \times 10^4 \, Pa$
ⓓ. $5.2 \times 10^4 \, Pa$
Correct Answer: $4.9 \times 10^4 \, Pa$
Explanation: $P = \rho g h = 1000 \times 9.8 \times 5 = 49000 Pa$.
658. A velocity of $10 \, m/s$ is observed at the throat of a Venturi meter with inlet velocity $5 \, m/s$. Find the pressure difference. ($\rho = 1000 \, kg/m^3$)
ⓐ. 25 kPa
ⓑ. 30 kPa
ⓒ. 37.5 kPa
ⓓ. 50 kPa
Correct Answer: 37.5 kPa
Explanation: $\Delta P = \tfrac{1}{2}\rho(v_2^2 – v_1^2) = 0.5 \times 1000 (100 – 25) = 37,500 Pa$.
659. A spherical raindrop of radius $1 \, mm$ falls at terminal velocity. Density of water $= 1000$, density of air negligible, viscosity of air $= 1.8 \times 10^{-5} Ns/m^2$. Find terminal velocity.
ⓐ. 0.30 m/s
ⓑ. 0.40 m/s
ⓒ. 0.50 m/s
ⓓ. 0.60 m/s
Correct Answer: 0.60 m/s
Explanation: $v_t = \frac{2r^2 g (\rho – \sigma)}{9\eta}$. Substituting: $v_t \approx 0.6 m/s$.
660. A spherical drop of radius $R$ is divided into $n$ smaller drops of equal size. If the initial surface energy was $E$, what is the final surface energy?
ⓐ. $E/n^{1/3}$
ⓑ. $n^{1/3}E$
ⓒ. $n^{2/3}E$
ⓓ. $nE$
Correct Answer: $n^{2/3}E$
Explanation: Surface area ∝ $r^2$, and radius of new drop = $R/n^{1/3}$. Total area increases by factor $n^{2/3}$, so energy becomes $n^{2/3}E$.
661. A horizontal pipe carries water with a speed of $2 \, m/s$ at pressure $3 \times 10^5 \, Pa$. At a constriction, speed rises to $5 \, m/s$. Find pressure at the constriction. ($\rho = 1000 \, kg/m^3$)
ⓐ. $2.8 \times 10^5 \, Pa$
ⓑ. $2.7 \times 10^5 \, Pa$
ⓒ. $2.6 \times 10^5 \, Pa$
ⓓ. $2.5 \times 10^5 \, Pa$
Correct Answer: $2.7 \times 10^5 \, Pa$
Explanation: From Bernoulli: $P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 – v_2^2) = 3 \times 10^5 + 0.5 \times 1000 (4 – 25) = 2.725 \times 10^5 \, Pa$.
662. A soap bubble of radius $3 \, mm$ is formed. Surface tension is $0.05 \, N/m$. Calculate excess pressure inside the bubble.
ⓐ. 50 Pa
ⓑ. 60 Pa
ⓒ. 66.7 Pa
ⓓ. 70 Pa
Correct Answer: 66.7 Pa
Explanation: $\Delta P = \tfrac{4T}{r} = \tfrac{4 \times 0.05}{0.003} \approx 66.7 \, Pa$.
663. A capillary of radius $0.25 \, mm$ is dipped in water at 20°C. Find the rise of water. (Surface tension = 0.072 N/m, density = 1000, g = 9.8, $\theta = 0^\circ$)
ⓐ. 2.5 cm
ⓑ. 3.0 cm
ⓒ. 5.8 cm
ⓓ. 6.0 cm
Correct Answer: 5.8 cm
Explanation: $h = \tfrac{2T \cos \theta}{\rho g r} = \tfrac{2 \times 0.072}{1000 \times 9.8 \times 0.00025} = 0.058 \, m = 5.8 \, cm$.
664. A tank of height $4 \, m$ is filled with water. Find the velocity of efflux from a hole $1 \, m$ above the bottom.
ⓐ. 5 m/s
ⓑ. 5.67 m/s
ⓒ. 4 m/s
ⓓ. 7.67 m/s
Correct Answer: 7.67 m/s
Explanation: From Torricelli’s theorem: $v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 3} \approx 7.67 \, m/s$.
665. A steel ball of radius $0.2 \, mm$ falls through glycerine (viscosity $= 0.8 \, Ns/m^2$) with terminal velocity $2 \times 10^{-3} \, m/s$. Find density of steel if glycerine density is $1200 \, kg/m^3$. (g = 9.8)
ⓐ. 7200 $kg/m^3$
ⓑ. 7800 $kg/m^3$
ⓒ. 8000 $kg/m^3$
ⓓ. 8500 $kg/m^3$
Correct Answer: 7800 $kg/m^3$
Explanation: Terminal velocity $v_t = \frac{2r^2 g (\rho_s – \rho)}{9\eta}$. Solving: $\rho_s = \rho + \frac{9 \eta v_t}{2 r^2 g} = 1200 + \frac{9 \times 0.8 \times 2 \times 10^{-3}}{2 \times (0.0002^2) \times 9.8} \approx 7800 \, kg/m^3$.
666. In a Venturi meter, diameters of inlet and throat are 0.3 m and 0.15 m. If pressure difference is $1.8 \times 10^4 \, Pa$, calculate flow rate of water. ($\rho = 1000 \, kg/m^3$)
ⓐ. 0.012 $m^3/s$
ⓑ. 0.024 $m^3/s$
ⓒ. 0.030 $m^3/s$
ⓓ. 0.036 $m^3/s$
Correct Answer: 0.024 $m^3/s$
Explanation: Use continuity: $A_1 v_1 = A_2 v_2$. Bernoulli: $\Delta P = 0.5 \rho (v_2^2 – v_1^2)$. Solving gives $Q = 0.024 \, m^3/s$.
667. A spherical water drop of radius $2 \, mm$ is broken into 8 equal droplets. Calculate the increase in surface energy if surface tension is $0.072 \, N/m$.
ⓐ. $2.0 \times 10^{-5} J$
ⓑ. $3.0 \times 10^{-5} J$
ⓒ. $4.0 \times 10^{-5} J$
ⓓ. $5.0 \times 10^{-5} J$
Correct Answer: $4.0 \times 10^{-5} J$
Explanation: Initial area = $4\pi R^2$. Final area = $8 \times 4\pi r^2$, where $r = R/2$. Increase in area × surface tension = $4.0 \times 10^{-5} J$.
668. The excess pressure inside a water drop of radius $1 \, mm$ is $150 \, Pa$. Find surface tension.
ⓐ. 0.0375 N/m
ⓑ. 0.050 N/m
ⓒ. 0.075 N/m
ⓓ. 0.100 N/m
Correct Answer: 0.075 N/m
Explanation: For droplet: $\Delta P = \frac{2T}{r}$. Thus $T = \frac{\Delta P r}{2} = \frac{150 \times 0.001}{2} = 0.075 \, N/m$.
669. A hydraulic lift has a piston of diameter 1.5 m used to lift a car of mass 1200 kg. If the smaller piston has area $0.01 \, m^2$, calculate the force required on it.
ⓐ. 75 N
ⓑ. 100 N
ⓒ. 120 N
ⓓ. 150 N
Correct Answer: 100 N
Explanation: Force ratio = area ratio. $F_1/F_2 = A_1/A_2$. Weight = $1200 \times 9.8 \approx 11760 N$. Required force = $11760 \times (0.01 / 1.77) \approx 100 N$.
670. A horizontal capillary tube of radius $0.25 \, mm$ is dipped in a liquid with surface tension $0.04 N/m$, density $800 kg/m^3$, and contact angle $0^\circ$. Calculate rise of the liquid.
ⓐ. 6.0 cm
ⓑ. 7.5 cm
ⓒ. 8.0 cm
ⓓ. 9.5 cm
Correct Answer: 7.5 cm
Explanation: $h = \tfrac{2T \cos \theta}{\rho g r} = \tfrac{2 \times 0.04}{800 \times 9.8 \times 2.5 \times 10^{-4}} \approx 0.075 \, m = 7.5 \, cm$.
671. A horizontal tube of cross-sectional area $4 \times 10^{-4} \, m^2$ carries water at a velocity of $2 \, m/s$. The tube narrows to a cross-sectional area of $10^{-4} \, m^2$. Find the pressure drop between the wide and narrow sections. ($\rho = 1000 \, kg/m^3$)
ⓐ. $1.5 \times 10^3 \, Pa$
ⓑ. $2.0 \times 10^3 \, Pa$
ⓒ. $3.0 \times 10^3 \, Pa$
ⓓ. $3.0 \times 10^4 \, Pa$
Correct Answer: $3 \times 10^4 \, Pa$
Explanation: Continuity: $A_1 v_1 = A_2 v_2 \implies v_2 = (4 \times 10^{-4} \times 2)/(10^{-4}) = 8 \, m/s$.
Bernoulli: $\Delta P = \tfrac{1}{2}\rho (v_2^2 – v_1^2) = 0.5 \times 1000 (64 – 4) = 30,000 \, Pa$.
672. A spherical raindrop of radius $2 \, mm$ falls through air with viscosity $1.8 \times 10^{-5} \, Ns/m^2$. Density of water = $1000 \, kg/m^3$, air density negligible. Find terminal velocity. (g = 9.8)
ⓐ. 0.4 m/s
ⓑ. 0.6 m/s
ⓒ. 0.8 m/s
ⓓ. 1.0 m/s
Correct Answer: 0.6 m/s
Explanation: Stokes’ law: $v_t = \frac{2 r^2 g (\rho – \sigma)}{9\eta}$. Substituting values: $v_t \approx 0.6 \, m/s$.
673. A U-tube manometer contains mercury ($\rho = 13.6 \times 10^3 \, kg/m^3$). The difference in mercury levels is $10 \, cm$. Find the pressure difference across the two arms. (g = 9.8)
ⓐ. $1.23 \times 10^4 \, Pa$
ⓑ. $1.33 \times 10^4 \, Pa$
ⓒ. $1.43 \times 10^4 \, Pa$
ⓓ. $1.53 \times 10^4 \, Pa$
Correct Answer: $1.33 \times 10^4 \, Pa$
Explanation: $\Delta P = \rho g h = 13.6 \times 10^3 \times 9.8 \times 0.10 = 1.33 \times 10^4 \, Pa$.
674. A water tank has a hole 2 m below the free surface. Find velocity of efflux. (g = 9.8)
ⓐ. 5.5 m/s
ⓑ. 6.2 m/s
ⓒ. 6.3 m/s
ⓓ. 7.0 m/s
Correct Answer: 6.3 m/s
Explanation: Torricelli’s theorem: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2} = 6.26 \, m/s$.
675. A spherical soap bubble of radius $2 \, cm$ requires work of $2.01 \times 10^{-3} \, J$ to be formed. Find surface tension.
ⓐ. 0.020 N/m
ⓑ. 0.030 N/m
ⓒ. 0.040 N/m
ⓓ. 0.050 N/m
Correct Answer: 0.020 N/m
Explanation: Work = surface energy = $8\pi r^2 T$.
$T = W/(8\pi r^2) = 2.01 \times 10^{-3}/(8\pi (0.02)^2) \approx 0.020 N/m$.
676. The radius of a capillary tube is $0.25 \, mm$. Water rises 6 cm in it. Calculate surface tension. ($\rho = 1000, g = 9.8, \theta = 0^\circ$)
ⓐ. 0.070 N/m
ⓑ. 0.072 N/m
ⓒ. 0.075 N/m
ⓓ. 0.080 N/m
Correct Answer: 0.072 N/m
Explanation: $T = \frac{h \rho g r}{2 \cos \theta} = (0.06 \times 1000 \times 9.8 \times 0.00025)/2 \approx 0.072 \, N/m$.
677. The excess pressure inside a soap bubble of radius $1 \, cm$ is 30 Pa. Find surface tension.
ⓐ. 0.06 N/m
ⓑ. 0.075 N/m
ⓒ. 0.080 N/m
ⓓ. 0.090 N/m
Correct Answer: 0.075 N/m
Explanation: $\Delta P = \frac{4T}{r} \implies T = \frac{\Delta P \cdot r}{4} = \frac{30 \times 0.01}{4} = 0.075 N/m$.
678. A liquid jet emerges from a hole at velocity $10 \, m/s$. Find maximum height the jet can reach. (g = 9.8)
ⓐ. 4 m
ⓑ. 5 m
ⓒ. 6 m
ⓓ. 7 m
Correct Answer: 5 m
Explanation: $h = v^2/(2g) = 100/19.6 \approx 5.1 \, m$.
679. A hydraulic lift has a large piston of area $1.5 \, m^2$ and a small piston of area $0.01 \, m^2$. If a car of mass 1500 kg is to be lifted, calculate minimum force on small piston.
ⓐ. 80 N
ⓑ. 90 N
ⓒ. 100 N
ⓓ. 120 N
Correct Answer: 100 N
Explanation: Weight = $1500 \times 9.8 = 14,700 N$. Force ratio = area ratio. Required force = $14,700 \times (0.01/1.5) \approx 98 N$.
680. A horizontal pipe carries oil ($\rho = 900 \, kg/m^3$). At one end velocity is $2 m/s$, pressure $2 \times 10^5 Pa$. At another end velocity is $4 m/s$. Find pressure.
ⓐ. $1.92 \times 10^5 Pa$
ⓑ. $1.80 \times 10^5 Pa$
ⓒ. $1.72 \times 10^5 Pa$
ⓓ. $1.60 \times 10^5 Pa$
Correct Answer: $1.80 \times 10^5 Pa$
Explanation: Bernoulli: $P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 – v_2^2)$.
$P_2 = 2 \times 10^5 + 0.5 \times 900 (4 – 16) = 1.80 \times 10^5 Pa$.
681. A tank is filled with water up to 4 m. A small hole is made at 1 m above the base. Find the velocity of efflux. ($g = 9.8$)
ⓐ. 5.4 m/s
ⓑ. 6.3 m/s
ⓒ. 7.7 m/s
ⓓ. 8.5 m/s
Correct Answer: 7.7 m/s
Explanation: Depth of hole = 3 m. From Torricelli: $v = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 3} \approx 7.67 \, m/s$.
682. A capillary tube of radius $0.25 \, mm$ is dipped in water. Water rises to $6 \, cm$. Find surface tension. ($\rho = 1000, g = 9.8, \theta = 0^\circ$)
ⓐ. 0.070 N/m
ⓑ. 0.072 N/m
ⓒ. 0.075 N/m
ⓓ. 0.080 N/m
Correct Answer: 0.072 N/m
Explanation: $T = \frac{h \rho g r}{2} = \frac{0.06 \times 1000 \times 9.8 \times 0.00025}{2} \approx 0.072 N/m$.
683. The radius of a water droplet is $2 \, mm$. Find excess pressure inside. ($T = 0.072 \, N/m$)
ⓐ. 36 Pa
ⓑ. 54 Pa
ⓒ. 72 Pa
ⓓ. 90 Pa
Correct Answer: 72 Pa
Explanation: For droplet: $\Delta P = \tfrac{2T}{r} = \tfrac{2 \times 0.072}{0.002} = 72 \, Pa$.
684. A Venturi meter has a throat diameter of 0.1 m and main diameter of 0.2 m. If pressure drop is $5 \times 10^3 Pa$, calculate discharge. ($\rho = 1000$)
ⓐ. 0.005 m³/s
ⓑ. 0.010 m³/s
ⓒ. 0.015 m³/s
ⓓ. 0.020 m³/s
Correct Answer: 0.010 m³/s
Explanation: Using continuity and Bernoulli, solving gives flow rate ≈ $0.010 m³/s$.
685. A steel ball of radius $1 \, mm$ falls in glycerin (η = 0.8 Ns/m², density of glycerin = 1200, density of steel = 7800). Find terminal velocity.
ⓐ. 0.018 m/s
ⓑ. 0.022 m/s
ⓒ. 0.026 m/s
ⓓ. 0.030 m/s
Correct Answer: 0.026 m/s
Explanation: $v_t = \frac{2 r^2 g (\rho_s – \rho)}{9\eta}$. Substituting values gives $0.026 \, m/s$.
686. A spherical soap bubble of radius $2 \, cm$ requires 0.01 J work to blow. Find surface tension.
ⓐ. 0.02 N/m
ⓑ. 0.04 N/m
ⓒ. 0.06 N/m
ⓓ. 0.08 N/m
Correct Answer: 0.04 N/m
Explanation: Work = $8\pi r^2 T$. $T = W/(8\pi r^2) = 0.01/(8 \pi (0.02)^2) \approx 0.04 N/m$.
687. A 4 cm radius pipe carries water at 2 m/s. It branches into two pipes each of radius 2 cm. Find velocity in each branch.
ⓐ. 4 m/s
ⓑ. 6 m/s
ⓒ. 8 m/s
ⓓ. 10 m/s
Correct Answer: 8 m/s
Explanation: Continuity: $A_1 v_1 = 2 A_2 v_2$. $ \pi (0.04)^2 \times 2 = 2 \times \pi (0.02)^2 v_2 \implies v_2 = 8 m/s$.
688. A capillary tube of radius $0.2 \, mm$ dipped in kerosene (T = 0.025 N/m, ρ = 800). Find height rise. ($g=9.8, \theta=0$)
ⓐ. 2 cm
ⓑ. 3 cm
ⓒ. 4 cm
ⓓ. 5 cm
Correct Answer: 5 cm
Explanation: $h = \tfrac{2T}{\rho g r} = (2 \times 0.025)/(800 \times 9.8 \times 2 \times 10^{-4}) ≈ 0.051 m = 5.1 cm$.
689. A spherical raindrop of radius 2 mm falls with terminal velocity 0.5 m/s. Find viscosity of air. (ρwater = 1000, ρair negligible)
ⓐ. $1.8 \times 10^{-5}$ Ns/m²
ⓑ. $2.0 \times 10^{-5}$ Ns/m²
ⓒ. $2.5 \times 10^{-5}$ Ns/m²
ⓓ. $3.0 \times 10^{-5}$ Ns/m²
Correct Answer: $1.8 \times 10^{-5}$ Ns/m²
Explanation: From Stokes’ law: $\eta = \tfrac{2 r^2 g \rho}{9 v}$. Substituting values ≈ $1.8 \times 10^{-5}$.
690. A pipe of diameter 0.2 m carries water at 1.5 m/s. It contracts to 0.1 m diameter. Find velocity in narrow section.
ⓐ. 3 m/s
ⓑ. 4 m/s
ⓒ. 5 m/s
ⓓ. 6 m/s
Correct Answer: 6 m/s
Explanation: $A_1 v_1 = A_2 v_2$. $v_2 = (0.2^2 / 0.1^2) \times 1.5 = 6 m/s$.
691. The height of water column in a barometer tube is 10 m. If replaced with mercury, what will be the height? ($\rho_w = 1000, \rho_{Hg} = 13.6 \times 10^3$)
ⓐ. 0.65 m
ⓑ. 0.74 m
ⓒ. 0.76 m
ⓓ. 0.80 m
Correct Answer: 0.76 m
Explanation: $h_{Hg} = h_w (\rho_w / \rho_{Hg}) = 10 \times (1000 / 13600) = 0.735 ≈ 0.76 m$.
692. A U-tube with one end open and the other sealed contains air trapped above mercury. When the open end is exposed to 76 cm pressure, mercury levels differ by 12 cm. Find pressure of trapped air.
ⓐ. 64 cm Hg
ⓑ. 70 cm Hg
ⓒ. 76 cm Hg
ⓓ. 80 cm Hg
Correct Answer: 64 cm Hg
Explanation: Trapped air pressure = 76 − 12 = 64 cm Hg.
693. A soap bubble of radius $2 \, mm$ has excess pressure of 100 Pa. Find surface tension.
ⓐ. 0.025 N/m
ⓑ. 0.030 N/m
ⓒ. 0.050 N/m
ⓓ. 0.060 N/m
Correct Answer: 0.050 N/m
Explanation: $\Delta P = 4T/r \implies T = \Delta P r/4 = (100 \times 0.002)/4 = 0.05 N/m$.
694. A cube of side 20 cm is immersed in water. Find difference in pressure between top and bottom faces. (ρ=1000, g=9.8)
ⓐ. 1000 Pa
ⓑ. 1200 Pa
ⓒ. 1500 Pa
ⓓ. 2000 Pa
Correct Answer: 2000 Pa
Explanation: ΔP = ρ g h = 1000 × 9.8 × 0.2 = 1960 ≈ 2000 Pa.
695. A raindrop of radius 1 mm is formed by condensation of small droplets each of radius 0.1 mm. Find number of droplets.
ⓐ. 1000
ⓑ. 500
ⓒ. 100
ⓓ. 10
Correct Answer: 1000
Explanation: Volume conservation: $N r^3 = R^3$. $N = (1/0.1)^3 = 1000$.
696. In a fluid, pressure at depth h is given by $P = \rho g h$. If depth = 500 m in sea water (ρ=1030), find pressure (in atm). (1 atm = 1.013×10^5 Pa).
ⓐ. 45 atm
ⓑ. 50 atm
ⓒ. 55 atm
ⓓ. 60 atm
Correct Answer: 50 atm
Explanation: P = 1030×9.8×500 ≈ 5.04×10^6 Pa. In atm = 5.04×10^6 / 1.013×10^5 ≈ 49.8 ≈ 50 atm.
697. A drop of mercury (T=0.465 N/m) of radius 1 mm splits into 1000 equal droplets. Calculate increase in surface energy.
ⓐ. 0.35 J
ⓑ. 0.29 J
ⓒ. 0.25 J
ⓓ. 0.20 J
Correct Answer: 0.29 J
Explanation: ΔE = T(ΔA). Final radius = R/ N^(1/3) = 0.001 / 10 = 10^-4 m. ΔA = 1000×4πr^2 − 4πR^2 ≈ 0.196 − 0.0126 ≈ 0.183 m². ΔE ≈ 0.465×0.183 ≈ 0.29 J.
698. A water jet of cross-section 2 cm² comes out at velocity 20 m/s. Find thrust on the wall if it strikes normally and comes to rest. (ρ=1000)
ⓐ. 400 N
ⓑ. 450 N
ⓒ. 500 N
ⓓ. 600 N
Correct Answer: 400 N
Explanation: Thrust = rate of change of momentum = ρ A v². =1000×2×10^-4×20²=400 N.
699. A spherical ball of radius 0.5 cm and density 8000 falls in water (ρ=1000). Terminal velocity = 5 cm/s. Find viscosity of water.
ⓐ. 0.95 Ns/m²
ⓑ. 0.90 Ns/m²
ⓒ. 0.85 Ns/m²
ⓓ. 0.80 Ns/m²
Correct Answer: 0.80 Ns/m²
Explanation: $v_t = \frac{2 r^2 g (\rho_s – \rho)}{9\eta}$. Solve: η ≈ 0.80 Ns/m².
700. A steel tank is filled with oil (ρ=850). Pressure gauge at depth 2 m reads how much (in kPa)? (g=9.8)
ⓐ. 15.7 kPa
ⓑ. 16.7 kPa
ⓒ. 17.0 kPa
ⓓ. 18.0 kPa
Correct Answer: 16.7 kPa
Explanation: P = ρ g h = 850×9.8×2 ≈ 16,660 Pa = 16.7 kPa.
This is the final section of Class 11 Physics MCQs – Chapter 10: Mechanical Properties of Fluids (Part 7). The chapter has been fully covered with a total of 700 MCQs with correct answers and explanations, divided into 7 structured parts for easy revision. Based on the NCERT/CBSE syllabus, these questions are extremely useful for board exams and competitive exams like JEE, NEET, and state-level entrance tests. In this last part, you will find the final 100 solved MCQs, making it perfect for last-minute revision and completing your preparation for this chapter. 🎉 Congratulations on reaching the end of this chapter — now continue exploring MCQs of other chapters through the navigation bar above!
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