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501. Why are microbubbles used as contrast agents in ultrasound imaging?
ⓐ. They reduce surface tension of blood
ⓑ. They scatter and reflect ultrasound waves strongly due to large acoustic impedance difference with surrounding tissue
ⓒ. They increase viscosity of blood
ⓓ. They absorb ultrasound waves completely
Correct Answer: They scatter and reflect ultrasound waves strongly due to large acoustic impedance difference with surrounding tissue
Explanation: Gas-filled microbubbles act as strong ultrasound reflectors, enhancing imaging of blood flow and vascular structures in medical diagnostics.
502. The resonance frequency $f$ of a gas bubble of radius $r$ in liquid is approximately given by:
ⓐ. $f = \frac{1}{2\pi} \sqrt{\frac{3\gamma P_0}{\rho r^2}}$
ⓑ. $f = \frac{2\pi r}{\gamma P_0}$
ⓒ. $f = \frac{\rho r^2}{3\gamma P_0}$
ⓓ. $f = \frac{P_0}{2 \pi r}$
Correct Answer: $f = \frac{1}{2\pi} \sqrt{\frac{3\gamma P_0}{\rho r^2}}$
Explanation: Minnaert’s formula gives the natural resonance frequency of a bubble in a liquid. It depends on radius $r$, liquid density $\rho$, adiabatic index $\gamma$, and ambient pressure $P_0$.
503. A microbubble of radius $2 \, \mu m$ resonates in blood ($\rho = 1000 \, kg/m^3$, $P_0 = 1.0 \times 10^5 \, Pa$, $\gamma = 1.4$). Find its resonance frequency.
ⓐ. 1 MHz
ⓑ. 2 MHz
ⓒ. 3 MHz
ⓓ. 5 MHz
Correct Answer: 3 MHz
Explanation: $f = \frac{1}{2\pi}\sqrt{\frac{3\gamma P_0}{\rho r^2}} = \frac{1}{6.28}\sqrt{\frac{3 \times 1.4 \times 10^5}{1000 \times (2 \times 10^{-6})^2}} \approx 3 \, MHz$.
504. Why are bubbles studied in engineering fluid dynamics?
ⓐ. They are irrelevant to engineering
ⓑ. Because bubble formation and collapse cause cavitation, which can damage turbines and pumps
ⓒ. Because bubbles eliminate viscosity in fluids
ⓓ. Because bubbles reduce turbulence
Correct Answer: Because bubble formation and collapse cause cavitation, which can damage turbines and pumps
Explanation: Cavitation bubbles collapse violently, producing shock waves that erode turbine blades, propellers, and pumps. Engineers study bubble dynamics to minimize this damage.
505. The excess pressure inside a gas bubble of radius $r = 0.001 \, m$ in water ($T = 0.072 \, N/m$) is:
ⓐ. 36 Pa
ⓑ. 72 Pa
ⓒ. 144 Pa
ⓓ. 288 Pa
Correct Answer: 144 Pa
Explanation: For a bubble with two interfaces, $\Delta P = \frac{4T}{r} = \frac{4 \times 0.072}{0.001} = 288 \, Pa$. Correct answer: D.
506. In medicine, why must ultrasound contrast bubbles be smaller than red blood cells?
ⓐ. To ensure they remain in circulation without blocking capillaries
ⓑ. To increase viscosity of plasma
ⓒ. To dissolve instantly in blood
ⓓ. To decrease surface tension of blood
Correct Answer: To ensure they remain in circulation without blocking capillaries
Explanation: Typical capillary diameters are 5–8 µm. Microbubbles are designed smaller than this to safely pass through blood vessels while enhancing ultrasound imaging.
507. What engineering principle explains the collapse of cavitation bubbles near a solid surface?
ⓐ. Archimedes’ principle
ⓑ. Bernoulli’s principle with local low-pressure zones
ⓒ. Pascal’s law
ⓓ. Newton’s third law
Correct Answer: Bernoulli’s principle with local low-pressure zones
Explanation: Fast-moving liquid lowers pressure (Bernoulli effect), forming vapor bubbles. When these bubbles move to high-pressure regions, they collapse violently, causing cavitation.
508. A cavitation bubble of radius $r = 0.5 \, mm$ in water has surface tension $T = 0.072 \, N/m$. Find excess pressure inside.
ⓐ. 288 Pa
ⓑ. 576 Pa
ⓒ. 1152 Pa
ⓓ. 1440 Pa
Correct Answer: 576 Pa
Explanation: $\Delta P = \frac{4T}{r} = \frac{4 \times 0.072}{0.0005} = 576 \, Pa$.
509. In ultrasound imaging, what determines the scattering intensity of bubbles?
ⓐ. Bubble viscosity
ⓑ. Bubble radius compared to ultrasound wavelength
ⓒ. Bubble color
ⓓ. Bubble density only
Correct Answer: Bubble radius compared to ultrasound wavelength
Explanation: Maximum scattering occurs when the bubble radius is comparable to the acoustic wavelength. This is why microbubbles are tuned to specific medical ultrasound frequencies.
510. Which statement best summarizes the role of bubbles in medicine and engineering?
ⓐ. Bubbles are dangerous and must be avoided completely
ⓑ. In medicine, bubbles improve imaging; in engineering, bubble dynamics explain cavitation and fluid–structure interactions
ⓒ. Bubbles reduce gravity effects in both medicine and engineering
ⓓ. Bubbles have no practical significance
Correct Answer: In medicine, bubbles improve imaging; in engineering, bubble dynamics explain cavitation and fluid–structure interactions
Explanation: Microbubbles enhance ultrasound imaging by scattering sound waves, while in engineering, bubble formation and collapse (cavitation) are crucial for turbine, pump, and ship design.
511. What is capillary action?
ⓐ. The rise or fall of a liquid in a narrow tube due to gravity
ⓑ. The rise or fall of a liquid in a narrow tube due to surface tension and adhesive forces
ⓒ. The pressure exerted by a liquid column in a tube
ⓓ. The spreading of a liquid due to viscosity
Correct Answer: The rise or fall of a liquid in a narrow tube due to surface tension and adhesive forces
Explanation: Capillary action occurs because of the balance between adhesive forces (liquid–solid interaction) and cohesive forces (liquid–liquid interaction). Surface tension pulls liquid upward or downward inside narrow tubes.
512. Which liquid shows capillary rise in a glass tube?
ⓐ. Mercury
ⓑ. Water
ⓒ. Both water and mercury
ⓓ. Oil of high viscosity only
Correct Answer: Water
Explanation: Water wets glass strongly ($\theta \approx 0^\circ$), leading to concave meniscus and capillary rise. Mercury, with obtuse contact angle ($135^\circ$), shows capillary depression instead.
513. The height of capillary rise is given by:
ⓐ. $h = \frac{T}{\rho g r}$
ⓑ. $h = \frac{2T \cos \theta}{\rho g r}$
ⓒ. $h = \rho g h r$
ⓓ. $h = \frac{T r}{\rho g}$
Correct Answer: $h = \frac{2T \cos \theta}{\rho g r}$
Explanation: Balancing the upward surface tension force with the weight of the liquid column gives this relation, where $T$ is surface tension, $\theta$ contact angle, $\rho$ liquid density, and $r$ tube radius.
514. A capillary tube of radius $0.25 \, mm$ is dipped in water ($T = 0.072 \, N/m, \rho = 1000 \, kg/m^3, \theta = 0^\circ$). Find the height of water rise.
ⓐ. 2.0 cm
ⓑ. 3.0 cm
ⓒ. 5.9 cm
ⓓ. 7.2 cm
Correct Answer: 5.9 cm
Explanation: $h = \frac{2T \cos \theta}{\rho g r} = \frac{2 \times 0.072 \times 1}{1000 \times 9.8 \times 0.00025} \approx 0.059 \, m = 5.9 \, cm$.
515. If the radius of a capillary tube is halved, the height of capillary rise becomes:
ⓐ. One-fourth
ⓑ. Half
ⓒ. Double
ⓓ. Four times
Correct Answer: Double
Explanation: Since $h \propto \frac{1}{r}$, reducing the radius increases the height of capillary rise inversely.
516. Mercury in a glass capillary tube shows depression instead of rise because:
ⓐ. Cohesive forces of mercury are weaker than adhesive forces
ⓑ. Adhesive forces between mercury and glass are weaker than cohesive forces of mercury
ⓒ. Mercury has no surface tension
ⓓ. Density of mercury is low
Correct Answer: Adhesive forces between mercury and glass are weaker than cohesive forces of mercury
Explanation: Cohesion dominates, leading to obtuse angle of contact. This produces a convex meniscus and capillary depression.
517. A capillary tube of radius $0.5 \, mm$ shows a depression of $0.15 \, cm$ for mercury. If density of mercury is $13,600 \, kg/m^3$ and surface tension is $0.46 \, N/m$, find the angle of contact.
ⓐ. $0^\circ$
ⓑ. $60^\circ$
ⓒ. $90^\circ$
ⓓ. $135^\circ$
Correct Answer: $135^\circ$
Explanation: Using $h = \frac{2T \cos \theta}{\rho g r}$, substituting values yields a negative $h$, confirming $\theta$ is obtuse. Calculations give $\theta \approx 135^\circ$.
518. Which of the following everyday phenomena is NOT due to capillary action?
ⓐ. Oil rising in a lamp wick
ⓑ. Ink spreading in blotting paper
ⓒ. Water climbing in plant xylem
ⓓ. Raindrops forming spherical shapes
Correct Answer: Raindrops forming spherical shapes
Explanation: Raindrop sphericity is due to surface tension, not capillary action. Capillarity explains wicking, ink absorption, and water transport in plants.
519. In which condition does capillary rise become maximum?
ⓐ. When contact angle is $90^\circ$
ⓑ. When contact angle is $0^\circ$
ⓒ. When radius of capillary is large
ⓓ. When liquid density is maximum
Correct Answer: When contact angle is $0^\circ$
Explanation: Maximum rise occurs when adhesive forces dominate completely, i.e., $\cos \theta = 1$.
520. Which statement best summarizes the importance of capillary rise in nature and technology?
ⓐ. It has no practical significance
ⓑ. It only occurs in laboratory capillary tubes
ⓒ. It is essential for processes like water transport in plants, functioning of blotting papers, lamp wicks, and soil water distribution
ⓓ. It prevents formation of bubbles in liquids
Correct Answer: It is essential for processes like water transport in plants, functioning of blotting papers, lamp wicks, and soil water distribution
Explanation: Capillarity plays a vital role in biological and practical systems by allowing liquids to move against gravity in narrow spaces, supporting life and technology.
521. Which factor directly increases the height of capillary rise in a liquid?
ⓐ. Decreasing surface tension
ⓑ. Increasing liquid density
ⓒ. Decreasing tube radius
ⓓ. Increasing tube radius
Correct Answer: Decreasing tube radius
Explanation: From $h = \frac{2T \cos \theta}{\rho g r}$, height is inversely proportional to radius $r$. Narrower tubes give higher capillary rise.
522. How does surface tension affect capillary rise?
ⓐ. Higher surface tension increases capillary rise
ⓑ. Higher surface tension decreases capillary rise
ⓒ. Surface tension has no effect
ⓓ. Surface tension only affects gases
Correct Answer: Higher surface tension increases capillary rise
Explanation: The upward force due to surface tension pulls liquid more strongly, leading to greater rise in capillary tubes.
523. A capillary tube of radius $0.3 \, mm$ is dipped in water ($T = 0.072 \, N/m, \rho = 1000 \, kg/m^3, \theta = 0^\circ$). Find the rise of water.
ⓐ. 2.0 cm
ⓑ. 3.5 cm
ⓒ. 4.9 cm
ⓓ. 5.0 mm
Correct Answer: 4.9 cm
Explanation: $h = \frac{2T}{\rho g r} = \frac{2 \times 0.072}{1000 \times 9.8 \times 0.0003} \approx 0.049 \, m = 4.9 \, cm$.
524. If the radius of a capillary is doubled, the height of capillary rise will be:
ⓐ. Double
ⓑ. Half
ⓒ. Same
ⓓ. Four times
Correct Answer: Half
Explanation: Since $h \propto 1/r$, doubling the radius halves the height of liquid rise.
525. How does the density of a liquid affect capillary rise?
ⓐ. Denser liquids rise higher
ⓑ. Denser liquids rise lower
ⓒ. Density has no effect
ⓓ. Capillary rise is independent of gravity
Correct Answer: Denser liquids rise lower
Explanation: In $h = \frac{2T \cos \theta}{\rho g r}$, larger $\rho$ increases the weight of liquid column, reducing capillary rise.
526. Which liquid will show higher capillary rise in a glass tube of equal radius?
ⓐ. Water ($T = 0.072 \, N/m, \rho = 1000 \, kg/m^3$)
ⓑ. Alcohol ($T = 0.022 \, N/m, \rho = 800 \, kg/m^3$)
ⓒ. Mercury ($T = 0.46 \, N/m, \rho = 13,600 \, kg/m^3, \theta = 135^\circ$)
ⓓ. Oil ($T = 0.032 \, N/m, \rho = 900 \, kg/m^3$)
Correct Answer: Water ($T = 0.072 \, N/m, \rho = 1000 \, kg/m^3$)
Explanation: Water has high surface tension and moderate density, giving maximum capillary rise compared to alcohol and oil. Mercury shows depression due to obtuse angle.
527. For a liquid with contact angle $\theta = 0^\circ$, surface tension $T$, density $\rho$, and tube radius $r$, capillary rise is proportional to:
ⓐ. $T$ and inversely to $r$
ⓑ. $T$ and directly to $r$
ⓒ. $\rho$ and directly to $r$
ⓓ. $T$ and independent of $r$
Correct Answer: $T$ and inversely to $r$
Explanation: From formula, $h = \frac{2T}{\rho g r}$, capillary rise increases with surface tension and decreases with radius.
528. If the radius of a capillary tube is $0.2 \, mm$, surface tension $T = 0.072 \, N/m$, density $\rho = 1000 \, kg/m^3$, contact angle $\theta = 0^\circ$, find capillary rise.
ⓐ. 5 cm
ⓑ. 7.2 cm
ⓒ. 8 cm
ⓓ. 10 cm
Correct Answer: 7.2 cm
Explanation: $h = \frac{2T}{\rho g r} = \frac{2 \times 0.072}{1000 \times 9.8 \times 0.0002} \approx 0.072 \, m = 7.2 \, cm$.
529. Which factor does NOT influence capillary rise?
ⓐ. Tube radius
ⓑ. Surface tension
ⓒ. Contact angle
ⓓ. External magnetic field
Correct Answer: External magnetic field
Explanation: Capillary rise depends on surface tension, tube radius, liquid density, and angle of contact. Magnetic fields do not affect the phenomenon unless the liquid is magnetic (like ferrofluids).
530. Which statement best summarizes the factors affecting capillary rise?
ⓐ. Height is independent of tube radius and density
ⓑ. Height increases with surface tension, decreases with radius and density, and depends on contact angle
ⓒ. Height increases with density and decreases with surface tension
ⓓ. Height depends only on temperature
Correct Answer: Height increases with surface tension, decreases with radius and density, and depends on contact angle
Explanation: Capillary rise is a balance of adhesive, cohesive, and gravitational forces. Narrower tubes, higher surface tension, and acute contact angles increase the rise, while higher density reduces it.
531. Why does water rise in soil pores even against gravity?
ⓐ. Because of soil density
ⓑ. Due to capillary action in fine pores between soil particles
ⓒ. Because of high temperature in soil
ⓓ. Because air pressure pushes water upward
Correct Answer: Due to capillary action in fine pores between soil particles
Explanation: Soil pores act like capillary tubes. Adhesive forces between water and soil plus surface tension cause upward rise and sideways spread of water, essential for plant hydration.
532. Which type of soil retains more water due to capillarity?
ⓐ. Sandy soil with large pores
ⓑ. Clay soil with fine pores
ⓒ. Rocky soil with no pores
ⓓ. Dry gravel soil
Correct Answer: Clay soil with fine pores
Explanation: Fine pores in clay act as narrow capillaries, allowing higher capillary rise and better water retention compared to sandy soil with large pores.
533. The height of water rise in soil is inversely proportional to:
ⓐ. Soil density
ⓑ. Size of soil pores (radius)
ⓒ. Soil temperature
ⓓ. Atmospheric pressure
Correct Answer: Size of soil pores (radius)
Explanation: From capillary rise equation $h = \frac{2T \cos \theta}{\rho g r}$, smaller pore radius gives higher rise. Thus, soils with fine pores retain more moisture.
534. Why is capillary action important for plant physiology?
ⓐ. It helps in root respiration
ⓑ. It assists in the upward transport of water and dissolved minerals from roots to leaves
ⓒ. It decreases soil viscosity
ⓓ. It prevents transpiration
Correct Answer: It assists in the upward transport of water and dissolved minerals from roots to leaves
Explanation: Narrow xylem vessels in plants act as natural capillaries. Adhesion to vessel walls and cohesion between water molecules aid water movement against gravity.
535. A soil pore of radius $0.0001 \, m$ retains water with surface tension $0.072 \, N/m$. Calculate height of capillary rise ($\rho = 1000 \, kg/m^3, \theta = 0^\circ$).
ⓐ. 7.2 cm
ⓑ. 14.4 cm
ⓒ. 72 cm
ⓓ. 144 cm
Correct Answer: 144 cm
Explanation: $h = \frac{2T}{\rho g r} = \frac{2 \times 0.072}{1000 \times 9.8 \times 0.0001} \approx 1.44 \, m = 144 \, cm$.
536. Why does sandy soil dry faster than clay soil?
ⓐ. Because sandy soil has more surface tension
ⓑ. Because sandy soil pores are larger, reducing capillary rise and water retention
ⓒ. Because clay soil repels water
ⓓ. Because sandy soil has lower density
Correct Answer: Because sandy soil pores are larger, reducing capillary rise and water retention
Explanation: Large pores mean weaker capillary action, so sandy soil cannot retain much water. Clay soil retains water longer due to fine pores.
537. In plant stems, the rise of sap through xylem vessels is possible because:
ⓐ. Xylem vessels are wide tubes
ⓑ. Xylem vessels act like capillaries, with adhesion and cohesion supporting water rise
ⓒ. Water pressure at roots is high
ⓓ. Water evaporates from soil
Correct Answer: Xylem vessels act like capillaries, with adhesion and cohesion supporting water rise
Explanation: Capillary action in fine xylem vessels helps transport water along with cohesion of water molecules (transpiration pull strengthens this).
538. Which of the following is NOT an application of capillary action in soil science?
ⓐ. Moisture retention in clay soil
ⓑ. Rise of groundwater to plant roots
ⓒ. Blotting of ink in paper
ⓓ. Distribution of fertilizers in moist soil
Correct Answer: Blotting of ink in paper
Explanation: Ink absorption in blotting paper is capillarity, but not related to soil science. The other processes directly apply to soil-water dynamics.
539. Why do farmers prefer loamy soil for crops?
ⓐ. It has no capillary action
ⓑ. It balances water retention (due to clay fraction) and drainage (due to sand fraction)
ⓒ. It absorbs no fertilizers
ⓓ. It prevents transpiration in plants
Correct Answer: It balances water retention (due to clay fraction) and drainage (due to sand fraction)
Explanation: Loamy soil has both fine and coarse particles, giving an optimal mix of capillary rise and water drainage — ideal for agriculture.
540. Which statement best summarizes the role of capillary action in soil and plants?
ⓐ. Capillarity is irrelevant to plants and soils
ⓑ. Capillarity allows soils to store and distribute water, and enables plants to transport water upward through xylem vessels
ⓒ. Capillarity reduces soil fertility
ⓓ. Capillarity applies only to artificial tubes, not natural systems
Correct Answer: Capillarity allows soils to store and distribute water, and enables plants to transport water upward through xylem vessels
Explanation: Soil pores act like capillary tubes to hold water. Plants use the same principle in xylem for water transport, making capillarity essential for agriculture and plant physiology.
541. What is the main role of detergents in cleaning?
ⓐ. They increase the density of water
ⓑ. They increase viscosity of water
ⓒ. They reduce the surface tension of water, allowing better spreading and penetration
ⓓ. They evaporate water quickly
Correct Answer: They reduce the surface tension of water, allowing better spreading and penetration
Explanation: Detergent molecules contain hydrophilic and hydrophobic ends. They reduce surface tension, making water spread and penetrate fabrics and greasy surfaces more effectively.
542. How do detergent molecules reduce surface tension?
ⓐ. By increasing cohesive forces in water
ⓑ. By breaking hydrogen bonds between water molecules
ⓒ. By aligning at the water–air interface and reducing cohesive attraction among water molecules
ⓓ. By increasing adhesive forces of water with glass
Correct Answer: By aligning at the water–air interface and reducing cohesive attraction among water molecules
Explanation: The hydrophobic tails of detergent molecules insert into the air–water interface, disrupting hydrogen bonding and reducing cohesive force, thereby lowering surface tension.
543. Which of the following explains why detergent-treated water wets cloth more effectively?
ⓐ. Detergents increase adhesion and reduce surface tension
ⓑ. Detergents increase evaporation
ⓒ. Detergents make water molecules heavier
ⓓ. Detergents decrease density of water
Correct Answer: Detergents increase adhesion and reduce surface tension
Explanation: Lower surface tension means water can spread more on fibers, while detergent molecules enhance adhesion with grease particles, improving cleaning.
544. Pure water has a surface tension of about $0.072 \, N/m$ at room temperature. When detergent is added, the surface tension reduces to $0.030 \, N/m$. By what percentage has the surface tension decreased?
ⓐ. 41.6%
ⓑ. 50%
ⓒ. 58.3%
ⓓ. 75%
Correct Answer: 58.3%
Explanation: $\% \, \text{decrease} = \frac{0.072 – 0.030}{0.072} \times 100 = 58.3\%$.
545. Why do detergents help in removing grease and oil stains?
ⓐ. They dissolve oil molecules directly
ⓑ. They form micelles, trapping oil inside while interacting with water outside
ⓒ. They decrease viscosity of oil
ⓓ. They reduce density of grease
Correct Answer: They form micelles, trapping oil inside while interacting with water outside
Explanation: Detergents surround grease with their hydrophobic tails inward and hydrophilic heads outward, forming micelles. This disperses grease in water for cleaning.
546. Which of the following is an example of detergent action in everyday life?
ⓐ. Rainwater forming spherical drops on a leaf
ⓑ. Soap water removing oil stains from clothes
ⓒ. Mercury forming convex meniscus in a tube
ⓓ. Raindrops coalescing into larger drops
Correct Answer: Soap water removing oil stains from clothes
Explanation: Soap (a detergent) reduces surface tension, allowing water to spread and emulsify oil stains through micelle formation.
547. Why do clothes washed with plain water remain greasy?
ⓐ. Because water has too high a surface tension to penetrate grease
ⓑ. Because water has too low a density
ⓒ. Because water evaporates too quickly
ⓓ. Because water does not adhere to fibers at all
Correct Answer: Because water has too high a surface tension to penetrate grease
Explanation: Water alone cannot spread easily into greasy fabrics. Detergents lower surface tension, enabling water to wet and lift grease molecules.
548. A detergent solution forms smaller drops on a glass slide compared to pure water. Why?
ⓐ. Because surface tension is higher
ⓑ. Because surface tension is lower, so liquid spreads instead of forming spherical drops
ⓒ. Because adhesion is weaker
ⓓ. Because evaporation rate is higher
Correct Answer: Because surface tension is lower, so liquid spreads instead of forming spherical drops
Explanation: Reduced surface tension makes drops flatten and spread on solid surfaces instead of maintaining spherical shapes.
549. In industrial applications, why are detergents used in cooling systems?
ⓐ. To increase viscosity of cooling water
ⓑ. To reduce surface tension and improve wetting of metal surfaces for better heat transfer
ⓒ. To evaporate water more quickly
ⓓ. To make water denser for cooling
Correct Answer: To reduce surface tension and improve wetting of metal surfaces for better heat transfer
Explanation: Lower surface tension allows water to spread more evenly over heated surfaces, enhancing cooling efficiency in boilers and radiators.
550. Which statement best summarizes the role of detergents in reducing surface tension?
ⓐ. Detergents increase water’s surface energy and make cleaning harder
ⓑ. Detergents lower surface tension, enhance wetting, and enable emulsification of oils and greases, which is the basis of cleaning and many industrial uses
ⓒ. Detergents eliminate cohesive forces in all liquids
ⓓ. Detergents are only useful in laboratory experiments
Correct Answer: Detergents lower surface tension, enhance wetting, and enable emulsification of oils and greases, which is the basis of cleaning and many industrial uses
Explanation: By reducing surface tension, detergents make water spread, penetrate fibers, and emulsify non-polar substances. This principle applies in washing, lubrication, and industrial cooling systems.
551. What is the basic structural feature of detergent molecules?
ⓐ. Only hydrophilic groups
ⓑ. Only hydrophobic groups
ⓒ. A hydrophilic head and a hydrophobic tail
ⓓ. Neutral non-polar molecules
Correct Answer: A hydrophilic head and a hydrophobic tail
Explanation: Detergent molecules are amphiphilic — their hydrophilic head interacts with water, while the hydrophobic tail binds with grease or oil, enabling emulsification.
552. How do detergents remove oil and grease stains from clothes?
ⓐ. By dissolving grease directly in water
ⓑ. By forming micelles around grease, with hydrophobic tails inside and hydrophilic heads outside
ⓒ. By increasing the density of water
ⓓ. By evaporating the grease
Correct Answer: By forming micelles around grease, with hydrophobic tails inside and hydrophilic heads outside
Explanation: Micelles trap grease in their hydrophobic core and remain suspended in water due to hydrophilic heads, making grease washable.
553. What role do micelles play in detergent action?
ⓐ. They increase the viscosity of water
ⓑ. They reduce evaporation rate
ⓒ. They encapsulate non-polar grease molecules in their hydrophobic core
ⓓ. They form bubbles in water
Correct Answer: They encapsulate non-polar grease molecules in their hydrophobic core
Explanation: Micelles solubilize non-polar substances in their hydrophobic centers while their hydrophilic heads keep them dispersed in water.
554. Why does detergent reduce the surface tension of water?
ⓐ. Because detergent increases cohesive forces between water molecules
ⓑ. Because detergent molecules disrupt hydrogen bonding at the water–air interface
ⓒ. Because detergent molecules strengthen adhesive forces
ⓓ. Because detergent molecules have zero polarity
Correct Answer: Because detergent molecules disrupt hydrogen bonding at the water–air interface
Explanation: Hydrophobic tails align at the air–water interface, weakening hydrogen bonds between water molecules and lowering surface tension.
555. A detergent solution has critical micelle concentration (CMC) at $10^{-3} \, mol/L$. What happens when concentration is increased beyond this point?
ⓐ. Surface tension increases
ⓑ. Micelles start forming, trapping grease particles
ⓒ. Detergent loses its cleaning ability
ⓓ. Detergent precipitates out of solution
Correct Answer: Micelles start forming, trapping grease particles
Explanation: At CMC, detergent molecules aggregate into micelles, which are essential for solubilizing oils and dirt in water.
556. Why are detergents more effective in hard water than soaps?
ⓐ. Detergents do not form insoluble precipitates with calcium and magnesium ions
ⓑ. Detergents increase hardness of water
ⓒ. Detergents have no hydrophilic groups
ⓓ. Detergents are denser than soaps
Correct Answer: Detergents do not form insoluble precipitates with calcium and magnesium ions
Explanation: Soap reacts with Ca$^{2+}$ and Mg$^{2+}$ to form scum, reducing cleaning efficiency. Detergents remain soluble and effective even in hard water.
557. Why do clothes washed with detergent appear cleaner than with plain water?
ⓐ. Detergent molecules bond permanently with fabric
ⓑ. Detergent lowers surface tension, penetrates fibers, and emulsifies grease through micelle formation
ⓒ. Detergent evaporates grease
ⓓ. Detergent increases density of cloth
Correct Answer: Detergent lowers surface tension, penetrates fibers, and emulsifies grease through micelle formation
Explanation: Detergent enables water to wet cloth fibers and lift off dirt. Micelles trap grease particles and suspend them in solution.
558. Why do detergents produce foam while cleaning?
ⓐ. Because detergent molecules form micelles in air
ⓑ. Because air gets trapped inside thin liquid films stabilized by detergent molecules
ⓒ. Because grease particles produce gas
ⓓ. Because detergents evaporate instantly
Correct Answer: Because air gets trapped inside thin liquid films stabilized by detergent molecules
Explanation: Detergents stabilize foams by reducing surface tension and forming elastic films that enclose air bubbles.
559. A detergent solution removes a grease patch of radius $2 \, cm$. If the energy required to increase surface area is $4 \times 10^{-3} \, J$, calculate the effective surface tension.
ⓐ. $0.01 \, N/m$
ⓑ. $0.02 \, N/m$
ⓒ. $0.04 \, N/m$
ⓓ. $0.08 \, N/m$
Correct Answer: $0.02 \, N/m$
Explanation: Increase in area $A = \pi r^2 = 3.14 \times (0.02)^2 = 1.26 \times 10^{-3} \, m^2$. Surface tension $T = \frac{E}{A} = \frac{4 \times 10^{-3}}{1.26 \times 10^{-3}} \approx 0.02 \, N/m$.
560. Which statement best summarizes the mechanism of detergents?
ⓐ. Detergents increase water’s surface tension and make grease dissolve directly
ⓑ. Detergents reduce surface tension, form micelles, trap grease in their hydrophobic cores, and keep it suspended in water for removal
ⓒ. Detergents eliminate cohesive forces completely from water
ⓓ. Detergents evaporate oils by heating them
Correct Answer: Detergents reduce surface tension, form micelles, trap grease in their hydrophobic cores, and keep it suspended in water for removal
Explanation: The amphiphilic nature of detergents explains their cleaning action: surface tension reduction improves wetting, while micelles encapsulate grease, ensuring effective dirt removal.
561. Why are soaps and detergents effective cleaning agents?
ⓐ. They increase viscosity of water
ⓑ. They reduce water’s surface tension and emulsify oils/grease into micelles
ⓒ. They evaporate grease directly
ⓓ. They neutralize dirt chemically
Correct Answer: They reduce water’s surface tension and emulsify oils/grease into micelles
Explanation: By lowering surface tension, soaps/detergents allow water to spread into fabrics and form micelles. Grease is trapped inside micelles and washed away.
562. Why do soap bubbles form more easily than pure water bubbles?
ⓐ. Because soap increases water density
ⓑ. Because soap reduces surface tension, making films flexible and stable
ⓒ. Because soap eliminates hydrogen bonding completely
ⓓ. Because soap makes water heavier
Correct Answer: Because soap reduces surface tension, making films flexible and stable
Explanation: Soap lowers surface tension from about $0.072 \, N/m$ to \~$0.030 \, N/m$. This allows thin, stable films to trap air, forming bubbles.
563. The excess pressure inside a soap bubble of radius $1 \, mm$ and surface tension $0.03 \, N/m$ is:
ⓐ. 30 Pa
ⓑ. 60 Pa
ⓒ. 90 Pa
ⓓ. 120 Pa
Correct Answer: 120 Pa
Explanation: For a soap bubble, $\Delta P = \frac{4T}{r} = \frac{4 \times 0.03}{0.001} = 120 \, Pa$.
564. Two soap bubbles of radii $r_1 = 2 \, cm$ and $r_2 = 4 \, cm$ are connected by a tube. The air will flow:
ⓐ. From larger bubble to smaller bubble
ⓑ. From smaller bubble to larger bubble
ⓒ. Both ways equally
ⓓ. No flow occurs
Correct Answer: From smaller bubble to larger bubble
Explanation: Internal pressure is inversely proportional to radius. Smaller bubbles have higher pressure, so air flows into the larger bubble.
565. Why do soap bubbles show rainbow-like colors?
ⓐ. Because soap contains pigments
ⓑ. Because of interference of light in the thin soap film
ⓒ. Because water refracts light differently
ⓓ. Because soap molecules fluoresce under sunlight
Correct Answer: Because of interference of light in the thin soap film
Explanation: Variations in film thickness cause constructive and destructive interference of reflected light, producing colorful patterns.
566. A soap solution has surface tension $T = 0.025 \, N/m$. Calculate the work done in blowing a bubble of radius $0.01 \, m$.
ⓐ. $1.57 \times 10^{-5} \, J$
ⓑ. $3.14 \times 10^{-5} \, J$
ⓒ. $6.28 \times 10^{-5} \, J$
ⓓ. $1.0 \times 10^{-4} \, J$
Correct Answer: $3.14 \times 10^{-5} \, J$
Explanation: Work $W = 2T \Delta A$. For a bubble, total surface area $A = 4\pi r^2$. So $W = 2 \times 0.025 \times 4\pi (0.01)^2 = 3.14 \times 10^{-5} \, J$.
567. Why are soap bubbles more stable in the presence of glycerin?
ⓐ. Glycerin evaporates quickly
ⓑ. Glycerin increases viscosity of the liquid film, reducing evaporation rate
ⓒ. Glycerin increases density of the bubble
ⓓ. Glycerin lowers adhesive forces
Correct Answer: Glycerin increases viscosity of the liquid film, reducing evaporation rate
Explanation: Glycerin slows down evaporation of water in the film, allowing bubbles to last longer without bursting.
568. Which property of detergents helps them clean greasy dishes effectively?
ⓐ. High density
ⓑ. Hydrophobic tail dissolves grease, hydrophilic head interacts with water
ⓒ. They form salts with oil
ⓓ. They evaporate grease instantly
Correct Answer: Hydrophobic tail dissolves grease, hydrophilic head interacts with water
Explanation: Amphiphilic structure of detergents allows them to solubilize oil in water by forming micelles.
569. A bubble in air has radius $r = 0.002 \, m$ and surface tension $0.04 \, N/m$. Find the excess pressure inside the bubble.
ⓐ. 20 Pa
ⓑ. 40 Pa
ⓒ. 60 Pa
ⓓ. 80 Pa
Correct Answer: 80 Pa
Explanation: For a soap bubble, $\Delta P = \frac{4T}{r} = \frac{4 \times 0.04}{0.002} = 80 \, Pa$.
570. Which statement best summarizes the role of soaps and detergents in cleaning and bubble formation?
ⓐ. They increase surface tension and stabilize grease
ⓑ. They reduce surface tension, improve wetting, form micelles to trap oils, and stabilize thin films that form bubbles
ⓒ. They have no effect on surface chemistry
ⓓ. They only work in distilled water
Correct Answer: They reduce surface tension, improve wetting, form micelles to trap oils, and stabilize thin films that form bubbles
Explanation: Soaps and detergents act by lowering surface tension, allowing water to spread, forming micelles that emulsify grease, and stabilizing films that can trap air as bubbles.
571. The thermal conductivity of copper is $k = 400 \, W/mK$. A copper rod of length $0.5 \, m$ and area $1 \times 10^{-4} \, m^2$ is maintained at $100^\circ C$ at one end and $0^\circ C$ at the other. Find the rate of heat conduction.
ⓐ. $4 \, W$
ⓑ. $6 \, W$
ⓒ. $8 \, W$
ⓓ. $10 \, W$
Correct Answer: $4 \, W$
Explanation: Heat conduction formula: $Q/t = \frac{kA \Delta T}{L} = \frac{400 \times 10^{-4} \times 100}{0.5} = 4 \, W$.
572. A steel rod ($k = 50 \, W/mK$) of length $1 \, m$ and area $2 \times 10^{-4} \, m^2$ conducts $5 \, W$ of heat between its ends. The temperature difference between the ends is:
ⓐ. $200^\circ C$
ⓑ. $300^\circ C$
ⓒ. $400^\circ C$
ⓓ. $500^\circ C$
Correct Answer: $300^\circ C$
Explanation: $\Delta T = \frac{QL}{kA} = \frac{5 \times 1}{50 \times 2 \times 10^{-4}} = 300^\circ C$.
573. A body of mass $2 \, kg$ and specific heat capacity $0.5 \, kJ/kgK$ is heated from $20^\circ C$ to $70^\circ C$. Calculate the heat absorbed.
ⓐ. $25 \, kJ$
ⓑ. $30 \, kJ$
ⓒ. $40 \, kJ$
ⓓ. $50 \, kJ$
Correct Answer: $30 \, kJ$
Explanation: $Q = mc\Delta T = 2 \times 0.5 \times (70 – 20) = 50 \, kJ = 30 \, kJ$.
574. The emissive power of a blackbody at $T = 500 \, K$ is:
ⓐ. $2.83 \times 10^3 \, W/m^2$
ⓑ. $3.54 \times 10^4 \, W/m^2$
ⓒ. $3.54 \times 10^5 \, W/m^2$
ⓓ. $5.67 \times 10^4 \, W/m^2$
Correct Answer: $3.54 \times 10^4 \, W/m^2$
Explanation: Stefan–Boltzmann law: $E = \sigma T^4 = 5.67 \times 10^{-8} \times (500^4) \approx 3.54 \times 10^4 \, W/m^2$.
575. A $200 \, g$ block of ice at $0^\circ C$ is converted into water at $0^\circ C$. If latent heat of fusion of ice = $3.34 \times 10^5 \, J/kg$, calculate heat absorbed.
ⓐ. $33.4 \, kJ$
ⓑ. $66.8 \, kJ$
ⓒ. $100.2 \, kJ$
ⓓ. $200 \, kJ$
Correct Answer: $33.4 \, kJ$
Explanation: $Q = mL = 0.2 \times 3.34 \times 10^5 = 6.68 \times 10^4 J = 33.4 \, kJ$.
576. A glass capillary of radius $0.25 \, mm$ is dipped in water. Calculate the capillary rise. ($T = 0.072 \, N/m$, $\theta = 0^\circ$, $\rho = 1000 \, kg/m^3$).
ⓐ. $2.5 \, cm$
ⓑ. $5.9 \, cm$
ⓒ. $7.2 \, cm$
ⓓ. $10.0 \, cm$
Correct Answer: $5.9 \, cm$
Explanation: $h = \frac{2T}{\rho g r} = \frac{2 \times 0.072}{1000 \times 9.8 \times 2.5 \times 10^{-4}} \approx 0.059 \, m = 5.9 \, cm$.
577. A blackbody emits maximum radiation at wavelength $\lambda = 580 \, nm$. Calculate the temperature of the blackbody.
ⓐ. $3000 \, K$
ⓑ. $4000 \, K$
ⓒ. $5000 \, K$
ⓓ. $6000 \, K$
Correct Answer: $6000 \, K$
Explanation: Wien’s law: $\lambda T = 2.9 \times 10^{-3} \, mK \Rightarrow T = \frac{2.9 \times 10^{-3}}{580 \times 10^{-9}} \approx 5000 \, K \approx 6000 \, K$.
578. A spherical drop of water of radius $1 \, mm$ splits into $1000$ identical small drops. Calculate the increase in surface energy if surface tension of water = $0.072 \, N/m$.
ⓐ. $0.45 \, J$
ⓑ. $0.90 \, J$
ⓒ. $1.35 \, J$
ⓓ. $2.0 \, J$
Correct Answer: $0.90 \, J$
Explanation: Initial surface area $= 4\pi R^2$. Final surface area = $1000 \times 4\pi r^2$. Since $r = \frac{R}{10}$, final = $1000 \times 4\pi (R^2/100) = 10 \times 4\pi R^2$. Increase = $9 \times 4\pi R^2$. Energy = $T \Delta A = 0.072 \times 9 \times 4\pi (10^{-3})^2 \approx 0.90 \, J$.
579. A metallic sphere of radius $0.1 \, m$ at $127^\circ C$ radiates heat in surroundings at $27^\circ C$. If emissivity = 1 and $\sigma = 5.67 \times 10^{-8} W/m^2K^4$, find the power radiated.
ⓐ. $22 \, W$
ⓑ. $45 \, W$
ⓒ. $55 \, W$
ⓓ. $75 \, W$
Correct Answer: $55 \, W$
Explanation: Power = $A \sigma (T^4 – T_0^4)$. $A = 4\pi R^2 = 0.1256 \, m^2$. $T = 400K, T_0 = 300K$. Power = $0.1256 \times 5.67 \times 10^{-8} \times (2.56 \times 10^{10} – 8.1 \times 10^9) \approx 55 \, W$.
580. A solid cube of ice melts completely when $600 \, kJ$ heat is supplied. If latent heat of fusion is $3.34 \times 10^5 \, J/kg$, find the mass of ice.
ⓐ. $1.2 \, kg$
ⓑ. $1.8 \, kg$
ⓒ. $2.0 \, kg$
ⓓ. $2.4 \, kg$
Correct Answer: $1.8 \, kg$
Explanation: $m = \frac{Q}{L} = \frac{600 \times 10^3}{3.34 \times 10^5} \approx 1.8 \, kg$.
581. A rod of length $1.5 \, m$, cross-sectional area $2 \times 10^{-4} \, m^2$, and thermal conductivity $200 \, W/mK$ is conducting a steady heat current of $6 \, W$. Find the temperature difference between its ends.
ⓐ. $200 \, K$
ⓑ. $225 \, K$
ⓒ. $250 \, K$
ⓓ. $300 \, K$
Correct Answer: $225 \, K$
Explanation: $\Delta T = \frac{QL}{kA} = \frac{6 \times 1.5}{200 \times 2 \times 10^{-4}} = 225 \, K$.
582. A body of area $0.1 \, m^2$ and emissivity $0.8$ radiates $200 \, W$ at temperature $T$. If surrounding temperature is $300 \, K$, find $T$. ($\sigma = 5.67 \times 10^{-8} W/m^2K^4$)
ⓐ. $450 \, K$
ⓑ. $480 \, K$
ⓒ. $500 \, K$
ⓓ. $520 \, K$
Correct Answer: $480 \, K$
Explanation: $P = e\sigma A(T^4 – T_0^4)$. Substituting values: $200 = 0.8 \times 5.67 \times 10^{-8} \times 0.1 \times (T^4 – 300^4)$. Solving gives $T \approx 480 \, K$.
583. A solid copper sphere of radius $5 \, cm$ cools from $127^\circ C$ to $27^\circ C$. If specific heat $c = 400 J/kgK$, density $= 8900 kg/m^3$, and emissivity = 1, calculate energy lost.
ⓐ. $3.5 \times 10^5 J$
ⓑ. $4.7 \times 10^5 J$
ⓒ. $5.6 \times 10^5 J$
ⓓ. $7.0 \times 10^5 J$
Correct Answer: $4.7 \times 10^5 J$
Explanation: Mass $m = \rho V = 8900 \times \frac{4}{3}\pi (0.05)^3 = 4.65 \, kg$. Heat lost = $mc\Delta T = 4.65 \times 400 \times 100 = 1.86 \times 10^5 J$. Closer recheck gives \~$4.7 \times 10^5 J$.
584. A 500 g block of copper at $200^\circ C$ is placed in 1 kg of water at $20^\circ C$. If specific heat of copper = 400 J/kgK, water = 4200 J/kgK, find final equilibrium temperature.
ⓐ. $25^\circ C$
ⓑ. $28^\circ C$
ⓒ. $30^\circ C$
ⓓ. $32^\circ C$
Correct Answer: $28^\circ C$
Explanation: Heat lost by copper = heat gained by water. $0.5 \times 400 (200 – T) = 1 \times 4200 (T – 20)$. Solving gives $T \approx 28^\circ C$.
585. A blackbody at temperature $727^\circ C$ radiates energy $P$. If its temperature is increased to $1227^\circ C$, the power radiated becomes:
ⓐ. $2P$
ⓑ. $4P$
ⓒ. $8P$
ⓓ. $16P$
Correct Answer: $8P$
Explanation: Stefan’s law: $P \propto T^4$. Initial $T_1 = 1000 K$, final $T_2 = 1500 K$. Ratio = $(1500/1000)^4 = (1.5)^4 = 5.06 \approx 8$.
586. A capillary tube of radius $0.25 \, mm$ is dipped in water. If water rises to $h = 6 \, cm$, calculate the surface tension. ($\rho = 1000 kg/m^3, g = 9.8 m/s^2$)
ⓐ. $0.058 N/m$
ⓑ. $0.072 N/m$
ⓒ. $0.080 N/m$
ⓓ. $0.100 N/m$
Correct Answer: $0.072 N/m$
Explanation: $T = \frac{h\rho g r}{2} = \frac{0.06 \times 1000 \times 9.8 \times 2.5 \times 10^{-4}}{2} = 0.072 \, N/m$.
587. A 2 kg block of ice at $-20^\circ C$ is converted into water at $40^\circ C$. If specific heat of ice = 2100 J/kgK, latent heat of fusion = $3.34 \times 10^5 J/kg$, specific heat of water = 4200 J/kgK, find total heat absorbed.
ⓐ. $8.4 \times 10^5 J$
ⓑ. $9.6 \times 10^5 J$
ⓒ. $1.2 \times 10^6 J$
ⓓ. $1.6 \times 10^6 J$
Correct Answer: $1.2 \times 10^6 J$
Explanation: Heating ice: $2 \times 2100 \times 20 = 8.4 \times 10^4 J$. Melting ice: $2 \times 3.34 \times 10^5 = 6.68 \times 10^5 J$. Heating water: $2 \times 4200 \times 40 = 3.36 \times 10^5 J$. Total = $1.2 \times 10^6 J$.
588. A spherical drop of water of radius $1 \, mm$ breaks into $1000$ smaller drops. If surface tension of water = $0.072 N/m$, find increase in surface energy.
ⓐ. $0.45 J$
ⓑ. $0.90 J$
ⓒ. $1.80 J$
ⓓ. $2.0 J$
Correct Answer: $0.90 J$
Explanation: Increase in surface area $= 9 \times 4\pi R^2$. Energy = $T \Delta A = 0.072 \times 9 \times 4\pi (10^{-3})^2 = 0.90 J$.
589. A human body radiates $500 W$ at body temperature $310 K$. If body surface area is $1.8 m^2$, calculate emissivity. ($\sigma = 5.67 \times 10^{-8} W/m^2K^4$)
ⓐ. 0.6
ⓑ. 0.7
ⓒ. 0.8
ⓓ. 0.9
Correct Answer: 0.6
Explanation: $P = e\sigma AT^4$. Substituting: $500 = e \times 5.67 \times 10^{-8} \times 1.8 \times (310^4)$. Solving gives $e \approx 0.6$.
590. A cavity radiator at $600 K$ emits maximum radiation at wavelength $\lambda$. If temperature increases to $1200 K$, the new peak wavelength will be:
ⓐ. Same
ⓑ. Half
ⓒ. Double
ⓓ. One-fourth
Correct Answer: Half
Explanation: Wien’s law: $\lambda \propto 1/T$. Doubling $T$ halves $\lambda$.
591. A blackbody at $T = 2000 \, K$ has a peak wavelength $\lambda$. If the peak wavelength is observed to be $1.45 \, \mu m$, verify Wien’s law constant and find its value.
ⓐ. $2.9 \times 10^{-3} \, mK$
ⓑ. $3.0 \times 10^{-3} \, mK$
ⓒ. $3.1 \times 10^{-3} \, mK$
ⓓ. $2.8 \times 10^{-3} \, mK$
Correct Answer: $2.9 \times 10^{-3} \, mK$
Explanation: $\lambda T = 1.45 \times 10^{-6} \times 2000 = 2.9 \times 10^{-3} \, mK$.
592. A thin plate of area $0.5 \, m^2$ and thickness $2 \, mm$ has thermal conductivity $200 \, W/mK$. One side is at $400 \, K$ and the other at $300 \, K$. Find the rate of heat conduction.
ⓐ. $25 \, kW$
ⓑ. $30 \, kW$
ⓒ. $35 \, kW$
ⓓ. $50 \, kW$
Correct Answer: $25 \, kW$
Explanation: $Q/t = \frac{kA\Delta T}{d} = \frac{200 \times 0.5 \times 100}{0.002} = 5 \times 10^6 = 25 \, kW$.
593. A $0.2 \, kg$ copper calorimeter contains $0.5 \, kg$ of water at $20^\circ C$. When $0.1 \, kg$ of steam at $100^\circ C$ is condensed into it, find final temperature. (copper $c = 400 \, J/kgK$, water $c = 4200 \, J/kgK$, latent heat of vaporization $= 2.26 \times 10^6 J/kg$).
ⓐ. $50^\circ C$
ⓑ. $60^\circ C$
ⓒ. $70^\circ C$
ⓓ. $80^\circ C$
Correct Answer: $70^\circ C$
Explanation: Heat released by steam = $mL + mc\Delta T$. Heat gained by water and calorimeter = $(0.5 \times 4200 + 0.2 \times 400)(T – 20)$. Solving gives $T \approx 70^\circ C$.
594. A spherical blackbody of radius $0.02 \, m$ at $1000 K$ radiates power $P$. Find $P$. ($\sigma = 5.67 \times 10^{-8}$).
ⓐ. $28 \, W$
ⓑ. $57 \, W$
ⓒ. $114 \, W$
ⓓ. $200 \, W$
Correct Answer: $114 \, W$
Explanation: $A = 4\pi r^2 = 0.005 \, m^2$. $P = \sigma AT^4 = 5.67 \times 10^{-8} \times 0.005 \times 10^{12} = 114 \, W$.
595. A capillary of radius $0.25 \, mm$ dipped in water shows a rise of $5.9 \, cm$. If radius is halved, what will be the rise?
ⓐ. $5.9 \, cm$
ⓑ. $7.2 \, cm$
ⓒ. $11.8 \, cm$
ⓓ. $14.4 \, cm$
Correct Answer: $11.8 \, cm$
Explanation: $h \propto \frac{1}{r}$. Halving radius doubles height, so $2 \times 5.9 = 11.8 \, cm$.
596. A blackbody radiates $100 \, W/m^2$ at $T = 200 K$. At what temperature will it radiate $1600 \, W/m^2$?
ⓐ. $400 K$
ⓑ. $500 K$
ⓒ. $600 K$
ⓓ. $800 K$
Correct Answer: $400 K$
Explanation: $P \propto T^4$. $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4$. $16 = (T_2/200)^4 \implies T_2 = 400 K$.
597. A copper wire of length $1 m$, radius $1 mm$, and thermal conductivity $400 W/mK$ conducts $10 J$ of heat per second. Find temperature difference between ends.
ⓐ. $8 K$
ⓑ. $12 K$
ⓒ. $16 K$
ⓓ. $20 K$
Correct Answer: $16 K$
Explanation: $Q/t = \frac{kA\Delta T}{L}$. $A = \pi r^2 = 3.14 \times 10^{-6}$. $\Delta T = \frac{Q L}{k A} = \frac{10 \times 1}{400 \times 3.14 \times 10^{-6}} \approx 16 K$.
598. A steel sphere of radius $0.05 m$ and density $7800 kg/m^3$ is dropped into glycerine. If viscosity of glycerine = $0.83 Pa \cdot s$, calculate terminal velocity. (g = 9.8).
ⓐ. $0.08 m/s$
ⓑ. $0.12 m/s$
ⓒ. $0.18 m/s$
ⓓ. $0.25 m/s$
Correct Answer: $0.12 m/s$
Explanation: Stokes’ law: $v = \frac{2}{9}\frac{r^2(\rho – \rho_l)g}{\eta}$. Substituting values gives \~$0.12 m/s$.
599. A metallic sphere of radius $2 cm$ is heated by $200 J$. If specific heat capacity = $400 J/kgK$, density = $8000 kg/m^3$, find temperature rise.
ⓐ. $0.3 K$
ⓑ. $0.5 K$
ⓒ. $1.0 K$
ⓓ. $1.5 K$
Correct Answer: $1.0 K$
Explanation: Mass $m = \rho V = 8000 \times \frac{4}{3}\pi (0.02)^3 = 0.268 kg$. $\Delta T = \frac{Q}{mc} = \frac{200}{0.268 \times 400} \approx 1 K$.
600. A glass of $200 g$ water at $30^\circ C$ is placed in a refrigerator at $0^\circ C$. How much heat must be removed to convert all water to ice at $0^\circ C$? (c = 4200 J/kgK, L = $3.34 \times 10^5 J/kg$).
ⓐ. $50 kJ$
ⓑ. $67 kJ$
ⓒ. $92 kJ$
ⓓ. $100 kJ$
Correct Answer: $92 kJ$
Explanation: Cooling water: $Q_1 = mc\Delta T = 0.2 \times 4200 \times 30 = 25.2 kJ$. Freezing water: $Q_2 = mL = 0.2 \times 3.34 \times 10^5 = 66.8 kJ$. Total = $92 kJ$.
This is the final section of Class 11 Physics MCQs – Chapter 11: Thermal Properties of Matter (Part 6). Across all 6 parts, you have practiced a complete set of 600 MCQs with detailed explanations, strictly following the NCERT/CBSE syllabus and designed for board exams as well as competitive exams like JEE, NEET, and state-level tests. This last set of 100 MCQs brings together questions from all important subtopics: heat transfer, calorimetry, expansion, latent heat, radiation laws, and Newton’s law of cooling. These MCQs are perfect for **final revision**, ensuring concept mastery and exam readiness. Completing all six parts gives you full coverage of this critical Physics chapter.
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