Physics MCQs

Question #541

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A machine gun fires 60 bullets per minute, each of mass 10 g with velocity 600 m/s. The power developed by the gun is:

A. 1800 W

B. 18000 W

C. 300 W

D. 3000 W

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Correct Answer: Option A

Explanation:
Power = work done per unit time = kinetic energy imparted per second. KE per bullet = ½mv² = ½×0.01×(600)² = 1800 J. Bullets per second = 60/60 = 1. Thus power = 1800 J/s = 1800 W. Note: Mass conversion: 10 g = 0.01 kg. This problem combines kinetic energy formula with power definition. Exam tip: Always convert to SI units first (grams to kg, minutes to seconds). Such numerical problems test multi-step calculation skills and unit conversion proficiency essential for competitive examinations.

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Question #542

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The kinetic energy of a body becomes four times its initial value. The new momentum will be:

A. Four times the initial

B. Same as initial

C. Half the initial

D. Twice the initial

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Correct Answer: Option D

Explanation:
Kinetic energy KE = p²/(2m), where p is momentum. Thus p = √(2m·KE). If KE becomes 4 times, p_new = √(2m·4KE) = 2√(2m·KE) = 2p_initial. Hence momentum doubles. This relationship shows momentum scales with square root of kinetic energy for constant mass. Memory tip: p ∝ √KE when mass constant. This conceptual link between momentum and kinetic energy is frequently tested in competitive exams to assess depth of understanding beyond rote formula memorization. Always derive relationships when direct formulas aren't recalled.

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Question #543

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A spring of force constant k is stretched by a length x. The work done in stretching it further by the same length x is:

A. ³/₂kx²

B. ½kx²

C. kx²

D. 2kx²

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Correct Answer: Option A

Explanation:
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.

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Question #544

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According to the law of conservation of energy, in an isolated system:

A. Total mechanical energy remains constant if only conservative forces act

B. Kinetic energy remains constant

C. Potential energy remains constant

D. Thermal energy is always zero

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Correct Answer: Option A

Explanation:
Law of conservation of energy states total energy (all forms) is conserved in isolated systems. For mechanical energy (KE + PE) to be conserved, only conservative forces (like gravity, spring) must act; non-conservative forces (friction) convert mechanical energy to heat. Option C precisely states this condition. Options A and B are incorrect as KE and PE individually vary. Option D is false as thermal energy can exist. Memory tip: 'Mechanical energy conserved ⇔ no non-conservative work'. This nuanced understanding is critical for energy problems in competitive exams, where distractors often omit the conservative force condition.

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Question #545

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The escape velocity from Earth's surface is approximately 11.2 km/s. If a planet has twice Earth's radius and same density, its escape velocity will be:

A. 15.8 km/s

B. 5.6 km/s

C. 11.2 km/s

D. 22.4 km/s

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Correct Answer: Option D

Explanation:
Escape velocity v_e = √(2GM/R). Mass M = density ρ × volume = ρ×(4/3)πR³. Thus v_e = √(2Gρ×4/3πR³ / R) = √(8GρπR²/3) ∝ R√ρ. Given same density and R_planet = 2R_earth, v_e ∝ R, so v_e_planet = 2 × 11.2 = 22.4 km/s. This derivation shows escape velocity scales linearly with radius for constant density. Memory tip: v_e ∝ √(M/R) and M ∝ R³ for constant ρ, so v_e ∝ R. Such proportional reasoning questions test conceptual grasp of gravitation formulas in competitive exams without heavy calculation.

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Question #546

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The acceleration due to gravity on Earth's surface is g. At a height equal to Earth's radius above the surface, g' will be:

A. g/9

B. 2g

C. g/2

D. g/4

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Correct Answer: Option D

Explanation:
Acceleration due to gravity at height h: g' = gR²/(R+h)², where R is Earth's radius. Given h = R, so g' = gR²/(2R)² = gR²/4R² = g/4. This inverse square law dependence arises from Newton's law of gravitation. At depth, variation differs, but for height, this formula applies. Memory aid: Double the distance from center ⇒ gravity becomes one-fourth. This standard result is frequently tested in gravitation sections of competitive exams. Always verify whether height or depth is specified, as formulas differ significantly.

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Question #547

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An artificial satellite orbits Earth in a circular path. If its orbital radius is increased, its orbital velocity:

A. Remains constant

B. Decreases

C. Increases

D. Becomes zero

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Correct Answer: Option B

Explanation:
Orbital velocity v = √(GM/r), where r is orbital radius from Earth's center. Thus v ∝ 1/√r. If r increases, v decreases. This follows from equating gravitational force to centripetal force: GMm/r² = mv²/r ⇒ v = √(GM/r). Higher orbits have slower speeds but longer periods. Memory aid: 'Higher orbit, slower speed' – counterintuitive but fundamental. This relationship is crucial for satellite motion questions in competitive exams. Always distinguish orbital velocity from escape velocity (which is √2 times orbital velocity at same radius).

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Question #548

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The universal law of gravitation states that the force between two masses is directly proportional to:

A. Difference of the masses

B. Square of the distance between them

C. Sum of the masses

D. Product of the masses

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Correct Answer: Option D

Explanation:
Newton's law of gravitation: F = G·m₁m₂/r². Thus force is directly proportional to the product of the masses (m₁m₂) and inversely proportional to square of distance (r²). Option D incorrectly states 'directly proportional to square of distance' – it's inverse square. This fundamental law governs celestial mechanics. Memory tip: 'F ∝ m₁m₂ and F ∝ 1/r²'. Competitive exams often test precise wording of physical laws; distractors may reverse proportionality or misstate dependencies. Always recall the exact mathematical form.

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Question #549

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A body floats in a liquid with one-third of its volume above the surface. The density of the body relative to the liquid is:

A. 3/2

B. 2/3

C. 3

D. 1/3

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Correct Answer: Option B

Explanation:
By Archimedes' principle, weight of body = weight of displaced liquid. Let V be total volume, ρ_b body density, ρ_l liquid density. Volume submerged = (2/3)V. Thus ρ_b·V·g = ρ_l·(2V/3)·g ⇒ ρ_b/ρ_l = 2/3. Fraction submerged equals density ratio. Memory tip: 'Fraction submerged = ρ_object / ρ_fluid'. This direct application of floatation condition is common in competitive exams. Always identify submerged fraction correctly: here 'one-third above' means two-thirds submerged. Such problems test conceptual clarity in buoyancy applications.

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Question #550

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The pressure at a point in a fluid at rest depends on:

A. Depth below the free surface

B. Shape of the container

C. Volume of the fluid

D. Area of the surface

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Correct Answer: Option A

Explanation:
Hydrostatic pressure P = P₀ + ρgh, where h is depth below free surface, ρ fluid density, g gravity. Pressure depends only on depth, not container shape, surface area, or total volume (Pascal's paradox). This is a fundamental result in fluid statics. Memory aid: 'Pressure increases linearly with depth'. Competitive exams often test this independence from container geometry to assess deep understanding versus misconceptions. Always recall that pressure is scalar and acts equally in all directions at a point in static fluid.

This question belongs to: Science Physics

Question #551

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Atmospheric pressure at sea level is approximately:

A. 10³ Pa

B. 10⁵ Pa

C. 10² Pa

D. 10⁷ Pa

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Correct Answer: Option B

Explanation:
Standard atmospheric pressure at sea level is 101,325 Pa, approximately 10⁵ Pa (or 1 bar, 760 mm Hg). This value is crucial for pressure calculations in fluids, thermodynamics, and meteorology. Option B (1000 Pa) is too low (about 1% of atmospheric), C and D are orders of magnitude off. Memory tip: 'Atmospheric pressure ≈ 10⁵ Pa = 100 kPa = 1 bar'. This factual knowledge is frequently tested in competitive exams as a baseline for pressure-related problems. Always use SI units (Pascal) unless specified otherwise.

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Question #552

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A block of wood floats in water with 60% of its volume submerged. In a liquid of density 0.8 g/cm³, the fraction of volume submerged will be:

A. 80%

B. 48%

C. 75%

D. 60%

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Correct Answer: Option C

Explanation:
From floatation: ρ_wood/ρ_water = fraction submerged in water = 0.6. Thus ρ_wood = 0.6 g/cm³ (since ρ_water = 1 g/cm³). In new liquid (ρ_l = 0.8 g/cm³), fraction submerged f = ρ_wood/ρ_l = 0.6/0.8 = 0.75 = 75%. This uses the principle that fraction submerged equals density ratio. Memory aid: 'f = ρ_object / ρ_fluid'. Such comparative buoyancy problems test application of Archimedes' principle across scenarios, common in competitive exams. Always maintain consistent units (g/cm³ here simplifies calculation).

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Question #553

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The specific heat capacity of a substance is defined as the amount of heat required to:

A. Vaporize 1 kg of the substance at its boiling point

B. Raise the temperature of 1 kg of the substance by 1°C

C. Raise the temperature of 1 g of the substance by 1°C

D. Melt 1 kg of the substance at its melting point

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Correct Answer: Option B

Explanation:
Specific heat capacity c is defined as heat required to raise temperature of unit mass (1 kg in SI) by 1°C (or 1 K). Formula: Q = mcΔT. Option D describes specific heat in cgs units (cal/g°C), but SI definition uses kg. Options A and C describe latent heats. Memory aid: 'Specific heat = per kg per degree'. This precise definition is crucial for calorimetry problems. Competitive exams often test unit awareness (kg vs g) to distinguish careful students. Always note the mass unit in the definition context.

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Question #554

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During the melting of ice at 0°C, the temperature remains constant because the heat supplied is used to:

A. Increase kinetic energy of molecules

B. Increase potential energy only

C. Break intermolecular bonds

D. Both (b) and (c)

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Correct Answer: Option D

Explanation:
During phase change (melting), heat energy (latent heat) is used to overcome intermolecular forces, increasing potential energy while kinetic energy (hence temperature) remains constant. Temperature reflects average kinetic energy; since it doesn't change, kinetic energy is unchanged. The energy breaks hydrogen bonds in ice, allowing molecules to move freely in liquid. Memory tip: 'Latent heat changes state, not temperature'. This conceptual question tests understanding of microscopic interpretation of phase changes, frequently examined in competitive tests to assess depth beyond formula recall.

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Question #555

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A metal rod of length L at temperature T₀ is heated to temperature T. If α is the coefficient of linear expansion, the new length is:

A. Lα(T - T₀)

B. L(1 + αT)

C. L(1 + αT₀)

D. L[1 + α(T - T₀)]

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Correct Answer: Option D

Explanation:
Linear expansion formula: ΔL = L₀αΔT, where ΔT = T - T₀. Thus new length L' = L₀ + ΔL = L₀[1 + α(T - T₀)]. Option A incorrectly uses absolute temperature T instead of temperature change. Option C gives only the change, not total length. Option D uses initial temperature incorrectly. Memory aid: 'Expansion depends on temperature change, not absolute value'. This standard formula application is common in thermal physics questions. Always identify initial length and temperature reference point to avoid sign errors.

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Question #556

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The mode of heat transfer that can occur in vacuum is:

A. Convection

B. Radiation

C. Both conduction and convection

D. Conduction

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Correct Answer: Option B

Explanation:
Radiation transfers heat via electromagnetic waves and requires no medium, thus works in vacuum (e.g., solar energy reaching Earth). Conduction needs direct contact between particles; convection requires fluid movement. Vacuum lacks matter for conduction/convection. Memory tip: 'Radiation = only mode in vacuum'. This fundamental distinction is frequently tested in competitive exams. Applications include thermos flasks (minimizing all three modes) and space technology. Always recall that radiation speed is light speed, while conduction/convection are much slower.

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Question #557

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The quantity of heat required to convert 1 kg of water at 100°C to steam at 100°C is called:

A. Thermal conductivity

B. Latent heat of vaporization

C. Latent heat of fusion

D. Specific heat capacity

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Correct Answer: Option B

Explanation:
Latent heat of vaporization is the heat required to change unit mass from liquid to vapor at constant temperature (boiling point). For water, it's approximately 2260 kJ/kg. Latent heat of fusion applies to solid-liquid transition. Specific heat relates to temperature change without phase change. Thermal conductivity measures heat transfer rate through conduction. Memory aid: 'Vaporization = liquid to gas; fusion = solid to liquid'. This definition-based question tests precise terminology, crucial for thermodynamics sections in competitive exams where distractors often mix phase change terms.

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Question #558

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A gas expands at constant pressure. The work done by the gas is given by:

A. PΔV

B. P/V

C. VΔP

D. Δ(PV)

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Correct Answer: Option A

Explanation:
For constant pressure (isobaric) process, work done by gas W = PΔV, where ΔV is change in volume. This derives from mechanical work W = F·d and pressure P = F/A, so F = PA, and dV = A·d, thus W = P·dV. Option D Δ(PV) applies to other contexts like enthalpy. Memory tip: 'Isobaric work = P times volume change'. This fundamental thermodynamics result is frequently used in first law applications. Competitive exams often combine this with ideal gas law for numerical problems; always verify process constraints (constant P here).

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Question #559

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The speed of sound in air at 0°C is approximately 332 m/s. At 27°C, it will be approximately:

A. 360 m/s

B. 332 m/s

C. 350 m/s

D. 340 m/s

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Correct Answer: Option C

Explanation:
Speed of sound in air v ∝ √T, where T is absolute temperature in Kelvin. T₀ = 0°C = 273 K, v₀ = 332 m/s. T = 27°C = 300 K. Thus v = v₀√(T/T₀) = 332 × √(300/273) ≈ 332 × √1.099 ≈ 332 × 1.048 ≈ 348 m/s ≈ 350 m/s. Approximate rule: speed increases by 0.6 m/s per °C rise, so 27×0.6≈16.2, 332+16.2≈348.2 m/s. Memory aid: 'v ∝ √T in Kelvin'. Such temperature-dependence problems test application of sound wave properties, common in competitive exams with emphasis on absolute temperature usage.

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Question #560

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The phenomenon of echo is based on which property of sound?

A. Reflection

B. Refraction

C. Diffraction

D. Interference

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Correct Answer: Option A

Explanation:
Echo is the reflection of sound waves from a rigid surface, returning to the listener after a time delay. For distinct echo, minimum distance to reflector is about 17 m (so time delay > 0.1 s, persistence of hearing). Refraction bends sound in media with varying density; interference combines waves; diffraction bends around obstacles. Memory tip: 'Echo = sound reflection like light in mirror'. This application-based question links wave properties to daily-life phenomena, frequently tested in competitive exams to assess conceptual clarity beyond definitions. Always distinguish echo from reverberation (multiple reflections).

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