Physics MCQs

Question #961

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Object floats with 40% volume submerged. Density ratio (object/liquid) is:

A. 0.4

B. 1.4

C. 0.6

D. 2.5

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Correct Answer: Option A

Explanation:
Floatation: weight = buoyant force ⇒ ρ_obj·V·g = ρ_liq·V_sub·g ⇒ ρ_obj/ρ_liq = V_sub/V = 0.4. Fraction submerged equals density ratio. Memory tip: 'Submerged fraction = ρ_object/ρ_fluid'. Direct Archimedes' principle application frequently tested in buoyancy sections of competitive exams.

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Question #962

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Temperature where Fahrenheit reading is double Celsius reading:

A. 40°C

B. 320°C

C. 160°C

D. 80°C

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Correct Answer: Option C

Explanation:
F = (9/5)C + 32. Given F = 2C: 2C = 1.8C + 32 ⇒ 0.2C = 32 ⇒ C = 160°. Verify: F = 2×160 = 320°; formula: (9/5)×160+32 = 288+32 = 320°. Memory aid: Solve linear equation for temperature conversion problems. Algebraic manipulation of scale relations frequently tested in competitive exams.

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Question #963

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Heat required to raise 2 kg water from 20°C to 100°C: (c_water = 4200 J/kg·K)

A. 336 kJ

B. 840 kJ

C. 168 kJ

D. 672 kJ

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Correct Answer: Option D

Explanation:
Q = mcΔT = 2×4200×(100-20) = 2×4200×80 = 672000 J = 672 kJ. Direct calorimetry formula application. Memory tip: 'Q = mcΔT; ΔT in Celsius or Kelvin same'. Basic numerical testing specific heat concept, frequently appearing in thermodynamics sections of competitive exams.

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Question #964

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During boiling, temperature remains constant because heat is used to:

A. Break intermolecular bonds

B. Expand volume

C. Increase kinetic energy

D. Increase pressure

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Correct Answer: Option A

Explanation:
At boiling point, supplied heat (latent heat) breaks intermolecular bonds for phase change, not increasing kinetic energy (temperature). Potential energy increases while kinetic energy (temperature) stays constant. Memory aid: 'Latent heat changes state, not temperature'. Conceptual question testing phase change understanding, common in competitive exams to assess microscopic interpretation.

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Question #965

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Linear expansion coefficient α = 2×10⁻⁵/°C. Length increase for 1 m rod heated 50°C:

A. 1 mm

B. 0.1 mm

C. 10 mm

D. 0.01 mm

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Correct Answer: Option A

Explanation:
ΔL = L₀αΔT = 1×2×10⁻⁵×50 = 100×10⁻⁵ = 10⁻³ m = 1 mm. Direct expansion formula application. Memory tip: 'ΔL = LαΔT; convert result to required units'. Tests thermal expansion calculation with unit conversion, frequently appearing in competitive exams.

This question belongs to: Science Physics

Question #966

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Heat transfer requiring medium and bulk motion is:

A. Radiation

B. Conduction

C. Convection

D. All three

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Correct Answer: Option C

Explanation:
Convection involves fluid motion carrying heat. Conduction needs medium but no bulk motion; radiation needs no medium. Memory aid: 'Convection = fluid flow + heat transfer; e.g., boiling water, weather'. Definition-based question testing heat transfer modes, frequently examined in competitive exams to distinguish mechanisms.

This question belongs to: Science Physics

Question #967

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Sound cannot travel through:

A. Vacuum

B. Gases

C. Liquids

D. Solids

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Correct Answer: Option A

Explanation:
Sound requires material medium for mechanical wave propagation. Vacuum lacks particles to transmit vibrations. Light (EM wave) travels through vacuum. Memory tip: 'Sound = mechanical wave ⇒ needs medium; light = EM wave ⇒ no medium needed'. Fundamental wave property frequently tested in competitive exams to assess medium-dependence understanding.

This question belongs to: Science Physics

Question #968

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Frequency of sound wave determines:

A. Pitch

B. Speed

C. Loudness

D. Timbre

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Correct Answer: Option A

Explanation:
Frequency → pitch perception: higher frequency = higher pitch. Loudness depends on amplitude; timbre on waveform; speed on medium properties. Memory aid: 'Frequency = pitch; Amplitude = loudness'. Basic sound property question frequently tested to verify perceptual attribute associations in competitive exams.

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Question #969

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Echo minimum distance for distinct hearing (speed sound = 340 m/s, persistence = 0.1 s):

A. 34 m

B. 17 m

C. 8.5 m

D. 68 m

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Correct Answer: Option B

Explanation:
Time for to-and-fro: t = 2d/v ≥ 0.1 s ⇒ d ≥ vt/2 = 340×0.1/2 = 17 m. Minimum distance for brain to distinguish echo from original sound. Memory tip: 'd_min = v×persistence/2'. Application of sound reflection with human perception limit, common in competitive exam numericals.

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Question #970

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SONAR uses waves of frequency:

A. > 20 kHz

B. Any frequency

C. 20 Hz - 20 kHz

D. < 20 Hz

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Correct Answer: Option A

Explanation:
SONAR uses ultrasonic waves (>20 kHz) for better directionality, resolution, and reduced absorption in water. Audible sound (20Hz-20kHz) diffracts more; infrasonic (

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Question #971

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Doppler effect for sound depends on:

A. Medium motion only

B. Relative motion between source and observer

C. Observer motion only

D. Source motion only

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Correct Answer: Option B

Explanation:
Apparent frequency change depends on relative velocity between source and observer with respect to medium. Both motions contribute: f' = f(v±v_o)/(v∓v_s). Memory tip: 'Approaching ⇒ higher pitch; receding ⇒ lower pitch'. Conceptual question testing Doppler effect fundamentals, frequently examined in competitive exams.

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Question #972

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Concave mirror forms real image when object is placed:

A. Beyond focus

B. Between pole and focus

C. At pole

D. At focus

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Correct Answer: Option A

Explanation:
Concave mirror: real, inverted image forms when object beyond focal point (u > f). Between pole and focus: virtual, erect image. At focus: image at infinity. Memory aid: 'Concave: object beyond F ⇒ real image; within F ⇒ virtual'. Ray optics concept frequently tested to verify mirror image formation understanding in competitive exams.

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Question #973

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Refractive index n = c/v. For glass n=1.5, light speed is:

A. 1.5×10⁸ m/s

B. 2×10⁸ m/s

C. 4.5×10⁸ m/s

D. 3×10⁸ m/s

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Correct Answer: Option B

Explanation:
v = c/n = 3×10⁸/1.5 = 2×10⁸ m/s. Direct refractive index formula application. Memory tip: 'Higher n ⇒ slower light; v = c/n'. Basic optics calculation frequently tested in competitive exams to verify formula recall and unit handling.

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Question #974

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Lens power P = -4 D. Focal length and type:

A. 25 cm, concave

B. 25 cm, convex

C. 40 cm, convex

D. 40 cm, concave

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Correct Answer: Option A

Explanation:
P = 1/f(m) ⇒ f = 1/P = 1/(-4) = -0.25 m = -25 cm. Negative focal length indicates diverging (concave) lens. Memory aid: 'Negative power = concave lens; f(cm) = 100/P(D)'. Direct lens power application frequently tested in competitive exams with sign convention awareness.

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Question #975

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Myopia correction uses:

A. Bifocal lens

B. Convex lens

C. Concave lens

D. Cylindrical lens

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Correct Answer: Option C

Explanation:
Myopia (nearsightedness): far point reduced; corrected by concave (diverging) lens to shift image to retina. Hypermetropia uses convex lens. Memory tip: 'Myopia = near-sighted ⇒ concave lens; Hyper = far-sighted ⇒ convex'. Vision defect correction frequently tested in competitive exams to assess application knowledge.

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Question #976

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Prism dispersion occurs because:

A. Total internal reflection

B. Refractive index depends on wavelength

C. Different colors have different frequencies

D. Prism absorbs colors selectively

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Correct Answer: Option B

Explanation:
Dispersion: n varies with λ (Cauchy's law), causing different refraction angles for colors. Frequency constant across media; wavelength changes. Memory aid: 'n(λ) ⇒ v(λ) ⇒ different bending; violet bends most'. Conceptual question testing dispersion mechanism, frequently examined in competitive optics sections.

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Question #977

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Rainbow formation involves:

A. Refraction only

B. Reflection only

C. Diffraction

D. Dispersion and total internal reflection

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Correct Answer: Option D

Explanation:
Rainbow: sunlight enters raindrop (refraction + dispersion), reflects internally (TIR), exits with further refraction. Primary rainbow: one TIR; secondary: two TIRs. Memory tip: 'Rainbow = refraction + dispersion + TIR in water droplets'. Application question testing atmospheric optics, frequently appearing in competitive exams.

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Question #978

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Resistance of wire: R = ρL/A. If length tripled and area doubled, new R is:

A. 1.5R

B. 6R

C. 3R

D. R/6

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Correct Answer: Option A

Explanation:
R' = ρ(3L)/(2A) = (3/2)(ρL/A) = 1.5R. Resistance scales directly with length, inversely with area. Memory aid: 'R ∝ L/A; changes multiply'. Proportional reasoning problem testing resistance formula, frequently appearing in competitive electricity sections.

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Question #979

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Ohm's law V = IR holds for:

A. Ohmic conductors at constant temperature

B. Insulators

C. Semiconductors only

D. All materials

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Correct Answer: Option A

Explanation:
Ohm's law applies to ohmic materials (metals) at constant temperature. Non-ohmic devices (diodes, filaments) have nonlinear V-I characteristics. Memory tip: 'Ohmic = linear V-I graph through origin; temperature must be constant'. Conceptual question testing Ohm's law limitations, frequently examined in competitive exams.

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Question #980

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Three 3Ω resistors in parallel: equivalent resistance is:

A. 9Ω

B. 0.33Ω

C. 1Ω

D. 3Ω

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Correct Answer: Option C

Explanation:
1/R_eq = 1/3 + 1/3 + 1/3 = 1 ⇒ R_eq = 1Ω. For n equal resistors in parallel: R_eq = R/n. Memory tip: 'Parallel: reciprocal sum; result < smallest resistor'. Basic circuit calculation frequently tested in competitive exams to verify parallel combination formula.

This question belongs to: Science Physics

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