Physics MCQs

Question #561

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The frequency of a sound wave determines its:

A. Speed

B. Pitch

C. Quality

D. Loudness

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Correct Answer: Option B

Explanation:
Frequency determines pitch: higher frequency ⇒ higher pitch (e.g., whistle vs drum). Loudness depends on amplitude; quality (timbre) on waveform/harmonics; speed on medium properties (density, elasticity). This is a fundamental characteristic of sound perception. Memory aid: 'Frequency → Pitch; Amplitude → Loudness'. Competitive exams often test these distinctions to identify misconceptions. Note: Speed of sound is independent of frequency in a given medium (non-dispersive for audible sound in air). Always link physical quantities to perceptual attributes correctly.

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Question #562

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SONAR technology utilizes which type of waves?

A. Radio waves

B. Visible light

C. Infrasonic waves

D. Ultrasonic waves

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Correct Answer: Option D

Explanation:
SONAR (Sound Navigation and Ranging) uses ultrasonic waves (frequency > 20 kHz) because they have high directionality, short wavelength for better resolution, and minimal absorption in water. Radio waves attenuate rapidly in water; infrasonic waves have long wavelengths unsuitable for detection; light doesn't propagate far in water. Memory tip: 'Ultra = beyond human hearing; used in SONAR/ultrasound imaging'. This application question tests knowledge of wave frequency ranges and their practical uses, common in competitive exams linking physics to technology. Always recall frequency thresholds: infrasonic 20kHz.

This question belongs to: Science Physics

Question #563

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When a source of sound moves towards a stationary observer, the apparent frequency:

A. Increases

B. Remains same

C. Decreases

D. Becomes zero

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Correct Answer: Option A

Explanation:
This is the Doppler effect: when source moves toward observer, wavefronts compress, decreasing wavelength, thus increasing apparent frequency (f' = f·v/(v - v_s), where v is sound speed, v_s source speed). Conversely, receding source decreases frequency. This applies to all waves (light, sound). Memory aid: 'Approaching = higher pitch (frequency); Receding = lower pitch'. Competitive exams often test qualitative understanding before formula application. Always identify motion direction relative to observer for sign conventions in numerical problems.

This question belongs to: Science Physics

Question #564

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The focal length of a concave mirror is 20 cm. An object is placed at 30 cm from the mirror. The image distance is:

A. 20 cm

B. 30 cm

C. 12 cm

D. 60 cm

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Correct Answer: Option D

Explanation:
Mirror formula: 1/f = 1/u + 1/v. Sign convention: f negative for concave mirror? Wait, standard Cartesian: f = -20 cm (concave), u = -30 cm (object real). Thus 1/(-20) = 1/(-30) + 1/v ⇒ -1/20 = -1/30 + 1/v ⇒ 1/v = -1/20 + 1/30 = (-3+2)/60 = -1/60 ⇒ v = -60 cm. Magnitude 60 cm, negative sign indicates real image in front of mirror. Many exams use magnitude-only options; here 60 cm is correct. Memory tip: 'Concave mirror: object beyond C (2f) ⇒ image between F and C, real, inverted'. This numerical tests mirror formula application with sign conventions, crucial for optics in competitive exams.

This question belongs to: Science Physics

Question #565

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A ray of light passes from air to glass. Which quantity remains unchanged?

A. Wavelength

B. Amplitude

C. Speed

D. Frequency

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Correct Answer: Option D

Explanation:
When light crosses media, frequency remains constant as it's determined by the source. Speed and wavelength change proportionally: v = fλ, and v = c/n, so λ decreases in denser medium (glass). Amplitude may change due to reflection/transmission coefficients. Memory aid: 'Frequency is source-dependent; speed and wavelength are medium-dependent'. This fundamental wave property is frequently tested in refraction questions. Competitive exams often use this to assess understanding of wave behavior at interfaces. Always recall that color (frequency) doesn't change during refraction, though speed and wavelength do.

This question belongs to: Science Physics

Question #566

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The power of a lens is -2 D. This lens is:

A. Concave lens of focal length 50 cm

B. Convex lens of focal length 200 cm

C. Convex lens of focal length 50 cm

D. Concave lens of focal length 200 cm

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Correct Answer: Option A

Explanation:
Power P = 1/f (in meters). P = -2 D ⇒ f = 1/(-2) = -0.5 m = -50 cm. Negative focal length indicates diverging (concave) lens. Convex lenses have positive power. Option A correctly states concave lens with |f| = 50 cm. Memory tip: 'Negative power = concave lens; f(in cm) = 100/P(in D)'. This direct application of lens power definition is common in competitive exams. Always convert diopters to focal length carefully, noting sign convention: negative for diverging lenses, positive for converging.

This question belongs to: Science Physics

Question #567

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A person cannot see objects clearly beyond 2 m. This defect of vision is:

A. Astigmatism

B. Presbyopia

C. Myopia

D. Hypermetropia

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Correct Answer: Option C

Explanation:
Myopia (nearsightedness) is the inability to see distant objects clearly, with far point less than infinity. Here, far point is 2 m. Corrected using concave lens. Hypermetropia (farsightedness) affects near vision; presbyopia is age-related loss of accommodation; astigmatism involves uneven corneal curvature. Memory aid: 'Myopia = near-sighted; far point reduced'. This application question tests knowledge of vision defects and corrections, frequently appearing in competitive exams. Always link symptom (blurred distant vision) to defect (myopia) and correction (diverging lens).

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Question #568

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When white light passes through a prism, it disperses into colors because:

A. Total internal reflection occurs

B. Prism absorbs some colors

C. Different colors have different frequencies

D. Different colors have different speeds in glass

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Correct Answer: Option D

Explanation:
Dispersion occurs because refractive index of glass varies with wavelength (color): n ∝ 1/λ approximately. Since v = c/n, different colors travel at different speeds in glass, causing different refraction angles (Snell's law). Frequency remains constant across media; wavelength changes. Option B is true but not the direct cause; speed difference causes angle difference. Memory tip: 'VIBGYOR: Violet bends most (slowest in glass), red least'. This conceptual question tests understanding of dispersion mechanism, crucial for optics in competitive exams. Always distinguish cause (speed variation) from property (frequency/wavelength).

This question belongs to: Science Physics

Question #569

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The refractive index of water is 4/3. The speed of light in water is:

A. 3×10⁸ m/s

B. 1.5×10⁸ m/s

C. 4×10⁸ m/s

D. 2.25×10⁸ m/s

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Correct Answer: Option D

Explanation:
Refractive index n = c/v, where c = 3×10⁸ m/s (speed in vacuum). Thus v = c/n = (3×10⁸)/(4/3) = 3×10⁸ × 3/4 = 9/4 × 10⁸ = 2.25×10⁸ m/s. This direct formula application is fundamental in optics. Memory aid: 'Higher n ⇒ slower light in medium'. Competitive exams frequently test this calculation with common refractive indices (water 4/3, glass ~1.5). Always use c = 3×10⁸ m/s unless specified otherwise, and ensure unit consistency (m/s here).

This question belongs to: Science Physics

Question #570

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Laws of reflection state that:

A. Light bends towards normal

B. Incident ray, reflected ray, and normal lie in same plane

C. Angle of incidence equals angle of reflection

D. Both (a) and (b)

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Correct Answer: Option D

Explanation:
Laws of reflection: (1) Angle of incidence (i) equals angle of reflection (r); (2) Incident ray, reflected ray, and normal at point of incidence all lie in the same plane. Option D describes refraction, not reflection. These laws hold for all reflecting surfaces (plane, curved). Memory tip: 'i = r and coplanar rays'. This definition-based question tests precise recall of fundamental optics laws, frequently examined in competitive exams. Always distinguish reflection laws (i=r) from refraction laws (Snell's law: n₁sin i = n₂sin r).

This question belongs to: Science Physics

Question #571

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A convex lens forms a real image of an object. If the lower half of the lens is covered, the image will:

A. Be half-formed

B. Be fainter but complete

C. Become virtual

D. Disappear completely

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Correct Answer: Option B

Explanation:
Covering part of a lens reduces the amount of light reaching the image, making it fainter, but the entire image is still formed because light from each object point can pass through the uncovered portion and converge to the corresponding image point. Image position, size, and nature remain unchanged. Memory aid: 'Partial lens coverage ⇒ reduced intensity, not partial image'. This conceptual question tests understanding of image formation principles, commonly featured in competitive exams to identify misconceptions about ray optics. Always recall that every point on lens contributes to entire image.

This question belongs to: Science Physics

Question #572

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The resistance of a wire is R. If its length is doubled and cross-sectional area halved, the new resistance is:

A. R/4

B. 2R

C. 4R

D. R

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Correct Answer: Option C

Explanation:
Resistance R = ρL/A, where ρ is resistivity (material property). New length L' = 2L, new area A' = A/2. Thus R' = ρ(2L)/(A/2) = ρ·2L·2/A = 4(ρL/A) = 4R. Resistance scales directly with length and inversely with area. Memory tip: 'Double length ⇒ double R; halve area ⇒ double R; combined ⇒ 4×'. This proportional reasoning problem is frequent in electricity sections of competitive exams. Always verify if resistivity changes (it doesn't here, same material).

This question belongs to: Science Physics

Question #573

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Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:

A. 6Ω

B. 0.5Ω

C. 1Ω

D. 11Ω

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Correct Answer: Option C

Explanation:
For parallel combination: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1. Thus R_eq = 1Ω. Note: Equivalent resistance in parallel is always less than the smallest individual resistance (here 2Ω). Memory tip: 'Parallel: reciprocal sum; result < min(R)'. This standard calculation tests parallel resistance formula application, frequently appearing in competitive exams. Always simplify fractions carefully; common denominator method avoids errors. Verify: 1Ω < 2Ω, consistent with parallel behavior.

This question belongs to: Science Physics

Question #574

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The heating effect of electric current is described by:

A. Ohm's law

B. Joule's law

C. Lenz's law

D. Faraday's law

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Correct Answer: Option B

Explanation:
Joule's law states that heat produced H = I²Rt, where I is current, R resistance, t time. This quantifies electrical energy converted to thermal energy in resistors. Ohm's law relates V, I, R; Faraday's and Lenz's laws concern electromagnetic induction. Memory aid: 'Joule heating ∝ I²R'. This definition-based question tests knowledge of fundamental laws in electricity, commonly examined in competitive exams. Applications include electric heaters, fuses, and incandescent bulbs. Always distinguish Joule's law (heating) from power formulas (P=VI=I²R=V²/R).

This question belongs to: Science Physics

Question #575

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A 100 W bulb operates at 220 V. The current drawn by it is approximately:

A. 0.22 A

B. 4.5 A

C. 2.2 A

D. 0.45 A

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Correct Answer: Option D

Explanation:
Electric power P = VI. Thus I = P/V = 100 W / 220 V ≈ 0.4545 A ≈ 0.45 A. This direct application of power formula is fundamental in electricity. Memory tip: 'I = P/V for resistive loads'. Competitive exams frequently test such calculations with household appliance ratings. Always use consistent units (watts, volts, amperes). Note: This assumes purely resistive load (valid for incandescent bulbs); for motors or electronics, power factor may matter, but not at this level.

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Question #576

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In a series circuit, which quantity remains the same across all components?

A. Resistance

B. Voltage

C. Current

D. Power

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Correct Answer: Option C

Explanation:
In series circuits, current is the same through all components because there's only one path for charge flow. Voltage divides across components proportional to resistance (V = IR). Resistance and power vary per component. Memory aid: 'Series: same current; Parallel: same voltage'. This fundamental circuit property is frequently tested in competitive exams to assess basic electronics understanding. Always apply Kirchhoff's current law: current entering a junction equals current leaving; in series, no junctions, so current constant.

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Question #577

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The magnetic field inside a long straight solenoid carrying current is:

A. Radially outward

B. Non-uniform and circular

C. Uniform and parallel to axis

D. Zero

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Correct Answer: Option C

Explanation:
Inside a long solenoid, magnetic field is uniform (constant magnitude and direction), parallel to the axis, given by B = μ₀nI, where n is turns per unit length. Outside, field is nearly zero. This uniformity makes solenoids useful for creating controlled magnetic fields. Memory tip: 'Solenoid: uniform B inside like bar magnet'. This conceptual question tests knowledge of magnetic field configurations, crucial for electromagnetism in competitive exams. Always distinguish solenoid field (uniform inside) from straight wire field (circular, non-uniform).

This question belongs to: Science Physics

Question #578

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The working principle of an electric generator is:

A. Magnetic effect of current

B. Chemical effect of current

C. Electromagnetic induction

D. Heating effect of current

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Correct Answer: Option C

Explanation:
Electric generators convert mechanical energy to electrical energy using electromagnetic induction: rotating a coil in magnetic field (or vice versa) induces EMF due to changing magnetic flux. This is Faraday's law application. Motors use the reverse principle (force on current-carrying conductor in field). Memory tip: 'Generator: motion ⇒ electricity (induction); Motor: electricity ⇒ motion (force)'. This application question tests understanding of device principles, common in competitive exams. Always distinguish generator (induction) from motor (Lorentz force) operation.

This question belongs to: Science Physics

Question #579

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A transformer has 100 turns in primary and 200 turns in secondary. If primary voltage is 120 V, the secondary voltage is:

A. 120 V

B. 240 V

C. 60 V

D. 480 V

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Correct Answer: Option B

Explanation:
Transformer voltage ratio: V_s/V_p = N_s/N_p, where N is number of turns. Thus V_s = V_p × (N_s/N_p) = 120 V × (200/100) = 240 V. This assumes ideal transformer (no losses). Step-up transformer (N_s > N_p) increases voltage. Memory aid: 'Turns ratio = voltage ratio'. This direct formula application is frequent in competitive exams. Always verify if transformer is step-up or step-down from turn counts. Note: Current ratio is inverse: I_s/I_p = N_p/N_s for power conservation (V_pI_p = V_sI_s).

This question belongs to: Science Physics

Question #580

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The magnetic field lines around a straight current-carrying conductor are:

A. Radial and outward

B. Straight and parallel to conductor

C. Circular and concentric around conductor

D. Elliptical

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Correct Answer: Option C

Explanation:
Magnetic field lines form concentric circles around a straight current-carrying wire, with direction given by right-hand thumb rule (thumb in current direction, fingers curl in field direction). This is derived from Biot-Savart law or Ampère's circuital law. Memory tip: 'Right-hand rule: thumb = current, fingers = field circles'. This fundamental field pattern is frequently tested in competitive exams. Always visualize field direction: for current toward you, field lines are counterclockwise. Distinguish from solenoid field (axial inside) or bar magnet (dipole pattern).

This question belongs to: Science Physics

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