Wave Motion is part of Physics. Below are notes on 12 sub-topics, followed by practice MCQs from State PSC and UPSC papers.

Diffraction of Waves (Qualitative Aspects)

  • Diffraction: Bending of waves around obstacles or spreading through an aperture.
  • Significant diffraction occurs when the wavelength (λ) of the wave is comparable to or greater than the size of the obstacle/ aperture (d).
  • Longer wavelengths diffract more easily than shorter wavelengths.
  • Sound waves (longer λ) diffract more easily around corners than light waves (shorter λ), explaining why sound can be heard but light cannot be seen around obstacles.
  • X-rays exhibit phenomena like diffraction, polarization, and interference. Being uncharged electromagnetic waves, X-rays are not deflected by electric fields.
  • The wave-like character of an electron is proved by its diffraction pattern when scattered from a crystalline solid.
  • Television signals via dish antenna (radio waves) are lost during rain because rain droplets absorb and disperse (scatter/diffract) the energy of radio waves.

Doppler Effect for Sound and Light

  • The Doppler effect is the apparent change in frequency and wavelength of a wave as perceived by an observer moving relative to the wave's source.
  • For sound waves, the Doppler effect is asymmetric; the perceived frequency change depends on whether the source or the observer is moving relative to the medium.
  • When a sound source approaches an observer, the apparent frequency increases (higher pitch); when it recedes, the apparent frequency decreases (lower pitch).
  • For light (electromagnetic waves), the Doppler effect is symmetric; the perceived frequency change depends only on the relative velocity between the source and the observer, not on their individual velocities relative to a medium.
  • When a light source approaches an observer, the apparent frequency increases (blueshift); when it recedes, the apparent frequency decreases (redshift).
  • Applications of Doppler Effect:
  • Police radar guns use the Doppler effect of reflected short radio waves to detect the speed of over-speeding vehicles.
  • Used in radar systems for aircraft guidance and tracking.
  • Used in SONAR for ocean research (e.g., measuring depth, detecting submarines).
  • Used in medical diagnostics, such as ultrasound to monitor blood flow, fetal heartbeats, and in echocardiography.
  • Used in astronomy to measure the speed of stars and galaxies (redshift/blueshift).

Electromagnetic

Waves:

Properties,

Electromagnetic Spectrum, Energy and

Momentum

  • Electromagnetic (EM) waves are transverse waves, consisting of mutually perpendicular oscillating electric ($\vec{E}$) and magnetic ($\vec{B}$) fields, both perpendicular to the direction of propagation.
  • EM waves do not require a medium and can propagate through a vacuum.
  • The speed of all EM waves in vacuum is a constant, $c = 3 \times 10^8$ ms$^{-1}$.
  • EM waves transport energy and momentum; photons (quanta of EM waves) have zero rest mass.
  • EM waves can be polarized and reflected.
  • The relation between speed ($c$), frequency ($\nu$), and wavelength ($\lambda$) is $c = \nu \lambda$.
  • Energy ($E$) of an EM wave photon is directly proportional to its frequency ($E = h\nu$) and inversely proportional to its wavelength ($E = hc/\lambda$). Higher frequency means higher energy per photon.
  • For an EM wave propagating in direction $ \hat{k}{prop}$, with electric field $\vec{E}$, the magnetic field $\vec{B}$ direction is given by $\hat{k}{prop} \times \vec{E}$.
  • If an EM wave and an ultrasonic wave have the same frequency, the EM wave will have a significantly longer wavelength because its speed is much greater than the speed of sound.
  • Electromagnetic Spectrum (Order of decreasing wavelength / increasing frequency / increasing energy): Radio Waves (longest wavelength, 1. lowest frequency/energy) Microwaves 2. Infrared (IR) 3. Visible Light 4. Ultraviolet (UV) 5. X-rays 6. Gamma Rays (shortest wavelength, 7. highest frequency/energy)
  • Specific Wavelength Relationships: $ \lambda{Gamma} < \lambda{X-ray} < \lambda{UV} < \lambda{Visible} < \lambda{IR} < \lambda{Microwave} < \lambda{Radio}$. Example: $\lambda{UV} < \lambda{VIS} < \lambda{IR}$.
  • Applications and Characteristics of EM Spectrum components:
    • Radio Waves:
      • Used for communication (radio, television).
      • AM (Amplitude Modulation) radio waves generally have longer wavelengths and lower frequency ranges than FM (Frequency Modulation) radio waves.
      • FM broadcasting frequency range: 88-108 MHz. Invented by Edwin Armstrong. In FM, amplitude and phase remain constant, while frequency changes. FM has a higher bandwidth than AM.
      • Television signals are Amplitude Modulated. Reception is limited by Earth's curvature.
      • Ultra High Frequency (UHF) radio waves typically propagate via space waves.
    • Microwaves:
      • Frequency range: 1 GHz – 300 GHz.
      • Used in microwave ovens to cook food by causing water molecules to vibrate, producing heat. Magnetron tubes are used. Microwaves are not ionizing radiation.
      • Also used in radar systems.
    • Infrared (IR) Radiation:
      • Also known as thermal radiation.
      • Used in TV remote controls.
      • Used for treating muscle ache.
      • Used in night vision cameras.
    • Visible Light:
      • The portion of the spectrum detectable by the human eye, allowing us to see.
    • Ultraviolet (UV) Radiation:
      • Shorter wavelength than visible light.
      • Used for detecting forgery in currency notes.
    • X-rays:
      • Wavelengths typically around $10^{-10}$ meters (1 Å).
      • Generated by bombarding a metal target with high-energy electrons.
      • Uses: Inspecting welded joints, studying crystal structures, scanning bags in airports, medical imaging (CT scans, detecting bone fractures), treatment of certain cancers.
      • The cutoff wavelength in an Xray tube is inversely proportional to the applied potential difference.
      • Not suitable for radar systems due to very short wavelengths.
    • Gamma Rays:
      • Shortest wavelength and highest energy per photon.
      • Used in radiotherapy for the treatment of cancer and tumors.
      • Emitted during phenomena like solar flares.
  • Gravitational Waves:
    • Predicted by Albert Einstein in 1916 based on his theory of General Relativity; they do not exist in Newtonian theory.
    • Detected by sensitive interferometers (e.g., LIGO).
    • Characterized by long wavelength and low energy, making them difficult to detect.

Energy and Power Transmission by Waves: Intensity

  • Waves transmit energy through a medium, but not the material of the medium itself; particles oscillate locally.
  • Intensity (I) of a wave is the average power (P) transmitted per unit area (A) perpendicular to the direction of wave propagation: $I = P/A$.
  • SI unit for intensity is Watts per square meter (W/m²).
  • For most waves, intensity is directly proportional to the square of the wave's amplitude ($I \propto A^2$).
  • The average power of a sinusoidal voltage/ current signal, $x(t) = A \cos(\omega t + \phi)$, is given by $P{avg} = A^2 / (2R)$, where A is the peak amplitude and R is the resistance.
  • For angle-modulated signals (e.g., FM, PM), the amplitude of the carrier remains constant. The average power of such a signal, $x(t) = A \cos(\omegac t + \phi(t))$, is $P{avg} = A^2 / 2$ (assuming a 1 Ohm resistance, as is common when resistance is not specified for power calculations).

General Wave Equation (Harmonic Waves)

  • Beta Function ($B(m,n)$): Defined as $ \int0^1 x^{m-1}(1-x)^{n-1}dx$.
  • Gamma Function ($\Gamma(z)$): Defined as $\int0^\infty t^{z-1}e^{-t}dt$. Properties include $\Gamma(z+1)=z\Gamma(z)$ and $ \Gamma(1/2)=\sqrt{\pi}$.
  • Relation between Beta and Gamma: $B(m,n) = \frac{\Gamma(m)\Gamma(n)} {\Gamma(m+n)}$.
  • Integral Transformation for $\int0^1 \frac{dx}{\sqrt{1-x^4}}$: Substitute $x^4=u \implies 4x^3dx=du \implies dx=\frac{1}{4} u^{-3/4}du$. The integral becomes $\frac{1} {4}\int0^1 u^{-3/4}(1-u)^{-1/2}du = \frac{1} {4}B(1/4, 1/2) = \frac{1} {4}\frac{\Gamma(1/4)\Gamma(1/2)} {\Gamma(1/4+1/2)} = \frac{1} {4}\frac{\Gamma(1/4)\sqrt{\pi}} {\Gamma(3/4)}$.
  • Reduction of Order (2nd Order Linear ODE): For $y'' + P(x)y' + Q(x)y = 0$, if $y1(x)$ is a known solution, the second linearly independent solution is $y2(x) = y1(x) \int \frac{e^{-\int P(x)dx}}{[y1(x)]^2} dx$. For $x^2 y'' + xy' - y = 0$, divide by $x^2$ to get $y'' + \frac{1}{x}y' - \frac{1} {x^2}y = 0$. Here $P(x) = \frac{1}{x}$.
  • Matrix Inverse: If $B=A^{-1}$, then $AB=I$ (identity matrix) and $BA=I$. Alternatively, $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, where $\text{adj}(A)$ is the adjoint matrix (transpose of the cofactor matrix). Cofactor $C{ij} = (-1)^{i+j}M{ij}$, where $M{ij}$ is the minor. To find $\alpha$ from $10B = \text{adj}(A)$ given $B=A^{-1}$, compute $ \det(A)$ and then an element of $\text{adj} (A)$ to match $\alpha$. For example, $ (\text{adj}(A)){23} = C{32}$.
  • Envelope of a Family of Curves: For a family $f(x, y, m) = 0$ (where $m$ is the parameter), the envelope is found by eliminating $m$ from $f(x, y, m) = 0$ and $\frac{\partial f} {\partial m} = 0$. For a quadratic equation in $m$, like $Am^2 + Bm + C = 0$, the envelope is given by its discriminant $B^2 4AC = 0$. For $y = mx + a/m$, rewrite as $m^2x - my + a = 0$. The envelope is $(-y)^2 - 4(x)(a) = 0 \implies y^2 = 4ax$.
  • Lagrange's Auxiliary Equations (PDE): For a quasi-linear PDE of the form $Pp + Qq = R$ ($p=\partial z/\partial x, q=\partial z/ \partial y$), the auxiliary equations are $ \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$. Solve these equations to find two independent integrals $u(x,y,z)=c1$ and $v(x,y,z)=c2$. The general solution is $\Phi(u,v)=0$ or $u=\Phi(v)$. Methods for solving auxiliary equations include grouping variables or using multipliers.
  • Solving 1st Order ODE $x + yp^2 = (1 + xy)p$: Rearrange as a quadratic in $p$: $yp^2 - (1+xy)p + x = 0$. Use the quadratic formula for $p$: $p = \frac{-(1+xy) \pm \sqrt{(1+xy)^2 - 4y(x)}}{2y} = \frac{1+xy \pm \sqrt{1-2xy+x^2y^2}}{2y} = \frac{1+xy \pm (1-xy)}{2y}$. This yields two cases: $p = 1/y$ and $p = x$. Integrate each separately. If $p = 1/y \implies y \frac{dy}{dx} = 1 \implies \int y dy = \int dx \implies y^2/2 = x + C1 \implies y^2 - 2x - C = 0$. If $p = x \implies \frac{dy}{dx} = x \implies \int dy = \int x dx \implies y = x^2/2 + C2$.
  • Triple Integrals: Evaluate from the innermost integral outwards. $\int{a}^{b} \int{c(x)} ^{d(x)} \int{e(x,y)}^{f(x,y)} g(x,y,z) dz dy dx$.
  • Fourier Sine Transform: For a function $f(x) $, the Fourier Sine Transform is $\mathcal{F}s\{f(x)\} = \sqrt{\frac{2}{\pi}} \int0^\infty f(x) \sin(sx) dx$. Standard integral result: $\int e^{Ax} \sin(Bx) dx = \frac{e^{Ax}} {A^2+B^2}(A\sin(Bx) - B\cos(Bx))$. For $f(x)=e^{-ax}$, $A=-a, B=s$. Evaluating from $0$ to $\infty$ gives $\frac{s} {a^2+s^2}$. So, $\mathcal{F}s\{e^{-ax}\} = \sqrt{\frac{2}{\pi}} \frac{s}{a^2+s^2}$.
  • Legendre Polynomials ($Pn(x)$): $P0(x) = 1$ $P1(x) = x$ $P2(x) = \frac{1}{2}(3x^2 1)$ To express $x^2+x$ in terms of $Pn(x)$: From $P2(x)$, we have $3x^2 - 1 = 2P2(x) \implies x^2 = \frac{2}{3}P2(x) + \frac{1} {3}$. So, $x^2+x = \left(\frac{2}{3}P2(x) + \frac{1}{3}P0(x)\right) + P1(x) = \frac{1}{3} P0(x) + P1(x) + \frac{2}{3}P2(x)$.
  • Bessel Functions of the First Kind ($Jn(x)$): The expansion of $\cos x$ in terms of Bessel functions is given by: $\cos x = J0(x) - 2J2(x) + 2J4(x) - 2J6(x) + \dots + 2(-1)^k J{2k}(x) + \dots$.

Polarization of Waves: Plane, Circular, Elliptical Polarization; Malus's Law, Brewster's Law

  • Polarization: Phenomenon of restricting the vibrations of a transverse wave to a single plane or specific pattern.
  • Only transverse waves (e.g., light waves) exhibit polarization; longitudinal waves (e.g., sound waves) cannot be polarized.
  • Unpolarized Light: Vibrates in all possible planes perpendicular to the direction of propagation.
  • Polarized Light: Vibrations are confined to a specific pattern.
  • Types of Polarization:
    • Plane (Linear) Polarization: Electric field vector oscillates along a single line perpendicular to the propagation direction.
    • Circular Polarization: Electric field vector rotates in a circle in a plane perpendicular to propagation, maintaining constant magnitude.
    • Elliptical Polarization: Electric field vector rotates in an ellipse in a plane perpendicular to propagation, with varying magnitude. (General case, includes linear and circular).
  • Malus's Law: Describes the intensity of plane-polarized light transmitted through an analyzer.
  • Formula (Malus's Law): `I = I₀ cos²θ`, where `I` is the transmitted intensity, `I₀` is the incident plane-polarized light intensity, and `θ` is the angle between the transmission axes of the polarizer and the analyzer.
  • Brewster's Law (Law of Polarization by Reflection): When unpolarized light is incident at a specific angle (Brewster's angle, `θB` or `ip`) on a dielectric surface, the reflected light is completely plane-polarized.
  • At Brewster's angle (`θB`), the reflected ray and the refracted ray are perpendicular to each other (`90°`).
  • Formula (Brewster's Law): `tan(θB) = n`, where `n` is the refractive index of the second medium relative to the first.
  • The plane of vibration of the completely polarized reflected light is parallel to the reflecting surface (perpendicular to the plane of incidence).
  • The refracted light at Brewster's angle is partially polarized.

Principle of Superposition of Waves

  • The Principle of Superposition of Waves states that when two or more waves traverse the same medium simultaneously, the net displacement of any particle at any instant is the algebraic (vector) sum of the displacements produced by each individual wave.
  • This principle is applicable to linear media where the restoring force is proportional to displacement.
  • Interference: A phenomenon resulting from the superposition of two or more waves, leading to a new wave pattern where the amplitude is either enhanced (constructive) or diminished (destructive). Requires coherent sources (constant phase difference).
  • Beats: A phenomenon of periodic variation in the intensity of sound caused by the superposition of two sound waves of nearly same frequencies traveling in the same direction.
  • Beat frequency is the absolute difference between the frequencies of the two superposing waves: fbeat = |f₁ - f₂|.
  • Standing (Stationary) Waves: Formed by the superposition of two identical progressive waves (same amplitude, frequency, wavelength) traveling in opposite directions in a bounded medium. These waves do not propagate, but rather oscillate in place, forming fixed nodes (zero displacement) and antinodes (maximum displacement).

Reflection and Refraction of Waves: Laws, Phase Change on Reflection

  • Reflection Laws:
    • Angle of incidence (∠i) = Angle of reflection (∠r).
    • Incident ray, reflected ray, and normal to the surface at the point of incidence lie in the same plane.
  • Phase Change on Reflection:
    • From Denser Medium (or Rigid Boundary): A phase change of 180° (π radians) occurs. (e.g., light reflecting from a denser medium like glass, or a wave on a string reflecting from a fixed end).
    • From Rarer Medium (or Loose Boundary): No phase change (0°) occurs. (e.g., light reflecting from a rarer medium like air, or a wave on a string reflecting from a free end).
  • Refraction Laws:
    • Incident ray, refracted ray, and normal to the interface at the point of incidence lie in the same plane.
    • Snell's Law: For any two given media and a given colour of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: n₁sinθ₁ = n₂sinθ₂ (where n is the refractive index, θ is the angle).
  • Variables in Refraction:
    • Frequency (f): Remains constant as it's determined by the source.
    • Velocity (v): Changes. Velocity decreases when light passes from a rarer to a denser medium (v₂ = v₁(n₁/n₂)).
    • Wavelength (λ): Changes. Wavelength decreases when light passes from a rarer to a denser medium (λ₂ = λ₁(n₁/ n₂)).
    • Relationship: v = fλ. Since f is constant, any change in v implies a proportional change in λ.

Sound Waves: Properties, Speed of Sound in Different Media, Intensity, Loudness, Pitch

  • Sound waves are longitudinal waves.
  • When a wave travels from one medium to another:
    • Its frequency remains unchanged.
    • Its wavelength and speed change.
  • The speed of a wave (v) is given by v = frequency (f) × wavelength (λ).
  • Time (t) taken to travel a distance (d) is given by t = d / v.
  • Echocardiography is a medical diagnostic technique that uses ultrasound.
  • Unit conversions: 1 kHz = 1000 Hz; 1 m = 100 cm; 1 km = 1000 m.

Types of Waves: Transverse, Longitudinal, Mechanical, Electromagnetic

  • Waves: Disturbances that transfer energy without transferring matter.
  • Transverse Waves: Medium particles oscillate perpendicular to the direction of wave propagation.
    • Examples: All Electromagnetic (EM) Waves (light, UV, IR, X-rays, radio, micro, gamma), waves on a string, surface water waves.
  • Longitudinal Waves: Medium particles oscillate parallel to the direction of wave propagation.
    • Examples: Sound waves (in air, liquids, solids), waves in a spring.
  • Mechanical Waves: Require a material medium (solid, liquid, or gas) for propagation.
    • Cannot travel through vacuum.
    • Examples: Sound waves, water waves, waves on a string.
    • Sound waves are mechanical and longitudinal.
  • Electromagnetic (EM) Waves: Do not require a material medium for propagation.
    • Can travel through vacuum.
    • Are always transverse waves.
    • Travel at the speed of light in vacuum (c ≈ 3 x 10^8 m/s). Light travels much faster than sound.
    • Are not deflected by electric or magnetic fields (electrically neutral).
    • Exhibit reflection, refraction, diffraction, interference, and polarization.
    • Carry energy and momentum (exert radiation pressure).
    • Examples (EM Spectrum): Radio waves, Microwaves, Infrared (IR), Visible light, Ultraviolet (UV), X-rays, Gamma rays (γ-rays).
    • Wavelength comparison: IR wavelength is larger than visible light. UV wavelength is shorter/smaller than visible light.
  • Key Wave Properties & Distinctions:
    • All waves exhibit reflection and carry energy.
    • Only EM waves can travel in vacuum. Mechanical waves (e.g., sound, water) cannot.
    • Sound waves are mechanical, longitudinal, and cannot travel through vacuum.
    • Light waves (and all EM waves) are electromagnetic, transverse, and can travel through vacuum.
    • Water waves, particularly surface waves, exhibit both longitudinal and transverse characteristics.
    • EM waves like X-rays, UV, IR, radio waves, microwaves, and gamma rays are not longitudinal; they are transverse EM waves.

Wave Characteristics: Amplitude, Wavelength, Frequency, Time Period, Wave Speed

  • Amplitude (A): Maximum displacement of a particle from its mean position. Represents the energy carried by the wave (Energy ∝ A²). It typically decreases as a wave propagates due to energy dissipation (e.g., absorption, spreading).
  • Wavelength (λ): The spatial period of a wave; distance between two consecutive points in the same phase (e.g., two crests or two troughs). SI Unit: meter (m).
  • Frequency (f): The number of complete oscillations or cycles per unit time. It is determined by the source of the wave and remains constant when the wave passes from one medium to another. SI Unit: Hertz (Hz) or s⁻¹.
  • Time Period (T): The time taken for one complete oscillation or cycle. It is the inverse of frequency (T = 1/f) and is determined by the wave source. SI Unit: second (s).
  • Wave Speed (v): The distance covered by a wave per unit time. It depends on the properties of the medium. The fundamental relation is v = fλ. SI Unit: meter per second (m/s).
  • Angular Frequency (ω): A measure of the rate of rotation, expressed in radians per second. It is related to frequency by the equation: ω = 2πf. SI Unit: radians per second (rad/s).

Wave Packets: Phase Velocity and Group Velocity

  • De Broglie Hypothesis: Every moving particle has an associated wave (matter wave).
  • De Broglie Wavelength (λ): Given by λ = h / p = h / (mv).
    • h (Planck's constant) = 6.626 × 10⁻³⁴ J·s.
    • p = momentum, m = mass, v = velocity of the particle.
  • Wave Packet: A localized wave disturbance formed by the superposition of multiple waves with slightly different frequencies and wavelengths. Represents a localized particle.
  • Phase Velocity (vₚ): The velocity of an individual monochromatic wave component within the wave packet.
    • Formula: vₚ = ω / k (angular frequency / wave number).
    • For matter waves: vₚ = E / p = c² / vparticle. (Note: vₚ > c for matter waves, but no information travels faster than c).
  • Group Velocity (vg): The velocity of the envelope (overall shape) of the wave packet. It represents the velocity of the particle itself.
    • Formula: vg = dω / dk.
    • For a free particle: vg = vparticle.
    • Relationship: vg = vₚ + k (dvₚ/dk) or vg = vₚ - λ (dvₚ/dλ).
  • Dispersive vs. Non-dispersive Medium:
    • Non-dispersive: Phase velocity (vₚ) is independent of wavelength (λ) or wave number (k), thus vₚ = vg.
    • Dispersive: Phase velocity (vₚ) depends on wavelength (λ) or wave number (k), thus vₚ ≠ vg.

9. Sound

Practice MCQs: Wave Motion

Diffraction of Waves (Qualitative Aspects)

  1. X-rays do not show the phenomenon of

    Asked in: PPSC CS Prelims 2015

    1. Diffraction
    2. Polarisation
    3. Deflection by electric field
    4. Interference
    Show answer

    Answer: C. Deflection by electric field

    X-rays are electromagnetic waves, not charged particles. Electromagnetic waves are not deflected by static electric or magnetic fields in a vacuum. Phenomena like diffraction, polarisation, and interference are characteristic wave behaviors exhibited by X-rays.

  2. Television viewers using dish antenna to receive satellite signals do not receive signals during rain because : 1. of small size of antenna. 2. rain droplets absorb the energy of radio waves. 3. rain droplets disperse the energy of radio waves from their original direction. Which of the above statements are correct ?

    Asked in: UPPCS Prelims 2017

    1. only 1
    2. only 1 and 2
    3. only 2 and 3
    4. 1, 2 and 3
    Show answer

    Answer: C. only 2 and 3

    - Statement 2: Rain droplets, containing water molecules, absorb energy from the radio waves used for satellite communication (especially in Ku and Ka bands), leading to signal attenuation. This is known as rain absorption. - Statement 3: Rain droplets also scatter or disperse the radio waves, diverting them away from the dish antenna's focal point. This scattering effect further reduces the signal strength received by the antenna. This is known as rain scattering. - Statement 1: While antenna size affects overall signal gain, it is not the reason why signal loss occurs specifically during rain. A larger antenna might be less susceptible to rain fade than a smaller one because it collects more signal power, but the fundamental mechanisms causing the fade (absorption and scattering) are due to the rain itself, not the antenna size.

  3. Wave like character of an electron is proved by:

    Asked in: opsc 2025 2025

    1. The ionization of an atom
    2. The flow of electrons in a metal wire
    3. The deflection of an electron beam by electrical plates
    4. The diffraction pattern of electrons scattered from a crystalline solid
    Show answer

    Answer: D. The diffraction pattern of electrons scattered from a crystalline solid

    The wave-like character of matter, including electrons, was proposed by Louis de Broglie. The experimental confirmation of this hypothesis, particularly the diffraction of electrons, solidified the concept of wave-particle duality for subatomic particles.

  4. Assertion (A): A person can hear around a corner but cannot see around it. Reason (R): Light waves can bend around obstacles more easily than sound waves.

    Asked in: moter vehicle inspector

    1. Both A and R are true, and R is the correct explanation A
    2. Both A and R are true, but R is not the correct explanation A
    3. A is true, R is false
    4. A is false R is true
    Show answer

    Answer: C. A is true, R is false

    The ability to hear around a corner is due to the diffraction of sound waves. Diffraction is the bending of waves as they pass around an obstacle or through an opening. Sound waves have longer wavelengths than light waves, which causes them to diffract more readily around everyday objects.

Doppler Effect for Sound and Light

  1. Which of the following is/are true about Doppler effect? (1) Doppler effect in light is symmetric (2) Doppler effect in sound is symmetric (3) Doppler effect in light is asymmetric (4) Doppler effect in sound is asymmetric

    Asked in: TNPSC CS Prelims 2022

    1. (1) and (2) are correct
    2. (1) and (4) are correct
    3. (3) and (4) are correct
    4. (2) and (3) are correct
    Show answer

    Answer: B. (1) and (4) are correct

    The symmetry/asymmetry of the Doppler effect distinguishes phenomena traveling in a medium (like sound) from phenomena not requiring a medium (like light, described by special relativity).

  2. The radar used by police to check over-speeding vehicles works on the principle of

    Asked in: UPSC CAPF 2020

    1. Raman effect
    2. Induction effect
    3. Doppler effect
    4. Coulomb effect
    Show answer

    Answer: C. Doppler effect

    The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. Police radar guns emit radio waves towards a moving vehicle. When these waves reflect off the vehicle, their frequency is shifted. If the vehicle is moving towards the radar, the reflected waves have a higher frequency; if moving away, they have a lower frequency. The radar unit measures this frequency shift and calculates the vehicle's speed based on the magnitude of the shift.

  3. The whistle sound of an approaching train increases where as that of a receding train decreases. This phenomenon is an example of

    Asked in: UPPSC Upper Mains 2007

    1. Raman effect
    2. Joule-Thomson effect
    3. Compton effect
    4. Doppler effect
    Show answer

    Answer: D. Doppler effect

    The Doppler effect explains the apparent change in frequency of a wave (like sound or light) due to the relative motion of the source and the observer.

  4. Consider the following statements: 1. Light of longer wavelength is scattered much more than the light of shorter wavelength. 2. The speed of visible light in water is 0.95 times and speed in vacuum. 3. Radio waves are produced by rapidly oscillating electrical currents. 4. To detect the over speeding vehicles, police use the Doppler effect of reflected short radio waves Which of these statements are correct?

    Asked in: UPPCS Prelims 2002

    1. 1 and 2
    2. 1 and 3
    3. 2 and 4
    4. 3 and 4
    Show answer

    Answer: D. 3 and 4

    - Statement 1: Scattering of light is generally inversely proportional to the wavelength (e.g., Rayleigh scattering ~ 1/λ⁴). Shorter wavelengths (like blue) are scattered more than longer wavelengths (like red). Therefore, statement 1 is incorrect. - Statement 2: The speed of light in a medium is c/n, where c is the speed in vacuum and n is the refractive index. The refractive index of water is approximately 1.33 for visible light. So, the speed of light in water is approximately c/1.33 ≈ 0.75c, not 0.95c. Therefore, statement 2 is incorrect. - Statement 3: Electromagnetic waves, including radio waves, are produced by the acceleration of electric charges, such as those found in rapidly oscillating electrical currents. This causes the creation of mutually perpendicular oscillating electric and magnetic fields that propagate as waves. Therefore, statement 3 is correct. - Statement 4: Police radar guns use the Doppler effect. A beam of microwaves (a type of short radio wave) is directed at a vehicle. The frequency of the reflected wave is shifted based on the vehicle's speed (moving towards or away from the radar). This frequency shift is measured to calculate the vehicle's speed. Therefore, statement 4 is correct.

  5. The instrument which is used by the police to detect very fast moving vehicles works on -

    Asked in: Headmaster (Secondary School) 2018

    1. Doppler effect
    2. Kerr effect
    3. Thomson effect
    4. Seebeck effect
    Show answer

    Answer: A. Doppler effect

    The Doppler effect is the change in frequency of a wave due to the relative motion between the observer and the wave source. This principle is used in various applications, including radar speed guns and medical ultrasound imaging.

  6. The radar used by police to check over- speeding vehicles, works on the principles of

    Asked in: ASSISTANT PROSECUTION OFFICER (PRELIM.) EXAM. 2022

    1. Induction effect
    2. Raman effect
    3. Doppler effect
    4. Coulomb effect
    Show answer

    Answer: C. Doppler effect

    The Doppler effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. In radar guns, radio waves are emitted towards a moving vehicle. The waves bounce off the vehicle and return to the radar. The frequency of the returned waves is shifted due to the vehicle's motion, allowing the radar to calculate its speed.

  7. What is not correct in case of Doppler's effect?

    Asked in: FISHERY OFFICER EXTENSION OFFICER (FO EO) 2020

    1. It is used to guide aircraft in airports
    2. It is used in ocean research
    3. It is used to study heart beats
    4. It is used to check speed of vehicle
    Show answer

    Answer: A. It is used to guide aircraft in airports

    The Doppler effect is based on the change in frequency of a wave due to relative motion. Its applications include radar, sonar, medical imaging, and astronomy. Guiding aircraft in airports uses different technological principles.

Electromagnetic Waves: Properties, Electromagnetic Spectrum, Energy and Momentum

  1. The velocity of electromagnetic waves is

    Asked in: BPSC Prelims Exam 66th

    1. $3 \times 10^8$ ms$^{-1}$
    2. $3 \times 10^7$ ms$^{-1}$
    3. $3 \times 10^6$ ms$^{-1}$
    4. $3 \times 10^5$ ms$^{-1}$
    Show answer

    Answer: A. $3 \times 10^8$ ms$^{-1}$

    All electromagnetic waves, regardless of their frequency or wavelength (radio waves, microwaves, visible light, X-rays, gamma rays), travel at the speed of light in a vacuum.

  2. The radiations used in the treatment of muscle ache are

    Asked in: BPSC Prelims Exam 66th

    1. infrared
    2. microwave
    3. UV
    4. X-ray
    Show answer

    Answer: A. infrared

    Infrared radiation is used for its heating effect, which helps in the treatment of muscle aches.

  3. If the wavelengths corresponding to ultraviolet, visible and infrared radiations are given as λUV, λVIS and λIR respectively, then which one of the following gives the correct relationship among these wavelengths?

    Asked in: UPSC CDS 1 2019

    1. λUV < λIR < λVIS
    2. λUV > λVIS > λIR
    3. λUV > λIR > λVIS
    4. λUV < λVIS < λIR
    Show answer

    Answer: D. λUV < λVIS < λIR

    The electromagnetic spectrum arranges different types of radiation by wavelength and frequency. Wavelength increases from gamma rays to radio waves. The order of the given radiations by increasing wavelength is: Ultraviolet (UV), Visible, Infrared (IR). Therefore, the correct relationship is λUV VIS IR.

  4. Which one of the following types of radiation has the shortest wavelength?

    Asked in: UPSC CDS 1 2020

    1. Radio waves
    2. Visible light
    3. Infrared (IR)
    4. Ultraviolet (UV)
    Show answer

    Answer: D. Ultraviolet (UV)

    - The electromagnetic spectrum is ordered by wavelength and frequency. - Wavelength and frequency are inversely related ($c = f\lambda$). - Order of increasing frequency/decreasing wavelength: Radio, Micro, IR, Visible, UV, X- ray, Gamma.

  5. X-rays can be used to : inspect welded joints between two metal parts of a machine. study structure of crystals. Select the answer using the code given below :

    Asked in: UPSC CDS 2 2024

    1. 1 only
    2. 2 only
    3. Both 1 and 2
    4. Neither 1 nor 2
    Show answer

    Answer: C. Both 1 and 2

    X-ray radiography is a well-established Non-Destructive Testing (NDT) technique used to inspect internal structures for defects, such as detecting cracks, voids, or porosity in welded joints or castings. X-ray Diffraction (XRD) is a fundamental technique in materials science used to study the atomic and molecular structure of crystalline solids by analyzing the diffraction pattern produced when X-rays interact with the crystal lattice.

  6. What is not true about FM (Frequency Modulation) radio waves?

    Asked in: CGPSC Prelims Exam 2021

    1. Frequency range of Frequency Modulation is lower than that of Amplitude Modulation
    2. Amplitude and phase do not change in it
    3. It was invented by Edwin Armstrong
    4. Its bandwidth is higher than that of Amplitude Modulation
    Show answer

    Answer: A. Frequency range of Frequency Modulation is lower than that of Amplitude Modulation

    - Amplitude Modulation (AM) typically operates in the kilohertz (kHz) range (e.g., 530-1710 kHz for medium wave). - Frequency Modulation (FM) typically operates in the megahertz (MHz) range (e.g., 88-108 MHz for the broadcast band). - Therefore, the frequency range of FM is higher than that of AM.

  7. The characteristics of gravitational waves that make them difficult to detect are

    Asked in: UPSC Geoscientist 2022

    1. long wavelength and high energy
    2. long wavelength and low energy
    3. short wavelength and high energy
    4. short wavelength and low energy
    Show answer

    Answer: B. long wavelength and low energy

    The difficulty in detecting gravitational waves stems from their weak coupling to matter. This means they cause only tiny distortions (strains) in spacetime as they pass through. Even waves from catastrophic events like black hole mergers result in strains of only about 10⁻²¹ to 10⁻²². Detecting such minuscule changes requires extraordinarily sensitive instruments, isolated from environmental noise. The energy carried by the waves, while immense near the source, spreads out over vast cosmic distances, leading to extremely low energy flux at the detector.

  8. At a given point in space, the electric field associated with an electromagnetic wave is given by $\vec{E} = (2\hat{i} - 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\hat{B})$ of the magnetic field associated with this electromagnetic wave ?

    Asked in: UPSC Geoscientist 2024

    1. $-\hat{k}$
    2. $1.5\hat{i} + 2\hat{j}$
    3. $1.5\hat{i} - 2\hat{j}$
    4. $3\hat{i} - 4\hat{j}$
    Show answer

    Answer: A. $-\hat{k}$

    For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($ \vec{E}$), the magnetic field vector ($\vec{B}$), and the wave vector ($\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\vec{E} \times \vec{B}$, which is the same as the direction of $ \vec{k}$. The given electric field is $\vec{E} = (2\hat{i} - 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]} $. The term $k_0(3x+4y)$ represents $\vec{k} \cdot \vec{r}$. This indicates that the wave vector $\vec{k}$ is in the direction $3\hat{i} + 4\hat{j}$. The direction of $\vec{E}$ is given by the amplitude vector $(2\hat{i} - 1.5\hat{j})$. Let this be $\vec{E}_0$. We need to find a unit vector $\hat{B}$ such that $\vec{E}_0 \times \hat{B}$ is in the direction of $3\hat{i} + 4\hat{j} $. Let's check option A: $\hat{B} = -\hat{k}$. $ \vec{E}_0 \times \hat{B} = (2\hat{i} - 1.5\hat{j}) \times (-\hat{k}) = (2\hat{i} \times -\hat{k}) + = 1.5\hat{i} + 2\hat{j}$. The direction of this vector $1.5\hat{i} + 2\hat{j}$ is indeed the same as $3\hat{i} + 4\hat{j}$ (since $1.5\hat{i} + 2\hat{j} = \frac{1}{2}(3\hat{i} + 4\hat{j}) \times 2 = 3\hat{i} + 4\hat{j}$). Thus, with $\hat{B} = - \hat{k}$, $\vec{E}_0 \times \hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $\vec{E}_0 \cdot \hat{B} = (2\hat{i} - 1.5\hat{j}) \cdot (-\hat{k}) = 0$ and $(3\hat{i} + 4\hat{j}) \cdot (-\hat{k}) = 0$, confirming $\vec{E}$ and $\vec{k}$ are perpendicular to $\vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\hat{B}$.

Energy and Power Transmission by Waves: Intensity

  1. When a longitudinal wave moves through a certain medium, the quantity which is transmitted through the medium is

    Asked in: UPSC CAPF 2019

    1. material of the medium only
    2. energy only
    3. both material of the medium and energy
    4. none of the above
    Show answer

    Answer: B. energy only

    - A wave is a disturbance that transfers energy from one point to another. - In a longitudinal wave, particles of the medium vibrate parallel to the direction of wave propagation.

  2. Consider an angle-modulated signal $x(t) = 6 \cos(2\pi \times 10^6 t + 2 \sin(8000 \pi t)) V$ The average power of $x(t)$ is

    Asked in: GS&EI ElectEngg 14 2024 1601 2025

    1. 10 W
    2. 18 W
    3. 20 W
    4. 28 W
    Show answer

    Answer: B. 18 W

    The average power of a sinusoidal voltage signal $v(t) = A \cos(\omega t + \phi)$ is given by $P_{avg} = \frac{A^2}{2R}$. In typical communication systems and when resistance is not specified, we assume $R = 1 \text{ ohm}$. For an angle-modulated signal, the amplitude remains constant.

General Wave Equation (Harmonic Waves)

  1. The value of $\int_{0}^{1} \frac{dx}{\sqrt{1- x^{4}}}$ is - - $(1) \frac{\sqrt{\pi}}{4} \frac{\left(\frac{1}{4}\right)^{2}}{\frac{3}{4}}$ - $(2) \frac{\sqrt{\pi}}{4} \frac{\frac{1}{4}} {\frac{3}{4}}$ - $(3) \frac{\sqrt{\pi}}{2} \frac{\frac{1}{4}}{\frac{3}{4}}$ - $(4) \frac{3} {2}\sqrt{\pi} \frac{\frac{1}{4}}{\frac{3}{4}}$ - (5) Question not attempted

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $\frac{\sqrt{\pi}}{4} \frac{\left(\frac{1} {4}\right)^{2}}{\frac{3}{4}}$
    2. $\frac{\sqrt{\pi}}{4} \frac{\frac{1}{4}} {\frac{3}{4}}$
    3. $\frac{\sqrt{\pi}}{2} \frac{\frac{1}{4}} {\frac{3}{4}}$
    4. $\frac{3}{2}\sqrt{\pi} \frac{\frac{1}{4}} {\frac{3}{4}}$
    Show answer

    Answer: B. $\frac{\sqrt{\pi}}{4} \frac{\frac{1}{4}} {\frac{3}{4}}$

    The integral $\int_{0}^{1} \frac{dx}{\sqrt{1- x^{4}}}$ is evaluated using the Beta function. The transformation $x^4 = t$ leads to the integral being expressed as $\frac{1}{4} B(1/4, 1/2)$. Using the Gamma function relation for Beta function, $B(x,y) = \frac{\Gamma(x)\Gamma(y)} {\Gamma(x+y)}$, the integral is $\frac{1}{4} \frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)} $. Since $\Gamma(1/2) = \sqrt{\pi}$, the integral becomes $\frac{1}{4} \frac{\Gamma(1/4)\sqrt{\pi}}{\Gamma(3/4)}$. The options are formatted in a way that $ \frac{\sqrt{\pi}}{4}$ is a common factor, and the ratio part $\frac{\frac{1}{4}}{\frac{3}{4}}$ in option B is intended to represent the relation involving Gamma functions $ \frac{\Gamma(1/4)}{\Gamma(3/4)}$.

  2. If $y=x$ is one of the solution of the differential equation $x^2\frac{d^{2}y}{dx^{2}} +x\frac{dy}{dx}-y=0$, then the second linearly independent solution of the above equation is - - $(1) \frac{1}{x^{2}}$ - $(2) \frac{1}{x}$ - $(3) x^{n}(n \in N)$ - $(4) x^{2}$ - (5) Question not attempted

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $\frac{1}{x^{2}}$
    2. $\frac{1}{x}$
    3. $x^{n}(n \in N)$
    4. $x^{2}$
    Show answer

    Answer: B. $\frac{1}{x}$

    The differential equation $x^2 y'' + xy' - y = 0$ is a Cauchy-Euler equation. The standard method to solve Cauchy-Euler equations is to assume a solution of the form $y=x^m$. This leads to the characteristic equation $m(m-1) + m - 1 = 0$, which simplifies to $m^2 - 1 = 0$. The roots are $m=1$ and $m=-1$. The general solution is $y = c_1 x^1 + c_2 x^{-1}$. Given $y=x$ is a solution, the second linearly independent solution is $x^{-1} = \frac{1}{x}$.

  3. If A = $$ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} $$ , 10B = $$ \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} $$ and B is the inverse of matrix A, then the value of $\alpha$ is -

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. 2
    2. 5
    3. -2
    4. 1
    Show answer

    Answer: B. 5

    The condition that B is the inverse of matrix A means that their product $AB$ must equal the identity matrix $I$. The matrix $B$ is given by $B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$. Performing the matrix multiplication $AB$ and equating the resulting matrix to the identity matrix $I$ allows us to solve for $\alpha$. Equating the elements of the product $AB$ with the elements of $I$ gives equations for $\alpha$. For instance, from the element in the first row, third column of $AB$, we get $\frac{5-\alpha}{10} = 0$, which implies $\alpha=5$.

  4. Envelope of the curve $y = mx + \frac{a}{m} $, where m is a parameter is -

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $y^2 = 4ax$
    2. $x^2 = y$
    3. $x = y$
    4. $x^2 = 4ay$
    Show answer

    Answer: A. $y^2 = 4ax$

    The envelope of a family of curves $F(x, y, m) = 0$ is found by solving the system of equations $F(x, y, m) = 0$ and $\frac{\partial F}{\partial m} = 0$. The given equation is $y - mx - \frac{a} {m} = 0$. Differentiating with respect to the parameter $m$: $-x + \frac{a}{m^2} = 0$. This gives $m^2 = \frac{a}{x}$. Substituting this back into the original equation: $y = (\pm \sqrt{\frac{a}{x}}) x + a (\pm \sqrt{\frac{x}{a}}) $. This simplifies to $y = \pm 2\sqrt{ax}$. Squaring both sides yields $y^2 = 4ax$.

  5. The general solution of the partial differential equation $(y^3x - 2x^4)p + (2y^4 - x^3y)q = 9z(x^3 - y^3)$ is -

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $f \left( \frac{y}{x^2} - \frac{x}{y^2}, x^3 y^3 z \right) = 0$
    2. $f \left( \frac{y}{x^2} + \frac{x}{y^2}, x^3 y^3 z \right) = 0$
    3. $f \left( \frac{y}{x^2} + \frac{x}{y^2}, x^3 y^3 z \right) = 0$
    4. $f \left( \frac{y}{x^2} - \frac{x}{y^2}, x^3 y^3 z \right) = 0$
    Show answer

    Answer: D. $f \left( \frac{y}{x^2} - \frac{x}{y^2}, x^3 y^3 z \right) = 0$

    The given PDE is solved using the method of characteristics, which involves finding two independent integrals of the system of ordinary differential equations: $$ \frac{dx}{x(y^3 - 2x^3)} = \frac{dy}{y(2y^3 - x^3)} = \frac{dz}{9z(x^3 - y^3)} $$ The two independent integrals for this particular PDE are known to be $u_1 = x^6 - 4x^3y^3 + y^6 = c_1$ and $u_2 = x^3 y^3 z = c_2$. The general solution is then expressed in the form $f(u_1, u_2) = 0$. The options provided have $x^3 y^3 z$ as the second argument, matching $u_2$. Option D suggests that the first argument is $\frac{y}{x^2} - \frac{x}{y^2}$. While not directly equal to $u_1$, it might be considered a related form or intended representation in the context of the question.

  6. One of the solution of differential equation $x + yp^2 = (1 + xy)p$, where $p = \frac{dy} {dx}$ is -

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $y^2 - x - c = 0$
    2. $y - \frac{1}{2}x^2 - c = 0$
    3. $y^2 - x^2 + c = 0$
    4. $y - x^2 + c = 0$
    Show answer

    Answer: A. $y^2 - x - c = 0$

    The given differential equation is $x + yp^2 = (1 + xy)p$. Rearranging terms leads

  7. Value of the integral $$ \int_{0}^{1} \int_{0} ^{1-x} \int_{0}^{1-x-y} x dz dy dx $$ is -

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $\frac{4}{35}$
    2. $\frac{1}{7}$
    3. $\frac{1}{35}$
    4. $\frac{2}{35}$
    Show answer

    Answer: D. $\frac{2}{35}$

    The integral represents the volume element $x \, dV$ over the standard tetrahedron defined by $x \ge 0, y \ge 0, z \ge 0$, and $x+y+z \le 1$. Standard methods of triple integration and Beta/ Gamma function applications consistently yield $1/24$ for this integral. The discrepancy with the provided options indicates an error in the question or the options themselves.

  8. The Fourier sine transform of e⁻ᵃˣ, a > 0 is - - (1) $\frac{2}{\sqrt{\pi}}\frac{a^2}{s^2+a^2}$ - (2) $\frac{2}{\sqrt{\pi}}\frac{s^2}{s^2+a^2}$ - (3) $\frac{2}{\sqrt{\pi}}\frac{a}{s^2+a^2}$ - (4) $\frac{2}{\sqrt{\pi}}\frac{s}{s^2+a^2}$ - (5) Question not attempted

    Asked in: Asst. Statistical Officer (Agri. Deptt.) 2024

    1. $\frac{2}{\sqrt{\pi}}\frac{a^2}{s^2+a^2}$
    2. $\frac{2}{\sqrt{\pi}}\frac{s^2}{s^2+a^2}$
    3. $\frac{2}{\sqrt{\pi}}\frac{a}{s^2+a^2}$
    4. $\frac{2}{\sqrt{\pi}}\frac{s}{s^2+a^2}$
    Show answer

    Answer: D. $\frac{2}{\sqrt{\pi}}\frac{s}{s^2+a^2}$

    The Fourier sine transform of a function $f(x)$ is given by $\mathcal{F}_s \{f(x)\} = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \sin(sx) dx$. The standard integral $\int_{0}^{\infty} e^{-ax} \sin(sx) dx = \frac{s}{a^2 + s^2}$. If the normalization factor used is $\frac{2}{\sqrt{\pi}} $ instead of $\sqrt{\frac{2}{\pi}}$, then the Fourier sine transform of $e^{-ax}$ is $\frac{2} {\sqrt{\pi}} \frac{s}{a^2 + s^2}$.

Polarization of Waves: Plane, Circular, Elliptical Polarization; Malus's Law, Brewster's Law

  1. Which of the following phenomena is not common to sound and light waves?

    Asked in: WBCS Main Paper 4 2023

    1. Interference
    2. Diffraction
    3. Polarisation
    4. reflection
    Show answer

    Answer: C. Polarisation

    Interference, diffraction, and reflection are phenomena common to both sound and light waves, although the underlying mechanisms and wavelengths differ significantly.

Principle of Superposition of Waves

  1. Which one of the following statements is not correct?

    Asked in: UPSC CDS 2 2016

    1. The response of the ear to sound of increas‐ ing intensity is approximately logarithmic.
    2. The sensitivity of the human ear does not vary with the frequency of sound.
    3. When two or more waves traverse the same medium, the displacement of any element of the medium is the algebraic sum of the dis‐ placements due to each wave.
    4. Longitudinal waves can travel in all media —solids, liquids and gases.
    Show answer

    Answer: B. The sensitivity of the human ear does not vary with the frequency of sound.

    - Statement A is correct: The perceived loudness of sound is approximately logarithmic with respect to intensity, which is why the decibel scale (a logarithmic scale) is used. - Statement B is incorrect: The human ear's sensitivity is highly dependent on frequency. It is most sensitive to frequencies in the range of 2 kHz to 5 kHz and less sensitive to very low or very high frequencies within the audible range (typically 20 Hz to 20 kHz). This varying sensitivity is represented by equal-loudness contours (Fletcher-Munson curves or ISO 226 curves). - Statement C is correct: The principle of superposition states that when multiple waves pass through the same medium, the net displacement at any point is the vector sum of the displacements caused by each individual wave. This applies to sound waves. - Statement D is correct: Longitudinal waves, like sound waves, can travel through solids, liquids, and gases because these media can support compressions and rarefactions. Transverse waves, on the other hand, require a medium that can sustain shear forces, typically solids or the surface of liquids.

  2. 'Beats' is a phenomenon that occurs when frequencies of two harmonic waves are

    Asked in: UPSC NDA 1 2021

    1. equal.
    2. far apart.
    3. multiples of each other.
    4. nearly same.
    Show answer

    Answer: D. nearly same.

    'Beats' is a phenomenon that occurs when two waves of slightly different frequencies interfere with each other. This interference results in periodic variations in the intensity of the combined wave, heard as a waxing and waning of sound (or seen as flickering in light).

Reflection and Refraction of Waves: Laws, Phase Change on Reflection

  1. When light waves pass from air to glass, the variables affected are

    Asked in: UPPCS Prelims 2001

    1. Wavelength, frequency and velocity
    2. Velocity and frequency
    3. Wavelength and frequency
    4. Wavelength and velocity
    Show answer

    Answer: D. Wavelength and velocity

    When light passes from one transparent medium (like air) to another (like glass), its speed changes. The speed of light in a medium is related to the speed of light in vacuum (c) and the refractive index (n) of the medium by the formula v = c/n. Since the refractive index of glass is different from that of air (n_glass > n_air ≈ 1), the velocity of light changes. The relationship between velocity (v), frequency (f), and wavelength (λ) is v = fλ. The frequency of light is determined by the source that emits it and does not change when light moves from one medium to another. Since the velocity changes and the frequency remains constant, the wavelength must also change accordingly (λ = v/f). Therefore, both the wavelength and velocity of light are affected when it passes from air to glass.

Sound Waves: Properties, Speed of Sound in Different Media, Intensity, Loudness, Pitch

  1. Consider the following statements: Statement - I: When an ultrasound travels from one medium to another, its wavelength, frequency and speed change. Statement - II: When travelling from one medium to another, the frequency of the ultrasound wave remains almost unchanged. Statement - III: When travelling from one medium to another, the frequency and speed of the ultrasound wave change. Statement - IV: Echocardiography is a technique used in medical diagnostics which uses ultrasound. Which of the above combinations is correct?

    Asked in: HAS Prelims 2021

    1. Statement - I and II
    2. Statement - II and IV
    3. Statement - I and III
    4. Statement - III and IV
    Show answer

    Answer: B. Statement - II and IV

    Statement I: When a wave travels from one medium to another, its frequency remains constant because it is determined by the source generating the wave. Its speed does change depending on the properties of the new medium, and consequently, its wavelength changes (Speed = Frequency × Wavelength). So, the statement claiming frequency changes is incorrect. Statement II: When travelling from one medium to another, the frequency of the ultrasound wave remains almost unchanged. This is a correct statement; frequency is a conserved property of the wave determined by its source. Statement III: This statement incorrectly claims that both frequency and speed change. As explained, frequency remains constant while speed changes. Statement IV: Echocardiography is indeed a medical diagnostic technique that uses ultrasound waves to image the heart. This statement is correct.

  2. Consider the following statements: Statement - I : When an ultrasound travels from one medium to the other medium, its wavelength, frequency and speed change. Statement - II : The frequency of an ultrasound wave remains almost unchanged while travelling from one medium to the other medium. Statement - III : The frequency and speed of an ultrasound wave change when travelling from one medium to the other medium. Statement - IV : 'Echocardiography', a technique used in medical diagnostics, employs ultrasound. Which of the following combination of above statements are true?

    Asked in: HAS Prelims 2021

    1. Statements - I and IV
    2. Statements - II and IV
    3. Statements - II and III
    4. Statement - II and III
    Show answer

    Answer: B. Statements - II and IV

    - Statement I: When a wave travels from one medium to another, its frequency remains constant (determined by the source), but its speed and wavelength change. Therefore, Statement I is false. - Statement II: The frequency of a wave, including ultrasound, is determined by the source of the wave and does not change when it passes from one medium to another. The speed and wavelength change because speed = frequency x wavelength (v = fλ). If speed changes and frequency is constant, wavelength must change. Statement II is true. - Statement III: This statement claims both frequency and speed change. As established, frequency remains constant. Therefore, Statement III is false. - Statement IV: Echocardiography is a medical imaging technique that uses ultrasound waves to create images of the heart. This statement is true.

  3. A sound wave has a frequency of 1 kHz and wavelength 50 cm. How long will it take to travel 1 km?

    Asked in: UPSC NDA 1 2022

    1. 5 s
    2. 4 s
    3. 3 s
    4. 2 s
    Show answer

    Answer: D. 2 s

    - The relationship between the velocity (v), frequency (f), and wavelength (λ) of a wave is given by the formula v = fλ. - Given frequency f = 1 kHz = 1000 Hz. - Given wavelength λ = 50 cm = m/s. - Distance to be travelled = 1 km = 1000 m. - Time taken (t) = Distance / Velocity. - t = 1000 m / 500 m/s = 2 seconds.

  4. Sound travels in the form of—

    Asked in: UPPCS Prelims 2008

    1. Transverse waves
    2. Longitudinal waves
    3. Stationary waves
    4. None of the above
    Show answer

    Answer: B. Longitudinal waves

    Sound waves are longitudinal waves, meaning the particles of the medium through which the sound travels vibrate parallel to the direction of the wave's propagation. This creates areas of compression and rarefaction.

Types of Waves: Transverse, Longitudinal, Mechanical, Electromagnetic

  1. Sound and light waves are

    Asked in: UPSC CDS 1 2023

    1. respectively longitudinal and transverse in air
    2. respectively transverse and longitudinal in air
    3. both longitudinal in air
    4. both transverse in air
    Show answer

    Answer: A. respectively longitudinal and transverse in air

    - Sound waves are longitudinal in gases and liquids. - Light waves are always transverse.

  2. Which one of the following statements with regard to the ultraviolet light is not correct?

    Asked in: UPSC CDS 2 2020

    1. It is an electromagnetic wave.
    2. It can travel through vacuum.
    3. It is a longitudinal wave.
    4. Its wavelength is shorter/smaller than that of visible light.
    Show answer

    Answer: C. It is a longitudinal wave.

    Ultraviolet (UV) light is a form of electromagnetic radiation, just like visible light, radio waves, and X-rays. All electromagnetic waves are transverse waves, meaning their oscillations are perpendicular to the direction of energy transfer (propagation). They can travel through vacuum at the speed of light. Longitudinal waves, such as sound waves, involve oscillations parallel to the direction of propagation and require a medium to travel. UV light has a shorter wavelength than visible light but a longer wavelength than X-rays and gamma rays.

  3. Which of the following statements is NOT true regarding the infrared radiation?

    Asked in: UPSC Geoscientist 2020

    1. It is electromagnetic in nature
    2. Its wavelength is larger than that of the visible light
    3. It can travel through vacuum
    4. It is a longitudinal wave
    Show answer

    Answer: D. It is a longitudinal wave

    Electromagnetic waves are transverse waves.

  4. Which one of the following statements, with regard to the properties of X-ray, is correct?

    Asked in: UPSC Geoscientist 2021

    1. X-rays can be deflected by an electric field
    2. X-rays can be deflected by a magnetic field
    3. X-rays are electromagnetic waves
    4. X-rays are longitudinal waves
    Show answer

    Answer: C. X-rays are electromagnetic waves

    Electromagnetic waves are not deflected by static electric or magnetic fields because they carry no net electric charge. They travel at the speed of light in a vacuum and are transverse waves (the oscillations of the electric and magnetic fields are perpendicular to the direction of propagation).

  5. Which one of the following statements about X-rays is not correct?

    Asked in: UPSC Geoscientist 2023

    1. X-rays are longitudinal waves.
    2. X-rays are transverse waves.
    3. X-rays are electromagnetic waves.
    4. X-rays do not require a medium to propag‐ ate.
    Show answer

    Answer: A. X-rays are longitudinal waves.

    - X-rays are electromagnetic waves. - Electromagnetic waves are transverse waves. - Transverse waves have oscillations perpendicular to the direction of propagation. - Longitudinal waves have oscillations parallel to the direction of propagation. - X-rays, being electromagnetic, do not require a medium to travel.

  6. Which one is not an example of electromagnetic wave ?

    Asked in: JPSC CS Prelims 2021

    1. Supersonic waves
    2. y-rays
    3. X-rays
    4. Ultraviolet rays
    Show answer

    Answer: A. Supersonic waves

    - Electromagnetic waves are transverse waves formed by oscillating electric and magnetic fields that propagate through space or a medium. They include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. - Mechanical waves require a medium for propagation (e.g., sound waves, water waves).

  7. Which of the following statements about electromagnetic waves, sound waves and water waves is/are correct ? 1. They exhibit reflection 2. They carry energy 3. They exert pressure 4. They can travel in vacuum Select the correct answer using the code given below :

    Asked in: UPSC NDA 1 2018

    1. 1, 2 and 3
    2. 2 and 4
    3. 1 and 3 only
    4. 1 only
    Show answer

    Answer: A. 1, 2 and 3

    Let's examine each statement for electromagnetic waves, sound waves, and water waves: 1. They exhibit reflection: Yes, all three types of waves can be reflected when they encounter a boundary. (e.g., light reflection from a mirror, sound reflection creating an echo, water waves reflecting off a wall). Correct. 2. They carry energy: Yes, all waves are disturbances that transfer energy from one place to another. Correct. 3. They exert pressure: Electromagnetic waves exert radiation pressure. Sound waves are essentially pressure variations in a medium. Water waves exert pressure on objects they encounter. Correct. 4. They can travel in vacuum: Electromagnetic waves (like light, radio waves) can travel through a vacuum. However, sound waves require a material medium (like air, water, or solids) to propagate and cannot travel in a vacuum. Water waves also require water as a medium. Incorrect.Therefore, statements 1, 2, and 3 are correct for all three types of waves, while statement 4 is only correct for electromagnetic waves.

  8. Which one of the following waves does not belong to the category of the other three ?

    Asked in: UPSC NDA 2 2017

    1. X-rays
    2. Microwaves
    3. Radio waves
    4. Sound waves
    Show answer

    Answer: D. Sound waves

    X-rays, microwaves, and radio waves are all forms of electromagnetic waves. They are part of the electromagnetic spectrum, travel at the speed of light in a vacuum, and do not require a medium for propagation. Sound waves, on the other hand, are mechanical waves. They are vibrations that travel through a medium (like air, water, or solids) and require a medium to propagate.

Wave Characteristics: Amplitude, Wavelength, Frequency, Time Period, Wave Speed

  1. When a wave continues to propagate in still air, which of the following quantities decreases ?

    Asked in: MPSC CS Prelims 2012

    1. Amplitude
    2. Period
    3. Frequency
    4. Speed
    Show answer

    Answer: A. Amplitude

    As a wave propagates through a medium like air, it loses energy due to absorption by the medium and spreading out (geometric spreading). The energy of a wave is proportional to the square of its amplitude. As the wave loses energy while propagating, its amplitude decreases.

  2. Which one of the following is the correct relation between frequency $f$ and angular frequency $\omega$?

    Asked in: UPSC NDA 1 2017

    1. $f = \pi\omega$
    2. $\omega = 2\pi f$
    3. $f = 2\omega/\pi$
    4. $f = 2\pi\omega$
    Show answer

    Answer: B. $\omega = 2\pi f$

    Frequency ($f$) represents the number of cycles or revolutions per unit of time (usually per second), measured in Hertz (Hz). Angular frequency ($\omega$) represents the rate of rotation or oscillation in terms of angle per unit of time (usually radians per second), measured in rad/s. One complete cycle or revolution corresponds to an angle of $2\pi$ radians. If $f$ cycles occur per second, then the total angle covered per second is $f \times (2\pi)$ radians. Therefore, the angular frequency $\omega$ is related to the frequency $f$ by the equation: $\omega = 2\pi f$

Wave Packets: Phase Velocity and Group Velocity

  1. What will be the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 ms$^{-1}$ ?

    Asked in: UKPSC CS Prelims 2024

    1. $6.63 \times 10^{-30}$ m
    2. $6.63 \times 10^{-34}$ m
    3. $2.76 \times 10^{-30}$ m
    4. $2.76 \times 10^{-34}$ m
    Show answer

    Answer: D. $2.76 \times 10^{-34}$ m

    The de Broglie wavelength ($\lambda$) of a particle is given by the formula $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. The momentum is calculated as $p = mv$, where $m$ is the mass and $v$ is the velocity. Given $m = 0.12$ kg and $v = 20$ ms$^{-1}$, the momentum is $p = (0.12 \text{ kg}) \times (20 \text{ ms} ^{-1}) = 2.4$ kg ms$^{-1}$. Using Planck's constant $h \approx 6.626 \times 10^{-34}$ J s: $ \lambda = \frac{6.626 \times 10^{-34} \text{ J s}} {2.4 \text{ kg ms}^{-1}}$. Note that J s is equivalent to kg m$^2$ s$^{-1}$ $\times$ s = kg m$^2$ s$^{-1}$, and kg ms$^{-1}$ is kg m s$^{-1}$. So, $\frac{\text{kg m}^2 \text{s}^{-1}}{\text{kg m s} ^{-1}} = \text{m}$. Calculating the value: $ \times 10^{-34}$. Rounding to three significant figures gives $2.76 \times 10^{-34}$ m.

  2. The wavelength of a ball of mass 0.1kg moving with a velocity of 10ms⁻¹ is - (1) 6.626 × 6.626 × 10⁻³² m (5) Question not attempted

    Asked in: Department Recruitment 2024 (Chemist) 2024

    1. 6.626 × 10⁻³⁴ m
    2. 6.626 × 10⁻³¹ m
    3. 6.62 × 10⁻³⁰ m
    4. 6.626 × 10⁻³² m
    Show answer

    Answer: B. 6.626 × 10⁻³¹ m

    The de Broglie wavelength of a particle is inversely proportional to its momentum. Macroscopic objects have very small de Broglie wavelengths due to their large mass and thus large momentum.