Q.Water waves produced by a motor boat sailing in water are

Concept and Intuition

Radio communication modes are classified by how the wave propagates: ground waves follow the earth's curvature (used for AM broadcasts), sky waves reflect off the ionosphere (used for long-distance shortwave), and space waves travel in a straight line (line-of-sight), used for TV, FM, and satellite communication because they penetrate the ionosphere rather than reflecting off it.

Step-by-Step Solution

1. Line-of-sight communication (like TV/FM/microwave links) requires waves that don't bend and travel in a straight path between transmitter and receiver.
2. Satellite communication requires waves that can pass through the ionosphere (not reflect off it), reaching the satellite in orbit.
3. Both requirements are met specifically by space waves.

Common Mistakes

• Confusing sky waves (used for long-distance shortwave via ionospheric reflection) with space waves (used for line-of-sight/satellite, which must pass through the ionosphere).

ANSWER: A

Concept and Intuition

The magnetron combines a static magnetic field (from permanent magnets) with a radial electric field in a set of resonant cavities. Electrons emitted from a central cathode spiral outward under the combined fields and interact with the cavity resonators, giving up energy to sustain oscillations at microwave frequencies (typically GHz range). This is exactly the working principle used in radar transmitters and household microwave ovens.

Step-by-Step Solution

1. Recall standard sources for different EM bands: X-rays from X-ray tubes/inner-shell transitions, gamma rays from nuclear transitions, radio waves from LC oscillators/antennas, microwaves from klystrons and magnetrons.
2. The magnetron is specifically identified (in NCERT's EM waves chapter) as a microwave-producing device.
3. Hence the magnetron valve produces microwaves.

Common Mistakes

• Confusing the magnetron with an X-ray tube (which needs high-energy electron bombardment of a metal target, not a magnetic-field cavity device).
• Thinking any "valve"/vacuum tube automatically means radio waves — the magnetron's cavity-resonator design specifically targets the microwave band.

ANSWER: B

Concept and Intuition

Solar radiation spans the EM spectrum (mostly visible and infrared, some UV) and, like all electromagnetic waves, consists of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation — making it transverse, not mechanical or longitudinal.

Step-by-Step Solution

1. Solar radiation travels through vacuum (space) to reach Earth, ruling out mechanical waves (which need a medium).
2. It's not stationary (it propagates outward from the sun).
3. All electromagnetic waves — visible light included — are transverse.

Common Mistakes

• Confusing "radiation" with sound or heat conduction (mechanical processes); solar radiation is purely electromagnetic.

ANSWER: C

Concept and Intuition

Radio communication can use ground waves (follow Earth's curvature, low frequency), sky waves (reflect off the ionosphere, medium/high frequency), or space waves (travel in a straight line directly between antennas, high frequency such as VHF/UHF/microwave). Because high-frequency waves are neither reflected by the ionosphere nor able to follow Earth's curvature well, they require direct line-of-sight propagation — hence the technique's name.

Step-by-Step Solution

1. LOS communication is used for high-frequency signals (VHF and above) which are not reflected by the ionosphere.
2. Such signals propagate directly (in a straight line) between transmitter and receiver — this is the definition of a space wave.

Common Mistakes

• Confusing space waves with sky waves (which rely on ionospheric reflection, used for longer-range non-LOS communication).

ANSWER: B

Concept and Intuition

A longitudinal wave (like sound) propagates by successive compressions (particles

bunched closer, so density is locally higher) and rarefactions (particles spread

farther apart, so density is locally lower). This periodic density variation is the

defining feature of how the disturbance travels through the medium.

Step-by-Step Solution

1. In a longitudinal wave, particle displacement is along the direction of wave propagation.
2. This causes local bunching (compression, higher density, higher pressure) and local spreading (rarefaction, lower density, lower pressure) that travel through the medium.
3. So density (and correspondingly pressure) at any point oscillates about the equilibrium value as the wave passes — density varies.

Common Mistakes

• Assuming pressure/density stay constant because "the medium as a whole" isn't compressed — but locally, at any instant, density does vary from point to point.

ANSWER: A

Concept and Intuition

A travelling (progressive) wave is defined by an equation of the form $y=f(kx\pm\omega t)$ — a single, finite, real-valued function of the combined variable $kx\pm\omega t$, valid for every $x$ and $t$. Two properties distinguish it from a standing wave:

1. It must genuinely be a function of the single combined argument (not separately of $x$ and $t$) — a product like $\cos(kx)\sin(\omega t)$ is a standing wave, formed by superposing two waves travelling in opposite directions, and every point oscillates "in place" with an amplitude that varies with position, rather than a disturbance that moves.
2. It must be finite and well-defined (real) for all values of $x$ and $t$ — a function that becomes undefined, infinite, or imaginary over part of the domain does not correctly describe a physical wave everywhere.

Step-by-Step Solution

1. (A) $y=2\cos3x\sin10t$: this is a product of a function of $x$ alone and a function of $t$ alone — the classic signature of a standing wave (e.g. a vibrating string fixed at both ends). Not progressive.
2. (B) $y=2\sqrt{x-vt}$: this is a function of the single variable $(x-vt)$, which looks promising. But $\sqrt{x-vt}$ is only real when $x\ge vt$; for $x<vt$ the expression is imaginary, so $y$ is not defined as a real physical displacement over the whole domain. A genuine progressive wave function must be finite and real everywhere — this one fails that test, so it is not accepted as representing a progressive wave.
3. (C) $y=3\sin(5x-0.5t)+4\cos(5x-0.5t)$: both terms are functions of the same combined argument $(5x-0.5t)$. Using $R\sin\theta+S\cos\theta = \sqrt{R^2+S^2}\sin(\theta+\phi)$, this combines into a single sinusoid $5\sin(5x-0.5t+\phi)$ (with $\tan\phi=4/3$) — a perfectly well-behaved, everywhere-finite function of $(5x-0.5t)$ alone. This is a genuine progressive (travelling) wave, moving with speed $0.5/5=0.1$.
4. (D) $y=\cos x\sin t+\cos2x\sin2t$: this is a sum of two different standing waves (different spatial and temporal frequencies), each individually a product $f(x)g(t)$. It is not a function of a single $(x\pm vt)$ variable and represents superposed standing-wave modes, not a travelling wave.
5. So only (C) qualifies as a progressive wave, matching option (B) of the answer choices.

Common Mistakes

• Accepting (B) just because $\sqrt{x-vt}$ "looks like" $f(x-vt)$, without checking that it must be finite/real for all $x,t$ — the domain restriction disqualifies it.
• Missing that (C)'s two terms share the exact same argument, and instead treating $\sin(\cdot)$ and $\cos(\cdot)$ as needing to be separately checked — together they still reduce to one clean travelling sinusoid.
• Confusing (D) (sum of two standing waves) with a travelling wave just because it 'isn't a single simple product' — each term individually is still $f(x)g(t)$ in form.

ANSWER: B

Concept and Intuition

Radio wave propagation mode depends strongly on frequency: low-frequency ground waves follow the Earth's curvature (only over shorter ranges, limited by frequency), sky waves (used by shortwave/HF) reflect off the ionosphere, but this reflection requires the wave frequency to be below the ionosphere's critical frequency — UHF is far above this. So UHF signals pass straight through the ionosphere and must be received along a direct, unobstructed line-of-sight path between transmitting and receiving antennas — this is called space-wave (or "line of sight") propagation, and is why UHF-band systems (TV, mobile telephony) need many closely-spaced towers/base stations or satellites.

Step-by-Step Solution

1. UHF range is roughly 300 MHz to 3 GHz.
2. At these frequencies, the ionosphere does not reflect the wave — it passes through, ruling out sky-wave propagation.
3. Ground/surface waves are heavily attenuated at high frequencies due to ground absorption, ruling them out too.
4. The only remaining and dominant propagation mechanism at UHF is direct line-of-sight travel between antennas — space wave propagation.

Common Mistakes

• Confusing UHF with the (much lower) MF/HF bands that use sky-wave propagation via ionospheric reflection.
• Thinking ground waves can carry UHF signals over long distances — ground-wave range shrinks drastically as frequency increases, and is negligible at UHF.

ANSWER: A

Concept and Intuition

Each region of the electromagnetic spectrum is suited to particular applications by its wavelength/frequency and energy. Microwaves, with wavelengths from about 1 mm to 30 cm, are used for radar, satellite communication, and microwave ovens because they can be focused into narrow beams by comparatively small antennas and reflect well off metallic/large objects.

Step-by-Step Solution

1. FM radio uses long-wavelength radio waves (tens of MHz), not microwaves.
2. Eye surgery (e.g. LASIK) uses UV/laser light, not microwaves.
3. Cancer treatment (radiotherapy) uses high-energy gamma rays or X-rays, not microwaves.
4. Radar uses microwaves — their short wavelength allows a narrow, directable beam and good reflection off aircraft/ships, giving precise range and direction.

Common Mistakes

• Confusing "microwave" (used for radar and ovens) with "radio wave" (used for FM broadcast) — both are electromagnetic but in very different bands.

ANSWER: D

Concept and Intuition

The electromagnetic spectrum is ordered by wavelength: radio waves (km–m), microwaves (mm–cm), infrared, visible (~400–700 nm), ultraviolet (~10–400 nm), X-rays (~0.01–10 nm), and gamma rays (<0.01 nm, overlapping with X-rays at the boundary). A wavelength of 1 nm sits squarely in the X-ray region.

Step-by-Step Solution

1. Given $\lambda = 1$ nm $=10^{-9}$ m.
2. Compare to standard band boundaries: UV extends down to about 10 nm; X-rays span roughly 0.01–10 nm; gamma rays are typically below 0.01 nm (though the X-ray/gamma boundary is more about origin than strict wavelength).
3. 1 nm falls within the X-ray range.

Common Mistakes

• Confusing X-rays and gamma rays — the boundary is fuzzy, but a wavelength of exactly 1 nm is conventionally classified as (soft) X-ray, not gamma ray, in standard NCERT-level classification.

ANSWER: C

Concept and Intuition

The speed of a mechanical wave is always $v=\sqrt{\text{elastic property}/\text{inertial property}}$, but which elastic property matters depends on the type of deformation the wave produces and the geometry of the medium (thin rod vs. bulk medium vs. surface).

Step-by-Step Solution

1. (a) Transverse wave in a steel rod: transverse waves in a solid are shear waves, restored by the rigidity (shear) modulus $\eta$: $v=\sqrt{\eta/\rho}$ — matches (ii).
2. (b) Longitudinal waves in the Earth's crust: the crust behaves as an extended (bulk) solid medium, not a thin rod, so the longitudinal wave speed involves both the bulk modulus and $\tfrac43$ of the shear modulus: $v=\sqrt{(B+\tfrac43\eta)/\rho}$ — matches (i).
3. (c) Longitudinal waves through a steel rod: in a thin rod (free lateral surfaces, no lateral constraint), the relevant modulus is Young's modulus: $v=\sqrt{Y/\rho}$ — matches (iv).
4. (d) Ripples: these are short-wavelength water waves dominated by surface tension $T$ rather than gravity, with speed $v=\sqrt{2\pi T/(g\lambda)}$ — matches (iii).
5. Combined: (a-ii), (b-i), (c-iv), (d-iii) — option (A).

Common Mistakes

• Using the bulk-modulus-plus-shear formula for a thin rod (that formula is only for an unbounded/extended medium like Earth's crust) instead of the simpler $\sqrt{Y/\rho}$.
• Confusing ripples (surface-tension dominated, short wavelength) with ordinary gravity waves (which instead use $v=\sqrt{g\lambda/2\pi}$).

ANSWER: A