Q.What physical quantity is the same for X-rays of wavelength $10^{-10}\ \text{m}$, red light of wavelength $6800\ \text{Å}$ and radiowaves of wavelength $500\ \text{m}$?

The key here is understanding what "bandwidth" means for a coaxial cable. Unlike an ideal wire that passes all frequencies, a real coaxial cable has frequency-dependent losses (skin effect, dielectric losses) that limit how high a frequency it can carry before the signal becomes too weak or distorted. For standard coaxial cables used in TV, internet, and radio frequency applications, the practical bandwidth is in the megahertz (MHz) range — not kilohertz (too low), not gigahertz (too high for typical long runs), and certainly not hertz (absurdly low).

Let’s reason through the options:

1. Eliminate obviously wrong orders of magnitude.

   • 750 Hz is audio frequency — a coaxial cable can easily carry that, but its bandwidth (the range of frequencies it can handle) is far larger. So (B) is wrong.
   • 750 kHz is AM radio range — again, too narrow for modern coaxial use. So (D) is wrong.

2. Consider the upper limit.

   • 750 GHz is in the terahertz range, far beyond what any conventional coaxial cable can transmit. At such frequencies, the cable acts like a waveguide with severe attenuation. So (A) is unrealistic.

3. Identify the correct range.

   • Coaxial cables are commonly used for cable TV, broadband internet, and RF signal distribution. Their bandwidth typically spans from a few MHz up to several hundred MHz (or a few GHz for premium cables like RG-6 or LMR-400).
   • 750 MHz is a standard figure: many cable TV systems operate up to around 750 MHz, and common coaxial cables are rated for frequencies up to 1 GHz or more. Thus, 750 MHz is a plausible and widely cited approximate bandwidth.

The klystron works on a beautiful principle: an electron beam is shot through a series of cavities, and its speed is varied (velocity modulation) by an input radio-frequency signal. This causes the electrons to bunch together as they drift, creating a strong, amplified signal at the output cavity. The key is that this process is designed for very high frequencies — specifically, the microwave region of the electromagnetic spectrum.

Why not the other options? Gamma rays and X-rays are produced by nuclear transitions or high-energy electron collisions with metal targets (like in an X-ray tube), not by velocity modulation in a vacuum tube. Infrared waves are typically generated by thermal sources or LEDs, not by klystrons. The klystron’s cavity dimensions and operating principles are tuned to wavelengths from about 1 mm to 30 cm — that’s the microwave band.

1. Understand the device’s purpose: A klystron is a specialized vacuum tube used as an amplifier or oscillator for radio frequencies. Its design — with resonant cavities and an electron gun — is optimized for frequencies above 1 GHz.

2. Identify the frequency range: The cavities in a klystron have physical dimensions that are comparable to the wavelength of the signal. For practical sizes (a few centimeters to millimeters), the corresponding frequencies fall in the microwave range (300 MHz to 300 GHz). This is far above the frequencies of ordinary radio waves but far below infrared.

3. Eliminate other options:

   • (A) Gamma rays: These have wavelengths less than 10⁻¹¹ m and are produced by nuclear decay or particle annihilation. A klystron cannot generate them.
   • (B) X-rays: Produced when high-speed electrons strike a metal target (Bremsstrahlung). A klystron’s electrons are deliberately kept from striking the output cavity wall — they are collected gently — so no X-rays are generated.
   • (D) Infrared waves: These have wavelengths from about 0.7 µm to 1 mm. While the longest infrared waves (far-infrared) border microwaves, klystrons are not designed for this range; they operate at longer wavelengths (microwaves).

4. Confirm the application: Klystrons are widely used in radar systems, satellite communication, and particle accelerators — all of which rely on microwave signals. For example, the klystron in a radar transmitter generates powerful pulses at several gigahertz.

The key idea here is that in free space, an electromagnetic wave's electric and magnetic fields are not independent — they are locked together by the speed of light. For a plane wave, the magnitudes satisfy $E = cB$ at every point and instant. This comes directly from Maxwell's equations: in a vacuum, the ratio of the field strengths is fixed by the universal constant $c$.

Let’s walk through it.

1. Recall the fundamental relation. For any electromagnetic wave travelling in free space (or vacuum), the magnitudes of the electric field $\vec{E}$ and magnetic field $\vec{B}$ are related by

where $c = 3 \times 10^8 \ \mathrm{m/s}$ is the speed of light. This is not an approximation — it follows from the wave solutions of Maxwell’s equations.

2. Identify the given data. The magnetic field is given as $\vec{B} = 2.1 \times 10^{-8} \ \hat{k} \ \mathrm{T}$. The direction ($\hat{k}$) tells us the field points along the z-axis, but for magnitude we only need the number:

3. Apply the relation. Substitute into $E = cB$:

Multiply the coefficients: $3 \times 2.1 = 6.3$. Multiply the powers of ten: $10^8 \times 10^{-8} = 10^0 = 1$. So

4. Check the options. The result $6.3 \ \mathrm{Vm}^{-1}$ matches option (D) exactly.

The key concept here is that in a plane electromagnetic wave in vacuum, the electric and magnetic fields are related by $E = cB$. But energy density depends on the square of the field, and the constants in the formulas exactly cancel this factor. Let’s see why.

1. Recall the energy density formulas

   The energy density stored in an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$.

   The energy density stored in a magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.

2. Use the wave relation between $E$ and $B$

   For an electromagnetic wave in vacuum, $E = cB$. Also, $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$.

3. Substitute $B = E/c$ into $u_B$

4. Replace $c^2$ with $1/(\varepsilon_0 \mu_0)$

5. Compare $u_E$ and $u_B$

They are identical. So the ratio $u_E : u_B = 1 : 1$.

The core idea is that audio signals (like speech or music) have low frequencies—typically 20 Hz to 20 kHz. Such low-frequency waves do not radiate efficiently from antennas of practical size, and they suffer from severe attenuation over distance. Modulation solves this by “riding” the audio signal onto a high-frequency carrier wave, which can be transmitted far more effectively.

Why not the other options?

• (A) & (B): Modulation does not primarily change the intensity (amplitude) of the audio signal; it can even reduce it. The goal is not to boost loudness.
• (D): Modulation does increase the frequency of the transmitted wave, but that is a means, not the need. The need is to enable long-distance travel, not just to raise frequency for its own sake.

Step-by-step reasoning:

1. Understand the limitation of raw audio signals

   Audio frequencies are low. For efficient transmission via an antenna, the antenna length should be comparable to the wavelength ($\lambda = c/f$). For a 1 kHz audio signal, $\lambda \approx 300$ km—impractical. Also, low-frequency waves are quickly absorbed by the atmosphere.

2. Recognize the purpose of modulation

   Modulation shifts the audio signal to a higher frequency band (e.g., radio frequencies). The high-frequency carrier wave can be transmitted with a reasonably sized antenna and travels much farther due to better propagation (e.g., via skywave or line-of-sight).

3. Evaluate each option

   • (A) Increase intensity: Modulation does not inherently amplify the audio signal; it only changes its form.
   • (B) Decrease intensity: Not the goal; we want to preserve or even improve signal strength at the receiver.
   • (C) Transmit to large distances: This is exactly why modulation is used—to overcome the range limitation of baseband audio.
   • (D) Increase frequency: While true that the carrier frequency is higher, this is a technique, not the fundamental need. The need is long-distance transmission.

4. Conclude

   The primary need for modulation in communication systems is to enable the audio signal to be sent over large distances.

The key idea is that Maxwell’s equations (Gauss’s law, Gauss’s law for magnetism, Faraday’s law, and Ampère’s law with Maxwell’s correction) are universal laws of nature. They describe how electric and magnetic fields are generated and interact, and they predict the existence of electromagnetic waves. These waves can have any frequency or wavelength — from radio waves to gamma rays — and the equations apply equally to all of them.

1. Understand what Maxwell’s equations govern

   They are a set of four differential or integral equations that relate electric fields $\mathbf{E}$, magnetic fields $\mathbf{B}$, charge density $\rho$, and current density $\mathbf{J}$. They do not contain any restriction on wavelength or frequency — they are scale-invariant in that sense.

2. Recall the prediction of electromagnetic waves

   Maxwell showed that a changing electric field produces a magnetic field, and a changing magnetic field produces an electric field. This mutual induction allows a self-sustaining wave to travel through space. The wave equation derived from Maxwell’s equations is:

and similarly for $\mathbf{B}$. The speed of these waves is $c = 1/\sqrt{\mu_0 \epsilon_0}$, which is constant — no wavelength dependence appears.

3. Consider the electromagnetic spectrum

   The spectrum includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. All are electromagnetic waves differing only in frequency $\nu$ and wavelength $\lambda = c/\nu$. Maxwell’s equations hold for every single one of them — they are the same physics.

4. Eliminate the incorrect options

   • (B) Ultraviolet only: No, because radio waves, visible light, etc., also obey Maxwell’s equations.
   • (C) Visible only: Same reasoning — visible light is just a tiny slice of the spectrum.
   • (D) Radio waves only: Again, too narrow; the equations apply to all frequencies.

The core idea is simple: when a low-frequency signal (the message) is superimposed on a high-frequency carrier wave in amplitude modulation, the resulting modulated wave contains three frequency components — the original carrier and two new frequencies called the upper sideband and lower sideband. These sidebands are the carrier frequency plus and minus the signal frequency.

Why does this happen? Mathematically, multiplying a carrier $\cos(\omega_c t)$ by a modulating signal $(1 + m \cos(\omega_m t))$ produces terms like $\cos(\omega_c t) \cos(\omega_m t)$, which expands to $\frac{1}{2}[\cos((\omega_c + \omega_m)t) + \cos((\omega_c - \omega_m)t)]$. So the sidebands appear naturally at $f_c \pm f_m$.

Now let’s apply this to the given numbers.

1. Identify the given frequencies.

   The carrier frequency is $f_c = 3\ \text{MHz}$. The modulating (signal) frequency is $f_m = 20\ \text{kHz}$.

   To avoid unit mismatch, convert everything to kHz: $f_c = 3000\ \text{kHz}$, $f_m = 20\ \text{kHz}$.

2. Compute the upper sideband frequency.

   Upper sideband = $f_c + f_m = 3000\ \text{kHz} + 20\ \text{kHz} = 3020\ \text{kHz}$.

3. Compute the lower sideband frequency.

   Lower sideband = $f_c - f_m = 3000\ \text{kHz} - 20\ \text{kHz} = 2980\ \text{kHz}$.

4. Match with the options.

   The pair (3020 kHz, 2980 kHz) appears exactly in option (D).