Past Questions — 2022

To identify the false statement, let's analyze each option based on measurement science (metrology): - **Option (a):** Reproducibility refers to the closeness of agreement between measurements of the same measurand when carried out under changed conditions (different observers, different instruments, or different locations). While the wording in the option is slightly simplified, it generally aligns with the concept of consistency across varied trials. - **Option (b):** Drift is the gradual change in the output of a measuring instrument over time while the input remains constant. Therefore, if the measured value does not vary with time for a given input, the instrument has zero drift. This is a true statement. - **Option (c):** In measurement and signal processing, noise is defined as any unwanted signal or interference that obscures the useful information of the primary signal. This is a true statement. - **Option (d):** Repeatability is the closeness of agreement between independent test results obtained with the same method on identical test items in the same laboratory. The variations in repeatability are caused by random errors (stochastic processes), not deterministic ones. Deterministic errors are systematic and predictable (like offset or scale factor errors), whereas repeatability deals with the random dispersion of measurements. Therefore, stating that the nature of repeatability is 'deterministic' is false. Since the question asks for the false statement, option (d) is the correct choice.

The circuit is a linear network, so the current varies linearly with the applied voltage. Given that when V = 0, the current is I = 3 A, this means the circuit has a 3 A offset current. When V is increased to 12 V, the current increases by another 3 A to become 6 A. Hence, the correct answer is 6 A. Why the others are incorrect: - a) 3 A: This is only the current at V = 0, not at V = 12 V. - c) 7 A: This does not match the linear change implied by the given condition. - d) 9 A: This would require a larger increase than what the circuit relation indicates.

To find the electric field strength 'E' on a spherical surface enclosing a charge 'q', we use Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε₀). 1. Formula for Flux: Φ = ∮ E · dA = q / ε 2. For a spherical surface of radius 'r', the electric field 'E' is constant in magnitude and perpendicular to the surface at every point. Therefore, the integral becomes: E × (Surface Area of Sphere) = q / ε 3. The surface area of a sphere is 4πr². 4. Substituting the area: E × (4πr²) = q / ε 5. Solving for E: E = q / (4πεr²) Comparing this result to the options: - Option (a) is missing the factor of π. - Option (b) is missing the permittivity constant ε. - Option (c) matches the derived formula exactly. - Option (d) is dimensionally incorrect.

In electrostatics, a conductor is a material that allows free movement of electrons. When a conductor is placed in an external electric field or given a net charge, the free electrons redistribute themselves almost instantaneously. If there were a non-zero electric field inside the conductor, the free charges would experience a force (F = qE) and move. This movement continues until the internal electric field created by the redistributed charges exactly cancels out the external electric field. Therefore, in a state of electrostatic equilibrium, the net electric field inside the bulk of a conductor is always zero. Analysis of other options: - Option (a) 'unity' is incorrect as there is no physical basis for the field to be exactly 1 V/m. - Option (b) 'always perpendicular to the surface' describes the electric field at the SURFACE of the conductor, not 'in' (inside) the conductor. - Option (c) 'always parallel to the surface' is incorrect because any parallel component of the field would cause charges to move along the surface until that component is neutralized.

In Amplitude Modulation (AM), the modulated signal consists of a carrier frequency (fc) and two sidebands: the Upper Sideband (USB) and the Lower Sideband (LSB). 1. **Bandwidth (BW)** of an AM signal is defined as the difference between the highest frequency component (f_max) and the lowest frequency component (f_min). For AM, BW = 2 * fm, where fm is the highest frequency of the modulating signal. Given: BW = 10 kHz. 2. **Highest Frequency Component (f_max)** is the upper edge of the Upper Sideband. Formula: f_max = fc + fm. Given: f_max = 705 kHz. 3. **Calculating fm**: Since BW = 2 * fm, then fm = BW / 2 = 10 kHz / 2 = 5 kHz. 4. **Calculating Carrier Frequency (fc)**: Using the formula f_max = fc + fm: 705 kHz = fc + 5 kHz fc = 705 kHz - 5 kHz fc = 700 kHz. Therefore, the carrier signal frequency is 700 kHz. Option (a) 695 kHz represents the lowest frequency component (f_min = fc - fm), and options (c) and (d) are incorrect based on the bandwidth and maximum frequency provided.

In ship-to-ship communication, the primary cause of fading is multipath propagation. Signals reflect off the sea surface, creating multiple paths that interfere with each other at the receiver (often resulting in Rayleigh fading). 1. **Space Diversity (Correct):** This involves using two or more antennas separated by a specific distance. Because the fading patterns are spatially dependent, it is highly unlikely that all antennas will experience a deep fade simultaneously. By combining the signals from multiple antennas, the receiver can maintain a stable connection, effectively reducing fading. This is the standard solution for maritime environments where the sea surface acts as a reflective plane. 2. **Directional Antenna (Incorrect):** While a directional antenna increases gain in one direction, it does not eliminate multipath fading caused by reflections from the water surface, and it is impractical for ships that are constantly moving and pitching/rolling. 3. **Broad band antenna (Incorrect):** Bandwidth refers to the range of frequencies an antenna can handle; it does not mitigate the physical phenomenon of multipath fading. 4. **Frequency Diversity (Incorrect):** While frequency diversity can reduce fading, it requires transmitting the same information on different frequencies. In maritime communication, space diversity is more efficient and commonly implemented via antenna arrays to combat the specific geometry of sea-surface reflections.

In a rectangular waveguide, the boundary conditions require that the tangential electric field must be zero at the conducting walls. For Transverse Electric (TE) modes, the electric field is entirely transverse to the direction of propagation (Ez = 0), and the boundary conditions are satisfied for various integer values of m and n (where m, n = 0, 1, 2...). For Transverse Magnetic (TM) modes, the magnetic field is entirely transverse (Hz = 0), but the boundary conditions for the electric field require that both m and n must be non-zero integers (m, n = 1, 2, 3...). Analyzing the options: - Option (a) 'TM': While TM modes can exist, they cannot exist for all indices (specifically, m=0 or n=0 is forbidden). However, TE modes are the primary and most common modes associated with rectangular waveguides. - Option (b) 'TE': This is the most general and correct answer as TE modes (including the dominant TE10 mode) always exist in rectangular waveguides. - Option (c) 'TM 10': This mode cannot exist because for TM modes, the index 'n' cannot be zero. If n=0, the electric field components vanish identically. - Option (d) 'TM 01': This mode cannot exist because for TM modes, the index 'm' cannot be zero. Therefore, while both TE and TM modes can exist in a rectangular waveguide, the specific TM options provided (TM 10 and TM 01) are physically impossible. Between 'TM' and 'TE', 'TE' is the definitive characteristic mode of rectangular waveguides (starting from the dominant TE10 mode).

In Delta Modulation (DM), slope overload distortion occurs when the rate of change of the input signal is faster than the maximum rate of change that the modulator can track. The maximum slope that a DM system can track is given by the formula: (Δ / Ts), where Δ is the step size and Ts is the sampling period. 1. To prevent or reduce slope overload distortion, the system's maximum slope (Δ / Ts) must be greater than or equal to the maximum slope of the input signal (max |dm(t)/dt|). 2. To increase the tracking capability (the maximum slope), one can either increase the step size (Δ) or decrease the sampling period Ts (which means increasing the sampling rate). Evaluating the options: - Option (a) Decreasing the step size: This would lower the maximum slope the system can track, actually increasing the likelihood of slope overload distortion. - Option (b) Decreasing the granular noise: Granular noise is the opposite of slope overload; it occurs when the step size is too large for a slowly varying signal. Reducing granular noise usually involves decreasing the step size, which worsens slope overload. - Option (c) Decreasing the sampling rate: Decreasing the sampling rate increases Ts, which reduces the maximum slope (Δ / Ts), thereby increasing slope overload distortion. - Option (d) Increasing the step size: This directly increases the maximum slope the system can track, thereby reducing slope overload distortion.

In ground wave propagation, the electromagnetic wave travels along the surface of the Earth. The Earth is not a perfect conductor; it has a finite conductivity (σ). Because the Earth has a certain thickness of conductive material over a non-conductive core (or simply due to the nature of the soil/sea interface), it acts as a dielectric medium with loss. Specifically, the Earth's surface acts as a capacitor (storing energy) but because it is not a perfect insulator, it allows some current to leak through (due to its finite conductivity). This combination of capacitive behavior and resistive loss is electrically modeled as a 'leaky capacitor'. This leakage causes the attenuation of the ground wave as it travels, as energy is dissipated as heat in the ground.

20 dB is the best answer because it directly satisfies the fact, principle, or condition asked in the question.

Let's trace the execution of the instructions step-by-step for the 8085 microprocessor: 1. **MVI A, A7H**: This instruction moves the immediate value A7H into the Accumulator (A). - Accumulator (A) = A7H = 1010 0111 in binary. 2. **ORA A**: This is a logical OR operation between the Accumulator and itself. Any value ORed with itself remains unchanged. - A = 1010 0111 OR 1010 0111 = 1010 0111 (A7H). - Note: This instruction is often used to clear the Carry flag and set the Zero flag based on the content of A. 3. **RLC**: This is 'Rotate Accumulator Left'. In this operation, the Most Significant Bit (MSB) is moved to the Least Significant Bit (LSB) position and also copied into the Carry flag. - Current A = 1010 0111 - MSB is '1'. - Shifting left: 0100 111_ - Moving MSB (1) to LSB: 0100 1111 - Binary 0100 1111 converted to Hexadecimal: - 0100 = 4 - 1111 = F - Result = 4FH. Wait, let's re-examine the options and the logic. If the result is 4FH, that matches option (b). However, let's double-check the binary for A7H: A=1010, 7=0111. Total: 10100111. RLC: 01001111 = 4FH. Wait, looking at the options provided in the prompt, if the intended answer is CEH, let's check if the input was different. If the input was A7H and we rotate left, we get 4FH. If we rotate right (RRC), we get 53H. Let's re-read the binary: A7H = 1010 0111. RLC (Rotate Left): Bit 7 -> Bit 0 Bit 6 -> Bit 7 Bit 5 -> Bit 6 Bit 4 -> Bit 5 Bit 3 -> Bit 4 Bit 2 -> Bit 3 Bit 1 -> Bit 2 Bit 0 -> Bit 1 1 0 1 0 0 1 1 1 ---> 0 1 0 0 1 1 1 1 (4FH). However, in many competitive exam keys for this specific question (MPSC/Technical), there is often a typo in the question's provided hex value or the options. If the starting value was 67H (0110 0111), RLC would be CEH (1100 1110). If the starting value was A7H, the result is 4FH. Given the options provided, if we strictly follow A7H, the answer is 4FH. But if the question intended a different starting value to reach CEH, we must check the logic. Let's re-verify A7H: 1010 0111. RLC -> 0100 1111 (4FH). Correction: Let's check if the instruction was 'RAL' (Rotate Left through Carry). If Carry was 1: 1(C) 1010 0111 -> 1010 0111 1 -> 0100 1111 (Still 4FH). Actually, looking at the binary for CEH: 1100 1110. If we rotate this RIGHT (RRC), we get 0110 0111 (67H). If the question is exactly as written: 1. A = A7H (1010 0111) 2. A = A OR A = A7H 3. A = RLC(A7H) = 4FH. Therefore, the correct mathematical result is 4FH (Option b).

To find the total power produced by an Amplitude Modulated (AM) transmitter, we use the standard power formula for AM: Total Power (Pt) = Pc * (1 + (m^2 / 2)) Where: - Pc = Carrier Power = 50 kW - m = Modulation Index = 80% = 0.8 Step-by-step calculation: 1. Square the modulation index: m^2 = (0.8)^2 = 0.64 2. Divide by 2: 0.64 / 2 = 0.32 3. Add 1 to this value: 1 + 0.32 = 1.32 4. Multiply by the carrier power: Pt = 50 kW * 1.32 5. Pt = 66 kW Therefore, the total power produced is 66 kW. Analysis of other options: - Option (a) 40 kW: This is less than the carrier power, which is impossible since modulation adds sideband power to the carrier. - Option (b) 50 kW: This would be the power if the modulation index were 0%. - Option (d) 100 kW: This would require a modulation index of approximately 1.73 (173%), which is over-modulation and not consistent with the given 80%.

Satellite communication requires frequencies that can penetrate the Earth's ionosphere without being reflected back to Earth. 1. **Ionospheric Penetration**: Lower frequency bands such as MF (Medium Frequency) and HF (High Frequency) are reflected by the ionosphere (sky-wave propagation), making them unsuitable for space communication. 2. **VHF (Very High Frequency)**: Ranges from 30 MHz to 300 MHz. While some VHF signals are reflected, the upper end of the VHF spectrum can penetrate the ionosphere, and it is used in various satellite services (e.g., some LEO satellites). 3. **UHF (Ultra High Frequency)**: Ranges from 300 MHz to 3 GHz. UHF signals easily penetrate the ionosphere and are widely used for satellite uplinks and downlinks (including the L-band and S-band). 4. **Higher Bands**: Most modern commercial satellites operate even higher in the SHF (Super High Frequency) range (C-band, Ku-band, Ka-band), but among the given options, the combination of VHF and UHF is the only one that allows for ionospheric penetration. Therefore, options (a), (b), and (d) are either incomplete or incorrect because MF and HF cannot reach space.

When designing or selecting a satellite system, the primary objective is to determine who needs to be served and where they are located. This is known as the 'Coverage Area' or 'Footprint'. 1. **Coverage Area (The First Factor):** Before calculating power or antenna sizes, the engineer must define the geographic region the satellite must cover. The coverage area determines the orbit (LEO, MEO, or GEO), the number of satellites required for a constellation, and the beamwidth of the antennas. 2. **Why other options are secondary:** - **Antenna Gain (c) and Antenna Size (a):** These are technical parameters derived from the required coverage area. Once the coverage area is defined, the beamwidth is calculated, which then determines the required antenna gain and, consequently, the physical size of the antenna. - **EIRP (b):** Effective Isotropic Radiated Power (EIRP) is a measure of the power transmitted. This is calculated based on the link budget to ensure that the signal reaching the ground (within the already defined coverage area) is strong enough to be detected by the receiver. Therefore, the logical flow of satellite system design is: Coverage Area $\rightarrow$ Orbit/Beamwidth $\rightarrow$ Antenna Gain/Size $\rightarrow$ Link Budget/EIRP.

In satellite communication, the C-band (approximately 4-8 GHz) was the first widely used frequency band. However, as the number of satellites increased, the C-band became heavily congested (overcrowded). Additionally, C-band shares frequencies with terrestrial microwave links, leading to significant interference. To overcome this 'overcrowding' and interference, the industry shifted toward the Ku-band (12-18 GHz). The Ku-band offers more available spectrum and allows for smaller antenna sizes due to the higher frequency. Evaluating other options: - (a) Lesser attenuation: Incorrect. Higher frequencies (Ku-band) actually suffer from GREATER attenuation, particularly rain fade, compared to C-band. - (b) Less power requirements: Incorrect. Higher frequencies generally require more sophisticated power management and can be more susceptible to signal loss. - (c) More bandwidth: While Ku-band does provide more bandwidth, the primary driver for the 'shift' from the established C-band was the saturation and interference (overcrowding) of the C-band spectrum.

A Klystron is a specialized linear-beam vacuum tube used as an amplifier for microwave frequencies. Its operation is based on the principle of velocity modulation. 1. **Velocity Modulation**: In a klystron, an electron beam is passed through a 'buncher cavity'. An RF input signal is applied to this cavity, creating an oscillating electric field. Electrons passing through the cavity are either accelerated or decelerated depending on the phase of the RF signal at the moment they enter. This creates variations in the velocities of the electrons, which is termed 'velocity modulation'. 2. **Bunching**: As these velocity-modulated electrons travel through a field-free drift space, the faster electrons catch up with the slower ones that preceded them, causing the electrons to form 'bunches'. 3. **Energy Conversion**: These bunches then pass through a 'catcher cavity', where they induce a high-frequency current, effectively converting the kinetic energy of the electron beam into microwave energy. Why other options are incorrect: - **Amplitude Modulation (b)**, **Frequency Modulation (c)**, and **Phase Modulation (d)** are methods of encoding information onto a carrier wave (modulation techniques) and are not the physical principles upon which the internal operation of a microwave vacuum tube like the Klystron is based.

1. Definitions: - Bit Rate (R): The number of bits transmitted per second (bps). - Baud Rate (S): The number of signal elements (symbols) transmitted per second (symbols/sec). 2. The Relationship: The relationship between bit rate and baud rate is given by the formula: Bit Rate = Baud Rate × log₂(L), where L is the number of levels (symbols) in the signal. 3. Binary Transmission: In binary transmission, there are only two possible signal levels (L = 2), such as 0 and 1. Substituting L = 2 into the formula: Bit Rate = Baud Rate × log₂(2) Since log₂(2) = 1, we get: Bit Rate = Baud Rate. 4. Analysis of Options: - Option (a) is correct because in a binary system, one symbol carries exactly one bit. - Option (b) and (d) refer to the Nyquist bandwidth (BW = S/2 for ideal low-pass channels), but the question asks specifically about the relationship between baud rate and bit rate in binary transmission, not the channel capacity limit. - Option (c) is incorrect because 'signalling rate' is another term for 'baud rate'; therefore, they are always equal by definition.

In a Pulse Code Modulation (PCM) system, the transmission bandwidth (B) is directly proportional to the bit rate (R). 1. The formula for the bit rate is: R = n * fs, where 'n' is the number of bits per sample and 'fs' is the sampling frequency. 2. The minimum bandwidth required for the transmission of a PCM signal is B = R / 2 (for baseband) or B = R (for certain line codes), but in all cases, B ∝ n. 3. Given: Initial bits per sample (n1) = 8 Final bits per sample (n2) = 16 4. The ratio of the new bandwidth (B2) to the old bandwidth (B1) is: B2 / B1 = n2 / n1 = 16 / 8 = 2. Therefore, the bandwidth increases by 2 times. Option (a) is incorrect because the bandwidth increases, not decreases. Option (c) and (d) are incorrect as they do not follow the linear relationship between the number of bits per sample and the required bandwidth.

48 H is correct based on the standard electrical/electronic relation governing the quantity in the question.

16 is correct based on the standard electrical/electronic relation governing the quantity in the question.

Purely resistive is correct based on the standard electrical/electronic relation governing the quantity in the question.