National Institute of Technology Mizoram Entrance Exam Set

The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the Discrete Fourier Transform (DFT) and its properties. The scenario involves a discrete-time signal \(x[n]\) and its DFT \(X[k]\). The core concept being tested is the linearity property of the DFT, which states that if \(x_1[n] \leftrightarrow X_1[k]\) and \(x_2[n] \leftrightarrow X_2[k]\), then \(a x_1[n] + b x_2[n] \leftrightarrow a X_1[k] + b X_2[k]\) for constants \(a\) and \(b\). In this problem, we are given a signal \(y[n] = 3x[n-2] + 5x[n+1]\). We need to find its DFT, \(Y[k]\). First, consider the time-shifting property of the DFT: if \(x[n] \leftrightarrow X[k]\), then \(x[n-n_0] \leftrightarrow e^{-j \frac{2\pi k n_0}{N}} X[k]\), where \(N\) is the DFT length. Applying this to the first term, \(3x[n-2]\): The DFT of \(x[n-2]\) is \(e^{-j \frac{2\pi k (2)}{N}} X[k] = e^{-j \frac{4\pi k}{N}} X[k]\). Therefore, the DFT of \(3x[n-2]\) is \(3 e^{-j \frac{4\pi k}{N}} X[k]\). Applying this to the second term, \(5x[n+1]\): The DFT of \(x[n+1]\) is \(e^{-j \frac{2\pi k (-1)}{N}} X[k] = e^{j \frac{2\pi k}{N}} X[k]\). Note that \(n+1\) is equivalent to \(n – (-1)\), so \(n_0 = -1\). Therefore, the DFT of \(5x[n+1]\) is \(5 e^{j \frac{2\pi k}{N}} X[k]\). Now, using the linearity property, the DFT of \(y[n] = 3x[n-2] + 5x[n+1]\) is the sum of the DFTs of the individual terms: \(Y[k] = 3 e^{-j \frac{4\pi k}{N}} X[k] + 5 e^{j \frac{2\pi k}{N}} X[k]\). Factoring out \(X[k]\), we get: \(Y[k] = \left(3 e^{-j \frac{4\pi k}{N}} + 5 e^{j \frac{2\pi k}{N}}\right) X[k]\). This result directly corresponds to one of the options. The question tests the candidate’s ability to recall and apply the time-shifting and linearity properties of the DFT, which are foundational concepts in digital signal processing, a key area of study in electrical engineering programs at institutions like the National Institute of Technology Mizoram. Understanding these properties is crucial for analyzing and manipulating signals in various applications, from communication systems to image processing, aligning with the interdisciplinary research strengths at NIT Mizoram.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in the context of signal reconstruction. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a signal containing frequencies up to 15 kHz is being sampled. To avoid aliasing and ensure faithful reconstruction, the sampling frequency must be at least twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the signal is sampled at 25 kHz, which is less than the Nyquist rate of 30 kHz, aliasing will occur. Aliasing is the phenomenon where higher frequencies in the analog signal are misrepresented as lower frequencies in the sampled digital signal. This distortion makes it impossible to accurately reconstruct the original analog signal from the sampled data. The frequencies above \(f_s/2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\) will fold back into the lower frequency range, corrupting the information content. Consequently, the reconstructed signal will not be identical to the original signal, and the fidelity will be compromised due to the presence of these aliased components. The core concept tested here is the direct application of the Nyquist criterion for preventing aliasing, a critical aspect of digital signal processing taught in introductory courses at institutions like NIT Mizoram, particularly within their Electrical and Electronics Engineering or Computer Science programs. Understanding this principle is vital for anyone working with analog-to-digital conversion and subsequent signal processing.

The question assesses understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for signal reconstruction. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling rate is known as the Nyquist rate. In the given scenario, a continuous-time signal containing frequencies up to 15 kHz is being sampled. To avoid aliasing, which is the distortion that occurs when the sampling frequency is too low, the sampling frequency must be greater than twice the maximum frequency. Therefore, the minimum required sampling frequency is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the signal is sampled at 25 kHz, which is less than the Nyquist rate of 30 kHz, aliasing will occur. This means that frequencies above \(f_s/2 = 25 \text{ kHz}/2 = 12.5 \text{ kHz}\) will be incorrectly represented as lower frequencies within the range of 0 to 12.5 kHz. Specifically, the 15 kHz component will be aliased. The aliased frequency (\(f_{aliased}\)) can be calculated using the formula \(f_{aliased} = |f – n \cdot f_s|\), where \(n\) is an integer chosen such that \(0 \le f_{aliased} \le f_s/2\). For the 15 kHz component and a sampling frequency of 25 kHz: \(f_{aliased} = |15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). This 10 kHz component will be indistinguishable from an original 10 kHz signal, leading to distortion. The question asks about the consequence of sampling at 25 kHz when the signal has a maximum frequency of 15 kHz. Since 25 kHz is less than the Nyquist rate (30 kHz), aliasing will occur. The highest frequency component (15 kHz) will be misrepresented. The correct option describes this phenomenon and its impact on the reconstructed signal’s fidelity. The National Institute of Technology Mizoram, with its focus on cutting-edge engineering and technology, emphasizes a deep understanding of such fundamental signal processing principles, which are crucial for fields like telecommunications, control systems, and embedded systems development, all areas of significant research and academic pursuit within the institute.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). If the sampling frequency is less than twice the highest frequency, aliasing occurs, where higher frequencies in the analog signal masquerade as lower frequencies in the sampled digital signal. In this scenario, the analog signal contains frequency components up to 15 kHz. The sampling is performed at 20 kHz. According to the Nyquist criterion, the minimum sampling frequency required to avoid aliasing for a signal with a maximum frequency of 15 kHz would be \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). Since the actual sampling frequency (20 kHz) is less than the required minimum (30 kHz), aliasing will occur. To prevent aliasing before sampling, an anti-aliasing filter is used. This is a low-pass filter that attenuates or removes frequencies in the analog signal that are above half the sampling frequency (the Nyquist frequency). The Nyquist frequency for a sampling rate of 20 kHz is \(20 \text{ kHz} / 2 = 10 \text{ kHz}\). Therefore, the anti-aliasing filter should be designed to effectively remove or significantly reduce any signal components at frequencies greater than 10 kHz. This ensures that the signal presented to the sampler does not contain frequencies that would cause aliasing when sampled at 20 kHz. The correct cutoff frequency for the anti-aliasing filter, to prevent aliasing of a signal with components up to 15 kHz when sampled at 20 kHz, is therefore 10 kHz.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its practical implications in preventing aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal. Mathematically, this is expressed as \(f_s \ge 2f_{max}\). In this scenario, the analog signal contains frequency components up to 15 kHz. Therefore, the minimum sampling frequency required to avoid aliasing, according to the Nyquist-Shannon theorem, is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The digital system samples this signal at 25 kHz. Since 25 kHz is less than the required minimum of 30 kHz, aliasing will occur. Aliasing is the phenomenon where high-frequency components in the original analog signal are misinterpreted as lower frequencies in the sampled digital signal, leading to distortion and loss of information. When aliasing occurs, the highest frequency that can be accurately represented without distortion is half the sampling frequency, which is known as the Nyquist frequency. In this case, the Nyquist frequency is \(25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). Any frequency component in the original signal above 12.5 kHz will be aliased. Specifically, the 15 kHz component will be reflected back into the spectrum below the Nyquist frequency. The aliased frequency of a signal with frequency \(f\) sampled at \(f_s\) is given by \(|f – k \cdot f_s|\), where \(k\) is an integer chosen such that the result is within the range \([0, f_s/2]\). For the 15 kHz component and a sampling frequency of 25 kHz, the aliased frequency is \(|15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). This 10 kHz frequency is within the representable range \([0, 12.5 \text{ kHz}]\). Therefore, the 15 kHz component will be incorrectly represented as 10 kHz in the digital signal. This demonstrates a critical failure in the sampling process, highlighting the importance of adhering to the Nyquist criterion for accurate signal digitization, a fundamental concept in electrical engineering and signal processing taught at institutions like NIT Mizoram.

The question tests the understanding of the fundamental principles of sustainable development and its application in the context of a developing region like Mizoram, which is known for its rich biodiversity and unique cultural heritage. The core concept is to balance economic growth with environmental protection and social equity. Option (a) directly addresses this by emphasizing community participation, resource conservation, and equitable benefit sharing, which are hallmarks of sustainable practices. Option (b) focuses solely on economic growth, neglecting the environmental and social dimensions crucial for sustainability. Option (c) prioritizes environmental preservation to the extent that it might hinder necessary economic development and social progress, creating an imbalance. Option (d) leans towards technological solutions without adequately considering the socio-cultural context and local participation, which are vital for long-term success in regions like Mizoram. Therefore, a strategy that integrates local knowledge, promotes ecological stewardship, and ensures inclusive development is the most aligned with the principles of sustainable development as expected in the academic discourse at National Institute of Technology Mizoram.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its practical implications in analog-to-digital conversion. The scenario describes a signal with a maximum frequency component of 15 kHz. According to the Nyquist-Shannon sampling theorem, to perfectly reconstruct an analog signal from its sampled digital representation, the sampling frequency (\(f_s\)) must be at least twice the maximum frequency (\(f_{max}\)) present in the signal. This minimum sampling rate is known as the Nyquist rate, given by \(f_{Nyquist} = 2 \times f_{max}\). In this case, \(f_{max} = 15 \text{ kHz}\). Therefore, the minimum sampling frequency required to avoid aliasing and ensure faithful reconstruction is \(f_s \ge 2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The question asks about the consequence of sampling at a rate *below* this minimum requirement. Sampling below the Nyquist rate leads to a phenomenon called aliasing. Aliasing occurs when higher frequencies in the analog signal “fold over” or masquerade as lower frequencies in the sampled digital signal. This distortion makes it impossible to recover the original signal accurately, as the aliased frequencies are indistinguishable from genuine lower frequencies. Consequently, the reconstructed analog signal will contain spurious frequency components that were not present in the original signal, and the original high-frequency components will be misrepresented or lost. This fundamentally compromises the integrity of the digital representation and its subsequent analog reconstruction. Therefore, sampling at 20 kHz for a signal with a maximum frequency of 15 kHz will result in aliasing.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal, i.e., \(f_s \ge 2f_{max}\). If the sampling frequency is less than twice the highest frequency, aliasing occurs, where high-frequency components masquerade as lower frequencies. Consider a signal with a maximum frequency component of \(f_{max} = 15\) kHz. If this signal is sampled at a rate of \(f_s = 20\) kHz, the Nyquist frequency is \(f_s/2 = 10\) kHz. Since \(f_{max} > f_s/2\), aliasing will occur. Specifically, frequencies above \(f_s/2\) will be reflected back into the baseband. For instance, a frequency of 12 kHz would appear as \(|12 \text{ kHz} – 20 \text{ kHz}| = 8\) kHz. A frequency of 15 kHz would appear as \(|15 \text{ kHz} – 20 \text{ kHz}| = 5\) kHz. To prevent aliasing, an anti-aliasing filter, which is a low-pass filter, must be applied to the analog signal *before* sampling. This filter attenuates or removes all frequency components above the Nyquist frequency (\(f_s/2\)). In this scenario, with a sampling rate of 20 kHz, the Nyquist frequency is 10 kHz. Therefore, the anti-aliasing filter must have a cutoff frequency at or below 10 kHz to ensure that no frequencies above 10 kHz are present in the signal when it is sampled. This guarantees that all sampled frequencies are within the range of 0 to 10 kHz, and no aliasing will occur. The correct cutoff frequency for the anti-aliasing filter should be set to \(f_s/2\), which is 10 kHz.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for signal reconstruction. The theorem states that to perfectly reconstruct a continuous-time signal from its samples, the sampling frequency \(f_s\) must be greater than twice the highest frequency component \(f_{max}\) present in the signal. This critical frequency, \(2f_{max}\), is known as the Nyquist rate. In this scenario, the analog signal has a maximum frequency component of 15 kHz. Therefore, the minimum sampling frequency required for perfect reconstruction, according to the Nyquist-Shannon sampling theorem, is \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). The provided sampling frequencies are 20 kHz, 30 kHz, 40 kHz, and 50 kHz. – A sampling frequency of 20 kHz is less than the Nyquist rate of 30 kHz, so aliasing will occur, and perfect reconstruction is not possible. – A sampling frequency of 30 kHz is exactly the Nyquist rate. While theoretically sufficient, in practice, it is often insufficient due to non-ideal filters and the presence of frequencies very close to \(f_{max}\). – A sampling frequency of 40 kHz is greater than the Nyquist rate of 30 kHz. This provides a margin of safety and allows for the use of practical anti-aliasing filters with a gradual roll-off, ensuring that frequencies above \(f_{max}\) are sufficiently attenuated before sampling, and that the signal can be accurately reconstructed. – A sampling frequency of 50 kHz is also greater than the Nyquist rate of 30 kHz and would also allow for reconstruction. However, the question asks for the *most appropriate* sampling frequency that balances fidelity with practical considerations like data storage and processing bandwidth. While higher frequencies are also valid, 40 kHz offers a good compromise, providing a sufficient guard band without excessive oversampling. Considering the practical aspects of signal reconstruction in digital systems, a sampling rate that provides a reasonable margin above the theoretical Nyquist rate is generally preferred. This margin helps mitigate the effects of non-ideal anti-aliasing filters and ensures robust reconstruction. Therefore, 40 kHz is the most appropriate choice among the given options for ensuring accurate reconstruction of a signal with a maximum frequency of 15 kHz.

The question probes the understanding of fundamental principles in materials science and engineering, specifically concerning the relationship between crystal structure, bonding, and macroscopic properties, a core area of study at NIT Mizoram. The scenario describes a novel alloy exhibiting exceptional tensile strength and ductility at elevated temperatures, alongside high electrical conductivity. These properties, particularly the combination of strength at high temperatures and good conductivity, are indicative of a material with a strong, yet somewhat flexible, metallic bonding structure, likely with a well-ordered crystalline lattice that facilitates electron movement. Considering the options: A) A face-centered cubic (FCC) lattice with predominantly covalent bonding would typically exhibit high hardness and brittleness, and while some FCC metals have good conductivity, the high-temperature strength and ductility combination points away from purely covalent character. Covalent bonds are directional and strong, leading to rigidity. B) A body-centered cubic (BCC) lattice with ionic bonding is highly improbable. Ionic bonding involves electron transfer and forms rigid, brittle structures with poor electrical conductivity, contradicting the observed properties. BCC structures can be strong but often lack the ductility of FCC at high temperatures. C) A hexagonal close-packed (HCP) lattice with metallic bonding, while offering good strength, can sometimes exhibit anisotropic properties and may not always provide the optimal balance of ductility and conductivity across all orientations compared to other structures. Metallic bonding is key for conductivity. D) A face-centered cubic (FCC) lattice with predominantly metallic bonding aligns best with the observed properties. FCC structures are known for their ductility due to the availability of multiple slip systems, which is crucial for the observed ductility at high temperatures. Metallic bonding inherently provides good electrical conductivity and can contribute to high tensile strength, especially in alloys where solute atoms might strengthen the lattice without completely sacrificing ductility. The combination of FCC and metallic bonding is a common foundation for many high-performance engineering alloys that balance strength, ductility, and conductivity. Therefore, the most fitting explanation for the observed properties of the new alloy is an FCC lattice with predominantly metallic bonding.

The question probes the understanding of the fundamental principles of digital logic design, specifically concerning the minimization of Boolean expressions and the implications of using different logic gates. The scenario describes a combinational logic circuit designed to control a traffic light system at an intersection near the National Institute of Technology Mizoram campus. The inputs represent sensor readings from different approaches, and the output dictates the state of the traffic light. The goal is to identify the most efficient implementation in terms of gate count and complexity, which is a core concern in digital system design. Let the input variables be A, B, C, and D, representing the sensor states. The desired output function, F, is given by the Boolean expression: \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) To simplify this expression, we can use Karnaugh maps or Boolean algebra. Let’s use Boolean algebra: \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) We can group terms that differ by only one variable. Consider the first two terms: \(\overline{A}B\overline{C} + A\overline{B}\overline{C}\). This doesn’t simplify directly. Consider the first and third terms: \(\overline{A}B\overline{C} + \overline{A}\overline{B}C = \overline{A}(B\overline{C} + \overline{B}C)\). This is the XOR operation, \(\overline{A}(B \oplus C)\). Consider the second and fourth terms: \(A\overline{B}\overline{C} + ABC\). This doesn’t simplify directly. Consider the third and fourth terms: \(\overline{A}\overline{B}C + ABC = C(\overline{A}\overline{B} + AB)\). This is \(C(\overline{A \oplus B})\), or \(C \cdot \overline{(A \oplus B)}\). Let’s re-examine the original expression and try a different grouping strategy or use a Karnaugh map. Using a Karnaugh map for 4 variables (A, B, C, D – though D is not in the expression, we can consider it as a don’t care or simply not present in the logic): | A\BC | 00 | 01 | 11 | 10 | |—|—|—|—|—| | 00 | 0 | 0 | 0 | 1 | (\(\overline{A}\overline{B}\overline{C}\) is 0, \(\overline{A}\overline{B}C\) is 1) | 01 | 0 | 0 | 0 | 1 | (\(\overline{A}B\overline{C}\) is 1) | 11 | 0 | 0 | 1 | 0 | (\(ABC\) is 1) | 10 | 0 | 0 | 0 | 1 | (\(A\overline{B}\overline{C}\) is 1) The minterms are: \(\overline{A}B\overline{C}\) (010), \(A\overline{B}\overline{C}\) (100), \(\overline{A}\overline{B}C\) (001), \(ABC\) (111). Let’s correct the Karnaugh map based on these minterms: | A\BC | 00 | 01 | 11 | 10 | |—|—|—|—|—| | 00 | 0 | 1 | 0 | 0 | (\(\overline{A}\overline{B}C\)) | 01 | 1 | 0 | 0 | 0 | (\(\overline{A}B\overline{C}\)) | 11 | 0 | 0 | 1 | 0 | (\(ABC\)) | 10 | 1 | 0 | 0 | 0 | (\(A\overline{B}\overline{C}\)) The minterms are: m1 (\(\overline{A}\overline{B}C\)), m4 (\(\overline{A}B\overline{C}\)), m8 (\(A\overline{B}\overline{C}\)), m15 (\(ABC\)). Let’s re-evaluate the expression and minterms: \(\overline{A}B\overline{C}\) = 010 (m2) \(A\overline{B}\overline{C}\) = 100 (m4) \(\overline{A}\overline{B}C\) = 001 (m1) \(ABC\) = 111 (m7) Corrected Karnaugh Map: | A\BC | 00 | 01 | 11 | 10 | |—|—|—|—|—| | 00 | 0 | 1 | 0 | 0 | (m1) | 01 | 0 | 1 | 0 | 0 | (m2) | 11 | 0 | 0 | 1 | 0 | (m7) | 10 | 0 | 1 | 0 | 0 | (m4) The ‘1’s are at positions (0,0,1), (0,1,0), (1,0,0), (1,1,1). Let’s map these to the standard 4-variable K-map structure (assuming C is the LSB for BC pair): A=0, B=0, C=1 -> 001 -> m1 A=0, B=1, C=0 -> 010 -> m2 A=1, B=0, C=0 -> 100 -> m4 A=1, B=1, C=1 -> 111 -> m7 Karnaugh Map (assuming A, B, C are the only variables): | BC\A | 0 | 1 | |—|—|—| | 00 | 0 | 1 | (m0, m4) -> m4 is present | 01 | 1 | 0 | (m1, m5) -> m1 is present | 11 | 0 | 1 | (m3, m7) -> m7 is present | 10 | 1 | 0 | (m2, m6) -> m2 is present | BC\A | 0 | 1 | |—|—|—| | 00 | 0 | 1 (m4) | | 01 | 1 (m1) | 0 | | 11 | 0 | 1 (m7) | | 10 | 1 (m2) | 0 | Grouping the ‘1’s: – m1 and m2: \(\overline{A}\overline{B}C + \overline{A}B\overline{C}\). Grouping these doesn’t yield a simple term. – m1 and m4: \(\overline{A}\overline{B}C + A\overline{B}\overline{C}\). No direct simplification. – m2 and m4: \(\overline{A}B\overline{C} + A\overline{B}\overline{C}\). No direct simplification. – m7: \(ABC\). Let’s use Boolean algebra on the original expression again, carefully: \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) Add redundant terms to facilitate grouping: \(A\overline{B}C\) and \(\overline{A}B C\). \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC + A\overline{B}C + \overline{A}BC\) (Adding terms that are already covered by other terms or are 0) This is not the correct way to add redundant terms. Let’s use the property \(X + \overline{X}Y = X + Y\). \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) Consider the terms with \(\overline{C}\): \(\overline{A}B\overline{C} + A\overline{B}\overline{C} = \overline{C}(\overline{A}B + A\overline{B}) = \overline{C}(A \oplus B)\). Consider the terms with \(C\): \(\overline{A}\overline{B}C + ABC\). This doesn’t simplify easily. Let’s try a different approach by factoring. \(F = \overline{A}B\overline{C} + A\overline{B}\overline{C} + \overline{A}\overline{B}C + ABC\) \(F = \overline{C}(\overline{A}B + A\overline{B}) + C(\overline{A}\overline{B} + AB)\) \(F = \overline{C}(A \oplus B) + C(\overline{A \oplus B})\) This expression is of the form \(\overline{C}X + C\overline{X}\), where \(X = A \oplus B\). This is the definition of the XOR operation between C and X. So, \(F = C \oplus X = C \oplus (A \oplus B)\). The XOR operation is associative, so \(F = A \oplus B \oplus C\). Let’s verify this: If A=0, B=0, C=0: \(0 \oplus 0 \oplus 0 = 0\). Original: \(\overline{0}0\overline{0} + 0\overline{0}\overline{0} + \overline{0}\overline{0}0 + 000 = 0 + 0 + 0 + 0 = 0\). Correct. If A=0, B=0, C=1: \(0 \oplus 0 \oplus 1 = 1\). Original: \(\overline{0}0\overline{1} + 0\overline{0}\overline{1} + \overline{0}\overline{0}1 + 001 = 0 + 0 + 1 + 0 = 1\). Correct. If A=0, B=1, C=0: \(0 \oplus 1 \oplus 0 = 1\). Original: \(\overline{0}1\overline{0} + 0\overline{1}\overline{0} + \overline{0}\overline{1}0 + 010 = 1 + 0 + 0 + 0 = 1\). Correct. If A=1, B=0, C=0: \(1 \oplus 0 \oplus 0 = 1\). Original: \(\overline{1}0\overline{0} + 1\overline{0}\overline{0} + \overline{1}\overline{0}0 + 100 = 0 + 1 + 0 + 0 = 1\). Correct. If A=0, B=1, C=1: \(0 \oplus 1 \oplus 1 = 0\). Original: \(\overline{0}1\overline{1} + 0\overline{1}\overline{1} + \overline{0}\overline{1}1 + 011 = 0 + 0 + 0 + 0 = 0\). Correct. If A=1, B=0, C=1: \(1 \oplus 0 \oplus 1 = 0\). Original: \(\overline{1}0\overline{1} + 1\overline{0}\overline{1} + \overline{1}\overline{0}1 + 101 = 0 + 0 + 0 + 0 = 0\). Correct. If A=1, B=1, C=0: \(1 \oplus 1 \oplus 0 = 0\). Original: \(\overline{1}1\overline{0} + 1\overline{1}\overline{0} + \overline{1}\overline{1}0 + 110 = 0 + 0 + 0 + 0 = 0\). Correct. If A=1, B=1, C=1: \(1 \oplus 1 \oplus 1 = 1\). Original: \(\overline{1}1\overline{1} + 1\overline{1}\overline{1} + \overline{1}\overline{1}1 + 111 = 0 + 0 + 0 + 1 = 1\). Correct. The simplified expression is \(A \oplus B \oplus C\). This expression can be implemented using two 2-input XOR gates. For example, \( (A \oplus B) \oplus C \). The original expression requires 4 product terms, each with 3 literals and 2 AND gates, plus an OR gate. This would be 4 * (2 AND gates + 3 literals) + 1 OR gate, which is significantly more complex. Alternatively, using the factored form \(\overline{C}(A \oplus B) + C(\overline{A \oplus B})\) requires two XOR gates for \(A \oplus B\) and \(\overline{A \oplus B}\), one XOR gate for the final output, and two AND gates and one OR gate. This is also more complex than two XOR gates. Therefore, the most efficient implementation in terms of gate count uses two 2-input XOR gates. The question asks about the most efficient implementation in terms of gate count. The simplified expression \(A \oplus B \oplus C\) can be realized using two 2-input XOR gates. For instance, one gate computes \(A \oplus B\), and the output of this gate is XORed with C using a second 2-input XOR gate. This is the minimal implementation. The other options represent less efficient implementations: – Using only AND and OR gates: This would require implementing the original sum-of-products form, which is significantly more complex. For example, each product term \(\overline{A}B\overline{C}\) requires two AND gates (one for \(\overline{A}\) and one for \(\overline{C}\)) and then a 3-input AND gate. This leads to a much higher gate count. – Using NAND gates only: While NAND gates are universal, converting the minimal XOR implementation to NAND gates would still likely result in more gates than the direct XOR implementation. For example, an XOR gate can be implemented with 4 NAND gates. Thus, two XOR gates would require 8 NAND gates. However, the question is about the most efficient implementation in terms of gate count, and the direct XOR implementation is the most straightforward and minimal in terms of gate types and count for this specific function. – Implementing the factored form \(\overline{C}(A \oplus B) + C(\overline{A \oplus B})\) directly without further simplification would involve more gates than the final XOR form. It would require gates to compute \(A \oplus B\), \(\overline{A \oplus B}\), two AND gates, and an OR gate, in addition to the final XOR. The most efficient implementation, in terms of gate count and simplicity for the function \(A \oplus B \oplus C\), is to use two 2-input XOR gates.

The question assesses understanding of the foundational principles of digital logic design, specifically related to combinational circuits and their implementation using basic gates. The scenario describes a system where a warning light activates under specific conditions related to three input sensors: a temperature sensor (T), a pressure sensor (P), and a fluid level sensor (L). The warning light (W) should illuminate when either the temperature is high (T=1) and pressure is normal (P=0), OR when the fluid level is low (L=0) and pressure is normal (P=0). This logic can be translated into a Boolean expression. Condition 1: Temperature high AND Pressure normal. This translates to \(T \cdot \bar{P}\). Condition 2: Fluid level low AND Pressure normal. This translates to \(\bar{L} \cdot \bar{P}\). The warning light activates if either Condition 1 OR Condition 2 is met. Therefore, the complete Boolean expression for the warning light (W) is: \(W = (T \cdot \bar{P}) + (\bar{L} \cdot \bar{P})\) To simplify this expression using Boolean algebra, we can factor out the common term \(\bar{P}\): \(W = \bar{P} \cdot (T + \bar{L})\) This simplified expression represents the most efficient way to implement the logic using the fewest gates. The original expression \( (T \cdot \bar{P}) + (\bar{L} \cdot \bar{P}) \) would require two AND gates and one OR gate. The simplified expression \( \bar{P} \cdot (T + \bar{L}) \) requires one NOT gate (for \(\bar{P}\)), one OR gate (for \(T + \bar{L}\)), and one AND gate. This is a more economical implementation. The question asks for the most efficient implementation, which corresponds to the simplified Boolean expression.

The question probes the understanding of the fundamental principles of sustainable development as applied to regional planning, a core concern for institutions like the National Institute of Technology Mizoram, which is situated in a region with unique ecological and socio-economic characteristics. The calculation involves a conceptual weighting of different sustainability pillars. Let’s assign hypothetical weights to the three pillars of sustainable development: Economic Viability (EV), Environmental Protection (EP), and Social Equity (SE). For a project to be considered truly sustainable, it must achieve a minimum threshold across all pillars, not just a high average. Suppose a project scores 80% on EV, 70% on EP, and 90% on SE. A common misconception is to average these scores: \(\frac{80 + 70 + 90}{3} = \frac{240}{3} = 80\%\). However, this average doesn’t account for the critical interdependence and minimum requirements of each pillar. A more robust approach, often implicitly used in advanced sustainability assessments, is to consider the *minimum* score achieved across the pillars, as a weakness in one area can undermine the entire effort. In this case, the minimum score is 70%. However, the question asks about the *most critical factor* for long-term viability in a region like Mizoram, which is known for its rich biodiversity and tribal heritage, alongside developmental aspirations. While economic growth is essential, and social well-being is paramount, the unique geographical and ecological context of Mizoram strongly emphasizes the foundational role of environmental integrity. Without a healthy ecosystem, economic activities can be unsustainable, and social well-being can be compromised by resource depletion or environmental degradation. Therefore, environmental protection, when considered as the bedrock upon which economic and social progress must be built, becomes the most critical factor for enduring success in such a setting. This is not a simple mathematical calculation but a reasoned prioritization based on regional context and the interconnectedness of sustainability dimensions. The “calculation” here is conceptual: identifying the pillar that, if neglected, poses the greatest existential threat to the other two in the specific context of Mizoram. Environmental degradation can lead to economic losses (e.g., loss of forest resources, impact on agriculture) and social disruption (e.g., displacement, health issues). Thus, prioritizing environmental protection ensures the long-term capacity for economic and social development.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation. Aliasing occurs when a continuous-time signal is sampled at a rate lower than twice its highest frequency component (Nyquist rate). This leads to the misrepresentation of higher frequencies as lower frequencies in the sampled signal. To prevent aliasing, an anti-aliasing filter, which is a low-pass filter, is applied to the analog signal *before* sampling. This filter attenuates frequencies above the desired bandwidth, ensuring that only frequencies below half the sampling rate are present in the signal when it is sampled. Consider a scenario where a signal contains frequency components up to \(f_{max}\). If this signal is sampled at a rate \(f_s\), aliasing will occur if \(f_{max} > f_s/2\). To avoid this, an anti-aliasing filter must be designed to have a cutoff frequency \(f_c\) such that \(f_c \le f_s/2\). Furthermore, the filter’s transition band should be sufficiently narrow to effectively remove frequencies above \(f_s/2\) without excessively distorting frequencies below it. The ideal scenario for anti-aliasing involves a sharp cutoff at \(f_s/2\). Therefore, the most effective strategy to prevent aliasing in digital signal processing at the National Institute of Technology Mizoram, which emphasizes rigorous theoretical understanding and practical application in its engineering programs, is to employ an anti-aliasing filter with a cutoff frequency precisely at half the sampling rate, ensuring that all frequencies above this threshold are attenuated before the analog-to-digital conversion process. This directly addresses the core cause of aliasing by limiting the spectral content of the signal to within the Nyquist limit.

The question revolves around the fundamental principles of digital logic design and the concept of Karnaugh maps (K-maps) for Boolean function minimization. The scenario describes a specific logic function with given input conditions and desired output. The task is to identify the most simplified Sum of Products (SOP) expression that satisfies these conditions, considering the potential for don’t care conditions. The given truth table implicitly defines the function. Let the inputs be A, B, and C. The output Y is 1 for the minterms 001, 011, 100, and 110. The minterms are represented by their decimal equivalents: Minterm 001 = 1 Minterm 011 = 3 Minterm 100 = 4 Minterm 110 = 6 So, the canonical SOP form is \(Y = \Sigma m(1, 3, 4, 6)\). To simplify this using a K-map, we construct a 3-variable K-map: “` BC 00 01 11 10 A 0 | 1 0 1 0 | 1 | 1 0 0 1 | “` Now, we group the adjacent 1s in powers of two. 1. Grouping the 1s at positions 1 (001) and 3 (011): These two 1s are adjacent. In the K-map, they share the same A=0 and C=1. The B variable changes, so it’s eliminated. This group represents \(\bar{A}C\). 2. Grouping the 1s at positions 4 (100) and 6 (110): These two 1s are adjacent. In the K-map, they share the same A=1 and C=0. The B variable changes, so it’s eliminated. This group represents \(AC\). The simplified SOP expression is the sum of these groups: \(Y = \bar{A}C + AC\). This expression can be further simplified using the distributive property of Boolean algebra: \(Y = C(\bar{A} + A)\). Since \(\bar{A} + A = 1\), the expression becomes \(Y = C(1)\), which simplifies to \(Y = C\). Therefore, the most simplified Sum of Products expression is \(C\). This question tests the ability to translate a functional requirement into a Boolean expression, represent it on a Karnaugh map, and apply Boolean algebra rules for minimization. Understanding how to group adjacent 1s in a K-map to eliminate variables is crucial. The simplification process demonstrates a core competency in digital logic design, essential for efficient circuit implementation, which is a foundational aspect of electrical and computer engineering programs at institutions like the National Institute of Technology Mizoram. The ability to derive the simplest form of a logic function directly impacts the number of gates and interconnections required, leading to cost savings and improved performance in digital systems. This skill is paramount for students aiming to excel in VLSI design, embedded systems, and computer architecture.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications for aliasing. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). This minimum sampling frequency is known as the Nyquist rate. Consider a scenario where a continuous-time signal with a maximum frequency component of 15 kHz is sampled at a rate of 25 kHz. According to the Nyquist criterion, the minimum sampling rate required to avoid aliasing would be \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). Since the actual sampling rate (25 kHz) is less than the required Nyquist rate (30 kHz), aliasing will occur. When aliasing occurs, frequencies above \(f_s/2\) (the Nyquist frequency) are folded back into the lower frequency range, appearing as lower frequencies that were not originally present in the signal. The aliased frequency (\(f_{alias}\)) of a frequency \(f\) sampled at \(f_s\) is given by \(f_{alias} = |f – k \cdot f_s|\), where \(k\) is an integer chosen such that \(0 \le f_{alias} \le f_s/2\). In this case, the Nyquist frequency is \(25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). A frequency component of 15 kHz, which is above the Nyquist frequency, will be aliased. To find the aliased frequency, we can use the formula. For \(f = 15 \text{ kHz}\) and \(f_s = 25 \text{ kHz}\): We need to find an integer \(k\) such that \(0 \le |15 \text{ kHz} – k \cdot 25 \text{ kHz}| \le 12.5 \text{ kHz}\). If \(k=1\), \(|15 \text{ kHz} – 1 \cdot 25 \text{ kHz}| = |-10 \text{ kHz}| = 10 \text{ kHz}\). Since \(10 \text{ kHz}\) is within the range \([0, 12.5 \text{ kHz}]\), the aliased frequency is 10 kHz. This understanding is crucial for students at the National Institute of Technology Mizoram, particularly in fields like Electronics and Communication Engineering, where signal processing is a core subject. Proper sampling is essential for accurate data acquisition, transmission, and analysis, preventing distortion and ensuring the integrity of information. Failure to adhere to sampling principles can lead to erroneous results in applications ranging from audio and video processing to medical imaging and telecommunications, all of which are areas of active research and development at NIT Mizoram.

The question probes the understanding of the ethical considerations in scientific research, specifically focusing on the principle of informed consent and its application in a hypothetical scenario involving vulnerable populations. The scenario describes a research project at the National Institute of Technology Mizoram aiming to understand the impact of traditional agricultural practices on local biodiversity. The researchers plan to interview elderly farmers in a remote village. The core ethical dilemma lies in ensuring genuine informed consent from individuals who may have limited formal education, potentially different cultural understandings of consent, and a power imbalance with the researchers. The correct approach, therefore, must prioritize the protection of these individuals. This involves not just obtaining a signature, but ensuring comprehension of the research purpose, risks, benefits, and the right to withdraw, all communicated in an understandable language and cultural context. The explanation of the research must be clear, avoiding jargon, and allowing ample time for questions. The researchers must also be sensitive to any potential coercion or undue influence, especially if they are perceived as authority figures or if there’s an expectation of benefit from participation. The principle of beneficence (doing good) and non-maleficence (avoiding harm) are paramount. The researchers must also consider the potential for secondary use of data and ensure anonymity and confidentiality are maintained throughout the study, aligning with the rigorous academic standards expected at the National Institute of Technology Mizoram.

The question assesses understanding of the fundamental principles of sustainable development and their application in the context of regional growth, a key focus for institutions like the National Institute of Technology Mizoram. The scenario describes a community in Mizoram facing resource depletion and environmental degradation due to rapid, unmanaged industrialization. The core challenge is to balance economic progress with ecological preservation and social equity. Sustainable development, as defined by the Brundtland Commission, is development that meets the needs of the present without compromising the ability of future generations to meet their own needs. This encompasses three interconnected pillars: economic viability, environmental protection, and social equity. In the given scenario, the unbridled industrial growth has led to significant environmental damage (deforestation, water pollution) and social strain (displacement, inequitable distribution of benefits). To address this, a strategy must integrate all three pillars. Option (a) proposes a multi-pronged approach: investing in renewable energy to reduce reliance on fossil fuels and mitigate pollution, implementing strict environmental regulations with robust enforcement to protect natural resources, and promoting community-based resource management to ensure equitable benefit sharing and local participation. This directly addresses the economic, environmental, and social dimensions of sustainability. Option (b) focuses solely on economic incentives for industries, neglecting environmental and social safeguards, which would likely exacerbate the existing problems. Option (c) emphasizes conservation efforts without considering the economic needs of the community or providing alternative livelihoods, which might lead to social unrest and hinder long-term adoption. Option (d) suggests technological solutions without addressing the underlying governance and community involvement aspects, which are crucial for the successful and equitable implementation of any development strategy. Therefore, the most comprehensive and effective approach, aligning with the principles of sustainable development and the mission of institutions like NIT Mizoram to foster responsible regional growth, is the one that integrates economic, environmental, and social considerations.

The question probes the understanding of fundamental principles in materials science and engineering, specifically concerning the behavior of crystalline solids under stress, a core area for students aspiring to engineering disciplines at National Institute of Technology Mizoram. The scenario describes a metallic crystal lattice. The critical aspect is identifying the primary mechanism responsible for plastic deformation in such structures. Plastic deformation, the permanent change in shape, in crystalline materials occurs through the movement of dislocations. Dislocations are line defects within the crystal lattice. Their movement, facilitated by shear stress, allows planes of atoms to slip past one another without breaking all atomic bonds simultaneously. This slip system, defined by a specific crystallographic plane and direction, is the most efficient path for dislocation motion. While other phenomena like twinning (a coordinated shear of a crystal across a specific plane, resulting in a mirror image lattice) and grain boundary sliding (at higher temperatures) also contribute to deformation, dislocation slip is the dominant mechanism for plastic deformation in most metals at typical engineering temperatures. Therefore, understanding dislocation theory and its role in slip systems is paramount. The question requires differentiating between these mechanisms based on their fundamental nature and prevalence in crystalline solids.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). If this condition is violated, higher frequencies in the analog signal masquerade as lower frequencies in the sampled digital signal, a distortion known as aliasing. Consider an analog signal with a maximum frequency component of 15 kHz. If this signal is sampled at a rate of 25 kHz, the Nyquist frequency, which is half the sampling rate, is \(f_{Nyquist} = f_s / 2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). Since the signal contains frequencies up to 15 kHz, which is greater than the Nyquist frequency of 12.5 kHz, aliasing will occur. Frequencies above the Nyquist frequency will be reflected back into the baseband. For instance, a frequency of 13 kHz would appear as \(25 \text{ kHz} – 13 \text{ kHz} = 12 \text{ kHz}\). A frequency of 15 kHz would appear as \(25 \text{ kHz} – 15 \text{ kHz} = 10 \text{ kHz}\). To prevent aliasing, an anti-aliasing filter, which is a low-pass filter, is applied to the analog signal *before* sampling. This filter attenuates or removes all frequency components above the Nyquist frequency. Therefore, to accurately sample a signal with a maximum frequency of 15 kHz without aliasing, the sampling frequency must be at least \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). If the sampling frequency is fixed at 25 kHz, the anti-aliasing filter must ensure that no frequencies above 12.5 kHz are present in the signal before it is sampled. Thus, the anti-aliasing filter must have a cutoff frequency at or below 12.5 kHz. The most appropriate cutoff frequency for the anti-aliasing filter in this scenario, to ensure no aliasing occurs when sampling at 25 kHz, would be the Nyquist frequency itself, or slightly below it to provide a guard band. Therefore, a cutoff frequency of 12.5 kHz is the critical requirement. The question asks about the *purpose* of the anti-aliasing filter in the context of sampling an analog signal at the National Institute of Technology Mizoram, implying a need to preserve signal integrity in digital systems. The core principle is to ensure that the sampled digital representation accurately reflects the original analog signal’s spectral content within the desired bandwidth. Without proper filtering, higher frequencies that exceed half the sampling rate will fold back into the lower frequency range, corrupting the signal and making accurate reconstruction impossible. This is a fundamental concept in any field involving analog-to-digital conversion, from telecommunications to sensor data acquisition, areas of study at NIT Mizoram. The filter’s role is to act as a spectral gatekeeper, ensuring that only frequencies within the valid range for the given sampling rate are allowed to pass through to the sampler, thereby preventing the introduction of spurious spectral components.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the Nyquist-Shannon sampling theorem and its implications in the context of signal reconstruction. The theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency \(f_s\) must be at least twice the highest frequency component \(f_{max}\) present in the signal, i.e., \(f_s \ge 2f_{max}\). Consider a scenario where a continuous-time signal \(x(t)\) contains frequency components up to 15 kHz. To avoid aliasing and ensure faithful reconstruction, the sampling frequency \(f_s\) must satisfy \(f_s \ge 2 \times 15 \text{ kHz}\), which means \(f_s \ge 30 \text{ kHz}\). If the sampling frequency used is 25 kHz, which is less than the required minimum of 30 kHz, aliasing will occur. Aliasing is the phenomenon where high-frequency components in the original signal are incorrectly interpreted as lower frequencies after sampling. This distortion makes it impossible to reconstruct the original signal accurately. Therefore, when a signal with a maximum frequency of 15 kHz is sampled at 25 kHz, the resulting samples will contain aliased components, rendering perfect reconstruction of the original 15 kHz signal impossible. The highest frequency that can be accurately represented without aliasing at a sampling rate of 25 kHz is \(f_{max} = f_s / 2 = 25 \text{ kHz} / 2 = 12.5 \text{ kHz}\). Any frequency component above 12.5 kHz in the original signal will be aliased. The question asks about the consequence of sampling a signal with a maximum frequency of 15 kHz at a rate of 25 kHz. Based on the Nyquist-Shannon theorem, this sampling rate is insufficient. The correct answer is that the original signal cannot be perfectly reconstructed because aliasing will occur, distorting the high-frequency components.

The question probes the understanding of the fundamental principles of data integrity and the role of hashing in ensuring it, particularly in the context of distributed systems and secure data storage, which are relevant to computer science and engineering curricula at institutions like NIT Mizoram. A cryptographic hash function takes an input (or ‘message’) of arbitrary size and produces a fixed-size string of characters, which is typically a hexadecimal number. This output is known as a hash value, message digest, or simply hash. Key properties of a good cryptographic hash function include: 1. **Pre-image resistance (One-way property):** It should be computationally infeasible to find a message \(m\) such that \(H(m) = h\) for a given hash value \(h\). 2. **Second pre-image resistance:** It should be computationally infeasible to find a different message \(m’\) such that \(H(m) = H(m’)\), given an initial message \(m\). 3. **Collision resistance:** It should be computationally infeasible to find two distinct messages \(m\) and \(m’\) such that \(H(m) = H(m’)\). The scenario describes a distributed ledger system where each block contains a hash of the previous block. This creates a chain, where any alteration to a previous block would change its hash, thus invalidating the hash stored in the subsequent block, and so on, breaking the chain. This mechanism is central to the immutability and integrity of blockchain technology. Let’s analyze the options: * **Option a) Collision resistance:** If a hash function lacks collision resistance, it means an attacker could find two different data sets that produce the same hash. In the context of a distributed ledger, if an attacker could find a different set of transactions (or even a single altered transaction) that produces the same hash as a legitimate block’s hash, they could potentially substitute a malicious block for a legitimate one without detection by the hashing mechanism. This directly undermines the integrity of the ledger. For instance, if \(H(\text{Block}_n) = H(\text{TamperedBlock}_n)\), then the link \(H(\text{Block}_n)\) in \(\text{Block}_{n+1}\) would still match \(H(\text{TamperedBlock}_n)\), allowing the tampered block to be accepted as valid. This is the most critical property for maintaining the integrity of a chained ledger. * **Option b) Pre-image resistance:** While important for security, pre-image resistance (making it hard to find the original message given the hash) is less directly related to detecting *alterations* within the chain itself. The primary threat in a chained ledger is modifying existing data, not necessarily creating a new block with a specific hash from scratch. * **Option c) Second pre-image resistance:** This property is also crucial, as it prevents an attacker from substituting a different valid block for an existing one. However, collision resistance is a stronger requirement that encompasses the ability to find *any* two different inputs that hash to the same output, making it more broadly applicable to detecting any form of data manipulation that results in a hash match. In many practical discussions of blockchain integrity, collision resistance is highlighted as the paramount property for preventing malicious data substitution. * **Option d) Reversibility:** Hash functions are intentionally designed to be irreversible (one-way). If a hash function were easily reversible, it would be trivial to reconstruct the original data from its hash, which would be a catastrophic security failure, but this is a property that *should not* exist, rather than a property that, if absent, compromises integrity. The question asks what property’s absence would compromise integrity. Therefore, the absence of collision resistance is the most direct and severe threat to the integrity of a distributed ledger system that relies on chaining hashes of previous blocks.

The question assesses understanding of fundamental principles in materials science and engineering, specifically concerning the relationship between crystal structure, mechanical properties, and processing methods, relevant to disciplines like Mechanical and Metallurgical Engineering at NIT Mizoram. Consider a hypothetical scenario involving the development of a novel alloy for high-temperature structural components, a common research area at NIT Mizoram. The alloy exhibits a face-centered cubic (FCC) crystal structure. During processing, it is observed that the alloy’s yield strength significantly increases after a specific heat treatment, accompanied by a subtle change in its diffraction pattern, suggesting the formation of a secondary phase or precipitation hardening. However, the ductility of the alloy decreases substantially post-treatment. To understand this phenomenon, we must analyze the underlying mechanisms. FCC structures are generally known for their good ductility due to the presence of multiple slip systems. The observed increase in yield strength and decrease in ductility after heat treatment strongly indicates a strengthening mechanism that impedes dislocation motion. Precipitation hardening, where fine, dispersed particles of a second phase form within the matrix, is a primary mechanism for strengthening alloys. These precipitates act as obstacles to dislocation movement, requiring higher stress to initiate plastic deformation. The decrease in ductility is a common consequence of significant strengthening. As dislocations become more difficult to move, the material’s ability to deform plastically before fracture is reduced. The subtle change in the diffraction pattern supports the formation of these precipitates, which would alter the overall diffraction peaks. Therefore, the most accurate explanation for the observed behavior, considering the FCC structure and the effects of heat treatment on mechanical properties, is that the heat treatment induced precipitation hardening within the FCC matrix, leading to increased yield strength by hindering dislocation movement, but consequently reducing the material’s overall ductility. This aligns with principles taught in materials science courses at NIT Mizoram, emphasizing the interplay between microstructure and mechanical performance.

The question probes the understanding of the fundamental principles of electromagnetic wave propagation and interaction with materials, a core concept in physics and engineering disciplines offered at National Institute of Technology Mizoram. Specifically, it tests the ability to discern the primary mechanism by which a dielectric material, characterized by its permittivity and permeability, influences the speed of an electromagnetic wave. When an electromagnetic wave encounters a dielectric medium, its electric field component interacts with the bound charges within the material. This interaction causes polarization, where the positive and negative charges within the dielectric are slightly displaced, creating induced dipoles. These induced dipoles then re-radiate electromagnetic waves, which interfere with the original wave. This interference effectively slows down the propagation of the wave through the medium. The extent of this slowing is quantified by the refractive index, \(n\), which is related to the material’s relative permittivity (\(\epsilon_r\)) and relative permeability (\(\mu_r\)) by the equation \(n = \sqrt{\epsilon_r \mu_r}\). The speed of the wave in the medium is then given by \(v = \frac{c}{n}\), where \(c\) is the speed of light in a vacuum. Therefore, the polarization of the dielectric material, driven by the electric field component of the wave, is the primary reason for the reduction in wave speed. While magnetic dipoles can also form in some materials (paramagnetic and ferromagnetic), for typical dielectrics encountered in introductory physics and engineering, the electric polarization is the dominant effect influencing wave propagation speed. The scattering of waves is a separate phenomenon, and absorption relates to energy dissipation, not the fundamental speed change.

The question probes the understanding of the fundamental principles of sustainable development and its application in the context of regional planning, a key area of focus for institutions like the National Institute of Technology Mizoram, which emphasizes balanced growth. The scenario involves a hypothetical development project in a region with unique ecological and socio-economic characteristics, mirroring the challenges faced in Mizoram. The core concept being tested is the integration of environmental preservation, economic viability, and social equity. To arrive at the correct answer, one must analyze the proposed actions against the pillars of sustainable development. The proposed establishment of a large-scale, monoculture agricultural plantation, while potentially boosting immediate economic output, carries significant risks. Monoculture farming often leads to soil degradation, increased reliance on chemical inputs (pesticides and fertilizers), and a reduction in biodiversity, all of which are detrimental to long-term ecological health. Furthermore, if this plantation displaces traditional, diverse farming practices or local communities without adequate compensation or alternative livelihoods, it would undermine social equity. The question requires evaluating which approach best aligns with the principles of sustainable development. Option (a) suggests a multi-pronged strategy that includes ecological restoration, diversification of economic activities, and community involvement. This approach directly addresses the environmental concerns (restoration, biodiversity), economic sustainability (diversification beyond a single crop), and social equity (community participation, preservation of traditional livelihoods). It acknowledges that development must be holistic and consider the long-term well-being of both the environment and its inhabitants. In contrast, other options might focus too narrowly on economic gains without sufficient environmental or social safeguards, or propose solutions that are not contextually appropriate for a region like Mizoram, which is known for its rich biodiversity and distinct cultural heritage. Therefore, a strategy that prioritizes ecological resilience, equitable benefit sharing, and the integration of local knowledge is paramount for true sustainable development.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). If the sampling frequency is less than twice the highest frequency, aliasing occurs, where higher frequencies masquerade as lower frequencies. In this scenario, the analog signal contains frequency components up to 15 kHz. The sampling is performed at 20 kHz. According to the Nyquist criterion, the minimum sampling frequency required to avoid aliasing would be \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). Since the actual sampling frequency (20 kHz) is less than the required minimum (30 kHz), aliasing will occur. An anti-aliasing filter is a low-pass filter placed before the sampler to remove or attenuate frequencies above half the sampling frequency (\(f_s/2\)). In this case, \(f_s/2 = 20 \text{ kHz} / 2 = 10 \text{ kHz}\). Therefore, an effective anti-aliasing filter for this system should have a cutoff frequency at or below 10 kHz. This ensures that any frequency components in the original analog signal that are above 10 kHz (and would otherwise cause aliasing when sampled at 20 kHz) are removed before sampling. The original signal’s highest frequency is 15 kHz, which is above the \(f_s/2\) of 10 kHz. Without an anti-aliasing filter, the 15 kHz component would alias. By using an anti-aliasing filter with a cutoff frequency of 10 kHz, the 15 kHz component is attenuated, preventing it from causing aliasing. The signal that can be perfectly reconstructed after sampling at 20 kHz will therefore be limited to frequencies up to 10 kHz.

The question assesses understanding of the fundamental principles of sustainable development and their application in the context of regional growth, a key focus for institutions like the National Institute of Technology Mizoram. The scenario describes a community in Mizoram aiming to leverage its unique biodiversity for economic advancement while preserving its ecological integrity and cultural heritage. The core concept here is the interconnectedness of economic, social, and environmental pillars of sustainability. A project that solely focuses on economic gain without considering environmental impact or community well-being would be unsustainable. Similarly, a purely conservation-focused approach that neglects economic opportunities might not be viable for the local population. The optimal approach, therefore, would integrate all three dimensions. This involves identifying economic activities that are inherently low-impact and regenerative, such as ecotourism, sustainable agriculture, or the ethical harvesting and processing of non-timber forest products. These activities should be developed in partnership with the local communities, ensuring they benefit directly and have a say in the management of resources. Furthermore, the project must incorporate robust environmental monitoring and conservation strategies to protect the biodiversity that forms the basis of these economic activities. This holistic approach aligns with the National Institute of Technology Mizoram’s commitment to fostering innovation that is both technologically advanced and socially responsible, contributing to the balanced development of the region.

The question probes the understanding of fundamental principles in digital logic design, specifically related to combinational circuits and their minimization. The scenario describes a scenario where a digital circuit needs to be designed to control a three-phase motor based on inputs from temperature sensors and a manual override. The truth table provided implicitly defines the desired output behavior. To determine the most efficient implementation, we need to derive the minimal Sum of Products (SOP) or Product of Sums (POS) expression. Let the inputs be \(A\) (high temperature), \(B\) (overload detected), and \(C\) (manual override). Let the output \(Y\) control the motor. From the truth table, the output \(Y\) is HIGH (1) for the following input combinations: – \(A=0, B=0, C=0\) (Normal, no override) – \(A=0, B=0, C=1\) (Normal, manual override) – \(A=1, B=0, C=1\) (High temp, manual override) – \(A=0, B=1, C=1\) (Overload, manual override) The minterms for which \(Y=1\) are: – \(A’B’C’\) (000) – \(A’B’C\) (001) – \(A B’C\) (101) – \(A’B C\) (011) The SOP expression is: \(Y = A’B’C’ + A’B’C + AB’C + A’BC\) Now, we can use a Karnaugh map (K-map) to simplify this expression. For a 3-variable K-map with variables A, B, C: “` BC A 00 01 11 10 —|—|—|—|— 0 | 1 | 1 | 1 | 0 1 | 0 | 0 | 1 | 0 “` Grouping the 1s: 1. Group the two 1s in the \(A’B’\) column: \(A’B’\) (covers 000 and 001). 2. Group the two 1s in the \(C\) column where \(A=0\) and \(A=1\): \(B’C\) (covers 001 and 101). Note that 001 is already covered. 3. Group the two 1s in the \(C\) column where \(A=0\) and \(B=1\): \(A’C\) (covers 011). Let’s re-examine the K-map and grouping for minimal SOP: “` BC A 00 01 11 10 —|—|—|—|— 0 | 1 | 1 | 1 | 0 (A’ row) 1 | 0 | 0 | 1 | 0 (A row) “` – Group 1: \(A’B’\) covers \(A’B’C’\) and \(A’B’C\). – Group 2: \(A’C\) covers \(A’B’C\) and \(A’BC\). – Group 3: \(AB’C\) covers \(AB’C\). The essential prime implicants are \(A’B’\) and \(AB’C\). The remaining minterm is \(A’BC\). This can be covered by \(A’C\). So, the minimal SOP expression is \(Y = A’B’ + AB’C + A’C\). Let’s check if \(A’C\) can be simplified further. Consider the terms \(A’B’\) and \(A’C\). They share \(A’\). Consider \(A’B’\) and \(AB’C\). No direct simplification. Consider \(A’C\) and \(AB’C\). They share \(C\). Let’s try another grouping strategy for the K-map: – Group 1: \(A’B’\) (covers 000, 001) – Group 2: \(A’C\) (covers 001, 011) – Group 3: \(AB’C\) (covers 101) The essential prime implicants are \(A’B’\) and \(AB’C\). The minterm \(A’BC\) (011) is covered by \(A’C\). So, \(Y = A’B’ + AB’C + A’C\). We can simplify \(A’C + AB’C\) using the consensus theorem or by observation. \(A’C + AB’C = C(A’ + AB’)\). Using \(A’ + AB’ = (A’ + A)(A’ + B’) = I(A’ + B’) = A’ + B’\). So, \(A’C + AB’C = C(A’ + B’) = A’C + B’C\). Therefore, \(Y = A’B’ + A’C + B’C\). This is a standard consensus form. Let’s re-evaluate the K-map and look for larger groups. The 1s are at (0,0,0), (0,0,1), (1,0,1), (0,1,1). “` BC A 00 01 11 10 —|—|—|—|— 0 | 1 | 1 | 1 | 0 1 | 0 | 0 | 1 | 0 “` – Group 1: \(A’B’\) covers (0,0,0) and (0,0,1). – Group 2: \(A’C\) covers (0,0,1) and (0,1,1). – Group 3: \(AB’C\) covers (1,0,1). Essential prime implicants are \(A’B’\) and \(AB’C\). The minterm (0,1,1) is covered by \(A’C\). So, \(Y = A’B’ + AB’C + A’C\). Let’s test the options. Option A: \(Y = A’B’ + A’C + B’C\) Let’s see if this is equivalent to our derived expression. \(A’B’ + A’C + B’C\) Using \(A’C + B’C = C(A’ + B’)\). So, \(Y = A’B’ + C(A’ + B’)\). Let’s expand \(A’B’ + AB’C + A’C\): \(A’B'(C+C’) + AB’C + A’C\) \(A’B’C + A’B’C’ + AB’C + A’C\) This matches the original minterms. Now let’s check if \(A’B’ + A’C + B’C\) is equivalent. \(A’B’ + A’C + B’C\) Consider the minterms: – \(A’B’\): \(A’B’C + A’B’C’\) – \(A’C\): \(A’B’C + A’BC\) – \(B’C\): \(AB’C + A’B’C\) Combining these: \(A’B’C + A’B’C’ + A’BC + AB’C\). This matches the required minterms. Therefore, \(Y = A’B’ + A’C + B’C\) is the correct minimal SOP expression. The question asks for the most efficient implementation. In digital logic, efficiency is often measured by the number of gates or the complexity of the logic. A minimal SOP or POS form generally leads to a more efficient implementation. The expression \(A’B’ + A’C + B’C\) is a simplified form. The core concept tested here is Boolean algebra simplification, specifically using K-maps or Boolean identities to arrive at a minimal Sum of Products (SOP) expression for a given set of minterms derived from a functional description. The scenario of controlling a motor based on sensor inputs is a common application of combinational logic. Understanding how to translate a functional requirement into a truth table, and then into a minimized Boolean expression, is crucial for efficient circuit design. The National Institute of Technology Mizoram Entrance Exam, with its focus on engineering principles, would expect candidates to demonstrate proficiency in these fundamental digital logic design skills. The ability to identify essential prime implicants and cover all minterms with the fewest terms and literals is key to optimizing hardware implementation, reducing gate count, and improving performance. The specific expression \(A’B’ + A’C + B’C\) is a classic example of a simplified Boolean function that can be implemented using basic logic gates, and its derivation requires careful application of simplification rules.

The question probes the understanding of the fundamental principles of digital signal processing, specifically concerning the aliasing phenomenon and its mitigation through anti-aliasing filters. The Nyquist-Shannon sampling theorem states that to perfectly reconstruct a signal from its samples, the sampling frequency (\(f_s\)) must be at least twice the highest frequency component (\(f_{max}\)) present in the signal, i.e., \(f_s \ge 2f_{max}\). If the sampling frequency is less than twice the highest frequency, aliasing occurs, where high-frequency components masquerade as lower frequencies. In this scenario, the analog signal contains frequency components up to 15 kHz. The sampling is performed at 20 kHz. According to the Nyquist criterion, the minimum required sampling frequency to avoid aliasing for a signal with a maximum frequency of 15 kHz would be \(2 \times 15 \text{ kHz} = 30 \text{ kHz}\). Since the actual sampling frequency (20 kHz) is less than the required minimum (30 kHz), aliasing will occur. Specifically, frequencies above \(f_s/2 = 20 \text{ kHz} / 2 = 10 \text{ kHz}\) will be aliased. The highest frequency component at 15 kHz will be aliased to \(|15 \text{ kHz} – 20 \text{ kHz}| = 5 \text{ kHz}\). An anti-aliasing filter is a low-pass filter placed before the sampler. Its purpose is to attenuate or remove frequency components in the analog signal that are above half the sampling frequency (\(f_s/2\)) to prevent aliasing. Therefore, to effectively prevent the 15 kHz component from causing aliasing, the anti-aliasing filter must have a cutoff frequency that is below 15 kHz but also above the highest desired frequency component that needs to be preserved without aliasing. Given the sampling rate of 20 kHz, the Nyquist frequency is 10 kHz. Any signal component above 10 kHz will alias. To prevent the 15 kHz component from aliasing, the anti-aliasing filter must attenuate frequencies above 10 kHz. Thus, the cutoff frequency of the anti-aliasing filter should be set at or below 10 kHz.

The question probes the understanding of fundamental principles in materials science and engineering, specifically concerning the behavior of crystalline solids under applied stress, a core area of study at National Institute of Technology Mizoram. The scenario describes a metal exhibiting plastic deformation, which is characterized by irreversible changes in shape. This deformation occurs when dislocations, which are line defects in the crystal lattice, move. The critical resolved shear stress (\(\tau_{CRSS}\)) is the minimum shear stress required to initiate dislocation motion on a specific slip system. The resolved shear stress (\(\tau\)) on a particular slip system is given by \(\tau = \sigma \cos\phi \cos\lambda\), where \(\sigma\) is the applied tensile stress, \(\phi\) is the angle between the tensile axis and the normal to the slip plane, and \(\lambda\) is the angle between the tensile axis and the slip direction. For plastic deformation to occur, \(\tau\) must reach \(\tau_{CRSS}\). The question asks which factor, when altered, would *most* directly influence the onset of plastic deformation in a single crystal oriented for slip. Let’s analyze the options: * **A) The crystallographic orientation of the single crystal relative to the applied stress:** This directly affects \(\phi\) and \(\lambda\). By changing the orientation, the \(\cos\phi \cos\lambda\) term changes, thus altering the resolved shear stress (\(\tau\)) for a given applied stress (\(\sigma\)). If the orientation is such that \(\tau\) reaches \(\tau_{CRSS}\) at a lower applied stress, plastic deformation will initiate sooner. This is a primary determinant of when plastic deformation begins. * **B) The magnitude of the applied tensile stress:** While the magnitude of applied stress is crucial because \(\tau\) is directly proportional to \(\sigma\), the question asks what *influences the onset* of plastic deformation. The onset is determined by when \(\tau\) reaches \(\tau_{CRSS}\). If the orientation is unfavorable, a very high \(\sigma\) might be needed, but the *onset* is still governed by the resolved shear stress reaching the critical value. Changing \(\sigma\) *causes* the onset if the orientation is favorable, but the orientation *determines* the \(\sigma\) required for onset. * **C) The elastic modulus of the material:** The elastic modulus (e.g., Young’s modulus) governs the material’s response in the elastic region, where deformation is reversible. Plastic deformation occurs after the elastic limit is reached and is related to dislocation movement, not directly to the elastic modulus. While a higher elastic modulus means less elastic strain for a given stress, it doesn’t directly dictate the stress required for dislocation motion. * **D) The density of grain boundaries within the material:** Grain boundaries act as barriers to dislocation motion, increasing the stress required for plastic deformation (Hall-Petch effect). However, the question specifies a *single crystal*. In a single crystal, grain boundaries are absent. Therefore, this factor is irrelevant to the scenario presented. Considering the formula \(\tau = \sigma \cos\phi \cos\lambda\) and the condition for plastic deformation \(\tau \ge \tau_{CRSS}\), altering the crystallographic orientation (\(\phi\) and \(\lambda\)) directly changes the resolved shear stress (\(\tau\)) for a given applied stress (\(\sigma\)), thereby most directly influencing the *onset* of plastic deformation by determining the required applied stress to reach \(\tau_{CRSS}\).