GATE IN 2018 question paper with solutions

GATE IN 2018 answer key

General Aptitude GA Q.1 to Q.10

1. GA Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   This is a homophone question. Stairs (steps) and stares (fixed looks) sound the same, so pick each word from its meaning. You fall down stairs, and onlookers give you stares. The word but sets the many looks against the little help she got.

   WATCH OUT FOR

   Picking by sound, or assuming the same word goes in both blanks (option B or D). Another slip is swapping the two words (option C). Test each blank on its own: can you fall down it, and can you receive many of it?

   Full solution →
2. GA Q.2MCQ · 1 markOfficial answer: C

   General Aptitude › Grammar & vocabulary

   Let the context decide whether the word is positive or negative. "Warned repeatedly" and "failed to correct" mean the behaviour was wrong, so the blank needs a negative adjective. Of the four options, only errant (straying from what is right, misbehaving) is negative.

   WATCH OUT FOR

   Choosing rational or reasonable because they sound formal, or because the student half-remembers them as words used about behaviour. Both are positive, and you do not correct sensible behaviour. Some students also mix up errant with arrogant or errand. Errant comes from the same root as err and error and means going wrong.

   Full solution →
3. GA Q.3MCQ · 1 markOfficial answer: B

   General Aptitude › Numerical

   Each function decreases where its derivative is negative: $\sin x$ where $\cos x<0$, which is $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$, and $\cos x$ where $\sin x>0$, which is $(0,\pi )$. The answer is where the two intervals overlap. You can also check each quadrant on the unit circle: the second quadrant is the only one where both coordinates are shrinking.

   WATCH OUT FOR

   Picking (C) because $\sin x$ and $\cos x$ are both negative in the third quadrant. Being negative is not the same as decreasing: in $\left(\pi ,\frac{3\pi }{2}\right)$, $\cos x$ climbs from $-1$ to $0$. Another slip is mixing up which derivative controls which function, for example deciding that $\sin x$ decreases wherever $\sin x>0$.

   Full solution →
4. GA Q.4MCQ · 1 markOfficial answer: C

   General Aptitude › Geometry

   The area of an equilateral triangle depends only on its side: $A=\frac{\sqrt{3}}{4}{a}^2$. Solve this for $a$, then multiply by 3 to get the perimeter.

   WATCH OUT FOR

   Stopping at the side length, which gives $a=2$ and wrongly picks (A) 2. Another mistake is using $A=\frac{\sqrt{3}}{2}{a}^2$ (mixing up the altitude $\frac{\sqrt{3}}{2}a$ with the area). That gives $a=\sqrt{2}$, and the perimeter $3\sqrt{2}$ is not among the options.

   Full solution →
5. GA Q.5MCQ · 1 markOfficial answer: D

   General Aptitude › Geometry

   Work out each volume with its standard formula: cuboid $lbh$, cube $a^3$, cylinder $\pi r^{2}h$, sphere $\frac{4}{3}\pi r^3$. The cylinder has $h=r=7$, so its volume is $\pi r^3$ and the sphere of the same radius is exactly $\frac{4}{3}$ times it. Both are more than $3\times 343=1029{\text{cm}}^3$, far above the cube (512) and the cuboid (480), so the only close comparison is cube against cuboid.

   WATCH OUT FOR

   Guessing that the cuboid is bigger than the cube because its longest side (10 cm) is longer than the cube's. In fact $10\times 8\times 6=480<512=8^3$. Another slip is using $\pi r^2$ or $\frac{4}{3}\pi r^2$ for the sphere and ending up with the sphere smaller than the cylinder. Also, the question asks for descending order, so reading it as ascending picks the reverse of the right sequence.

   Full solution →
6. GA Q.6MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   Average speed means total distance divided by total time. When two legs cover the same distance at speeds $v_1$ and $v_2$, the average is their harmonic mean, $\frac{2v_1{v}_2}{v_1+v_2}$. The car spends longer on the slower leg, so that speed counts for more.

   WATCH OUT FOR

   Taking the arithmetic mean of the two speeds, $\frac{60+90}{2}=75$ km/h, and picking option (D). That would only be right if the car spent equal TIME at each speed. Here it covers equal DISTANCES, so it spends 1.5 times as long at 60 km/h as at 90 km/h, and the true average falls below 75.

   Full solution →
7. GA Q.7MCQ · 2 marksOfficial answer: C

   General Aptitude › Numerical

   A parallelogram is fixed by choosing its two sides from each set of parallel lines, so the count is $\left(\genfrac{}{}{0ex}{}{m}{2}\right)\times \left(\genfrac{}{}{0ex}{}{n}{2}\right)$ for $m$ lines crossing $n$ lines. This counts every parallelogram of every size, not only the smallest cells of the grid.

   WATCH OUT FOR

   Counting only the smallest cells of the grid, $(4-1)\times (5-1)=12$, which leaves out the larger parallelograms made up of several cells. Another mistake is multiplying $4\times 5=20$ (option A), which is the number of intersection points, not the number of parallelograms.

   Full solution →
8. GA Q.8MCQ · 2 marksOfficial answer: D

   General Aptitude › Numerical

   Three-set inclusion-exclusion with one unknown. Call the all-three count $x$; it is also the none count. Writing the union two ways, as $N-x$ and as singles minus pairs plus $x$, fixes $x$. Passing means exactly two correct plus all three correct, and exactly two correct is the sum of the pairwise counts minus $3x$.

   WATCH OUT FOR

   There are three common traps. (1) Taking the passers to be the sum of the pairwise counts (2,70,000, option B). That counts the all-three group three times. (2) Counting only the exactly-one group as failures (3,90,000, option C) and forgetting the 30,000 who got none right. (3) Reporting $x=30,000$ (option A), which is the all-three count and not the number who failed.

   Full solution →
9. GA Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Numerical

   Divide the quadratic by $x$ to get $x-\frac{1}{x}$, then square twice. Each squaring turns $x^n\pm \frac{1}{x^n}$ into $x^{2n}+\frac{1}{x^{2n}}$ plus or minus 2, because $x^n\cdot \frac{1}{x^n}=1$. You never have to solve for $x$.

   WATCH OUT FOR

   Getting the sign of $x-\frac{1}{x}$ wrong. Dividing by $x$ gives $x+1-\frac{1}{x}=0$, so $x-\frac{1}{x}=-1$, not $x+\frac{1}{x}=-1$. If you take $x+\frac{1}{x}=-1$, the first squaring gives $x^2+\frac{1}{x^2}=1-2=-1$, which is impossible for real $x$ and leads to $x^4+\frac{1}{x^4}=-1$, not any of the options. A second slip is forgetting the $\pm 2$ cross term when squaring, which gives 1 or 9 (options A and D) instead of 7.

   Full solution →
10. GA Q.10MCQ · 2 marksOfficial answer: A

    General Aptitude › Data interpretation

    When two series move up and down together on a graph, that shows correlation and nothing more. It cannot show that either one causes the other, in either direction, and it gives no rule for predicting one from the other. Here the two series also come from unrelated studies, which makes this even clearer.

    WATCH OUT FOR

    Picking (B) or (C) because the two curves match so closely. That reads a cause-and-effect link, or a rule for predicting one series from the other, into what is only co-movement between two series from unrelated studies.

    Full solution →

Instrumentation Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   If $Nv=\lambda v$, then $N^{k}v={\lambda }^{k}v$. So a matrix with $N^k=0$ (a nilpotent matrix) can only have the eigenvalue $0$, repeated $n$ times for an $n\times n$ matrix, whatever the size of $k$.

   WATCH OUT FOR

   One mistake is to think $N^2=0$ forces $N=0$ and then stop. Another is to allow eigenvalues of $1$ because $N$ is not the zero matrix. $N$ need not be zero (for example, a single $1$ above the diagonal), but every eigenvalue still satisfies ${\lambda }^2=0$. A nonzero nilpotent matrix still has only the eigenvalue $0$.

   Full solution →
2. Q.2MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Complex analysis

   A function is analytic in a region when it satisfies the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$ there and its partial derivatives are continuous. Equivalently, it must not depend on $\bar{z}$: $\partial f/\partial \bar{z}=0$. Every polynomial in $z$ is entire. $\bar{z}$ is the standard example of a function that is continuous everywhere but differentiable nowhere.

   WATCH OUT FOR

   Assuming $\bar{z}$ is analytic because it is continuous and "simple" (it is linear in $x$ and $y$). Continuity is not enough. For $\bar{z}=x-iy$, $u_x=1$ while $v_y=-1$, so the Cauchy-Riemann equations fail at every point. Some students also get a sign wrong when expanding $z^2$ and wrongly conclude that $u_y=-v_x$ fails for $z^2$.

   Full solution →
3. Q.3MCQ · 1 markOfficial answer: D

   Engineering Mathematics › Probability & statistics

   When two random variables are independent, their variances add, whether you add or subtract the variables: $\mathrm{Var}(X\pm Y)=\mathrm{Var}(X)+\mathrm{Var}(Y)$. The coefficient $-1$ is squared, so the minus sign disappears.

   WATCH OUT FOR

   Subtracting the variances ($1-2=-1$, or $|1-2|=1$, which gives option B) because $Z$ is a difference. Variance cannot be negative. The coefficient of $Y$ is squared ($(-1{)}^2=1$), so the variances add. Only the means subtract: $E[Z]=E[X]-E[Y]$.

   Full solution →
4. Q.4NAT · 1 markOfficial answer: 1

   Engineering Mathematics › Calculus

   Where two curves meet, their values are equal, so setting $f(x)=g(x)$ gives the quadratic $x^2-6x+5=0$. It has two real roots, $1$ and $5$, and the question asks for the smaller one.

   WATCH OUT FOR

   Three slips are common. Some students answer $5$, the larger root. Others answer $3$, which is where $f(x)-g(x)$ is smallest (the vertex of $x^2-6x+5$), not where it equals zero. Others expand $(x-2{)}^2$ wrongly, as $x^2+4$ or $x^2-4x-4$, and end up with the wrong quadratic.

   Full solution →
5. Q.5MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Probability & statistics

   For the $r$-th success to land exactly on trial $n$, trial $n$ must be a success and the first $n-1$ trials must hold exactly $r-1$ successes. That gives the negative binomial probability $\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)p^r(1-p{)}^{n-r}$. Here it is $\left(\genfrac{}{}{0ex}{}{4}{2}\right){\left(\frac{1}{2}\right)}^5=\frac{3}{16}$.

   WATCH OUT FOR

   Using $\left(\genfrac{}{}{0ex}{}{5}{3}\right){\left(\frac{1}{2}\right)}^5=\frac{10}{32}=\frac{5}{16}$ (option A). That is the chance of exactly 3 heads somewhere in 5 tosses. It includes sequences such as HHHTT, where the third head comes before the fifth toss. The fifth toss has to be fixed as the third head, so only the first four tosses are free to arrange, giving $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$.

   Full solution →
6. Q.6MCQ · 1 markOfficial answer: A

   Networks (Electric Circuits) › AC, power & resonance

   With $R$ fixed and $C$ varied, the impedance $Z=R-j{X}_C$ moves along the vertical half-line $\text{Re}Z=R$ below the real axis. Inverting a straight line that does not pass through the origin gives a circle that does pass through the origin, so $\underline{I}=V/Z$ traces a circle through the origin with diameter $1/R$ on the real axis. Because the circuit is capacitive, the current leads the voltage, so only the upper half ($y>0$) is traced.

   WATCH OUT FOR

   Picking (B) by mixing up leading and lagging. The fourth-quadrant semicircle is the current locus of a series R-L circuit with $L$ varied, where the current lags. A related slip is to sketch the impedance locus, which for R-C lies below the real axis, and take it for the current locus: inverting $Z$ flips the sign of the angle. Options (C) and (D) need a negative real part of the current, which a positive resistance cannot give.

   Full solution →
7. Q.7MCQ · 1 markOfficial answer: C

   Networks (Electric Circuits) › Theorems

   $V_{th}$ is the open-circuit voltage. No current flows in the resistor leading to the open terminal, so it is just the divider voltage across the shunt resistor, $4\times \frac{1}{2}=2\text{V}$. For $R_{th}$, short the voltage source: $(100\parallel 100)+100=150\text{k}\Omega$. A Thevenin equivalent is always a source in series with a resistance.

   WATCH OUT FOR

   Three slips are common. (1) Treating the resistor leading to p as part of the divider, which gives $V_{th}=4/3\text{V}$ or a similar wrong value, even though it carries no current when p-q is open. (2) Forgetting that same series resistor when finding $R_{th}$, which gives 50 kΩ. (3) Opening the voltage source instead of shorting it, which gives $100+100=200\text{k}\Omega$ and makes (D) look plausible. Separately, any option with a voltage source in parallel with a resistor has the wrong form for a Thevenin equivalent.

   Full solution →
8. Q.8NAT · 1 markOfficial answer: 1009 to 1011

   Networks (Electric Circuits) › AC, power & resonance

   A lossy coil (R in series with L) in parallel with a capacitor resonates when the capacitor's susceptance cancels the coil's susceptance. This is the unity-power-factor condition: total admittance purely real. Only the coil's conductance $R/(R^2+X_L^2)$ is left, so the circuit looks like the dynamic resistance $Z_D=(R^2+X_L^2)/R=R(1+Q^2)$. For a high-Q coil this is roughly $L/(CR)$.

   WATCH OUT FOR

   Treating resonance as $X_C=X_L=100\Omega$ and then combining $(10+j100)$ with $-j100$ in parallel gives $1000-j100\Omega$. That is not purely resistive, so it is not resonance. With the coil resistance present, the capacitor has to cancel the coil's susceptance ($\omega C=100/10100$ S), not its reactance. A second slip is answering $10\Omega$, as if the circuit were series resonant: at parallel resonance the impedance is at its largest, not its smallest.

   Full solution →
9. Q.9MCQ · 1 markOfficial answer: D

   Signals and Systems › Signals basics

   Time reversal ($t\to -t$) and time shift ($t\to t+t_0$) leave the period unchanged. Time scaling $t\to at$ divides it by $|a|$. When the ratio of two periods is rational, the sum is periodic with period equal to the LCM of the two periods. That holds unless parts of the two terms cancel exactly, and the question gives no relation between x and y that could make them cancel.

   WATCH OUT FOR

   Answering 1.5 s (B). Some students apply the compression by 2 to the whole of z(t), or take the smaller of the two periods instead of their LCM. Only the y term is compressed. Its 1.5 s repetition does not make x(-t) repeat, so the sum repeats only at 3 s. Another slip is thinking the +1 in y(2t+1) or the time reversal in x(-t) changes the period. Neither does.

   Full solution →
10. Q.10MCQ · 1 markOfficial answer: A

    Signals and Systems › Signals basics

    Sifting property: an impulse at $t_0$ inside the limits of integration picks out the value of the multiplying function at $t_0$. Here that gives $2x(t-3)$ evaluated at $t=4$, which is $2x(1)$.

    WATCH OUT FOR

    Substituting the impulse location into the shift the wrong way, for example evaluating $x(4)$ or $x(4+3)$ instead of $x(4-3)=x(1)$. Since $|4|>2$ and $|7|>2$, this gives 0 (option C). The other slip is dropping the factor 2 and answering 1 (option B).

    Full solution →
11. Q.11MCQ · 1 markOfficial answer: C

    Signals and Systems › Fourier series

    A zero-mean square wave with a 50% duty cycle has half-wave symmetry, so it contains only odd harmonics of its fundamental: $f_0,3f_0,5f_0,\dots$ With $f_0=50\text{Hz}$ these are $50,150,250,\dots \text{Hz}$. An ideal LPF cutting off at $120\text{Hz}$ passes only the $50\text{Hz}$ fundamental, so the output is a single sinusoid.

    WATCH OUT FOR

    Assuming every harmonic is present and counting $100\text{Hz}$ as a component below the cut-off, which leads to (B). A symmetric square wave has no even harmonics. A second trap is (D): reasoning that the square wave's fundamental is below the cut-off, so the square wave passes. A square shape needs the higher odd harmonics, and the filter removes all of them.

    Full solution →
12. Q.12NAT · 1 markOfficial answer: 90, or -270

    Control Systems › Bode & margins

    For a stable LTI system, a sinusoid of frequency $\omega$ comes out scaled by $|G(j\omega )|$ and shifted by $\angle G(j\omega )$. $\frac{s-1}{s+1}$ is an all-pass network with phase $180^{\circ }-2{\tan }^{-1}\omega$, so at $\omega =1\text{rad/s}$ you evaluate $G(j1)=j$ and read off $90^{\circ }$.

    WATCH OUT FOR

    Taking the angle of the right-half-plane zero factor $(j\omega -1)$ as ${\tan }^{-1}(\omega /(-1))=-45^{\circ }$ from a calculator, when $-1+j$ actually sits in the second quadrant at $135^{\circ }$. That slip gives a total of $-90^{\circ }$, which becomes $270^{\circ }$ in the requested range, and that is wrong. Another error is to treat the numerator like a left-half-plane zero ($+{\tan }^{-1}\omega$), which gives a phase of $0^{\circ }$.

    Full solution →
13. Q.13MCQ · 1 markOfficial answer: A

    Control Systems › Bode & margins

    When factors are multiplied, their phases add. A pure delay $e^{-sT}$ has magnitude 1 but phase $-\omega T$, which is a straight falling line on a linear frequency axis. A second-order pair with $\zeta \approx 0$ has a phase that jumps almost vertically from $0^{\circ }$ to $-180^{\circ }$ at ${\omega }_n$. The answer is the delay's line with a $-180^{\circ }$ step cut into it at ${\omega }_n=100\text{rad/s}$.

    WATCH OUT FOR

    The most common mistake is leaving out the delay because $|e^{-j\omega T}|=1$, which leads to (B). The delay's magnitude is 1, but its phase is not zero. Two other slips: taking the delay's phase as $+\omega T$, which gives the rising slope of (C); and evaluating ${\tan }^{-1}\frac{0.2\omega }{100^2-{\omega }^2}$ on a calculator above ${\omega }_n$. There the denominator is negative, so the angle lies in the second quadrant and the phase is near $-180^{\circ }$, not near $0^{\circ }$.

    Full solution →
14. Q.14NAT · 1 markOfficial answer: 180, or -180

    Control Systems › Bode & margins

    Phase margin is measured where $|G(j\omega )|=1$. Here the DC gain $\frac{2}{1\cdot 2}$ is exactly $1$ and the magnitude only falls from there, so the gain crossover sits at $\omega =0$, where the phase is $0^{\circ }$. That gives $\text{PM}=180^{\circ }+0^{\circ }=180^{\circ }$.

    WATCH OUT FOR

    Seeing that $|G(j\omega )|$ never goes above $0\text{dB}$ and deciding there is no gain crossover, then answering "infinite" or "undefined". The magnitude equals $1$ at $\omega =0$, and that point is the gain crossover. Another mistake is to mix up phase margin with gain margin. The phase only reaches $-180^{\circ }$ as $\omega \to \infty$, so there is no finite phase crossover and the gain margin is infinite, but that is a separate quantity from the phase margin. A third slip is dropping the constant term when expanding $({\omega }^2+1)({\omega }^2+4)=4$, which gives a spurious non-zero crossover frequency.

    Full solution →
15. Q.15NAT · 1 markOfficial answer: 0

    Analog Electronics › Biasing

    Check the sign of the feedback before assuming $V_{-}=V_{+}$. The CE stage inverts from base to collector, and the feedback goes to the (−) input, which inverts again. Two inversions make the feedback positive, so the op-amp saturates. The consistent state has the collector at 5 V, above the 2 V reference, with the op-amp output at its negative limit and the transistor cut off. So $I_o=0$.

    WATCH OUT FOR

    Using the virtual short straight away ($V_C=2$ V, so $I_C=285.7\mu$A and $I_o=I_E=300\mu$A) without checking that the feedback is negative. That answer holds only if the op-amp inputs are swapped. As drawn, the loop is regenerative and the transistor latches off.

    Full solution →
16. Q.16NAT · 1 markOfficial answer: 200

    Analog Electronics › Diode circuits

    The left half is a positive clamper: its capacitor charges to $V_m$, and node p rides on the source at $v+V_m$ (between $0$ and $2V_m$). The right half is a negative peak detector: its capacitor holds q at $-V_m$. So $V_{pq}=v+2V_m$. At $t=60\text{ms}=3T$ the source is exactly zero, which leaves $V_{pq}=2V_m=200\text{V}$.

    WATCH OUT FOR

    The usual slip is treating p as a peak detector held at a constant value. p is a clamper output that swings from $0$ to $200\text{V}$, so its value depends on the instant. Two common wrong answers come from evaluating at a source peak instead of at $t=60\text{ms}$: $300\text{V}$ (positive peak) and $100\text{V}$ (negative peak). The other slip is giving q the wrong sign: taking $V_q=+100\text{V}$ gives $0\text{V}$, and computing $V_q-V_p$ instead of $V_p-V_q$ gives $-200\text{V}$.

    Full solution →
17. Q.17NAT · 1 markOfficial answer: 10

    Digital Electronics › Sequential

    Work out the counter state during the asked-for clock cycle, then use the diode directions to find which branches conduct. The MSB and LSB diodes can only source current from a HIGH output. The middle diode can only sink current into a LOW middle bit. In state 101 two 1 V sources, each behind $100\Omega$, act in parallel through $50\Omega$ into 0 V.

    WATCH OUT FOR

    Taking the state during the 5th clock cycle to be $(100{)}_2$ (the count before the 5th edge) instead of $(101{)}_2$. That leaves only the MSB branch conducting and gives $1/150\text{A}=6.7\text{mA}$. A second slip is misreading the middle diode's direction. It points INTO the counter, so current flows only when the middle bit is 0; if you treat it as a source like the other two you get the wrong network.

    Full solution →
18. Q.18MCQ · 1 markOfficial answer: C

    Digital Electronics › Number systems

    Convert the integer part and the fractional part separately. For the integer part, divide repeatedly by 2 and read the remainders from the last one up. For the fractional part, multiply repeatedly by 2 and read the carries from the first one down. $0.625=\frac{1}{2}+\frac{1}{8}$ terminates in binary, so the result is exact.

    WATCH OUT FOR

    The usual slip is reading the fractional carries in the wrong order. Reading them bottom to top gives $.101$ again only because it is a palindrome, but on other fractions it gives a wrong answer. The distractors come from these slips: (A) $11011.110=27.75$ gets the fraction wrong, (B) $11101.101=29.625$ swaps bits in the integer part, and (D) $10111.110=23.75$ reverses the integer remainders (reading them top to bottom) and also gets the fraction wrong. A quick place-value check of the chosen option catches all three.

    Full solution →
19. Q.19MCQ · 1 markOfficial answer: C

    Digital Electronics › ADC & DAC

    A flash ADC uses one comparator for each boundary between adjacent output codes. An $n$-bit converter has $2^n$ codes and so $2^n-1$ boundaries, which means it needs $2^n-1$ comparators. That number doubles with every extra bit, which is why flash converters are fast but large.

    WATCH OUT FOR

    Picking 256 ($2^n$) by counting output levels or ladder resistors when you should count the boundaries between levels. The other trap is picking 8 ($n$) by confusing a flash ADC with a successive-approximation ADC, which uses a single comparator for $n$ clock cycles.

    Full solution →
20. Q.20MCQ · 1 markOfficial answer: D

    Electrical and Electronic Measurements › Time, frequency & scopes

    On a CRO, volts per division = voltage ÷ vertical divisions it covers, and time per division = time ÷ horizontal divisions it covers. Take the peak (or peak-to-peak) voltage and the period $T=2\pi /\omega$ from the signal, count their divisions on the graticule, and divide.

    WATCH OUT FOR

    Three common slips. (1) Dividing the peak voltage by the peak-to-peak height: $6\text{V}/6\text{div}$ gives the wrong answer of 1 V/div. (2) Taking $100\pi$ as the frequency in hertz, or using $f=100\text{Hz}$ ($T=10\text{ms}$, which gives 1 ms/div), when in fact $\omega =2\pi f$ gives $f=50\text{Hz}$. (3) Thinking that because the input is a cosine, the trace cannot start at a zero crossing. Where the trace starts depends only on the trigger point.

    Full solution →
21. Q.21NAT · 1 markOfficial answer: 200

    Electrical and Electronic Measurements › V, I, power & energy

    For an LPF wattmeter, full-scale reading = rated $V\times$ rated $I\times$ rated power factor. Here that is $300\times 5\times 0.2=300\text{W}$ across 150 divisions, so each division is worth $2\text{W}$.

    WATCH OUT FOR

    Leaving out the rated power factor and taking full scale as $VI=1500\text{W}$. That gives $10\text{W/div}$ and a wrong answer of $1000\text{W}$. A second slip is to multiply the reading by 0.2 again. The rated power factor only sets the full-scale value; it is not applied to the reading.

    Full solution →
22. Q.22NAT · 1 markOfficial answer: 95 to 97
23. Q.23MCQ · 1 markOfficial answer: A
24. Q.24NAT · 1 markOfficial answer: 0.3
25. Q.25NAT · 1 markOfficial answer: 75
26. Q.26MCQ · 2 marksOfficial answer: D

    Engineering Mathematics › Vector calculus

    To evaluate a line integral along a given path, parametrise the path, here $x=y=z=t$ for $0\le t\le 1$. Write $\vec{F}\cdot d\vec{l}=F_{x}dx+F_{y}dy+F_{z}dz$ in terms of $t$ and integrate over the parameter range. Because $\nabla \times \vec{F}\ne \vec{0}$, no potential function exists, so the integral has to be taken along this particular path.

    WATCH OUT FOR

    Dropping the minus signs in $-2y$ or $-4yz$. Both give $t$-terms of the same sign, and losing either one moves the result to a distractor. A second mistake is to assume $\vec{F}$ is conservative and try to evaluate a potential at the endpoints: the curl is not zero, so the result depends on the path.

    Full solution →
27. Q.27MCQ · 2 marksOfficial answer: B

    Engineering Mathematics › Probability & statistics

    When equal-sized groups are pooled, the proportion of a colour in the mixture is the plain average of the group proportions: $\frac{1}{2}(20\%+30\%)=25\%$. This is the total probability theorem with $P(A)=P(B)=\frac{1}{2}$. The yellow balls in bag $B$ make no difference, because only the red fraction of each bag is used.

    WATCH OUT FOR

    Adding the two percentages ($20\%+30\%=50\%$) without dividing by the doubled total, which gives an option that is not even listed. Another mistake is letting the green or yellow split change the answer. Note too that the simple average works only because the bags hold equal numbers of balls. With unequal bags the average must be weighted by bag size.

    Full solution →
28. Q.28MCQ · 2 marksOfficial answer: C

    Engineering Mathematics › Linear algebra

    A $2\times 2$ system has infinitely many solutions only when $\rho (A)=\rho ([A|B])<2$. Put another way, the coefficient ratios must all be equal: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$. Here that reads $\frac{3}{k}=\frac{2k}{6}=\frac{-2}{2}=-1$, and both equalities give $k=-3$.

    WATCH OUT FOR

    Stopping at $\det A=0$ and choosing $k=3$ (option A). A zero determinant only rules out a unique solution. At $k=3$ the two lines are parallel and distinct ($3x+6y=-2$ against $3x+6y=2$), so the system has no solution. You have to check that the augmented matrix has the same rank before calling the solutions infinite.

    Full solution →
29. Q.29MCQ · 2 marksOfficial answer: D

    Engineering Mathematics › Calculus

    To find $V$ from its two partial derivatives, integrate one of them, letting the 'constant' of integration be a function of the other variable. Then differentiate that result with respect to the other variable and match it to the second given partial to fix that function. For an MCQ, the quickest check is to differentiate each option and compare with both given partials.

    WATCH OUT FOR

    One mistake is to integrate each equation on its own and add the results. That counts the shared cross terms twice, giving $2x^{2}y+2x{y}^2$ instead of $x^{2}y+x{y}^2$. Another is to treat the $2xy$ term as if it came from an $xy$ (or $2xy$) term in $V$, which leads to (A) or (C). The right source is $x^{2}y+x{y}^2$: differentiating it in $x$ gives $2xy+y^2$, and in $y$ gives $x^2+2xy$. A third mistake is to use a true constant instead of $g(y)$ as the constant of integration after integrating in $x$.

    Full solution →
30. Q.30NAT · 2 marksOfficial answer: 0.95 to 1.05

    Networks (Electric Circuits) › Theorems

    When one source is set to zero, the resistor in that source's branch ends up connected between node A and the reference. The one branch current you are given then fixes the node voltage $V_A$ ($I_2^{\prime }=V_A^{\prime }/2$, $I_1^{\prime \prime }=V_A^{\prime \prime }/3$). The $6\Omega$ current in each sub-circuit is $V_A/6$, and superposition adds the two. You never need the source values or their polarities.

    WATCH OUT FOR

    One mistake is to read each given current as a current flowing into node A. All three reference arrows point away from A, so $I_2^{\prime }=-6\text{A}$ makes $V_A^{\prime }=-12\text{V}$, not $+12\text{V}$. Another is to try to find $V_1$ and $V_2$ first when their polarities are not shown. A third is to add the given currents $I_2^{\prime }$ and $I_1^{\prime \prime }$ directly instead of first working out each sub-circuit's own $I_3$.

    Full solution →
31. Q.31NAT · 2 marksOfficial answer: 66 to 68

    Networks (Electric Circuits) › AC, power & resonance

    In parallel branches across one voltage source, reactive power adds branch by branch (complex power is conserved). The inductor absorbs $|I_1{|}^2{X}_L=V^2{X}_L/|Z_1{|}^2$ and the capacitor supplies $V^2/X_C$. Their difference is what the source must provide or absorb.

    WATCH OUT FOR

    Writing $Q_L=V^2/X_L=230^2/162.6\approx 325.4\text{VAr}$, as if the inductor had the full 230 V across it. It is in series with the resistor, so only part of the voltage appears across it, and you must use $|I_1{|}^2{X}_L$ with $|I_1|=V/|Z_1|$. That shortcut gives $|Q|\approx 95.4\text{VAr}$, which is wrong. A second slip is to report $-67.33$ when the question asks for a magnitude.

    Full solution →
32. Q.32MCQ · 2 marksOfficial answer: A

    Networks (Electric Circuits) › Laws & node/mesh

    When a current source is the only element in a branch shared by two meshes, it gives a constraint between their mesh currents, here $I_3-I_1=2$. The KVL equation that replaces those two meshes goes round their combined outer boundary (the supermesh) and leaves out the source, because the voltage across the source is unknown.

    WATCH OUT FOR

    There are two common mistakes. The first is to write KVL round mesh 1 or mesh 3 alone as if the current source had zero voltage across it. The second is to test the options only against the source constraint. Every option satisfies $I_3-I_1=2$, and every option also satisfies the mesh-2 equation, so only the supermesh equation $8I_1-7I_2+2I_3=0$ picks out (A). Watch the sign of the 5 V source too: going clockwise in mesh 2 you meet − then +, which is a rise.

    Full solution →
33. Q.33MCQ · 2 marksOfficial answer: C

    Signals and Systems › Fourier transform

    Multiplying by $e^{j{\omega }_{0}t}$ in time shifts the spectrum to the right by ${\omega }_0$: $e^{j{\omega }_{0}t}x(t)\leftrightarrow X(j(\omega -{\omega }_0))$. So $Y(j\omega )=X(j\omega )+X(j(\omega -1))$. To evaluate the shifted copy at $\omega =0.5$, read the original spectrum at $0.5-1=-0.5$.

    WATCH OUT FOR

    Shifting the wrong way, i.e. reading $X(j(\omega +1))$ at $\omega =0.5$. That point, $1.5$, is outside the band, so the second term is taken as $0$ and the answer comes out as $1$ (option B). Another slip is reading the flat value $1$ at $\omega =-0.5$ as well, which gives $2$, not one of the options. The ramp gives $0.5$ at $\omega =-0.5$.

    Full solution →
34. Q.34MCQ · 2 marksOfficial answer: D

    Signals and Systems › z-transform

    Convolution in time becomes multiplication in the z-domain, and a delay of $k$ samples multiplies by $z^{-k}$. Delaying either operand of a convolution by 2 samples delays the output by 2 samples: $x[n]\ast h[n-2]=y[n-2]$, so $P(z)=z^{-2}Y(z)$.

    WATCH OUT FOR

    Using $z^{+2}$ for the delay $h[n-2]$, which gives $z^{2}Y(z)=2z^2+3z+1$ (option C). A delay (shift right) multiplies by $z^{-2}$, not $z^2$. Another slip is to shift only part of the expression, which produces mixed-power options such as (A) or (B).

    Full solution →
35. Q.35MCQ · 2 marksOfficial answer: C

    Signals and Systems › DTFT & DFT

    $x[n]=(-1{)}^n=e^{j\frac{2\pi }{N}\cdot \frac{N}{2}n}$ is one DFT basis vector, the one at the Nyquist bin $k=N/2$. Because the DFT basis vectors are orthogonal, all its energy falls in that one bin: $X(k)=N\delta [k-N/2]$. Here that is $4\delta [k-2]$.

    WATCH OUT FOR

    Picking $k=0$ because it is the DC term, when in fact $X(0)=\sum x[n]=0$ for this zero-mean sequence. A second mistake is getting the sign of $e^{-j\pi n/2}$ wrong at $n=1,3$, which makes $X(1)$ or $X(3)$ look non-zero. Remember $e^{-j\pi /2}=-j$ and $e^{-j3\pi /2}=+j$.

    Full solution →
36. Q.36NAT · 2 marksOfficial answer: 0

    Control Systems › Nyquist

    The zeros $-0.1\pm j9.9995$ are the mirror images of the poles $0.1\pm j9.9995$ across the $j\omega$ axis, so $G(s)$ is all-pass: $|G(j\omega )|=1$ at every frequency. Multiplying by $H=\frac{1}{2}$ puts the entire Nyquist plot on a circle of radius $0.5$. The point $-1$ lies outside that circle and cannot be encircled.

    WATCH OUT FOR

    Students see two open-loop RHP poles ($P=2$) and write $N=-2$ or $N=2$, as if a stable closed loop were being assumed, or they take the count from $P$ alone. The encirclement count depends only on where the plot goes, and this plot stays on a circle of radius $0.5$, so $N=0$. That gives $Z=N+P=2$: the closed loop is unstable, but the answer asked for is still $0$. A second slip is writing the angle as $2{\tan }^{-1}\left(\frac{0.2\omega }{100-{\omega }^2}\right)$ with the principal-value arctangent. That formula goes wrong for $\omega >10$ because it drops the quadrant.

    Full solution →
37. Q.37MCQ · 2 marksOfficial answer: D

    Control Systems › State space

    For a diagonal $A$, $e^{At}$ is diagonal with entries $e^{{\lambda }_{i}t}$, because every power of a diagonal matrix is diagonal. Each state decays on its own, $x_i(t)=e^{{\lambda }_{i}t}{x}_i(0)$, and there are no cross terms.

    WATCH OUT FOR

    Using $\Phi (0)=I$ to eliminate options. Here every option equals $I$ at $t=0$, including (C), whose $-t^2{e}^{-2t}$ term vanishes at $t=0$, so that check eliminates nothing. The other mistake is misreading $e^{-2t}$ as $e^{2t}$, which is option (B). The mode for an eigenvalue at $-2$ is $e^{-2t}$.

    Full solution →
38. Q.38NAT · 2 marksOfficial answer: 8.9 to 9.1

    Control Systems › Routh–Hurwitz

    Write the closed-loop characteristic polynomial $s(s+1)(s+2)+(s+\alpha )$, build its Routh array, and find the value of $\alpha$ that makes a whole row zero. The $s^1$ row vanishes at $\alpha =9$. The auxiliary polynomial from the row above it ($3s^2+9$) then gives the imaginary-axis poles $\pm j\sqrt{3}$ directly.

    WATCH OUT FOR

    Dropping the $s$ in the denominator of $H(s)$ and writing $(s+1)(s+2)+(s+\alpha )=0$, a second-order equation that can never put poles on the $j\omega$ axis. Another mistake is forgetting the extra $s$ contributed by $H(s)$'s numerator, which gives $s^3+3s^2+2s+\alpha$ and the wrong answer $\alpha =6$. A third is giving $\alpha =0$ just because it also zeroes a first-column entry: that only puts a cancelled root at the origin.

    Full solution →
39. Q.39MCQ · 2 marksOfficial answer: D

    Control Systems › Time response

    With zero initial conditions, the transfer function is the Laplace transform of the output divided by the Laplace transform of the input. For a unit-step input that means $G(s)=sY(s)$. The same result follows from differentiating the step response to get the impulse response, $h(t)=2e^{-2t}u(t)$, whose transform is $\frac{2}{s+2}$.

    WATCH OUT FOR

    Reading $1-e^{-2t}$ as a pole at $s=-2$ and choosing $\frac{1}{s+2}$ (option C) without working out the gain. That system settles at $0.5$, not $1$. The numerator must be $2$ so that $G(0)=1$. A second mistake is taking the transform of $y(t)$ itself as the transfer function, which forgets to divide by the step input's $\frac{1}{s}$.

    Full solution →
40. Q.40NAT · 2 marksOfficial answer: Marks to all

    Analog Electronics › Filters & waveforms

    A square/triangle generator needs an odd number of inversions round its loop: the integrator must ramp $v_1$ TOWARD the comparator's current threshold, not away from it. Check this before using any frequency formula. Once it oscillates, $T/2=\Delta V/|d{v}_1/dt|$, where $\Delta V$ is the gap between the two thresholds and $|d{v}_1/dt|=V_A/RC$.

    WATCH OUT FOR

    Writing $f=1/(4\beta RC)=200\text{kHz}$ without checking the sign of the loop. As printed, both stages invert, so the integrator ramps away from the threshold and the circuit latches. It is also wrong to assume the frequency comes out the same however the drawing is corrected. The threshold swing is $\pm 2.5\text{V}$ under one correction and $\pm 5\text{V}$ under the other, which halves the frequency. The $50\text{k}\Omega$ at the integrator's virtual ground is a distractor: it carries no current.

    The question and its official answer →
41. Q.41NAT · 2 marksOfficial answer: 17.5 to 17.9

    Analog Electronics › Op-amps

    The slew rate limits the output of the op-amp, not its input. First find the output peak, $V_m=A_v{V}_{in}$, and check that it stays inside the supply rails so clipping is ruled out. Then set the sinusoid's largest slope, $2\pi f{V}_m$, equal to the slew rate: $f_{\max }=SR/(2\pi V_m)$.

    WATCH OUT FOR

    Using the input amplitude of 0.3 V in place of the output peak of 4.5 V gives about 265 kHz, which is 15 times too high. A second mistake is to use the supply voltage of 5 V as $V_m$, which gives about 15.9 kHz. The supply only tells you whether the output clips. The slew-rate limit depends on the actual output swing.

    Full solution →
42. Q.42NAT · 2 marksOfficial answer: 0.98 to 1.00

    Analog Electronics › Op-amps

    The two cascaded inverting stages give a net gain of $+2$, so the right end of the $10.1\text{ k}\Omega$ resistor sits at $2v_s$. That resistor returns almost all of the current that the $10.0\text{ k}\Omega$ resistor draws into the virtual ground, and the source supplies only the small difference $v_s\left(\frac{1}{10\text{ k}\Omega }-\frac{1}{10.1\text{ k}\Omega }\right)$. The circuit makes the source see a very large input resistance of about $1.01\text{ M}\Omega$, and this is the principle behind bootstrapping and negative-impedance input stages.

    WATCH OUT FOR

    Stopping at $100\mu \text{A}$ (only the $10.0\text{ k}\Omega$ branch), or taking $V_{o2}=-2\text{ V}$ by forgetting that the second inverting stage flips the sign again. The second error gives $I=1/10+3/10.1\approx 397\mu \text{A}$. Another slip is rounding $1/10.1\text{ mA}$ to $0.099\text{ mA}$ too early. The answer is a small difference of two nearly equal currents, so keep at least five significant figures until the final subtraction.

    Full solution →
43. Q.43MCQ · 2 marksOfficial answer: B

    Digital Electronics › Boolean logic

    In a canonical POS, each sum term is a maxterm and marks one input where $F=0$. Read off its index by setting plain variables to 0 and complemented ones to 1. The canonical SOP is the sum of the minterms with every other index: $\prod M(S)=\sum m(\text{all indices not in }S)$.

    WATCH OUT FOR

    Using the maxterm indices directly as minterms, which gives $\sum m(1,3,4,7)$ = option (A). A second mistake is reading a maxterm's index the minterm way ($\overline{X}\to 0$). That turns $(A+B+\overline{C})$ into index 6 instead of 1, and the whole set comes out wrong.

    Full solution →
44. Q.44MCQ · 2 marksOfficial answer: C

    Digital Electronics › Sequential

    In a synchronous counter, work out every J and K from the present state, then apply the J-K rules (00 hold, 01 reset, 10 set, 11 toggle) to both flip-flops at once. You can also check it in one line with $Q^{+}=J\overline{Q}+\overline{K}Q$. Here the inputs simplify to $Q_1^{+}=Q_2$ and $Q_2^{+}=\overline{Q_1}$, a 2-bit twisted-ring counter, and that gives the Gray sequence 00, 01, 11, 10.

    WATCH OUT FOR

    Updating $Q_1$ first and then using the new $Q_1$ to work out $J_2$ and $K_2$ within the same clock edge. That is ripple-style thinking, and it is wrong for a synchronous counter, where every input comes from the old state. Other slips: mixing up $JK=01$ (reset) with $10$ (set), and missing an overbar on the input labels. For example, $J_1=Q_1+Q_2$ has no bars, while $K_1=\overline{Q_1}+\overline{Q_2}$ has two.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 279 to 282

    Digital Electronics › Processors & memory

    For a delay loop with an initial count $N$, the loop body runs $N$ times, but the conditional jump is taken only $N-1$ times and falls through once, at its shorter not-taken time. The total T-states are $N_T=T_{\text{init}}+N{T}_{\text{DCR}}+(N-1)T_{\text{JNZ,taken}}+T_{\text{JNZ,not taken}}$, and the delay is $N_T/f$. Convert a hexadecimal count such as $64\text{H}$ to decimal first.

    WATCH OUT FOR

    Reading $64\text{H}$ as decimal 64 gives $7+256+630+7=900$ T-states, or $180\mu \text{s}$. Another error is to halve the 5 MHz clock, as if it were an 8085 crystal frequency, which doubles the answer to $561.6\mu \text{s}$. The question states that the processor itself is driven by a 5 MHz clock, so $T=0.2\mu \text{s}$. Counting the jump as taken all 100 times is also wrong, even though it happens to land inside the key's range here.

    Full solution →
46. Q.46MCQ · 2 marksOfficial answer: A

    Digital Electronics › Boolean logic

    This is the standard four-NAND XOR. The first NAND makes $\overline{XY}$. Each input is then NANDed with it, and the last NAND turns the two terms into $X\overline{XY}+Y\overline{XY}=(X+Y)(\overline{X}+\overline{Y})$, which is $X\oplus Y$.

    WATCH OUT FOR

    Treating the output NAND as an AND or an OR of the two middle outputs and dropping a complement. Another slip is treating the gates as AND gates because the bubbles are small, which gives $XY$ or a wrong option such as (D). The quickest way to catch these is a four-row truth table with every gate treated as a NAND. It also separates XOR (A) from XNOR (B).

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 2.49 to 2.53

    Electrical and Electronic Measurements › Errors & standards

    An instrument's accuracy stated as a percentage of full scale is a fixed absolute error, so convert it to a relative error at the actual reading. For a product or quotient, the limiting (worst-case) relative errors add.

    WATCH OUT FOR

    One mistake is to use the full-scale percentages directly ($1\%+0.5\%=1.5\%$) and forget that the error is a percentage of full scale, not of the reading. The other is to subtract the ammeter's relative error because $I$ is in the denominator. Worst-case errors of a quotient always add.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 0.9 to 1.1

    Electrical and Electronic Measurements › Bridges

    A coil's self-capacitance $C_d$ adds to the Q-meter's tuning capacitance, so ${\omega }^{2}L(C+C_d)=1$ at every resonance. Two resonances give two equations. Equating them removes $L$ and gives $C_d$. For a frequency ratio of 2, $C_d=(C_1-4C_2)/3$. Put $C_d$ back into either equation to get $L$.

    WATCH OUT FOR

    The usual slip is the powers of ten in the last step. $1000\text{pF}$ is $10^{-9}\text{F}$, not $1\text{F}$, so ${\omega }_1^2(C_1+C_d)=10^{12}\times 10^{-9}=10^3$. That gives $L=1\text{mH}$, not $1\text{H}$ (1000 mH). Two other errors: leaving out $C_d$ altogether, which gives $L=1/(10^{12}\times 990\times 10^{-12})\approx 1.01\text{mH}$ from the first resonance and a different value from the second, and subtracting $C_d$ from $C$ when it should be added.

    Full solution →
49. Q.49NAT · 2 marksOfficial answer: 221 to 223

    Electrical and Electronic Measurements › Bridges

    This is a Maxwell inductance-capacitance bridge. At balance, $Z_x=R_2{R}_4{Y}_1$. The parallel $R_1{C}_1$ arm contributes an admittance, so its $j\omega C_1$ term turns directly into $j\omega L_x$. That gives $L_x=R_2{R}_4{C}_1$, independent of frequency and of $R_1$.

    WATCH OUT FOR

    One mistake is using the impedance of the $R_1{C}_1$ arm instead of its admittance. Another is pairing the wrong arms, for example writing $Z_1{Z}_4=Z_2{Z}_x$ when the detector's position makes $Z_1$ opposite $Z_x$. A third is bringing $R_1$ into the inductance formula. Each of these gives a wrong or frequency-dependent $L_x$. The unit prefixes also cause slips: $22.2\text{k}\Omega \times 10\text{k}\Omega \times 1\text{nF}=0.222\text{H}$, which is 222 mH, not 222 H or 22.2 mH.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: 3.5 to 5.7
51. Q.51NAT · 2 marksOfficial answer: 0.48 to 0.51
52. Q.52NAT · 2 marksOfficial answer: 10.6 to 11.0

    Analog Electronics › Noise & shielding (IN)

    Only a stray capacitance that ends on a high-impedance node lets interference through. The inverting input of an ideal op-amp is a virtual ground, so the 3 pF capacitor passes a current $V_n\omega C$ that must all flow through $R_f$. Capacitors that end on low-impedance nodes (the ideal dc source, or the op-amp output) send their current into those sources and have no effect on $V_o$.

    WATCH OUT FOR

    Adding all three capacitances (4.6 pF) and getting about 16.6 mV. The 1.3 pF current goes into the ideal dc source and the 0.3 pF current goes into the zero-impedance op-amp output, so neither reaches $V_o$. Other slips: treating the 1.3 pF current as an input through the 10 kΩ resistor with gain $-R_f/R_1$, or forgetting that the answer is asked for in mV.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 352000
54. Q.54NAT · 2 marksOfficial answer: 249 to 251
55. Q.55NAT · 2 marksOfficial answer: 29 to 31

    Electromagnetics › Waveguides & fibres

    For a fibre launched from air, the acceptance angle satisfies $\sin {\theta }_A=\text{NA}=\sqrt{n_1^2-n_2^2}$. This comes from total internal reflection at the core–cladding boundary ($\sin {\theta }_c=n_2/n_1$) together with Snell's law at the end face. Here $n_2\approx \sqrt{2}$, so $\text{NA}=\sqrt{2.25-2}=0.5$.

    WATCH OUT FOR

    Answering with the critical angle ${\theta }_c={\sin }^{-1}(n_2/n_1)={\sin }^{-1}(0.9428)\approx 70.5^{\circ }$, or its complement inside the core, $90^{\circ }-{\theta }_c\approx 19.5^{\circ }$, instead of the angle in air. The core-side angle has to be carried through Snell's law at the end face, $\sin {\theta }_A=n_1\cos {\theta }_c$. Another slip is to take $\text{NA}=n_1-n_2$ instead of $\sqrt{n_1^2-n_2^2}$.

    Full solution →