GATE IN 2023 question paper with solutions

GATE IN 2023 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   When a sentence places something relative to two separate things joined by "and" ("______ X and Y"), the preposition is between. Among is for a group of three or more, and at, in and through do not take a two-part "X and Y" object of place.

   WATCH OUT FOR

   Picking "in" because "nestled in" appears earlier in the sentence. That "in" belongs to "a green spot". The blank has to describe where the spot sits relative to the ocean and the hills, and "in the ocean and the hills" would put land inside the ocean.

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2. Q.2MCQ · 1 markOfficial answer: C

   General Aptitude › Reasoning

   Put the first pair into a sentence ("When you disagree, you protest"), then keep the relation and swap the attitude ("When you agree, you ______"). Protest is the outward, active expression of disagreement, so the answer has to be the outward, active expression of agreement: recommend (to endorse or speak in favour of something).

   WATCH OUT FOR

   Choosing Refuse or Refute because they feel close to "Protest". This matches the answer to the second word of the first pair when it should match the relation. Both words express disagreement, so they have the wrong polarity for "Agree". Pretext is a noun that has nothing to do with either attitude.

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3. Q.3MCQ · 1 markOfficial answer: D

   General Aptitude › Numerical

   Count position by position. The first digit has 5 odd choices. Each later digit must differ only from the digit right before it, so it has 4 choices. The multiplication principle then gives $5\times 4\times 4$.

   WATCH OUT FOR

   Reading "no two adjacent digits the same" as "all three digits different" gives $5\times 4\times 3=60$, option (C). That wrongly rules out numbers like $131$, where only the non-adjacent first and last digits match. Ignoring the adjacency rule altogether gives $5^3=125$, option (A).

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4. Q.4MCQ · 1 markOfficial answer: C

   General Aptitude › Statistics & probability

   The median is defined by position: at least half of the ordered scores sit at or below it, so no more than half can be strictly above it. The mean depends on the size of the scores, not their position. Depending on skew it can lie above or below the median, and more than half the scores can exceed it.

   WATCH OUT FOR

   Choosing (A) or (B) from a remembered rule such as "mean > median for skewed data". That rule depends on the direction of skew, so neither direction holds for every set of scores. A second mistake is treating the mean as a "middle" value and picking (D). A single left-skewed set like $1,8,9$ puts two of three scores above the mean.

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5. Q.5MCQ · 1 markOfficial answer: B

   General Aptitude › Spatial

   The top view of a hill is its horizontal cross-sections seen from above, which is a contour map. Each oval in the side view becomes a closed curve covering the same distances. Where the hill is steep, one cross-section ends close to the next, so the contours bunch together. Where it is gentle, they spread out. This hill is steep on the 0 km side and has a long gentle tail to the right. So its contours must be bunched on the left and spread on the right, with the summit contour off-centre towards 0 km.

   WATCH OUT FOR

   Choosing P, the mirror image, by ignoring the 0 km mark that fixes left and right in every option. Other traps: picking R because a hill 'should' give concentric rings, which is true only for a symmetric hill. Reading crowded contours as a gentle slope instead of a steep one. Not noticing that the touching contours in S would mean vertical cliffs, which this hill does not have.

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6. Q.6MCQ · 2 marksOfficial answer: Marks to all

   General Aptitude › Reasoning

   Turn each sentence into a set relation: $W$ is not empty, $W\subseteq A$, and more than half of $W$ lies in $B$. Then test every option against those relations and nothing else. Work out what the demonstrative "these" points back to before you start, because the whole question turns on it. Remember that "most of a non-empty set" guarantees "some".

   WATCH OUT FOR

   Reading "All well-established residents are academicians" backwards as "All academicians are well established" and choosing (B). A subtler slip is rejecting (A) by quietly taking "these academicians" to mean every academician in the complex, when "these" points back to the well-established residents just described. A third is forgetting that a "some" conclusion needs at least one member to exist; here "many well-established individuals" supplies it.

   The question and its official answer →
7. Q.7MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   Look at the last move. The final move onto stair $N$ is either a 1-stair step from $N-1$ or a 2-stair step from $N-2$. Those two cases are disjoint and cover every way, so $F(N)=F(N-1)+F(N-2)$, the Fibonacci recurrence. Starting from $F(1)=1$ and $F(2)=2$, the values run 1, 2, 3, 5, 8.

   WATCH OUT FOR

   One common mistake is to count only the combinations of step sizes (all 1s; three 1s and one 2; one 1 and two 2s), which gives 3, instead of counting orderings. For example, 1+2+2, 2+1+2 and 2+2+1 are three different ways. Another is to use the standard Fibonacci sequence 1, 1, 2, 3, 5 and pick 5, option (D). The question's own values $F(1)=1$, $F(2)=2$ shift the sequence by one place, so $F(5)=8$.

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8. Q.8MCQ · 2 marksOfficial answer: D

   General Aptitude › Reading

   An inference 'with certainty' must be forced by the passage alone. Being 'non-coding' (not read in triplets into amino acids) is a narrower claim than having 'no role in protein synthesis', and it says nothing at all about whether a region's function is 'understood'.

   WATCH OUT FOR

   Picking (A): reading 'only about 2% is coding' as 'the other 98% has no role in making proteins'. The passage says only that the 98% is not read into amino acids, not that it plays no part. Picking (B) or (C) is also a mistake, because it brings in the familiar outside idea that non-coding DNA is poorly understood, which the passage never states. Answer from the passage alone, not from what you already know about biology.

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9. Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   Break a sum of moduli at the points where each inner expression is zero ($x=-2$ and $x=1$), write the function piece by piece, and apply the outer modulus last. A quick check: $|x+2|-|x-1|$ is (distance from $x$ to $-2$) minus (distance from $x$ to $1$). The triangle inequality keeps that difference between $-3$ and $3$, so $0\le f(x)\le 3$ for every $x$. That alone rules out Q (it goes negative) and R and S (they reach about 9), leaving only P.

   WATCH OUT FOR

   Forgetting the outer modulus gives the ramp from $-3$ to $3$, which is figure Q. Using the middle-interval formula $|2x+1|$ (or $-(2x+1)$) for all $x<-2$ instead of the constant 3 produces the rising left arm of R or S. Always evaluate each piece only on its own interval.

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10. Q.10MCQ · 2 marksOfficial answer: D

    General Aptitude › Spatial

    Under a parallel beam, the shadow is an orthographic projection. For a cylinder, that is always one end-ellipse (minor axis $d\cos \theta$ along the projected axis, major axis $d$ across it) swept a distance $L\sin \theta$ along the projected axis. The result is a circle ($\theta =0^{\circ }$), a capsule ($0^{\circ }<\theta <90^{\circ }$) or a rectangle ($\theta =90^{\circ }$). Every one of them is symmetric about a line parallel to its long edges, and any straight end meets those edges at a right angle.

    WATCH OUT FOR

    Picturing a tilted cylinder the way it looks in a perspective or oblique drawing, and assuming the tilt skews the rectangle into a parallelogram. Under parallel light, tilting the axis only curves the ends into half-ellipses; it never slants them. The opposite slip is to reject Q because the ends of a cylinder are flat. Those flat ends are circles seen at an angle, so they cast rounded shadows.

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Instrumentation Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   For a homogeneous system $A\mathbf{x}=0$ in $n$ unknowns, the solution set is a subspace of dimension $n-\text{rank}(A)$. Here that is $3-2=1$, so $S$ is a line spanned by one vector. Two independent equations in three unknowns leave one free parameter. Any proposed spanning vector must satisfy every equation.

   WATCH OUT FOR

   Choosing (D) because $(1,0,1)$ satisfies the second equation $x-z=0$, without checking the first equation, where it gives $2\ne 0$. A second mistake is choosing (B) or (C) because they contain the correct vector $(1,1,1)$. Adding a second independent vector turns the set into a plane, and that plane includes non-solutions such as $(1,0,1)$ or $(2,1,2)$.

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2. Q.12MCQ · 1 markOfficial answer: A

   Electromagnetics › Maxwell's equations

   A non-magnetic conductor near an AC coil behaves like a shorted secondary winding. Its eddy currents oppose the coil's flux (Lenz's law) and reflect a negative reactance into the coil, $-{\omega }^2{M}^2{L}_2/(R_2^2+{\omega }^2{L}_2^2)$. The measured inductance falls and the measured resistance rises. Only a magnetic material (${\mu }_r\gg 1$) would raise $L$.

   WATCH OUT FOR

   Many students pick (C) because they assume any metal near a coil increases its inductance, as an iron core does. Copper has ${\mu }_r\approx 1$, so it adds no magnetic path, and its eddy currents cancel part of the flux. Option (D) is also wrong: the reduction is a steady-state AC effect with no later rise above 10 mH.

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3. Q.13MCQ · 1 markOfficial answer: D
4. Q.14MCQ · 1 markOfficial answer: C
5. Q.15MCQ · 1 markOfficial answer: A

   Electromagnetics › Electrostatics

   $C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$ assumes a uniform field confined between infinitely large plates. Real plates have fringing flux at the edges and on the outer faces, and that flux stores extra charge at the same voltage. So the real capacitance is always a little above the formula value. With meter and cable errors excluded, a correct reading is slightly more than 20 pF.

   WATCH OUT FOR

   Choosing 20 pF because the meter is called ideal. An ideal meter shows the true capacitance of the structure, and the formula itself is the approximation because it leaves out fringing. The second trap is 19.5 pF, which treats the difference as random scatter. Fringing always adds capacitance, so the error has a known direction.

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6. Q.16MCQ · 1 markOfficial answer: A

   Digital Electronics › Sequential

   Look for the pattern that links the input to the change of state. A T flip-flop holds when $T=0$ and toggles when $T=1$, so $Q(t+1)=T\oplus Q(t)$. A D flip-flop copies its input, so $Q(t+1)=D$.

   WATCH OUT FOR

   Picking D because the input equals $Q(t+1)$ in the first two rows. Rows 3 and 4 break that pattern. The input actually equals $Q(t)\oplus Q(t+1)$, which means it tells the flip-flop whether to change state. That is the T flip-flop's excitation rule.

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7. Q.17MCQ · 1 markOfficial answer: D

   Digital Electronics › Boolean logic

   XOR outputs $1$ exactly when its inputs differ. So a signal XORed with itself gives $0$, and XORed with its complement gives $1$. XOR with a constant makes a controlled inverter: a control of $0$ passes $X$ through unchanged, and a control of $1$ inverts it to $\overline{X}$.

   WATCH OUT FOR

   Mixing up the two constant cases: taking $X\oplus 1=1$ or $X\oplus 0=0$ by treating XOR like OR or AND. Another mistake is swapping (i) and (ii), writing $X\oplus X=1$ because "a signal matches itself". Options (A), (B) and (C) each contain one of these slips.

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8. Q.18MCQ · 1 markOfficial answer: B
9. Q.19MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Complex analysis

   Shift to the new centre with $w=z-2$ and rewrite $1-z$ as $-(1+w)$. That turns $F$ into $-\frac{1}{1+w}$, a geometric series that converges for $|w|<1$. You can confirm the result in one line with $a_0=F(2)=-1$.

   WATCH OUT FOR

   The usual slip is dropping the minus sign in $1-z=-(1+w)$. That gives $a_k=(-1{)}^k$, which is option (A). A second trap is mixing up the centre: the familiar $\frac{1}{1-z}=\sum z^k$ is centred at $z=0$, and forcing a factor of $\frac{1}{2}$ into the expansion leads to options (C) or (D), which belong to a different expansion with a different ROC. Checking $a_0=F(2)=-1$ catches all three.

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

    Engineering Mathematics › Differential equations

    The equation involves only $\dot{x}$ and $\ddot{x}$, so substitute $v=\dot{x}$ to get the first-order equation $\dot{v}=-kv$. Its solution is $v(t)=v(0)e^{-kt}$. With $v(0)=0$ the velocity is identically zero, so $x(t)$ is the constant $x(0)$. Equivalently, the roots $m=0,-k$ give $x=A+B{e}^{-kt}$, and $\dot{x}(0)=0$ forces $B=0$.

    WATCH OUT FOR

    Checking only the initial conditions. Option (A) satisfies both $x(0)=1$ and $\dot{x}(0)=0$ and is picked for that reason alone, but it does not satisfy $\ddot{x}=-k\dot{x}$. Every option also has to be substituted into the differential equation. A second slip is to read $\ddot{x}=-k\dot{x}$ as $\ddot{x}=-kx$ (a harmonic oscillator), which gives oscillating answers.

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11. Q.21MCQ · 1 markOfficial answer: C

    Control Systems › Bode & margins

    In sinusoidal steady state a stable LTI system scales the amplitude by $|H(j\omega )|$ and shifts the phase by $\angle H(j\omega )$, and it never changes the frequency. To find the input from the output, you undo that shift. Here $(s-\pi )/(s+\pi )$ is all-pass: the gain is 1 at every frequency, and at $\omega =\pi$ the phase is exactly $+90^{\circ }$. So the input has to lag the output by $\pi /2$.

    WATCH OUT FOR

    The usual slip is getting the sign of the phase wrong. The numerator $j\pi -\pi$ sits in the second quadrant, at $135^{\circ }$ and not $45^{\circ }$. People who treat $(s-\pi )/(s+\pi )$ as a lag, or read the numerator's angle as $-45^{\circ }$, conclude that the system shifts the phase by $-\pi /2$. That leads them to pick (B), which actually produces $-\sin \pi t$. A second mistake is to pick (A) on the grounds that $|H|=1$, so the input must equal the output. That forgets that an all-pass network still shifts the phase.

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12. Q.22MCQ · 1 markOfficial answer: B

    Digital Electronics › Logic families

    ECL is the fastest family because its current-switch transistors never saturate, so there is no storage delay. It also uses a small swing (about 0.85 V) and low-resistance emitter-follower outputs, so the load capacitance charges quickly. The price is the highest static power dissipation of the common families.

    WATCH OUT FOR

    Picking CMOS because it is the dominant modern technology. CMOS's strength is very low static power, not speed: standard CMOS gates are slower than TTL, and high-speed CMOS only matches standard TTL. A second mistake is confusing 'fastest' with 'best speed-power product'. ECL wins on raw delay but not on power.

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13. Q.23MCQ · 1 markOfficial answer: A

    Engineering Mathematics › Calculus

    A function that tends to zero, multiplied by a bounded function, tends to zero. Here $|x\sin (1/x)|\le |x|$, and the squeeze theorem finishes it. The oscillation of $\sin (1/x)$ does not matter, because its amplitude is being shrunk to zero by the factor $x$.

    WATCH OUT FOR

    Picking (D) because $\sin \frac{1}{x}$ oscillates and has no limit. That is true of $\sin \frac{1}{x}$ by itself, but not of $x\sin \frac{1}{x}$, whose swings are held inside $\pm |x|$. Picking (B) by rewriting the function as $\frac{\sin (1/x)}{1/x}$ and applying the standard limit $\frac{\sin t}{t}\to 1$. That standard limit needs $t\to 0$, but here $t=\frac{1}{x}\to \infty$, and $\frac{\sin t}{t}\to 0$ as $t\to \infty$. ($x\sin \frac{1}{x}\to 1$ is correct only as $x\to \infty$.)

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14. Q.24NAT · 1 markOfficial answer: 2

    Control Systems › Routh–Hurwitz

    In the Routh array, the number of sign changes down the first column equals the number of roots in the right-half plane. For a cubic $s^3+a_2{s}^2+a_{1}s+a_0$ with positive coefficients, the one test that matters is whether $a_2{a}_1>a_0$. When it fails, as it does here, the $s^1$ entry turns negative and two roots lie in the RHP.

    WATCH OUT FOR

    Seeing that every coefficient is positive and answering 0. Positive coefficients are necessary but not sufficient for stability once the order is 3 or more. A second mistake is to spot that the system is unstable and answer 1. The number of sign changes is the number of RHP roots, and here there are two, $2\to -35$ and $-35\to 80$.

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15. Q.25NAT · 1 markOfficial answer: 2

    Control Systems › Nyquist

    When a question asks about encircling the origin, not the point $-1$, apply the argument principle to $G(s)H(s)$ itself. Then $N=Z-P$ uses the open-loop right-half-plane zeros and poles. Here the two RHP zeros at $s=1,2$ give $N=2$ clockwise. The all-pass magnitude of $\frac{1}{2}$ and the $720^{\circ }$ total phase swing confirm it directly.

    WATCH OUT FOR

    Answering 1 by tracing only the $\omega =0$ to $\infty$ half of the plot, which is one circle. The mirror half from $-\infty$ to $0$ adds a second turn. Other traps: counting zeros of $1+G(s)H(s)$ (the closed-loop rule for the $-1$ point) in place of the zeros of $G(s)H(s)$, and deciding there is no encirclement because the magnitude is constant.

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16. Q.26NAT · 1 markOfficial answer: 1110

    Analog Electronics › Op-amps

    The input offset voltage is amplified by the noise gain $1+R_f/R_1$, whatever configuration the circuit is drawn in. When the + input is grounded directly, the bias current into the − input is forced through $R_f$ alone, because the virtual short leaves no voltage across $R_1$. That gives $I_B{R}_f$. Superpose the two and add magnitudes for the worst case.

    WATCH OUT FOR

    Multiplying $V_{OS}$ by the inverting gain $R_f/R_1=100$ gives 1000 µV instead of 1010 µV, which would be the correct figure only for a signal applied through $R_1$. Other errors are dropping the bias-current term (answer 1010), or letting part of $I_B$ flow through the 1 kΩ resistor, when that resistor has 0 V across it.

    Full solution →
17. Q.27NAT · 1 markOfficial answer: 4.9 to 5.1

    Electromagnetics › Magnetostatics

    The force per unit length between two long parallel currents is $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$. The field of a long straight wire falls off as $1/d$, not $1/d^2$, so with the currents unchanged the force is inversely proportional to the separation.

    WATCH OUT FOR

    Treating the force like Coulomb's law between point charges, as $1/d^2$, which gives 2.5 μN/m. The field of an infinitely long line current falls off as $1/d$, so doubling the gap only halves the force. A second error is trying to find the actual currents: they are not needed, because only the ratio matters.

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18. Q.28NAT · 1 markOfficial answer: -3

    Signals and Systems › Signals basics

    Find the switching point of each step. $u[-n+5]$ is 1 for $n\le 5$ (a reflected step, so it is on to the left), and $u[n+3]$ is 1 for $n\ge -3$ (on to the right). These two breakpoints cut the time axis into three regions. The difference is zero only where the two steps overlap, $-3\le n\le 5$, and the left end of that overlap is the answer.

    WATCH OUT FOR

    Getting the direction of the reflected step wrong. Some students read $u[-n+5]$ as 1 for $n\ge 5$ or $n\ge -5$, when it is actually 1 for $n\le 5$. Two other slips: answering $-4$, the last point where $x[n]=1$, instead of $-3$, the first point where $x[n]=0$; and assuming that $x[n]=0$ again for $n\ge 6$, when it is $-1$ there.

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19. Q.29NAT · 1 markOfficial answer: 25

    Signals and Systems › Signals basics

    Time scaling $y(t)=x(at)$ with $a>0$ divides the fundamental period by $a$: $T_y=T_x/a$. Compression ($a>1$) shortens the period and expansion ($0<a<1$) lengthens it.

    WATCH OUT FOR

    Multiplying instead of dividing, which gives $4\times 100=400\text{s}$. Replacing $t$ by $4t$ compresses the signal in time, so each cycle takes a quarter as long and the period is $100/4=25\text{s}$.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 0.99 to 1.01

    Electrical and Electronic Measurements › Bridges

    In a balanced Kelvin double bridge the galvanometer current is zero. The inner arms $R_a$ and $R_b$ therefore carry the same current and act as one series path of $R_a+R_b$, in parallel with the link $R_c$. The $3\text{mA}$ in the arm feeding the link splits by a simple current divider: $I_{R_a}=3\times \frac{10}{10+20}=1\text{mA}$.

    WATCH OUT FOR

    One mistake is to put $R_a$ in parallel with $R_c$ alone and forget that $R_b$ is in series with it, which gives $1.5\text{mA}$. Another is to flip the divider and use the path's own resistance in the numerator, $3\times \frac{20}{30}=2\text{mA}$, which is actually the current through $R_c$. A third is trying to solve the whole network with the $1\text{V}$ source and the $100\text{k}\Omega$ galvanometer. At balance neither affects the answer, and the outer arms have no values printed anyway.

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21. Q.31NAT · 1 markOfficial answer: 1.0

    Networks (Electric Circuits) › Theorems

    With only independent sources present, $R_{th}$ is found by deactivating them: short every voltage source and open every current source. Any resistor in series with a current source then carries no current and drops out, and any resistor directly across a voltage source is shorted out.

    WATCH OUT FOR

    Keeping the $1\Omega$ resistors that sit in series with the $1\text{A}$ sources in the network. Once a current source is opened, its whole branch carries no current, so those resistors do not count (keeping them gives about $0.46\Omega$). A second mistake is opening the $1\text{V}$ source instead of shorting it. That leaves the left $1\Omega$ in the path and gives $3\parallel 2=1.2\Omega$.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: -0.5

    Engineering Mathematics › Probability & statistics

    First find the missing probability from $\sum P=1$. Then use $E[g(X)]=\sum g(x)P(X=x)$: take $g(x)=x$ for the mean and $g(x)=x^2$ for the second moment. Linearity also gives the answer in one line as $E[X(1-X)]$, and only $x=2$ adds anything to that sum.

    WATCH OUT FOR

    There are two common slips. One is leaving out the $x=2$ term because its probability is not printed, when it must be found as $1-0.25-0.5=0.25$. The other is computing $E[X^2]$ as $(E[X]{)}^2=1$, which gives $0$ instead of $-0.5$. $E[X^2]$ is the second moment, not the square of the mean. The two differ by the variance, which here is $0.5$.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 0.95 to 1.05

    Analog Electronics › Diode circuits

    The ideal diode shorts R whenever the source current flows in the diode's forward direction, so R carries only the other half-cycles: a half-wave-rectified sine of peak $\pi$ mA. Its average over a full period is $I_m/\pi =1$ mA, whatever the value of R, because a current source drives the circuit.

    WATCH OUT FOR

    Averaging over only the half-cycle in which R conducts gives $2I_m/\pi =2$ mA. The average must be taken over the full period, which gives $I_m/\pi$. A second trap is to think the answer needs the value of R: with a current source and the diode open, all of $i_s$ goes through R, whatever its resistance.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 0

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

    A wattmeter shows a reading only when current flows through its current coil. Here the current coil is the only conductor between the primary circuit and the secondary circuit, which the transformer isolates from each other. KCL gives no return path, so $I_{CC}=0$ and the reading is $0\text{W}$ whatever the potential-coil voltage.

    WATCH OUT FOR

    Working out the resistor's power, $\frac{100^2}{100}=100\text{W}$, and answering 100 W, the meter's full scale. That assumes the current coil carries the secondary load current, but it does not: it only bridges the grounded primary return to the secondary loop and forms no closed path.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 0

    Electrical Machines › DC machines

    A DC shunt generator self-excites only if the field-current mmf produced by the small residual-flux EMF reinforces (adds to) the residual flux. This "sense of reinforcement" depends on both the direction of rotation and the fixed physical connection of the field winding to the armature; reversing only the rotation direction reverses the EMF polarity and hence the field-current direction, which flips the field mmf from reinforcing to opposing the residual flux — killing self-excitation.

    WATCH OUT FOR

    Assuming the output voltage stays at 200 V (or drops to some other nonzero value) because "the machine is unchanged" — forgetting that self-excitation depends on the relative direction between the residual-flux EMF and the fixed field-winding connection, so reversing rotation reverses this relative sense and can convert a cumulatively-connected (self-exciting) machine into a differentially-connected (de-exciting) one.

    Full solution →
26. Q.36MCQ · 2 marksOfficial answer: A

    Signals and Systems › LTI & convolution

    An impulse at $t_0$ in $h(t)$ adds a copy of the input delayed by $t_0$. Here the delay of 4 s is a phase lag of $\frac{7\pi }{4}\times 4=7\pi$ rad, an odd multiple of $\pi$. That turns the delayed copy into $-0.5\cos \left(\frac{7\pi }{4}t\right)$, which cancels half of the direct term. Put another way, $|H(j\omega )|=|1+0.5e^{-j4\omega }|$ is at its minimum of $0.5$, with zero phase, at this frequency.

    WATCH OUT FOR

    Adding the two impulse strengths without accounting for the delay ($1+0.5=1.5$) gives option (B). That is only right if the phase lag $4\omega$ is an even multiple of $\pi$. A second trap is thinking the delay turns cosine into sine (options C and D). A phase shift of $7\pi$ only flips the sign; a sine would need an odd multiple of $\pi /2$.

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

    Signals and Systems › Laplace

    The time-shift property $g(t-t_0)u(t-t_0)⟷e^{-s{t}_0}G(s)$ only works when the whole function is shifted, not just the step. So first write $e^{-3t}=e^{-15}{e}^{-3(t-5)}$. The constant $e^{-15}$ multiplies the transform, and the ROC stays $\mathrm{Re}\{s\}>-3$. The same result follows from shifting in $s$: $e^{-at}f(t)⟷F(s+a)$, where $u(t-5)⟷\frac{e^{-5s}}{s}$ gives $\frac{e^{-5(s+3)}}{s+3}$.

    WATCH OUT FOR

    The usual mistake is to apply the time shift to the step alone and write $\frac{e^{-5s}}{s+3}$ (option A). That drops the factor $e^{-15}$, because $e^{-3t}u(t-5)$ is not the same as $e^{-3(t-5)}u(t-5)$. Getting the sign of the exponential wrong leads to (B) or (D).

    Full solution →
28. Q.38MCQ · 2 marksOfficial answer: A
29. Q.39MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Complex analysis

    Invert the bilinear map and substitute a real value. For real $u$, the numbers $j-u$ and $j+u$ have the same modulus $\sqrt{u^2+1}$, so their ratio always has modulus 1. Put the other way round: $f$ is a Cayley-type transform that takes the unit circle onto the real axis, so $f^{-1}$ takes the real axis back onto the unit circle.

    WATCH OUT FOR

    Using the forward map $f$ instead of $f^{-1}$. For real $x$, $f(x)=j\frac{1-x}{1+x}$ is purely imaginary, so $f$ maps the real axis to the imaginary axis, which leads to the wrong option (C). A second slip is getting the image circle right but not checking its centre. The points $1$, $j$ and $-1$ are all at distance 1 from the origin, which fixes the centre at $0$ and rules out (B).

    Full solution →
30. Q.40MCQ · 2 marksOfficial answer: C

    Digital Electronics › Boolean logic

    Plot the 8 minterms on a 4-variable K-map. They form exactly two non-overlapping quads: $\{4,5,12,13\}$ gives $X\overline{Y}$ and $\{10,11,14,15\}$ gives $WY$. Both are essential prime implicants, and together they cover every minterm. The third prime implicant, $WX$, is the consensus of these two and is redundant.

    WATCH OUT FOR

    Choosing (B) because the quad $\{12,13,14,15\}$ ($WX$) is a valid prime implicant. It is the consensus term of $X\overline{Y}$ and $WY$, so it adds no coverage. Choosing (A) is the same kind of error: it describes the correct function but is not the minimal form the question asks for.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: A

    Digital Electronics › Sequential

    A closed loop with an odd number of inversions is a ring oscillator. Its period is twice the total propagation delay round the loop, and only gates whose inputs actually change count towards that delay. Here the constant inputs A and B make the OR and AND gates act as plain delays, so the loop is effectively one inverter plus 20 ns of extra delay.

    WATCH OUT FOR

    Taking the period as a single trip round the loop gives $25\text{ns}$ and 40 MHz (option C). You need two trips, one for each edge. Counting only the OR and AND delays and missing the NOT gate in the loop gives $20\text{ns}$, which students often misread as the period and so pick 50 MHz (option D). Adding the NOT gate on $B$ is also wrong: $B$ is constant, so that gate is not in the loop.

    Full solution →
32. Q.42MCQ · 2 marksOfficial answer: C

    Digital Electronics › Sequential

    A 4-bit shift register fully empties in four shifts. So B ends up holding A's original contents (1101), and A ends up holding the four feedback bits $D=A_4\oplus \overline{A_2}$ it produced, the newest on the left. Each feedback bit is worked out from the register contents just before its clock edge.

    WATCH OUT FOR

    Treating A as a plain ring counter, with $A_4$ fed straight back to $A_1$. That brings A back to 1101 after four shifts and gives option (A), but it ignores the XOR and the inverter. Other slips are dropping the inverter on $A_2$, taking the tap from the wrong cell, and working out $D$ from contents that have already shifted instead of the contents just before the edge.

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

    Control Systems › Bode & margins

    A pure delay $e^{-sT}$ does not change $|G(j\omega )|$. It adds phase $-\omega T$, which is a straight line on a linear frequency axis and has no lower limit. A second-order rational transfer function on its own can never go below $-180^{\circ }$. So the magnitude plot gives the quadratic (${\omega }_n$ from where the peak sits, $\zeta$ from how high it is), and the extra linear phase gives the delay $T$.

    WATCH OUT FOR

    Choosing (A) because its magnitude plot matches exactly, without noticing that its phase stops at $-180^{\circ }$ while the plot goes to $-720^{\circ }$. Another trap is choosing (C) because it 'has a delay' without checking the size: $T=0.5\times 10^{-12}\text{s}$ adds only about $10^{-10}\text{rad}$ of phase here. Some students also take ${\omega }_n=10000$ instead of ${\omega }_n^2=10000$, or pick (D) by mistaking $100$ for ${\omega }_n$ when it is ${\omega }_n^2$ in that option. Finally, a straight phase line on a linear frequency axis points to a delay, not to extra poles; pole and zero asymptotes are straight on a log axis.

    Full solution →
34. Q.44MSQ · 2 marksOfficial answer: B, C

    Signals and Systems › Fourier transform

    For a real signal, $x(-t)\leftrightarrow X^{\ast }(j\omega )$. So the even part $x(t)+x(-t)$ has transform $2\text{Re}\{X(j\omega )\}$, which is real, and the odd part $x(t)-x(-t)$ has transform $j2\text{Im}\{X(j\omega )\}$, which is imaginary. Multiplying a signal by $j$ multiplies its transform by $j$, which turns real into imaginary and imaginary into real. Causality with positive energy makes sure the even and odd parts are both nonzero, so none of the four transforms is trivially zero.

    WATCH OUT FOR

    Marking only (B) because you remember "real and odd gives imaginary" and forgetting that the factor $j$ in (C) turns the real transform of the even part into an imaginary one. The opposite mistake is marking (D) because it "has a $j$ in it": there $j\cdot j=-1$, so its transform comes out real.

    Full solution →
35. Q.45NAT · 2 marksOfficial answer: 0.620 to 0.640

    Electromagnetics › Waveguides & fibres

    The NA is set by whatever medium directly surrounds the core. Remove the glass cladding and put the core in water, and the water becomes the cladding: $\text{NA}=\sqrt{n_{\text{core}}^2-n_{\text{surround}}^2}$ with $n_{\text{surround}}=1.33$.

    WATCH OUT FOR

    Using the original cladding index 1.44, which gives about 0.295. Another slip is taking the square root of the index difference, $\sqrt{1.47-1.33}\approx 0.374$, when the formula needs the difference of the squares. Some students also divide by $n_{\text{water}}$ as if light were launched from water. The NA is defined for launch from air ($n_0=1$); water here only forms the cladding.

    Full solution →
36. Q.46NAT · 2 marksOfficial answer: 1.40 to 1.50

    Networks (Electric Circuits) › AC, power & resonance

    Work with admittances, because parallel admittances simply add. Set the imaginary part of $Y_1+Y_2$ to zero. The condition factorises into $X_C=X_L$ or $X_C{X}_L=R^2$ (that is, $C=L/R^2$). Here $\omega L\approx R$, so the network is almost the classic constant-resistance circuit ($R_1=R_2=\sqrt{L/C}$). That is why the two roots nearly coincide at about 1.45 mF.

    WATCH OUT FOR

    Setting $\omega L=1/(\omega C)$ straight away, as if this were a series or pure-LC resonance, without deriving the condition. That happens to land in the range here, but only because of the numbers. A second common slip is writing $Y_2=1/(R+j{X}_C)$ with the wrong sign on the capacitive reactance. That makes the two imaginary parts add instead of cancel, so no real solution exists.

    Full solution →
37. Q.47NAT · 2 marksOfficial answer: 0.43 to 0.45

    Networks (Electric Circuits) › Transients

    This is a first-order RL circuit, so $i_L(t)=i_L(\infty )+[i_L(0^{+})-i_L(\infty )]e^{-t/\tau }$ with $\tau =L/R$, where $R$ is the resistance the inductor sees with the current source open-circuited. The non-zero initial current matters: the gap still to close is only $\frac{4}{5}{I}_0$. The 99% point is where that gap has shrunk to $0.01I_0$, so $e^{-t/\tau }=\frac{0.01}{0.8}=\frac{1}{80}$.

    WATCH OUT FOR

    Ignoring the initial current and using the zero-state rule $t=\tau \ln 100=0.46\mu \text{s}$, which falls outside the key. A second mistake is shorting the current source when finding $\tau$ (that would give $\tau =0$). An ideal current source is replaced by an open circuit, so the inductor sees $R$ alone.

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

    Control Systems › Compensators & PID

    Read the controller's zeros off the root locus. A PID adds a pole at the origin and two zeros. The whole real axis left of $-1$ is on the locus, which by the odd-count rule is only possible if both zeros sit at $-1$. So $C(s)=K_D(s+1{)}^2/s$, and at $\omega =1$ the proportional, integral and derivative terms combine into the real number $2K_D$.

    WATCH OUT FOR

    Taking the single 'o' marker as a single zero, $C(s)=K_P(s+1)/s$. That leaves $K_D$ undetermined (or wrongly zero), and it contradicts the plot, because a single zero at $-1$ would take the real axis left of $-1$ off the locus. Overlapping zeros are drawn as one marker, so count them with the real-axis rule, not by eye. A second slip is using $|j\omega +1|$ instead of $|j\omega +1{|}^2$ in the magnitude, which gives $K_D=\sqrt{2}$.

    Full solution →
39. Q.49NAT · 2 marksOfficial answer: 240

    Engineering Mathematics › Probability & statistics

    There are 5 places and only 4 digits, and just one digit may repeat, so the digit pattern has to be $2+1+1+1$: all four digits appear and one of them appears twice. Choose the repeated digit (4 ways), then count the orders of a multiset, $\frac{5!}{2!}=60$. That gives $4\times 60=240$.

    WATCH OUT FOR

    One slip is to forget to choose which digit repeats, which gives $\frac{5!}{2!}=60$. Another is to count $4\times 5!=480$ without dividing by $2!$ for the two identical digits. A third is to pick the repeated digit and then choose 3 of the remaining digits as $\left(\genfrac{}{}{0ex}{}{3}{3}\right)$ while also multiplying by $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$ or a similar factor, which counts the same number more than once. Subtracting orders with a leading zero is also wrong here, because $0$ is not one of the digits.

    Full solution →
40. Q.50NAT · 2 marksOfficial answer: -2 to 2

    Control Systems › Bode & margins

    A right-half-plane zero $(1-s)$ has the same magnitude as the left-half-plane zero $(1+s)$, but its phase is the opposite: it contributes $-{\tan }^{-1}\omega$, a lag. Get the gain crossover from the magnitude, then add up the phase with the correct sign for each factor.

    WATCH OUT FOR

    Treating the zero $(1-s)$ as if it were $(1+s)$ and giving it a phase lead of $+{\tan }^{-1}\omega$. That makes $\angle G(j\sqrt{3})=60^{\circ }-120^{\circ }=-60^{\circ }$ and a wrong PM of $120^{\circ }$. Another slip is to read the leading minus in $-(s-1)$ as an extra $180^{\circ }$. Written as $2(1-s)$ the gain is $+2$ and adds no phase.

    Full solution →
41. Q.51NAT · 2 marksOfficial answer: 1.77 to 1.78
42. Q.52NAT · 2 marksOfficial answer: 34.5 to 35.5
43. Q.53NAT · 2 marksOfficial answer: 0.98 to 1.02

    Engineering Mathematics › Probability & statistics

    The sample standard deviation divides the sum of squared deviations by $n-1$ (Bessel's correction), not by $n$: $s=\sqrt{\sum (x_i-\bar{x}{)}^2/(n-1)}$. Here the mean is 80 kg, the squared deviations add up to 4, and $s=\sqrt{4/4}=1$.

    WATCH OUT FOR

    Dividing by $n=5$ gives the population standard deviation, $\sqrt{4/5}\approx 0.89\text{kg}$. That falls outside the key's range of 0.98 to 1.02. The word "sample" in the question means you divide by $n-1=4$.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 4.00 to 4.10
45. Q.55NAT · 2 marksOfficial answer: 17.00 to 18.00

    Networks (Electric Circuits) › Coupled circuits

    The energy stored in coupled inductors is $w=\frac{1}{2}{L}_1{i}_1^2+\frac{1}{2}{L}_2{i}_2^2\pm M{i}_1{i}_2$, with $M=k\sqrt{L_1{L}_2}$. Use the plus sign when both currents enter (or both leave) their dotted ends. Here the two currents are in phase, so all three terms peak together, when both currents are at their peak values.

    WATCH OUT FOR

    Using the minus sign for the mutual term gives $12-5.37=6.63\mu \text{J}$. The right sign comes from reading the dots correctly: both currents enter dotted ends here. Other slips: leaving out the mutual term entirely (12 μJ), or taking $M=k{L}_1{L}_2$ or $k(L_1+L_2)/2$ instead of $k\sqrt{L_1{L}_2}$. Some also use rms currents, but the question asks for the peak energy, which needs the peak currents.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 0.64 to 0.70

    Analog Electronics › Amplifiers

    The drain bias of 4 V fixes $I_D=(5-4)/1\text{k}\Omega =1\text{mA}$ straight away. That gives $V_S=1\text{V}$, $V_{GS}=2\text{V}$ and $g_m=2\text{mA/V}$. Because the output is taken at the source, the stage is a source follower with gain $g_m{R}_S/(1+g_m{R}_S)$. $R_D$ only sets the bias and plays no part in the signal gain.

    WATCH OUT FOR

    Taking the output as if it came from the drain and using the degenerated common-source gain $-g_m{R}_D/(1+g_m{R}_S)=-0.67$, which has the right magnitude but the wrong sign and the wrong physical meaning. Other slips: dropping the factor $\frac{1}{2}$ in the square law so that $g_m$ comes out wrong, or adding a body-effect term $g_{mb}$ even though the body is tied to the source.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 0.9 to 1.1

    Engineering Mathematics › Calculus

    The graph of $\max \{f,h\}$ is the upper boundary of the two curves. Its minimum is either at the lowest point of a piece inside that piece's own interval, or at a crossing point where the upper boundary switches from one curve to the other. Here the line is on top on $[-1,3]$ and keeps falling up to $x=3$. After $x=3$ the parabola is on top and keeps rising. So the minimum is at the crossing $x=3$.

    WATCH OUT FOR

    Taking the vertex of the parabola, $x=2$, $g=0$, as the answer. At $x=2$ the parabola is not the larger piece: the line gives $3$, so $g(2)=3$. The opposite slip is to minimise $\min \{f,h\}$ instead of $\max \{f,h\}$. The vertex value $0$ is only the minimum of the parabola alone, and $g$ never gets that low.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 0.95 to 1.05

    Electrical Machines › Transformers

    A short-circuit test gives the transformer's series (leakage) impedance as seen from the primary; when a capacitor is placed in series with that impedance and its reactance is chosen to exactly cancel the leakage reactance, the circuit becomes purely resistive (series resonance), so the current is simply the source voltage divided by the short-circuit resistance.

    WATCH OUT FOR

    Forgetting that with the secondary shorted, the transformer itself still contributes its short-circuit impedance $R_{sc}+j{X}_{sc}$ to the new circuit (treating the capacitor as connected to an ideal short instead of to the transformer's leakage impedance), or making a sign/algebra slip when combining $-j{X}_C$ with $+j{X}_{sc}$ and not noticing they are equal in magnitude (resonance), leading to a wrong non-real or mismatched impedance.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 0.49 to 0.51

    Analog Electronics › Op-amps

    This is the input stage of a three-op-amp instrumentation amplifier. The virtual shorts put $V_{S1}$ and $V_{S2}$ across the middle resistor, and the same current flows through all three resistors of the left chain, so $V_{o1}=2V_{S1}-V_{S2}$ and $V_{o2}=2V_{S2}-V_{S1}$. The difference between the outputs is amplified, but their average stays equal to the input common-mode voltage. The equal-resistor divider on the right picks out that average: $V_O=(V_{S1}+V_{S2})/2$. Only the DC parts of the inputs survive averaging, so the answer is $(3-2)/2=0.5\text{V}$.

    WATCH OUT FOR

    A common error is to treat the circuit as a differential amplifier and compute $V_{o1}-V_{o2}=3(V_{S1}-V_{S2})$, which averages to $15\text{V}$. But $V_O$ is measured from the midpoint of the right divider to ground, not across the two outputs, so it is the common-mode average $(V_{o1}+V_{o2})/2$. Other slips: getting the sign of $IR$ wrong for the bottom op-amp ($V_{o2}=V_{S2}-IR$, because the current flows from B towards the bottom output), and averaging the sine amplitudes into the answer when a sinusoid averages to zero.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 99 to 101

    Analog Electronics › Op-amps

    This circuit is a switched-capacitor inverting integrator. In every clock period the 1 nF capacitor picks up $C_1{V}_{in}$ from the input and dumps it at the virtual ground, onto the feedback capacitor, so the output steps by $-\frac{C_1}{C_f}{V}_{in}=-10\text{mV}$ each period. The 1 nC starting charge fixes the starting point: because the inverting input is a virtual ground, $V_o(0)=+Q_0/C_f=+100\text{mV}$. After 20 periods, $V_o=100-200=-100\text{mV}$.

    WATCH OUT FOR

    1. Ignoring the 1 nC initial charge. That gives $|V_o|=200\text{mV}$.
    2. Reading the polarity backwards, so $V_o(0)=-100\text{mV}$. That gives $-300\text{mV}$, a magnitude of $300\text{mV}$.
    3. Treating the stage as non-inverting and adding the steps: $100+200=300\text{mV}$.
    4. Counting 19 or 40 transfers instead of 20. There is exactly one charge transfer per 1 ms period, at each High-to-Low edge.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 2.1 to 2.5

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

    The multiplier's $V_x$ input is proportional to the load current and its $V_y$ input is proportional to the load voltage shifted by $90^{\circ }$. That makes the circuit a varmeter: the average output is $V_{x,\text{rms}}{V}_{y,\text{rms}}\cos (90^{\circ }+\varphi )$, whose magnitude is $VI\sin \varphi \times R_{\text{shunt}}\times A_1$ for a load with power-factor angle $\varphi$. Here that is $230\times 1\times \sin 45^{\circ }\times 1\times \frac{1}{50\sqrt{2}}=2.3\text{V}$.

    WATCH OUT FOR

    Ignoring the $+90^{\circ }$ phase shifter and computing active power, $P\times R\times A_1=162.63/70.71=2.3\text{V}$, reaches the right number only by accident, because $\cos 45^{\circ }=\sin 45^{\circ }$. The method would fail for any other power factor. Two other slips: using peak values instead of rms (which gives $4.6\text{V}$), and taking $|I|=230/(230/\sqrt{2})=\sqrt{2}\text{A}$ by dividing by one component of $Z$ instead of $|Z|=230\Omega$.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 1.9 to 2.1

    Analog Electronics › Op-amps

    The output feeds back to both inputs: negative feedback through the $3R$–$3R$ divider (fraction $\frac{1}{2}$) and positive feedback through the R-network (fraction $\frac{1}{3}$). Net feedback is negative, so the virtual short holds. It gives $V_{+}=V_o/2$, and nodal analysis of the non-inverting network then fixes $V_o$.

    WATCH OUT FOR

    One mistake is to ignore the $R$ from node X to the output and treat the non-inverting side as a plain divider of the 1 V source. That gives $V_{+}=1\cdot \frac{2R}{5R}=0.4\text{V}$ and $V_o=0.8\text{V}$. Another is to see positive feedback and assume the op-amp saturates, without checking that the negative feedback ($\frac{1}{2}$) is stronger than the positive feedback ($\frac{1}{3}$).

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: -8

    Engineering Mathematics › Linear algebra

    A rank-one matrix has every row proportional to one nonzero row. That is the same as saying every $2\times 2$ minor vanishes. Each unknown sits in a row with one known entry, and that known entry fixes the scale factor for its row.

    WATCH OUT FOR

    Using the wrong entry to fix a row's multiplier, for example taking row 3 as $6\times$ row 1 instead of $\frac{6}{4}\times$ row 1, which gives $\beta =6$ and a ratio of $-2$. Another slip is computing $\beta /\alpha =-0.125$ and rounding it to $0$. Also, setting only one minor to zero is not enough: you have to use both the row-2 and row-3 conditions.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 0.009 to 0.011

    Electrical and Electronic Measurements › Digital instruments

    The meter measures $V=I_{s}R$ across the CT burden. For the worst case, add the limiting relative errors of the CT ratio (Class 1, so $\pm 1\%$) and the burden ($\pm 0.5\%$), turn that into volts at the $0.5\text{V}$ reading, then add the meter's own error: $0.1\%$ of reading plus 2 counts of $0.001\text{V}$ each on the $1.999\text{V}$ range.

    WATCH OUT FOR

    The usual mistakes are leaving out the digit error or giving it the wrong size. On a 3½-digit meter, $\pm 2$ digits on the $1.999\text{V}$ range is $\pm 0.002\text{V}$, and that alone is 20 % of the answer. Other slips: forgetting the burden tolerance, or combining the errors as a root-sum-square when the question asks for the worst case, which is a straight sum.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 99.0 to 101.0

    Networks (Electric Circuits) › AC, power & resonance

    The element values are chosen so that $\omega =\frac{1}{\sqrt{LC}}$: here $\sqrt{LC}=\frac{1}{20\pi }\times 10^{-3}\text{s}$, so $\frac{1}{\sqrt{LC}}=20000\pi \text{rad/s}$, which is exactly the source frequency. At series resonance the reactances cancel and the current is set only by the total loop resistance, and that total includes the source's internal resistance.

    WATCH OUT FOR

    Leaving out the 50 Ω internal resistance because it is not drawn, which gives $10/50=200\text{mA}$. Using the peak voltage $10\sqrt{2}\text{V}$ in place of the rms value, which gives $141.4\text{mA}$. Reading mF as μF, which makes $X_C=1000\Omega$ and takes the circuit far from resonance.

    Full solution →