GATE IN 2026 question paper with solutions

GATE IN 2026 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   To turn a quote into indirect speech with a past reporting verb, make four changes together: keep \"said\", change the pronoun to fit the speaker (I becomes he), move the tense one step back (simple past becomes past perfect), and shift the time word (yesterday becomes the day before).

   WATCH OUT FOR

   Making only some of the changes. (A) leaves the pronoun and the tense as they were. (C) shifts the time word to \"the day before\" but switches the reporting verb to \"says\" and keeps \"did\", which mixes present-tense reporting with past-tense shifts. The correct answer has to make every change at once.

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

   General Aptitude › Numerical

   In a tile series, ignore the pictures at first and just count what is on each tile. Here the counts 0, 1, 1, 2, 3, 5 form the Fibonacci sequence, where each term is the sum of the two before it. So the next term is 3 + 5 = 8, and only one option has 8 dots.

   WATCH OUT FOR

   Tile 6 looks like the five on a die, so it is easy to assume the tiles are die faces counting up and pick (B), the six. That rule breaks at the start: the tiles begin 0, 1, 1, so a count does repeat. The other mistake is picking (D), the full 3 x 3 grid, because it looks like the natural way to finish the pattern. It has 9 dots, not 8.

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

   General Aptitude › Reasoning

   Treat the walkways as the edges of a 5-cycle and the junctions as its vertices. Each guard covers only the 2 edges at his junction, so covering 5 edges needs at least $\lceil \frac{5}{2}\rceil =3$ guards, and 3 well-placed guards (apex plus both bottom corners) do the job.

   WATCH OUT FOR

   Answering 2 by putting guards at two opposite-looking corners, such as the two upper side junctions, and forgetting that the bottom walkway is then unwatched. Two guards can cover at most 4 walkways, and there are 5. The other trap is thinking a guard is needed at every junction (5), when a guard at one junction also covers both walkways ending there.

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

   General Aptitude › Numerical

   Count the days between the two dates and divide by 7. Only the remainder shifts the weekday: here 31 days is 4 full weeks plus 3 days, so Thursday moves forward 3 days to Sunday.

   WATCH OUT FOR

   Off-by-one counting. Some students count 2 June as day 1 and get 32 days (remainder 4, Monday, which is not even an option). Others take June as 31 days and get 32 days again. The gap is 3 July minus 2 June: 28 days to the end of June plus 3 days of July, which is 31.

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5. Q.5MCQ · 1 markOfficial answer: C

   General Aptitude › Reasoning

   Count inversions. An adjacent swap reverses the order of exactly one H–T pair, so the minimum number of adjacent swaps equals the number of pairs that must change order: 3 heads × 3 tails = 9.

   WATCH OUT FOR

   Answering 3 by thinking of each H trading places with a T in one move. A step only swaps neighbours, so every H must pass each of the 3 T's one step at a time, which gives 3 × 3 = 9. Answering 6 is another common slip, from counting only how far the blocks shift and not the number of crossings.

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6. Q.6MCQ · 2 marksOfficial answer: A

   General Aptitude › Reasoning

   In an analogy, first name the relation in the given pair (here antonyms, 'worsen : lessen'). Then check each option for both the relation and the order. A quick test is to swap in synonyms: aggravate is close to exacerbate, and alleviate is close to mitigate.

   WATCH OUT FOR

   Picking (B) because it pairs a 'lessen' word with a 'bring on' word. That misses the order: the given pair starts with the worsening word, so (B) is reversed even on the loosest reading. Another trap is choosing a pair of words that merely look alike (Emancipate : Exonerate) without checking their relation.

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7. Q.7MCQ · 2 marksOfficial answer: B

   General Aptitude › Spatial

   In a cross-shaped net, squares one apart in a straight line become opposite faces, and every other pair shares an edge. Fix one face as the front, fold the net, and note which edge or corner each marking touches. A drawn cube is possible only if all three visible faces can meet at one corner, each marking touches the right edge or corner, and the faces go round the corner in the right order (no mirror image).

   WATCH OUT FOR

   Accepting (i) because its three faces (L, S, H) can meet at a corner and the two big markings, L's white quarter and H's grey half, are in the right places, without checking which way the thin stripe on S runs. That stripe has to cross the S–L edge at a right angle, so it cannot be parallel to the front edge. A second common slip is forgetting that H turns over when it folds onto S: its lower half in the net becomes the half of the back face next to T.

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8. Q.8MCQ · 2 marksOfficial answer: B

   General Aptitude › Reasoning

   Write "all X are Y" as $C\subseteq B$ and "no one is both" as $C\cap D=\varnothing$. Then no one has all three conditions, and the high-BP circle splits into three separate parts: cholesterol + BP, only BP, and BP + diabetes. Subtract the two known parts from $|B|$ to get the third. The total of 75 is only needed to check that the numbers fit.

   WATCH OUT FOR

   Using the total of 75 to find the overlap. For example, $75-45-10=20$ gives option (C), but that counts the cholesterol group twice: it lies inside the BP group and is already part of the 45. Another error is to subtract only the 20 "only high BP" patients from 45 and answer 25, which forgets that the 10 cholesterol patients are also in the high-BP group.

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9. Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Reasoning

   Turn each statement into an age inequality and chain them. $P<S$ and $R<P$ give $R<P<S$, so neither P nor S can be the youngest. Q rules itself out, which leaves R. You do not need the full order to answer.

   WATCH OUT FOR

   Misreading R's statement "P is older than me" as $R>P$. It means $P>R$, so R is younger than P. Some candidates also think Q's statement makes the order ambiguous. Q can sit second or third, but R is youngest either way, so the question still has one answer.

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

    General Aptitude › Geometry

    When circles touch externally, the distance between two centres is the sum of their radii. That gives all three sides of triangle $O_1{O}_3{O}_2$ in terms of $r_3$: $2+r_3$, $1+r_3$ and $3$. The right angle at $O_3$ makes $O_1{O}_2=3$ the hypotenuse, and Pythagoras then gives a quadratic in $r_3$.

    WATCH OUT FOR

    The usual slip is to put the hypotenuse in the wrong place. The right angle is at $O_3$, so the hypotenuse is $O_1{O}_2=3$, not one of the sides that involve $r_3$. Other errors are forgetting that $C_1$ and $C_2$ also touch (so $O_1{O}_2=r_1+r_2$), and picking option (B), $\frac{1}{2}(3+\sqrt{17})$, after a sign slip in the quadratic formula. Option (B) is about 3.56 cm, larger than both given radii. That contradicts both the figure and Pythagoras, since the legs $2+r_3$ and $1+r_3$ could not then fit a hypotenuse of 3.

    Full solution →

Instrumentation Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   If a polynomial $p$ satisfies $p(P)=0$, then every eigenvalue $\lambda$ of $P$ satisfies $p(\lambda )=0$, because $p(P)\mathbf{x}=p(\lambda )\mathbf{x}$ for an eigenvector $\mathbf{x}$. So the eigenvalues can only be roots of $p$. The given polynomial need not be the characteristic polynomial: $P=I$ and $P=2I$ also satisfy it.

   WATCH OUT FOR

   Choosing (A) on the assumption that a $3\times 3$ matrix needs three different eigenvalues, when the cubic has a repeated root and $3$ is not a root at all. A second error is to treat the given cubic as the characteristic polynomial. Here that happens to give the same set, but the reasoning is wrong, since $P=I$ also satisfies the identity and has only the eigenvalue $1$.

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2. Q.12MCQ · 1 markOfficial answer: D

   Engineering Mathematics › Calculus

   Split $|x|$ by sign to get $f(x)=x^2$ for $x\ge 0$ and $-x^2$ for $x<0$. At $0$, both the one-sided limits and the one-sided difference quotients $\frac{h|h|}{h}=|h|$ go to $0$, so $f$ is continuous and differentiable there, with $f^{\prime }(0)=0$. Multiplying $|x|$ by $x$ smooths out the corner of $|x|$ once: $f^{\prime }(x)=2|x|$ still has a corner, so $f^{\prime \prime }(0)$ does not exist.

   WATCH OUT FOR

   Seeing $|x|$ and picking C by reflex, because $|x|$ itself is not differentiable at $0$. The extra factor $x$ makes the difference quotient $|h|$, which tends to $0$ from both sides. A second slip is to confuse "differentiable once" with "differentiable twice": $f^{\prime \prime }(0)$ does not exist, but the question asks only about the first derivative.

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3. Q.13MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Complex analysis

   Cauchy integral formula: if $f$ is analytic on and inside a simple closed contour $\gamma$ traversed anticlockwise, and $z_0$ lies inside $\gamma$, then ${\oint }_{\gamma }\frac{f(z)}{z-z_0}dz=2\pi jf(z_0)$. Here $f$ is a polynomial, so the integral is simply $2\pi j$ times $f(1)$, whatever shape $\gamma$ has.

   WATCH OUT FOR

   Leaving out the factor $2\pi$. Writing $\pi jf(1)=3\pi j$ gives option (B), and dropping the factor 3 gives $\pi j$, option (D). Another slip is evaluating $f$ at $z=0$ or at the wrong point, or squaring twice, which can lead to $12\pi j$ (C). The integral is $2\pi j$ times $f$ evaluated at the enclosed pole, $z_0=1$.

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4. Q.14MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   When a square $A$ is singular ($\det A=0$), $Ax=b$ has either no solution or infinitely many. It has infinitely many exactly when $b$ satisfies the same linear dependence as the rows of $A$, i.e. when $\rho (A)=\rho ([A|b])$. Here $R_3=R_1+R_2$ but $b_3\ne b_1+b_2$, so there are no solutions.

   WATCH OUT FOR

   Seeing $\det A=0$ and choosing "infinitely many" (D) straight away. A zero determinant only rules out a unique solution. You still have to check $b$ against the row dependence, and here $b_3=0\ne b_1+b_2=4$. A sign slip in the cofactor $(-1\cdot 1-1\cdot 0)=-1$ can also make $\det A$ look nonzero, which leads to the wrong answer (B).

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5. Q.15MCQ · 1 markOfficial answer: Marks to all

   Electromagnetics › Electrostatics

   $\mathbf{E}=-\nabla V$. For $V=r=\sqrt{x^2+y^2+z^2}$, $\nabla r={\hat{a}}_r$, so the field has magnitude 1 V/m everywhere and points toward the origin. The force on the charge is $\mathbf{F}=q\mathbf{E}$, and at $(1,1,1)$ each component is $-\frac{1}{\sqrt{3}}$.

   WATCH OUT FOR

   There are two. The first is dropping the minus sign in $\mathbf{E}=-\nabla V$, which gives option (A) exactly but puts the force on a positive charge pointing toward higher potential. The second is carrying over the inverse-square habit from a point charge and dividing by $r^2$, which gives (B). Here the potential grows linearly with $r$, so the field has the same magnitude everywhere; it does not fall off as $1/r^2$.

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6. Q.16MCQ · 1 markOfficial answer: D

   Networks (Electric Circuits) › AC, power & resonance

   To classify a filter, see what each element becomes at the frequency extremes and at resonance. At DC a capacitor is an open circuit, and at very high frequency an inductor is an open circuit. A series L-C branch has zero impedance at ${\omega }_0=1/\sqrt{LC}$. When that branch is the shunt element and the output is taken across it, the input passes unchanged at both ends of the frequency range and is shorted out at ${\omega }_0$. That is a band stop (notch) response.

   WATCH OUT FOR

   Answering band pass. That answer is right for two other circuits, not this one: the same circuit with the output taken across $R$, or a circuit whose shunt element is a parallel L-C tank, which is an open circuit at resonance. A series L-C branch is the opposite of a tank: it is a short circuit at resonance, so across the output it removes the band around ${\omega }_0$ instead of passing it. A second slip is to check only one end of the range. Seeing that $C$ blocks DC and stopping there can lead to high pass or low pass, when the gain is actually 1 at both ends.

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7. Q.17MCQ · 1 markOfficial answer: C

   Electrical Machines › Transformers

   When an impedance is referred from one side of an ideal transformer to the other, it scales by the square of the turns ratio between the side it is being referred TO and the side it originally sits on: $Z_2^{\prime }=a^2{Z}_2$, where $a=N_1/N_2$.

   WATCH OUT FOR

   Confusing which winding has more turns in a step-down transformer (thinking the primary of a step-down transformer has fewer turns because "step-down" sounds like a reduction), or forgetting the transformation is by $a^2$, not $a$, and concluding $Z_2^{\prime }=Z_2$ or $Z_2^{\prime }<Z_2$.

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8. Q.18MCQ · 1 markOfficial answer: A

   Signals and Systems › Fourier series

   A periodic signal is a sum of sinusoids at whole-number multiples of its fundamental. An ideal LPF keeps only the components below its cut-off. If the cut-off lies between $f_0$ and the next harmonic that is present, only the fundamental sinusoid gets through, whatever the input's wave shape.

   WATCH OUT FOR

   Choosing (B) because the frequency is unchanged. The fundamental frequency does survive, but a square shape needs its higher harmonics (60 kHz, 100 kHz, ...) and the filter removes all of them. Another mistake is choosing (C), from the fact that integrating a square wave gives a triangular wave. An ideal LPF is not an integrator. It removes harmonics completely instead of weighting them by $1/n$.

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9. Q.19MCQ · 1 markOfficial answer: D

   Signals and Systems › Signals basics

   A sum of sinusoids is periodic only if the ratio of their periods is rational. When it is, the fundamental period is the least common multiple of the individual periods. Equivalently, the fundamental frequency is the GCD of the individual frequencies. The amplitudes (4 and 7), and whether each term is a sine or a cosine, have no effect on the period.

   WATCH OUT FOR

   Taking the LCM of the coefficients of $\pi t$ (15 and 4) and getting 60, which is option (B). Another version of the same slip is to multiply by the wrong factor and get 15 or 30. Those numbers are angular-frequency coefficients, not periods. Convert each term to its period first ($2/15\text{s}$ and $1/2\text{s}$), then take the LCM of the periods. That gives $2\text{s}$.

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10. Q.20MCQ · 1 markOfficial answer: C

    Signals and Systems › LTI & convolution

    Linearity depends on how the dependent variable $x(t)$ and its derivatives appear. Each must appear only to the first power, with no products or nonlinear functions of them. A coefficient that depends on time, like $2t$, makes the system time-varying but leaves it linear.

    WATCH OUT FOR

    Marking (i) as nonlinear because of the $2t$ coefficient, which leads to option (A). A time-dependent coefficient breaks time-invariance, not linearity. Superposition still holds for (i).

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

    Analog Electronics › Amplifiers

    Identify a single-transistor stage by what its input and output terminals look like. Common source (input at the high-resistance gate, output at the high-resistance drain) is a VCCS or transconductance stage. Common drain (high $R_{in}$, low $R_{out}\approx 1/g_m$) is a VCVS or voltage follower. Common gate (low $R_{in}\approx 1/g_m$, high $R_{out}$, $A_i\approx 1$) is a CCCS or current follower.

    WATCH OUT FOR

    Picking (C), voltage controlled current source, because the MOSFET law $i_d=g_m{v}_{gs}$ is a transconductance. That describes the device, and it fits the common-source stage, whose input is the high-resistance gate. In the common-gate stage the signal goes into the low-resistance source terminal ($R_{in}=1/g_m$), so the input is naturally a current, and the same current comes out of the drain.

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

    Analog Electronics › Biasing

    This is voltage-divider bias. Replace the divider with its Thevenin equivalent ($V_{TH}=2.5\text{V}$, $R_{TH}=2.5\text{k}\Omega$). The emitter resistor looks $(\beta +1)$ times larger from the base, so $I_B=(V_{TH}-V_{BE})/[R_{TH}+(\beta +1)R_E]$, and then $I_C=\beta I_B$. Always confirm active mode afterwards: $V_C>V_B$.

    WATCH OUT FOR

    One common slip is treating the divider as stiff: ignoring $R_{TH}$ and taking $I_C\approx I_E=(2.5-0.65)/5=0.37\text{mA}$. That is close but is not one of the options. Another is forgetting the $(\beta +1)$ factor that reflects $R_E$ into the base loop, which gives a base current, and so a collector current, that is much too large.

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

    Digital Electronics › ADC & DAC

    For a full-scale sinusoid into an ideal $n$-bit quantizer, $\text{SQNR}=\frac{3}{2}{2}^{2n}$, which is $1.76+6.02n\text{dB}$. Each extra bit adds about $6\text{dB}$.

    WATCH OUT FOR

    Using $20{\log }_{10}(2^n)=6.02n$ alone gives $96.3\text{dB}$ because it leaves out the $1.76\text{dB}$ from the sinusoid's power relative to the ${\Delta }^2/12$ noise. Here (D) is still the nearest option, but the correct maximum is $98.1\text{dB}$. A second slip is to use $10{\log }_{10}(2^n)\approx 3n$, which gives about $48\text{dB}$ and leads to the distractor (A).

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14. Q.24MCQ · 1 markOfficial answer: B

    Electrical and Electronic Measurements › Bridges

    For a low resistance, lead and contact resistance are comparable to the resistance being measured. The Kelvin double bridge removes them in two ways. The second pair of ratio arms, set so that $P/Q=p/q$, cancels the link resistance, leaving $R=(P/Q)S$. Four-terminal (current + potential) construction of $R$ and $S$ keeps the contact drops outside the measured section.

    WATCH OUT FOR

    Mixing up the two terminal techniques. Four-terminal (Kelvin) connection handles LOW resistances by excluding lead and contact drops. Three-terminal (guard) connection handles HIGH resistances by bypassing surface-leakage current. The word "double" may also suggest two measurements or a high range, which pulls students toward (C) or (D).

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15. Q.25MCQ · 1 markOfficial answer: B
16. Q.26MCQ · 1 markOfficial answer: D
17. Q.27MCQ · 1 markOfficial answer: D
18. Q.28MCQ · 1 markOfficial answer: C
19. Q.29MSQ · 1 markOfficial answer: A, C

    Electrical and Electronic Measurements › Errors & standards

    When quantities are multiplied or divided, their limiting (worst-case) relative errors add in magnitude. For $P=VI$: $\pm (5\%+2.5\%)=\pm 7.5\%$ of $500\text{mW}$, which is $\pm 37.5\text{mW}$. An MSQ can list the same result twice, once as a percentage and once as an absolute error, and then both options are correct.

    WATCH OUT FOR

    Picking only one of (A) and (C) when they state the same error in two forms. Another mistake is subtracting the relative errors ($5\%-2.5\%=2.5\%$, giving (B)/(D)) or taking the root-sum-square ($\approx 5.59\%$). A limiting error is worst-case, so the magnitudes add.

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20. Q.30MSQ · 1 markOfficial answer: A
21. Q.31NAT · 1 markOfficial answer: 1

    Engineering Mathematics › Probability & statistics

    The median is the middle value of the data once it is sorted. For an odd count $n$ it is the value at position $\frac{n+1}{2}$ in the sorted list, and repeated values each keep their own place.

    WATCH OUT FOR

    Taking the middle entry of the list as printed, without sorting it first. The 5th printed entry is $-3$, which is wrong. Two other slips: collapsing repeated values into one (the distinct values $-3,-1,0,1,2$ have middle $0$, also wrong), and giving the mean ($3/9\approx 0.33$, which rounds to $0$) when the question asks for the median.

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22. Q.32NAT · 1 markOfficial answer: 5

    Networks (Electric Circuits) › Laws & node/mesh

    One volt is one joule per coulomb. The potential difference $v_A-v_B$ is the work needed per coulomb to move a charge from $B$ to $A$, so $W_{B\to A}=q(v_A-v_B)$.

    WATCH OUT FOR

    Getting the direction backwards. Some students use $v_B-v_A$ and answer $-5$. Others think the charge goes "downhill" from $B$ to $A$. $A$ is 5 V above $B$, so moving a positive charge from $B$ to $A$ needs $+5\text{J}$ of work.

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23. Q.33NAT · 1 markOfficial answer: 25

    Networks (Electric Circuits) › AC, power & resonance

    In a series RL circuit the current lags the source voltage by ${\tan }^{-1}(\omega L/R)$. A 45° lag means $X_L=R$, so $f=\frac{R}{2\pi L}$. Here the π in $R=100\pi \Omega$ cancels the π in $2\pi$.

    WATCH OUT FOR

    Stopping at $\omega =R/L=50\pi \approx 157$ and entering that as the answer. That number is the angular frequency in rad/s, not the frequency in Hz. Another slip is to drop the π from $R$ and get $f=100/(4\pi )\approx 8$ Hz.

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24. Q.34NAT · 1 markOfficial answer: 11

    Analog Electronics › Diode circuits

    The Zener is in series with the $8\Omega$ resistor and does not sit directly across the output, so it does not clamp $v_{out}$ at $3\text{V}$. Once it breaks down it behaves like a $3\text{V}$ battery. The output is then that $3\text{V}$ plus the drop across the $8\Omega$ resistor, which keeps rising with the source: $v_{out}=\frac{2v_s}{3}+1$.

    WATCH OUT FOR

    One mistake is to treat the Zener as a clamp across the output and answer $3\text{V}$, which ignores the $8\Omega$ drop in series with it. Another is to apply the voltage divider to the whole peak, $\frac{8}{12}\times 15=10\text{V}$, which leaves the Zener's $3\text{V}$ out of the loop. That value is actually the size of the negative-half-cycle extreme, not the positive maximum.

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25. Q.35NAT · 1 markOfficial answer: 63

    Digital Electronics › ADC & DAC

    An $n$-bit flash ADC compares the input against every decision threshold at once, with one comparator per threshold. $2^n$ levels have $2^n-1$ thresholds between them, so it needs $2^n-1$ comparators. This count doubles with every added bit, which is why flash converters are fast but expensive at high resolution.

    WATCH OUT FOR

    Answering $2^6=64$ (counting levels instead of the thresholds between them), or answering 6 (mixing up the flash ADC with a successive-approximation ADC, which uses a single comparator for $n$ clock cycles). The resistor ladder has $2^n$ resistors, but it needs only $2^n-1$ comparators.

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26. Q.36MCQ · 2 marksOfficial answer: C

    Engineering Mathematics › Probability & statistics

    Drawing in succession means drawing without replacement, so the events are dependent. For a specified order, multiply the probability of the first draw by the conditional probability of the second draw given the first: $P(A_1\cap O_2)=P(A_1)P(O_2\mid A_1)$. Counting ordered pairs ($6$ favourable out of $5\times 4=20$) gives the same answer.

    WATCH OUT FOR

    Each wrong option matches a specific slip. (A) $\frac{3}{5}$ is $P(\text{first is an apple})$ on its own. It is also the probability of one apple and one orange in either order, $2\times \frac{3}{10}$, so it is what you get if you ignore the order the question asks for. (D) $\frac{2}{5}$ is the probability of drawing an orange from the full box. Treating the draws as independent (with replacement) gives $\frac{3}{5}\cdot \frac{2}{5}=\frac{6}{25}$, which is not an option. That is a sign the second draw has to use the $4$ fruits left in the box.

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

    Electromagnetics › Electrostatics

    In free space, the point form of Gauss's law gives ${\rho }_v={ϵ}_0\nabla \cdot \mathbf{E}$. Only the diagonal terms count: $\partial E_x/\partial x$, $\partial E_y/\partial y$ and $\partial E_z/\partial z$. The off-diagonal coefficients (5, 6, 10) do not affect the charge density. Here they are symmetric, which is exactly the condition for $\nabla \times \mathbf{E}=0$.

    WATCH OUT FOR

    Adding every coefficient, or picking up the off-diagonal ones (for example $\partial E_y/\partial z=10$), gives distractors such as $10{ϵ}_0$. The divergence takes only the $x$-derivative of $E_x$, the $y$-derivative of $E_y$ and the $z$-derivative of $E_z$. Another slip is dividing by ${ϵ}_0$ instead of multiplying: ${\rho }_v={ϵ}_0\nabla \cdot \mathbf{E}$, not $\nabla \cdot \mathbf{E}/{ϵ}_0$.

    Full solution →
28. Q.38MCQ · 2 marksOfficial answer: C

    Electromagnetics › Magnetostatics

    For a straight wire in a uniform field, $\mathbf{F}/L=I\hat{n}\times \mathbf{B}$. The force is perpendicular to both the wire and the field, and its size can never be more than $I|\mathbf{B}|$. Those two facts rule out wrong options before you finish the cross product.

    WATCH OUT FOR

    Common mistakes are reversing the order to $\mathbf{B}\times \hat{n}$, which flips the sign to $-\frac{4}{\sqrt{6}}\hat{i}+\frac{4}{\sqrt{6}}\hat{k}$, and dropping or mis-handling the $1/\sqrt{6}$ factor in $\hat{n}$, which gives a vector in the right direction with the wrong size, like option (A). A quick check: the force must be perpendicular to $\mathbf{B}$, and its magnitude cannot be more than $I|\mathbf{B}|=2\sqrt{2}\text{N/m}$.

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29. Q.39MCQ · 2 marksOfficial answer: B

    Networks (Electric Circuits) › Two-port

    Set $R_L$ equal to the Thevenin impedance at port 2. That impedance is the network's output impedance with the source's own $5\Omega$ resistance ($R_s$) loading port 1: $Z_{out}=z_{22}-\frac{z_{12}{z}_{21}}{z_{11}+R_s}$.

    WATCH OUT FOR

    Leaving out the source resistance: $z_{22}-\frac{z_{12}{z}_{21}}{z_{11}}=10-2.5=7.5\Omega$. Taking $R_L=z_{22}=10\Omega$, which is the output impedance only when port 1 is open. Adding the correction term instead of subtracting it, which gives $11.67\Omega$. Another mistake is killing the 5 V source by open-circuiting it; a voltage source is killed by shorting it, so the $5\Omega$ resistor stays connected across port 1.

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30. Q.40MCQ · 2 marksOfficial answer: B

    Networks (Electric Circuits) › Laws & node/mesh

    Reduce the ladder step by step, starting at the passive left end and working toward the output. Parallel branches collapse by Millman's theorem (a single source transformation does the same job), and series resistors simply add. The output is open-circuited, so $V_0$ is the node voltage at C. The passive left section draws current from the sources, which pulls the output below 4 V.

    WATCH OUT FOR

    Seeing two equal 4 V sources and deciding that no current flows, which gives $V_0=4\text{V}$ (option C). The two left-hand 2 Ω resistors are a load of $1\Omega$ at A, and that load draws current through the 1 Ω resistors, so the output settles at 3 V. A second trap is adding the two sources to get 8 V (option D), but they are in parallel branches, not in series. Misreading the two left-hand 2 Ω resistors as being in series gives about 3.48 V, which matches none of the options. Their tops are joined by a plain wire, so they are in parallel.

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31. Q.41MCQ · 2 marksOfficial answer: B

    Signals and Systems › Signals basics

    Do the time operation on the argument first and the amplitude operations last. For $x(4-t)$, the value $x(\tau )$ lands at $t=4-\tau$, which reflects $x(t)$ about $t=2$ and reverses the order of the breakpoints. Then negate and add 1. One point check, such as $y(0)=1-x(4)=2$, separates the options quickly.

    WATCH OUT FOR

    Shifting in the wrong direction is the usual error. Writing $x(4-t)$ as $x(-t)$ advanced by 4, i.e. $x(-(t+4))$, puts the breakpoints near $t=-7\dots -1$ instead of $-1\dots 7$. Other slips: shifting without reversing, which gives (C), $x(t+1)$. Letting the minus sign act on the constant as well, which gives (A), $-1-x(4-t)$. Adding 1 without negating first.

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

    Signals and Systems › LTI & convolution

    Test the amplitude and the time argument separately. A non-linear operation on the amplitude, such as squaring or taking the modulus, makes a system non-linear. Anything done to the time argument other than a pure shift, such as the reversal $t\to -t$, makes it time-varying. $x(t-3)$ is only a shift, so it is LTI. $x(-t)$ and $|x(3-t)|$ both reverse time, so both are time-varying. $x^2(t)$ and $|\cdot |$ fail scaling, so both are non-linear.

    WATCH OUT FOR

    Getting the time-invariance test wrong for $x(-t)$ by substituting $t\to t-t_0$ everywhere in $y(t)$. That only produces $y(t-t_0)$ and compares it with itself. The response to the delayed input must be found by delaying $x$ first, which gives $x(-t-t_0)$. A second mistake is to read $x(3-t)$ as a shift by 3 and so call (O) time-invariant, when $x(3-t)=x(-(t-3))$ contains a reversal. A third is to treat the modulus as linear because it keeps the shape of a positive signal.

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

    Signals and Systems › Laplace

    Convolution in time becomes multiplication in $s$. The pulse $u(t-2)-u(t-3)$ transforms to $\frac{e^{-2s}-e^{-3s}}{s}$ by the time-shift property, and $e^{-4t}u(t)$ transforms to $\frac{1}{s+4}$. Taking out $e^{-3s}$ as a common factor puts the product in the printed form. Quick check: $x(t)\ge 0$ and $y(t)\ge 0$, so $z(t)\ge 0$ and its area $Z(0)$ must be positive. The pulse has area $1$ and $y$ has area $\frac{1}{4}$, so $Z(0)=\frac{1}{4}$. Option (A) tends to $+\frac{1}{4}$ as $s\to 0$ and (D) tends to $-\frac{1}{4}$.

    WATCH OUT FOR

    Getting the sign wrong when taking out the common factor. $e^{-2s}-e^{-3s}$ equals $e^{-3s}\left(e^s-1\right)$, not $e^{-3s}\left(1-e^s\right)$, and the wrong sign leads to (D). A second mistake is to shift the $\frac{1}{s+4}$ factor and get something like $(s+3)$ in the denominator, as in (B) and (C). Only the step functions pick up $e^{-s{t}_0}$. The exponential is not shifted, so its transform stays $\frac{1}{s+4}$.

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

    Control Systems › State space

    For a state-space model the poles are the roots of $|sI-A|=0$, the eigenvalues of $A$. Check them quickly with trace $A$ = sum of the eigenvalues and $\det A$ = their product. A pole drops out of the transfer function only if a zero of $C\text{adj}(sI-A)B$ cancels it. Here the only zero is at $s=1$, so both eigenvalues are poles.

    WATCH OUT FOR

    Getting the sign of the $-10$ entry wrong when forming $sI-A$. Writing $s(s+7)-10$ gives $s^2+7s-10$, whose roots are not integers and match no option. Another mistake is to take the roots of $s^2-7s+10$ (that is, $\det (A-sI)$ with the signs slipped) and pick $2,5$, option (B). Two eigenvalues whose sum is the negative trace $-7$ and whose product is $+10$ must both be negative.

    Full solution →
35. Q.45MCQ · 2 marksOfficial answer: C

    Analog Electronics › Op-amps

    This is the input stage of the three-op-amp instrumentation amplifier. The two virtual shorts force the full input difference across $R$. Since no current enters either op-amp input, the same current $V_{in}/R$ flows through the string $R_f$–$R$–$R_f$ between the two outputs. That makes the differential gain $1+2R_f/R$.

    WATCH OUT FOR

    Using the single non-inverting gain $1+R_f/R=11$ gives $1.1\text{V}$, option (D). That forgets the second $R_f$, which carries the same current. Another mistake is to see the $\pm 15\text{V}$ supply and assume the output saturates, which leads to (A) or (B). In fact each output moves only about $1\text{V}$ from its input, so both op-amps stay linear.

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

    Analog Electronics › Filters & waveforms

    Treat the three op-amps as one feedback loop, not as a cascade. Write each stage with the virtual-ground rule (inverter: $-1$; integrator: $-1/sRC$; lossy summing integrator: KCL with $R\parallel C$ feedback). Then close the loop through the top resistor. The result is a second-order low-pass biquad, and at $\omega ={\omega }_0$ its gain magnitude is $|\text{DC gain}|\times Q$.

    WATCH OUT FOR

    The usual mistake is to ignore the global feedback resistor from $V_{out}$ back to op-amp 1 and multiply the three stage gains as if they were a simple cascade. At $sRC=j$ that gives $\left|\frac{-1}{1+j}\right|\times 1\times 1=0.707$, which is option (B). The loop through the top $R$ changes the denominator to $(sRC{)}^2+sRC+1$, and that raises the gain at ${\omega }_0$ to $Q=1$. A second slip is getting the sign around the loop wrong: the integrator and the inverter together give $V_{out}=+V_1/(sRC)$, not $-V_1/(sRC)$.

    Full solution →
37. Q.47MCQ · 2 marksOfficial answer: C

    Digital Electronics › Sequential

    In a synchronous circuit, write $Q^{+}=D$ for every flip-flop. Then work out every $D$ from the present state, update all flip-flops together on each clock edge, and repeat once per edge. Here the circuit is a right-shift register with XOR feedback: $D_3=Q_1\oplus Q_0$, $D_2=Q_3$, $D_1=Q_2$, $D_0=Q_1$.

    WATCH OUT FOR

    The most common slip is stopping after one clock edge. That gives $0101$, which is option (B). A second mistake is updating the flip-flops one at a time, so that an already-changed $Q$ feeds the next $D$ within the same edge. A third is reading the XOR gate as an XNOR: the first edge then gives $1101$, which is option (A). A fourth is shifting in the wrong direction. Each wire goes from $Q_k$ to $D_{k-1}$, so the bits move from $Q_3$ towards $Q_0$.

    Full solution →
38. Q.48MCQ · 2 marksOfficial answer: D

    Electrical and Electronic Measurements › Time, frequency & scopes

    For a stationary closed Lissajous figure, $f_y/f_x$ equals the number of tangencies on a horizontal edge divided by the number on a vertical edge. The spot reaches its highest point once per Y cycle and its rightmost point once per X cycle.

    WATCH OUT FOR

    Inverting the ratio, i.e. taking $f_y/f_x=2/3$, gives $400\text{Hz}$, option (B). Horizontal tangencies are set by the vertical (Y) deflection, because the spot touches the top edge once per Y cycle, so they go in the numerator with $f_y$. Another slip is to add the top and bottom tangencies together, when the count should be taken along one edge only.

    Full solution →
39. Q.49MCQ · 2 marksOfficial answer: B

    Electrical and Electronic Measurements › Errors & standards

    In the 2019 SI each base unit is defined by fixing an exact value of a constant: $\Delta {\nu }_{\text{Cs}}$ (second), $c$ (metre), $h$ (kilogram) and $k$ (kelvin). To see what a unit depends on, write the constant's unit in base units: $\text{Hz}={\text{s}}^{-1}$, ${\text{m s}}^{-1}$, $\text{J s}={\text{kg m}}^2{\text{s}}^{-1}$, ${\text{J K}}^{-1}={\text{kg m}}^2{\text{s}}^{-2}{\text{K}}^{-1}$. The dependencies line up as second → metre → kilogram → kelvin.

    WATCH OUT FOR

    Going by the old definitions or the usual listing order (metre, kilogram, second, ...) and putting the metre before the second. Since 1983 the metre has been defined through the second, and the kilogram is no longer a platinum-iridium artefact: since 2019 it is defined through $h$, which brings in the metre and the second. A second trap is thinking the kelvin is independent (the old triple-point-of-water definition). Because $k$ is in ${\text{J K}}^{-1}$, the kelvin now depends on the kilogram, metre and second.

    Full solution →
40. Q.50MCQ · 2 marksOfficial answer: C
41. Q.51MCQ · 2 marksOfficial answer: A
42. Q.52MCQ · 2 marksOfficial answer: A
43. Q.53MSQ · 2 marksOfficial answer: A, D

    Engineering Mathematics › Calculus

    $x^2-2xy+y^2$ is the perfect square $(x-y{)}^2$, so a level set $(x-y{)}^2=k$ with $k>0$ is two parallel lines $x-y=\pm \sqrt{k}$, not a closed curve. The discriminant $B^2-4AC=0$ gives the same answer: it rules out circles and ellipses.

    WATCH OUT FOR

    Keeping only the positive root and marking (A) alone. The complete contour is both lines, because $(x-y{)}^2=1$ also holds when $x-y=-1$. Another trap is to see squared terms with equal coefficients and guess a circle. That ignores the $-2xy$ cross term, which turns the expression into a perfect square.

    Full solution →
44. Q.54MSQ · 2 marksOfficial answer: B

    Engineering Mathematics › Differential equations

    $\ddot{x}+x=0$ is simple harmonic motion, and $x^2+{\dot{x}}^2$ is its conserved energy. Multiply the equation by $\dot{x}$ and you get $\frac{d}{dt}\left[\frac{1}{2}(x^2+{\dot{x}}^2)\right]=0$. In the phase plane the trajectory is a circle of radius $\sqrt{c}$, and $x(0)\ne 0$ means that radius cannot be zero.

    WATCH OUT FOR

    Picking (C) because the characteristic roots are $\pm j$ and the solutions contain $e^{\pm jt}$. That mixes up the form of $x(t)$ with the value of $|x{|}^2+|\dot{x}{|}^2$, which is a real constant. Another slip is choosing (A) because the equation is homogeneous and has no forcing. The sum is zero only for the trivial solution, and $x(0)\ne 0$ rules that out. Some students also expand the squares, forget that the $\pm 2k_1{k}_2\sin t\cos t$ cross terms cancel, and wrongly decide the sum depends on $t$.

    Full solution →
45. Q.55MSQ · 2 marksOfficial answer: Marks to all

    Digital Electronics › Logic families

    For a static CMOS gate, read the NMOS pull-down network (series = AND, parallel = OR) and complement the result: here $f=\overline{A+BC}$. Then use De Morgan to spot options that are the same function written another way. Finally, check that the pull-up network is the dual. When it is not, as here, test each option only on inputs where the circuit actually gives a defined 0 or 1.

    WATCH OUT FOR

    Swapping series and parallel in the pull-down network gives $\overline{A(B+C)}$, so options (A) and (D) get ticked. Reading the function from the drawn pull-up (C in series with A parallel to B) gives $\overline{C+AB}$, which matches no option. A third slip in this MSQ is ticking only (B) and missing that (C) is the same function after De Morgan's theorem.

    Full solution →
46. Q.56MSQ · 2 marksOfficial answer: B, C

    Digital Electronics › Processors & memory

    An $n$-bit field in a fixed instruction format can encode at most $2^n$ distinct values. So a 7-bit opcode gives 128 opcodes, a 3-bit register field gives 8 registers, and $7+3\times 3=16$ bits fills the instruction exactly.

    WATCH OUT FOR

    One mistake is to count the $16-7=9$ non-opcode bits and pick $2^9=512$ as the number of opcodes (D). Those 9 bits hold the three operand fields, not opcode bits. Another is to assume that a 16-bit processor must have 16 registers (A). The register count depends only on the width of the register field (3 bits), not on the processor's word length.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 2.4 to 2.6

    Engineering Mathematics › Calculus

    For a piecewise function, the integral over the whole real line is the total signed area under its graph. Here that is a rectangle on $[0,2]$ (area $2$) plus a right triangle on $[2,3]$ (area $0.5$). Where $f=0$, nothing is added.

    WATCH OUT FOR

    Integrating $-\tau +3$ over $[0,3]$ as if it applied everywhere gives $4.5$. Using $\tau$ in place of $3-\tau$ on $[2,3]$ gives $2.5$ for that piece and a total of $4.5$. Both come from misreading which formula belongs to which interval. Some students also drop the rectangle and report only the triangle, $0.5$.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 2490 to 2510

    Electrical Machines › Transformers

    In an open-circuit test performed from a given winding, the no-load current I0 splits into an in-phase (core-loss) component Iw = P0/V0 and a quadrature (magnetizing) component Im = sqrt(I0² − Iw²); the magnetizing reactance on the test side is V0/Im, and to refer any impedance/reactance from the LV side to the HV side it is multiplied by the square of the turns ratio a = (HV rated voltage / LV rated voltage).

    WATCH OUT FOR

    Forgetting to refer the LV-side magnetizing reactance to the HV side (i.e., stopping at 100 Ω), or applying the turns ratio the wrong way (dividing by a² instead of multiplying), since the OC test was performed from the LV side but the question asks for Xm referred to the HV side.

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

    Control Systems › Time response

    In a unity-feedback loop, the ramp error depends only on the velocity constant: $e_{ss}=1/K_v$ with $K_v={\lim }_{s\to 0}sG(s)$. That formula is only valid once you have confirmed the closed loop is stable, which here takes a quick Routh check on $s^3+2s^2+15s+10$.

    WATCH OUT FOR

    One slip is to cancel the $s$ wrongly and use the zero at $s=-1$ or the numerator gain alone, giving $K_v=10$ and $e_{ss}=0.1$. The correct value is $K_v=10\cdot 1/5=2$, because the constant term $5$ of $s^2+2s+5$ divides the gain. Another mistake is to skip the stability check. If the loop were unstable, $1/K_v$ would give a meaningless number.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 0.39 to 0.41

    Control Systems › Bode & margins

    Gain margin is read at the phase-crossover frequency. With four identical poles at $s=-1$, each pole adds $-45^{\circ }$ at $\omega =1\text{rad/s}$, so the total phase is $-180^{\circ }$ there. A 20 dB gain margin means $|G(j{\omega }_{pc})|=0.1$, which gives $K/4=0.1$.

    WATCH OUT FOR

    Setting $|G(j{\omega }_{pc})|=10$ instead of $0.1$ (sign slip in $GM=-20{\log }_{10}|G|$), which gives $K=40$. Another slip is writing $(\sqrt{2}{)}^4$ as $2$ instead of $4$, which gives $K=0.2$. Some students also take the gain-crossover frequency ($|G|=1$) in place of the phase-crossover frequency.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 512

    Analog Electronics › Filters & waveforms

    When a PLL is locked, both inputs of the phase detector run at the same frequency. With a divide-by-M counter on the reference path and a divide-by-N counter in the feedback path, $f_{\text{in}}/M=f_{\text{out}}/N$, so $f_{\text{out}}=(N/M)f_{\text{in}}$. This is how an integer-N frequency synthesiser works.

    WATCH OUT FOR

    Swapping the two dividers, which gives $(20/1024)\times 10\text{kHz}\approx 0.2\text{kHz}$. The divider in the feedback path multiplies the output frequency and the reference divider divides it. Other slips: ignoring the $÷20$ counter and answering $1024\times 10\text{kHz}=10240\text{kHz}$, or giving the answer in Hz (512000) when the blank asks for kHz.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 1

    Electrical and Electronic Measurements › Time, frequency & scopes

    When a counter measures over one input period, it counts $N=T/T_c$ clock ticks and carries the usual $\pm 1$ count error. That makes the relative error $1/N$, and because $f=1/T$ the frequency has the same relative error: $\Delta f\approx f/N=f^2{T}_c$.

    WATCH OUT FOR

    Using the frequency-mode formula (error $=\pm 1/T_{\text{gate}}$ with a 0.1 μs gate), which gives an absurd $\pm 10\text{MHz}$. Another slip is writing the error as $\pm T_c=0.1\mu \text{s}$ and never converting it from time to frequency. Here the counter counts the clock during one input period, so the $\pm 1$ count is a $\pm 1\%$ error on $T$, and so on $f$.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.25 to 0.27
54. Q.64NAT · 2 marksOfficial answer: 537.00 to 539.00
55. Q.65NAT · 2 marksOfficial answer: 128