GATE EE 2026 question paper with solutions

GATE EE 2026 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   A simple sentence divides into a subject (who does the action) and a predicate (what is said about the subject). A transitive verb inside the predicate takes a direct object, which answers the question "subject + verb + what?". The shopkeeper sells what? Lemons, so ‘lemons’ is the object.

   WATCH OUT FOR

   Picking (C) 'predicate' because ‘lemons’ sits in the predicate half of the sentence. The predicate is the whole verb phrase ‘sells lemons’. A single noun inside it that receives the action is the object, not the predicate.

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2. Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Spatial

   Treat the '?' box's border as a set of checkpoints. First mark every point where a visible shape crosses into the box. Then carry each visible straight side on into the box until it meets its partner: that meeting point is the hidden vertex. The correct option must have black on the border exactly at those crossings and must put its vertices at those points. Test one shape at a time, and drop an option as soon as one checkpoint fails.

   WATCH OUT FOR

   Judging the options by overall look instead of by where they touch the border. R's top-right piece looks like a triangle tip, but it never reaches the box's right edge, so the triangle stays broken. S's large D shape looks like 'more oval', but the oval only crosses the left edge down to 37% and crosses the top edge, which S leaves white. Q copies the correct lower triangle but puts it on the wrong side.

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3. Q.3MCQ · 1 markOfficial answer: A

   General Aptitude › Numerical

   Set the two expressions for $y$ equal to each other. The number of real roots of the resulting quadratic, read from its discriminant, is the number of intersection points. Here $2x^2+2x+1=0$ has $D=-4<0$, so there are none. Another way to see it: $y=x^2$ opens upward with its vertex at $(0,0)$, and $y=-(x+1{)}^2$ opens downward with its vertex at $(-1,0)$. One curve lies on or above the $x$-axis and the other on or below it, and they touch the axis at different points.

   WATCH OUT FOR

   Making a sign error while moving terms across. For example, writing $2x^2+2x-1=0$ gives $D=12>0$ and the wrong answer (C) 2. The other trap is noticing that both parabolas touch the $x$-axis, assuming they touch each other there, and choosing (B) 1. They touch the axis at $x=0$ and $x=-1$, which are different points.

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4. Q.4MCQ · 1 markOfficial answer: C

   General Aptitude › Reading

   Read both the counterfactual and the superlative strictly. “Would have been made captain” means Anish is not captain now. “Would have become the youngest captain in the team’s history” means he is younger than every captain so far, and that includes the current one. The comparison covers captains only, not every player.

   WATCH OUT FOR

   Choosing (D): stretching “youngest captain” into “youngest player”. The passage compares Anish only with the team’s captains, so it tells us nothing about the ages of players who never captained. Some readers also pick (B) because they miss that someone can only be “made” captain if he is not captain already.

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5. Q.5MCQ · 1 markOfficial answer: B

   General Aptitude › Spatial

   A rotation is rigid: every motif turns through the same angle and keeps its place and its contacts with its neighbours. Each option differs from the true rotation in just one or two motifs. Check the four vertex hooks first, since they come from four identical corner hooks and must all look alike. Then check the centre, and the one option with no odd motif is the answer.

   WATCH OUT FOR

   Stopping once the outline and the four vertices look right. P passes that check and differs from Q only in its central motif, which is turned 180°, so a quick comparison picks P. The centre has to be checked too, for example by seeing which hooks it touches.

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6. Q.6MCQ · 2 marksOfficial answer: A

   General Aptitude › Grammar & vocabulary

   In a matching question, fix the one or two pairs you are sure of first. Each sure pair rules out every option that disagrees with it. Then check the remaining pairs of the option that survives, because the last option standing is not proof on its own. Here Lonely = Solitary, Literal = Verbatim, Lousy = Terrible and Lethal = Deadly.

   WATCH OUT FOR

   Matching by sound rather than meaning. All four Column I words begin with "L", which pushes you to guess. Two traps: reading "lousy" as something harmful and pairing it with "deadly", which is option B's (iii)-(r); and pairing "literal" with "terrible". Lousy means poor or bad (terrible). Lethal is the one that means deadly.

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7. Q.7MCQ · 2 marksOfficial answer: B

   General Aptitude › Geometry

   Join $RQ$. The angle in a semicircle makes $\angle TRQ=90^{\circ }$. The inscribed angle $\angle RQS$ is half the central angle $\angle ROS$, so it is $40^{\circ }$. The third angle of triangle $RQT$ is then $50^{\circ }$. Another way: the angle between two secants from an outside point is half the difference of the arcs they cut off, $\frac{180^{\circ }-80^{\circ }}{2}$.

   WATCH OUT FOR

   Picking $40^{\circ }$ (A) confuses the inscribed angle $\angle RQS$ with the angle at $T$. Picking $80^{\circ }$ (D) copies the central angle across. Some students use the rule for chords that cross inside a circle, half the SUM of the arcs, which gives $\frac{180^{\circ }+80^{\circ }}{2}=130^{\circ }$. When the lines meet outside the circle, the rule is half the DIFFERENCE. Measuring the figure does not work either, because it is not drawn to scale.

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8. Q.8MCQ · 2 marksOfficial answer: A

   General Aptitude › Reasoning

   The number inside each polygon is the factorial of its number of sides: $3!=6$, $4!=24$, $5!=120$, so the 6-sided hexagon holds $6!=720$.

   WATCH OUT FOR

   Looking only at the numbers and ignoring the shapes. From 6, 24, 120 alone you might guess "double the last number" ($120\times 2=240$, option D) or reuse an earlier number (24, option C). The multiplier is not fixed: it goes up by one each time because it is the next polygon's side count, $\times 4$ then $\times 5$ then $\times 6$.

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9. Q.9MCQ · 2 marksOfficial answer: B

   General Aptitude › Reasoning

   Update every bulb at the same time from the previous Step's states, and first check that your reading of the rules reproduces the given Step 1 and Step 2 rows. Then keep stepping until the pattern stops changing. Here it settles by Step 4 at $1010101$, where every OFF bulb has two ON neighbours (not exactly one) and every ON bulb has no ON neighbours, so nothing can change after that.

   WATCH OUT FOR

   (1) Swapping the legend. Hollow is ON and filled is OFF. With the colours swapped, the initial row is $1110111$ and the rules give $1010101$ as the next row, which does not match the figure's Step 1. (2) Reading "exactly one ON neighbour" as "at least one". Bulb 4 then turns ON in Step 3, and the sequence drifts to a different count. (3) Updating bulbs one at a time from left to right instead of all together from the previous Step's states. (4) Stopping at Step 3, which has 6 bulbs ON.

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

    General Aptitude › Numerical

    Factor $P^2-Q^2=(P-Q)(P+Q)$. Since 13 is prime, the two positive integer factors can only be 1 and 13. That fixes $P-Q=1$ and $P+Q=13$, so $P=7$ and $Q=6$ are the only solution.

    WATCH OUT FOR

    Seeing 13 in the question and picking option (A) 13 or (B) 26 without solving for $P$ and $Q$. Another slip is to set $P-Q=13$ and $P+Q=1$, which makes $Q$ negative. $P+Q$ must be the larger factor.

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Electrical Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: C

   Signals and Systems › z-transform

   $0.9^{|n|}$ is two-sided. Its causal half $0.9^{n}u[n]$ converges outside its pole ($|z|>0.9$) and its anti-causal half $0.9^{-n}u[-n-1]$ converges inside its pole ($|z|<1/0.9$). The Z-transform of the whole sequence exists only on the intersection of the two regions, which is a ring between the two poles. The ring is non-empty because the base satisfies $|0.9|<1$.

   WATCH OUT FOR

   Reading $0.9^{|n|}$ as the causal $0.9^{n}u[n]$ and choosing (A). Another mistake is taking the union of the two half-ROCs instead of their intersection, or inverting the left-sided condition, which leads to (D). An ROC is always one connected ring, never two separate pieces.

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2. Q.12MCQ · 1 markOfficial answer: D

   Signals and Systems › Signals basics

   A discrete-time period must be a whole number of samples. So $x[n]$ repeats only when some integer number of samples $N$ spans a whole number of continuous-time periods: $N{T}_s=mT$, which means $T/T_s=N/m$ is rational. If $T/T_s=p/q$ in lowest terms, then $N=p$ samples is a period.

   WATCH OUT FOR

   Picking A because the samples per period $T/T_s$ looks like the discrete period. That ratio need not be an integer. For $T/T_s=3/2$ the period is $3$ samples, not $1.5$, and for an irrational ratio (for example $T_s=T/\sqrt{2}$) a sampled sinusoid never repeats. Another trap is to think sampling a periodic signal always gives a periodic sequence.

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3. Q.13MCQ · 1 markOfficial answer: A

   Control Systems › Time response

   Remove the $\frac{1}{s}$ step factor and what is left, $\frac{1}{1+sT}$ with $T=0.01\text{s}$, is a first-order system. Every time-domain figure is then a fixed multiple of $T$: the 10–90% rise time is $T\ln 9\approx 2.2T$, and the 2% settling time is $T\ln 50\approx 4T$.

   WATCH OUT FOR

   Reading $T_c$ as 100 s from the pole at $s=-100$, which makes every value 10,000 times too large. The time constant is the reciprocal of the pole magnitude. A second slip is taking the rise time as $T\ln 10=2.3T$ (from 0 to 90%) and not from 10% to 90%. A third is using $\ln 4$ or the second-order formula $\frac{4}{\zeta {\omega }_n}$ for settling time, when this is a first-order response.

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4. Q.14MCQ · 1 markOfficial answer: A

   Signals and Systems › LTI & convolution

   Linearity depends only on how the dependent variable $y$ and its derivatives appear: first power, no products of them with each other, no nonlinear functions of them. Coefficients that depend on $t$ do not affect linearity. What they do affect is time-invariance: a system described by a differential equation is time-invariant only when every coefficient is constant.

   WATCH OUT FOR

   Reading the term $8ty$ (or $t^2{y}^{\prime \prime }$) as a product of two variables and calling the equation nonlinear. $t$ is the independent variable, so $8t$ is just a coefficient. The equation would be nonlinear only if $y$ were multiplied by $y$ or by one of its derivatives. The opposite slip is to see "linear" and assume "LTI": coefficients that vary with $t$ make the system time-varying, so (C) is false even though (A) is true.

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5. Q.15MCQ · 1 markOfficial answer: C

   Electromagnetics › Electrostatics

   Electric field is a vector. At the centre of a uniformly charged ring, each charge element is cancelled by the element diametrically opposite it, so $\mathbf{E}=0$ and $\mathbf{F}=q\mathbf{E}=0$. Potential is a scalar, so nothing cancels: $V=\frac{Q}{4\pi ϵR}$ at the centre is not zero.

   WATCH OUT FOR

   Mixing up potential and field. Students pick (A), which is the potential energy $qV$ at the centre and not a force, or (B), which adds the magnitudes of the elemental forces as if all of them pointed the same way. (D) looks convincing because it does have the units of force, but symmetry forces the vector sum to be exactly zero.

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6. Q.16MCQ · 1 markOfficial answer: B

   Electromagnetics › Magnetostatics

   The Lorentz force is $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. The magnetic term is always perpendicular to both $\mathbf{v}$ and $\mathbf{B}$. Here $\mathbf{v}\times \mathbf{B}$ comes out along $+{\hat{a}}_y$, the same direction as $\mathbf{E}$, so the two forces add and the total force is parallel to $\mathbf{E}$ and perpendicular to $\mathbf{B}$.

   WATCH OUT FOR

   Getting the cross-product sign wrong. ${\hat{a}}_x\times {\hat{a}}_z=-{\hat{a}}_y$, not $+{\hat{a}}_y$. With the minus sign on $\mathbf{B}$ as well, the result is $+30{\hat{a}}_y$. If you drop either sign, the magnetic force appears to oppose $\mathbf{E}$, but that still leaves the force along the $y$-axis, so the answer is (B) either way. The other trap is picking (C) because the magnetic force alone is perpendicular to $\mathbf{B}$ and $\mathbf{v}$, and then wrongly assuming it must also be perpendicular to $\mathbf{E}$.

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7. Q.17MCQ · 1 markOfficial answer: C

   Electrical Machines › Transformers

   Converting a two-winding transformer into an autotransformer (by electrically joining the two windings in series) multiplies its VA rating by the factor $\frac{V_H}{V_H-V_L}$, because the same physical windings now also carry a conductively-transferred (not just transformed) component of power.

   WATCH OUT FOR

   Assuming the autotransformer rating equals the original two-winding rating (15 kVA) because "it's the same transformer," ignoring that autotransformer action lets the common winding carry more net through-current than pure transformer action alone.

   Full solution →
8. Q.18MCQ · 1 markOfficial answer: B
9. Q.19MCQ · 1 markOfficial answer: D
10. Q.20MCQ · 1 markOfficial answer: B

    Electrical Machines › Synchronous machines

    A synchronous motor and a synchronous generator on the same mechanical shaft run at the same speed $N$; each machine's own pole count then converts that common $N$ to its own electrical frequency via $N=\frac{120f}{P}$.

    WATCH OUT FOR

    Assuming the frequency ratio equals the pole ratio directly in the same sense (i.e. picking $P_{motor}:P_{gen}=f_{motor}:f_{gen}$) instead of inversely — since speed is fixed, a lower output frequency needs FEWER poles on the generator, not more, so $P_{gen}/P_{motor}=f_{gen}/f_{motor}$.

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

    Electrical Machines › Induction machines

    In a double-squirrel-cage rotor, the outer cage (small bars near the air gap) has high resistance and low leakage reactance, while the inner cage (large bars deep in the slot) has low resistance and high leakage reactance. Since rotor reactance scales with slip ($X=s{X}_{20}$), at starting ($s=1$) the reactances are at their supply-frequency maximum, so the high-reactance inner cage is effectively "choked off" and current is forced into the low-reactance outer cage — giving high effective rotor resistance and hence high starting torque, exactly like an external rotor-resistance starter. Near synchronous speed, slip is small, both reactances collapse, and the division of current is governed by resistance alone, so current concentrates in the low-resistance inner cage — giving low running losses and good efficiency.

    WATCH OUT FOR

    Confusing which cage dominates at which condition — assuming the low-resistance (inner) cage always carries most of the current (true only near rated speed, not at starting), or believing the purpose of the double cage is to reduce rotor resistance at starting rather than to raise it for better starting torque.

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

    Electrical Machines › Synchronous machines

    On an infinite bus (fixed V, f), real power P is set by the prime mover and stays constant when only field excitation changes. Since $P=\frac{EV}{X_s}\sin \delta$ with V and $X_s$ fixed, $E\sin \delta$ stays constant — so increasing E forces $\sin \delta$ (hence $\delta$) to fall. Reactive power $Q=\frac{EV\cos \delta -V^2}{X_s}$ starts at zero (unity PF) and rises positive as E increases, making the machine over-excited: it now supplies lagging VARs, so $|I_s|=\sqrt{P^2+Q^2}/V$ grows and the terminal power factor becomes lagging.

    WATCH OUT FOR

    Confusing the machine-side excitation change with the load angle direction: many students assume increasing E always increases δ (as it does at constant P and Q on a V-curve sketch), forgetting that on an infinite bus P is the fixed, externally-set quantity while δ is the variable that adjusts — so δ must fall, not rise, to keep $E\sin \delta$ constant as E grows. Another common slip is stopping at "generator becomes over-excited, so PF is leading" — for a generator, over-excitation (supplying VARs to the bus) gives a lagging power factor, the opposite convention from a motor.

    Full solution →
13. Q.23MCQ · 1 markOfficial answer: B

    Networks (Electric Circuits) › Laws & node/mesh

    Linearity belongs to each element, not to the circuit it sits in. An element is linear when its $v$–$i$ relation obeys superposition. A constant-valued R, L or C does, because scaling and differentiation are both linear operations. An ideal diode's characteristic has a corner at the origin, so homogeneity fails and it is nonlinear.

    WATCH OUT FOR

    There are two common traps. (1) Calling L and C nonlinear because their relations contain derivatives, or because they store energy. Differentiation is a linear operation, so L and C with constant values are linear. (2) Calling the ideal diode linear because each of its two pieces is a straight line. Piecewise linear is not linear: scaling a forward-conducting point by $-1$ gives a reverse current the diode cannot carry. The diode being in the loop, or the source being DC, does not change whether R, L or C are linear.

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

    Networks (Electric Circuits) › Theorems

    If a chain made only of ideal voltage sources connects the two terminals, the open-circuit voltage is just the KVL sum of those sources along the chain (here $20+100+20=140\text{V}$), whatever the resistors are. Killing the sources turns that chain into a short across the terminals, so $R_{TH}=0$ and the Thevenin equivalent is an ideal voltage source.

    WATCH OUT FOR

    There are two traps. (1) Reading the visible $10\text{k}\Omega$ between Y and Z as $R_{TH}$ (options A and D). Once the sources are shorted, that resistor is in parallel with a zero-resistance path through X and W, so it adds no resistance. (2) Misreading the polarity of the right-hand 20 V source, which has its $+$ toward W. If it is taken as opposing the other two sources, the sum becomes $20+100-20=100\text{V}$ (options A and C) instead of the correct $140\text{V}$.

    Full solution →
15. Q.25MCQ · 1 markOfficial answer: A

    Networks (Electric Circuits) › Laws & node/mesh

    Ideal voltage sources in parallel must have the same voltage (KVL), and ideal current sources in series must carry the same current (KCL). A voltage source paired with a current source, in either arrangement, can never conflict, so it puts no condition on its constant. Only circuits 1 and 4 constrain anything: $k_1=1$ and $3k_4=-1$.

    WATCH OUT FOR

    Reading the two upward arrows in circuit 4 as "the same direction" and setting $I_X=I_D$, which gives $k_4=+\frac{1}{3}$. Around the single loop, the two upward arrows actually drive current in opposite senses, so KCL at the top node needs $I_X=-I_D$. A second trap is thinking the voltage-source/current-source pairs (circuits 2 and 3) fix $k_2$ or $k_3$. They don't: the voltage source takes any current, and the current source takes any voltage.

    Full solution →
16. Q.26MCQ · 1 markOfficial answer: B

    Networks (Electric Circuits) › AC, power & resonance

    When the voltage is purely sinusoidal, only the fundamental current carries real power, but every harmonic adds to the rms current. So PF = P/S = cos φ₁ × (I₁/I_rms). Here that is the displacement factor (1) times the distortion factor (12/15), which gives 0.80.

    WATCH OUT FOR

    One mistake is taking the power factor as cos φ₁ = 1 because the fundamental is in phase, which ignores the harmonic and gives option (A). Another is adding the current amplitudes directly, 12 + 9 = 21, instead of in quadrature, which gives 12/21 ≈ 0.57, option (D). The same 0.57 also comes from wrongly using 1/(1 + THD) with THD = 0.75.

    Full solution →
17. Q.27MCQ · 1 markOfficial answer: C
18. Q.28MCQ · 1 markOfficial answer: B
19. Q.29MCQ · 1 markOfficial answer: A

    Control Systems › Bode & margins

    Read an asymptotic Bode magnitude plot one slope at a time. The starting slope counts the poles or zeros at the origin ($-20\text{dB/dec}$ means one pole). Each change of slope at a corner frequency is a simple pole ($-20$) or a simple zero ($+20$). The gain comes from where the low-frequency asymptote crosses $0\text{dB}$: for $K/s$ that crossing is at $\omega =K$.

    WATCH OUT FOR

    Two traps. First, choosing (B) because ${\omega }_0$ looks like a single corner, without checking which way the low-frequency line slopes. (B) rises at $+20\text{dB/dec}$ and the figure falls. Second, thinking the $1-s/{\omega }_0$ factor in (C) or (D) changes the magnitude. It does not: $|1-j\omega /{\omega }_0|=|1+j\omega /{\omega }_0|$, so both of those are all-pass with a flat $0\text{dB}$ magnitude. The sign affects only the phase.

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20. Q.30MSQ · 1 markOfficial answer: A, B, D

    Engineering Mathematics › Linear algebra

    $\lambda$ is an eigenvalue of $M$ exactly when $\det (M-\lambda I)=0$. If $A$ and $B$ share the eigenvector $x$ with eigenvalues $\lambda$ and $\mu$, then $(A+B)x=(\lambda +\mu )x$, so $A+B$ has eigenvalue $\lambda +\mu$ (here $2+2=4$). Eigenvalues add only because the eigenvector is common.

    WATCH OUT FOR

    Ticking (C) because $2$ is an eigenvalue of both $A$ and $B$, as if it must then be an eigenvalue of $A+B$ too. The shared eigenvector gives $A+B$ the eigenvalue $4$, not $2$. $\det (A+B-2I)$ can be zero for some particular matrices (for example $A=B=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]$ gives $A+B=\left[\begin{array}{cc}4& 0\\ 0& 2\end{array}\right]$), but it fails for $A=B=2I$, so (C) is not guaranteed. A second trap is assuming eigenvalues of $A+B$ are always sums of eigenvalues of $A$ and $B$. That holds only for a shared eigenvector, which is given here.

    Full solution →
21. Q.31MSQ · 1 markOfficial answer: A

    Electrical Machines › Transformers

    Same physical transformer means same turns ratio, same core, same magnetizing inductance and same core-loss constants. Only the applied V and f change, so every core-dependent quantity (peak flux density, magnetizing current, hysteresis loss, eddy-current loss) scales with the ratio V/f, while copper loss for a given load current is unaffected.

    WATCH OUT FOR

    Assuming the transformer is "under-utilized" or "over-fluxed" in the USA just because the voltage is lower, without actually computing V/f; or forgetting that eddy current loss scales as $f^2{B}_m^2$ (not $f{B}_m^2$) and hysteresis loss as $f{B}_m^{1.6}$, which is needed to see that both core losses actually fall in the USA rather than rise.

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22. Q.32NAT · 1 markOfficial answer: 0

    Engineering Mathematics › Vector calculus

    For a closed surface, use the divergence theorem to turn the surface integral into ${\iiint }_V\nabla \cdot \vec{F}dV$. Each component here depends only on the other coordinates ($F_x$ has no $x$, $F_y$ has no $y$, $F_z$ is constant), so the divergence is zero and the flux through any closed surface is zero.

    WATCH OUT FOR

    Working out the surface integral in spherical coordinates and getting stuck on terms like $\sin (\sin \theta \sin \varphi )$. Another mistake is to treat $\sin (y)$ or $\cos (x)$ as if it depended on its own coordinate, for example writing ${\partial }_x\sin (y)=\cos (y)$, which gives a nonzero divergence. Some students also multiply the constant $5\hat{z}$ by the sphere's area ($5\times 4\pi$) and forget that the flux in through the lower hemisphere cancels the flux out through the upper one.

    Full solution →
23. Q.33NAT · 1 markOfficial answer: 47.5 to 48.0

    Analog Electronics › Diode circuits

    In a series-pass (emitter-follower) linear regulator, the same load current flows from the input to the output through the pass transistor. With bias currents ignored, the efficiency is therefore just $V_{out}/V_{in}$. Here $V_{out}=V_Z-V_{BE}=4.8\text{V}$, which gives $4.8/10=48\%$.

    WATCH OUT FOR

    One mistake is to take $V_{out}=5.4\text{V}$, the Zener voltage, and forget the $V_{BE}$ drop, which gives 54 %. Another is to use $V_Z+V_{BE}=6.0\text{V}$, which gives 60 %. A third is to include the Zener or resistor current even though the question says to ignore it: that gives about 47.6 %. That value happens to fall inside the key range, but it is not the "maximum possible" efficiency the question asks for.

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24. Q.34NAT · 1 markOfficial answer: 32.3 to 32.5
25. Q.35NAT · 1 markOfficial answer: 0

    Engineering Mathematics › Linear algebra

    A quadratic form ${\mathbf{x}}^{\mathrm{T}}A\mathbf{x}$ is a scalar, so it equals its own transpose ${\mathbf{x}}^{\mathrm{T}}{A}^{\mathrm{T}}\mathbf{x}$. For a skew-symmetric matrix that transpose is the negative of the form, so the form must be zero. Only the symmetric part $\frac{1}{2}(A+A^{\mathrm{T}})$ of any matrix contributes to a quadratic form, and a skew-symmetric matrix has no symmetric part.

    WATCH OUT FOR

    Treating ${\mathbf{x}}^{\mathrm{T}}\mathbf{x}=1$ as important and trying to get an answer from it, for example by guessing 1 (as for the identity matrix) or bringing in eigenvalues. The eigenvalues of a real skew-symmetric matrix are zero or purely imaginary, and none of that is needed here. The unit-length condition is a distractor: the form is zero for every real vector. A second slip is forgetting that $({\mathbf{x}}^{\mathrm{T}}A\mathbf{x}{)}^{\mathrm{T}}$ reverses the order of the product.

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

    Signals and Systems › Fourier series

    Summing $g(x+2\pi m)$ over all $m$ is simply the $2\pi$-periodic extension of $g$, which here is an odd square wave. Odd symmetry makes every cosine coefficient zero. The sine series of this square wave is standard: $b_n=\frac{4k}{n\pi }$ for odd $n$ and $0$ for even $n$. The \"sum of the third-harmonic coefficients\" is therefore just $b_3$.

    WATCH OUT FOR

    Using the peak-to-peak value $2k$ as the amplitude, which gives $b_3=\frac{8k}{3\pi }$ and $k=\frac{1}{4}$ (option D). Another slip is dropping the factor 2 from the half-range integral, which gives $b_3=\frac{2k}{3\pi }$ and $k=1$ (option A). A third is assuming $a_3\ne 0$ without checking the symmetry. The waveform swings between $-k$ and $+k$, so its amplitude is $k$, and it is odd, so all $a_n=0$.

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

    Analog Electronics › Op-amps

    When the open-loop gain is finite, drop the virtual-ground assumption and use $V_{out}=A_0(v_{+}-v_{-})$ together with KCL at the inverting node. For an inverting amplifier this gives $A_{CL}=\frac{-R_f/R_1}{1+(1+R_f/R_1)/A_0}$. The noise gain $1+R_f/R_1=21$ (that is, $1/\beta$) is what gets divided by $A_0$.

    WATCH OUT FOR

    Assuming an ideal op-amp (virtual ground) and writing $-R_f/R_1=-20.00$, which is option (B). This is the answer you also get if you read the printed 105 as $10^5$, since $-20/(1+21\times 10^{-5})\approx -20.00$. The paper prints 105, and with a loop gain of only $A_0\beta =5$ the finite gain matters. Another slip is using the non-inverting formula $-(1+R_f/R_1)=-21.00$ (option C), which mixes up the noise gain with the signal gain.

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28. Q.38MCQ · 2 marksOfficial answer: D

    Electrical Machines › Transformers

    For a delta-delta bank where both HV and LV windings have their dotted terminal wired straight to the line, write each winding's dot-to-finish equation from how the diagram actually interconnects it (not by assumption), use the transformer dot rule (same-core EMFs in phase, scaled by real turns ratio k) to relate LV winding phasors to the corresponding HV line phasors, then solve the delta loop KVL (with one node as reference) for the LV line-to-line phasor in terms of the HV line phasor to read off the clock number.

    WATCH OUT FOR

    Assuming the LV delta closes in the same left-to-right order (a\to b\to c) as it visually appears, without noticing the LV block is drawn mirrored — this silently reverses the internal loop to a\to c\to b\to a. Missing that mirroring flips the sign in the loop solution and typically leads to answering Dd0, Dd2 or Dd6 instead of the correct Dd10.

    Full solution →
29. Q.39MCQ · 2 marksOfficial answer: D
30. Q.40MCQ · 2 marksOfficial answer: A
31. Q.41MCQ · 2 marksOfficial answer: D
32. Q.42MCQ · 2 marksOfficial answer: C

    Networks (Electric Circuits) › AC, power & resonance

    For sinusoidal $v$ and $i$ at the same frequency, $p(t)$ is a constant $\frac{V_m{I}_m}{2}\cos \theta$ plus a ripple at $2\omega$. The ripple's period is $\frac{\pi }{\omega }$, exactly one half cycle of $\omega$. So averaging over a half cycle gives the same result as averaging over a full cycle: $P=V_{\text{rms}}{I}_{\text{rms}}\cos \theta$.

    WATCH OUT FOR

    Three traps. Picking (B) means dropping the $\frac{1}{2}$ from the product-to-sum identity, which is the same as treating peak values as rms. Picking (A) means assuming anything sinusoidal averages to zero over a half cycle; $v$ and $i$ on their own do, but their product carries a constant $\cos \theta$ term. Picking (D) means thinking a half cycle halves the average. It does not, because $p(t)$ repeats every half cycle.

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

    Networks (Electric Circuits) › Transients

    For a single capacitor, find three things: the initial value from the DC circuit before the switching event (capacitor open, and its voltage is continuous), the final value from the DC circuit after it, and $\tau =R_{th}C$, where $R_{th}$ is seen from the capacitor terminals with the independent sources killed. Then $v_C(t)=v_C(\infty )+[v_C(0^{+})-v_C(\infty )]e^{-t/\tau }$. A source transformation makes both the final value and $R_{th}$ one-line calculations here.

    WATCH OUT FOR

    Taking $v_C(0^{+})=10\text{V}$ by assuming the capacitor charges to the full source voltage and forgetting the 100 kΩ divider arm. This leads to (C) or (D). The second trap is swapping the last two values (A/D). The response rises monotonically towards 28 V, so its value at 0.5 s (2.5 time constants) must be below 28 V. Some students also use only 100 kΩ or only 25 kΩ for $\tau$ instead of their parallel combination, 20 kΩ.

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

    Networks (Electric Circuits) › Theorems

    A linear one-port has a straight-line terminal characteristic $v_t=V_{th}+R_{th}{i}_t$ (with $i_t$ entering the + terminal). Two measured points fix it: the slope is $R_{th}=R_N$, the intercept at $i_t=0$ is the open-circuit voltage $V_{th}$, and the Norton current is the short-circuit current $I_N=V_{th}/R_{th}$. The third row serves as a check.

    WATCH OUT FOR

    Reading $i_t$ as leaving terminal X and writing $v_t=V_{th}-R_{th}{i}_t$. That gives $R_{th}=-12\Omega$, which is physically impossible and should tell you the sign is wrong. Two other slips: taking $V_{th}=24$ as the Norton resistance (that gives option D), or swapping the two values (that gives option B).

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

    Digital Electronics › Boolean logic

    The NOR output $\bar{x}\bar{y}\bar{z}$ is 1 only when $y=z=0$. The OR output $\bar{x}(y+z)$ needs $y$ or $z$ to be 1. The final AND gate needs both to be 1 at once, which never happens, so OUT is the constant 0.

    WATCH OUT FOR

    Stopping at the OR gate and choosing (C) $\bar{x}(y+z)$, even though that signal still passes through the final AND gate with the NOR output. A second mistake is misusing De Morgan and writing $\overline{x+y+z}$ as $\bar{x}\overline{yz}$ (option A); the correct form is $\bar{x}\bar{y}\bar{z}$. A third is treating the wire jumps as connections, which puts the wrong input on the AND gates.

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

    Digital Electronics › Logic families

    Treat each resistive-load NMOS stage as a pull-down network: series switches give AND, parallel switches give OR, and the output is the complement of that. The right stage makes $X=\overline{C}$, which drives $M_X$ in parallel with the $A$–$B$ series pair, so $\text{OUT}=\overline{AB+\overline{C}}=\overline{AB}C$. With ideal switches, a branch dissipates $V^2/R$ only while its pull-down path conducts. Both branches can conduct at once only when $C=1$ and $AB=1$, which gives $5+2.5=7.5\text{mW}$.

    WATCH OUT FOR

    The usual slip is to stop once the Boolean expression is found. Option (A) is the same expression as (D), written before applying De Morgan, so you can only separate them by working out the power. The other trap is the power itself: taking the $C=0$ case, where $M_X$ is on, gives only 5 mW, because the right branch is then off. The maximum needs $C=1$, which turns the right branch on and $M_X$ off, together with $A=B=1$ so that the left branch still conducts through the series pair.

    Full solution →
37. Q.47MCQ · 2 marksOfficial answer: B
38. Q.48MCQ · 2 marksOfficial answer: C
39. Q.49MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Linear algebra

    The $p$ vectors are the columns of an $n\times p$ matrix, and $\text{rank}\le \min (n,p)$. Independence needs $\text{rank}=p$, which is impossible when $p>n$: ${\mathbb{R}}^n$ cannot hold more than $n$ independent vectors. The reverse does not hold. $p\le n$ allows independence but never guarantees it, and independence is a much weaker condition than orthogonality.

    WATCH OUT FOR

    Treating a necessary condition as a sufficient one. $p\le n$ is necessary for independence but not sufficient, so (D) fails: a vector and a multiple of it are dependent in any dimension. Another mistake is confusing independence with orthogonality in (C). Orthogonal nonzero vectors are always independent, but independent vectors need not be orthogonal.

    Full solution →
40. Q.50MCQ · 2 marksOfficial answer: A

    Engineering Mathematics › Differential equations

    For $y^{\prime \prime }+a{y}^{\prime }+by=0$ with complex roots $\alpha \pm j\beta$, the solution is $e^{\alpha x}(A\cos \beta x+B\sin \beta x)$. Then $y(0)$ fixes $A$, and $y^{\prime }(0)=\alpha A+\beta B$ fixes $B$. All four options share the same envelope, the same frequency and $A=1$, so only the $y^{\prime }(0)$ condition decides between them.

    WATCH OUT FOR

    Leaving out the $\alpha A$ term that the product rule gives to $y^{\prime }(0)$, and writing $\beta B=1$ instead of $\alpha A+\beta B=1$. That gives $B=\frac{2}{\sqrt{3}}$, which is not an option. Another slip is dropping the sign of $\alpha$ and writing $+\frac{A}{2}$, which gives $\frac{\sqrt{3}}{2}B=\frac{1}{2}$, $B=\frac{1}{\sqrt{3}}$, and so the wrong option (B).

    Full solution →
41. Q.51MCQ · 2 marksOfficial answer: A

    Electromagnetics › Maxwell's equations

    A loop in the XY plane links flux only through $B_z$. Faraday's law turns each patch's flux into a sinusoidal emf. Differentiating $\cos (2\omega t+\pi /4)$ doubles the amplitude of the second term. Because the two emfs are at different frequencies ($\omega$ and $2\omega$), their rms values combine as $E_{rms}=\sqrt{E_{1,rms}^2+E_{2,rms}^2}$, not by adding peaks or phasors.

    WATCH OUT FOR

    Using the full field magnitude, including the in-plane $\hat{x}$ or $\hat{y}$ component, instead of only $B_z$. Other mistakes: forgetting the factor $2$ that the derivative of $\cos (2\omega t+\pi /4)$ brings out, which gives option (D)'s $2a_2^2$ coefficient; forgetting to divide the squared peaks by 2 when converting to rms, which leads to option (B); and treating the two emfs as phasors at one frequency and adding them.

    Full solution →
42. Q.52MCQ · 2 marksOfficial answer: D

    Control Systems › State space

    The denominator of the transfer function is $|sI-A|$. Set it equal to the polynomial built from the required poles, $(s+2{)}^2+9=s^2+4s+13$, and compare coefficients. Then check that the numerator does not cancel a pole.

    WATCH OUT FOR

    The usual slip is a sign error in the determinant. Writing $s(s-a)-ab$ instead of $s(s-a)+ab$ (the off-diagonal product is $(-b)(a)=-ab$, and it is subtracted) gives $ab=-13$ and points to (A) or (B). A second slip is matching $-a$ to $+4$ as $a=4$, which leads to (A) or (C). Remember that a pole at $-2$ needs a positive coefficient of $s$ in the characteristic polynomial.

    Full solution →
43. Q.53MCQ · 2 marksOfficial answer: C

    Control Systems › State space

    For a change of state $\mathbf{z}=T\mathbf{x}$, the new model is $\left(TA{T}^{-1},TB,C{T}^{-1},D\right)$. The eigenvalues, and so the trace and determinant, of the system matrix do not change. That gives a quick check that removes wrong options without doing all the multiplication.

    WATCH OUT FOR

    Forgetting to pre-multiply by $T$ and stopping at $A{T}^{-1}=\left[\begin{array}{cc}2& 3\\ 0& 3\end{array}\right]$ gives the system matrix printed in (B). Mixing up the direction of the transform is the other trap. Because the question defines $\mathbf{z}=T\mathbf{x}$, the new matrix is $TA{T}^{-1}$, not $T^{-1}AT$. The two have the same trace and determinant, so the invariance check cannot catch this mistake. Only reading the definition carefully does. The same applies to the output matrix: it is $C{T}^{-1}$, not $CT$.

    Full solution →
44. Q.54MSQ · 2 marksOfficial answer: A, D
45. Q.55MSQ · 2 marksOfficial answer: A, B, D

    Engineering Mathematics › Linear algebra

    Every property here follows from $A^{T}A=I$. Taking determinants gives $(\det A{)}^2=1$. Writing ${\mathbf{x}}^T{A}^{T}A\mathbf{y}={\mathbf{x}}^T\mathbf{y}$ shows that inner products are preserved, and so lengths are too. Applying the length result to a possibly complex eigenvector, using the conjugate transpose, gives $|\lambda |=1$.

    WATCH OUT FOR

    One mistake is to read (B) as saying the eigenvalues are $\pm 1$ and reject it because rotation matrices have complex eigenvalues. The option only claims modulus 1, and $e^{\pm j\theta }$ has modulus 1. Another is to accept (C) because its first half ($\Vert A\mathbf{x}\Vert =\Vert \mathbf{x}\Vert$) is true, without noticing that its second half contradicts inner-product preservation. A third is to prove $|\lambda |=1$ with $v^T$ in place of $v^{\ast }$. That argument fails when the eigenvector is complex.

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

    Engineering Mathematics › Linear algebra

    For a square system, a unique solution exists exactly when $A$ is non-singular: $\text{rank}(A)=\text{rank}([A\mid \mathbf{b}])=n$, which is the same as $\det A\ne 0$. Every power of $A$ is then non-singular too, because $\det (A^m)=(\det A{)}^m\ne 0$. So $A^m$ keeps full rank $n$, and $A^m\mathbf{x}=\mathbf{b}$ also has exactly one solution.

    WATCH OUT FOR

    Picking (D) because "rank" and "augmented matrix" sound like the uniqueness test. $\text{rank}(A)<\text{rank}([A\mid \mathbf{b}])$ actually means the system is inconsistent, with no solution at all. A second trap is rejecting (C) on the idea that powers of a matrix can lose rank. That only happens when $A$ is singular (a nilpotent matrix, for example), and a unique solution rules that out.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 750

    Electrical and Electronic Measurements › Bridges

    A Wheatstone bridge is balanced when the products of opposite arms are equal. Here P is opposite S and Q is opposite R, so $PS=QR$. This condition does not depend on the supply voltage, the source's internal resistance or the galvanometer's resistance.

    WATCH OUT FOR

    Pairing the wrong arms. Writing $P/Q=S/R$ gives $S=PR/Q=12000\Omega$. Always find which arms sit opposite each other in the figure: they are the arms that share neither a supply node nor a galvanometer node. A second mistake is trying to use the 1 Ω internal resistance, the 50 Ω galvanometer resistance or E = 50 V. They are distractors, because no current flows through G at balance.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: Marks to all
49. Q.59NAT · 2 marksOfficial answer: 0.09

    Electrical Machines › Induction machines

    For a constant-torque load driven at low slip, torque is approximately proportional to s/R₂ (the sX₂ term in the torque expression is negligible at low slip), so equal torque before and after adding rotor resistance gives s₁/R₂ = s₂/(R₂+r), which can be solved directly for the unknown rotor resistance R₂.

    WATCH OUT FOR

    Forgetting to neglect the sX₂ term (i.e., trying to use the full torque expression with an unknown X₂, which the problem gives no way to determine) or mixing up which resistance (R₂ vs R₂+r) goes with which slip, both of which make the problem unsolvable or give a wrong sign on r.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 2.81 to 2.91

    Networks (Electric Circuits) › Theorems

    Match $R_L$ to the Thevenin resistance at port 2. When the ideal voltage source is replaced by a short, it shorts out the 5 Ω shunt at port 1. That leaves the 10 Ω series arm in parallel with the 4 Ω shunt at port 2.

    WATCH OUT FOR

    Open-circuiting the voltage source instead of shorting it keeps the 5 Ω in the path and gives $(10+5)\parallel 4=60/19\approx 3.16\Omega$, which is wrong. Another mistake is working out two-port parameters (Z or Y) of the pi network alone. That ignores the fact that the ideal source fixes $v_1$ and so shorts port 1 when it is deactivated.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 1.04 to 1.14
52. Q.62NAT · 2 marksOfficial answer: 25.00 to 25.60

    Engineering Mathematics › Complex analysis

    Find where each pole sits relative to the circle $|z-(2+i)|=3/2$ by working out its distance from the centre $2+i$. Only $z=2$ (distance 1) is inside, so the integral is $2\pi i$ times that one residue, $\frac{9}{2-i}$. Its magnitude is $2\pi \cdot \frac{9}{|2-i|}=\frac{18\pi }{\sqrt{5}}$.

    WATCH OUT FOR

    Including the pole at $z=i$ as well. It is at distance 2 from the centre $2+i$, so it lies outside the radius-$3/2$ circle. Counting both residues gives $2\pi i[{\text{Res}}_{z=i}+{\text{Res}}_{z=2}]$, which is a different number. A second slip is to treat $|z-2-i|$ as a circle centred at $2-i$ or $-2-i$ instead of $2+i$. A third is to report $2\pi \cdot 9/5$ after dropping the modulus of $(2+i)$ from the residue.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.38 to 0.42

    Electromagnetics › Electrostatics

    Outside a uniformly charged sphere, the field and potential are those of a point charge $Q$ at the centre. Electrostatic work depends only on the end points, $W=q(V_B-V_A)$, so only the two radial distances matter (5 m and 4 m). The point charge is given as $\pi {ϵ}_0$ so that $q/(4\pi {ϵ}_0)=1/4$ exactly.

    WATCH OUT FOR

    Three common mistakes: (1) using the $z$-coordinates or the straight-line distance between the points, 5 m apart, instead of each point's distance from the origin; (2) using the interior potential formula when both points are outside the sphere ($r_A=5$, $r_B=4>R=2$); (3) writing $V_A-V_B$, which gives $-0.40\mu \text{J}$ (the work done by the field) instead of the work done by the external agent in moving the charge.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 0.1

    Engineering Mathematics › Probability & statistics

    $E\left((X-\alpha Y{)}^2\right)=E(X^2)-2\alpha E(XY)+{\alpha }^{2}E(Y^2)$ is a quadratic in $\alpha$ with a positive leading coefficient. Its vertex is at ${\alpha }^{\ast }=E(XY)/E(Y^2)$. This is the least-mean-square (projection) coefficient of $X$ onto $Y$, and only raw second moments enter it.

    WATCH OUT FOR

    Using central moments by mistake, for example computing $\text{Cov}(X,Y)/\text{Var}(Y)=(0.9-2)/(9-4)=-0.22$. That ratio minimises $E\left((X-{\mu }_X-\alpha (Y-{\mu }_Y){)}^2\right)$, which is a different expression. Other slips are dropping the factor 2 on the cross term, which gives $0.2$, and writing $E(X)E(Y)$ where the question needs $E(XY)$.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 0.25

    Engineering Mathematics › Calculus

    On $(0,\infty )$ the substitution $x\to 1/x$ turns $\frac{x^n}{1+x^n}$ into $\frac{1}{1+x^n}$ and leaves $\frac{dx}{1+x^2}$ unchanged. Adding the two forms, the $x^n$ part cancels and only ${\int }_0^{\infty }\frac{dx}{1+x^2}=\frac{\pi }{2}$ remains, so $I=\pi /4$ whatever $n$ is.

    WATCH OUT FOR

    Trying partial fractions or an asymptotic estimate on $x^{2026}$. Another mistake is forgetting the $1/\pi$ outside and entering $\pi /4\approx 0.79$ in place of $0.25$. A third is dropping the $1/t^2$ from $dx$ during the substitution, which leaves a spurious $t^2$ factor and breaks the symmetry.

    Full solution →