GATE EE 2022 question paper with solutions

GATE EE 2022 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   This is a homophone pair: work out which part of speech each blank needs. After the possessive your the blank needs a noun (heel, part of the foot). After the infinitive to it needs a verb (heal, to recover).

   WATCH OUT FOR

   Picking by sound or by the \"healing\" theme of the sentence and choosing (C) heal / heal. Heal is a verb and cannot follow your, which needs a noun. Choosing (B) gets both blanks backwards.

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

   General Aptitude › Numerical

   If two runners run for the same length of time, the distances they cover are in the same ratio as their speeds. Only count the stretch where both are running, which starts when Q sets off after P has already covered 140 m.

   WATCH OUT FOR

   Using the full 500 m for P (Q would then cover about 666.7 m, which is impossible), or subtracting the head start twice. Another slip is answering 140 m, which is P's head start and not the gap at the finish.

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

   General Aptitude › Numerical

   Events that repeat every a, b and c minutes coincide again after LCM(a, b, c) minutes. Find it from prime factorisation by taking the highest power of each prime: LCM(20, 30, 50) = 2^2 × 3 × 5^2 = 300 minutes = 5 hours.

   WATCH OUT FOR

   Using the HCF (10 minutes) in place of the LCM. Multiplying the periods (20 × 30 × 50 = 30000 minutes), which gives a common multiple but not the first one. Taking the LCM of only two bells, for example LCM(30, 50) = 150 minutes gives 2:30 PM, which does not suit P because 150 is not a multiple of 20. Adding 300 minutes as if it were 3 hours.

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

   General Aptitude › Reasoning

   'Some B are C' + 'All C are K' gives 'Some B are K'. 'All C are K' converts only to 'Some K are C' (here cups exist because Statement 1 names one), never to 'All K are C'. A conclusion follows only if it is true in every diagram consistent with both statements, so one diagram that meets both statements but breaks the conclusion is enough to reject it.

   WATCH OUT FOR

   Reading a statement as reversible. Students take 'All cups are knives' to mean 'All knives are cups' (IV), or take 'Some bottles are cups' to mean 'All cups are bottles' (III). A universal affirmative converts only to a particular ('Some knives are cups'), and a particular gives only a particular.

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

   General Aptitude › Spatial

   Looking at the disc from behind is the same as turning it $180^{\circ }$ about the vertical axis 1-1. A flip about 1-1 therefore just swaps the two identical views, and nothing changes. A flip about a slanted axis mirrors the disc in that slanted line. That swings the big shaded sector from pointing down to pointing sideways, so the views change. Two of the three axes do this, which gives $\frac{2}{3}$.

   WATCH OUT FOR

   One mistake runs: 'both faces carry the same pattern, so any flip just swaps identical faces', giving probability 0, option (A). That misses that a flip about a slanted axis mirrors the disc in that slanted line, which moves the big sector away from the vertical. The opposite mistake comes from forgetting that the rear view is drawn as seen from behind (mirrored left-right). A student who does that thinks even the flip about 1-1 changes the picture and answers 1, option (D).

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

   General Aptitude › Reading

   Split the passage into two lists: causes it accepts (social bonding, familiarity, group identification) and causes it rejects (empathy, guilt). An option is a valid inference only if every cause it accepts is on the first list and every cause it rules out is on the second.

   WATCH OUT FOR

   Picking (A) or (B) because each rejects one of empathy and guilt, and missing that each also accepts the other rejected cause. Another trap is checking only the first half of an option. (D)'s \"but not familiarity\" contradicts the passage as directly as its \"due to empathy\" does.

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

   General Aptitude › Statistics & probability

   "Only Q, U and V" asks that every die shows a letter from that set. For independent dice, multiply the single-die probability: ${\left(\frac{3}{6}\right)}^2=\frac{1}{4}$. "Identical" dice are still two separate dice, so count the 36 ordered outcomes, which are equally likely.

   WATCH OUT FOR

   Counting only pairs with two different letters from {Q, U, V} (QU, QV, UV and their reverses, 6 outcomes) gives $\frac{6}{36}=\frac{1}{6}$, which is trap option (C). The wording "any combination" allows repeats like QQ. A second trap is treating the "identical" dice as giving 21 unordered outcomes: those outcomes are not equally likely, and the count gives $\frac{6}{21}$, which is not an option. Option (B), $\frac{3}{4}$, is the complement: the probability that at least one die shows R, S or T.

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

   General Aptitude › Numerical

   "10% cheaper at S than at M" means the 10% is taken on M's price, so S's price is $0.9m$. The saving is M's price minus everything paid at S, delivery included: $0.1m-150=100$.

   WATCH OUT FOR

   Taking the 10% on the wrong price. If you write M's price as $1.1s$, you get $0.1s-150=100$ and $s=2500$, which is option (A). That is actually M's price, not S's. Two other slips: forgetting the Rs. 150 delivery charge ($0.1m=100$ gives Rs. 900 at S, which is not an option), and answering with M's price of 2500 when the question asks for S's.

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

   General Aptitude › Reasoning

   In a regular hexagon with vertices numbered 1 to 6, the chord joining vertices i and j has a direction fixed by (i+j) mod 6: equal values mean parallel, values 3 apart mean perpendicular. The only way a longer segment can be perpendicular to a shorter one on four distinct vertices is a long diagonal perpendicular to a short diagonal. That fixes R and S up to order, and the parallel condition then pins the whole labelling down to one arrangement and its mirror image.

   WATCH OUT FOR

   Drawing one arrangement that satisfies the statements and reading the answer off it, without checking that it is forced. Here that works only because the two surviving labellings are mirror images. Students also often let RS be a side perpendicular to a short diagonal PQ, forgetting that such a pair always shares a vertex, which the four distinct letters rule out.

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

    General Aptitude › Spatial

    If the distance to the top-right corner must strictly decrease, every move has to be up or to the right. A left or down move takes the ant further away. So a valid trajectory is a staircase that goes only up and right. It never revisits a point, never comes back to P, and never has two branches leaving P.

    WATCH OUT FOR

    Picking (A) or (B) because most of their segments go up or right. Both end with a downward segment, and both have two segments meeting at P, which the ant could only use by coming back to its starting point. Another slip is choosing (D) because a square is made of grid lines, forgetting that going around it means moving left and down back to P.

    Full solution →

Electrical Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: B or D

   Control Systems › Bode & margins

   A strictly proper transfer function (numerator degree lower than the denominator's) always rolls off to zero at high frequency, so it can never pass high frequencies.

   WATCH OUT FOR

   Thinking the zero at $s=-B/A$ lets high frequencies through. One zero offsets only one pole's roll-off; with two poles the magnitude still falls at 20 dB per decade as $\omega \to \infty$.

   Full solution →
2. Q.12MCQ · 1 markOfficial answer: D

   Analog Electronics › Amplifiers

   In an ideal MOSFET the insulated gate draws no current, so every configuration that takes its input at the gate (common source and common drain) has zero input current and infinite current gain. The common drain stage has a voltage gain close to 1, not a current gain of 1.

   WATCH OUT FOR

   Choosing (B) 1 because the source follower is remembered as a unity-gain stage. That value is its voltage gain. Its current gain is infinite because the gate current is zero. For the common gate stage, where input current enters at the source, the current gain is 1.

   Full solution →
3. Q.13MCQ · 1 markOfficial answer: A
4. Q.14MCQ · 1 markOfficial answer: D
5. Q.15MCQ · 1 markOfficial answer: B
6. Q.16MCQ · 1 markOfficial answer: B

   Analog Electronics › Op-amps

   An ideal op-amp integrator has no resistor across its capacitor, so it has no DC feedback. Any nonzero DC input is integrated into a ramp that only stops at a supply rail. Because the integrator is inverting, a positive input drives the output to the negative rail. At DC the capacitor is open, the op-amp runs open-loop, and the sign of $V_{+}-V_{-}$ decides which rail.

   WATCH OUT FOR

   Choosing (A) by forgetting that the integrator inverts: the ramp slope is $-V_{in}/(R_1{C}_1)$. Choosing (C) or (D) by assuming the virtual short still holds at steady state, or by treating the circuit as a unity-gain follower or inverter once the capacitor is open. When the capacitor is open there is no feedback at all, so the virtual short no longer holds and the output saturates.

   Full solution →
7. Q.17MCQ · 1 markOfficial answer: A

   Control Systems › Bode & margins

   A pole and a zero at mirror-image points, $s=-a$ and $s=+a$, have equal distances to every point $j\omega$, so the magnitude stays flat. Their angles do not cancel, though. Each one contributes $-{\tan }^{-1}(\omega /a)$, which gives the first-order all-pass $K\frac{a-s}{a+s}$ a phase that runs from $0^{\circ }$ to $-180^{\circ }$. A flat magnitude with a moving phase is the signature of a non-minimum-phase (RHP) zero.

   WATCH OUT FOR

   Picking (B) because a pole and a zero at the same frequency give a flat magnitude. With both in the LHP they cancel completely, so the phase is also constant at $0^{\circ }$ and can never reach $-180^{\circ }$. A second error is to assume an RHP zero behaves like an LHP zero in phase. At high frequency an LHP zero adds $+90^{\circ }$, but an RHP zero adds $-90^{\circ }$.

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

   Electrical and Electronic Measurements › Errors & standards

   At balance, the products of opposite arms of a Wheatstone bridge are equal: $R_{AB}{R}_{CD}=R_{BC}{R}_{DA}$. When the unknown is a product or quotient of the other arms, its limiting percentage error is the sum of their percentage errors. You then turn that percentage into ohms using the computed value.

   WATCH OUT FOR

   Pairing the wrong arms. Writing $R_{CD}=R_{AB}{R}_{DA}/R_{BC}=3000\Omega$ gives options (C) or (D). Another trap is reading the $\pm 3\%$ relative error as $\pm 3\Omega$, which gives option (A). A third is subtracting the error of the divisor arm, $2.1-0.5-0.4=1.2\%$: in a limiting-error calculation, relative errors always add.

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

   Control Systems › Routh–Hurwitz

   Stability depends on the closed-loop poles, the roots of $1+G(s)=0$, not on the open-loop poles. The open-loop plant has a right-half-plane pole at $s=1$, since $s^2+4s-5=(s+5)(s-1)$. Feedback gain $k$ moves the constant term from $-5$ to $k-5$. For a quadratic, all coefficients positive is both necessary and sufficient for stability.

   WATCH OUT FOR

   One mistake is to apply the Routh test to the open-loop denominator $s^2+4s-5$ and conclude the system is unstable for every $k$. Another is dropping the sign of the $-5$ and writing $k+5>0$. A third is treating $k=5$ as stable, which gives $k\ge 5$: the characteristic equation is then $s(s+4)=0$, a pole at the origin, which is only marginal stability. Options (A) and (B) use 3, a number that has no role in this characteristic equation.

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

    Engineering Mathematics › Linear algebra

    If the $(i,j)$ entry is an odd function of $(i-j)$, swapping $i$ and $j$ flips its sign, so $A^T=-A$ and the matrix is skew-symmetric. An even function of $(i-j)$, such as $(i-j{)}^2$ or $|i-j|$, would make it symmetric instead. You never need to write the matrix out. Looking only at the parity of the power is enough.

    WATCH OUT FOR

    Treating $(i-j{)}^3$ as if it were $(i-j{)}^2$ or $|i-j|$, which leads to 'symmetric'. The cube keeps the sign, so $a_{21}=+1$ while $a_{12}=-1$. A second trap is choosing 'null' because every diagonal entry is $0$: a zero diagonal is only what skew-symmetry forces, and the off-diagonal entries are non-zero.

    Full solution →
11. Q.21MCQ · 1 markOfficial answer: C

    Networks (Electric Circuits) › Three-phase (EE)

    In a three-wire star system, the load neutral sits at the source neutral potential exactly when the Millman numerator $\sum E_k/Z_k$ is zero. The same condition says the currents $E_k/Z_k$ satisfy KCL at $n^{\prime }$. The load does not have to be balanced for this: it is enough that the three phase currents, each set by its own phase voltage, add to zero.

    WATCH OUT FOR

    Assuming $V_{n^{\prime }n}=0$ needs a balanced load and picking $Z_C=Z_B=20\angle 60^{\circ }\Omega$ (option D). That cannot be right, since $Z_A=10\Omega$ already differs from $Z_B$. Two other slips: using $E_C=100\angle -120^{\circ }$ (the $ACB$ sequence) instead of $100\angle 120^{\circ }$, and dropping the minus sign in $I_C=-(I_A+I_B)$, which gives $20\angle 120^{\circ }$, not one of the options.

    Full solution →
12. Q.22MCQ · 1 markOfficial answer: C
13. Q.23MCQ · 1 markOfficial answer: C

    Electromagnetics › Magnetostatics

    When the current density varies with $r$, find the enclosed current by integrating over thin rings of area $2\pi r^{\prime }d{r}^{\prime }$. Then apply Ampère's law on a circle of radius $r$: $H_{\varphi }(2\pi r)=I_{\text{enc}}$. A current density that goes as $r^n$ gives $H_{\varphi }=\frac{J_a{r}^{n+1}}{n+2}$, so here ($n=3$) the result is $\frac{J_a{r}^4}{5}$.

    WATCH OUT FOR

    The first mistake is treating $J$ as uniform and writing $I_{\text{enc}}=J\pi r^2$. That gives $\frac{J_a{r}^4}{2}$. The second is leaving out the factor $r^{\prime }$ in the ring area $dS=2\pi r^{\prime }d{r}^{\prime }$, so that $\int r^{\prime 3}d{r}^{\prime }$ is integrated instead of $\int r^{\prime 4}d{r}^{\prime }$. That gives $\frac{J_a{r}^3}{4}$. Neither result is an option, and option (A) pairs the power $r^4$ with the wrong factor $\frac{1}{4}$. Option (D) is just the magnitude of $\mathbf{J}$, which mixes up the current density with the field it produces.

    Full solution →
14. Q.24MCQ · 1 markOfficial answer: D

    Electrical Machines › Induction machines

    Running power factor in a single-phase induction motor depends on what remains permanently connected in the auxiliary-winding circuit during steady running, not on the starting arrangement. Only a permanently connected running capacitor, sized to bring the auxiliary winding current close to quadrature with the main winding at rated load, improves the running power factor toward that of a balanced two-phase machine.

    WATCH OUT FOR

    Confusing "best starting torque/performance" with "best running power factor" — a capacitor-start motor has excellent starting torque but its capacitor is switched OUT after starting, so it gives no running power-factor benefit; only a permanently connected running capacitor (as in capacitor-start-capacitor-run) improves the running power factor.

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

    Analog Electronics › Diode circuits

    $D_1$, $R_1$ and $D_2$ are all in one series loop with the source, and both diodes point the same way round it. With ideal diodes the loop is either a short circuit through $R_1$ ($V_{in}>0$, so $V_{out}=V_{in}$) or open ($V_{in}<0$, so $V_{out}=0$). It is a half-wave rectifier, and the second diode only adds another series switch in the same direction.

    WATCH OUT FOR

    Reading $D_2$ as pointing against $D_1$ because it is drawn facing left. Trace the loop instead: after $R_1$ the current enters $D_2$ from the right, which is its anode, so both diodes conduct in the same loop direction. A second mistake is treating the two-diode circuit as a bridge or full-wave rectifier and choosing (D). With only one closed loop, nothing can flip the polarity, so the output is zero for $V_{in}<0$.

    Full solution →
16. Q.26NAT · 1 markOfficial answer: 10

    Digital Electronics › Sequential

    When counters are cascaded, their moduli multiply: ${\text{MOD}}_{\text{total}}=M_1\times M_2$. The later stage completes one cycle only after the earlier stage has completed $M_1$ cycles.

    WATCH OUT FOR

    Adding the moduli ($2+5=7$) instead of multiplying them. Another mistake is counting flip-flops instead of states: MOD 2 needs 1 flip-flop and MOD 5 needs 3, so there are 4 flip-flops, but the question asks for the modulus, which is the number of states, $10$.

    Full solution →
17. Q.27NAT · 1 markOfficial answer: 48

    Networks (Electric Circuits) › AC, power & resonance

    In a series circuit at resonance both elements carry the same current and have the same reactance $X$. Each loss resistance is $X/Q$, and these resistances add, so the reciprocals of the Q-factors add: $\frac{1}{Q}=\frac{1}{Q_L}+\frac{1}{Q_C}$. The rule works like resistors in parallel.

    WATCH OUT FOR

    Adding the Q-factors (60 + 240 = 300), averaging them (150), or multiplying them. The losses add, not the Q-factors, so the overall Q has to come out smaller than the smaller of the two (60). Another mistake is to take the lossless capacitor's Q as infinite and answer 60, even though the capacitor's Q is given as 240.

    Full solution →
18. Q.28NAT · 1 markOfficial answer: 250

    Networks (Electric Circuits) › AC, power & resonance

    For any resonant network, Q = f_0 / BW, where BW is the half-power bandwidth. When f_0 and BW are given, you get Q straight from them. You don't need the element values or the circuit topology (series, parallel or tank).

    WATCH OUT FOR

    Mixing units: dividing 150 by 600 (kHz over Hz) gives 0.25 instead of 250. A second mistake is to multiply or divide by 2π when one quantity is in rad/s and the other in Hz. Here both are in Hz, so no 2π appears. A third is to try to find Q as ω_0 L / R or R√(C/L). That can't be done without element values and isn't needed.

    Full solution →
19. Q.29NAT · 1 markOfficial answer: 12.3 to 12.7

    Digital Electronics › Sequential

    In a ripple counter, the clock of each stage comes from the output of the stage before it. The worst-case settling time is therefore the sum of all the flip-flop delays, $n{t}_{pd}$, and the clock period must be at least that long, so $f_{\max }=1/(n{t}_{pd})$.

    WATCH OUT FOR

    Using the single flip-flop delay, $f=1/t_{pd}=50\text{MHz}$. That is the synchronous-counter result, where every flip-flop shares the clock and switches at the same time. A ripple counter's delays add in series, so the period must cover all four stages.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 13

    Electrical and Electronic Measurements › Errors & standards

    For a power ratio, decibels are $10{\log }_{10}(P_{in}/P_{out})$. If $5\%$ gets through, the ratio is $20$, and $10{\log }_{10}20=10(1+0.301)\approx 13\text{dB}$.

    WATCH OUT FOR

    Using $20{\log }_{10}$, which is the voltage or current formula, and getting $26\text{dB}$. The other common slip is taking the log of the lost fraction ($95\%$) instead of the ratio of input to output, which gives about $0.22\text{dB}$. A third is entering $-13$: the question asks for a loss, so the answer is positive.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 4.95 to 5.05

    Networks (Electric Circuits) › Transients

    In DC steady state a capacitor carries no current, so it acts as an open circuit. Any resistor in series with it (here the $10\Omega$) carries no current either, and the $RC$ branch drops out. The circuit is then just the source, $2\Omega$ and $6\parallel 3=2\Omega$ in one loop, and the voltage splits equally, giving $5\text{V}$.

    WATCH OUT FOR

    Counting the $10\Omega$ resistor in the steady-state network. Putting it in parallel with $6\parallel 3$ gives $R_{eq}=1.667\Omega$ and about $4.55\text{V}$. A second mistake is to find the current in the $6\Omega$ resistor ($5/6\approx 0.83\text{A}$) and report that instead of the voltage. A third is to use the initial ($t=0^{+}$) condition, where the uncharged capacitor acts as a short.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 795 to 805
23. Q.33NAT · 1 markOfficial answer: 0.43 to 0.47
24. Q.34NAT · 1 markOfficial answer: 20

    Networks (Electric Circuits) › Three-phase (EE)

    Replace the balanced delta by its star equivalent ($Z_1/3$). Two balanced star loads fed from the same supply have their star points at neutral potential, so the per-phase impedances are in parallel. Here $Z_1/3=Z_2=(6+j8)\Omega$, so the per-phase impedance is $(3+j4)\Omega$, $|Z|=5\Omega$, and the line current is $100/5=20\text{A}$.

    WATCH OUT FOR

    Three errors are common here. (1) Leaving out the factor of 3 in the delta-to-star conversion, or dividing the delta's phase current by $\sqrt{3}$ instead of multiplying by it. Either slip gives the wrong delta line current. (2) Applying the line voltage $100\sqrt{3}$ V across a star impedance, which gives 17.3 A per branch. (3) Reporting the current of only one load (10 A). Adding the two line-current magnitudes arithmetically is valid here only because both loads have the same impedance angle, $53.13^{\circ }$. In general the currents must be added as phasors.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 580 to 590

    Electrical Machines › Induction machines

    In an induction motor, the rotor current frequency equals slip times the stator (supply) frequency: $f_r=s{f}_s$. Once slip is found, the actual rotor speed is obtained from the synchronous speed as $N=N_s(1-s)$.

    WATCH OUT FOR

    Confusing rotor frequency relation direction (using $f_s=s{f}_r$ instead of $f_r=s{f}_s$), or forgetting to compute $N_s$ correctly for an 8-pole machine (using $P$ instead of pole pairs correctly in $N_s=120f_s/P$).

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

    Analog Electronics › Op-amps

    Voltage-sampling negative feedback divides the op-amp's open-loop output resistance by $1+A(f)\beta$. For a follower $\beta =1$, so $Z_{\text{out}}=r_o/(1+A(f))$. The open-loop gain rolls off above $f_b$, so $|Z_{\text{out}}|$ rises from $r_o/(1+A_0)$ to $r_o$ between $f_b$ and $f_T$: a zero at $f_b$ and a pole at about $f_T$.

    WATCH OUT FOR

    Treating the op-amp as ideal, so that $Z_{\text{out}}=0$ at every frequency, and picking the flat plot (A). The question says the op-amp is non-ideal. Another mistake is sketching the shape of the open-loop gain curve (flat, falling, flat) and picking (D). $Z_{\text{out}}$ varies inversely with the loop gain, so it rises where the gain falls.

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

    Control Systems › Bode & margins

    For a stable LTI system, a DC-plus-sinusoid input gives a DC-plus-sinusoid steady-state output. The DC level is scaled by $G(0)$ and the sinusoid by $|G(j\omega )|$ at its own frequency. When a lightly damped second-order system $\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$ is driven exactly at $\omega ={\omega }_n$, the real parts of the denominator cancel and $|G(j{\omega }_n)|=\frac{1}{2\zeta }$, so a small input sinusoid is amplified many times over.

    WATCH OUT FOR

    There are two traps. The first is using one gain for both terms: the DC term needs $G(0)$ and the sinusoid needs $|G(j10)|$, and on a sharp resonance these can differ by a factor of 100. The second is forcing an answer when the numbers you get match no option. Here, picking (B) because its $a$ agrees ignores that its $b$ is wrong by a factor of about nine. Before choosing, check whether a printed constant looks inconsistent with the options.

    The question and its official answer →
28. Q.38MCQ · 2 marksOfficial answer: B

    Control Systems › Nyquist

    Where the contour is indented decides $P$. Here the indentation goes into the left half plane, so the pole at the origin is inside the contour and $P=1$. The closed loop is stable, so $Z=0$. Then $N=Z-P=-1$: one counter-clockwise encirclement of $-1$. On the plot, the small left-side arc maps to an infinite arc swept counter-clockwise on the left, and that arc wraps round $-1$.

    WATCH OUT FOR

    Using the usual textbook contour, which indents to the right of the origin. That leaves the origin pole outside, gives $P=0$, and leads to $N=0$ (option A). With the indentation drawn to the left, $P=1$. The small arc then maps to an infinite arc on the left, passing through $-\infty$, instead of one on the right, and it encircles $-1$ once. A second trap is the sign: $N=-1$ means one counter-clockwise encirclement. The number of encirclements is still $1$.

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

    Control Systems › Time response

    Reduce nested loops from the inside out. The inner unity-feedback loop around an integrator $\frac{K}{s}$ becomes a first-order lag $\frac{K}{s+K}$. Closing the outer loop then gives a standard second-order denominator, and you read ${\omega }_n$ from the constant term and $\zeta$ from the $s$ coefficient.

    WATCH OUT FOR

    Ignoring the inner loop and treating the forward path as $\frac{100}{s^2}$. That gives $\frac{100}{s^2+100}$, so $\zeta =0$, which is not an option. Another slip is taking ${\omega }_n$ as 100 (the value of ${\omega }_n^2$) or setting $\zeta {\omega }_n=10$ instead of $2\zeta {\omega }_n=10$, which gives $\zeta =1$.

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

    Engineering Mathematics › Linear algebra

    From $Av=\lambda v$ you get $A^{k}v={\lambda }^{k}v$ for every $k$. Put that into the series $e^A=\sum A^k/k!$ and it gives $e^{A}v=e^{\lambda }v$. More generally, $f(A)$ keeps the eigenvectors of $A$ and maps each eigenvalue $\lambda$ to $f(\lambda )$.

    WATCH OUT FOR

    Thinking the function changes the eigenvectors along with the eigenvalues, and so picking (A). Applying a polynomial or power series to $A$ changes only the eigenvalue ($\lambda \to f(\lambda )$). The eigenvector $v$ stays the same. A second slip is to treat $e^A$ as the entry-wise exponential of $A$. It is the matrix power series, not $e^{a_{ij}}$ taken entry by entry.

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

    Engineering Mathematics › Calculus

    If $f(x)={\int }_a^{x}g(t)dt$ with $g$ continuous, then $f^{\prime }(x)=g(x)$. So $f$ is increasing or decreasing according to the sign of the integrand at $t=x$. You never need to evaluate the integral.

    WATCH OUT FOR

    Integrating $e^t(t-1)(t-2)$ by parts to find $f(x)$ explicitly wastes time and invites sign errors. A second trap is reading the sign of $(x-1)(x-2)$ backwards. The quadratic opens upward, so it is negative between its roots, not outside them. That backwards reading picks (B) or (C) instead.

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

    Engineering Mathematics › Linear algebra

    By the Cayley-Hamilton theorem, $A$ satisfies its characteristic equation ${\lambda }^3-6{\lambda }^2+11\lambda -6=0$. Multiplying by $A^{-1}$ gives $6A^{-1}$ as a quadratic in $A$. Because $A$ is block diagonal, the characteristic polynomial is just $(1-\lambda )$ times the $2\times 2$ block's polynomial, ${\lambda }^2-(\mathrm{tr}B)\lambda +\det B$.

    WATCH OUT FOR

    Getting the sign of $\det B$ wrong: $4\cdot 1-(-2)\cdot 1=6$, not $2$. With $\det B=2$ you get the wrong cubic and the wrong constant term, and a coefficient such as $-6$ (option C) or $11$ (option D) looks like the answer. Those are the separate values of $c$ and $d$, not their sum. Another slip is dropping the $-6I$ term when multiplying by $A^{-1}$, which loses the factor $6$ in front of $A^{-1}$.

    Full solution →
33. Q.43MCQ · 2 marksOfficial answer: B
34. Q.44MCQ · 2 marksOfficial answer: C

    Analog Electronics › Op-amps

    The "ideal current amplifier" is an ideal op-amp with capacitive feedback. Virtual ground at the inverting input forces all of $I_{in}$ through $C_f$, which sets $V_o=I_{in}/(s{C}_f)$. The zero-resistance output then drives $C_c$ to ground, giving $I_{out}=s{C}_c{V}_o$. The $s$ terms cancel, so the gain is the capacitor ratio $C_c/C_f$. Its sign depends only on the reference directions drawn for $I_{in}$ and $I_{out}$.

    WATCH OUT FOR

    Using the inverting-amplifier rule "gain $=-Z_f/Z_{in}$" out of habit and picking a negative option, (B) or (D). The sign has to come from the arrows. The source draws $I_{in}$ out of the virtual-ground node, so $V_o$ comes out positive ($V_o=+I_{in}/(s{C}_f)$). $I_{out}$ is defined in the same direction as the current $V_o$ drives through $C_c$ to ground, so the ratio is $+C_c/C_f$. Another slip is inverting the ratio to $C_f/C_c$, which confuses the impedance $1/(sC)$ with the admittance $sC$.

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

    Electromagnetics › Magnetostatics

    For a static field, the point form of Ampere's circuital law is $\mathbf{J}=\nabla \times \mathbf{H}$. Take the curl symbolically first, substitute the point only after that, and then find the magnitude from all three components.

    WATCH OUT FOR

    Substituting the point into $\mathbf{H}$ before differentiating, which makes every derivative zero. Another mistake is getting a curl component's sign wrong: $J_y=\partial H_x/\partial z-\partial H_z/\partial x$, not the reverse. Adding the components ($4+8+8=20$, option D) instead of taking the root of the sum of their squares, or dropping $J_x$ ($\sqrt{128}\approx 11.3$), also gives a wrong answer.

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

    Signals and Systems › LTI & convolution

    For a first-order system $y+\tau \frac{dy}{dt}=Kx$, the transfer function is $H(s)=\frac{K}{1+s\tau }=\frac{K/\tau }{s+1/\tau }$, so the impulse response is $h(t)=\frac{K}{\tau }{e}^{-t/\tau }u(t)$. Here $K=2$ and $\tau =\frac{1}{4}$, so the gain is $8$ and the decay rate is $4$.

    WATCH OUT FOR

    Reading the $\frac{1}{4}$ coefficient as the decay rate, which gives $e^{-t/4}$ (options A and C), or keeping the right-hand-side gain 2 without dividing by the time constant (options A and B). You have to normalise the derivative coefficient to 1 first: $\frac{dy}{dt}+4y=8x$.

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

    Signals and Systems › Fourier transform

    $h(t)$ is an ideal low-pass impulse response, $\frac{\sin (10\pi t)}{\pi t}$, multiplied by $\cos (40\pi t)$. That multiplication shifts the low-pass band to $\pm 40\pi$, so the system is an ideal band-pass filter over $30\pi <|\omega |<50\pi$. The factor 2 cancels the $\frac{1}{2}$ from modulation, so the gain is exactly 1. An LTI system passes each sinusoid on its own, scaled by $H({\omega }_k)$, so we only need to check which input frequencies fall inside the band.

    WATCH OUT FOR

    The usual slip is to ignore the $\cos (40\pi t)$ factor and treat $h(t)$ as a low-pass filter with cutoff $10\pi$. That keeps the low tones and points to option (A), which is wrong. A second slip is to drop the $\frac{1}{2}$ of the modulation property. That gives a passband gain of 2, and the resulting $14\sin (42\pi t)+8\cos (45\pi t)$ is not among the options. A third is to mix up rad/s and Hz and read the band as 30 to 50 Hz. The input tones are at 5, 7.5, 21 and 22.5 Hz, and none of those would fall inside a 30 to 50 Hz band.

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

    Signals and Systems › LTI & convolution

    A system that only changes the time argument, $y(t)=x(f(t))$, is always linear, because the amplitude of $x$ passes through unchanged. It is causal only if $f(t)\le t$ for every $t$. Here $f(t)=e^t>t$ for every real $t$, so the output always depends on a future input value.

    WATCH OUT FOR

    Mixing up $x(e^t)$ with $e^{x(t)}$. The second one really is non-linear, and that confusion leads to option (C) or (D). A second trap is checking causality only for large $t$, or assuming negative $t$ maps to the past. It does not: $y(-1)=x(e^{-1})\approx x(0.368)$, which is still later than $t=-1$.

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

    Signals and Systems › Fourier series

    DTFS coefficients are periodic in $k$ with period $N$, so an index outside $0$ to $N-1$ maps back into that range: $a_{-3}=a_0$ and $a_4=a_1$. The half-angle factorisation $1+e^{j\theta }=2e^{j\theta /2}\cos (\theta /2)$ then turns $2+e^{j\frac{2\pi }{3}n}$ into the form used in the options.

    WATCH OUT FOR

    Treating $a_{-3}$ and $a_4$ as harmonics at $-3{\omega }_0$ and $4{\omega }_0$ that are separate from $a_0$ and $a_1$. A common slip is to add $a_{-3}+a_4$ as the DC term, or to read $2e^{j\frac{2\pi }{6}n}\cos \left(\frac{2\pi }{6}n\right)$ as one term instead of $1+e^{j\frac{2\pi }{3}n}$. Either way the constant comes out as $2$, which leads to option (D). Expand each option back into exponentials and check it at one or two values of $n$.

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

    Engineering Mathematics › Vector calculus

    The direction of fastest increase of a scalar field is the direction of its gradient at the point, and the maximum rate of increase is the gradient's magnitude. Here $\nabla f(1,0,2)=20\hat{i}+7\hat{j}+12\hat{k}$ and the maximum rate is $\sqrt{593}$.

    WATCH OUT FOR

    The usual slip is to drop a component because a coordinate of $P$ is zero. $\partial f/\partial y=7x$ does not contain $y$, so it equals 7 at $P$ even though $y=0$ there. That slip gives option (C). Another slip is differentiating $3x{z}^2$ with respect to $z$ as $3z^2$ instead of $6xz$, or forgetting the $3z^2$ term in $\partial f/\partial x$, which gives 8 instead of 20.

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

    Engineering Mathematics › Vector calculus

    By Green's theorem, ${\oint }_C(Pdx+Qdy)$ taken counter-clockwise equals the area of $R$ exactly when $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$. That holds for $xdy$ and for $\frac{1}{2}(xdy-ydx)$. For $ydx$ it is $-1$, which gives the negative of the area.

    WATCH OUT FOR

    Forgetting the minus sign in Green's theorem, $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$, and so treating ${\oint }_{C}ydx$ as the area. Counter-clockwise, the area is ${\oint }_{C}xdy=-{\oint }_{C}ydx$. A second mistake is to pick (D) because of its factor $\frac{1}{2}$. The two terms each contribute half the area, so together they give the whole area.

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

    Engineering Mathematics › Vector calculus

    Work out the divergence $\vec{\nabla }\cdot \vec{E}=4x+8$ and integrate it over the unit cube. You can check the result with the divergence theorem: the volume integral of the divergence equals the net outward flux through the cube's six faces, and here that is $2+5+3=10$.

    WATCH OUT FOR

    Treating the divergence as the constant $2+5+3=10$ reaches the right number for the wrong reason. The $x$-term is $4x$, not a constant, and it only averages to $2$ over $0\le x\le 1$. A second mistake is to differentiate $2x^2$ and then set $x=1$, giving $4+5+3=12$, which is not an option. A third is to add only the constant terms $5+3=8$ and drop the $x$-term, which picks the distractor (B).

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

    Electromagnetics › Electrostatics

    No charge sits between the shells, so $V$ obeys Laplace's equation. With spherical symmetry the general solution is $V(r)=\frac{A}{r}+B$, and the two shell potentials fix $A$ and $B$. In an exam, the fastest route is to put $r=c$ and $r=d$ into each option. Only the correct expression gives $V_1$ and $V_2$.

    WATCH OUT FOR

    Checking only one boundary. Option (A) also gives $V_1$ at $r=c$, and fails only at $r=d$, where it gives $2V_1-V_2$. Check both walls. Other slips are using the cylindrical solution $A\ln r+B$ instead of the spherical $A/r+B$, and losing a sign when solving for $B$.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 0.99 to 1.01

    Engineering Mathematics › Probability & statistics

    A PDF must have total area 1. Because $e^{-2|x|}$ is even, the area is twice the area over $x\ge 0$, which is $2\times \frac{1}{2}=1$. So $a=1$. In general, $a{e}^{-\lambda |x|}$ is a valid PDF only when $a=\lambda /2$.

    WATCH OUT FOR

    Integrating only from $0$ to $\infty$ and forgetting the negative half of the line gives $a/2=1$, so $a=2$. That constant belongs to the one-sided exponential $2e^{-2x}$ for $x\ge 0$, not to this two-sided density. Another slip is dropping the absolute value and integrating $e^{-2x}$ over all of $\mathbb{R}$, which diverges.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 98 to 102

    Networks (Electric Circuits) › Laws & node/mesh

    The short in the middle holds the node after the CCVS at ground, so the circuit splits into two loops. Only the dependent sources connect them. The right loop is a series path driven by the current source, which gives $V_1=8000I$. Put that into the left loop's KVL and you have one equation in $I$.

    WATCH OUT FOR

    One mistake is to treat the middle wire as if it carried the $8\text{k}\Omega$ current, or to think the short "kills" the right loop. The short only fixes a node voltage. The current source still pushes $I$ through the $8\text{k}\Omega$ resistor. A second mistake is to get the CCVS polarity wrong, adding $0.5V_1$ as a rise instead of a drop when the loop is traversed from $+$ to $-$. That gives $75=2000I-4000I$, so $I$ comes out negative and $|V_1|=300\text{V}$.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 188 to 192, or 198 to 202
47. Q.57NAT · 2 marksOfficial answer: 74 to 76
48. Q.58NAT · 2 marksOfficial answer: 3170 to 3190
49. Q.59NAT · 2 marksOfficial answer: 17.3 to 18
50. Q.60NAT · 2 marksOfficial answer: 0.46 to 0.48
51. Q.61NAT · 2 marksOfficial answer: 165 to 171
52. Q.62NAT · 2 marksOfficial answer: 480 to 485

    Electrical Machines › DC machines

    With constant field flux, back-EMF is proportional to speed (Eb = k·N) and developed torque is proportional to armature current (T = k·Ia). At steady state the developed torque equals the load torque, and since load torque is proportional to speed here, armature current also becomes proportional to speed (Ia = k'·N). Writing the voltage equation V = Eb + Ia·R for both operating points (before and after adding series resistance) and using the two proportionality constants found from the first operating point gives a single linear equation in the new speed.

    WATCH OUT FOR

    Assuming armature current stays the same (30 A) after adding the series resistance, or assuming speed stays constant and only current changes — both ignore that the load torque itself changes with speed, which is what links Ia to N in this problem (T ∝ Ia and T_L ∝ N together force Ia ∝ N at every steady-state operating point, not just current or speed alone).

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 675 to 700, or 1725 to 1740

    Electrical Machines › Induction machines

    During the linear speed ramp, the motor's electrical output energy (losses neglected) must equal the sum of (1) the mechanical work done against the constant load torque over the angle turned, and (2) the increase in the rotor's stored kinetic energy: $E=T_L\theta +\frac{1}{2}J({\omega }_2^2-{\omega }_1^2)$, where $\theta$ is the angle turned only during the 4 s ramp.

    WATCH OUT FOR

    Forgetting that the motor must supply BOTH the energy absorbed by the constant load torque (work done against the load, $T_L\theta$) AND the change in the rotor's kinetic energy ($\Delta KE$) during the acceleration — computing only one of the two terms is the usual error. A second common slip is using the wrong duration (4 s, the ramp portion only, not 8 s, the full graph) for $\theta$.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 240 to 248

    Electrical Machines › Synchronous machines

    For a synchronous motor with negligible armature resistance, the phasor equation is V = E + jXsI (current taken as flowing into the motor), so E = V - jXsI. Input electrical power equals shaft output plus internal losses, and for a leading power factor the current phasor leads the reference voltage.

    WATCH OUT FOR

    Forgetting to add the 2 kW loss to the 10 kW shaft output before computing the line current (using 10 kW instead of 12 kW input power), or using the wrong sign convention for a leading power factor (taking the current phase angle as -36.87° instead of +36.87°, since for a motor a leading power factor means the current phasor leads the voltage phasor).

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 0.61 to 0.65

    Electrical Machines › Induction machines

    An induction motor at starting behaves like a fixed impedance (magnetizing branch neglected), so its starting current is directly proportional to the terminal voltage applied to it. An ideal autotransformer of transformation ratio x reduces the motor's terminal voltage to xV and also reduces the line (supply) current by the same factor x relative to the motor's own current — so the current is stepped down twice: once by voltage reduction and once by the autotransformer's current ratio.

    WATCH OUT FOR

    Applying the ratio x only once (treating supply current = x × DOL current) instead of x² × DOL current, which gives x = 0.4 instead of the correct x ≈ 0.63 — forgetting that the autotransformer transforms current a second time between motor winding and supply line.

    Full solution →