GATE IN 2020 question paper with solutions

GATE IN 2020 answer key

General Aptitude GA Q.1 to Q.10

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

   General Aptitude › Grammar & vocabulary

   "Crocodile tears" is a fixed idiom for fake sorrow. The noun crocodile describes tears the way an adjective would, and the idiom never uses the possessive form.

   WATCH OUT FOR

   Picking (B) crocodile’s because the tears seem to "belong" to the crocodile. The idiom is always written crocodile tears, with no possessive. The other trap is picking a fox option because "unscrupulous" suggests cunning, but no idiom pairs fox with tears.

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2. GA Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   When you compare two qualities of the same subject, use "more + positive adjective ... than ...", even for a short adjective like "fast". The "-er" form (faster) is for comparing one quality across two different subjects.

   WATCH OUT FOR

   Choosing (B) "faster", because "fast" normally takes "-er". That form is right for "Archer is faster than Broad" but wrong when one person's speed is compared with his own accuracy. (D) "more faster" is a double comparative and is never correct.

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3. GA Q.3MCQ · 1 markOfficial answer: C

   General Aptitude › Reasoning

   Find the grammatical relation in the first pair (verb → the noun formed from that verb), then pick the option with the same relation to "grow". The distractors are all verb forms of "grow" (past tense, past participle) or a non-word.

   WATCH OUT FOR

   Reading "Building" only as the "-ing" form of "build" and then looking for another inflected verb form of "grow" (Grown or Grew). Other students pick "Growed" because it copies the regular "-ed" pattern. In this analogy "building" works as a noun, so the answer must be the noun "growth".

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4. GA Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Grammar & vocabulary

   "Having said that" is a concessive connector, like "nevertheless" or "even so". The speaker stands by the earlier statement and then adds a point that runs against it. Check an option by putting it in the phrase's place: the right one leaves the meaning unchanged.

   WATCH OUT FOR

   Picking (D) "contrary to" or (A) "as opposed to" because the two sentences contrast. Both of those mean the speaker is contradicting or withdrawing the first statement, but the speaker keeps it. Picking (C) "in addition to" misses the contrast altogether.

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5. GA Q.5MCQ · 1 markOfficial answer: D

   General Aptitude › Geometry

   $[x]$ is a step function, constant between consecutive integers. Split $[1,4]$ at the integers. The area is then the sum of rectangles 1 unit wide whose heights are the step values 1, 2 and 3.

   WATCH OUT FOR

   Counting the value at $x=4$ as a fourth step and getting $1+2+3+4=10$, when that point has zero width. Another slip is treating $[x]$ as $x$ and using the trapezoid ${\int }_1^{4}xdx=7.5$. A third is taking just one step, or the width 3 alone, and picking 3.

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6. GA Q.6MCQ · 2 marksOfficial answer: D

   General Aptitude › Reading

   In the passage, \"willing to invest\" means the same as \"voluntary\". All four options repeat \"on web-based platforms\", so the only thing that tells them apart is the word describing the contributions. Test that word against the passage: willing means voluntary, not unwilling or coerced, and \"smaller amounts\" rules out large.

   WATCH OUT FOR

   Choosing (B) because the passage says \"large number of people\". \"Large\" describes how many people contribute, not how much each one gives. The passage says each person gives \"smaller amounts\".

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7. GA Q.7MCQ · 2 marksOfficial answer: D

   General Aptitude › Reasoning

   Unique coding means one code per letter with no repeats. Two symbols over two positions give only $2\times 2=4$ codes. Once P takes $\alpha \alpha$ and Q takes $\alpha \beta$, R and S must share out the two codes that are left, $\beta \alpha$ and $\beta \beta$.

   WATCH OUT FOR

   Checking only that R and S get different codes from each other, and forgetting to check them against P and Q. Option (A) looks fine on that test because $\beta \alpha \ne \alpha \beta$, but it gives S the code $\alpha \beta$, which Q already has.

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8. GA Q.8MCQ · 2 marksOfficial answer: D

   General Aptitude › Numerical

   Write each term as $\frac{8}{9}(10^k-1)$, since a string of $k$ eights is $\frac{8}{9}$ of a string of $k$ nines. The sum then splits into a geometric series in powers of 10 minus $n$ copies of 1. The fastest way to decide between options like these is to put in $n=1$ and $n=2$ and check against 8 and 96.

   WATCH OUT FOR

   Forgetting to subtract the $n$ (the "$-1$" inside each $10^k-1$) gives a plus sign and points to (C). Starting the GP at $10^0=1$ instead of $10^1=10$ loses the factor 10 and gives a coefficient of $\frac{8}{81}$ instead of $\frac{80}{81}$. Mixing up $\frac{8}{9}$ and $\frac{9}{8}$ leads to (A) or (B). Putting in $n=1$ (the sum must be 8) rules out every wrong option at once.

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9. GA Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Numerical

   On $0<x<1$, raising a fraction to a power greater than 1 makes it smaller and raising it to a power less than 1 makes it larger, so $x^m<x<x^{1/m}$. $x^m$ is convex with slope 0 at the origin, $x^{1/m}$ is concave with a vertical tangent there, and as inverse functions their graphs are mirror images in $y=x$.

   WATCH OUT FOR

   Carrying over the intuition from $x>1$, where $x^m$ is the larger of the two, and so choosing (B), which has the right shapes but swapped labels. Another mistake is to check only the endpoints $(0,0)$ and $(1,1)$, which every option satisfies, and never test a point in between such as $x=\frac{1}{4}$.

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10. GA Q.10MCQ · 2 marksOfficial answer: C

    General Aptitude › Data interpretation

    "Average of success rates" means the simple mean of each school's own rate (passed / appeared x 100), with every school weighted equally. It is not the success rate of all four schools pooled together.

    WATCH OUT FOR

    Pooling the totals: total passed / total appeared = $\frac{1305}{2200}\times 100\approx 59.3\%$. That gives option (D), a distractor. Pooling weights each school by how many students it has, so the large School R (65 %) pulls the figure up. The question asks for the average of the four rates.

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Instrumentation Engineering Q.1 to Q.55

1. Q.1MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Vector calculus

   Two non-parallel vectors in the plane $z=0$ span that plane. The only direction perpendicular to both is the plane's normal, which is $\hat{k}$. The cross product shows this: $\vec{a}\times \vec{b}=(a_1{b}_2-a_2{b}_1)\hat{k}$.

   WATCH OUT FOR

   Picking an in-plane option such as $\hat{ı}$ or $-\hat{ȷ}$ because it is perpendicular to one particular vector you have in mind. The question asks for a vector perpendicular to BOTH of two non-parallel vectors in the plane, and no nonzero vector in the plane can be. Another slip is to overlook the condition $\vec{a}\ne \alpha \vec{b}$. It is what makes $\vec{a}\times \vec{b}$ nonzero, so the perpendicular direction is exactly $\hat{k}$.

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2. Q.2MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Numerical methods (IN)

   A recursion of the form $X_{n+1}=X_n+hf(X_n)$ is forward Euler for $\dot{X}=f(X)$. Rearrange the given update until it has exactly that shape, then read off $f$. The starting value $X_0$ gives the initial condition.

   WATCH OUT FOR

   Getting the sign wrong on the $X_n$ term inside the bracket. $-h(F-X_n)$ expands to $-hF+h{X}_n$, so $f=-F+X$, not $-F-X$. Students who read only the $-hF(X_n)$ part pick (A); students who drop the minus sign in front of $h$ pick (C) or (D).

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3. Q.3MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Linear algebra

   For an $m\times n$ matrix $A$ with $m<n$, the $n\times n$ matrix $A^{T}A$ has rank $\rho (A)\le m<n$, so it is always singular. An underdetermined system (fewer equations than unknowns) can never have a unique solution, and the normal equations cannot change that.

   WATCH OUT FOR

   Carrying over the least-squares recipe $x=(A^{T}A{)}^{-1}{A}^{T}b$ without checking when it applies. It gives a unique answer only when $A$ has full column rank, which needs at least as many rows as columns ($m\ge n$). A second trap is picking (C) or (D) because they sound like numerical-analysis caveats: conditioning describes how sensitive an invertible matrix is to small errors, but here $A^{T}A$ is exactly singular, with infinite condition number, for every $2\times 4$ matrix $A$.

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4. Q.4MCQ · 1 markOfficial answer: C

   Networks (Electric Circuits) › Laws & node/mesh

   In a series circuit every element carries the same current. Work out each resistor's own current limit, $I_{\max }=\sqrt{P_{\text{rated}}/R}$. The safe current is the smallest of these limits, because the most restrictive element decides it.

   WATCH OUT FOR

   Adding the power ratings (1.75 W) and the resistances (103 Ω) and computing $\sqrt{1.75/103}\approx 0.13\text{A}$. That current would overload the 100 Ω resistor, since each rating applies only to its own resistor. Another mistake is taking the largest limit (0.5 A, option B) when the smallest one governs.

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5. Q.5MCQ · 1 markOfficial answer: C

   Signals and Systems › Fourier transform

   Differentiating in time multiplies the transform by $s$ (by $j\omega$ on the frequency axis). So an ideal differentiator has $H(j\omega )=j\omega$. Its magnitude $\omega$ grows in direct proportion to frequency, and its phase stays fixed at $+90^{\circ }$.

   WATCH OUT FOR

   Mixing up magnitude and phase and picking (A). The phase of $j\omega$ is a constant $+90^{\circ }$, not linear in $\omega$; linear phase belongs to a pure time delay $e^{-j\omega t_0}$. The other common slip is swapping differentiator and integrator: $|H|\propto 1/\omega$ (option D) is the integrator $1/s$.

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6. Q.6MCQ · 1 markOfficial answer: A

   Control Systems › Compensators & PID

   For a lead network $\frac{1+Ts}{1+\alpha Ts}$ with $\alpha <1$, the phase peaks at the geometric mean of the zero and pole corner frequencies: ${\omega }_m=\sqrt{{\omega }_z{\omega }_p}=\frac{1}{T\sqrt{\alpha }}$. On a log-frequency axis this is the midpoint between the two corners.

   WATCH OUT FOR

   Picking the pole corner, $1/0.05=20\text{rad/s}$ (option B), or taking $1/(0.2\times 0.05)=100$ (option C), which is ${\omega }_m^2$ and needs a square root. The other slip is using the arithmetic mean $(5+20)/2=12.5$ in place of the geometric mean.

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7. Q.7MCQ · 1 markOfficial answer: A

   Analog Electronics › Diode circuits

   The Zener's breakdown voltage (5 V) equals the battery voltage (5 V). When the series diode is off, the battery sets the node at 5 V through the 100 Ω resistor with no current flowing. When the diode is on, the ideal Zener breaks down and clamps the node at 5 V, taking all the source current. Either way the voltage across the 100 Ω resistor is zero, so it carries no current.

   WATCH OUT FOR

   Ignoring the Zener clamp and putting the source peak minus the battery, $(10-5)=5\text{V}$, across the $100\Omega$ resistor gives $25/100\text{W}$ (option C). Reading the battery polarity backwards, $(10+5)=15\text{V}$, gives $225/100\text{W}$ (option D). The Zener pins the node at exactly the battery voltage, so the resistor never sees any voltage across it.

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8. Q.8MCQ · 1 markOfficial answer: B

   Digital Electronics › Boolean logic

   Put the don't-cares into the K-map as 1s wherever that makes a group bigger. Here the only 0s are $4,5,12,13$, and all four have $B=1$ and $C=0$. So $\overline{f}=B\overline{C}$, and $f=\overline{B}+C$.

   WATCH OUT FOR

   Forgetting the don't-cares, especially 3, 7, 14 and 15. Without them the octets break into quads and pairs, which gives a longer expression that is not in the options. Another mistake is picking (A) $A\overline{B}+CB$ because it looks close. It misses minterms 0, 1 and 2.

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9. Q.9MCQ · 1 markOfficial answer: B

   Electrical and Electronic Measurements › Bridges

   A Q meter is a series-resonant circuit made of the coil under test and a calibrated tuning capacitor. At resonance $V_C=QE$, so it reads a coil's $Q$ directly. Resonating the same coil at $f_1$ and at $2f_1$ gives its distributed capacitance from two dial readings: $C_d=(C_1-4C_2)/3$.

   WATCH OUT FOR

   Picking (A) because the instrument is named after the quality factor. A Q meter reads a coil's Q directly. A low-loss capacitor has a much higher Q, which the meter can only find indirectly from the small difference between two readings, and that result is much less accurate. The standard "direct" uses of a Q meter are a coil's Q, its inductance and its distributed capacitance.

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10. Q.10MCQ · 1 markOfficial answer: B
11. Q.11MCQ · 1 markOfficial answer: A

    Digital Electronics › Boolean logic

    Shading "A only" plus "B only" is the exclusive-OR, $A\overline{B}+\overline{A}B$. The options are in product-of-sums form, so build the POS from the regions that are NOT shaded: outside both circles gives the maxterm $(A+B)$ and the overlap gives $(\overline{A}+\overline{B})$. Their product is the answer.

    WATCH OUT FOR

    Picking option (B), $(A+\overline{B})(\overline{A}+B)$, because it looks like a "mixed" XOR pattern. Multiplied out it is $AB+\overline{A}\overline{B}$, the XNOR, which is exactly the unshaded area. The other trap is putting the maxterms for the shaded regions into the POS instead of the maxterms for the unshaded ones.

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

    Digital Electronics › Combinational

    A MUX with fixed data inputs is just a truth table. Input $I_k$ is the output for select combination $k$ (with $S_1$ as MSB), so once you know the four input values you can read the function straight off. Here the inputs are 0, 1, 1, 0 for select values 00, 01, 10, 11, which is XOR.

    WATCH OUT FOR

    Losing track of the inverter count on each wire. $I_0$ goes through two inverters, so it is back to 0, not 1. $I_1$ and $I_2$ come off the middle node after one inverter, so they are 1. If you treat $I_0$ as 1, you get $0,1,1,1$ and pick OR, option (C). Swapping the MSB and LSB would not change this answer, because XOR is symmetric and $I_1=I_2$, but it does matter in general.

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

    Signals and Systems › Signals basics

    Check the samples before applying any periodicity rule. At integer $n$, $\sin (2\pi n)=0$, so $x[n]=\sin (2\pi n)u[n]$ is identically zero. A discrete-time period must be a positive integer, and the zero sequence satisfies $x[n+N]=x[n]$ for every $N$, so the fundamental period is the smallest positive integer, $N_0=1$. The rule ${\omega }_0/2\pi =1=1/1$ gives the same $N=1$.

    WATCH OUT FOR

    The usual mistake is to call the signal aperiodic because $u[n]$ switches it on at $n=0$. That argument is right for a one-sided sinusoid with a non-zero sample somewhere, for example $\sin (\pi n/2)u[n]$. It does not apply here: $\sin (2\pi n)$ is zero at every integer, so $x[n]\equiv 0$ and $u[n]$ has no effect. Another slip is to reason about the continuous curve $\sin (2\pi t)$ without looking at the samples, which hides the fact that the sequence is all zeros.

    The question and its official answer →
14. Q.14MCQ · 1 markOfficial answer: D

    Control Systems › Bode & margins

    For a stable LTI system, a sinusoid at $\omega$ produces a steady-state sinusoid at the same $\omega$. Its amplitude is multiplied by $|G(j\omega )|$ and its phase is shifted by $\angle G(j\omega )$. Here $G(j1)=\frac{1}{1+j}$, so the amplitude becomes $\frac{1}{\sqrt{2}}$ and the phase shifts by $-\frac{\pi }{4}$.

    WATCH OUT FOR

    Getting the sign of the phase wrong and picking (C). A pole in the denominator makes the phase negative: $\angle \frac{1}{1+j}=-{\tan }^{-1}1=-\frac{\pi }{4}$, so the output lags the input. Picking (A) is another mistake: it uses the DC gain $G(0)=1$, which only applies to a step input, not to a sinusoid.

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15. Q.15MCQ · 1 markOfficial answer: B

    Engineering Mathematics › Complex analysis

    The only pole, $z=-a$, is the centre of the circle, so it is enclosed for every radius. Its residue is 1, and an anti-clockwise loop gives $2\pi i\times 1$. This is the standard result $\oint \frac{dz}{z-z_0}=2\pi i$ for any positively oriented circle around $z_0$.

    WATCH OUT FOR

    Two traps. Some candidates pick 0 because they take the circle to be centred at the origin, or think a small $R$ could leave the pole outside; the pole is the centre, so it is always inside. Others pick $-2\pi i$ by getting the orientation wrong: anti-clockwise is the positive direction and gives $+2\pi i$.

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16. Q.16NAT · 1 markOfficial answer: 0.09

    Engineering Mathematics › Probability & statistics

    Read "only in the second attempt" as one specific sequence: a miss on the first throw, then a basket on the second. Independence lets you multiply the two probabilities: $P(F_1)P(S_2)=(1-p)p$. This is the geometric probability that the first success comes on trial 2.

    WATCH OUT FOR

    Answering $0.1$ by ignoring the requirement that the first throw misses. Another error is answering $0.18$, which is $P(\text{exactly one success in two throws})=2\times 0.9\times 0.1$ and wrongly includes the case where the ball goes in on the first throw only. A third is answering $0.01$, which is the probability of scoring on both throws.

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17. Q.17NAT · 1 markOfficial answer: 7

    Analog Electronics › Op-amps

    An ideal op-amp with negative feedback forces its (−) input to the voltage on its (+) input. That fixes node A at 3 V and node B at 2 V. With both ends of every resistor at node A known except $V_1$, KCL at node A alone gives $V_1$. The 5 kΩ and 3 kΩ resistors only decide the right op-amp's output ($-4\text{V}$).

    WATCH OUT FOR

    There are three common mistakes. (1) Reading the left stage as a plain non-inverting amplifier with gain $1+1/1=2$, which gives $V_1=6\text{V}$. This ignores the top 1 kΩ, which carries 1 mA from A to B. (2) Treating node B as a virtual ground (0 V) instead of 2 V, which gives $V_1=9\text{V}$. (3) Writing KCL at the $V_1$ node, where the op-amp's output current is unknown, instead of at the op-amp's input node, where the input current is zero.

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18. Q.18NAT · 1 markOfficial answer: 1.7 to 1.8

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

    In a balanced delta load, each phase is connected across a line-to-line voltage, so $V_{ph}=V_L$. Each line current is the phasor difference of two phase currents that are $120^{\circ }$ apart, which makes it $\sqrt{3}$ times as large: $I_L=\sqrt{3}{I}_{ph}$.

    WATCH OUT FOR

    Using the star relations for a delta load. Some candidates divide the voltage by $\sqrt{3}$ (taking $V_{ph}=400/\sqrt{3}=230.9\text{V}$, which gives $I_{ph}=0.577\text{A}$). Others forget the $\sqrt{3}$ factor and report the phase current of $1\text{A}$ as the line current. Both answers fall outside the key.

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19. Q.19NAT · 1 markOfficial answer: 2

    Signals and Systems › Fourier transform

    A Fourier transform at zero frequency is the area under the signal: $X(j0)={\int }_{-\infty }^{\infty }x(t)dt$. For the two-sided exponential $e^{-|t|}$, that area is $2$. The pair $e^{-a|t|}\leftrightarrow \frac{2a}{a^2+{\omega }^2}$ confirms it.

    WATCH OUT FOR

    Integrating only over $t\ge 0$, as if the signal were $e^{-t}u(t)$. That gives $1$ instead of $2$. The absolute value makes the signal two-sided, so the left half adds an equal area. A second slip is using the one-sided pair $\frac{1}{1+j\omega }$, which also gives $1$ at $\omega =0$.

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20. Q.20NAT · 1 markOfficial answer: 0.78 to 0.79

    Control Systems › Time response

    Peak time depends only on the imaginary part of the closed-loop poles: $t_p=\pi /{\omega }_d$. For poles at $-3\pm j4$, ${\omega }_d$ is simply 4, so there is no need to compute $\zeta$ or ${\omega }_n$ first.

    WATCH OUT FOR

    Using ${\omega }_n=5$ in place of ${\omega }_d=4$ gives $\pi /5=0.63\text{s}$, which is wrong. The other common slip is using the real part: $\pi /3=1.05\text{s}$ is also wrong. Only the damped frequency ${\omega }_d$, the imaginary part of the pole, sets the peak time.

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21. Q.21NAT · 1 markOfficial answer: -4

    Analog Electronics › Op-amps

    With an ideal op-amp in linear operation, the input node sits at $V_x$ and the non-inverting gain fixes $V_o=(1+2/1)V_x=3V_x$. The $8\text{k}\Omega$ feedback resistor is the only path for $I_x$. The far end of that resistor is driven to three times the input voltage, so current flows out of the source terminal instead of in, and the resistance seen at the input is negative: $R_{in}=-R_f/(A-1)$.

    WATCH OUT FOR

    Seeing an 8 kΩ resistor from the input to the output and treating it as a load to ground, which gives $+8\text{k}\Omega$. Another slip is to use $V_o=-2V_x$, the inverting gain, because the divider looks like an inverting amplifier. The source drives the $+$ input, so the gain is $+3$, the resistor current is $(V_x-3V_x)/8\text{k}\Omega$, and the sign of the answer must be negative.

    Full solution →
22. Q.22NAT · 1 markOfficial answer: 25

    Signals and Systems › Sampling

    Sampling a tone $f_0$ at rate $f_s$ creates lines at every $|k{f}_s\pm f_0|$, not only at $f_0$. The ideal LPF then passes every line with $|f|\le f_c$. Here the lines are 5, 10, 20, 25, 35, ... kHz. The highest one the filter lets through is $f_s+f_0=25$ kHz, which falls exactly on the cut-off, and the cut-off is part of the passband.

    WATCH OUT FOR

    There are three common mistakes. (1) Answering 10 kHz, as if the filter recovers the original tone. That only happens when $f_s\ge 2f_0$ and $f_c$ lies between $f_0$ and $f_s-f_0$, and neither is true here. (2) Answering 5 kHz, keeping only the alias $f_s-f_0$ and forgetting that the 25 kHz cut-off also passes the higher copies. (3) Answering 20 kHz by rejecting the 25 kHz line because it sits exactly on the cut-off. An ideal LPF passes $|f|\le f_c$, the edge included, and the official key (25) uses this convention.

    The question and its official answer →
23. Q.23NAT · 1 markOfficial answer: 223.3 to 223.5

    Digital Electronics › ADC & DAC

    A dual-slope ADC integrates the input for a fixed time $T_1$, then integrates the opposite-polarity reference until the integrator output returns to zero. Because both ramps cover the same voltage, $T_2=T_1{V}_{in}/V_{ref}$, and $RC$ drops out. In continuous mode, one conversion lasts the whole cycle $T_1+T_2$, so the conversion time depends on the input voltage.

    WATCH OUT FOR

    Giving only the run-down time $T_2=123.4\text{ms}$, or only the fixed run-up $T_1=100\text{ms}$, as the conversion time. A conversion needs both phases, so the answer is $T_1+T_2=223.4\text{ms}$. A second error is using the full-scale value (200 mV) in place of $V_{ref}$ (100 mV) in $T_2=T_1{V}_{in}/V_{ref}$, which gives $T_2=61.7\text{ms}$ and a wrong total of 161.7 ms. Another is taking the worst case, $T_1+T_{2,\max }=300\text{ms}$, when the question asks about one particular input.

    Full solution →
24. Q.24NAT · 1 markOfficial answer: 5
25. Q.25NAT · 1 markOfficial answer: 2.5, or 3.5

    Electrical and Electronic Measurements › Errors & standards

    For a sum of quantities, the absolute limiting errors add: 3 Ω + 4 Ω = 7 Ω on a 200 Ω total. So the relative error of a series combination is the resistance-weighted average of the individual tolerances, not their sum. Treating the errors as independent and combining them by root-sum-square (5 Ω) gives 2.5 % instead, and the key accepts both.

    WATCH OUT FOR

    Adding the percentages directly (3 % + 4 % = 7 %). Percentage errors add only for a product or a quotient. For a sum you add the absolute errors in ohms and then divide by the total resistance, which gives 7/200 = 3.5 %.

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

    Engineering Mathematics › Linear algebra

    Test each option directly against $Mv=\lambda v$: one value of $\lambda$ has to work in every entry. A useful shortcut: if every row of a matrix has the same sum $s$, then ${\left[\begin{array}{cccc}1& 1& \cdots & 1\end{array}\right]}^T$ is an eigenvector with eigenvalue $s$. Here $s=0$.

    WATCH OUT FOR

    Stopping once one entry of $Mv$ matches a multiple of $v$. For option (A) the first and third entries of $Mv=(2,4,2{)}^T$ fit $\lambda =2$, but the middle entry does not. Every entry has to give the same $\lambda$. A second slip is to reject (D) because $Mv=0$. The zero vector is not allowed as an eigenvector, but $\lambda =0$ is a perfectly valid eigenvalue, and it occurs exactly when $M$ is singular.

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

    Engineering Mathematics › Differential equations

    In an autonomous equation $\frac{dx}{dt}=f(x)$, every root of $f(x)=0$ gives a constant solution. If the initial value sits on such a root and $f$ is Lipschitz, uniqueness says the solution never leaves it. Here $\sin 0=0$, so $x(t)\equiv 0$.

    WATCH OUT FOR

    Reading $\sin (x)$ as $\sin (t)$ and integrating gives $x=1-\cos t$. That is not an option, and it tempts students to pick (B) or (C) by pattern. A second trap: separating variables as $\frac{dx}{\sin x}=dt$ divides by $\sin x$, which is zero at $x=0$. That throws away exactly the equilibrium solution the initial condition picks out, and $\ln \left|\tan \frac{x}{2}\right|=t+C$ then has no finite $C$ for $x(0)=0$.

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

    Engineering Mathematics › Calculus

    A line with x-intercept $a$ and y-intercept $b$ (both non-zero) is $\frac{x}{a}+\frac{y}{b}=1$. Its slope is $-\frac{b}{a}$ and the y-intercept is $b$ itself. Here the slope is $-\frac{1}{-0.5}=2$ and $c=1$.

    WATCH OUT FOR

    Taking the slope as $b/a$ or $a/b$ without the minus sign, or using the x-intercept value $-0.5$ as the slope. Both lead to option (A) $y=-0.5x+1$. That option has the right y-intercept, but it cuts the x-axis at $x=2$. Always check both intercepts of the option you pick.

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

    Control Systems › Nyquist

    The pole at $s=0$ maps to an infinite semicircle, and here that semicircle sweeps clockwise through the left half-plane, from $-j\infty$ via $-\infty$ to $+j\infty$. The finite branches never go left of $\text{Re}=-\frac{1}{4}$, so $-1+j0$ sits between them and the infinite arc and is enclosed once, clockwise. You can confirm this from $N=P-Z$ with $P=1$ (the pole at $s=2$) and $Z=2$ (the closed-loop double pole at $s=1$).

    WATCH OUT FOR

    The usual slip is to draw only the $\omega >0$ polar plot. It never crosses the negative real axis, so people answer (C) "does not encircle". Another route to (C) is closing the plot through the right half-plane: for this type-1 system with a right-half-plane pole, the indentation maps to an arc through $-\infty$, not $+\infty$. Others see an open-loop right-half-plane pole ($P=1$) and assume one counterclockwise encirclement (B). That is only what stability would need. The actual count is $N=P-Z=1-2=-1$.

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

    Networks (Electric Circuits) › Laws & node/mesh

    When every mesh current in a resistive network without dependent sources is drawn in the same direction, you can write the mesh equations by inspection. The resistance matrix is symmetric. Each diagonal entry is the total resistance around that mesh and is always positive. Each off-diagonal entry is minus the resistance the two meshes share. Each right-hand-side entry is the net source voltage rise met when you go around the mesh in the direction of its current.

    WATCH OUT FOR

    Writing $+V_3$ instead of $-V_3$. The clockwise current $I_3$ goes down through $V_3$ from its + terminal to its − terminal, so $V_3$ is a drop in that mesh and appears as $-V_3$ on the source side. Another mistake is accepting a negative diagonal entry (options B and C). A self-resistance can never be negative, and the matrix must be symmetric. Those two checks rule out B and C straight away.

    Full solution →
31. Q.31NAT · 2 marksOfficial answer: 0.5

    Engineering Mathematics › Calculus

    A constrained extremum on a line becomes a one-variable problem once you substitute the constraint ($y=1-x$), which gives a parabola whose vertex is the minimum. Lagrange multipliers give the same point. Geometrically, $x^2+y^2$ is the squared distance from the origin, so the answer is the square of the perpendicular distance from the origin to the line, ${\left(\frac{1}{\sqrt{2}}\right)}^2=0.5$.

    WATCH OUT FOR

    Giving the distance $\frac{1}{\sqrt{2}}\approx 0.7$ instead of its square: $f$ is the squared distance, so the answer is $0.5$. Another slip is to find the critical point $x=y=\frac{1}{2}$ and then enter the coordinate $0.5$ without checking that it is the value of $f$ there. It happens to be the same number here, but only by coincidence.

    Full solution →
32. Q.32NAT · 2 marksOfficial answer: 0.7

    Engineering Mathematics › Probability & statistics

    This is Bayes' theorem. The two bags are equally likely to be picked, so the priors cancel. The posterior $P(B_1\mid R)$ is then bag B1's share of all the red balls: $\frac{7}{7+3}=0.7$.

    WATCH OUT FOR

    The common slip is to answer $P(R)=0.5$, the overall chance of drawing a red ball, or to multiply out only the numerator, $P(B_1)P(R\mid B_1)=0.35$, and forget to divide by $P(R)$. Both confuse the joint or total probability with the conditional probability $P(B_1\mid R)$.

    Full solution →
33. Q.33NAT · 2 marksOfficial answer: 1

    Networks (Electric Circuits) › AC, power & resonance

    At $\omega =1000\text{rad/s}$ the series $L$ and $C$ have equal reactances ($100\Omega$ each), so the circuit is at series resonance. The two reactances cancel, only the $100\Omega$ resistor limits the current, and $I_{\text{rms}}=V_{\text{rms}}/R=100/100=1\text{A}$.

    WATCH OUT FOR

    Using the peak voltage $100\sqrt{2}\text{V}$ in place of the rms value gives $1.414\text{A}$, which is the peak current, not the rms current asked for. A second slip is adding the reactances ($X_L+X_C$) when they should be subtracted. That gives $|Z|=\sqrt{100^2+200^2}$ and a current that is far too small.

    Full solution →
34. Q.34NAT · 2 marksOfficial answer: 115 to 116

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

    In an unbalanced star load with no neutral, the star point floats. Find its shift from the source neutral with Millman's theorem, then subtract it from the phase voltage. Here the R arm is 75 Ω against 300 Ω in each of the other two arms, which puts the star point exactly halfway to V_R, so the R arm gets V_R/2.

    WATCH OUT FOR

    The usual error is to assume the star point stays at the source neutral and answer 230.94 V (the full phase voltage). That ignores the neutral shift an unbalanced, ungrounded star produces. A second error is to put only part of the arm voltage across the 100 Ω by treating it as in series with the 300 Ω, when the two resistors are actually in parallel.

    Full solution →
35. Q.35NAT · 2 marksOfficial answer: 0

    Signals and Systems › LTI & convolution

    $g[n]$ is the unit impulse $\delta [n]$, and convolving any sequence with $\delta [n]$ gives the sequence back: $h[n]\otimes \delta [n]=h[n]$. So $y[2]$ is simply $h[2]$, which is 0 because 2 is not a multiple of 3.

    WATCH OUT FOR

    Not noticing that $g[n]$ is just $\delta [n]$ and working through the convolution sum term by term, which risks errors with the flipped index. A second mistake is misreading $h[n]$ as 1 at every $n$ up to 9, or as periodic in both directions, and answering 1. It is non-zero only at $n=0,3,6,9,\dots$

    Full solution →
36. Q.36NAT · 2 marksOfficial answer: 5.0 to 5.1

    Control Systems › Root locus

    Find the breakaway point from $dK/ds=0$, where $K$ comes from the characteristic equation. Keep only the root that lies on a real-axis segment of the locus. Then get the gain there from the magnitude condition $K=\prod |s_b-p_i|$.

    WATCH OUT FOR

    Stopping at the breakaway point $s_b\approx -0.90$ and giving that as the answer, when the question asks for $K$. Another slip is picking the root $s=-4.43$, which lies between $-2$ and $-6$. That segment is not on the locus, and that root gives a negative $K$.

    Full solution →
37. Q.37NAT · 2 marksOfficial answer: 4

    Control Systems › Time response

    With unity feedback around a type-0 first-order plant, the closed-loop DC gain is $\frac{K}{1+K}$. The step response settles at $10\cdot \frac{K}{1+K}$, not at $10K$. Set this equal to the asymptote in the plot, 8, and solve for $K$. The time constant $\tau$ only changes how fast the response rises, not where it settles.

    WATCH OUT FOR

    Using the open-loop gain as though the loop were open, $10K=8$ giving $K=0.8$, which ignores the feedback path. Another error is reading the 8 as the steady-state error rather than the output, which gives $\frac{10}{1+K}=8$ and $K=0.25$. Some candidates also try to find $\tau$ from the curve, but the plot has no time scale and $\tau$ is not asked for.

    Full solution →
38. Q.38NAT · 2 marksOfficial answer: 1

    Analog Electronics › Op-amps

    This is an inverting precision half-wave rectifier. In one half-cycle the output is held at 0 V. In the other it is $-(R_f/R_1)v_i$, with gain $10/20=0.5$. The diode drops sit inside the feedback loop, so they have no effect. A PMMC meter with a 1 Hz bandwidth reads only the DC (average) value, $V_m/\pi$ for a half-wave rectified sine.

    WATCH OUT FOR

    Common mistakes: forgetting the gain of 0.5 and using the source peak $2\pi \text{V}$, which gives 2 V; using the full-wave average $2V_m/\pi$ (2 V); or taking the RMS value $V_m/2$ ($\pi /2\approx 1.57\text{V}$). A PMMC meter reads the average, and a half-wave output averages to $V_m/\pi$.

    Full solution →
39. Q.39NAT · 2 marksOfficial answer: 0.45 to 0.55

    Analog Electronics › Op-amps

    An integrator inside a negative-feedback loop drives its own input to zero at DC. That input is op-amp 1's output, so op-amp 1's inputs end up at the same voltage: $V_{+}=V_{-}=0$. The divider voltage $V_O/1000$ must therefore exactly cancel $V_X$, so $V_O=1000V_X$. The integrator enforces this even when $A$ is finite, so the result does not depend on $A$.

    WATCH OUT FOR

    One mistake is reading the source polarity backwards. Its $-$ sign is drawn as a short vertical bar because the source is drawn horizontally, and a backwards reading gives $V_O=-0.50\text{V}$. Another is treating op-amp 1 as an open-loop comparator that saturates. A third is assuming a DC input makes the integrator output ramp forever, when in fact the loop drives the integrator's input to zero. Students also sometimes load the divider with a current through $V_X$, but that current is zero because $V_X$ feeds an op-amp input. Always check that the overall loop is negative feedback before using the virtual short.

    Full solution →
40. Q.40NAT · 2 marksOfficial answer: 7

    Digital Electronics › Sequential

    Write down the excitation equations from the gates ($T_A=\overline{xB}$, $T_B=x$). Then step through a state table one clock at a time, working each T input out from the present state and that cycle's $x$, and using $Q^{+}=T\oplus Q$. The output $y=A+B$ is combinational, so read it from the state after the 3rd clock.

    WATCH OUT FOR

    The usual slips: (i) reading the gate as an AND and forgetting the bubble, which gives $T_A=xB$ and a different sequence; (ii) working a cycle's $T_A$ out from the value B will have after the edge instead of its present value (at clock 1, for example, $T_A$ uses $B=1$, not the new $B=0$); (iii) taking $y$ from the state before the 3rd clock, or leaving $y$ out of the 3-bit word.

    Full solution →
41. Q.41NAT · 2 marksOfficial answer: 3.5 to 4.5

    Digital Electronics › ADC & DAC

    The two DAC LSBs ($A_1$, $A_0$) are grounded and the processor drives the upper eight bits, so the 10-bit code is always $4\times$ the data word. A one-count change in the data therefore moves the DAC by $2^2=4$ steps of $1.023/1023\text{V}=1\text{mV}$ each.

    WATCH OUT FOR

    Treating the change as one DAC step (1 mV), which ignores that $D_0$ drives $A_2$, not $A_0$. Another slip is to work out the step from an 8-bit DAC ($1.023/255\text{V}\approx 4.01\text{mV}$). That lands inside the key range by luck, but it describes the wrong hardware.

    Full solution →
42. Q.42NAT · 2 marksOfficial answer: 199 to 201

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

    In the two-wattmeter method, $W_1=V_L{I}_L\cos (30^{\circ }+\varphi )$ and $W_2=V_L{I}_L\cos (30^{\circ }-\varphi )$. If one meter reads zero, then $\varphi =60^{\circ }$ (power factor 0.5), and the other meter carries all the power: $W_2=V_L{I}_L\cos 30^{\circ }=\sqrt{3}{V}_L{I}_L\cos 60^{\circ }$.

    WATCH OUT FOR

    One common mistake is using the phase voltage $400/\sqrt{3}$ in the meter formula instead of the line voltage. Each pressure coil sees a line-to-line voltage, and using phase voltage gives about $115.5\text{W}$. Another is forgetting that one meter at zero means $\varphi =60^{\circ }$ and the other meter reads the whole power. That leads to wrongly halving the total, or using $\cos \varphi =1$ and getting $400\text{W}$.

    Full solution →
43. Q.43NAT · 2 marksOfficial answer: 999 to 1001

    Electrical and Electronic Measurements › Time, frequency & scopes

    A timer-counter shows its reading as a number in the selected range unit. Turn the period reading into seconds, take the reciprocal to get the frequency, then write that frequency in the new range unit (Hz).

    WATCH OUT FOR

    Using the displayed number without its unit and writing $f=1/10^6=10^{-6}$. Another error is treating ns as $10^{-6}\text{s}$ (that is microseconds), which gives 1 Hz. The reading $10^6\text{ns}$ is $10^{-3}\text{s}$, so the frequency is $1000\text{Hz}$.

    Full solution →
44. Q.44NAT · 2 marksOfficial answer: 3.4 to 3.6

    Analog Electronics › Op-amps

    This is a charge amplifier. With the + input biased at $V_{CC}/2=2.5\text{ V}$, the virtual short puts the sensor's own capacitance and the cable capacitance ($10+100\text{ pF}$) at zero volts across, so they carry no current. All of the sensor charge goes onto $C_f$, and $v_o=V_{CC}/2-q_p/C_f$. Biasing at mid-supply lets a single-supply op-amp handle a bipolar charge signal. The peak is the bias plus $q_{p,\max }/C_f=1\text{ V}$.

    WATCH OUT FOR

    Reading the ground symbol under the op-amp as the + input's reference. That ground is the op-amp's negative supply pin; the + input is wired to the 2.5 V rail. With the + input taken as 0 V you get $v_o=\mp \sin (10000\pi t)$, which clips at the 0 V rail and gives a wrong 1.0 V. Another mistake is dividing by $C_f+110\text{ pF}$: the input capacitors have zero voltage across them under the virtual short, so they take no charge. The phrase '10 cycles' is a distractor. With zero-mean charge nothing accumulates, and the peak is the same in every cycle.

    Full solution →
45. Q.45NAT · 2 marksOfficial answer: 0.95 to 1.05
46. Q.46NAT · 2 marksOfficial answer: 27.5 to 28.5

    Electromagnetics › Waveguides & fibres

    For maximum coupling, every ray of the focused cone must lie inside the fibre's acceptance cone. The steepest ray is the marginal ray, at $\tan \theta =a/f$ where $a$ is the beam radius. Setting $\theta ={\theta }_a$ with $\sin {\theta }_a=\text{NA}=\sqrt{n_1^2-n_2^2}$ gives the shortest focal length that still couples the whole beam: $f_{\min }=a/\tan {\theta }_a$.

    WATCH OUT FOR

    Using the full diameter (10 mm) where the geometry needs the radius (5 mm). That doubles the answer to about 56 mm. Other slips: forgetting the square root in $\text{NA}=\sqrt{n_1^2-n_2^2}$; writing $\text{NA}=n_1-n_2$; treating the lens index 1.5 as part of the NA. Setting $\sin {\theta }_a=a/f$ instead of $\tan {\theta }_a=a/f$ gives about 28.4 mm, which happens to fall inside the key range, but the ray geometry is a tangent, not a sine.

    Full solution →
47. Q.47NAT · 2 marksOfficial answer: 200

    Electromagnetics › Plane waves

    When the slab has the same medium (air) on both sides, only one of the two reflections picks up a $\pi$ phase shift. That shift already makes the two rays opposite in phase, so for destructive interference the round trip $2n_{2}t$ must be a whole number of wavelengths: $2n_{2}t=m\lambda$. The smallest non-zero thickness is a half-wave layer, $t=\lambda /(2n_2)$.

    WATCH OUT FOR

    One mistake is to forget the $\pi$ shift at the top face, or to count a $\pi$ shift at both faces. Either way you get $2n_{2}t=\lambda /2$ and the wrong answer $t=\lambda /(4n_2)=100\text{nm}$, which is the condition for constructive reflection here. Another is to use the free-space path $2t$ instead of the optical path $2n_{2}t$, which gives $300\text{nm}$.

    Full solution →
48. Q.48NAT · 2 marksOfficial answer: 0

    Signals and Systems › DTFT & DFT

    A flat $N$-point spectrum $X(k)=1$ is the DFT of the unit impulse $\delta (n)$, so its IDFT is $(1,0,\dots ,0)$. Worked out term by term, $x(2)$ is $\frac{1}{3}$ times the sum of the three cube roots of unity, and that sum is zero.

    WATCH OUT FOR

    Leaving out the $\frac{1}{N}$ factor or using the forward-DFT sign $e^{-j2\pi kn/N}$. For this spectrum neither changes $x(2)$, but both give the wrong $x(0)$ (3 instead of 1 without the factor). Another mistake is to assume that a constant spectrum gives a constant sequence $(1,1,1)$, or $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. A constant sequence has a spectrum that is an impulse. A constant spectrum belongs to an impulse sequence.

    Full solution →
49. Q.49NAT · 2 marksOfficial answer: 14.5 to 15.0

    Networks (Electric Circuits) › Theorems

    Short the independent DC source, reduce the passive ladder (7 µF parallel with 3 µF, then in series with 4 µF, giving 40/14 µF), and use Miller's theorem on the feedback capacitor: at the input side it looks like $C(1-G)$ to ground, which is 12 µF for G = −5. The two capacitances at A are in parallel.

    WATCH OUT FOR

    Three slips are common. (1) Getting the Miller sign wrong. Writing $C(1+G)$ with G = −5 gives a negative capacitance, and using the output-side value $C(1-1/G)=2.4\mu \text{F}$ gives about 5.3 µF. (2) Treating the source as an open circuit, so the 7 µF branch drops out and the passive part becomes $4\times 3/7=1.71\mu \text{F}$, for a total of about 13.7 µF. (3) Putting the 7 µF and 3 µF in series instead of in parallel once the source is shorted.

    Full solution →
50. Q.50NAT · 2 marksOfficial answer: -0.13 to -0.11

    Control Systems › State space

    With no input, the state response is $X(s)=(sI-A{)}^{-1}x(0)$, or equivalently $x(t)=e^{At}x(0)$. Since $x(0)=[01{]}^T$, $x(t)$ is just the second column of the state-transition matrix $\Phi (t)$. The eigenvalues $-2$ and $-3$ (the roots of $s^2+5s+6$) tell you in advance which exponentials will appear.

    WATCH OUT FOR

    The common mistakes are these: picking the first column of $(sI-A{)}^{-1}$ instead of the second, which gives $x_2=-6/((s+2)(s+3))$ and $x_2(1)\approx -0.51$; getting a sign wrong in the adjugate (the off-diagonal entries must become $+1$ and $-6$); and reporting $x_1(1)=e^{-2}-e^{-3}\approx 0.09$ when the question asks for $x_2(1)$. Also note that $x_2(1)$ comes out negative even though $x_2(0)=1$. The fast $e^{-3t}$ term dies away before the slower $-2e^{-2t}$ term does, so the answer really is negative. Do not drop the sign.

    Full solution →
51. Q.51NAT · 2 marksOfficial answer: 1.8

    Analog Electronics › Diode circuits

    On the right, the three diodes form an OR circuit that holds node Y at the highest source, 15 V. On the left, guess which diodes are ON, solve KCL at node X, then check every diode: an ON diode must carry forward current, and an OFF diode must be reverse biased. With both left diodes ON, the 12 V branch would need negative current. So only the 6 V diode conducts, $V_X=9.6\text{V}$ and $I_x=(15-9.6)/3=1.8\text{mA}$.

    WATCH OUT FOR

    Assuming both left-hand diodes conduct and stopping at $V_X=10.5\text{V}$, which gives $I_x=1.5\text{mA}$. That state needs $-0.75\text{mA}$ through the 12 V diode, which an ideal diode cannot carry. A second mistake is misreading the diode directions: the right-hand diodes pass current INTO node Y (so $V_Y$ is set by the highest source, 15 V, not the lowest), and the left-hand diodes pass current OUT of node X.

    Full solution →
52. Q.52NAT · 2 marksOfficial answer: 0.95 to 1.05

    Electrical and Electronic Measurements › Instrument transformers

    A UPF burden with a resistive secondary gives $\delta =0$. The reflected secondary current $n{I}_s$ then lies along the core-loss component $I_e$, and only the magnetizing component $I_m$ sits in quadrature. So the phase error fixes $I_m$ directly: $I_m=I_p\sin \theta \approx n{I}_s\theta$. The ratio error fixes $I_e$. Remember that $3.438$ minutes is exactly $1\text{mrad}$.

    WATCH OUT FOR

    The usual slips are computing $I_e$ from the ratio error and reporting it as the magnetizing current, and plugging $3.438$ minutes (or degrees) straight into the formula without converting to radians. With $\delta =0$, the ratio error comes from $I_e$ and the phase error comes from $I_m$. The two values are nearly equal here (about $1\text{A}$ each), which hides the mix-up but does not justify it.

    Full solution →
53. Q.53NAT · 2 marksOfficial answer: 4.3 to 4.6
54. Q.54NAT · 2 marksOfficial answer: 10

    Electrical and Electronic Measurements › Bridges

    At balance in a Maxwell-Wien (inductance-capacitance) bridge, the coil's $L_x/R_x$ ratio equals $C_1{R}_1$ of the parallel RC arm opposite it. So $Q=\omega L_x/R_x=\omega C_1{R}_1$ depends only on $\omega$, $C_1$ and $R_1$, and the ratio arms $R_2$, $R_4$ cancel.

    WATCH OUT FOR

    Using $\omega =2\pi \times 1000$, which gives $Q\approx 62.8$. The source is written as $\cos (1000t)$, so $1000$ is already $\omega$ in rad/s. A second mistake is to treat $R_1$ and $C_1$ as a series combination instead of a parallel one, which gives $Q=1/(\omega C_1{R}_1)=0.1$.

    Full solution →
55. Q.55NAT · 2 marksOfficial answer: 65.4 to 65.6

    Control Systems › Bode & margins

    The double pole at the origin holds the phase at exactly $-180^{\circ }$, so the phase margin is simply the phase lead of the zero, ${\tan }^{-1}{\omega }_{gc}$, measured at the gain crossover frequency. Setting $|GH|=1$ gives a quadratic in ${\omega }^2$.

    WATCH OUT FOR

    Students often misplace the zero's phase lead, or read $\angle GH=-180^{\circ }+{\tan }^{-1}\omega$ as meaning the system sits at the stability limit and write a phase margin of $0^{\circ }$. Some also take the negative root of the quadratic in ${\omega }^2$, or leave the calculator in radians and write ${\tan }^{-1}(2.197)=1.14$.

    Full solution →