GATE ECE 2026 question paper with solutions

GATE ECE 2026 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: B

   General Aptitude › Grammar & vocabulary

   A true antonym is the opposite on the same scale of meaning. Nocturnal is about the time of day (night), so its antonym must also be about time of day: diurnal (by day). The Latin roots help here: nox = night, dies = day.

   WATCH OUT FOR

   Seeing the shared ending "-nal" and choosing 'normal', or treating nocturnal as meaning "unusual" and then picking its opposite. Options (A), (C) and (D) are about how ordinary something is, not about night and day, so none of them can be the antonym.

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

   General Aptitude › Reasoning

   Knowing the highest and the fourth highest score boxes the top four students into an interval whose width is their difference. Here that width is $100-76=24\le 25$, so (S3) is forced. Each of the other implications falls to a single counterexample.

   WATCH OUT FOR

   Treating an implication as true because it holds in one convenient example (for example, choosing (C) or (D) because some score list fits), instead of asking whether it holds in every case. The other slip is reading (S3) as a claim about the highest score. (S3) fixes no particular value, so on its own it can never tell you the highest or the fourth highest score.

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

   General Aptitude › Spatial

   Each line has to pass straight through the gap. Join every stub to the stub on another edge with the same slope, which gives three lines: top 31% to the right edge near Y, top 57% to the bottom edge near Z, and left 70% to right 54%. Then check each option at the edges. P, Q and R each get exactly one of the three lines wrong. Only S has all three.

   WATCH OUT FOR

   Choosing P because its two top-edge lines and their crossing look right, without noticing that its near-horizontal line slopes the wrong way (high on the left, low on the right). Another trap is choosing R, whose single triple crossing looks neat, without checking that its rising line leaves through the right edge near X instead of the top edge.

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

   General Aptitude › Numerical

   Taking the $n$th root of a log's base multiplies the log by $n$: ${\log }_{a^{1/n}}x=n{\log }_{a}x$. Each factor therefore gains a factor of $n$, and ${\log }_{p}y\cdot {\log }_{y}p=1$. The whole product is $n^2$, so $n^2=16$.

   WATCH OUT FOR

   Treating ${\log }_{p^{1/n}}y$ as $\frac{1}{n}{\log }_{p}y$ (a root in the base multiplies the log by $n$, it does not divide it). That gives $\frac{1}{n^2}=16$, which no option satisfies. The other common slip is stopping at $n=16$ and forgetting that each of the two factors contributes an $n$, so the product holds $n^2$, not $n$.

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

   General Aptitude › Reasoning

   Negate the quantifiers one at a time. "For every student there exists a classmate within 10 marks" becomes "there exists a student for whom every classmate is more than 10 marks away". Then one class that makes the statement false (99 students on 50, one on 80) is enough to rule out every option that is stronger than the negation.

   WATCH OUT FOR

   Swapping the quantifiers on only half the sentence. Keeping "for each" and negating just the inner part gives option (A), which is far stronger than the negation. A single isolated student is enough to make the original statement false.

   The question and its official answer →
6. Q.6MCQ · 2 marksOfficial answer: A

   General Aptitude › Grammar & vocabulary

   Solve each clue from both its meaning and its letter pattern (length plus the fixed letters), take the first letter of each keyword, then keep only the option that uses exactly those four letters, each once.

   WATCH OUT FOR

   Stopping after two or three clues and guessing. Every wrong option except NOSE shares three letters with the answer (COIN has C, I, N; ITCH has I, C, H), so you need all four keywords to decide. Another slip is taking TRIPLE for Clue 3 because it fits _ _ _ PLE. The clue says two, so the word is COUPLE.

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

   General Aptitude › Reasoning

   Use rule (i) first, because it removes candidates directly: anyone who has partnered a grown-up cannot be that grown-up's child. Partnering P and R leaves X only one possible child. Then use rule (ii) on a match that already has one parent-child pair: it cannot hold a second one, and that decides Y's child.

   WATCH OUT FOR

   Using only rule (i). It fixes X as Q's parent, but Y could still seem to be the parent of P or R, which leaves a guess between (A) and (C). Only rule (ii) applied to Match 3, where X already faces Q, rules out Y being R's parent. Another slip is reading 'X and Y are parents to two of the children' as both grown-ups sharing the same children and so choosing (D). The question says the child of X is not the child of Y.

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

   General Aptitude › Numerical

   Write each term in terms of $k$ ($P_k=4k$), take the common factor 4 out, and use $1+2+\cdots +n=\frac{n(n+1)}{2}$.

   WATCH OUT FOR

   Summing the side lengths and stopping there gives $1+2+\cdots +10=55$ (option A). Using $2k$ per square, i.e. only two sides or a semi-perimeter, gives 110 (option B). Using the square's area-like factor or doubling the perimeter by mistake gives 440 (option D).

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

   General Aptitude › Geometry

   A ring of inner radius $r_1$ and outer radius $r_2$ has area $\pi (r_2^2-r_1^2)$. Work out each boundary radius by adding the widths from the centre outward. The factor $\pi$ cancels in a ratio of areas, so only the differences of squared radii matter. A quick check is that all the pieces together must add up to the area of the full disc.

   WATCH OUT FOR

   Comparing widths instead of areas: land widths $2.5+2+3=7.5$ against water widths $1+2+3=6$ gives $1.25$, but outer rings of the same width cover much more area. Another slip is leaving the Inner Island out of the land total, which gives $72/104\approx 0.69$. Others stop the city at E and drop the outer water ring EF, or divide land by the whole city ($78.25/182.25\approx 0.43$) instead of by the water.

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

    General Aptitude › Spatial

    A water image (reflection in a horizontal line) swaps top and bottom and turns each character upside down. It never changes left-to-right order. A left-right mirror reverses the order and keeps the lines where they are. A $180^{\circ }$ rotation does both: it reverses the order and swaps the lines.

    WATCH OUT FOR

    Seeing upside-down text and assuming a $180^{\circ }$ rotation, then choosing an option with the word order reversed (S, or P or Q, which each reverse one line). Check the reference: \"Ref\" still comes before \"Fig\" and \"12\" before \"34\", so the reading order is kept and only the vertical direction changes.

    Full solution →

Electronics and Communication Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Differential equations

   For $\dot{\vec{w}}=A\vec{w}$, each term $e^{\lambda t}\vec{v}$ of the solution is a mode, and a mode only works when $A\vec{v}=\lambda \vec{v}$. Here the modes lie along ${\vec{u}}_x$ (with $\lambda =1$) and ${\vec{u}}_y$ (with $\lambda =-2$). The eigenvectors are the standard basis vectors, so $A$ is simply $\text{diag}(1,-2)$: its eigenvalues sit on the diagonal in the same order as the axes.

   WATCH OUT FOR

   One mistake is picking (B) by reading the exponent $e^{-2t}$ as a pole at $+2$, which flips the signs. The eigenvalue is the coefficient of $t$ in the exponent itself, so it is $-2$ and not $+2$. Another is choosing (C) or (D) because they contain the right numbers $1$ and $-2$ in off-diagonal places. But those matrices mix $x$ and $y$, and their eigenvalues $\pm j\sqrt{2}$ would give oscillation, not $e^t$ and $e^{-2t}$.

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

   Engineering Mathematics › Vector calculus

   Write the surface as a level surface $\varphi (x,y,z)=0$. The gradient $\nabla \varphi$ at a point on it is normal to the surface there, and $\pm \nabla \varphi /|\nabla \varphi |$ are the two unit normals. An option has to pass two tests: its components must be in the same ratio as $\nabla \varphi$, and its magnitude must be 1.

   WATCH OUT FOR

   Stopping at $\nabla \varphi =4\hat{ı}-2\sqrt{2}\hat{k}$ and choosing an option by direction alone, without dividing by $|\nabla \varphi |=2\sqrt{6}$. The question asks for a unit normal, and (A) and (C) are not unit vectors. Other slips are reading the point as $(1,\sqrt{2},0)$ instead of $(1,0,\sqrt{2})$, and getting the sign of the $z$ term wrong when moving $z^2$ across. Either one puts a spurious $\hat{ȷ}$ component into the normal or changes its direction.

   Full solution →
3. Q.13MCQ · 1 markOfficial answer: D

   Signals and Systems › Laplace

   Convolution in time becomes a product in $s$, and a time delay of $t_0$ becomes the factor $e^{-s{t}_0}$. Here $\frac{e^{-2s}}{s}\cdot \frac{1}{s^2}=\frac{e^{-2s}}{s^3}$. In the time domain, convolving with a delayed step integrates the ramp once and delays the result: $x(t)=\frac{(t-2{)}^2}{2}u(t-2)$.

   WATCH OUT FOR

   Mixing up the two shifting properties. A delay in time, $x(t-2)$, multiplies the transform by $e^{-2s}$. A shift in $s$, such as $(s-2)$ or $e^{-2(s-2)}$, comes from multiplying by $e^{2t}$, and this signal has no such factor. That confusion produces options (A), (B) and (C). A second slip is to multiply the two signals in time instead of their transforms.

   Full solution →
4. Q.14MCQ · 1 markOfficial answer: C

   Electronic Devices › P-N junction

   A Zener diode is deliberately made with heavily doped $p$ and $n$ regions, so its depletion width is extremely thin; under reverse bias this thin barrier develops a very high electric field, which allows electrons to tunnel directly from the valence band on the $p$-side to the conduction band on the $n$-side — this is the Zener (tunneling) breakdown mechanism, distinct from avalanche breakdown (drift + impact ionization) that dominates in lightly doped, lightly biased junctions.

   WATCH OUT FOR

   Confusing the Zener (tunneling) breakdown mechanism with avalanche breakdown (impact ionization due to carrier drift), or assuming all diode breakdown is "avalanche" regardless of doping level.

   Full solution →
5. Q.15MCQ · 1 markOfficial answer: D

   Signals and Systems › Sampling

   A sinusoid at $f_0$ sampled at $F_s$ gives the same samples as one at $f_0-k{F}_s$ for any integer $k$, because the discrete-time frequency $2\pi f_0/F_s$ is only defined modulo $2\pi$. Here $50\text{Hz}-40\text{Hz}=10\text{Hz}$, so $x_2$ aliases exactly onto $x_1$ and every sample matches.

   WATCH OUT FOR

   Comparing the continuous-time waveforms instead of the samples. $\cos (100\pi t)$ oscillates five times faster than $\cos (20\pi t)$, so at most instants the two signals differ, and that tempts students towards (A), (B) or (C). At the sampling instants $t=n/40$, though, the extra $80\text{Hz}$ ($2\pi n$ in phase) adds a whole number of cycles, and the samples are identical. A second slip is counting the fourth sample as $n=4$; since sampling starts at $n=0$, the fourth to seventh samples are $n=3$ to $6$. Here that makes no difference because every sample is equal.

   Full solution →
6. Q.16MCQ · 1 markOfficial answer: A

   Signals and Systems › z-transform

   Find $Y(z)=H(z)X(z)$. The input's zero at $z=\frac{1}{3}$ cancels one system pole, which leaves one pole at $z=\frac{1}{2}$. A recursion run forward from rest gives a right-sided output, so the ROC is $|z|>\frac{1}{2}$. That ROC contains the unit circle, so $y[n]={\left(\frac{1}{2}\right)}^{n}u[n]$ is both causal and absolutely summable.

   WATCH OUT FOR

   Taking the left-sided ROC $|z|<\frac{1}{2}$ gives $y[n]=-{\left(\frac{1}{2}\right)}^{n}u[-n-1]$, which grows like $2^{|n|}$ as $n\to -\infty$. That points to option (D), "unstable and non-causal". But a difference equation computed forward in time from rest always gives a right-sided response. A second slip is to judge the response by the system's poles alone and miss the pole-zero cancellation. Here both poles ($\frac{1}{2}$ and $\frac{1}{3}$) lie inside the unit circle anyway, so the conclusion is the same, but the cancellation is what makes $y[n]$ a single geometric sequence.

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

   Analog Electronics › Op-amps

   For an ideal non-inverting amplifier, $V_o=\left(1+\frac{R_f}{R_1}\right)V_i$. Closing S shorts the $1\text{k}\Omega$ feedback resistor, so $R_f$ drops from $3\text{k}\Omega$ to $2\text{k}\Omega$ and the gain drops from 4 to 3. The ratio of the two outputs is the ratio of the two gains, $3/4$.

   WATCH OUT FOR

   Taking the ratio of the feedback resistances ($R_f^{\prime }/R_f=2/3$, option B) instead of the ratio of the gains. A non-inverting gain is $1+R_f/R_1$, not $R_f/R_1$, so the '1 +' term has to be kept. Another slip is using the inverting-amplifier formula $-R_f/R_1$, which also gives $2/3$ for the ratio.

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

   Analog Electronics › Biasing

   This is an op-amp driving a BJT, the standard voltage-to-current converter. Negative feedback through the emitter forces $V_E=V_i$. The op-amp raises its output to $V_i+V_{BE}$, which cancels $V_{BE}$. So the emitter current is set only by $V_i/R_E=3\text{mA}$, whatever $\beta$ or $V_{BE}$ are. The op-amp only has to supply the base current, $I_E/({\beta }_{dc}+1)$. The $2\text{k}\Omega$ collector resistor and $V_{CC}$ play no part in $I_o$; they matter only for checking that the transistor stays active.

   WATCH OUT FOR

   The first mistake is to treat the circuit as a plain emitter follower and write $V_E=V_i-V_{BE}=2.3\text{V}$. The feedback loop puts the base at $3.7\text{V}$, so the emitter sits at exactly $V_i=3\text{V}$. The second is to take $I_o$ as the emitter or collector current (about $3\text{mA}$) rather than the base current the op-amp supplies. The third is to divide by $\beta$ instead of $\beta +1$: that gives $30.3\mu \text{A}$, which is close but not exact. With ${\beta }_{dc}=99$, $\beta +1=100$ gives exactly $30\mu \text{A}$.

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

   Control Systems › Block diagrams & SFG

   Work out what signal actually reaches each summing junction before you reach for a reduction formula. Summing junction 2 subtracts two copies of the same signal, $E(s)/(s+4)$, so the feedback is exactly zero. The loop is effectively open and $C/R$ is just the forward block $\frac{1}{s+4}$.

   WATCH OUT FOR

   Treating the two $\frac{1}{s+4}$ blocks as a parallel pair whose outputs add, which gives $\frac{2}{s+4}$ (option C). Only the top block feeds $C(s)$. The other common error is using the standard $\frac{G}{1+GH}$ with $H=\frac{1}{s+4}$ and ignoring the $+C$ input at summing junction 2. That misses the cancellation, and the resulting denominator is not in the options.

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

    Digital Electronics › Logic families

    In a static CMOS gate, the NMOS pull-down network directly implements the complement of the output. Parallel NMOS devices give OR, and series NMOS devices give AND. The PMOS pull-up network is the dual: series where the NMOS are parallel, and parallel where they are in series. Parallel NMOS with series PMOS gives $Y=\overline{A+B}$, which is NOR.

    WATCH OUT FOR

    Reading the output from the pull-down function without inverting it. The parallel NMOS pair computes $A+B$, but it pulls $Y$ LOW, so the answer is NOR, not OR. A second slip is mixing up the dual pair: series PMOS with parallel NMOS gives NOR, while parallel PMOS with series NMOS gives NAND. It also helps to notice the hop on the $B$ wire, which shows $B$ does not connect to the output line.

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

    Electromagnetics › Antennas

    Subtracting one dBm value from another gives a dB ratio. Here $P_R-P_T=-100\text{dB}$ is the free-space path loss with unity-gain antennas, and ${\left(\frac{\lambda }{4\pi D}\right)}^2=10^{-10}$. Because Friis loss depends on the square of distance, 100 dB of power loss means $\frac{4\pi D}{\lambda }=10^5$, not $10^{10}$.

    WATCH OUT FOR

    The usual mistakes are about the square. One is forgetting the square root, which leaves $\frac{4\pi D}{\lambda }=10^{10}$ and a distance far larger than any option. Another is converting the 100 dB power ratio with $10^{-100/20}$ and then taking a square root as well, which applies the halving twice. A third is leaving $D$ in metres ($\frac{30000}{4\pi }$ m) and not turning it into km before comparing with the options. It also helps to remember that dBm minus dBm is a plain dB ratio, so there is no need to convert to watts.

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

    Networks (Electric Circuits) › Transients

    A current impulse $q\delta (t)$ into a capacitor dumps a charge $q$ on it instantly, so the voltage jumps by $q/C$. The impulse can be removed and replaced by the initial condition $V_0+q/C$. In the Laplace domain this is just moving $q$ into the initial-condition term: $C[sV(s)-V_0]=q-M(s)$ is the same as $C[sV(s)-(V_0+q/C)]=-M(s)$.

    WATCH OUT FOR

    Being put off by the label $t=0^{-}$ in option (A) and expecting $V_0+q/C$ to belong only at $0^{+}$. In the equivalent system there is no impulse, so $V$ is continuous and its $0^{-}$ and $0^{+}$ values are the same. Other traps: picking (B) or (C), which add a current $m(0^{-})$ to a voltage (a quick unit check rules them out), or picking (D), which forgets the voltage $V_0$ the capacitor already had.

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

    Electromagnetics › Plane waves

    Equal amplitudes with a $90^{\circ }$ phase difference give circular polarization. To find the sense, fix one plane (say $z=0$) and follow $\mathbf{E}$ as time increases. If the right-hand fingers curl the way $\mathbf{E}$ turns and the thumb points along the direction of travel, the wave is right-handed (IEEE convention). Equivalently, for travel along $+z$ the phasor $({\hat{a}}_x-j{\hat{a}}_y)e^{-jkz}$ is RHCP and $({\hat{a}}_x+j{\hat{a}}_y)e^{-jkz}$ is LHCP.

    WATCH OUT FOR

    Two errors lead to (B). (1) Tracking the rotation in space at a fixed instant (letting $z$ vary) rather than in time at a fixed plane: along $z$ the vector turns the opposite way, so the sense comes out reversed. (2) Using the classical optics convention, which names the sense as seen by an observer facing the oncoming wave. Under that convention the same wave is called left-handed. GATE uses the IEEE (engineering) convention, where the sense is judged with the thumb along the direction of travel. The stem's mention of a right-handed co-ordinate system points to that convention.

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

    Electronic Devices › P-N junction

    Under forward bias $V$, the depletion region is no longer in equilibrium, so a single Fermi level no longer describes it; separate quasi-Fermi levels $E_{fn}$ (electrons) and $E_{fp}$ (holes) are needed. In the quasi-neutral bulk on each side, each carrier type stays close to local equilibrium, so $E_{fn}$ and $E_{fp}$ are each nearly flat there. The applied bias does work $qV$ on carriers crossing the junction, and this work appears entirely as the split between the two (flat, bulk-side) quasi-Fermi levels: $E_{fn}-E_{fp}=qV$. With $V=2\text{V}$, the magnitude is $2\text{eV}$.

    WATCH OUT FOR

    Reporting the answer as "2 V" (option C) or "1 V" (option D) by treating the quasi-Fermi level separation as a voltage rather than an energy — a dimensional slip, since $E_{fn}-E_{fp}$ is an energy and must be quoted in eV (numerically equal to the bias in volts only because of the eV/V unit correspondence, $E=qV$). A second common mistake is halving the bias (giving 1 eV, option B) by mistakenly splitting the applied voltage symmetrically between the two quasi-Fermi levels relative to the equilibrium Fermi level, instead of recognizing that the full $qV$ appears as their mutual separation.

    Full solution →
16. Q.26MSQ · 1 markOfficial answer: B, D

    Engineering Mathematics › Linear algebra

    Expand along the row or column with the most zeros to get $\det (M)$ as a linear expression in $a$ and $b$, here $a+5b+3$. Then substitute each option. Every option that makes the expression zero gives a singular matrix and must be rejected.

    WATCH OUT FOR

    Getting a cofactor sign wrong. In the $2\times 2$ minor $\left|\begin{array}{cc}1& 3\\ -1& a\end{array}\right|$, the $-1$ makes it $a+3$, not $a-3$. The sign pattern on the middle element also matters. Either slip changes which options give zero. For example, $a-3+5b$ would wrongly reject (B) instead of (C). Since this is an MSQ, stopping after finding one option that works also loses the mark.

    Full solution →
17. Q.27MSQ · 1 markOfficial answer: B

    Digital Electronics › Sequential

    A counter with $N$ distinct states needs the smallest $n$ with $2^n\ge N$. Counting from 0 to $M$ gives $N=M+1$ states, so the number of flip-flops equals the number of bits in the largest count $M$: here $(64{)}_{10}=(1000000{)}_2$ has 7 bits.

    WATCH OUT FOR

    Taking $N=64$ instead of 65 and picking 6 because $2^6=64$. Six flip-flops count 0 to 63, which is 64 states, and cannot hold the state 64 itself. When counting starts at 0, the final count needs one more bit than you might first expect.

    Full solution →
18. Q.28MSQ · 1 markOfficial answer: A, B

    Electromagnetics › Plane waves

    Put $ϵ={ϵ}^{\prime }-j{ϵ}^{\prime \prime }$ into $\nabla \times \mathbf{H}=\sigma \mathbf{E}+j\omega ϵ\mathbf{E}$ and split the current into a part in phase with $\mathbf{E}$, $(\omega {ϵ}^{\prime \prime }+\sigma )\mathbf{E}$, and a part in quadrature, $j\omega {ϵ}^{\prime }\mathbf{E}$. The loss tangent is their ratio, $\tan \delta =(\omega {ϵ}^{\prime \prime }+\sigma )/(\omega {ϵ}^{\prime })$. That single formula is both option (B) and the wording of option (A).

    WATCH OUT FOR

    Rejecting (A) because the word 'displacement' seems to leave out conduction, or because it names 'real to imaginary' rather than 'imaginary to real'. With $\mathbf{E}$ as the phase reference, the lossy part of $\sigma \mathbf{E}+j\omega ϵ\mathbf{E}$ is the real part and the stored-energy part is the imaginary part, so real/imaginary is exactly $\tan \delta$. Another trap is picking (D) because $P/D$ looks like a dimensionless material ratio. It is $({ϵ}_r-1)/{ϵ}_r$, which measures polarisation, not loss.

    Full solution →
19. Q.29MSQ · 1 markOfficial answer: C, D

    Electronic Devices › BJT

    In the Early-effect model, $I_C=I_{C0}\left(1+\frac{V_{CE}}{V_A}\right)$ at fixed $I_B$, so each output curve's active-region straight-line segment has slope $\frac{I_{C0}}{V_A}$; extrapolated backward, that line meets the $V_{CE}$-axis at $V_{CE}=-V_A$. Since both curves in the figure branch from the same knee point (same $I_{C0}$), a flatter active-region slope means a larger $|V_A|$, and a steeper slope means a smaller $|V_A|$.

    WATCH OUT FOR

    Assuming a nearly-flat curve on a printed figure must mean $V_A\to \infty$, and so marking (B) or dismissing (D). The figure clearly shows BJT 2's line still has a small positive slope (it is not perfectly horizontal), so $|V_{A2}|$ is large but finite, not infinite; only a truly horizontal line would signal $V_A\to \infty$.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 0.95 to 1.05

    Networks (Electric Circuits) › Laws & node/mesh

    The four equal resistors form a balanced bridge, with the battery across the middle arm. By symmetry the battery shifts $V_A$ and $V_B$ equally in opposite directions, so it adds nothing to the rail-to-rail voltage. With the battery shorted, A and B become one node and the source sees $(1\Vert 1)+(1\Vert 1)=1\text{k}\Omega$. That gives $V_o=1\text{mA}\times 1\text{k}\Omega =1\text{V}$. The supernode equations confirm it directly: $V_A+V_B=1$ and $2V_o-(V_A+V_B)=1$.

    WATCH OUT FOR

    Assuming the 6 V battery must appear in $V_o$ and adding or subtracting it (for example, answering 4 V or 7 V). In a balanced bridge the battery only splits the middle nodes to $V_A=3.5\text{V}$ and $V_B=-2.5\text{V}$, and its effect cancels between the two rails. Another slip is to write a KCL at A or at B alone and then try to include the battery current, which is unknown. Enclose the battery in a supernode instead.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 71.50 to 72.50

    Networks (Electric Circuits) › AC, power & resonance

    An element connected straight across an ideal voltage source always has the source voltage across it. Here R is in parallel with the source, so resonance changes only the current the source supplies, not the power in R: $P=V_m^2/(2R)$ at any frequency.

    WATCH OUT FOR

    Using the peak value as if it were rms, $12^2/1000=144\text{mW}$, which is twice the correct answer. A second trap is to assume resonance must change the answer and start computing currents through an equivalent impedance. The parallel LC only changes the source current, while the voltage across R stays fixed by the ideal source.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 0.95 to 1.05
23. Q.33NAT · 1 markOfficial answer: 20.50 to 21.50

    Electromagnetics › Waveguides & fibres

    For ${\text{TE}}_{mn}$ in an air-filled rectangular guide, $f_{c,mn}=\frac{c}{2}\sqrt{{\left(\frac{m}{a}\right)}^2+{\left(\frac{n}{b}\right)}^2}$. With $m=1$, $n=0$ this becomes $f_{c,10}=c/2a$, which depends only on the broad dimension $a$. The whole problem comes down to converting $a$ from inches to metres correctly.

    WATCH OUT FOR

    Using the narrow dimension $b=0.14$ inch, or writing $f_c=c/a$ without the factor 2. Both give about $42.18\text{GHz}$, which is the ${\text{TE}}_{01}$/${\text{TE}}_{20}$ cut-off, not ${\text{TE}}_{10}$. Another slip is forgetting to convert inches to metres, which is off by a factor of about 39. (Taking $c=2.998\times 10^8\text{m/s}$ gives $21.08\text{GHz}$, which is still inside the key range.)

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 20.00 to 21.00

    Electromagnetics › Transmission lines

    S-parameter magnitudes squared are power fractions. In a passive two-port driven at port 1 with port 2 matched, reflected power is $|S_{11}{|}^2$ and transmitted power is $|S_{21}{|}^2$. Any shortfall from 1 is power lost inside the network. At a crossing point both curves have the same dB value, so both power fractions are $10^{\text{dB}/10}$.

    WATCH OUT FOR

    A common slip is converting $-4$ dB with $10^{-4/20}=0.631$. That is the magnitude $|S|$, not the power fraction. Adding the two magnitudes, or subtracting magnitudes from 1, gives a meaningless result. Another slip is to count only one of the two terms and write $1-0.398=60.2\%$. The reflected and transmitted powers must both be subtracted.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 4.90 to 5.10

    Digital Electronics › Sequential

    With $J=K=1$, a JK flip-flop toggles on every active clock edge ($Q_{n+1}=\overline{Q_n}$). A full output cycle needs two toggles, so the flip-flop works as a divide-by-2 counter: $f_Q=f_{clk}/2$. The triggering edge (rising or falling) only moves where the output transitions fall in time. It does not change their rate.

    WATCH OUT FOR

    One mistake is to count each toggle as a full output cycle and answer 10, which is the clock frequency. Another is to think negative-edge triggering changes the frequency, for example by counting both clock edges and answering 10 or 20. An edge-triggered flip-flop responds to only one edge per clock cycle, and it takes two toggles to complete one cycle of $Q$.

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

    Engineering Mathematics › Calculus

    An exponential in the denominator beats any power of $n$ in the numerator: for $\sum n^k/r^n$ with $r>1$ the ratio test gives the limit $1/r<1$, so the series converges. A series of positive terms, each a distinct term of a convergent geometric series, has bounded increasing partial sums and therefore converges, however irregular the pattern of left-out terms.

    WATCH OUT FOR

    Two traps. First, seeing $n^2$ in the numerator of $S_A$ and deciding the terms grow, when $2^n$ grows much faster and the terms tend to $0$. Second, deciding that $S_B$ is not geometric because of the gaps ($\frac{1}{4},\frac{1}{32},\dots$ are missing) and so it cannot be judged. Leaving out positive terms from a convergent geometric series can only make the sum smaller, so it still converges.

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

    Signals and Systems › Fourier series

    Shifting by half a period multiplies the $n$-th coefficient by $e^{-jn\pi }=(-1{)}^n$. With half-wave symmetry, $x\left(t-\frac{T}{2}\right)=-x(t)$, you need $(-1{)}^n{a}_n=-a_n$, so only odd harmonics survive. Every even-indexed coefficient is zero, and that includes $a_0$.

    WATCH OUT FOR

    Students often mix up half-wave symmetry, $x\left(t-\frac{T}{2}\right)=-x(t)$, with odd symmetry, $x(-t)=-x(t)$. Odd symmetry only makes the coefficients purely imaginary. It does not remove the even harmonics. Another slip is to use a shift of a full period $T$ instead of $T/2$, which gives $e^{-j2\pi n}=1$ and seems to tell you nothing. Some students also forget that $m=0$ is included, so the DC value $a_0$ is zero as well.

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

    Engineering Mathematics › Probability & statistics

    Expand each correlation with linearity of expectation. Independence makes the cross term zero, $E[XN]=E[X]E[N]=0$. A uniform distribution symmetric about zero has every odd moment zero, so $E[XY]=E[X^2]=\frac{1}{3}$ and $E[XZ]=E[X^3]=0$.

    WATCH OUT FOR

    Thinking that because $Z$ is built from $X$, the two must be correlated. $Z$ depends on $X$ only through $X^2$, and $E[X\cdot X^2]=E[X^3]=0$ for a symmetric $X$. So they are uncorrelated but not independent. A second trap is computing the normalised coefficient ${\rho }_{XY}=\frac{1}{2}$ and finding it in no option. In this question "correlation" means $E[XY]$.

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

    Networks (Electric Circuits) › AC, power & resonance

    In an RC branch and an RL branch connected in parallel, the two branch currents are $90^{\circ }$ apart exactly when $\tan {\theta }_1\tan {\theta }_2=1$, i.e. $\frac{X_C{X}_L}{R_C{R}_L}=1$. Since $X_L{X}_C=\omega L\cdot \frac{1}{\omega C}=\frac{L}{C}$, the condition becomes $R_C{R}_L=L/C$, and the frequency drops out.

    WATCH OUT FOR

    Setting ${\theta }_1={\theta }_2=45^{\circ }$ (taking $R_C=X_C$ and $R_L=X_L$ separately) is only one possible case. The question fixes only the sum of the angles, so only the product $R_C{R}_L$ is determined. Another slip is to write the quadrature condition as $\tan {\theta }_1+\tan {\theta }_2=1$ or ${\theta }_1={\theta }_2$, when the correct relation is $\tan {\theta }_1\tan {\theta }_2=1$. Some students also spend time computing $\omega$ exactly, even though it cancels.

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

    Control Systems › Compensators & PID

    For a second-order closed loop, the complex poles $-\sigma \pm j{\omega }_d$ fix the characteristic polynomial exactly as $s^2+2\sigma s+({\sigma }^2+{\omega }_d^2)$. The compensator gives two free parameters, one for each coefficient. Here $K$ alone sets the $s$ coefficient ($3+K$), and once $K$ is known, $\alpha$ sets the constant term ($2+K\alpha$).

    WATCH OUT FOR

    Taking the constant term of the desired polynomial as $3^2-5=4$ or as $9$, when it is $3^2+(\sqrt{5}{)}^2=14$. The other mistake is forgetting that $K$ multiplies $\alpha$ in the constant term, which gives $2+\alpha =14$. Every distractor is close. Option (D), $(3,3)$, gets $K$ right but gives $s^2+6s+11$, whose poles are $-3\pm j\sqrt{2}$. Options (A) and (C) have $K=2$, which gives an $s$ coefficient of 5 and a real part of $-2.5$.

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

    Control Systems › State space

    For a strictly proper state-space model, $G(s)=C\text{adj}(sI-A)B/\det (sI-A)$. The denominator is the characteristic polynomial of $A$, and for a $2\times 2$ system you can read it off directly as $s^2-(\text{trace }A)s+\det A=s^2+4s+6$. The numerator needs only one column of the adjugate, because $B$ has a single non-zero entry.

    WATCH OUT FOR

    Sign errors in $\det (sI-A)$. The off-diagonal product is $(1.5)(-4)=-6$, and subtracting it gives $+6$. Writing $-6$ leads to option (B). The other common slip is dropping the $0.625\times 8=5$ term: the second state contributes to the output through $C$ even though $u$ does not drive it directly, and missing that term leads to option (D). Also watch the adjugate: swap the diagonal entries and negate the off-diagonal ones. Transposing without negating gives the wrong numerator.

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

    Digital Electronics › Processors & memory

    A byte-addressable block of $N$ locations that starts at address $A$ occupies $A$ to $A+N-1$. Here $256\text{KB}=2^{18}=(40000{)}_H$, so the last address is $(2500{)}_H+(3FFFF{)}_H=(424FF{)}_H$.

    WATCH OUT FOR

    One mistake is to add the full size, $(2500{)}_H+(40000{)}_H=(42500{)}_H$. That is the first address after the memory, so you still have to subtract 1. The other is to write $256\text{KB}$ as $(10000{)}_H$ or $(30000{)}_H$ instead of $(40000{)}_H$ ($2^{16}$ is only 64 KB). That mistake leads to distractors such as (B) or (D).

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

    Signals and Systems › Signals basics

    Energy scales with the square of the amplitude and inversely with the magnitude of the time-scale factor: $E[Ax(at+b)]=\frac{A^2}{|a|}E[x(t)]$. A time shift or a time reversal leaves the energy unchanged.

    WATCH OUT FOR

    Three common slips: scaling energy by $A$ instead of $A^2$ (this gives $E=4$ and a ratio of 1, option B); multiplying by $|a|$ instead of dividing (compression by 3 makes the pulse narrower, so it has less energy, not more; this gives a ratio of 1/27, which is not an option); and thinking the shift $+5$ or the minus sign on $t$ changes the energy.

    Full solution →
34. Q.44MCQ · 2 marksOfficial answer: B
35. Q.45MCQ · 2 marksOfficial answer: B

    Digital Electronics › Number systems

    In base $r$, the $r$'s complement of an $n$-digit number $N$ is $r^n-N$. It equals the $(r-1)$'s complement, $(r^n-1)-N$, plus 1. For $N=47$: $100-47=53$, or $99-47=52$ and then $52+1=53$.

    WATCH OUT FOR

    Stopping at the 9's complement (52, option A) and forgetting to add 1 is the usual slip. A second one is using the wrong power of the base, for example $10^3-47=953$. With $n=2$ digits the power must be $10^2$.

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

    Signals and Systems › Fourier transform

    Use Rayleigh's (Parseval's) theorem. The total energy is $1/4$ and $|X(f){|}^2=\frac{1}{4(1+{\pi }^2{f}^2)}$. Integrating this over the two-sided band $|f|\le B$ gives the energy fraction $\frac{2}{\pi }{\tan }^{-1}(\pi B)$. Setting that equal to $0.99$ gives $\pi B=\cot (0.005\pi )\approx 63.66$.

    WATCH OUT FOR

    Working in rad/s, getting ${\omega }_B\approx 127.3\text{rad/s}$, then dividing by $\pi$ instead of $2\pi$. That gives $B\approx 127.3/\pi$ Hz, which is the trap option (C). A second slip is integrating only $0\le f\le B$ without doubling for the negative frequencies. The fraction then tops out at $50\%$ and can never reach $99\%$.

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

    Signals and Systems › LTI & convolution

    Delays, adders, pick-offs and LTI blocks joined without a feedback loop always make an LTI system. Here the overall impulse response is fixed: $h[n]=h_1[n]-h_2[n-1]+(h_1\ast h_2)[n-2]$, or $H(z)=H_1-z^{-1}{H}_2+z^{-2}{H}_1{H}_2$. Whether the system is causal depends on whether $h_1$ and $h_2$ are causal, and the question does not tell us.

    WATCH OUT FOR

    Picking (A) because the diagram contains only unit delays $z^{-1}$, which makes it look causal. A delay after an anti-causal block does not remove its anti-causality: with $h_1[n]=\delta [n+1]$ the whole system reduces to $y[n]=x[n+1]$ for any $h_2$. Another mistake is to read the cross-coupled branches as feedback that makes the system time-varying. There is no loop, and a fixed interconnection of LTI blocks stays LTI.

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

    Digital Electronics › Combinational

    A 4:1 MUX outputs the data input chosen by its select lines, $f={\sum }_k{m}_k(S_1,S_0)I_k$. Put in $S_1=x$, $S_0=z$ and the inputs read from the figure ($I_0=1$, $I_1=0$, $I_2=1$, $I_3=y$), then simplify. The two constant-1 inputs together cover every case with $z=0$.

    WATCH OUT FOR

    Most errors come from misreading the figure. If you treat the hop on the 0 line as a connection, you tie 0 to the vertical wire and get $I_2=0$ (and a 1-to-0 short). Mistaking which variable drives which select line is the other trap: $x$ drives $S_1$ and $z$ drives $S_0$, so $y$ is not a select input at all. Some students also take the select order as $(x,y)$ just because $x$ is the MSB. That pairs the terms wrongly and leads to options such as (D).

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

    Digital Electronics › Sequential

    XOR with a constant 1 is an inverter, so the serial input is $\overline{Q}$. $Q$ samples $b_7$, which stays 0 until a 1 has travelled all 8 cells, so for the first 8 edges $Q=0$ and a 1 shifts in at $b_0$ on every edge.

    WATCH OUT FOR

    Taking the serial input as $Q$ instead of $Q\oplus 1=\overline{Q}$ leaves the register at all zeros, which is not an option. Filling from the $b_7$ end, as if it were a shift-right register, gives $(11111000{)}_2$, also not an option. Counting one edge too many (for example, treating the initial clear as a clock) gives six 1s, $(00111111{)}_2$, which is the trap option (C).

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

    Analog Electronics › Amplifiers

    Bias with the full 1.5 kΩ in the emitter ($I_C\approx 1$ mA, so $g_m=40$ mS). Then check that both tones are in the mid-band: at $10^5$ rad/s the 100 nF capacitors are 100 Ω, which is tiny next to a 14.5 kΩ input and an open output, and the 10 µF is 1 Ω, which leaves only the 500 Ω unbypassed. With the parasitics neglected, both tones see one real gain, $-g_m{R}_C^{\prime }/(1+g_m{R}_E)\approx -9$.

    WATCH OUT FOR

    Treating the emitter as fully bypassed and taking the gain as $-g_m(R_C\parallel r_o)\approx -190$ (option C). The 10 µF bypasses only the 1 kΩ, so the 500 Ω still degenerates the gain by a factor of $1+g_m{R}_E=21$. Another trap is assuming the $10^7$ rad/s tone picks up a different phase and flips sign (options B and D). With the parasitics neglected and the coupling capacitors already short at $10^5$ rad/s, both tones sit in the flat mid-band. Also watch the bias: the DC emitter resistance is the full 1.5 kΩ, not 500 Ω.

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

    Networks (Electric Circuits) › Two-port

    Find the ABCD parameters from two port-2 tests. With port 2 open ($I_2=0$) you get $A=V_1/V_2$ and $C=I_1/V_2$. With port 2 shorted ($V_2=0$) you get $B=V_1/(-I_2)$ and $D=I_1/(-I_2)$. For a T-network with series arms $Z_1$, $Z_2$ and shunt arm $Z_3$ this gives $A=1+Z_1/Z_3$, $B=Z_1+Z_2+Z_1{Z}_2/Z_3$, $C=1/Z_3$ and $D=1+Z_2/Z_3$. Then check with $AD-BC=1$ (reciprocal) and $A=D$ (symmetrical).

    WATCH OUT FOR

    Doing the whole calculation in kΩ and never converting back. That gives $B=3$ and $C=1$, and writing these as $3\Omega$ and $1{\Omega }^{-1}$ lands on option (B). A second slip is to find $V_2/V_1$ in place of $V_1/V_2$, which gives $A=D=1/2$ as in option (A). Option (A) also fails the reciprocity check, since $AD-BC=1/4-3\ne 1$.

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

    Electromagnetics › Transmission lines

    A diametrically opposite point on the same $|\Gamma |$ circle means ${\Gamma }_{in}=-{\Gamma }_L$. That is a $180^{\circ }$ turn of the chart, which is a $\lambda /4$ lossless line, so the normalised impedance inverts: $z_{in}=1/z_L$, or $Z_{in}=Z_0^2/Z_L$.

    WATCH OUT FOR

    Option (D), $97.5+j65.0\Omega$, is the load impedance itself. You get it by converting ${\Gamma }_L$ and forgetting to move to the opposite point. Option (C) is its conjugate, which you get by reflecting the point across the real axis (flipping the sign of the phase) instead of turning it through $180^{\circ }$. A further slip is adding $180^{\circ }$ and then taking $\cos 210^{\circ }$ and $\sin 210^{\circ }$ with the wrong signs. Both parts of ${\Gamma }_{in}$ are negative.

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

    Analog Electronics › Op-amps

    With a virtual ground at the inverting input, the input current $V_i/1\text{ k}\Omega =2\text{ mA}$ has to pass through the feedback $1\text{ k}\Omega$ resistor, which puts node X at $-2\text{ V}$. The op-amp output then has to sink this $2\text{ mA}$ as well as the $1\text{ mA}$ drawn from ground through the $2\text{ k}\Omega$ load (KCL at X).

    WATCH OUT FOR

    Picking 2.0 mA by assuming the output resistor carries only the feedback current, which ignores the $1\text{ mA}$ in the grounded $2\text{ k}\Omega$ resistor at node X. Another mistake is to treat the circuit as a plain inverting amplifier with gain $-1$, so that $V_X=-2\text{ V}$ is taken to be the output voltage. The op-amp output is actually on the far side of the $2\text{ k}\Omega$ resistor, at $-8\text{ V}$.

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

    Electronic Devices › Optoelectronics

    A direct-bandgap LED emits photons with energy essentially equal to the bandgap energy, so its emission is concentrated at a single wavelength ${\lambda }_g=hc/E_g$, not spread over the full visible-to-IR spectrum; the LED "does not emit" in any stated window that does not bracket ${\lambda }_g$.

    WATCH OUT FOR

    Trying to match the bandgap to a "range" of wavelengths rather than computing a single sharp emission wavelength ${\lambda }_g=hc/E_g$ and checking which option windows exclude it; also common arithmetic slip: using $q$ in the wrong power of ten when converting J to eV, giving a ${\lambda }_g$ that falls in the wrong window.

    Full solution →
45. Q.55MSQ · 2 marksOfficial answer: C
46. Q.56MSQ · 2 marksOfficial answer: A, D

    Digital Electronics › Boolean logic

    Draw the K-map. The four corners give $\overline{x}\overline{z}$ and the centre quad gives $xz$. After that, $m_{14}$ is the only 1 left, and it has two adjacent 1s ($m_{15}$ and $m_{10}$). That gives two equally minimal covers, $wxy$ or $wy\overline{z}$. A function can have more than one minimal SOP. To test any option, list the minterms its terms cover and compare that set with the given one.

    WATCH OUT FOR

    Stopping at the first minimal cover you find ($\overline{x}\overline{z}+xz+wxy$) and marking only (A), without noticing that $m_{14}$ can also pair with $m_{10}$, which makes (D) valid too. Another mistake is to judge (B) and (C) only by whether they cover every 1. Both also include $\overline{w}xy\overline{z}=m_6$, where $f=0$.

    Full solution →
47. Q.57MSQ · 2 marksOfficial answer: A, B, C
48. Q.58MSQ · 2 marksOfficial answer: A, C, D

    Signals and Systems › Fourier series

    Orthogonality depends on the interval. Harmonically related sinusoids are orthogonal over any whole period of the fundamental, because every product-to-sum term then completes a whole number of cycles. Over a fraction of that period the inner product is generally non-zero. Orthonormal is stricter still: each signal must also have unit energy over the interval.

    WATCH OUT FOR

    Assuming a sine and a cosine are always orthogonal and so rejecting (D). That is true only over a whole number of periods; over half a period of $x_1$ the inner product is $4/(3\pi n)\ne 0$. Other traps: marking (B) by treating orthogonal and orthonormal as the same thing (the energies are $1/(2n)$ and $2/n$, not 1), and rejecting (C) because $x_2$ is a sine with amplitude 2. A harmonic is defined by its frequency being an integer multiple of the fundamental, whatever its amplitude and phase.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 1386

    Control Systems › Routh–Hurwitz

    Marginal stability happens at the gain that makes one row of the Routh array all zeros. The row above it then gives the auxiliary equation, and its roots are the pair on the $j\omega$ axis. For a cubic $a_0{s}^3+a_1{s}^2+a_{2}s+a_3$ this reduces to $a_1{a}_2=a_0{a}_3$.

    WATCH OUT FOR

    Expanding $(s+7)(s+11)$ wrongly (for example as $s^2+18s+18$, or leaving out the factor $s$) gives the wrong cubic. Another mistake is to report a range such as $0<K<1386$, or the oscillation frequency $\sqrt{77}$, when the question asks for the single marginal gain $K$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 2.50 to 2.60

    Analog Electronics › Biasing

    When the gate is tied to the drain, $V_{DS}=V_{GS}$, so an n-channel MOSFET that is on must be in saturation. Set the square-law current $\frac{1}{2}\mu C_{ox}\frac{W}{L}(V_G-V_{TH}{)}^2$ equal to the resistor current $(V_{DD}-V_G)/R$. Solve the quadratic and keep the root with $V_{GS}>V_{TH}$.

    WATCH OUT FOR

    Writing $I_D=\mu C_{ox}\frac{W}{L}(V_{GS}-V_{TH}{)}^2$ without the factor $\frac{1}{2}$. That gives $2x^2+x-4=0$, so $x=1.19\text{V}$ and $V_G\approx 2.19\text{V}$, which is outside the key. The key uses the standard $\frac{1}{2}\mu C_{ox}\frac{W}{L}$ form. Other mistakes: keeping the negative root (it makes $V_G<V_{TH}$, so the device would be off), and checking for the triode region, which a diode-connected device can never enter.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 0.60 to 0.70

    Engineering Mathematics › Calculus

    Put the origin at the centre of the square ($u=x-1$). The integrand becomes $u^2+2u+y^2$. The odd term $2u$ integrates to zero by symmetry, and the $-1$ cancels against the $+1$ from $(1+u{)}^2$. What remains, $\iint (u^2+y^2)dA$, is the polar moment of a square about its centre, $a^4/6$ with $a=\sqrt{2}$, which gives $2/3$.

    WATCH OUT FOR

    Taking the region as an axis-aligned square such as $[0,2]\times [-1,1]$ or $[0,1]\times [0,1]$. The figure shows a square standing on a corner, $|x-1|+|y|\le 1$, with side $\sqrt{2}$ and area $2$, not side $2$. A second slip is subtracting only $1$ for the constant term instead of $1\times \text{area}=2$. That gives $\frac{8}{3}-1=\frac{5}{3}$ instead of $\frac{8}{3}-2=\frac{2}{3}$.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 2.90 to 3.10

    Analog Electronics › Op-amps

    This is an inverting amplifier whose feedback resistor has been replaced by a T-network. Virtual ground at the (−) input fixes the voltage at the T-node, $v_X=-(R_2/R_1)v_i$. KCL at the T-node then gives $v_o$. The general result is $v_o/v_i=-\left(R_2+R_3+R_2{R}_3/R_4\right)/R_1$: the shunt resistor $R_4$ adds the term $R_2{R}_3/R_4$, which raises the gain above what $R_2+R_3$ alone would give.

    WATCH OUT FOR

    One mistake is to treat the feedback as a simple series resistor $R_2+R_3$ and ignore $R_4$, which gives a gain of magnitude 2. Another is to treat $R_4$ as a load that can be dropped. $R_4$ takes current away from the T-node, so the output has to swing further to push the same current through $R_2$. It adds the term $R_2{R}_3/R_4$ and brings the gain to 3. A third error is to take the voltage at the (−) input as $v_i$, when with the (+) input grounded it is at virtual ground (0 V).

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.10 to 0.20
54. Q.64NAT · 2 marksOfficial answer: -36.50 to -34.50, or 34.50 to 36.50

    Electronic Devices › Carrier transport

    Diffusion current density is $J_{n,\text{diff}}=q{D}_n\frac{dn}{dx}$, with $D_n$ obtained from the mobility via the Einstein relation $D_n={\mu }_n{V}_T$; only the magnitude and unit conversion (cm² to mm²) need care here.

    WATCH OUT FOR

    Forgetting that $1{\text{cm}}^2=100{\text{mm}}^2$, so converting A/cm² to A/mm² requires DIVIDING by 100 (not multiplying), and mixing up the sign convention or the direction of the gradient (which point's concentration is subtracted from which).

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 3.70 to 4.30

    Analog Electronics › Diode circuits

    A diode clipper with a resistor in series with the diode and its reference battery does not clip flat. Once D1 turns on, the series 10 kΩ and the branch 10 kΩ form a voltage divider between $v_i$ and the 2 V battery, so above 2 V the output follows $v_o=(v_i+2)/2$ and reaches 4 V at the 6 V input peak. D2, with its anode at −4 V, affects only the negative half-cycle.

    WATCH OUT FOR

    Treating D1 as a hard clamp and answering 2 V, which ignores the 10 kΩ resistor in D1's own branch. Another mistake is misreading the battery symbols (long plate = +). Taking D1's cathode at −2 V gives the wrong divider result, and taking D2's anode at +4 V wrongly brings D2 into the positive half-cycle.

    Full solution →