GATE ECE 2025 question paper with solutions

GATE ECE 2025 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   Both chains list the same kind of action in decreasing intensity: Group-I is disapproval (abuse > insult > ridicule) and Group-II is approval. The missing word must be a stronger form of praise, which is extol.

   WATCH OUT FOR

   Picking a word because of how it looks or sounds rather than what it means. Prize resembles praise and Appropriate resembles appreciate, but neither expresses approval more strongly than praise. Another mistake is reading the chain in the wrong direction and looking for something milder than appreciate.

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2. Q.2MCQ · 1 markOfficial answer: D

   General Aptitude › Grammar & vocabulary

   A clause that starts with \"Had + subject + past participle\" is an inverted third conditional: an unreal past condition. Its result clause must be would / could / might + have + past participle. No option with a future or present verb can complete it.

   WATCH OUT FOR

   Missing the inversion. With no \"if\" in sight, some candidates read \"Had I learnt\" as a plain past statement, or match it to a first-conditional result like \"will be\". Another trap is to pick \"can be\" because it expresses ability, but a present modal cannot give the result of a condition in the past.

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3. Q.3MCQ · 1 markOfficial answer: B

   General Aptitude › Numerical

   The notes form a geometric progression with common ratio $2^{1/12}$. Two notes that are $n$ steps apart have a frequency ratio of $2^{n/12}$. $F^{\#}$ is 6 steps above $C$ (half of the 12-step octave), so the ratio is $2^{6/12}=\sqrt{2}$. The actual frequency of $C$ does not matter.

   WATCH OUT FOR

   Counting notes instead of steps. $C$ to $F^{\#}$ covers 7 notes but only 6 multiplications, and $2^{7/12}$ is not among the options. Other slips: stopping at $F$ (5 steps, giving $2^{5/12}$); treating the progression as arithmetic; or working out $130.8\sqrt{2}\approx 185\text{ Hz}$ when the question asks for a ratio.

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4. Q.4MCQ · 1 markOfficial answer: B

   General Aptitude › Numerical

   When every segment is replaced by k copies, each scaled by a factor r, the total length is multiplied by k times r at every iteration. The length therefore grows geometrically, L_n = (kr)^n × L_0. Here k = 5 and r = 1/3, so L_n = (5/3)^n.

   WATCH OUT FOR

   Checking only n = 1. Option (D) also gives 5/3 at n = 1, because n(2n-1) = 1 there. Only n = 2 (25/9, against (5/3)^6 for D) separates it from (B). A second mistake is counting the segments of Iteration 2 as 5 × 3 = 15 instead of 5 × 5 = 25, or being put off by bumps that point in different directions. Orientation does not change length.

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5. Q.5MCQ · 1 markOfficial answer: A

   General Aptitude › Numerical

   For any non-zero $x$, $\frac{|x|}{x}$ is the sign of $x$: it is $+1$ when $x>0$ and $-1$ when $x<0$. So $f(x)=-\frac{|x|}{x}$ is a step function. It is constant on each side of the origin, undefined at $x=0$, and the leading minus sign puts the $+1$ level on the negative side.

   WATCH OUT FOR

   Dropping the leading minus sign and choosing (B), which is the graph of $\frac{|x|}{x}$ and not of $-\frac{|x|}{x}$. Another error is treating $\frac{|x|}{x}$ like $\frac{1}{x}$ and choosing the hyperbola-like curves in (D). Test one positive and one negative value of $x$ to rule out both.

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6. Q.6MCQ · 2 marksOfficial answer: B

   General Aptitude › Reading

   Look for connecting words. \"Such adaptations\" (P) and \"This process of adaptation\" (R) both need an earlier sentence to refer to, so neither can open. Q is the only sentence that states the general theme with no back-reference, so it opens. P's \"Over time\" gives the long-term consequence, so it closes.

   WATCH OUT FOR

   Opening with S because it reads smoothly on its own (\"such changes\" refers back to \"As environments change\" in the same sentence). S describes one particular case of a general claim, so it belongs after Q. Starting with S, as option (D) does, leaves the general opening statement Q stranded at the end.

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7. Q.7MCQ · 2 marksOfficial answer: D

   General Aptitude › Geometry

   When three lengths add up to $L$, the triangle inequality reduces to one simple test: every side must be shorter than $L/2$. Here $L=1$, so each of $b_1$, $b_2-b_1$ and $1-b_2$ must be less than $0.5$. Write out these three inequalities and match them to the options. To show that a condition is not necessary, one counter-example is enough: a valid triangle that breaks that condition.

   WATCH OUT FOR

   Confusing 'necessary' with 'sufficient'. Each of (A), (B) and (C) alone is not enough to guarantee a triangle, so students sometimes wrongly mark one of them as the answer. The question asks which condition a triangle does not REQUIRE. Another trap is assuming $b_1+b_2<1$ from the figure, or from a symmetry argument that puts both cuts near the middle. The figure is only representative. Cuts at $0.45$ and $0.7$ give a valid triangle with $b_1+b_2>1$.

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8. Q.8MCQ · 2 marksOfficial answer: Marks to all

   General Aptitude › Reasoning

   The students move clockwise, so 'behind' means counting anticlockwise. Musical chairs never changes the students' order round the circle, so each round just removes one name from a cyclic list, and you count only the students still playing. Rounds 1 to 3 remove S, U and R. Once U and R are gone, options (A) and (C) are ruled out before round 4.

   WATCH OUT FOR

   The first common mistake is counting 'behind' clockwise, in the direction of motion. That removes S, R, P, V and leaves T, Q, U, W, which is not an option. The second is counting the empty chairs of students who have left as if someone were still sitting there. A third is not noticing that U has already left when round 4 asks for the student behind U. The question is flawed as printed, and that is why the official key is MTA.

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9. Q.9MCQ · 2 marksOfficial answer: C

   General Aptitude › Data interpretation

   Add each row across to get every nation's total medal count, then test each statement against those totals. For a "less than 50%" claim, compare the part with half the grand total (here 188). That settles it without working out an exact percentage.

   WATCH OUT FOR

   Using only the Gold column, or reading the question as "which statement is correct". On gold alone, USA + Canada have 79 of 132 golds, which is also more than 50%, but options (A) and (D) need the totals across all three columns. Candidates also make arithmetic slips in the grand total (376); a slip there can make (C) look true.

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

    General Aptitude › Numerical

    First work out which slab applies. Even at the top of the first slab (fee Rs. 5,00,000) the client pays only Rs. 7,08,000, so the fee must be in the upper slab. There the overhead is 0.1F + 50,000, so the client pays 1.18(1.1F + 50,000). Set that equal to Rs. 10,00,000 and work backwards: take out the 18% GST first, then the overhead.

    WATCH OUT FOR

    One mistake is to charge the 10% on the whole fee instead of only on the part above Rs. 5,00,000, which gives an overhead of 1,00,000 + 0.1F. Another is to use the 20% slab rate for the whole fee. A third is to take 18% off the Rs. 10,00,000 (multiplying by 0.82) instead of dividing by 1.18. Each of these moves the answer away from (B) and can land near one of the other options.

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Electronics and Communication Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Linear algebra

   $\rho (A)\ge r$ exactly when some $r\times r$ minor is non-zero. Here the first two rows always give rank at least 2, so you only need to find one non-zero $3\times 3$ minor for each case. For an upper triangular matrix, $\det A$ is the product of the diagonal entries, $12\alpha \gamma$. That settles rank 4 straight away, and a zero diagonal entry does not by itself push the rank below 3.

   WATCH OUT FOR

   Counting non-zero diagonal entries and taking that as the rank. In case (i) the diagonal is $2,6,0,\gamma$, which looks like rank 3 for the wrong reason, and a student can easily talk themselves into "$\alpha =0$ breaks rank 3" and reject (i), which leads to option (D). Rank comes from independent rows or non-zero minors, not from the diagonal. In (i) the minor on columns 1, 2, 4 is $12\beta \ne 0$, so the rank is 3. Rows 3 and 4 are identical, so it cannot be 4.

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2. Q.12MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Calculus

   Use a standard test for each series. For $\sum 1/n^p$, check whether $p>1$. For $\sum 1/(n(n+1))$, split into partial fractions so the sum telescopes. For $\sum 1/n!$, the ratio test works well because consecutive factorials cancel down to $1/(n+1)$.

   WATCH OUT FOR

   Deciding that (i) converges because $1/\sqrt{n}\to 0$. A term going to zero is necessary for convergence but not sufficient. $\sum 1/\sqrt{n}$ is a p-series with $p=\frac{1}{2}\le 1$, so it diverges, and choosing option (D) this way is wrong. A second slip is to confuse $p=\frac{1}{2}$ with the convergent case and treat every series whose terms shrink as convergent.

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3. Q.13MCQ · 1 markOfficial answer: A

   Engineering Mathematics › Probability & statistics

   Without replacement, the first ball can be anything. What matters is the second draw: 3 balls are left and 2 of them are the other colour, so the probability is $\frac{2}{3}$. You get the same answer by counting unordered pairs, $\frac{2\cdot 2}{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}=\frac{4}{6}$.

   WATCH OUT FOR

   One mistake is to treat the draws as if they were with replacement, which gives $2\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{2}$, option (C). Another is to use the ordered count (4 ordered red-blue pairs, only one colour order) over the unordered total, or to leave out the factor 2 for the two orders in the sequential method ($\frac{2}{4}\cdot \frac{2}{3}=\frac{1}{3}$, option (B)).

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4. Q.14MCQ · 1 markOfficial answer: A
5. Q.15MCQ · 1 markOfficial answer: A
6. Q.16MCQ · 1 markOfficial answer: D

   Control Systems › Bode & margins

   On a Nyquist plot, read each crossover from where the positive-frequency locus is, not from the angle marks. It crosses the unit circle at the gain crossover ($|GH|=1$) and the negative real axis at the phase crossover ($\angle GH=-180^{\circ }$).

   WATCH OUT FOR

   Swapping the two definitions, which gives (C). Another error is to treat $R$ as special because it sits on the marked $45^{\circ }$ ray. It looks like a $45^{\circ }$ phase margin, but $R$ has $|GH|<1$, so it is neither crossover. $P$ is on the other $45^{\circ }$ ray at phase $-225^{\circ }$, not $-180^{\circ }$. Some students also assume $S$ is on the $-135^{\circ }$ ray and write the phase margin as $45^{\circ }$. In the figure $S$ is below that ray, so its phase is less lagging than $-135^{\circ }$.

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7. Q.17MCQ · 1 markOfficial answer: D

   Signals and Systems › LTI & convolution

   Reduce the block diagram to $y[n]=x[n]\ast h_1[n]\ast h_2[n]+b\left(h_2[n]-\delta [n]\right)$. That is an LTI response plus a fixed, nonzero signal that does not depend on the input. Because the added signal is nonzero, zero input does not give zero output, so the system is nonlinear. Because the added signal is tied to $n=0$ and is not a constant, it does not move when the input is delayed, so the system is time variant.

   WATCH OUT FOR

   Three traps. (1) Choosing (A) because every block is LTI, without noticing that the $\delta [n]$ sources inside the dashed box are not part of the input. (2) Thinking $+b\delta [n]$ and $-b\delta [n]$ cancel. They do not: the first one passes through $h_2[n]$ and becomes $b{h}_2[n]$, so they cancel only when $h_2[n]=\delta [n]$, which the question rules out. (3) Choosing (C) by treating the added term like a constant bias. A constant offset keeps a system time invariant, but $b(h_2[n]-\delta [n])$ is a signal fixed to $n=0$, not a constant.

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8. Q.18MCQ · 1 markOfficial answer: A

   Signals and Systems › Fourier transform

   $|F(\omega )|=\left|\int f(t)e^{-j\omega t}dt\right|\le \int |f(t)||e^{-j\omega t}|dt=\int |f(t)|dt$. The area under $|f(t)|$ bounds the spectrum's magnitude at every frequency. The bound is reached at $\omega =0$ when $f(t)\ge 0$, which is why the strict inequality in (B) can never hold.

   WATCH OUT FOR

   Choosing (C) or (D) because $F(0)=\int f(t)dt$. The signed area only gives the spectrum's value at $\omega =0$. It bounds $|F(\omega )|$ in neither direction: it can be zero or negative while $|F(\omega )|>0$ elsewhere, which breaks (C), and $|F(\omega )|$ can fall to zero at a spectral null, which breaks (D). A bound that holds at every frequency needs the absolute value inside the integral.

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9. Q.19MCQ · 1 markOfficial answer: A

   Networks (Electric Circuits) › Laws & node/mesh

   You only need the part of the network next to $X$. Ohm's law on the $3\Omega$ resistor gives the junction voltage, $V_N=9-1\times 3=6\text{V}$. Node $X$ has just two branches, so the $2\Omega$ and the $1\Omega$ carry the same current and act as a voltage divider between 8 V and 6 V. The delta and the dotted part never enter the calculation.

   WATCH OUT FOR

   Each wrong option matches a specific slip. (B) $32/3$: taking the 1 A as flowing up, out of $N$, which gives $V_N=9+3=12\text{V}$ instead of 6 V. (C) $22/3$: swapping the divider weights, i.e. putting the larger drop across the $1\Omega$ instead of the $2\Omega$. (D) $2/3$: giving the series current, $2/3\text{A}$, instead of the node voltage. Some students also waste time trying to reduce the delta, which the given 1 A makes unnecessary.

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

    Networks (Electric Circuits) › AC, power & resonance

    In a parallel circuit every branch has the same voltage. When branch values are given as admittances, each branch current is $I=YV$. In polar form the magnitudes multiply and the angles add. The source angle of $90^{\circ }$ gives $90^{\circ }+90^{\circ }=180^{\circ }$ for the capacitor, $90^{\circ }-90^{\circ }=0^{\circ }$ for the inductor and $90^{\circ }+0^{\circ }=90^{\circ }$ for the resistor.

    WATCH OUT FOR

    Treating the siemens values as impedances and computing $I=V/Y$, which gives the magnitudes 4, 10 and 5 in options (B) and (D). A second slip is to take the source as the $0^{\circ }$ reference and forget the $90^{\circ }$ source angle, or to add the angles with the wrong sign. That produces phase sets like the ones in (C) and (D).

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

    Analog Electronics › Amplifiers

    The source resistance $R_S$ and $r_{\pi }$ are in series, so $i_b=v_s/(R_S+r_{\pi })$. The dependent source forces $\beta i_b$ through $R_L$, which gives $|v_o|=\beta i_b{R}_L$. The source arrow points away from the output node, so the output voltage is negative.

    WATCH OUT FOR

    One mistake is to drop $r_{\pi }$ from the input loop and write $i_b=v_s/R_S$, which leads to option (C). The other is to misread the source's direction and lose the minus sign, which leads to (D). The downward arrow means the source pulls current OUT of the $v_o$ node through $R_L$ from ground, so $v_o$ is negative.

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

    Analog Electronics › Biasing

    This is collector-to-base feedback bias. The base current is drawn from the collector node through $R_B$, so the collector resistor carries $I_C+I_B=I_E$, the same current as the emitter resistor. That gives $V_{CE}=V_{CC}-I_E(R_C+R_E)$ directly, and once $I_B$ is given you do not need $R_B$ or $V_{BE}$.

    WATCH OUT FOR

    The wrong options are the other node voltages in this circuit: 4.95 V is $V_C$, 3.03 V is $V_E$, and 3.73 V is $V_B=V_E+0.7$. Picking one of them means you answered for a node voltage instead of the difference $V_C-V_E$. A second trap is putting only $I_C=1\text{mA}$ through the $5\text{k}\Omega$ resistor and forgetting the base current that also leaves the collector node through $R_B$. That gives 1.97 V, or 2.00 V if $I_C$ is used for both resistors, and neither is an option.

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

    Digital Electronics › Boolean logic

    A 3-input majority function is 1 when at least two inputs are 1, so it is $\sum m(3,5,6,7)$. Each 1 in that set pairs with $m_7=XYZ$, so the minimal sum of products is one AND term per pair of inputs: $F=XY+YZ+ZX$. This is also the carry output of a full adder.

    WATCH OUT FOR

    Picking $X\oplus Y\oplus Z$ because it also comes from the full adder. That expression is the SUM output (odd parity), which is 1 when exactly one input is 1. The majority function is the CARRY output, $XY+YZ+ZX$. Testing any single-1 input such as $XYZ=100$ rules out both XOR and OR at once.

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

    Digital Electronics › Combinational

    An XOR gate with one input tied to a control line $M$ passes $Y$ through when $M=0$ and inverts it when $M=1$. If the same $M$ also drives the carry-in, the full adder computes $X+Y$ for $M=0$ and $X+\overline{Y}+1=X-Y$ (2's complement) for $M=1$. This is the standard adder-subtractor, and the difference bit is $X\oplus Y$ while the borrow is $\overline{C_{out}}$.

    WATCH OUT FOR

    Answering "an adder" because the block is labelled full adder, without noticing that the XOR gate complements $Y$ when $Z=1$ and that $Z$ also supplies the carry-in of 1. A second slip is reading $C_{out}$ itself as the borrow. In 2's-complement subtraction the borrow is the complement of the carry-out: here $C_{out}=X+\overline{Y}$, and borrow $=\overline{X}Y$.

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15. Q.25MSQ · 1 markOfficial answer: A, B

    Engineering Mathematics › Calculus

    A local extremum is global only if the function never goes past it anywhere else in the domain. A polynomial of odd degree on all of $\mathbb{R}$ goes to $+\infty$ at one end and $-\infty$ at the other, so it never has a global maximizer or a global minimizer, whatever its local extrema are. To classify the stationary points of $f^{\prime }(x)=6(x-2)(x+1)$, use the sign of $f^{\prime \prime }(x)=12x-6$.

    WATCH OUT FOR

    Swapping the local classifications. For a cubic with positive leading coefficient the smaller critical point ($x=-1$) is the local maximum and the larger one ($x=2$) is the local minimum. Another mistake is to call a local extremum global without checking the ends of the domain: on a closed interval the cubic would have global extrema, but on $\mathbb{R}$ it has none.

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16. Q.26MSQ · 1 markOfficial answer: C

    Control Systems › Root locus

    A point is a closed-loop pole for some $K\ge 0$ only if it lies on the root locus. That is decided by the angle condition, $\angle G(s)=\pm 180^{\circ }$, not by the magnitude condition. A quicker check is to substitute the point into $1+KG(s)=0$ and solve for $K$. The point is reachable only if $K$ comes out real and non-negative; here $K=-j/5$. For two real poles and two real zeros, the complex part of the locus is a circle, here centred at $-2.5$ with radius $0.866$, and $-1+j1$ is $1.80$ from that centre.

    WATCH OUT FOR

    Using only the magnitude condition. $|G(-1+j1)|=5$ suggests $K=1/5$, which is option (B), or $K=5$ if the ratio is inverted, which is option (A). Both skip the angle test, which gives $-90^{\circ }$. Another trap is judging from the drawing: the stretched vertical scale makes the circle look as if it reaches $j1$, but its true radius is $0.866$ and it never gets near $-1+j1$.

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17. Q.27MSQ · 1 markOfficial answer: C, D

    Signals and Systems › z-transform

    The DTFT is $X(z)$ evaluated on $|z|=1$. The RoC is defined by absolute convergence, so a unit circle inside the RoC means ${\sum }_n|x[n]|<\infty$, and then the DTFT exists. The RoC can never contain a pole, but it can contain zeros. A signal that is nonzero only at $n=0$ has a constant $X(z)$, so its RoC is the whole $z$-plane with no exception at $0$ or $\infty$.

    WATCH OUT FOR

    Marking (B) true by misremembering the rule. The RoC contains no poles, but zeros can lie inside it. Marking (A) true on the assumption that any signal with a $z$-transform also has a DTFT; a growing signal such as $2^{n}u[n]$ has one and not the other. In (D), excluding $z=0$ or $z=\infty$ by habit. Those exclusions apply to a shifted impulse $\delta [n-k]$ with $k\ne 0$, not to $\delta [n]$.

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18. Q.28MSQ · 1 markOfficial answer: A, D
19. Q.29MSQ · 1 markOfficial answer: A, C

    Analog Electronics › Op-amps

    The ideal op-amp is an ideal voltage-controlled voltage source that responds only to the difference between its inputs. Its input resistance and differential gain are infinite; its output resistance and common-mode gain are zero, so CMRR is infinite.

    WATCH OUT FOR

    Marking every option with the word "infinite". Output resistance is zero, not infinite: an ideal voltage source must not depend on the load. Common-mode gain is zero, not infinite: it is the CMRR that is infinite. Students often mix up $A_{cm}$ and CMRR.

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

    Analog Electronics › Amplifiers

    Look at which terminal the signal enters. If it enters at the gate (common source or common drain), the input resistance is infinite because an ideal gate draws no current. If it enters at the source (common gate), the input resistance is only $1/g_m$. For phase, look at the sign of the gain: common source gives $-g_m{R}_D$ (inverting), while common drain gives $+\frac{g_m{R}_S}{1+g_m{R}_S}$ and common gate gives $+g_m{R}_D$ (both non-inverting).

    WATCH OUT FOR

    Assuming every MOSFET amplifier has infinite input resistance because the gate is insulated. In the common-gate stage the signal enters at the source, not the gate, and the whole channel current flows through the input, so $R_{in}=1/g_m$. A second slip is treating the common-source stage as non-inverting. Its gain $-g_m{R}_D$ carries a $180^{\circ }$ phase shift.

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21. Q.31MSQ · 1 markOfficial answer: A, D

    Electronic Devices › Semiconductors

    n-type doping of a Group IV semiconductor (silicon) requires pentavalent (Group V) donor impurities, which contribute a loosely bound extra electron to the conduction band; p-type doping requires trivalent (Group III) acceptor impurities.

    WATCH OUT FOR

    Confusing donor and acceptor group numbers — assuming Group III elements donate electrons, or forgetting that silicon (Group IV) needs a 5-valence-electron dopant to supply the extra "donor" electron for n-type conduction.

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22. Q.32NAT · 1 markOfficial answer: 0.25

    Engineering Mathematics › Differential equations

    This is a Cauchy-Euler equation, where each derivative $y^{(k)}$ is multiplied by $t^k$. Substituting $y=t^m$ turns it into the algebraic equation $m(m-1)-2m+2=0$. Its roots $m=1,2$ give the polynomial solution $y=c_{1}t+c_2{t}^2$, and the two conditions on $y^{\prime }$ fix $c_1$ and $c_2$. What is left is an ordinary maximisation over a closed interval: check the stationary point and both endpoints.

    WATCH OUT FOR

    The most common slip is reading the conditions as $y(0)=1$ and $y(1)=-1$ when they are on $y^{\prime }$. This cannot even work, because every solution has $y(0)=0$. The second is writing the auxiliary equation as $m^2-2m+2=0$, which drops the $-m$ from $m(m-1)$ and gives false complex roots. The third is forgetting to compare the stationary value with the endpoint values $y(0)$ and $y(1)$.

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23. Q.33NAT · 1 markOfficial answer: 3
24. Q.34NAT · 1 markOfficial answer: 50

    Analog Electronics › Filters & waveforms

    The input network (R in series with 10C) is a high-pass section with corner 1/(2π·10RC). The feedback network (2R in parallel with 0.1C) is a low-pass section with corner 1/(2π·0.2RC). The midband gain is -2R/R = -2. The -3 dB points, taken 3 dB below that midband gain, fall at 1.0004 and 49.98 times the lower corner. That makes f_H/f_L = 49.96, which is essentially the corner ratio 10RC/0.2RC = 50, so f_H is about 50 MHz.

    WATCH OUT FOR

    Measuring -3 dB from the actual peak gain (about 1.96, at the geometric-mean frequency) instead of from the midband gain -2R/R = 2. That moves the band edges apart and gives 54 MHz, which is wrong. A second mistake is pairing the wrong components: using R with 0.1C, or 2R with 10C, for the corners gives the wrong ratio (100 or 25 instead of 50).

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25. Q.35NAT · 1 markOfficial answer: 250

    Digital Electronics › ADC & DAC

    In a weighted-resistor DAC with a virtual-ground summing node, $V_O=-V_{REF}\frac{R_f}{R}\left(b_3+\frac{b_2}{2}+\frac{b_1}{4}+\frac{b_0}{8}\right)$. The output is proportional to the binary value of the input code, so any change in output equals (change in code) $\times$ (one LSB step). Here one LSB is $V_{REF}{R}_f/(8R)=0.25\text{V}$, and $14\to 13$ is a change of one step.

    WATCH OUT FOR

    Two bits flip ($b_1$: $1\to 0$, $b_0$: $0\to 1$), and a common error is to add the two contributions as magnitudes ($0.5+0.25=0.75\text{V}$) when they have opposite signs and partly cancel, leaving a net $0.25\text{V}$. Two other slips: pairing the LSB $b_0$ with the smallest resistor $R$ instead of $8R$, which reverses the bit weights, and taking the step as $V_{REF}/2^4=0.125\text{V}$ as if this were an R-2R ladder with full-scale $V_{REF}$. In this circuit the feedback resistor equals the MSB resistor, so the MSB alone gives $V_{REF}$ and the LSB step is $V_{REF}/8$.

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

    Control Systems › Compensators & PID

    A double integrator $\frac{1}{10s^2}$ has no $s^1$ term in its characteristic polynomial, so it has no damping. Stabilising it needs a controller that adds that term. A PD controller adds $K_{D}s$ and gives $10s^2+K_{D}s+K_P$, which is stable for all positive gains. An integral controller raises the order to $10s^3+K_I$ and still leaves the $s^2$ and $s$ coefficients at zero. That fails the necessary coefficient condition, whatever the gain.

    WATCH OUT FOR

    One mistake is misreading the wording of the statements. Statement (i) is a claim that no stabilising controller exists. The integral loop is unstable, so (i) is TRUE, not FALSE. A second mistake is stopping the Routh test for $10s^3+K_I$ at the zero in the first column of row $s^2$ and calling the system marginally stable. Completing the array with $\epsilon$ shows two sign changes, so two roots are in the right half-plane and the loop is unstable. A third is thinking that (ii) needs a particular tuning of $K_P$ and $K_D$. For a second-order polynomial, positive coefficients alone guarantee stability.

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27. Q.37MCQ · 2 marksOfficial answer: A

    Control Systems › Routh–Hurwitz

    A whole row of zeros in the Routh array means $p(s)$ has roots placed symmetrically about the origin, and those roots are the roots of the auxiliary polynomial built from the row above. After replacing the zero row with $dA/ds$, the total number of sign changes in the first column counts all right-half-plane roots. The sign changes from the auxiliary row downwards count only the auxiliary polynomial's right-half-plane roots. Here that count is zero, so both roots of $A(s)=-40(s^2+1)$ lie on the $j\omega$ axis.

    WATCH OUT FOR

    Stopping at the zero row, or putting a small $ϵ$ in place of the whole zero row (the $ϵ$ method is only for a single zero in the first column). That hides the $\pm j$ pair and leads to option (B) or (C). A second trap is reading the one sign change as the only non-left-half-plane root and forgetting to count the imaginary pair: that gives $L=4$, which matches no option.

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

    Signals and Systems › Sampling

    Shifting the sampling grid by $\tau$ multiplies the $k$-th spectral copy by $e^{-jk{\omega }_s\tau }$. The baseband copy ($k=0$) gets $e^0=1$, so the offset has no effect on what the lowpass filter recovers. The sampled spectrum carries a factor $1/T_s$, which the passband gain $T_s$ cancels. Nyquist (${\omega }_s>2{\omega }_c$) gives $T_s<\pi /{\omega }_c$.

    WATCH OUT FOR

    Picking (B) on the belief that the offset $\tau$ delays the output. Here $f$ is multiplied by the shifted train, so each impulse carries $f$'s true value at its own instant, and the only $\tau$-dependent phase lands on the $k\ne 0$ copies, which the filter removes. Other traps: writing the Nyquist condition as ${\omega }_s>{\omega }_c$, which gives $2\pi /{\omega }_c$ as in (C) and (D), and forgetting that the gain $T_s$ cancels the $1/T_s$ from sampling, as in (D).

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

    Networks (Electric Circuits) › AC, power & resonance

    The meters measure the load branch directly, so the unknown LTI box does not matter. The only thing the box sets is the frequency, which stays at $\omega =200\text{rad/s}$. The readings give $|Z|=V_{\text{rms}}/I_{\text{rms}}=5\Omega$, which equals the branch resistance, so the net reactance must be zero (series resonance) and $C=1/({\omega }^{2}L)$.

    WATCH OUT FOR

    Picking (D) because the contents of the box are unknown. The box only has to be LTI, which fixes the frequency at 200 rad/s, and the voltmeter and ammeter already give the branch's voltage and current. A second mistake is to use the 100 V source amplitude, or to mix peak and rms values, instead of the ratio of the two meter readings. A third is to use $f=200\text{Hz}$ instead of $\omega =200\text{rad/s}$, which gives the wrong capacitance.

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

    Networks (Electric Circuits) › Two-port

    Open-circuit one port at a time. With a port open, its series resistor carries no current, so it drops out of the input impedance and adds no drop to the output voltage. That makes $z_{11}$ a series-parallel reduction and $z_{21}$ a single current division. Reciprocity (the network has only resistors) and mirror symmetry then give $z_{12}=z_{21}$ and $z_{22}=z_{11}$ with no extra work.

    WATCH OUT FOR

    Each wrong option matches a specific slip. Adding all five resistors in series, as if the shunts were series elements, gives 10 and 2 (option C). Writing the divided current $I_1/3$ as $z_{21}$ and forgetting to multiply by the $2\text{Ω}$ shunt at B gives 1/3 (option D). Swapping the driving-point and transfer entries gives option B: $z_{11}$ must be the larger one, since $V_1$ includes the drop across the first series resistor. Another trap is counting the open-circuited right-hand series resistor in $z_{21}$. With $I_2=0$ it has no drop across it.

    Full solution →
31. Q.41MCQ · 2 marksOfficial answer: D
32. Q.42MCQ · 2 marksOfficial answer: A
33. Q.43MCQ · 2 marksOfficial answer: D

    Analog Electronics › Amplifiers

    In a symmetric differential pair driven by a purely differential signal, the tail current splits equally at DC and the common source node acts as an AC ground. Each drain therefore behaves like a common-source stage driven by half the differential input. At drain 1 the output is $V_{DD}-\frac{I_{SS}}{2}{R}_D$ at DC, plus an AC part of $-g_m\frac{v_d}{2}{R}_D$.

    WATCH OUT FOR

    One mistake is to use the full differential input $v_d=0.02\sin \omega t$ rather than $v_d/2$ at each gate, which gives $0.1\sin \omega t$ (options B, C). Another is to drop the inversion at the drain of M${}_1$: $V_1$ rising makes $V_3$ fall. A third is to put the whole 2 mA tail current through one resistor, which gives a DC level of 3 V instead of 4 V.

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

    Digital Electronics › ADC & DAC

    An ideal $n$-bit ADC driven by a full-scale sinusoid has ${\text{SNR}}_{\max }=\frac{3}{2}\cdot 2^{2n}$, which is $6.02n+1.76$ dB. This depends only on the resolution, not on $V_{FS}$ or the signal frequency. The output bit rate is resolution times sampling rate, $R_b=n{f}_s$.

    WATCH OUT FOR

    Using the 500 kHz input frequency instead of the 1 MHz sampling frequency for the data rate gives $10\times 0.5=5$ Mbps (options B and D). Another mistake is thinking the 3.3 V full-scale voltage changes the SNR. It cancels, because the signal and the quantisation step both scale with $V_{FS}$. Some students also drop the $+1.76$ dB term, or use $10{\log }_{10}$ where $20{\log }_{10}$ belongs.

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

    Digital Electronics › Sequential

    The XOR is fed by both complemented outputs, so $Y=\overline{Q_1}\oplus \overline{Q_2}=Q_1\oplus Q_2$. At each rising edge, read SEL and S. When $SEL=1$ the two flip-flops act as a 2-bit shift register ($Q_1\leftarrow S$, $Q_2\leftarrow$ old $Q_1$). When $SEL=0$ they load the constants $Q_1{Q}_2=01$, whatever their earlier state. Each $T_k$ is a falling edge, so it shows the state loaded at the rising edge just before it.

    WATCH OUT FOR

    One mistake is feeding FF1's new value into FF2 in the same clock edge, as if the flip-flops rippled. At the third edge this gives $Q_2=1$ and $Y(T_2)=0$, so the sequence becomes 1001, which is not an option. The other is reading SEL at the dashed line $T_2$, where it is still 1, instead of at the next rising edge, where it is already 0. That gives $Q_1{Q}_2=11$ at $T_3$ and the sequence 1010. Taking the XOR inputs as $Q$ instead of $\overline{Q}$ does not change the answer, because complementing both inputs leaves the XOR unchanged.

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

    Electronic Devices › Carrier transport

    In an intrinsic semiconductor, electrons and holes exist in equal concentration $n_i$, so both carrier types contribute to conductivity: ${\sigma }_i=e{n}_i({\mu }_n+{\mu }_p)$, and resistivity is simply the reciprocal, ${\rho }_i=1/{\sigma }_i$.

    WATCH OUT FOR

    Using only the electron mobility (or only the hole mobility) instead of the sum ${\mu }_n+{\mu }_p$, or forgetting to convert the final answer from $\Omega \cdot \text{m}$ to $\text{k}\Omega \cdot \text{m}$ (giving 1250 instead of 1.25).

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

    Analog Electronics › Biasing

    The diode-connected Q2 and the Zener form a fixed voltage stack below the 20 V rail. That stack pins Q1's base at $20-0.7-5=14.3\text{V}$, and so its emitter at $15\text{V}$. $R_2$ then sets the emitter current $I_E=(1+\beta )I_L/\beta$ across a fixed 5 V drop. $R_1$ carries the Zener current plus the PNP base current that flows out of Q1's base.

    WATCH OUT FOR

    A common slip is measuring the Zener voltage from ground and putting Q1's base at 5 V. The Zener actually sits under Q2, so what $R_1$ sees is $20-0.7-5=14.3\text{V}$. The other trap is units: volts divided by milliamps gives kilo-ohms directly, and a slip of a factor of 10 lands on option (D). The base current (0.1 mA) barely changes either value at one decimal place, but its direction matters. For a PNP it flows out of the base, so it adds to the Zener current in $R_1$.

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

    Electronic Devices › Carrier transport

    The Einstein relation links diffusivity and mobility: $D_n/{\mu }_n=V_T=k_{B}T/q$, the thermal voltage.

    WATCH OUT FOR

    Forgetting to convert from m²/s to cm²/s (multiply by 10⁴), which would give an answer near 0.0098 and no matching option, or misreading μn's units.

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

    Electromagnetics › Electrostatics

    Total charge is the surface integral of ${\rho }_s$ over the sheet. Here ${\rho }_s=4|y|$ does not depend on $x$, so the integral splits into (width in $x$) $\times$ $\int |y|dy$. Because $|y|$ is even, ${\int }_{-2}^2|y|dy$ is twice the integral from $0$ to $2$.

    WATCH OUT FOR

    Dropping the modulus and integrating $4y$ from $-2$ to $2$ gives $Q=0$. The absolute value makes the density positive on both halves, so the two halves add instead of cancelling. The other common slip is to treat ${\rho }_s$ as uniform, using its edge value $4\times 2=8\mu {\text{C/m}}^2$ times the area $16{\text{m}}^2$, which gives $128$. The density rises linearly from zero at $y=0$, so its average over the sheet is only $4\mu {\text{C/m}}^2$.

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

    Electromagnetics › Electrostatics

    Use Ohm's law in point form, $\mathbf{J}=\sigma \mathbf{E}$. With a steady field the current density is uniform across the wire, so the current is just $J$ times the cross-sectional area: $I=\sigma E\pi r^2$.

    WATCH OUT FOR

    Putting the diameter in place of the radius gives $A=\pi (10^{-3}{)}^2$ and a current four times too large, $1.82\text{A}$, which is option (B). Forgetting the factor $\pi$ gives $0.145\text{A}$, and taking the area as $d^2$ gives $0.58\text{A}$, which is option (C). Make sure mm is converted to m before you square it.

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

    Engineering Mathematics › Calculus

    Bound the integral using the interval length: $(b-a)m\le {\int }_a^{b}f(x)dx\le (b-a)M$, which here is $6m\le {\int }_2^{8}f\le 6M$. A constant function reaches both bounds, so a lower bound $pm$ needs $p\le 6$ and an upper bound $qM$ needs $q\ge 6$. The step from $6m$ down to $5m$ and from $6M$ up to $7M$ is valid only because $f\ge 0$.

    WATCH OUT FOR

    Using the upper limit $8$ (or the lower limit $2$) as the multiplier instead of the interval length $8-2=6$. Another mistake is dropping the non-negativity condition: if $f$ could be negative, then $5m>6m$ when $m<0$, and even (A) fails. For example, $f(x)=-1$ gives an integral of $-6$, which is less than $5m=-5$. A third mistake is marking (C) because $\alpha =6M$ is a valid upper bound, while missing that $\beta =7m$ breaks the lower bound. Both halves must hold.

    Full solution →
42. Q.52MSQ · 2 marksOfficial answer: A, B

    Engineering Mathematics › Complex analysis

    By Cauchy's theorem, a contour integral is zero whenever the integrand is analytic on and inside the contour. So first find where the singularities are. For powers of $z$, use the standard result ${\oint }_{|z|=1}{z}^{n}dz=2\pi j$ if $n=-1$ and $0$ for every other integer $n$. A pole inside the contour does not by itself make the integral nonzero; only the residue counts.

    WATCH OUT FOR

    Two traps. First, treating (B) as false because a negative even power such as $z^{-2}$ has a pole inside $C$: that pole's residue is $0$, so the integral is still $0$. Second, marking (D) true because $\sec z$ has poles at all: they sit at $\pm \pi /2,\pm 3\pi /2,\dots$ (all real, since $\cos z$ has no non-real zeros), which is outside the unit circle. Also watch the $\ne$ sign in (C) and (D): both integrals are $0$, so those statements are false.

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

    Control Systems › State space

    With a diagonal $A$, the states decouple into first-order systems $\frac{1}{s-{\lambda }_i}$. Every $b_i=1$, so every mode is driven by the input, and each state's boundedness is decided by its own eigenvalue alone. $\lambda =2$ (right half-plane) and $\lambda =0$ (a pole at the origin, whose step response is a ramp) can be made unbounded by a bounded input. $\lambda =-2$ cannot. So the option you need is "at least one state is unbounded", not "all states are unbounded".

    WATCH OUT FOR

    Treating the whole system as one unit and choosing (C) because it is unstable, without noticing that $x_2$ ($\lambda =-2$) stays bounded for every bounded input. Another mistake is to count the $\lambda =0$ state as marginally stable and so bounded: under a constant input, a pole at the origin integrates the input into a ramp.

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

    Signals and Systems › Fourier series

    Time scaling $y(t)=f(\alpha t)$ compresses the period to $T_0/\alpha$ and raises the fundamental frequency to $\alpha {\omega }_0$, but the Fourier series coefficients stay the same. The same harmonic weights now sit at harmonics of a higher fundamental frequency. In the analysis integral, the $1/\alpha$ from the change of variable cancels the $\alpha$ in $1/(T_0/\alpha )$.

    WATCH OUT FOR

    Carrying over the Fourier transform scaling rule $f(\alpha t)\leftrightarrow \frac{1}{|\alpha |}F(\omega /\alpha )$ and picking (C). The $1/\alpha$ appears in the transform because the transform integrates over all time. The series averages over one period, and the shorter period cancels it. A second slip is thinking compression lengthens the period and choosing (B).

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

    Control Systems › Block diagrams & SFG

    In a signal flow graph, a feedback branch must start at a node that carries exactly the signal the block diagram feeds back. Here the feedback takes $G_1$'s output before $G_2$'s output is added. The graph therefore needs a node that holds $G_{1}E$ alone, separate from a later node where $G_{2}E$ joins. Only graph B has one, so its only loop is $-G_1$ and $G_2$ stays a feedforward path.

    WATCH OUT FOR

    Treating the take-off point after $G_1$ and the second summing junction as one node. Graphs A and D do this. The fed-back signal then becomes $(G_1+G_2)E$, which adds a second loop $-G_2$ and gives $\Delta =1+G_1+G_2$. A second slip is missing the sign change in C: a $+1$ feedback branch turns the minus signs in $\Delta$ into plus signs.

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

    Analog Electronics › Diode circuits

    A full-wave bridge rectifies what the LOAD sees, not what the SOURCE sees. The current through R always flows the same way ($Q\to P$), so $V_O=|V_I|$. The current at the source terminals still reverses with $V_I$. With ideal diodes the source sees the resistance R connected one way round or the other, so $I_I=V_I/R$ is linear and odd.

    WATCH OUT FOR

    Picking (C) by assuming that since the output is rectified, the input current is rectified too. The load current $V_O/R=|V_I|/R$ is always positive, but for $V_I<0$ the source current flows against the marked $I_I$ arrow, so $I_I$ is negative. Picking (B) is the other slip: it treats the bridge as a straight-through connection and forgets that the two diodes conducting for $V_I<0$ swap which end of R connects to which terminal.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 6.95 to 7.05

    Engineering Mathematics › Probability & statistics

    Linearity of expectation: the mean of a sum is the sum of the means, $E[X_1+X_2]=E[X_1]+E[X_2]$, so there is no need to build the distribution of the sum. One fair die has mean $3.5$, so two dice have mean $7$.

    WATCH OUT FOR

    Averaging only the 11 possible sums, $2$ to $12$, as if each were equally likely. The mean happens to come out as $7$ here, but that is luck from symmetry: the sums are not equally likely ($P(X=7)=\frac{6}{36}$ while $P(X=2)=\frac{1}{36}$), and the same method fails for any non-symmetric case. A second mistake is thinking linearity needs the dice to be independent. It does not; independence matters for the variance, not the mean.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 2.95 to 3.05

    Engineering Mathematics › Linear algebra

    The smallest value of $\Vert \mathbf{a}x-\mathbf{b}{\Vert }_2$ is the distance from $\mathbf{b}$ to the line spanned by $\mathbf{a}$. It is reached at the orthogonal projection, $x^{\ast }=\frac{{\mathbf{a}}^T\mathbf{b}}{{\mathbf{a}}^T\mathbf{a}}$. The minimum distance squared is $\Vert \mathbf{b}{\Vert }_2^2-\frac{({\mathbf{a}}^T\mathbf{b}{)}^2}{{\mathbf{a}}^T\mathbf{a}}$.

    WATCH OUT FOR

    Students often report the minimiser $x^{\ast }=\frac{3\sqrt{2}}{2}\approx 2.12$ when the question asks for the minimum value of the norm, which is 3. Another slip is to report the squared norm, 9. A third is to forget to divide by ${\mathbf{a}}^T\mathbf{a}=2$, which gives $x^{\ast }=3\sqrt{2}$ and a residual of norm $3\sqrt{2}\approx 4.24$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 0.0
51. Q.61NAT · 2 marksOfficial answer: 13.32 to 13.34

    Analog Electronics › Diode circuits

    An ideal diode conducts whenever its anode is above its cathode. Read the battery polarity from the figure: the long plate is +, and here it is grounded. That puts the cathode at −5 V, so the diode conducts whenever V_I > −5 V. Turn this into a phase condition, sin θ > −1/2, which holds for 4π/3 of every 2π rad. The diode therefore conducts for 2/3 of the 20 ms period.

    WATCH OUT FOR

    The usual slip is misreading the battery polarity and taking the cathode as +5 V. That leads to the condition sin θ > 1/2, conduction for only 1/3 of the cycle, and an answer of 6.67 ms. The other slip is dropping the battery altogether and treating the circuit as a plain half-wave rectifier, which gives 10 ms.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 200

    Digital Electronics › Sequential

    Check every path from one flip-flop to the next with $T_{clk}\ge t_{clk\text{-}Q}+t_{comb}+t_{setup}$. The path with the most logic in it (here the feedback loop through the AND gate) sets the minimum clock period, and $f_{max}=1/T_{min}$.

    WATCH OUT FOR

    One mistake is checking only the direct FF1 → FF2 wire, which gives 4 ns and 250 MHz. Another is adding the hold time into the set-up equation, or adding the delays of both paths together (2 + 2 + 1 + 2 + 2 = 9 ns) as if the data had to cross both flip-flops in one clock cycle.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.23 to 0.24

    Electronic Devices › P-N junction

    In strong forward bias, $I\gg I_S$, so the diode equation $I=I_S(e^{V/V_T}-1)$ reduces to $I\approx I_S{e}^{V/V_T}$, giving $V=V_T\ln (I/I_S)$ with $V_T=k_{B}T/q$.

    WATCH OUT FOR

    Rounding $\ln (10^4)$ too early (e.g. using $\ln 10\approx 2.3$ instead of $2.302585$) or misplacing a decimal in $V_T$ shifts the final answer's second decimal place, since the true value (0.2383 V) sits close to the 0.235 rounding boundary.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 6.3 to 6.5

    Electromagnetics › Transmission lines

    On a lossless line the load receives the incident power minus the reflected power: $P_L=P_{inc}(1-|{\Gamma }_L{|}^2)$. Only $|{\Gamma }_L|$ matters. Here $|{\Gamma }_L{|}^2=75^2/(100^2+75^2)=0.36$, so the load takes 64% of the incident power. The phase of ${\Gamma }_L$ does not enter.

    WATCH OUT FOR

    Using $|{\Gamma }_L|$ in place of $|{\Gamma }_L{|}^2$ gives $10(1-0.6)=4\text{mW}$, which is wrong because power goes as the square of the voltage reflection coefficient. Other errors: reading the reflected power (3.6 mW) as the answer, or thinking the reactive part $-j75$ takes power. Only the resistive part of $Z_L$ absorbs power, and the reactance only sets how much power is reflected.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 0.62 to 0.63

    Electromagnetics › Maxwell's equations

    A uniform flux density that rises linearly with time gives a constant motional-free (transformer) emf, $|\mathcal{E}|=A\frac{dB}{dt}$. Every element in a single closed loop carries the same current, so the emf drives that current through the total series resistance.

    WATCH OUT FOR

    Treating the two $2\Omega$ resistors as parallel because they are drawn on opposite sides, which gives $R=1\Omega$ and $I=2.5\text{A}$. There is only one loop, so they are in series. A second slip is using $B=0.5t$ itself as the flux and forgetting to multiply by the $5{\text{m}}^2$ area, which gives $0.125\text{A}$.

    Full solution →