GATE EE 2024 question paper with solutions

GATE EE 2024 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Reasoning

   In an intensity analogy, every word in a chain names the same act at a different strength. Pin down the mild end and the extreme end first. The blank is then the option that means the same act at a strength between the two. Here that act is pleasing someone: please (mild), flatter (stronger, insincere praise), pander (extreme, improper indulgence).

   WATCH OUT FOR

   Choosing by sound rather than meaning. flatter, flutter, fritter and frizzle look and sound alike, and 'flutter' in particular is easily mistaken for 'flatter' under exam pressure. Only flatter is about pleasing a person. The other three have nothing to do with the please–pander scale.

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2. Q.2MCQ · 1 markOfficial answer: A

   General Aptitude › Reasoning

   Turn the story into firm conditions: the table must be at the window, and Bed Q must be close to the fan. Then strike out every plan that breaks either one. The almirah, the door and Bed P don't affect the answer.

   WATCH OUT FOR

   Checking only one condition, or checking the wrong bed. (B) puts the table at the window but leaves Bed Q far from the fan. (D) puts Bed Q next to the fan but gives the window to Bed Q instead of the table. (C) has a bed next to the fan, but it is Bed P, not Bed Q.

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3. Q.3MCQ · 1 markOfficial answer: D

   General Aptitude › Numerical

   In octal the place values are powers of 8 ($8^0=1$, $8^1=8$, $8^2=64$). To convert, divide by 8 repeatedly and read the remainders from bottom to top. Then expand the result in powers of 8 to confirm it gives back the decimal number.

   WATCH OUT FOR

   Reading the remainders top to bottom gives 451 instead of 154. Another slip is choosing 168 or 108 without noticing that 168 uses the digit 8, which octal does not have, and that 108 is just the decimal numeral copied unchanged.

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4. Q.4MCQ · 1 markOfficial answer: C

   General Aptitude › Numerical

   Profit and tax are successive percentage increases, so their multipliers multiply: final price $=1.2\times 1.18\times$ cost $=1.416\times$ cost. To go back to the cost, divide by each multiplier in reverse order (tax first, then profit). Never subtract the percentages from the final price.

   WATCH OUT FOR

   Removing the percentages from the final price as if they were shares of it. For example, taking 20% off Rs. 3,186 gives $3186\times 0.8=2548.80$, which is option (D). That treats the profit as 20% of the price the customer paid, when it is 20% of the cost, and it forgets the tax completely. A 20% markup is undone by dividing by 1.2, not by multiplying by 0.8, and the 18% tax is undone by dividing by 1.18.

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5. Q.5MCQ · 1 markOfficial answer: A

   General Aptitude › Numerical

   A ratio fixes $x$ and $y$ only up to a common factor. Write $x=k(a+1)$ and $y=k(a-1)$. Both $x^2-y^2$ and $x^2+y^2$ are homogeneous of degree 2, so $k^2$ cancels. You can also use componendo-dividendo: from $x^2:y^2=(a+1{)}^2:(a-1{)}^2$ you get $(x^2-y^2):(x^2+y^2)=4a:(2a^2+2)=2a:(a^2+1)$.

   WATCH OUT FOR

   Forgetting the cross terms: $(a+1{)}^2+(a-1{)}^2$ is $2a^2+2$, because the $+2a$ and $-2a$ cancel. It is not $2a^2$ and it is not $(2a{)}^2$. Another slip is dropping the factor 2 in $x-y=2k$, which leads to option (B), $a:(a^2+1)$. Mixing up $x^2-y^2$ with $x^2+y^2$ (or $x+y$ with $x-y$) leads to the $a^2-1$ options, (C) and (D).

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6. Q.6MCQ · 2 marksOfficial answer: C

   General Aptitude › Grammar & vocabulary

   "Row" is used in two senses. "A row with someone over something" is a quarrel: "with" names the other person and "over" names the cause. "Stand in a row" is a line of people, and "at" marks a point location such as an ATM booth.

   WATCH OUT FOR

   Reading both uses of "row" as a quarrel, or both as a line, and then picking "at a row" or "over a row" for blank (iii). Another mistake is putting "over" in blank (i) as in (D). "Over" names what the quarrel is about, not who it is with.

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7. Q.7MCQ · 2 marksOfficial answer: Marks to all

   General Aptitude › Geometry

   In a rhombus all sides are equal (so AD = CD) and the diagonals bisect each other at right angles (so BF = DB). Pythagoras in right triangle DAE gives the hypotenuse DE, and subtracting BE along the same line leaves DB.

   WATCH OUT FOR

   Answering 6, which is the whole diagonal DF rather than half of it (BF), or 4, which is AB rather than BF. Some solvers also miss that AD = CD because both are sides of the rhombus. Separately, the data over-determine the figure and contradict each other: triangle ABE forces AB² = 44, but AB is a leg of right triangle ABD with hypotenuse AD = 5.

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8. Q.8MCQ · 2 marksOfficial answer: C

   General Aptitude › Data interpretation

   This is a dual-axis chart. Read the bars on the right axis (people buying cars) and the fuel line on the left axis (Million liters). Then check each option on exactly the years and the group it names. A percentage change is always the change divided by the value at the start of the period.

   WATCH OUT FOR

   The usual slip is mixing up the periods. Options (B) and (D) deliberately set one group's change over 2010 to 2015 against the other group's change over 2015 to 2020, and students often compute both over the same period. The second slip is dividing by the end value in (A). That gives 16.7% against 33.3%, which happens to point the same way here but is wrong in method. The third is reading the fuel line on the buyers axis.

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9. Q.9MCQ · 2 marksOfficial answer: B

   General Aptitude › Spatial

   Meshing gears turn in opposite directions. A rack moves with the point of the pinion it touches: the top of a clockwise wheel moves right and its bottom moves left. Follow the one chain of contacts from A to B: P clockwise, Q anticlockwise, R clockwise, so B moves left.

   WATCH OUT FOR

   The usual slip is to count only the gear-to-gear meshes (two, so R turns the same way as P) and then say B moves the same way as A. That misses the extra reversal from A sitting on top of its pinion while B sits underneath its pinion. Another slip is to assume P and R mesh directly because they sit close together in the drawing. They do not touch, so the motion has to go through Q.

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

    General Aptitude › Geometry

    Equal base angles make triangle ABC isosceles. The surveyor stands at the midpoint M of AB, so CM is both the median and the altitude. That gives a right triangle with a 100 m leg next to the 87.8° angle, so $CM=100\tan 87.8^{\circ }=100/\tan 2.2^{\circ }$.

    WATCH OUT FOR

    The usual slip is working out the distance from an endpoint instead of from the surveyor. The slant side CA is the hypotenuse: $CA=100/\cos 87.8^{\circ }=100/\sin 2.2^{\circ }\approx 2605.0\text{m}$, which rounds to the distractor 2606 (B). The required distance is the altitude CM, the leg opposite the 87.8° angle. The approximation needs care as well. Taking $\tan 2.2^{\circ }\approx 2.2^{\circ }$ in radians gives about 2604.4, which is only just closer to 2603 than to 2606. Keep the $x^3/3$ term, or use a table value $\tan 2.2^{\circ }\approx 0.03842$, to get a clear answer of 2603.

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Electrical Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: C

   Engineering Mathematics › Linear algebra

   A square matrix is invertible exactly when its determinant is non-zero. A quick way to rule out options is to look for a row (or column) that is a scalar multiple of another, because that forces $\det =0$. For the one option left, compute the determinant to confirm it is non-zero.

   WATCH OUT FOR

   Picking (C) only because none of its rows is a multiple of another. Having no proportional pair of rows does not prove invertibility, since a row can still be a combination of two others. Always compute $\det$ for the matrix you choose. Another slip is the sign pattern $+,-,+$ in the cofactor expansion. Getting the middle term wrong gives $12-8-32=-28$, which happens to be non-zero as well, so the error goes unnoticed.

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2. Q.12MCQ · 1 markOfficial answer: A

   Networks (Electric Circuits) › Laws & node/mesh

   A junction (principal node) is a point where three or more elements meet. First merge every set of points joined only by bare wire into one node, then count the elements at each node. Nodes with only two elements, such as corners and the midpoint of a series pair, are not junctions.

   WATCH OUT FOR

   There are two common mistakes. The first is counting the ends of a bare wire as separate junctions: O and E, T and T', H and H' are three pairs, and each pair is one node. The second is counting every node rather than only junctions. The circuit has 9 nodes, which is option (D), but A, F and the point between the capacitor and the resistor in the middle branch join only two elements each. A third slip is counting only the solid dot, since the drawing marks just one of the six junctions with a dot.

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3. Q.13MCQ · 1 markOfficial answer: A

   Networks (Electric Circuits) › Laws & node/mesh

   The ideal current source sets the current in its own branch, and the $2\Omega$ resistor in series with it has no effect on that current. If the voltage source carried no current, the whole $10\text{A}$ would pass through the $1\Omega$ resistor and put exactly $10\text{V}$ on node A. That matches the source voltage, so KCL is satisfied with zero current in $\alpha$ for any value of $\alpha$. Because the voltage source carries no current, its power is zero.

   WATCH OUT FOR

   Picking (D) on the grounds that α is unknown, without checking whether it cancels. It does: $V_A(1+\alpha )=10(1+\alpha )$. A second mistake is to add the $2\Omega$ resistor into the node equation as $(V_A-V_x)/2$ with an unknown $V_x$, or to charge the current source's $10\text{A}$ to the voltage source and get $100\text{W}$, option (C). The current source's current goes entirely into node A and then down the $1\Omega$ resistor.

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4. Q.14MCQ · 1 markOfficial answer: A

   Networks (Electric Circuits) › Transients

   To find a time constant, set the independent sources to zero. A current source becomes an open circuit, so the left R–L branch ends up in series with the right-hand path. That leaves one R–L loop with R_eq = 2 Ω and L_eq = 1 + 1 + (1 ∥ 1) = 2.5 H, so τ = L_eq/R_eq = 1.25 s.

   WATCH OUT FOR

   Replacing the current source with a short circuit, as you would a voltage source. The short then bypasses the left 1 Ω + 1 H branch and leaves R = 1 Ω and L = 1 + 0.5 = 1.5 H, giving τ = 1.5 s, which is the distractor (D). The same wrong answer comes from ignoring the left branch because it 'belongs to the source'. A second mistake is adding the two right-hand 1 H inductors in series (as 2 H) when they are actually in parallel (0.5 H).

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5. Q.15MCQ · 1 markOfficial answer: A

   Signals and Systems › LTI & convolution

   $x(t)\ast x(-t)=\int x(\tau )x(\tau -t)d\tau$ is the autocorrelation of $x(t)$ (for real $x$). Putting $t\to -t$ and shifting the dummy variable returns the same integral, so the result is always even. In the frequency domain, $X(j\omega )X(-j\omega )$ does not change when $\omega \to -\omega$.

   WATCH OUT FOR

   Picking (C) or (D) because one of the two signals is time-reversed. Time-reversal flips where each signal is nonzero, but the convolution spreads over both sides of $t=0$: a pulse on $[0,1]$ convolved with its mirror image on $[-1,0]$ gives a triangle on $[-1,1]$. Some students also choose (B), treating reversal as a sign flip, but $z(0)$ is the signal's energy, which is positive, so the result cannot be odd.

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6. Q.16MCQ · 1 markOfficial answer: A

   Signals and Systems › Laplace

   How fast a signal grows only matters when the signal lasts forever. A bounded signal of finite duration has a Laplace transform that converges for every $s$, so its ROC is the whole $s$-plane. This holds even for something like $e^{t^2}$, which on its own grows faster than any exponential and has no Laplace transform at all.

   WATCH OUT FOR

   There are two traps. The first is taking the window edges $t=1$ and $t=10$ as limits on $\text{Re}(s)$ and picking (D) $1\le \text{Re}(s)\le 10$. Time limits never become ROC boundaries: those come from exponential growth rates or pole locations. The second is seeing $e^{t^2}$ and deciding the transform cannot exist, or that it needs a large $\text{Re}(s)$ as in (B). Because of the window, the integral runs over a finite interval of a bounded function, so it converges everywhere.

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7. Q.17MCQ · 1 markOfficial answer: C

   Electrical Machines › Losses & efficiency

   When stator copper loss, core loss, and rotational (mechanical) loss are all neglected, the only loss remaining is the rotor copper loss, which equals the slip fraction of the air-gap power. Since there are no stator losses, the input power equals the air-gap power, and the developed mechanical output power equals (1-s) times the air-gap power. Hence efficiency = 1 - s exactly.

   WATCH OUT FOR

   Confusing slip loss fraction with efficiency directly, or computing efficiency as Pmech/Pinput using an incorrect power balance (e.g., forgetting that with stator losses neglected, input power to the rotor equals air-gap power, so efficiency simplifies neatly to (1-s)).

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8. Q.18MCQ · 1 markOfficial answer: Marks to all

   Electrical Machines › Transformers

   Check three things for each option. On the primary, $V_p\approx E_p$ because the leakage drop is small. On the open-circuit secondary, $V_s=E_s$. And because one mutual flux induces both windings, the induced voltages at the two dotted terminals have the same instantaneous polarity. Only (B) passes all three when $E$ is read as the voltage across the winding.

   WATCH OUT FOR

   One mistake is to apply the dot rule to $V_p$ alone and ignore how the $E$ markings relate to the $V$ markings. Another is to mix conventions: drawing $E_p$ as a counter-emf opposing $V_p$ while treating $E_s$ as the secondary's terminal voltage. That can make (D) look right. Fix one sign convention first, then apply it to both windings.

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9. Q.19MCQ · 1 markOfficial answer: C
10. Q.20MCQ · 1 markOfficial answer: D

    Control Systems › Block diagrams & SFG

    The minus sign on the link between the two summing junctions lies on both the forward path and the outer loop. It makes the forward gain negative ($-G$). In the outer loop it cancels the minus at the first junction, so that loop is effectively positive ($+G$), the same as the inner positive-feedback loop. Two touching $+G$ loops give $\Delta =1-2G$.

    WATCH OUT FOR

    Counting the minus sign between the junctions only once. If you apply it to the outer loop but forget it on the forward path, you get $+G/(1-2G)$, which is option (C). If you ignore it in the outer loop and read that loop as ordinary negative feedback, the loop signs come out wrong and you land on a $1+2G$ denominator (options A and B) or on a result that is none of the options. That minus has to be counted in every path and loop that passes through it.

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

    Control Systems › Time response

    Settling time is set by the real part of the poles, $\sigma =\zeta {\omega }_n$. In this figure that is ${\omega }_n\cos \theta$: the horizontal distance of the poles from the $j\omega$ axis. The system whose poles sit closer to the imaginary axis settles more slowly, whatever its ${\omega }_n$ or $\theta$ is on its own.

    WATCH OUT FOR

    Comparing only one parameter. Some students see that System 2 has the smaller ${\omega }_n$, or the larger $\theta$, and stop there. Others think settling time depends on $\zeta$ alone and so pick (D). What decides it is the product $\zeta {\omega }_n={\omega }_n\cos \theta$. A related slip is measuring $\theta$ from the imaginary axis and writing $\zeta =\sin \theta$. It happens to give the same option here, but it is wrong in general.

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

    Control Systems › Time response

    To reduce the order, keep the dominant pole (the one closest to the origin, here $s=-1$) and drop the far pole and zero. Then pick the gain so the DC gains match, because by the final value theorem the unit-step steady state equals $G(0)$.

    WATCH OUT FOR

    There are three common traps. (1) Dropping the second block $\frac{s+40}{s+20}$ entirely and choosing (A) $\frac{1}{s+1}$. This ignores the block's DC gain of $\frac{40}{20}=2$ and halves the steady-state value. (2) Keeping the pole at $-20$ because it 'looks bigger', which gives (C) or (D). The pole to keep is the slow one closest to the origin, not the fast one. (3) Expanding in partial fractions, $G(s)=\frac{39/19}{s+1}-\frac{20/19}{s+20}$, and simply deleting the fast term. That leaves $\frac{39/19}{s+1}$, whose DC gain of $\frac{39}{19}\approx 2.05$ does not equal 2. The question asks for the steady states to match, so the gain has to be reset to $G(0)=2$.

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

    Electrical and Electronic Measurements › Instrument transformers

    The two transformers are connected differently, and that decides everything. A CT is in series with the line, so the line fixes its primary current, and an open secondary drives the core into saturation and puts a dangerous voltage on the secondary. A PT is in parallel across the grid, so its primary current is set by what the secondary draws, and a larger burden means a larger primary current.

    WATCH OUT FOR

    Giving S to the CT on the grounds that 'secondary burden affects accuracy'. The burden does change a CT's ratio and phase-angle errors, but it cannot change the CT's primary current, because the line fixes that. Only in the PT does the burden change the primary current. A second slip is thinking 'open secondary not desirable' applies to both. It is the CT that must never be open-circuited. A PT's secondary normally feeds a high-impedance load.

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

    Digital Electronics › Boolean logic

    Expand the function to its minterm list, $\sum m(0,1,2,3,5,7,8,10)$ in the order $PQRS$, and group the 1s on a 4-variable K-map. The $\overline{P}S$ quad $m(1,3,5,7)$ and the four-corner quad $m(0,2,8,10)=\overline{Q}\overline{S}$ cover every 1. Algebraically, $\overline{P}\overline{Q}$ splits into $\overline{P}\overline{Q}S$ and $\overline{P}\overline{Q}\overline{S}$. The first part is absorbed by $\overline{P}S$, and the second combines with $P\overline{Q}\overline{S}$ to give $\overline{Q}\overline{S}$.

    WATCH OUT FOR

    The usual mistake is to miss that the four corner cells $m(0,2,8,10)$ are adjacent (the K-map wraps around both ways), so $\overline{Q}\overline{S}$ never gets formed. Another is to keep $\overline{P}\overline{Q}$ as a prime implicant and pick (B). But (B) leaves out $m5$ and $m7$, which come from $\overline{P}QS$. Always check a chosen option against the full minterm list.

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15. Q.25MCQ · 1 markOfficial answer: C

    Digital Electronics › Sequential

    A D flip-flop latches the value at $D$ on the clock edge ($Q_{n+1}=D$). Here $D=S$ is computed from the present state $Z$. Because the combinational circuit has no delay, $S$ is then recomputed straight away from the new $Z$. With $X\oplus Y=1$, $S=\overline{Z}$ (overline = complement), so the circuit toggles $Z$.

    WATCH OUT FOR

    Reporting the value of $S$ from before the clock ($S=0$) together with the new $Z=0$, which gives option (A). The question asks for both values after the clock. Once $Z$ changes to $0$, the delay-free combinational circuit makes $S=1$ immediately.

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16. Q.26MCQ · 1 markOfficial answer: B

    Digital Electronics › Combinational

    In an $n$-select MUX used as a function generator, data input $I_k$ is simply the value of $F$ at the minterm whose binary code, with the MSB select line as the leading bit, equals $k$. Here $X$ drives the MSB, so the input order $P,Q,R,S$ matches $XY=00,01,10,11$.

    WATCH OUT FOR

    Mixing up which select line is the MSB. If you take $Y$ as the MSB, inputs 1 and 2 swap places. That gives the same pattern here only because $F$ is symmetric in $X$ and $Y$, so it will not work in general. A second slip is to read only the $X\overline{Y}$ term, mark just input 2 as 1, and forget that $\overline{X}$ also covers inputs 0 and 1.

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17. Q.27MCQ · 1 markOfficial answer: D
18. Q.28MCQ · 1 markOfficial answer: B
19. Q.29MSQ · 1 markOfficial answer: B, D

    Engineering Mathematics › Probability & statistics

    When $X$ is uniform on $N$ points, $P(g(X)=y)$ equals (the number of support points that $g$ sends to $y$) divided by $N$. So $g(X)$ is uniform exactly when every value in its range has the same number of preimages. Any one-to-one map keeps the distribution uniform: a strictly monotone function such as $x^3$, or squaring on a set whose values all have the same sign, as with $X+10\ge 0$. A square of a set that contains both $k$ and $-k$ for some $k$, next to values that have no partner, is not uniform.

    WATCH OUT FOR

    (1) Rejecting (D) because "a square is never one-to-one". Squaring is one-to-one on $\{0,1,\dots ,20\}$ because every value is non-negative. (2) Thinking the shift in (C) removes the $\pm$ pairing. Since $X-5$ still runs from $-15$ up to $+5$, the values $\pm 1,\dots ,\pm 5$ still pair up. (3) Treating "uniform" as meaning evenly spaced values and rejecting (B) or (D) because $-1000,\dots ,1000$ or $0,1,4,\dots ,400$ are unequally spaced. Uniformity is about equal probabilities, not about spacing.

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

    Engineering Mathematics › Complex analysis

    A function built from $z$ alone, such as a polynomial in $z$, is entire. Functions built from $\text{Re}(z)$, $\text{Im}(z)$ or $|z|$ depend on $\bar{z}$ as well as $z$, and they fail the Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$. A non-constant real-valued function can never be analytic.

    WATCH OUT FOR

    Checking only the first Cauchy-Riemann equation. In (A) and (B), $u_x=v_y=0$ holds trivially, and a student who stops there wrongly marks them analytic. Another mistake is to accept $e^{|z|}$ because the exponential is entire: $e^w$ is analytic in $w$, but $w=|z|$ is not an analytic function of $z$, so the composition is not analytic either.

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21. Q.31NAT · 1 markOfficial answer: 0.0

    Engineering Mathematics › Complex analysis

    $\cos z$ and $e^{z^2}$ are both even functions of $z$, so their Maclaurin series contain only even powers. The coefficient of any odd power, including $z^5$, is therefore zero. You do not need to expand anything once you see this.

    WATCH OUT FOR

    Expanding $e^z$ in place of $e^{z^2}$ picks up the odd term $\frac{z^5}{5!}\approx 0.0083$. Mixing $e^{z^2}$ up with $\sin z$ picks up $\frac{z^5}{5!}$ in the same way. Some students also report the $z^4$ coefficient ($\frac{13}{24}\approx 0.5$) by mistake. The printed function is $e^{z^2}$, which is even.

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22. Q.32NAT · 1 markOfficial answer: 29

    Engineering Mathematics › Linear algebra

    The sum of the eigenvalues equals the trace, so you only need the diagonal of $B^2$. As a check, if $B$ has eigenvalues ${\lambda }_i$, then $B^2$ has eigenvalues ${\lambda }_i^2$, and $\sum {\lambda }_i^2=(\mathrm{tr}B{)}^2-2\det B$ for a $2\times 2$ matrix.

    WATCH OUT FOR

    Squaring the entries one by one (getting diag $1,16$ and trace $17$) instead of multiplying the matrix by itself. A related slip is squaring the trace of $B$ ($5^2=25$) and forgetting the $-2\det B$ term: the eigenvalues of $B^2$ add up to $\sum {\lambda }_i^2$, which is not $(\sum {\lambda }_i{)}^2$.

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23. Q.33NAT · 1 markOfficial answer: 0.0

    Signals and Systems › Fourier transform

    $x(t)=e^{-t^4}\cos t$ is real and even, so $X(\omega )$ is real and even. The derivative of an even function is odd, and an odd function is zero at the origin. Equivalently, $X^{\prime }(0)=-j\int tx(t)dt$, which is the integral of an odd function over a symmetric range, so it is $0$. You never need to compute the transform of $e^{-t^4}$, which has no closed form.

    WATCH OUT FOR

    Trying to find $\mathcal{F}\{e^{-t^4}\}$ explicitly, which has no elementary closed form, or treating it as a Gaussian $e^{-t^2}$. Another mistake is to answer with the value $X(0)=\int e^{-t^4}\cos tdt$, which is not zero, when the question asks for the derivative $X^{\prime }(0)$. A third is to think the modulation by $\cos t$ destroys the symmetry: the spectrum shifts to $\pm 1$, but it shifts symmetrically, so $X(\omega )$ stays even.

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24. Q.34NAT · 1 markOfficial answer: 35
25. Q.35NAT · 1 markOfficial answer: 50
26. Q.36MCQ · 2 marksOfficial answer: C

    Engineering Mathematics › Vector calculus

    The directional derivative is the gradient dotted with the UNIT vector in the given direction: $D_{\hat{u}}f=\nabla f\cdot \frac{\bar{u}}{|\bar{u}|}$. Splitting the logarithms first turns $f$ into $5\ln x+3\ln y+4\ln z$, and its gradient can then be read off directly.

    WATCH OUT FOR

    Forgetting to normalise $\bar{u}$ and using $\nabla f\cdot \bar{u}=21$ gives option (D), which is exactly three times the correct value because $|\bar{u}|=3$. A second trap is collecting the log coefficients wrongly. For example, $y$ appears in both $2\ln (xy)$ and $\ln (yz)$, so its coefficient is $2+1=3$, not $2$ or $1$.

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

    Signals and Systems › LTI & convolution

    Move $e^{-t}$ inside the integral to get ${\int }_{-\infty }^t{e}^{-(t-\tau )}x(\tau )d\tau$. The kernel depends only on $t-\tau$, so this is the convolution of $x(t)$ with $h(t)=e^{-t}u(t)$. A system that is a convolution is always LTI, and here it is also causal because $h(t)=0$ for $t<0$.

    WATCH OUT FOR

    Seeing the explicit $e^{-t}$ multiplying the integral and deciding the system is time-varying, which leads to (C). The factor $e^{-t}$ is cancelled by the $e^{\tau }$ inside the integral: combined, they give $e^{-(t-\tau )}$, which depends only on the elapsed time $t-\tau$. A time-dependent factor makes a system time-varying only when nothing cancels it. Test with a delayed input; do not judge from the look of the formula.

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

    Signals and Systems › LTI & convolution

    To show a system is BIBO unstable, you only need one bounded input that gives an unbounded output. A constant fed into a running sum gives $n+1$. To show it is stable, bound $|y[n]|$ for every bounded input. For $T_2$, the weights $(1/2{)}^k$ add up to a finite total of 2, so $|y_2[n]|\le 2B$ for every $n$. This works directly even though $T_2$ is time-varying, so no impulse-response test is needed.

    WATCH OUT FOR

    Calling $T_2$ unstable because the number of terms grows with $n$. What decides stability is whether the weights have a finite total, and $\sum (1/2{)}^k=2$. Another mistake is to treat $T_2$ as LTI and use $\sum |h[n]|$. $T_2$ is time-varying, because its weights depend on $k$ and not on $n-k$. That test does not apply, even though here it happens to give the right answer. A third mistake is calling $T_1$ stable because every term $x[k]$ is bounded. A sum of more and more bounded terms can still grow without limit.

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29. Q.39MCQ · 2 marksOfficial answer: C

    Signals and Systems › z-transform

    Downsampling by 2 ($y[n]=x[2n]$) keeps only the even-indexed samples. Pick them out with the selector $\frac{1+(-1{)}^m}{2}$. With $z^{-n}={\left(z^{1/2}\right)}^{-m}$, that selector turns into the average of $X$ evaluated at $z^{1/2}$ and at $-z^{1/2}$.

    WATCH OUT FOR

    Mixing up downsampling with upsampling. $X(z^2)$, option (A), is the transform of the expanded signal $x[n/2]$, not of $x[2n]$. Option (B) comes from carrying the minus sign of $z^{-n}$ into the argument and writing $z^{-1/2}$. Remember that $X(\cdot )$ already contains the negative powers, so the argument is $z^{1/2}$. Option (D) applies the right averaging structure to the wrong operation.

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

    Electrical Machines › Synchronous machines

    When a generator supplies a lagging load through a purely reactive line, the terminal voltage phasor is found by adding the line's reactive drop ($j{X}_{line}I$) to the load voltage phasor; since only the terminal current and terminal voltage magnitude are given, the generator's own synchronous reactance (30 Ω) is irrelevant to finding the terminal power factor — it would only be needed to find the internal EMF.

    WATCH OUT FOR

    Using the generator's synchronous reactance (30 Ω) instead of, or in addition to, the line reactance (5 Ω) when computing the terminal voltage phasor — the synchronous reactance relates internal EMF to terminal voltage and plays no role in a question that already gives both rated terminal voltage and rated current.

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

    Analog Electronics › Biasing

    Swap the voltage divider for its Thevenin equivalent ($V_{TH}=4\text{V}$, $R_{TH}=33.33\text{k}\Omega$). Then write KVL around the base loop with $R_E$ reflected as $(\beta +1)R_E$ to find $I_B$. After that, $I_C=\beta I_B$, and KVL around the collector loop gives $V_{CE}$. Here $\beta R_E$ is only about $3R_{TH}$, so the exact Thevenin method is needed.

    WATCH OUT FOR

    Treating the divider as stiff, i.e. ignoring base current and taking $V_B=4\text{V}$ straight from the divider. That gives $I_E=3.3/1=3.3\text{mA}$ and $V_{CE}\approx 12-3.3\times 3=2.1\text{V}$, which is about 34% too much current and matches no option, because $\beta R_E=100\text{k}\Omega$ is only about $3R_{TH}$. A second slip is leaving out the $I_E{R}_E$ drop, or using $R_C$ alone, when writing the collector-loop KVL.

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

    Engineering Mathematics › Calculus

    $f^{\prime \prime }>0$ everywhere means $f^{\prime }$ is strictly increasing, so it has at most one zero. At any zero, $f^{\prime }$ changes sign from negative to positive. So $f$ has at most one stationary point, and that point can only be a minimum. $f^{\prime \prime }>0$ does not guarantee that a stationary point exists ($e^t$ has none), and it says nothing about the vertical position of the graph ($t^2-1$ dips below zero).

    WATCH OUT FOR

    Picturing only a U-shaped parabola and marking (A), or (A) and (D). A convex function can be strictly monotonic with no minimum at all ($e^t$, $e^{-t}$). A constant shift keeps $f^{\prime \prime }>0$ but can push the minimum value below zero ($t^2-1$).

    Full solution →
34. Q.44MSQ · 2 marksOfficial answer: B

    Engineering Mathematics › Calculus

    Write $f$ piecewise ($0$ for $t<0$, $t^2$ for $t\ge 0$) and test the junction $t=0$ with the limit definition, first for $f$ and then for $f^{\prime }$. Squaring the ramp $\max (0,t)$ smooths its corner by one order: $f^{\prime }=2\max (0,t)$ is continuous, but $f^{\prime }$ still has the ramp's corner, so $f^{\prime \prime }(0)$ does not exist.

    WATCH OUT FOR

    Seeing $\max (\cdot )$ and assuming a corner, which gives (A), or seeing the smooth-looking $t^2$ and ticking (D). The corner of $\max (0,t)$ is removed from $f$ by squaring, but it reappears one level down in $f^{\prime }=2\max (0,t)$. The one-sided second derivatives at $0$ are $0$ and $2$, so $f^{\prime }$ is not differentiable there.

    Full solution →
35. Q.45MSQ · 2 marksOfficial answer: B, D

    Engineering Mathematics › Differential equations

    Linearity is judged only by how the dependent variable $x$ and its derivatives appear. They must enter to the first power, never multiplied together and never inside a function such as $e^{(\cdot )}$ or $\sin (\cdot )$. Coefficients that depend on the independent variable $t$ (like $t$, $\cos t$, $\sin t$) keep an equation linear.

    WATCH OUT FOR

    One mistake is marking (A) or (C) as nonlinear because of the products $tx$, $x\cos t$ or $x^{\prime }\sin t$. Those factors depend only on $t$, so the equations stay linear. The other is confusing degree with linearity and passing over (B): its derivative appears to the first power, but the product $x{x}^{\prime }$ makes it nonlinear.

    Full solution →
36. Q.46MSQ · 2 marksOfficial answer: A, D
37. Q.47MSQ · 2 marksOfficial answer: B, D
38. Q.48NAT · 2 marksOfficial answer: -19.90 to -19.70

    Networks (Electric Circuits) › Two-port

    For networks in cascade, convert each one to its ABCD (transmission) matrix and multiply them in the order the signal passes through: $T_{\mathbf{R}}=T_{\mathbf{P}}{T}_{\mathbf{Q}}$. Then compute only the entry you need, here $\beta =A_{\mathbf{P}}{B}_{\mathbf{Q}}+B_{\mathbf{P}}{D}_{\mathbf{Q}}$. You get the Z→ABCD and Y→ABCD formulas by solving the port-2 equation for the port-1 variable. Checking $AD-BC$ ($=z_{12}/z_{21}$ for P and $=1$ for reciprocal Q) is a quick test of each conversion.

    WATCH OUT FOR

    (1) Multiplying in the wrong order: $T_{\mathbf{Q}}{T}_{\mathbf{P}}$ gives $\beta =A_{\mathbf{Q}}{B}_{\mathbf{P}}+B_{\mathbf{Q}}{D}_{\mathbf{P}}=-4+0.5=-3.5\Omega$. (2) Swapping $z_{12}$ and $z_{21}$ for the non-reciprocal P, i.e. dividing by 60 instead of 80, which gives about $-26.4\Omega$. (3) Dropping the minus signs in the Y→ABCD formulas. Because $y_{21}$ is negative here, the signs of every entry of $T_{\mathbf{Q}}$ then come out wrong.

    Full solution →
39. Q.49NAT · 2 marksOfficial answer: 5.32 to 5.34

    Networks (Electric Circuits) › Coupled circuits

    The secondary loop is nearly resonant ($Z_{22}=j1\Omega$), so the impedance it reflects into the primary, $(\omega M{)}^2/Z_{22}=-j625\Omega$, exactly cancels $j\omega L_1=j625\Omega$. The whole coupled branch is then a short circuit, and $Z_{th}$ reduces to a resistor ladder, $4+(4\parallel 2)$. Thevenin impedance is measured looking into the network the arrow points at, so the source and $Z_s$ are left out.

    WATCH OUT FOR

    (1) Getting the sign of $1/j$ wrong, so $625/j1$ is written as $+j625$ instead of $-j625$. That doubles the primary reactance instead of cancelling it. (2) Leaving the capacitor out of $Z_{22}$ (using $j5$), which gives $Z_r=-j125\Omega$ and a complex $Z_{th}$. (3) Adding $Z_s$ or the source to $Z_{th}$, although the arrow points away from them. (4) Thinking the answer depends on the dot convention: the reflected impedance depends on $M^2$, so the dots do not change it.

    Full solution →
40. Q.50NAT · 2 marksOfficial answer: 692 to 693

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

    First find where the load actually is. The third side of the "delta" is a wattmeter pressure coil, which carries no current, so the load is only $Z_1$ (R–B) and $Z_2$ (R–Y). Then get each reading from $W=\text{Re}\{V_{pc}{I}_{cc}^{\ast }\}$, taking the pressure-coil voltage from its $+$ end to its $-$ end and the current entering the $+$ end of the current coil. $I_R=10\angle 30^{\circ }\text{A}$ is at exactly $90^{\circ }$ to $V_{RB}$, so $W_1=0$ and the whole $692.82\text{W}$ taken by $Z_2$ shows on $W_2$.

    WATCH OUT FOR

    Treating the bottom side of the triangle as a third load impedance, or assuming the load is balanced and using $W_1-W_2=\sqrt{3}{V}_L{I}_L\sin \varphi$. That formula holds only for a balanced load, and this load is not balanced: nothing is connected between Y and B, and $Z_1$ and $Z_2$ differ. Two other slips are common. One is using a phase voltage of $400/\sqrt{3}$ V across the delta-connected impedances instead of the 400 V line voltage. The other is dropping the $V_{RY}/Z_2$ part of $I_R$, which leaves $W_1$ correct but only by luck and gets the current in line R wrong.

    Full solution →
41. Q.51NAT · 2 marksOfficial answer: -1.7 to -1.5

    Electromagnetics › Electrostatics

    Add the fields by superposition. Because the two equal charges $Q$ sit symmetrically about the $y$-axis, their $x$-components cancel at $P$ and only a net $+y$ field is left. The third charge lies on the $y$-axis, so its field is purely along $y$. Setting the $y$-components to sum to zero gives $\alpha$, and $k$ and $Q$ cancel out.

    WATCH OUT FOR

    The usual slip is adding the full magnitudes $kQ/1.25$ of the two symmetric charges instead of only their $y$-components $kQ(0.5)/(1.25{)}^{3/2}$. That gives $\alpha =-3.6$. The other common error is measuring the third charge's distance as $1$ or $0.5$ instead of $1.5$, or dropping the negative sign. A positive $\alpha$ would push $P$ further in $+y$ and could never cancel the field.

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

    Electromagnetics › Magnetostatics

    A proposed field is a physically valid magnetic field only if its divergence is zero ($\nabla \cdot \mathbf{B}=0$, so $\nabla \cdot \mathbf{H}=0$ in a linear homogeneous medium). Setting the divergence to zero fixes the unknown constant. In a current-free region $\nabla \times \mathbf{H}=0$ gives an independent check. The result, $2\cos \theta \hat{r}+\sin \theta \hat{\theta }$ over $r^3$, is the magnetic-dipole field.

    WATCH OUT FOR

    Using the Cartesian-style divergence $\partial H_r/\partial r+\partial H_{\theta }/\partial \theta$ and leaving out the $r^2$ and $\sin \theta$ factors. That gives $-3P\cos \theta /r^4+\cos \theta /r^3$, which cannot vanish for every $r$ and $\theta$, or leads to a wrong value such as $P=1$ or $P=3$. Another slip is dropping the $\sin \theta$ inside the $\theta$-derivative: differentiating $\sin \theta$ alone gives $\cos \theta$ and so $P=1$, when the correct derivative of ${\sin }^2\theta$ gives the factor of 2.

    Full solution →
43. Q.53NAT · 2 marksOfficial answer: 2.0

    Signals and Systems › Signals basics

    For $y(t)=Ax(at-b)$, energy scales as $E_y=\frac{A^2}{|a|}{E}_x$. Amplitude enters squared. Compressing time by $|a|$ divides the energy by $|a|$, and a time shift changes nothing.

    WATCH OUT FOR

    Students often multiply by $|a|$ instead of dividing: compressing the signal makes it shorter, so it holds less energy. That mistake gives $4\times 2=8$. Others forget to square the amplitude and get $2/2=1$, or think the shift of $-1$ changes the energy.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 0.26 to 0.29

    Electrical Machines › Induction machines

    For a slip-ring induction motor with the magnetizing branch neglected, the per-phase equivalent circuit is a series combination of $R_s$, $X_s$, $R_r^{\prime }/s$ and $X_r^{\prime }$ (all referred to the stator). Torque is maximum (as a function of variable rotor resistance) when $\frac{R_r^{\prime }}{s}=\sqrt{R_s^2+(X_s+X_r^{\prime }{)}^2}$. At starting, $s=1$, so this directly gives the total stator-referred rotor resistance needed for maximum starting torque.

    WATCH OUT FOR

    Forgetting to convert the required EXTRA resistance back from the stator-referred frame to the actual rotor-side value using $R_{\text{actual}}=R^{\prime }/a^2$ (with $a=3$, the stator:rotor turns ratio) — the question explicitly asks for the resistance "referred to the rotor side," not the stator-referred value.

    Full solution →
45. Q.55NAT · 2 marksOfficial answer: 27.19 to 27.39

    Electrical Machines › Losses & efficiency

    Get the constant rotational loss from the no-load test: $P_{rot}=V{I}_{a0}-I_{a0}^2{R}_a$. At rated load, the power converted in the armature, $V{I}_a-I_a^2{R}_a$, must equal the rated output plus $P_{rot}$. This gives a quadratic in $I_a$ with two roots. The smaller root is the one with the best efficiency. Add the field current to get the line current.

    WATCH OUT FOR

    Reading "best possible efficiency" as the maximum-efficiency condition (variable loss = constant loss). That calculation ignores the fixed 5 kW rated load and gives 30.60 A, or 37.28 A if the field copper loss is counted as a constant loss. Neither current delivers 5 kW. The phrase only tells you which root of the rated-load quadratic to pick. Other slips: forgetting to subtract the field current from the no-load line current, and giving the armature current (26.29 A) instead of the line current.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 0.10 to 0.12
47. Q.57NAT · 2 marksOfficial answer: 4 to 8

    Control Systems › Root locus

    For a complex pole with negative feedback and $K\ge 0$, the departure angle is ${\varphi }_d=180^{\circ }+\sum \angle (\text{zeros}\to \text{pole})-\sum \angle (\text{other poles}\to \text{pole})$. Measure each vector from the zero or other pole to the pole in question, and use its true quadrant. Then reduce the result into $[0^{\circ },360^{\circ })$.

    WATCH OUT FOR

    The usual slip is to take a zero's angle as a plain ${\tan }^{-1}$ value (for example $-26.57^{\circ }$ instead of $153.43^{\circ }$), which ignores that the vectors $-2+j1$ and $-2+j3$ lie in the second quadrant. Another is to point the vectors from the pole to the zeros instead of from the zeros to the pole. A third is to stop at $367^{\circ }$ without subtracting $360^{\circ }$, even though the question asks for a value between $0^{\circ }$ and $360^{\circ }$. Some students also put a minus sign on the zero terms, which gives a very different answer.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 0.09 to 0.11

    Control Systems › Bode & margins

    A $-20$ dB/dec asymptote with $-90^{\circ }$ phase that starts at low frequency means one integrator, so the system is type 1. For a type-1 system, the initial $-20$ dB/dec line crosses $0$ dB at $\omega =K_v$. Here nothing changes the slope before $100$ rad/s, so that crossing is the gain crossover itself, and $K_v={\omega }_{gc}=10$. For unity feedback, $e_{ss,\text{ramp}}=1/K_v$.

    WATCH OUT FOR

    Setting $e_{ss}=1/{\omega }_{gc}$ is only valid because the slope is still $-20$ dB/dec at crossover. If a corner frequency lay below ${\omega }_{gc}$, $K_v$ would come from the $0$ dB crossing of the extended initial line, not from ${\omega }_{gc}$. Other slips are using the type-0 formula $1/(1+K_p)$ (which gives $1/11\approx 0.09$ only by coincidence and uses the wrong reasoning), or answering that the error is zero or infinite without first reading the system type from the $-90^{\circ }$ phase.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: 1.11 to 1.13

    Control Systems › Bode & margins

    The integrator $K_I/s$ adds a fixed $-90^{\circ }$ at every frequency, and $K_I$ changes only the magnitude, never the phase. So you find the gain crossover frequency from the phase alone: $\angle G=-180^{\circ }+\text{PM}+90^{\circ }=-40^{\circ }$, which the table gives at $\omega =0.5\text{rad/s}$. Then you pick $K_I$ so that $|L|=0\text{dB}$ at that frequency: $K_I={\omega }_{gc}/|G(j{\omega }_{gc})|$.

    WATCH OUT FOR

    Forgetting the integrator's own $-90^{\circ }$ and $1/\omega$. Taking $\angle G=-130^{\circ }$ gives $\omega =2\text{rad/s}$, which is wrong. Dropping the $1/\omega$ factor gives $K_I=10^{7/20}\approx 2.24$, and that is wrong too. Another slip is turning $-7\text{dB}$ into a ratio with $10^{-7/10}$, the power formula, instead of $10^{-7/20}$.

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 1.64 to 1.70

    Analog Electronics › Diode circuits

    Assume a state for each ideal diode (ON means short circuit, OFF means open circuit), solve the linear circuit, then confirm the state: every ON diode must carry forward current and every OFF diode must be reverse-biased. Here $D1$ ON pins node B at 3 V. That turns node A into a single-node KCL problem, and $I$ follows from KCL at B.

    WATCH OUT FOR

    Taking $I$ as the full 3 mA that flows from node B to ground, which forgets that $D2$ already supplies 4/3 mA of it. Another mistake is working out the left branch as an unloaded divider ($V_A=5$ V, so $I_{D2}=2$ mA and $I=1$ mA), which ignores that $D2$ draws current from node A.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 39.50 to 41.50

    Analog Electronics › Op-amps

    With an ideal op-amp, the output is $A_1{v}_{in1}+A_2{v}_{in2}$, where $A_1=-R_f/R_1$ and $A_2=(1+R_f/R_1)R_4/(R_3+R_4)$. The two gains are equal and opposite only when $R_f/R_1=R_4/R_3$. Any mismatch leaves a common-mode gain $A_c=A_1+A_2$ alongside the differential gain $A_d=(A_2-A_1)/2$, and $\text{CMRR}=|A_d/A_c|$, expressed in dB as $20{\log }_{10}$ of that ratio. Here the op-amp is ideal, so the finite CMRR comes entirely from the mismatched resistors.

    WATCH OUT FOR

    Three common slips. (1) Taking $A_d$ as just $R_f/R_1=9.619$ (or $R_4/R_3$), which only holds for a matched bridge, and so getting a wrong ratio. (2) Leaving out the $(1+R_f/R_1)$ factor in the non-inverting path. (3) Rounding $A_1$ and $A_2$ to two or three digits before subtracting them. $A_c$ is the small difference of two nearly equal numbers, so early rounding throws the CMRR off badly. Keep exact fractions over the common denominator $1165.5$ until the last step. Using $10{\log }_{10}$ instead of $20{\log }_{10}$ for this voltage-gain ratio is another way to land outside the key.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 0.940 to 0.970
53. Q.63NAT · 2 marksOfficial answer: 50
54. Q.64NAT · 2 marksOfficial answer: 13
55. Q.65NAT · 2 marksOfficial answer: 113.00 to 116.00