GATE IN 2024 question paper with solutions

GATE IN 2024 answer key

General Aptitude Q.1 to Q.10

1. Q.1MCQ · 1 markOfficial answer: A

   General Aptitude › Grammar & vocabulary

   In an intensity analogy, every word in a chain names the same kind of thing at a different strength. First find what the other two words have in common (here, disagreement). Then pick the option that belongs to that same kind and sits at the weakest end, below 'quarrel'.

   WATCH OUT FOR

   Choosing a word that merely sounds like conflict (such as 'dodge', which is about avoiding something) or that suggests weakness without being about arguing (such as 'dither', which is about indecision). The blank has to be the same kind of thing as 'quarrel' and 'feud', only milder.

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

   General Aptitude › Reasoning

   Chain the two universal statements into Heroes ⊆ Winners ⊆ Lucky people. An "All A are B" statement can never be reversed to "All B are A". It can be converted by limitation to "Some B are A", since GATE treats every named class as non-empty.

   WATCH OUT FOR

   Students go wrong in two ways. Some reverse "All heroes are lucky people" into "All lucky people are heroes" and accept I, which leads to (A) or (C). Others argue that "All" never implies "Some" because the Heroes set might be empty, and choose (D) or reject II. GATE follows the traditional convention that universal statements are about classes with members, so both II and III follow.

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

   General Aptitude › Numerical

   Percentage error is always taken relative to the correct value. Here the ratio of the student's answer to the correct answer is $\frac{p/q}{pq}=\frac{1}{q^2}$, and $p$ cancels out. An 80% error means the student got 20% of the correct value, so $\frac{1}{q^2}=0.2$ and $q^2=5$.

   WATCH OUT FOR

   Dividing by the student's (wrong) answer $\frac{p}{q}$ instead of the correct answer $pq$. That gives $|q^2-1|=0.8$, so $q^2=1.8$ or $0.2$, and neither matches an option. Some students also read an 80% error as "the student got 80% of the correct value", which gives $q^2=1.25$. It actually means the answer is off by 80%, so the student got 20%.

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

   General Aptitude › Numerical

   The sum of the first $n$ positive odd numbers is always $n^2$, because $1+3+\cdots +(2n-1)=\frac{n}{2}(1+(2n-1))=n^2$. Here $n=20$, so the sum is exactly $20^2$ and the ratio is 1.

   WATCH OUT FOR

   Taking the 20th odd number to be 41 (confusing $2n+1$ with $2n-1$) or including 0, which gives a wrong sum. Another slip is using the formula for the sum of the first $n$ natural numbers, $\frac{n(n+1)}{2}=210$, in place of the sum of odd numbers. Both lead to a ratio that is not 1.

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

   General Aptitude › Numerical

   Use one variable for the girls, since their number is the same in both classes. Write each class's boys as total minus girls and set the two ratios equal. Because the ratios are reversed (girls : boys against boys : girls), the $g^2$ terms cancel when you cross-multiply, and what is left is a linear equation.

   WATCH OUT FOR

   Reading both ratios as girls : boys and writing $\frac{g}{450-g}=\frac{g}{360-g}$. That has no positive solution, because it forces $450=360$. Another slip is expanding $(450-g)(360-g)$ with the wrong sign on the $810g$ term. Testing the options is a fast check: with 200 girls, class VIII is $200:250$ and class IX is $160:200$, and both are $4:5$.

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

   General Aptitude › Grammar & vocabulary

   All four blanks are phrasal verbs built on stand, and the particle changes the meaning completely: stand out (be distinguished), stand down (resign a post), stand by (keep supporting one's words), stand for (represent or uphold). Pin down the blank with the strongest clue first. Here that is (ii), because a fellowship is a post you step down from. Then eliminate options one blank at a time.

   WATCH OUT FOR

   Picking (A) because (i), (ii) and (iv) are all right, and missing that stood in her writings means nothing. Stand in means to act as a substitute. The idiom for refusing to take back what you wrote is stand by. A second trap is swapping out and down for (i) and (ii). Stands down as an author would mean she quit writing, which contradicts a sentence praising her as an author.

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

   General Aptitude › Geometry

   The container and the sticks have the same length, so the volume ratio equals the ratio of cross-sectional areas. In the hexagonal packing of seven equal circles, the outer centres are 2r from the middle, so the enclosing circle has radius 3r. The area is then 9 small-circle areas: 7 are filled and 2 are empty.

   WATCH OUT FOR

   Answering 9/2 (C) or 7/9, which is occupied to total space rather than occupied to empty space. A second slip is taking the container's radius as 2r, the distance to the outer centres, and forgetting to add the outer stick's own radius.

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

   General Aptitude › Data interpretation

   The risk reduction over an interval is the vertical drop of the curve across it. This curve falls steeply at first and then flattens, so equal 5000-step intervals give smaller and smaller drops. The first interval, 0 to 5000, gives by far the largest one.

   WATCH OUT FOR

   Confusing the risk itself with the risk reduction. The risk is lowest between 15000 and 20000, but the curve is nearly flat there, so that interval gives the smallest reduction and (D) is wrong. The other mistake is comparing the curve's heights at the two ends of an interval instead of the drop across it.

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9. Q.9MCQ · 2 marksOfficial answer: A

   General Aptitude › Spatial

   An orthographic view flattens the solid along the line of sight. Each column of the image is as tall as the tallest part behind it, and a nearer object's outline appears inside the outline of a farther object it overlaps. Fix left and right from where the viewer stands. Facing the lower-right face, the front corner is on your left and the outer back-right edge is on your right.

   WATCH OUT FOR

   Getting left and right the wrong way round, as if the face were seen from behind, gives the mirror image (C). Confusing view Y with the view of the other vertical face puts the small cube outside the tall column (D). Forgetting that the small cube is only half as tall draws it as a full-height strip (B).

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

    General Aptitude › Spatial

    Three edges meet at each end of a body diagonal, and the six remaining corners form two equilateral triangles at heights $\frac{1}{3}$ and $\frac{2}{3}$ of the diagonal. A turn about a vertical axis cannot change a corner's height, so each triangle has to land on itself. That means the angle must be a multiple of $120^{\circ }$, and $120^{\circ }$ itself works: it is the cyclic swap $(x,y,z)\to (z,x,y)$.

    WATCH OUT FOR

    Choosing $60^{\circ }$ because the six middle corners look like a regular hexagon from above. They are not all at the same height: alternate corners sit at $\frac{1}{3}$ and $\frac{2}{3}$ of the diagonal, so a $60^{\circ }$ turn puts a corner where the cube has none. Choosing $90^{\circ }$ mixes up this axis with the face-centre axis, which is 4-fold.

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Instrumentation Engineering Q.11 to Q.65

1. Q.11MCQ · 1 markOfficial answer: B

   Engineering Mathematics › Complex analysis

   Substitute $z=x+iy$ and turn the complex equation into a real equation in $x$ and $y$. A sum of conjugates is real: $z^2+{\bar{z}}^2=2\text{Re}(z^2)=2(x^2-y^2)$. Then read off the conic from its Cartesian form: $x^2-y^2=1$ is a hyperbola.

   WATCH OUT FOR

   Treating $z^2+{\bar{z}}^2$ as if it were $z\bar{z}=|z{|}^2=x^2+y^2$. That gives $x^2+y^2=1$ and the wrong answer, circle (D). Another slip is to expand $z^2$ as $x^2+y^2$, forgetting that $(iy{)}^2=-y^2$. Squaring flips the sign of $y^2$, and that sign is what makes the curve a hyperbola and not a circle or an ellipse.

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

   Control Systems › Process control (IN)

   Sensitivity is the slope $df/dx$ of the inherent characteristic, so 'decreasing sensitivity' means a concave curve ($d^{2}f/d{x}^2<0$). Quick opening ($\sqrt{x}$) is concave, linear ($x$) has constant slope, and equal percentage (${\alpha }^{x-1}$) and $x^2$ are convex, so their sensitivity increases.

   WATCH OUT FOR

   Picking equal percentage (A) because each equal increment of stem gives an equal percentage change in flow. Read as a relative change, $\frac{1}{f}\frac{df}{dx}=\ln \alpha$, that looks constant or 'gentle', but the absolute slope $\frac{df}{dx}=(\ln \alpha ){\alpha }^{x-1}$ grows with opening. The other trap is mixing up which curve is concave: quick opening rises steeply and then flattens, so its sensitivity is highest near closure.

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3. Q.13MCQ · 1 markOfficial answer: D

   Signals and Systems › LTI & convolution

   At zero lag the sequence is multiplied by an unshifted copy of itself, so $R_{xx}[0]={\sum }_n{x}^2[n]$, which is the signal's energy. It is also the largest value the autocorrelation takes at any lag: $R_{xx}[0]\ge |R_{xx}[l]|$.

   WATCH OUT FOR

   Adding the samples instead of their squares gives $1+2+3+4=10$, option (B), which is a trap. Dividing by the number of samples (average power, $30/4=7.5$) is also wrong here: the question asks for the autocorrelation of an energy signal, not a time-averaged one.

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

   Electromagnetics › Electrostatics

   When dielectric layers are stacked between the plates and each covers the full plate area, they are capacitors in series. The total capacitance is $C=\frac{{\epsilon }_{0}A}{{\sum }_i{d}_i/{\epsilon }_{ri}}$: each layer adds its "equivalent air thickness" $d_i/{\epsilon }_{ri}$. Moving the plate changes only the air-gap term.

   WATCH OUT FOR

   Dividing the wrong thickness by ${\epsilon }_2$: the air gap (${\epsilon }_1=1$) enters as plain $d_1$ and only the slab enters as $d_2/{\epsilon }_2$, so options (B) and (C) have the roles reversed. Another slip is writing $d_1+\delta d_1$ (option D), which is an increase in the gap, not a decrease. A third is treating the stacked layers as capacitors in parallel.

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6. Q.16MCQ · 1 markOfficial answer: A

   Digital Electronics › Boolean logic

   Use the redundancy theorem $X+\overline{X}Y=X+Y$ again and again. It first turns $A+\overline{A}B$ into $A+B$. After $\overline{A}A=0$ removes one product, it takes $A+B+\overline{A}\overline{B}C$ down to $A+B+C$. A quick way to check: $F$ is 0 only when every input is 0, and that function is the OR of all the inputs.

   WATCH OUT FOR

   Losing the leading $A$ partway through. Redundancy on $B$ alone correctly gives $B+\overline{A}\overline{B}C=B+\overline{A}C$, which is option (C). But $F$ still has the $A$ term, and $A+\overline{A}C=A+C$ turns it into $A+B+C$. Other mistakes are keeping $\overline{A}AC$ as a real term instead of setting it to 0, or thinking the whole second product is absorbed just because it contains $A$.

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

   Control Systems › State space

   The poles of $\dot{x}=Ax+Bu$ are the eigenvalues of $A$. The product of the eigenvalues equals $\det A$, so the system has a pole at the origin exactly when $\det A=0$. You only need to find the one singular matrix, and you never have to solve a quadratic.

   WATCH OUT FOR

   One mistake is picking (A) or (D) because they have a zero on the diagonal. A zero entry says nothing about the eigenvalues; both of those matrices have $\det A=2$. The other is a sign slip in (B): $(-1.5)(-2)=+3$, not $-3$. With that slip you get $\det A=6$ and throw away the correct option.

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

   Control Systems › Bode & margins

   Split $G(j\omega )$ into real and imaginary parts and match them to the plot. A vertical line at $\text{Re}=1$ means the real part does not depend on $\omega$, so $G$ must be $1$ plus a purely imaginary term. The direction the line runs (down to $-j\infty$) fixes the sign of that term as $-j\omega T$. Neither rational option can do this: they stay bounded, tracing the unit circle (all-pass) or a semicircle into the origin (lag).

   WATCH OUT FOR

   Picking (C), $1+j\omega T$, because it gives the same vertical line. It runs upward to $+j\infty$, while the plot runs down. The other trap is choosing the all-pass (B) because it also has a right-half-plane zero and starts at $1$. Its magnitude stays at $1$, so it traces a semicircle ending at $-1$ and never heads off to infinity.

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

   Engineering Mathematics › Linear algebra

   A square $n\times n$ matrix has full rank $n$ exactly when its columns are linearly independent, i.e. when its determinant is nonzero. To rule out an option quickly, look for an obvious dependence: a zero column, or one column equal to a sum or multiple of the others. Then confirm the remaining option with a determinant, expanding along the row or column with the most zeros.

   WATCH OUT FOR

   Seeing that no two columns are parallel and concluding the rank is 3. Pairwise non-parallel columns are not enough: in (A) and (D) no column is a multiple of another, yet $v_3$ is a combination of $v_1$ and $v_2$, so the rank is only 2. A second slip is dropping the cofactor sign $(-1{)}^{i+j}$ in a row expansion. Here it does not change whether $\det M$ is zero, but it does give the wrong determinant.

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

    Electromagnetics › Electrostatics

    Assume charges $\pm Q$ on the two shells. Gauss's law gives $E=Q/(4\pi ϵr^2)$ in the gap. Integrate from $x$ to $y$ to get $V=\frac{Q}{4\pi ϵ}\left(\frac{1}{x}-\frac{1}{y}\right)$, then $C=Q/V=4\pi ϵ\frac{xy}{y-x}$. To check the answer quickly, let the outer radius go to infinity: the result must become the isolated-sphere value $4\pi ϵx$.

    WATCH OUT FOR

    Students often slip when combining $\frac{1}{x}-\frac{1}{y}$ into one fraction. Some write $\frac{x-y}{xy}$, which is negative, or $\frac{y-x}{y^2}$, and then pick (C) or (B). Others use the coaxial-cable result $2\pi ϵL/\ln (y/x)$, which applies to cylinders, not spheres. Putting $y\to \infty$ exposes all of these errors at once.

    Full solution →
11. Q.21NAT · 1 markOfficial answer: 0.15 to 0.17
12. Q.22NAT · 1 markOfficial answer: 0.70 to 0.80

    Signals and Systems › FIR & IIR filters

    Delays in the difference equation become factors $e^{-j\omega k}$ in the frequency response. At $\omega =\pi /2$, $e^{-j2\omega }=-1$ and $e^{-j4\omega }=1$, so the input delay $x[n-4]$ has magnitude 1 and drops out, and the denominator becomes $1+0.5=1.5$.

    WATCH OUT FOR

    Taking $e^{-j2\omega }$ as $+1$ at $\omega =\pi /2$ (it is $e^{-j\pi }=-1$). That gives a denominator of $0.5$ and the wrong answer $a=0.25$. Another mistake is thinking the delay $x[n-4]$ changes the magnitude. It only adds phase, because $|e^{-j4\omega }|=1$ at every frequency.

    Full solution →
13. Q.23NAT · 1 markOfficial answer: 1.6

    Analog Electronics › Diode circuits

    The open-loop op-amp is a comparator, so its high output is the $+5\text{V}$ rail. The CMOS input draws no current, so all of $I_{R_Z}=(5-1.8)/R_Z$ goes into the Zener. Setting this current to at least $I_{Z,\min }=2\text{mA}$ gives an upper limit on $R_Z$.

    WATCH OUT FOR

    Using the Zener voltage alone, $1.8/2\text{mA}=0.9\text{k}\Omega$, when the drop across $R_Z$ is actually $5-1.8=3.2\text{V}$. A second slip is to treat the $-5\text{V}$ state as the limiting case, or to subtract load current into the CMOS gate, which draws none.

    Full solution →
14. Q.24NAT · 1 markOfficial answer: 498 to 502

    Analog Electronics › Biasing

    The gate draws no DC current, so the 100 kΩ feedback resistor has no voltage across it. This forces $V_{GS}=V_{DS}$, which keeps the transistor in saturation (a diode-connected bias). Solve the square law and the drain-resistor KVL together, then keep only the root with $V_{GS}>V_T$.

    WATCH OUT FOR

    One common slip is including a voltage drop across the 100 kΩ resistor, or letting it share current with the drain resistor, when it carries no DC current. Another is forgetting the factor $\frac{1}{2}$ in the square law, which gives $k=1000\mu {\text{A/V}}^2$ and the wrong current. A third is keeping the negative root of the quadratic, which puts $V_{GS}$ below $V_T$ where the square law does not hold.

    Full solution →
15. Q.25NAT · 1 markOfficial answer: 32

    Signals and Systems › DTFT & DFT

    A radix-2 FFT of length $N=2^m$ has ${\log }_{2}N$ stages, each with $N/2$ butterflies. Each DIT butterfly needs one complex multiplication, by the twiddle factor $W_N^k$, and two complex additions. That gives $\frac{N}{2}{\log }_{2}N$ multiplications and $N{\log }_{2}N$ additions in total.

    WATCH OUT FOR

    There are three usual slips. The first is counting two multiplications per butterfly (one for each output), which gives $N{\log }_{2}N=64$, but the product $W_N^{k}B$ is computed once and shared by both outputs. The second is answering with the direct-DFT count $N^2=256$. The third is counting only the non-trivial twiddle products, skipping every $W_N^k$ equal to $1$ or $-j$. That gives a smaller number, but it is not the standard count the question asks for.

    Full solution →
16. Q.26NAT · 1 markOfficial answer: -2

    Engineering Mathematics › Linear algebra

    $|P|$ is the product of the eigenvalues, and the eigenvalues of $P^{-1}$ are their reciprocals. So $|P^{-1}|=\frac{1}{|P|}=\frac{1}{\frac{1}{4}\cdot 1\cdot (-2)}=-2$.

    WATCH OUT FOR

    Reading $|P^{-1}|$ as an absolute value and answering $2$ (it is the determinant, and it is negative here). Another slip is to give $|P|=-\frac{1}{2}$ and round it to $0$ or $-1$, forgetting that the question asks for the determinant of the inverse.

    Full solution →
17. Q.27NAT · 1 markOfficial answer: 400

    Signals and Systems › Sampling

    Before you apply $f_s=2f_{max}$ to a non-linear expression, first rewrite it as a sum of pure sinusoids. The identity ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$ turns $10{\sin }^2(200\pi t)$ into a DC term plus a cosine at $400\pi \text{rad/s}$ ($200\text{Hz}$), so $f_{max}=200\text{Hz}$ and the Nyquist rate is $400\text{Hz}$.

    WATCH OUT FOR

    Reading $f_{max}$ straight off the argument $200\pi t$ gives $100\text{Hz}$ and the wrong answer $200\text{Hz}$, because it misses that squaring doubles the frequency. A second slip is dividing by $2\pi$ incorrectly and treating $400\pi$ as $400\text{Hz}$, which gives $800\text{Hz}$.

    Full solution →
18. Q.28NAT · 1 markOfficial answer: 22

    Electrical and Electronic Measurements › Errors & standards

    If the arithmetic mean equals the true value, the deviations of the readings from the true value must add up to zero. Equivalently, the sum of all n readings must be n times the true value, and the missing reading is whatever is left over.

    WATCH OUT FOR

    Averaging only the five known readings (98/5 = 19.6 kΩ) and reporting that, or dividing by 5 instead of 6 when setting up the mean. The condition covers all six readings, so the total must be 6 × 20 = 120 kΩ.

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19. Q.29NAT · 1 markOfficial answer: -3

    Analog Electronics › Op-amps

    This is a charge amplifier. The virtual ground holds the inverting node at 0 V, so the sensor's shunt capacitance (2 µF) takes no charge and all of $q_p$ goes onto the feedback capacitors. Since nothing loads $V_x$, the two series capacitors carry the same charge. So $V_x$ gives you $q_p$ through the 100 pF capacitor, and $q_p$ then gives you the extra drop across the 200 pF capacitor.

    WATCH OUT FOR

    Taking $V_x$ as the output, or splitting the voltage the wrong way. With equal charge, the drop across each series capacitor goes as $1/C$, so the 200 pF capacitor drops half as much as the 100 pF one (1 V, not 4 V). Another mistake is letting the 2 µF sensor capacitance share the charge. It sits at virtual ground and takes none.

    Full solution →
20. Q.30NAT · 1 markOfficial answer: 200

    Networks (Electric Circuits) › AC, power & resonance

    Voltage and current components at different frequencies (a DC voltage paired with an AC current counts as different) transfer no average power over a common period. Only same-frequency pairs contribute, each giving $\frac{V_m{I}_m}{2}\cos ({\theta }_v-{\theta }_i)$. Here the only such pair is the 60 Hz one: $\frac{200\times 4}{2}\cos 60^{\circ }=200\text{W}$.

    WATCH OUT FOR

    Students often multiply the DC voltage by an AC current (writing $95\times 4$ or similar), or pair the 90 V (180 Hz) voltage with the 1.5 A (120 Hz) current. Both give zero average power, because the frequencies differ. A second common slip is using $V_m{I}_m\cos \varphi$ with peak values and forgetting the factor $\frac{1}{2}$, which gives 400 W. Using the wrong sign in the phase difference is harmless here, since $\cos (\pm 60^{\circ })=0.5$.

    Full solution →
21. Q.31NAT · 1 markOfficial answer: 27.0 to 30.0

    Signals and Systems › Signals basics

    For a sum of a dc term and sinusoids at distinct frequencies, the mean squares add because the cross terms average to zero: $I_{\text{rms}}=\sqrt{I_0^2+{\sum }_k{I}_{mk}^2/2}$. First merge the $\cos$ and $\sin$ terms at each frequency into one sinusoid of peak $\sqrt{a^2+b^2}$. The dc term counts at full value, not halved.

    WATCH OUT FOR

    Common mistakes: (1) halving the dc term, i.e. using $4^2/2$ instead of $4^2$; (2) adding peak or rms amplitudes directly instead of their squares, which gives $4+30/\sqrt{2}+\dots$; (3) treating the $\cos$ and $\sin$ terms at the same frequency as if their peaks add linearly ($30+20$) instead of in quadrature ($\sqrt{30^2+20^2}$). The negative signs on the sine terms have no effect on the rms value.

    Full solution →
22. Q.32NAT · 1 markOfficial answer: 10.1 to 10.3
23. Q.33NAT · 1 markOfficial answer: 15

    Networks (Electric Circuits) › Transients

    The inductor current cannot change instantly. Once the switch opens, the capacitor is the only path left for that current, so $i_C(0^{+})=i_L(0^{-})$ and $d{v}_c/dt{|}_{0^{+}}=i_L(0^{-})/C$.

    WATCH OUT FOR

    Some students treat the capacitor current as zero at $t=0^{+}$ because it was zero in steady state, and so answer 0. Only the capacitor voltage is continuous, not its current. Others keep the lower 4 Ω resistor in the circuit after the switch opens, splitting the current and getting a smaller slope. A third error is to take $i_L(0^{-})=12/4=3\text{A}$ by forgetting the second resistor, which gives 30.

    Full solution →
24. Q.34NAT · 1 markOfficial answer: 1.10 to 1.12

    Engineering Mathematics › Numerical methods (IN)

    Improved Euler predicts with a plain Euler step, then corrects by averaging the slope at the start, $f(t_0,y_0)$, with the slope at the predicted end point, $f(t_1,y_1^{\ast })$: $y_1=y_0+\frac{h}{2}\left[f(t_0,y_0)+f(t_1,y_1^{\ast })\right]$. The predicted $y_1^{\ast }$ must go into the second slope evaluation, not $y_0$.

    WATCH OUT FOR

    Stopping at the plain Euler step gives $1.10$, which only just touches the edge of the key range and is the wrong method. Another slip is to evaluate the second slope at $(t_1,y_0)$ instead of $(t_1,y_1^{\ast })$, which gives $f(0.1,1)=1.19$ and $y_1=1.1095\approx 1.11$: the right number by luck, from the wrong working. Also, some textbooks use 'modified Euler' for the midpoint rule. Here that also gives $1.11475\approx 1.11$, but Heun's form is the standard reading of 'improved Euler'.

    Full solution →
25. Q.35NAT · 1 markOfficial answer: 0.5

    Engineering Mathematics › Probability & statistics

    The two finalists come from different pools, so the events are independent. Split "one blue, one red" into its two exclusive cases (A blue with B red, and A red with B blue), multiply within each case and add the two. Checking with the complement (both the same colour) confirms the result.

    WATCH OUT FOR

    Counting only one case (for example A blue, B red $=\frac{12}{64}\approx 0.2$) and forgetting the reverse case. Another mistake is pooling all 16 teams and using $\frac{9}{16}\cdot \frac{7}{16}\cdot 2$ or $\left(\genfrac{}{}{0ex}{}{16}{2}\right)$. That ignores the fact that exactly one finalist comes from each pool.

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

    Electromagnetics › Magnetostatics

    When the current density varies across the cross-section, the total current is the surface integral $I={\int }_S\mathbf{J}\cdot d\mathbf{S}$, not $J\times \text{area}$. For $J$ that depends only on $s$, use rings of area $2\pi sds$, so $I={\int }_0^{a}J(s)2\pi sds$.

    WATCH OUT FOR

    Dropping the $s$ in the area element $sdsd\varphi$ gives $2\pi k{a}^3/3$, which is option (B). Taking the ring area as $\pi sds$ in place of $2\pi sds$ gives $\pi k{a}^4/4$, which is option (D). Multiplying $J(a)=k{a}^2$ by $\pi a^2$ gives $\pi k{a}^4$, which is double the right value.

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

    Electrical and Electronic Measurements › Errors & standards

    Always measure errors from the true value. The bias is the mean of the readings minus the true value. The maximum error is the largest single-reading deviation from the true value. How far the readings spread around their own mean is precision, not accuracy.

    WATCH OUT FOR

    The usual mistake is measuring from the wrong reference. The spread of the readings, 0.39 - 0.35 = 0.04, describes precision, not maximum error, and taking it for the maximum error leads to option (D). Using the reading nearest the true value (0.35 - 0.25 = 0.10) as the bias leads to option (C). A third slip is dividing by the 0 to 2 full-scale range, which gives a percentage of full scale when the question asks for flow units.

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

    Signals and Systems › z-transform

    A finite-length sequence has a finite sum for its $z$-transform, so the ROC is the whole $z$-plane with at most two points taken out: $z=0$ is excluded if the sequence has any non-zero sample at $n>0$ (a $z^{-n}$ term), and $z=\infty$ is excluded if it has any non-zero sample at $n<0$ (a $z^{|n|}$ term). Here all samples are at $0\le n\le 7$, so only $z=0$ is excluded.

    WATCH OUT FOR

    Writing the geometric-series closed form $\frac{1-(0.2{)}^8{z}^{-8}}{1-0.2z^{-1}}$, seeing the denominator vanish at $z=0.2$, and choosing (A) or setting the ROC to $|z|>0.2$ as for the infinite sequence $(0.2{)}^{n}u[n]$. That pole is cancelled by a zero of the numerator. The other common slip is choosing (B) because 'finite sequences converge everywhere', forgetting that the $z^{-n}$ terms blow up at $z=0$.

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

    Electrical and Electronic Measurements › Bridges

    An AC bridge balances when the products of the complex impedances of opposite arms are equal, $Z_1{Z}_4=Z_2{Z}_3$. The real parts and the imaginary parts must each match. The inductive arm does not balance on its own, so the answer depends on frequency. Take $\omega =5000\text{rad/s}$ straight from $5\sin (5000t)$.

    WATCH OUT FOR

    One mistake is to read 5000 as the frequency in hertz and use $\omega =2\pi \times 5000$. The argument of the sine is $\omega t$, so $\omega =5000\text{rad/s}$. Another is to pair adjacent arms instead of opposite ones, for example $Z_4=Z_1{Z}_2/Z_3$. That makes $Z_4$ inductive, which a series RC arm cannot be.

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

    Signals and Systems › Laplace

    When the ROC is a strip between poles, a pole to the right of the strip gives a left-sided (anti-causal) term, $-e^{-at}u(-t)$, and a pole to the left gives a right-sided (causal) term, $e^{-at}u(t)$. The ROC of the whole signal is where the individual ROCs overlap.

    WATCH OUT FOR

    Treating every term as causal (the one-sided Laplace habit), or dropping the minus sign in the left-sided pair $-e^{-at}u(-t)\leftrightarrow \frac{1}{s+a}$. Quick check on the other options: in (A) the term $-e^{-6t}u(t)$ needs $\text{Re}\{s\}>-6$ and $-e^{-7t}u(-t)$ needs $\text{Re}\{s\}<-7$, and no $s$ satisfies both. (C) has the same empty ROC. (D) is purely left-sided with ROC $\text{Re}\{s\}<-7$, and its transform is $\frac{1}{s+6}+\frac{1}{s+7}$, which is not $X(s)$.

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

    Analog Electronics › Op-amps

    A series offset source moves the node it is connected to. The virtual short pins the op-amp's inverting input at 0 V, so the junction node sits at $+V_{os}=1\text{mV}$ instead of at virtual ground. That node sees two resistors to 0 V and one feedback resistor, so it is amplified by the noise gain $1+R/(R\parallel R)=3$. The second op-amp simply inverts $V_X$ with gain $-1$.

    WATCH OUT FOR

    A common mistake is to treat node $X$ as a virtual ground, which gives $V_{out2}=0$ (options B or D). Another is to count only the source resistor when working out the gain, giving $1+R/R=2$ and $V_{out1}=2\text{mV}$ (option D). The resistor to the virtual ground of $A_2$ also loads node $X$, so the gain is 3. Reading the battery polarity backwards flips both signs.

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

    Digital Electronics › Sequential

    Each MUX in front of a D flip-flop picks between holding and shifting. With $S=1$ each stage loads the stage to its left (DIN for stage 1), so the circuit is a right-shift register. With $S=0$ each stage reloads its own $Q$ and nothing changes. The NAND of the two AND terms reduces to $Y=\overline{\overline{Q_1}{Q}_2{Q}_3}$, so you only need to find when the register sits in state 011. Take S and DIN at the instant of each rising edge.

    WATCH OUT FOR

    The usual slip is to miss the hold on clock 4. S is 0 at that edge, but students often shift DIN = 1 in anyway, which gives states 011, 101, 110 and $Y_3{Y}_4{Y}_5=011$, the trap option (D). A second slip is reading S or DIN just after an edge instead of at it. Both change shortly after edge 3, so at edge 3 they are still S = 1 and DIN = 0. A third is misreading the left AND input as $Q_1$ when the wire actually comes from $\overline{Q_1}$.

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

    Analog Electronics › Filters & waveforms

    In an inverting amplifier with $R_f\parallel C$ in the feedback path, the gain is $-\frac{R_f}{R}\cdot \frac{1}{1+j\omega R_{f}C}$. The dc gain $-\frac{R_f}{R}$ sets $R_f$, and the single pole sits at $f_{3\text{dB}}=\frac{1}{2\pi R_{f}C}$. The input resistor $R$ sets the gain but plays no part in the cut-off.

    WATCH OUT FOR

    Using the input resistor in the cut-off formula. $\frac{1}{2\pi RC}$ gives $1591.5\text{Hz}$ (option B), and $\frac{1}{2\pi (R\parallel R_f)C}$ gives $1750.7\text{Hz}$ (option C). A second trap: strict rounding of $159.155\text{Hz}$ gives $159.2\text{Hz}$, which is not printed. Candidates who look only for $159.2$ can wrongly reject (A), the only option that comes from $\frac{1}{2\pi R_{f}C}$.

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

    Control Systems › Time response

    With non-unity feedback, the error $r-y$ is not the signal leaving the summing junction. Find the closed-loop transfer function $T(s)=\frac{1}{s^2+s+K}$. For a unit step, $e_{ss}=1-T(0)=1-\frac{1}{K}$, provided the loop is stable ($K>0$).

    WATCH OUT FOR

    Picking option (C), 0, because the forward path $\frac{1}{s(s+1)}$ is type 1. The type-1 integrator drives the actuating signal $r-Ky$ to zero, not $r-y$. It forces $y(\infty )=1/K$, so the output tracks the step only when $K=1$. A sign slip in $1-\frac{1}{K}$ gives option (D), $\frac{1-K}{K}$.

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

    Control Systems › Time response

    For a complex pole pair of the standard second-order system, $|s|={\omega }_n$, so circles centred at the origin are lines of constant ${\omega }_n$. The angle from the negative real axis is ${\cos }^{-1}\xi$, so rays from the origin are lines of constant $\xi$. A wedge of half-angle $\theta$ around the negative real axis therefore means $\xi \ge \cos \theta$, and the region outside a circle of radius $r$ means ${\omega }_n\ge r$.

    WATCH OUT FOR

    One mistake is to use the marked $120^{\circ }$ directly, which gives $\cos 120^{\circ }=-\frac{1}{2}$. The damping angle has to be measured from the NEGATIVE real axis, which makes it $60^{\circ }$ and gives $\xi \ge \frac{1}{2}$. The other mistake is to take the corner point's imaginary part $\sqrt{3}$, from $-1+j\sqrt{3}$, as the bound on ${\omega }_n$ (options C and D). The bound is the radius, 2, not the height. The value $\xi \ge \frac{1}{4}$ in option B matches no boundary in the figure: ${\cos }^{-1}\frac{1}{4}\approx 75.5^{\circ }$, which lies outside the wedge.

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

    Engineering Mathematics › Vector calculus

    Green's theorem turns a counterclockwise line integral around a closed curve into the double integral of $\partial Q/\partial x-\partial P/\partial y$ over the region inside it. Here the integrand $e^x+e^y$ splits into two parts, and over a rectangle each part is a product of two one-variable integrals.

    WATCH OUT FOR

    Losing the minus sign on $P=-e^y$. Doing so gives the integrand $e^x-e^y$ and the value $e^2-2e+1$, which is not among the options. A second mistake is using the wrong side lengths: the $e^y$ term has to be multiplied by the $x$-width (2), and the $e^x$ term by the $y$-height (1). Swapping them gives $2e^2+e-3$.

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

    Electrical and Electronic Measurements › Instrument transformers

    Draw every phasor relative to the load voltage. The CT and PT each rotate their secondary quantity by their own phase angle, which can go either way, hence the $\pm$. The pressure-coil inductance makes $I_p$ lag its voltage by $\beta$. So the angle between the coil currents, $\gamma$, is the load angle $\alpha$ reduced by $\beta$ and shifted by the two transformer angles: $\alpha =\gamma +\beta \pm \delta \pm \theta$.

    WATCH OUT FOR

    Getting the sign of $\beta$ wrong, which leads to option (D). Because the pressure-coil current lags its voltage by $\beta$, it swings toward the lagging load current, so the gap between the coil currents is smaller than the load angle: $\gamma =\alpha -\beta$ (ignoring transformer errors), not $\alpha +\beta$. Options (A) and (B) also flip the sign of $\gamma$. That would make the coil currents close to $-\alpha$ apart for a lagging load, which is impossible.

    Full solution →
38. Q.48MCQ · 2 marksOfficial answer: A
39. Q.49MSQ · 2 marksOfficial answer: A, C

    Engineering Mathematics › Complex analysis

    Write the Cauchy-Riemann equations for $f=u+iv$ and for $\overline{f}=u+i(-v)$. Each pair of equations has the same left side and right sides of opposite sign, so every first partial derivative of $u$ and $v$ must be zero. That makes $f^{\prime }(z)=u_x+i{v}_x=0$, and on a connected domain $f$ is constant.

    WATCH OUT FOR

    Using $+v$ instead of $-v$ as the imaginary part of $\overline{f}$ in its Cauchy-Riemann equations. Doing that just repeats the equations for $f$ and gives no new information. Another trap is picking (B) because it looks like the standard CR equation $u_y=-v_x$. The equation does hold, but both sides equal zero, so the "$\ne 0$" part of (B) is false.

    Full solution →
40. Q.50NAT · 2 marksOfficial answer: 0.450 to 0.550

    Engineering Mathematics › Probability & statistics

    $R^2$ is the fraction of the spread in $y$ that the line accounts for. For a least-squares line it equals $1-S{S}_E/S{S}_T$, which for a straight line simplifies to $S_{xy}^2/(S_{xx}{S}_{yy})$ (the square of the correlation coefficient). First confirm that the given line really is the least-squares fit; here $b=S_{xy}/S_{xx}=0.04$ and $a=\bar{y}-b\bar{x}=10.32$.

    WATCH OUT FOR

    Common mistakes: reporting the correlation coefficient $r=\sqrt{0.5}\approx 0.707$ in place of $R^2$; using raw sums of $y^2$ instead of deviations from the mean, which loses the small differences (of order $0.01$) to rounding; and dividing by $n-1$ in only one of $S_{xy}$, $S_{xx}$ or $S_{yy}$. The ratio stays correct only when all three are treated the same way.

    Full solution →
41. Q.51NAT · 2 marksOfficial answer: 4

    Networks (Electric Circuits) › Laws & node/mesh

    Seen from the detector diagonal (A-B), with the excitation source shorted, a Wheatstone bridge is two parallel pairs in series: $R_{AB}=(R_1\parallel R_2)+(R_3\parallel R_4)$. Each pair is the two arms that meet at one detector node. An ideal INA draws no input current, so it does not appear in the result.

    WATCH OUT FOR

    Opening the voltage source instead of shorting it. The two paths from A to B are then each $(R-\Delta R)+(R+\Delta R)=2R$, and in parallel they give $R=4.5\text{k}\Omega$, which rounds to 5, not the key's 4. An ideal voltage source must be replaced by a short circuit. A second slip is to include a finite INA input resistance in parallel with the bridge. An ideal INA has infinite input resistance.

    Full solution →
42. Q.52NAT · 2 marksOfficial answer: 6.00 to 6.30

    Analog Electronics › Filters & waveforms

    In a symmetric op-amp astable, the capacitor swings between $\pm \beta V_{sat}$, charging each time toward $\pm V_{sat}$. Each half-period is $RC\ln \frac{1+\beta }{1-\beta }$, so $T=2RC\ln \frac{1+\beta }{1-\beta }=2RC\ln \left(1+\frac{2R_1}{R_2}\right)$. Here $\beta =\frac{R_1}{R_1+R_2}$, $R_1$ is the resistor from the $+$ input to ground (10 kΩ) and $R_2$ is the one from the $+$ input to the output (16 kΩ). The value of $V_{sat}$ cancels, so the ±13 V only confirms that the thresholds are ±5 V.

    WATCH OUT FOR

    Swapping the divider resistors. $\beta =\frac{16}{26}$, or $T=2RC\ln \left(1+\frac{2\cdot 16}{10}\right)=2RC\ln 4.2$, gives about 3.48 kHz. Taking $\beta =\frac{10}{16}$ (the resistor ratio instead of the divider fraction) gives about 3.41 kHz. A third mistake is timing the start-up interval (0 to $+V_P$, which is $RC\ln \frac{13}{8}$) as if it were a half-period; the figure marks $T$ in steady state, where $V_C$ swings the full $-V_P$ to $+V_P$.

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

    Signals and Systems › DTFT & DFT

    A circular convolution of length $N$ becomes a product of the $N$-point DFTs, $Z[k]=X[k]Y[k]$. Setting $n=0$ in the inverse DFT turns every exponential into 1, so $z[0]=\frac{1}{N}{\sum }_{k}Z[k]$. You never have to recover $x[n]$ or $y[n]$.

    WATCH OUT FOR

    Leaving out the $\frac{1}{N}$ in the inverse DFT gives $z[0]=8$. Because the forward DFT given here has no scale factor, the whole $\frac{1}{4}$ belongs to the inverse. The other slip is a sign error in $j\cdot j$: $(-j)(3j)=-3j^2=+3$, not $-3$. With that error $Z[1]$ and $Z[3]$ come out as $-3$, and the sum gives $z[0]=-1$.

    Full solution →
44. Q.54NAT · 2 marksOfficial answer: 58 to 62

    Networks (Electric Circuits) › Laws & node/mesh

    With zero galvanometer current, nodes $T$ and $C$ are at the same potential. The lower network ($40\Omega$ and $60\Omega$ from $L$ to $R$ through both $C$ and $B$) is itself balanced, so the $60\Omega$ resistor from $C$ to $B$ carries no current. $V_C$ is then just the $40:60$ divider, and the balance reduces to the ratio $R_x/90=40/60$.

    WATCH OUT FOR

    Setting up the ratio with the arms swapped, for example $R_x/90=60/40$, which gives $135\Omega$. Another mistake is assuming the bottom node $B$ is wired to the source, or trying to solve the full five-node network when the $C$–$B$ resistor carries no current. The arms that must be in the same ratio are the ones that meet at the source terminals: $R_x$ and $40\Omega$ both touch $L$, and $90\Omega$ and $60\Omega$ both touch $R$.

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

    Control Systems › Root locus

    A point on the $j\omega$-axis lies on the root locus exactly where the closed loop is marginally stable. Make the $s^1$ row of the Routh array zero to get the critical gain, then solve the auxiliary equation from the $s^2$ row to get the crossing frequency. As a quick cross-check, the pole angles at $s=j2$ ($90^{\circ }+63.43^{\circ }+26.57^{\circ }$) add up to exactly $180^{\circ }$.

    WATCH OUT FOR

    Stopping at the critical gain and entering $K=20$ when the question asks for the crossing frequency $\alpha =2$. Another slip is expanding $s(s+1)(s+4)$ as $s^3+5s^2+5s$ (the $s$ coefficient is $1\cdot 4=4$), which moves the crossing to $j\sqrt{5}$. Some students also treat the trivial root $\omega =0$ as the answer, but it corresponds to $K=0$ and is ruled out by $\alpha >0$.

    Full solution →
46. Q.56NAT · 2 marksOfficial answer: 2.81 to 2.85

    Analog Electronics › Noise & shielding (IN)

    The buffer holds the shield at the inner wire's voltage, so $C_{\text{stray}}$ has no voltage across it and carries no current. The ideal amplifier inputs draw no current either, so $R_{\text{wire}}$ and the $2\Omega$ input resistor drop nothing. The op-amp therefore sees the unloaded inductive-divider voltage $V_m=\frac{V_{s1}{L}_2+V_{s2}{L}_1}{L_1+L_2}=2\sin (2000\pi t)$. A non-inverting gain of 2 makes the peak 4 V, and $4/\sqrt{2}\approx 2.83\text{V}$ RMS.

    WATCH OUT FOR

    Swapping the divider weights. The junction voltage weights $V_{s1}$ by the FAR inductor $L_2$, not by $L_1$. Writing $\frac{6(2)+(-6)(4)}{6}$ gives $-2$, which has the same magnitude only by coincidence, so check the sign with KVL. Other slips: using an inverting gain of $-R_f/R_g=-1$ instead of the non-inverting $1+R_f/R_g=2$ (answer $1.41$); loading the divider with $R_{\text{wire}}$, the $2\Omega$ resistor or $C_{\text{stray}}$ even though no current flows; and reporting the peak (4 V) instead of the RMS value.

    Full solution →
47. Q.57NAT · 2 marksOfficial answer: 9.5 to 10.5

    Analog Electronics › Diode circuits

    A fixed current in a diode whose saturation current is held constant gives $V_D=\frac{T}{\alpha }\ln (I_D/I_S)$, which is proportional to absolute temperature. The non-inverting amplifier multiplies it by $1+5R/R=6$, so $d{V}_{out}/dT=\frac{6}{\alpha }\ln (I_D/I_S)$.

    WATCH OUT FOR

    (1) Using the familiar $-2{\text{mV/}}^{\circ }\text{C}$ rule for a real diode. That rule comes from $I_S$ rising steeply with temperature, and this question fixes $I_S$ at a constant, so here the voltage rises with temperature. (2) Taking the gain as $5R/R=5$, the inverting-amplifier gain, when the circuit is non-inverting with gain $1+5=6$; this gives 8.3 mV. (3) Dropping the $8\ln 10$ term or using ${\log }_{10}$ in place of $\ln$.

    Full solution →
48. Q.58NAT · 2 marksOfficial answer: 3

    Digital Electronics › ADC & DAC

    The number of bits per sample comes from the full-scale voltage and the resolution: $1.4/0.2=7$ steps, which means 8 levels and so 3 bits. The Nyquist rate for a 500 MHz band is $10^9$ samples/s. The serial bit rate is the bits per sample times the sampling rate: $3\times 1\text{GHz}=3\text{Gbps}$.

    WATCH OUT FOR

    Using the bandwidth (500 MHz) as the sampling rate instead of the Nyquist rate $2\times 500\text{MHz}$, which gives 1.5 Gbps. Another mistake is taking ${\log }_{2}7\approx 2.81$ as the bit count and giving about 2.8 Gbps. The bit count has to be a whole number, and 7 steps means 8 levels, so it needs 3 bits.

    Full solution →
49. Q.59NAT · 2 marksOfficial answer: -2.6 to -2.4

    Analog Electronics › Op-amps

    This is an inverting integrator. Its output equals minus the area under $V_{in}$ divided by $RC$, starting from the initial capacitor voltage. While $V_{in}=0$ there is no current, so the output holds its last value. Here the area up to 2 ms is $5\text{V}\times 1\text{ms}$ and $RC=2\text{ms}$, so $V_{out}=-2.5\text{V}$.

    WATCH OUT FOR

    (1) Treating the circuit as a passive RC charging exponentially with $\tau =RC$, e.g. $-5(1-e^{-0.5})\approx -2.0\text{V}$. The virtual ground keeps the resistor current constant, so the output ramp is linear. (2) Assuming the output decays back to 0 during the 1–2 ms gap. With zero input there is no current, so the capacitor holds its charge. (3) Dropping the minus sign of the inverting integrator and answering $+2.5$. (4) Unit slips in $RC$ ($1\text{k}\Omega \times 2\mu \text{F}=2\text{ms}$, not 2 s).

    Full solution →
50. Q.60NAT · 2 marksOfficial answer: 0

    Electrical and Electronic Measurements › V, I, power & energy

    The comparator, the gain $-1$ amplifier and the switch together multiply $V_{\text{sense}}$ by a $\pm 1$ square wave locked to $V_{\text{ref}}$. This is a phase-sensitive detector. A PMMC meter with a $1\text{Hz}$ bandwidth reads only the DC part of the product. That DC part is non-zero only when $V_{\text{sense}}$ has a component at the reference frequency or at one of its odd harmonics. Here $2\text{kHz}$ is an even harmonic of $1\text{kHz}$, so the reading is $0$.

    WATCH OUT FOR

    One error is to treat the circuit as a full-wave rectifier, taking $V_{\text{out}}=|V_{\text{sense}}|$ and reading $\frac{2V_m}{\pi }=\frac{4}{\pi }\approx 1.27\to 1\text{V}$. The switch follows $V_{\text{ref}}$, not $V_{\text{sense}}$, so the two signals are at different frequencies and the product is not rectified. Another error is to read an rms value, $\frac{2}{\sqrt{2}}\approx 1.41\to 1\text{V}$. A PMMC meter responds to the average value, not the rms value.

    Full solution →
51. Q.61NAT · 2 marksOfficial answer: 95.2 to 95.4

    Electrical Machines › Transformers

    With core loss negligible, all loss at any load is copper loss, which scales as the square of the per-unit load. Full-load unity-pf efficiency fixes the full-load copper loss; scale it to 75% load, recompute output at the new load fraction and power factor, and take the ratio.

    WATCH OUT FOR

    Forgetting that copper loss scales with the square of the load fraction (using 0.75x instead of 0.75 squared), or re-deriving core loss instead of taking it as negligible as stated.

    Full solution →
52. Q.62NAT · 2 marksOfficial answer: 17.00 to 17.30

    Electrical Machines › Induction machines

    With stator impedance neglected, the torque-slip relation about the maximum-torque point is T/Tmax = 2/(s/s_maxT + s_maxT/s). Evaluating at starting slip (s=1) with the given ratio Tst/Tmax = 1/3 gives a quadratic in s_maxT; the physically valid root (0 < s_maxT < 1) is the answer.

    WATCH OUT FOR

    Picking the wrong root of the quadratic (taking s = 3 + 2√2 ≈ 5.83, which is unphysical since slip cannot exceed 1 on this branch) instead of s = 3 - 2√2 ≈ 0.1716.

    Full solution →
53. Q.63NAT · 2 marksOfficial answer: 0.25

    Networks (Electric Circuits) › Coupled circuits

    Aiding minus opposing inductance is $4M$, and aiding plus opposing is $2(L_1+L_2)$. Once you have $M$ and the two self-inductances, $k=M/\sqrt{L_1{L}_2}$. You never need $L_1$ and $L_2$ separately unless they are needed for $\sqrt{L_1{L}_2}$. Here they are equal, so $\sqrt{L_1{L}_2}=L_1$.

    WATCH OUT FOR

    Writing $2M=L_{aid}-L_{opp}$ instead of $4M$ gives $M=100\text{mH}$ and $k=0.5$. A second slip is to divide $M$ by $L_1+L_2=400\text{mH}$ instead of by $\sqrt{L_1{L}_2}=200\text{mH}$, which gives $0.125$.

    Full solution →
54. Q.64NAT · 2 marksOfficial answer: 33

    Engineering Mathematics › Differential equations

    The operator is $(D+1{)}^3$, so substituting $y=u{e}^{-t}$ turns the ODE into $u^{\prime \prime \prime }=30$. That fixes $c_3=5$ straight away. The three initial conditions, taken through $y^{\prime }=(u^{\prime }-u)e^{-t}$ and $y^{\prime \prime }=(u^{\prime \prime }-2u^{\prime }+u)e^{-t}$ at $t=0$, then give $c_0$, $c_1$ and $c_2$ one after another.

    WATCH OUT FOR

    Missing the printed minus sign in front of $c_2{t}^2$. With it, $u^{\prime \prime }(0)=-2c_2$ and $c_2=25$. If you drop the sign you get $c_2=-25$ and a sum of $-17$. Another error is to try to find $c_3$ from the initial conditions: they only fix $c_0$, $c_1$ and $c_2$, while $c_3$ comes from the forcing term.

    Full solution →
55. Q.65NAT · 2 marksOfficial answer: 0.130 to 0.140

    Engineering Mathematics › Probability & statistics

    For a continuous random variable, $P(X>a)={\int }_a^{\infty }{f}_X(x)dx$. This pdf is the exponential distribution with $\lambda =1$, whose survival function is $P(X>a)=e^{-\lambda a}$, so the answer is $e^{-2}$ with no further working.

    WATCH OUT FOR

    Three common slips. (1) Integrating over $[0,2]$ and reporting $P(X\le 2)=1-e^{-2}\approx 0.865$ instead of the right tail. (2) Evaluating the density at the point, $f_X(2)=e^{-2}$, and getting the right number for the wrong reason. For a continuous variable a probability is an area, and the two only agree here because $\lambda =1$. (3) Rounding $e^{-2}$ badly, for example writing 0.14 or 0.13 instead of 0.135 to three decimal places.

    Full solution →