Free GATE ECE mock test — 65 questions, GATE marking 🇮🇳

8 original practice questions written by Myndaq in the GATE EC style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1General AptitudeOne correct answer

   Pipe P alone can fill a tank in 12 hours and pipe Q alone can fill it in 18 hours. Both pipes are opened together on an empty tank, and after 3 hours pipe P is shut. How many more hours does Q alone need to fill the tank completely?

   1. A. 9
   2. B. 10
   3. C. 10.5
   4. D. 12
   Show answer and explanation

   Answer: C

   In one hour P fills 1/12 and Q fills 1/18 of the tank. In the first 3 hours together they fill 3/12 + 3/18 = 1/4 + 1/6 = 5/12 of the tank, so 7/12 remains.

   Q alone fills 1/18 per hour, so it needs (7/12) ÷ (1/18) = 18 × 7/12 = 10.5 hours. 12 hours would be right only if the first 3 hours had filled one third of the tank (18 × 2/3 = 12), but they fill 5/12. In 10 hours Q fills only 10/18 = 5/9 of the tank, which is less than the 7/12 still needed. 9 hours would fill only half the tank.

2. Question 2Engineering MathematicsType a numeric answer

   The solution of the differential equation y'' + 4y' + 4y = 0 with y(0) = 1 and y'(0) = 0 is evaluated at t = 1. The value of y(1), rounded to two decimal places, is ______.

   Show answer and explanation

   Answer: 0.40 to 0.41

   The characteristic equation r² + 4r + 4 = (r + 2)² = 0 has the repeated root r = -2, so y(t) = (A + Bt)e^(-2t).

   y(0) = A = 1. Differentiating, y'(t) = Be^(-2t) - 2(A + Bt)e^(-2t), so y'(0) = B - 2A = 0 and B = 2. Hence y(t) = (1 + 2t)e^(-2t).

   Check: y' = -4t e^(-2t) and y'' = (-4 + 8t)e^(-2t); then y'' + 4y' + 4y = (-4 + 8t - 16t + 4 + 8t)e^(-2t) = 0, and y'(0) = 0.

   y(1) = 3e^(-2) = 3 × 0.1353 ≈ 0.406.

3. Question 3Networks, Signals and SystemsType a numeric answer

   A DC circuit has a node X and a ground node. Between X and ground there are three branches: (i) an ideal 24 V source in series with a 6 Ω resistor, with the positive terminal of the source towards X; (ii) a 3 Ω resistor; (iii) an ideal 2 A current source that drives current from ground into node X. A variable load resistor RL is then connected between X and ground.

   The maximum power, in watts, that can be delivered to RL is ______.

   Show answer and explanation

   Answer: 18

   Find the Thevenin equivalent seen by RL. With RL removed, KCL at X (currents leaving X) gives (V - 24)/6 + V/3 - 2 = 0. Multiplying by 6: V - 24 + 2V - 12 = 0, so 3V = 36 and Vth = 12 V.

   To find Rth, replace the voltage source by a short and the current source by an open circuit: Rth = 6 || 3 = 18/9 = 2 Ω.

   Maximum power transfer occurs at RL = Rth = 2 Ω, giving Pmax = Vth²/(4Rth) = 144/8 = 18 W. Check: the load current is 12/4 = 3 A and 3² × 2 = 18 W. Leaving out the current source would give Vth = 8 V and only 8 W.

4. Question 4Networks, Signals and SystemsOne or more correct answers

   A continuous-time system has input x(t) and output y(t) = t·x(t). Which of the following properties does the system have?

   1. A. Linear
   2. B. Time-invariant
   3. C. Causal
   4. D. BIBO stable
   5. E. Memoryless
   Show answer and explanation

   Answer: A, C, E

   Linear: an input a·x1(t) + b·x2(t) gives t(a·x1 + b·x2) = a·y1(t) + b·y2(t), so superposition holds.

   Not time-invariant: the input x(t - t0) gives t·x(t - t0), but the shifted output is y(t - t0) = (t - t0)·x(t - t0). These differ, because the gain t depends on time.

   Causal and memoryless: y at time t depends only on x at the same instant t.

   Not BIBO stable: the bounded input x(t) = 1 gives y(t) = t, which grows without bound. So A, C and E are correct.

5. Question 5Analog CircuitsOne correct answer

   An NMOS common-source amplifier has its source grounded, drain resistor RD = 5 kΩ to a 12 V supply and a DC gate-source voltage of 2 V. The transistor has threshold voltage VT = 1 V, obeys ID = (k/2)(VGS - VT)² in saturation with k = 2 mA/V², and channel-length modulation is negligible. The small-signal voltage gain from gate to drain, with no external load, is

   1. A. -5 V/V
   2. B. -10 V/V
   3. C. -20 V/V
   4. D. +10 V/V
   Show answer and explanation

   Answer: B

   Bias: ID = (2/2)(2 - 1)² = 1 mA, so VDS = 12 - 1 × 5 = 7 V. Since 7 V is greater than VGS - VT = 1 V, the transistor is in saturation and the square-law model applies.

   Transconductance: gm = k(VGS - VT) = 2 × 1 = 2 mA/V (equivalently 2ID/(VGS - VT) = 2 mA/V).

   Gain = -gm × RD = -2 mA/V × 5 kΩ = -10 V/V. The common-source stage inverts, so +10 is wrong; -5 comes from wrongly taking gm = ID/(VGS - VT), and -20 from doubling gm.

6. Question 6Digital CircuitsOne correct answer

   A four-variable function is given as F(W, X, Y, Z) = Σm(1, 3, 4, 6, 9, 11, 12, 14), with W as the most significant bit. Its minimal sum-of-products form is

   1. A. X'Z + XZ'
   2. B. X'Z' + XZ
   3. C. W'X'Z + XZ'
   4. D. X'Y'Z + XZ'
   Show answer and explanation

   Answer: A

   Write each minterm in binary as WXYZ. Minterms 1, 3, 9 and 11 are 0001, 0011, 1001 and 1011: in all four X = 0 and Z = 1 while W and Y take every combination, so together they reduce to X'Z.

   Minterms 4, 6, 12 and 14 are 0100, 0110, 1100 and 1110: in all four X = 1 and Z = 0, which reduces to XZ'.

   So F = X'Z + XZ', the XOR of X and Z. Option B is the XNOR, the complement of F. Option C misses minterms 9 and 11 (it forces W = 0), and option D misses minterms 3 and 11 (it forces Y = 0).

7. Question 7Control SystemsOne or more correct answers

   A unity negative-feedback system has open-loop transfer function G(s) = K / [s(s + 2)(s + 5)] with K > 0. Which of the following statements are TRUE?

   1. A. The closed-loop system is stable for K = 50.
   2. B. At K = 70 the closed-loop system has a pair of poles on the imaginary axis at s = ±j√10.
   3. C. The closed-loop system is stable for every K > 0.
   4. D. For K = 80, exactly one closed-loop pole lies in the right half of the s-plane.
   Show answer and explanation

   Answer: A, B

   The characteristic equation is s(s + 2)(s + 5) + K = s³ + 7s² + 10s + K = 0. Routh array: s³ row 1, 10; s² row 7, K; s¹ row (70 - K)/7; s⁰ row K. All first-column entries are positive only when 0 < K < 70, so K = 50 gives a stable system (A true) and stability does not hold for every K (C false).

   At K = 70 the s¹ row vanishes; the auxiliary equation 7s² + 70 = 0 gives s = ±j√10. Indeed s³ + 7s² + 10s + 70 = (s + 7)(s² + 10), so B is true.

   At K = 80 the first column is 1, 7, -10/7, 80. There are two sign changes, so two poles lie in the right half-plane, not one. D is false.

8. Question 8CommunicationsType a numeric answer

   A conventional AM (DSB with full carrier) transmitter is modulated by a single sinusoidal tone. The unmodulated carrier power is 500 W and the total transmitted power is 580 W. The modulation index, rounded to two decimal places, is ______.

   Show answer and explanation

   Answer: 0.56 to 0.57

   For single-tone AM the total power is Pt = Pc(1 + μ²/2). So 580 = 500(1 + μ²/2), which gives μ²/2 = 80/500 = 0.16 and μ² = 0.32.

   μ = √0.32 ≈ 0.566. A common slip is to take μ²/2 = 0.16 as μ = 0.16 or to forget the factor of 2 (which would give μ = 0.4); neither satisfies the power equation.