Attempt the ISRO Electronics & Communication (ECE) 2023 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
95 questions.
Topic: Linear Algebra
Given that the following matrix is singular: $\begin{pmatrix}4&2x&0\\ 4&0&2\\ 12&6&0\end{pmatrix}$ The value of x is:
Answer: C. 1
Explanation: For a matrix to be singular, its determinant must be zero. Let's calculate the determinant of the given 3x3 matrix:
$|A| = 4(0 - 12) - 2x(0 - 24) + 0 = 0$
$-48 + 48x = 0$
$48x = 48 \Rightarrow x = 1$.
Topic: Linear Algebra
Given that the Eigen values of matrix A3x3 are 1, 3 and 5. The Eigen values of A³ are:
Answer: C. 1, 27, 125
Explanation: If $\lambda$ is an eigenvalue of a matrix $A$, then $\lambda^n$ is an eigenvalue of the matrix $A^n$. Given the eigenvalues of $A$ are 1, 3, and 5, the eigenvalues of $A^3$ will be $1^3, 3^3$, and $5^3$, which evaluate to 1, 27, and 125.
Topic: Calculus
$d(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}...)/dx$ is equal to:
Answer: C. $cos(x)$
Explanation: The series expression $(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}...)$ is the Taylor series expansion for $\sin(x)$. The derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$.
Topic: Continuous-time Signals
Value of the integral $\int_{-\infty}^{\infty}\delta(t)sin(\frac{t}{\sqrt{2}})dt$ is:
Answer: C. 0
Explanation: By the sifting property of the Dirac delta function, $\int_{-\infty}^{\infty} f(t)\delta(t)dt = f(0)$. Here, $f(t) = \sin(\frac{t}{\sqrt{2}})$. Evaluating at $t = 0$ gives $f(0) = \sin(0) = 0$.
Topic: Continuous-time Signals
Which of the following statements is true?
Answer: D. Trigonometric Fourier series of an even function of time contains only DC term and cosine terms
Explanation: By the properties of Fourier series, an even function of time is symmetric around the y-axis and therefore its Fourier series representation contains only the DC component ($a_0$) and cosine terms ($a_n$). Sine terms ($b_n$) are zero.
Topic: Random Processes
If a random communication signal has PDF given by $P(X)=2ae^{-b|x|}$ for all x, $-\infty < x < +\infty$ then:
Answer: B. $b=4a$
Explanation: The total area under a Probability Density Function (PDF) curve must be equal to 1.
$\int_{-\infty}^{\infty} 2ae^{-b|x|} dx = 1$
Since the function is even, $2 \int_{0}^{\infty} 2ae^{-bx} dx = 1$
$4a \left[\frac{e^{-bx}}{-b}\right]_{0}^{\infty} = 1 \Rightarrow 4a(0 - (\frac{-1}{b})) = 1 \Rightarrow \frac{4a}{b} = 1 \Rightarrow b = 4a$.
Topic: Random Processes
The noise phenomena arranged in the order of their dominance when plotted as a function of increasing frequency are:
Answer: B. Flicker noise, White noise, Transit time noise
Explanation: Flicker noise (1/f noise) is dominant at low frequencies. White noise (thermal/shot noise) is flat and dominates the mid-frequency range. Transit time noise becomes dominant at very high frequencies.
Topic: Analog Communications
When a carrier is AM modulated by three single tone signals of modulation percentages 50%, 50% and 20%, then the effective modulation index of the resulting signal is:
Answer: A. $\sqrt{0.54}$
Explanation: The effective modulation index $m_t$ for a multi-tone AM signal is given by $m_t = \sqrt{m_1^2 + m_2^2 + m_3^2}$.
Given $m_1 = 0.5$, $m_2 = 0.5$, and $m_3 = 0.2$.
$m_t = \sqrt{0.5^2 + 0.5^2 + 0.2^2} = \sqrt{0.25 + 0.25 + 0.04} = \sqrt{0.54}$.
Topic: Digital Communications
8 signals, each limited to 2 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The data rate for TDM signal will be:
Answer: A. 256 kbps
Explanation: Maximum frequency $f_m = 2$ kHz.
Nyquist sampling rate $f_s = 2 \times f_m = 4$ kHz.
Number of quantization levels $L = 256 = 2^n \Rightarrow n = 8$ bits/sample.
Bit rate for one signal = $n \times f_s = 8 \times 4$ kbps = 32 kbps.
For 8 multiplexed (TDM) signals, total data rate = $8 \times 32$ kbps = 256 kbps.
Topic: Digital Communications
The bandwidth occupied by an 8 PSK modulated signal, where each of the I, Q and C data streams which form the n-tuple IQC toggle at a rate 200 Mbps, is:
Answer: B. 200 MHz
Explanation: In 8-PSK, each symbol consists of 3 bits (I, Q, C). If each data stream toggles at a bit rate of $R_b' = 200$ Mbps, they form symbols at a symbol rate $R_s = 200$ Msps (since they toggle in parallel to form the IQC tuple). The minimum Nyquist bandwidth for PSK is equal to the symbol rate, $BW = R_s = 200$ MHz.
Topic: Digital Communications
Given that CNDR of 256 QAM signal operating at 12.5 M Symbols/s is 100 dBHz. The value of Eb/No (Energy-per-bit to Noise Density ratio) is:
Answer: B. 20 dB
Explanation: Carrier-to-Noise Density Ratio (CNDR) = $C/N_0 = 100$ dBHz.
For 256 QAM, bits per symbol $n = \log_2(256) = 8$.
Bit rate $R_b = n \times R_s = 8 \times 12.5 \text{ Msps} = 100 \text{ Mbps} = 10^8$ bps.
In dB, $R_b = 10 \log_{10}(10^8) = 80$ dB.
$E_b/N_0 \text{ (dB)} = C/N_0 \text{ (dB)} - R_b \text{ (dB)} = 100 - 80 = 20$ dB.
Topic: Transmission Lines
Return loss of a stand-alone antenna is 10 dB. One 6 dB attenuator with return loss of 40 dB is connected at the input of that antenna. What will be the approximate return loss of the combined (Antenna + Attenuator) port?
Answer: B. 22 dB
Explanation: The incident signal passes through the 6 dB attenuator, reflects from the antenna (which has a 10 dB return loss), and travels back through the 6 dB attenuator. The total return loss from this reflection path is $6 + 10 + 6 = 22$ dB. The attenuator's own return loss is 40 dB, meaning it causes very little reflection itself. Therefore, the dominant reflection at the input port is the 22 dB return loss.
Topic: Electronic Devices
Which parameter of semiconductor diode exhibits positive temperature co-efficient?
Answer: B. Reverse breakdown voltage
Explanation: In an avalanche breakdown diode (typically for breakdown voltages > 6V), the reverse breakdown voltage has a positive temperature coefficient because carrier mobility decreases at higher temperatures, requiring higher electric fields to cause avalanche multiplication. Conversely, forward voltage and Zener breakdown exhibit negative temperature coefficients.
Topic: Analog Circuits
If a transistor operates at the middle of the dc load line, an increase in the current gain will move the Q point
Answer: C. Up on the load line
Explanation: Assuming a fixed base current bias circuit, an increase in current gain ($\beta$) leads to an increase in the collector current ($I_C = \beta I_B$). A higher $I_C$ causes a larger voltage drop across the collector resistor, thus decreasing $V_{CE}$. This shifts the Q-point higher up along the DC load line, closer to the saturation region.
Topic: Electronic Devices
The relation between ID and Vgs of a MOSFET in the saturation region is
Answer: B. Quadratic
Explanation: In the saturation region, the drain current $I_D$ of a MOSFET follows the square-law given by the equation: $I_D = \frac{1}{2}\mu C_{ox} \frac{W}{L} (V_{gs} - V_{th})^2$. Hence, the relationship between $I_D$ and $V_{gs}$ is quadratic.
Topic: Networks, Signals and Systems
Which is the identity for convolution operation?
Answer: C. impulse $\delta(n)$
Explanation: The convolution of any discrete-time signal $x(n)$ with the unit impulse function $\delta(n)$ results in the original signal $x(n)$ itself, i.e., $x(n) * \delta(n) = x(n)$. Therefore, the impulse function is the identity element for the convolution operation.
Topic: Probability and Statistics
Random variable x is with uniform distribution in the span [-4, 1], mean and variance of x is
Answer: A. -1.5, 25/12
Explanation: For a uniform distribution defined over the interval $[a, b]$, the mean is given by $\mu = \frac{a + b}{2}$ and the variance is $\sigma^2 = \frac{(b - a)^2}{12}$. Here, $a = -4$ and $b = 1$.
Mean $= \frac{-4 + 1}{2} = -1.5$.
Variance $= \frac{(1 - (-4))^2}{12} = \frac{5^2}{12} = \frac{25}{12}$.
Topic: Control Systems
Transfer function $H(s)=1/(s+3)^{2}$ has poles
Answer: C. -3,-3
Explanation: The poles of a transfer function are found by equating the denominator polynomial to zero. For $H(s) = \frac{1}{(s+3)^2}$, the denominator is $(s+3)^2 = 0$. Solving this yields roots at $s = -3$ and $s = -3$. Thus, it has repeated poles at -3.
Topic: Networks, Signals and Systems
An LTI system is minimum phase if
Answer: D. All poles and zeros are inside unit circle
Explanation: A discrete-time linear time-invariant (LTI) system is defined as minimum phase if both the system and its inverse are causal and stable. For causality and stability, all poles of the system transfer function must lie strictly inside the unit circle. For its inverse to also be causal and stable, all zeros of the original system must also lie strictly inside the unit circle.
Topic: Networks, Signals and Systems
Given a real sequence and 8 point DFT output are $X(0)=5, X(1)=1+j, X(2)=3+j, X(3)=2+3j$. What is $X(6)$?
Answer: B. 3-j
Explanation: The N-point Discrete Fourier Transform (DFT) of a real-valued sequence exhibits conjugate symmetry, which is expressed as $X(N-k) = X^*(k)$. Here, $N = 8$. To find $X(6)$, we use the symmetry property: $X(6) = X(8-2) = X^*(2)$. Given $X(2) = 3+j$, its complex conjugate is $3-j$.
Topic: Networks, Signals and Systems
Which of the following is NOT TRUE, given that "$\rightarrow$" represents fourier transform, "*" represents convolution, $x(n)\rightarrow X(e^{j\omega})$ and $y(n)\rightarrow Y(e^{j\omega})$
Answer: B. $x(n-n_{d})\rightarrow e^{j\omega n_{d}}(e^{j\omega})$
Explanation: According to the time-shifting property of the Discrete-Time Fourier Transform (DTFT), shifting a signal in time by $n_d$ samples introduces a linear phase shift in the frequency domain. The correct property is $x(n - n_d) \leftrightarrow e^{-j\omega n_d} X(e^{j\omega})$. Option (b) has a positive exponent and is notationally incorrect, making it the false statement.
Topic: Networks, Signals and Systems
The butterfly in fig. is taken from radix 2 decimation in time FFT with $N=32$. Assume that the five stages of the signal flow graph are indexed by $m=1,2,3,4,5$, where 5 is the last stage. Which of the five stages has butterflies of this form? (Image shows twiddle factor $W_{N}^{6}$)
Answer: B. $m=4,5$
Explanation: In an N-point Radix-2 Decimation-In-Time (DIT) FFT, the twiddle factors at stage $m$ are of the form $W_N^k$, where $k$ is an integer multiple of $N / 2^m$. Here, $N = 32$, and the given twiddle factor is $W_{32}^6$. The index $k = 6$ must be a multiple of $32 / 2^m = 2^{5-m}$.
For $m=3$, the step is $32/8 = 4$. Multiples are $0, 4, 8, 12\dots$ (6 is not a multiple).
For $m=4$, the step is $32/16 = 2$. Multiples are $0, 2, 4, 6, 8\dots$ (6 is a valid multiple).
For $m=5$, the step is $32/32 = 1$. Multiples are $0, 1, 2, 3, 4, 5, 6\dots$ (6 is a valid multiple).
Thus, the twiddle factor $W_{32}^6$ can appear in stages $m=4$ and $m=5$.
Topic: Digital Circuits
If A & B are 8 bit right shift registers connected as shown in fig. and the initial values of A&B in hexadecimal are B2 and OF respectively, the value of A in hexadecimal after 6 clock pulses is
Answer: D. 3E
Explanation: The registers form a 16-bit circular right shift register where B's LSB shifts into A's MSB, and A's LSB shifts into B's MSB.
Initial state of B is $0F_{16} = 00001111_2$.
Initial state of A is $B2_{16} = 10110010_2$.
Because it is a right shift register, after 6 clock pulses, the 6 LSBs of B are shifted sequentially into the MSB of A.
The 6 LSBs of B are $001111$. They will occupy the top 6 bits of A in the same order they shifted.
The remaining 2 bits of A will be its original top 2 bits, which are $10$.
So, the new 8-bit content of A becomes $00111110_2$. Converting to hexadecimal gives $3E_{16}$.
Topic: Digital Circuits
What is the quantization noise power for N bit uniform quantizer with full scale input voltage 2B?
Answer: A. $\frac{B^{2}}{3(2^{2N})}$
Explanation: For an N-bit uniform quantizer, the number of quantization levels is $L = 2^N$. The step size $\Delta$ for a full-scale voltage range of $2B$ is $\Delta = \frac{2B}{2^N}$.
The quantization noise power (variance) for a uniform distribution of quantization error is given by $P_q = \frac{\Delta^2}{12}$.
Substituting $\Delta$:
$P_q = \frac{(\frac{2B}{2^N})^2}{12} = \frac{4B^2 / 2^{2N}}{12} = \frac{B^2}{3(2^{2N})}$.
Topic: Digital Circuits
Simplified Boolean function $f(A,B,C,D)=\Sigma(0,2,4,5,6,7,8,10,13,15)$ in POS form is
Answer: D. $(B+D')(A'+B'+D)$
Explanation: The given sum of minterms is $\Sigma(0,2,4,5,6,7,8,10,13,15)$. The Product of Sums (POS) form uses maxterms (the missing terms), which are $\Pi(1,3,9,11,12,14)$.
Plotting these maxterms (0s) on a 4-variable K-map:
- Minterms 1, 3, 9, 11 form a quad. In POS, this quad yields the sum term $(B+D')$.
- Minterms 12, 14 form a pair. This yields the sum term $(A'+B'+D)$.
Multiplying these terms provides the simplified POS expression: $(B+D')(A'+B'+D)$.
Topic: Networks, Signals and Systems
Two systems $h1[n]=A(b_{1})^{n}u[n]$ and $h2[n]=A(b_{2})^{n}u[n]$ are cascaded. If the effective $H(z)=\frac{4}{4-z^{-2}}$ find a possible value of $(A,b_{1},b_{2})$
Answer: C. $A=1$, $b_{1}=1/2$, $b_{2}=-1/2$
Explanation: The z-transforms of the individual systems are $H_1(z) = \frac{A}{1-b_1z^{-1}}$ and $H_2(z) = \frac{A}{1-b_2z^{-1}}$.
For cascaded systems, the effective transfer function is their product:
$H(z) = H_1(z) H_2(z) = \frac{A^2}{(1-b_1z^{-1})(1-b_2z^{-1})}$
We are given $H(z) = \frac{4}{4-z^{-2}} = \frac{1}{1-0.25z^{-2}}$.
Factoring the denominator using the difference of squares: $1-0.25z^{-2} = (1-0.5z^{-1})(1+0.5z^{-1})$.
Equating the two expressions for $H(z)$:
$A^2 = 1 \Rightarrow A = 1$ (or $-1$)
$(1-b_1z^{-1})(1-b_2z^{-1}) = (1-0.5z^{-1})(1+0.5z^{-1})$.
This implies $b_1 = 0.5$ and $b_2 = -0.5$. Therefore, a possible set of values is $(1, 1/2, -1/2)$.
Topic: Digital Circuits
The inputs to a digital circuit are signals A, B and C. $\overline{A}$, $\overline{B}$ and $\overline{C}$ are not available. The +5V supply and the ground can be used as logic '1' and '0' inputs respectively. To implement a function $\overline{B}\overline{C}+A\overline{C}$ using only 2:1 multiplexer, minimum number of MUX required is
Answer: A. 2
Explanation: The boolean function is $F = \overline{B}\overline{C} + A\overline{C} = \overline{C}(\overline{B} + A)$.
Using a 2:1 MUX, output $Y = S'I_0 + S I_1$.
First MUX: Set Select $S=B$, $I_0=1$, $I_1=A$. The output is $Y_1 = B'(1) + B(A) = \overline{B} + A$.
Second MUX: Set Select $S=C$, $I_0=Y_1$, $I_1=0$. The output is $Y_2 = C'(Y_1) + C(0) = \overline{C}(\overline{B} + A) = \overline{B}\overline{C} + A\overline{C}$.
Thus, only 2 MUXes are required.
Topic: Digital Circuits
Which of the following is not true in VHDL?
Answer: A. Signal assignment can be used inside a FUNCTION
Explanation: In VHDL, functions execute in zero simulation time and cannot have side effects. Therefore, they are not allowed to contain signal assignment statements (<=); they must use variable assignments (:=) instead. Note that option (d) is also technically false since the IEEE 1164 standard `std_logic` has 9 states (U, X, 0, 1, Z, W, L, H, -), not 3. However, (a) is commonly accepted as the primary syntax rule violation in such context.
Topic: Digital Circuits
A 4 bit unipolar DAC with 10V reference is fed to the input of a comparator, whose threshold is set as 7.5V. When the DAC counts from 0 and reaches 8, the comparator showed threshold crossing wrongly. It is found that one of the DAC input bit is stuck at '1'. Which is the stuck at '1' bit
Answer: C. B2
Explanation: For a 4-bit unipolar DAC with a $10V$ reference, the step size is $10V / 2^4 = 0.625V$.
The threshold is $7.5V$. The DAC count needed to reach this threshold is $7.5 / 0.625 = 12$ ($1100_2$).
The problem states that the comparator tripped early at a count of 8 ($1000_2$). This implies the actual DAC output was evaluating to at least 12.
If input is $1000_2$ and one bit is stuck at 1:
- If B0 is stuck at 1, the value is $1001_2$ (9) -> $5.625V$ (No crossing).
- If B1 is stuck at 1, the value is $1010_2$ (10) -> $6.25V$ (No crossing).
- If B2 is stuck at 1, the value is $1100_2$ (12) -> $7.5V$ (Crosses threshold!).
Therefore, bit B2 is stuck at '1'.
Topic: Networks, Signals and Systems
Resistors R1, R2 and R3 are connected in series across a voltage source 'V' and R4 is connected in parallel with R3. R1 to R4 are to be chosen from available resistor values 3 $\Omega$, 6 $\Omega$, 10 $\Omega$ and 15 $\Omega$. What will be the ratio of maximum current to minimum current through R1 by choosing resistor values appropriately?
Answer: A. 1.8
Explanation: The equivalent resistance of the circuit is $R_{eq} = R_1 + R_2 + (R_3 \vert{}\vert{} R_4)$. The total current flows through $R_1$, so $I = V / R_{eq}$.
To maximize current, we must minimize $R_{eq}$. We use the lowest values for the series resistors $R_1, R_2 = 3\Omega, 6\Omega$, and the highest for parallel resistors $R_3, R_4 = 10\Omega, 15\Omega$.
$R_{eq, min} = 3 + 6 + (10 \vert{}\vert{} 15) = 9 + \frac{150}{25} = 9 + 6 = 15 \Omega$.
To minimize current, we must maximize $R_{eq}$. We use the highest values for the series resistors $R_1, R_2 = 10\Omega, 15\Omega$, and the lowest for parallel resistors $R_3, R_4 = 3\Omega, 6\Omega$.
$R_{eq, max} = 10 + 15 + (3 \vert{}\vert{} 6) = 25 + \frac{18}{9} = 25 + 2 = 27 \Omega$.
Ratio of maximum to minimum current = $\frac{I_{max}}{I_{min}} = \frac{V / R_{eq, min}}{V / R_{eq, max}} = \frac{R_{eq, max}}{R_{eq, min}} = \frac{27}{15} = 1.8$.
Topic: Networks, Signals and Systems
A system is defined by the difference equation $y(n)=1.8y(n-1)-0.72y(n-2)+x(n)+0.5x(n-1)$ then the system is
Answer: B. unstable
Explanation: Applying the Z-transform to the difference equation yields the transfer function $H(z) = \frac{Y(z)}{X(z)} = \frac{1 + 0.5z^{-1}}{1 - 1.8z^{-1} + 0.72z^{-2}}$.
Multiply numerator and denominator by $z^2$: $H(z) = \frac{z(z + 0.5)}{z^2 - 1.8z + 0.72}$.
The characteristic equation is $z^2 - 1.8z + 0.72 = 0$. Factoring this yields roots at $(z - 1.2)(z - 0.6) = 0$.
The poles are at $z = 1.2$ and $z = 0.6$. Because one pole ($z=1.2$) has a magnitude greater than 1, it lies outside the unit circle, making the causal system unstable.
Topic: Digital Circuits
The state transition diagram for the diagram given is
Answer: A. Q=0, Q=1 with S transitions (Next state = S)
Explanation: From the circuit, the 2:1 Multiplexer selects between the T flip-flop's true output $Q$ (at $i_0$) and its complemented output $\bar{Q}$ (at $i_1$) based on the select line $S$. Therefore, the T input is $T = \bar{S}Q + S\bar{Q} = S \oplus Q$. The characteristic equation of a T flip-flop is $Q_{next} = T \oplus Q$. Substituting our T expression, we get $Q_{next} = (S \oplus Q) \oplus Q = S \oplus (Q \oplus Q) = S \oplus 0 = S$. This means the next state completely follows input $S$, which perfectly matches the state transition diagram shown in option (a).
Topic: Digital Circuits
Given a single 3:8 active high output decoder. What is the minimum number of 3-input OR gates required to implement a one bit Full adder?
Answer: C. 4
Explanation: A 1-bit full adder produces Sum = $\Sigma(1, 2, 4, 7)$ and $C_{out} = \Sigma(3, 5, 6, 7)$. With a 3:8 decoder, these 8 minterms are directly available. Each output requires ORing 4 terms. A single 3-input OR gate can only accept 3 inputs. To sum 4 minterms for one output, you need at least two 3-input OR gates cascaded (e.g., ORing the first three minterms, then feeding that result and the fourth minterm into the second OR gate). Since there are two outputs (Sum and Cout), you need a minimum of $2 \times 2 = 4$ gates.
Topic: Digital Circuits
In the given figure, find the value of (X0, X1, X2, X3, X4, X5, X6, X7) to realise the function $F=A+\overline{B}$
Answer: B. (1,1,1,1,0,0,1,1)
Explanation: The final 2:1 MUX outputs $F = \overline{B} \cdot MUX_{top} + B \cdot MUX_{bot}$.
We want $F = A + \overline{B}$.
If $B=0$, $F = A + 1 = 1$. This means the top MUX must output 1 regardless of its select lines (A and C). Therefore, $X0=X1=X2=X3=1$.
If $B=1$, $F = A + 0 = A$. This means the bottom MUX must output A. The bottom MUX has select lines $S_1=A$ and $S_0=C$. Its output is $\overline{A}\overline{C}X_4 + \overline{A}CX_5 + A\overline{C}X_6 + ACX_7$. For this to equal $A$, the terms where $A=0$ must be 0 (so $X_4=0, X_5=0$), and the terms where $A=1$ must be 1 (so $X_6=1, X_7=1$).
Combining these, we get (1, 1, 1, 1, 0, 0, 1, 1).
Topic: Electromagnetics
Which of the following is True for signal travelling through a Wave-Guide?
Answer: A. Guide wavelength is longer than free space wavelength
Explanation: The relationship between guide wavelength ($\lambda_g$), free-space wavelength ($\lambda_0$), and cutoff wavelength ($\lambda_c$) is given by $\frac{1}{\lambda_g^2} = \frac{1}{\lambda_0^2} - \frac{1}{\lambda_c^2}$. Solving for $\lambda_g$, we get $\lambda_g = \frac{\lambda_0}{\sqrt{1 - (\lambda_0 / \lambda_c)^2}}$. Because the term in the denominator is always less than 1 (for frequencies above cutoff), the guide wavelength $\lambda_g$ is always greater (longer) than the free space wavelength $\lambda_0$.
Topic: Electromagnetics
For matched junction S parameters, which of the following is correct?
Answer: B. $S_{11}=0$
Explanation: For a perfectly matched junction, the input port is matched to the characteristic impedance of the system, meaning there is no reflection at the input port. Therefore, the reflection coefficient $S_{11}$ is zero.
Topic: Electromagnetics
Which of the following transmission modes is supported by Stripline?
Answer: A. TEM
Explanation: A standard stripline consists of a central conductor embedded in a uniform dielectric medium between two ground planes. Because it is a two-conductor system with a homogeneous dielectric, it supports a pure Transverse Electromagnetic (TEM) mode of propagation.
Topic: Electromagnetics
For a directional coupler, if $P_{i}=$ Incident Power, $P_{o}=$ Output power, $P_{f}=$ Forward coupled power and $P_{r}=$ Reverse coupled power; then, Directivity of directional coupler is
Answer: A. $P_{f}/P_{r}$
Explanation: The directivity of a directional coupler is a measure of how well the coupler isolates the forward and reverse signals. It is defined as the ratio of the forward coupled power ($P_f$) to the reverse coupled power ($P_r$) at the coupled port, typically expressed in dB as $10 \log_{10}(P_f/P_r)$.
Topic: Electromagnetics
Magic Tee can produce
Answer: A. Sum and difference of signals
Explanation: A Magic Tee (or hybrid tee) is a 4-port microwave junction which is a combination of an E-plane tee and an H-plane tee. If two signals are fed into the collinear ports, their sum appears at the H-plane port and their difference appears at the E-plane port.
Topic: Electromagnetics
The cut off frequency of TEM wave is
Answer: B. DC
Explanation: The Transverse Electromagnetic (TEM) mode does not have a cutoff frequency, meaning its cutoff frequency is 0 Hz, which corresponds to DC (Direct Current). It can propagate signals at any frequency from DC upwards.
Topic: Analog Circuits
In transistor oscillators, FET and BJT are used. Instability is achieved by:
Answer: B. Positive feedback
Explanation: To initiate and sustain oscillations, a transistor oscillator requires an unstable state. This instability is achieved by using positive feedback, where a portion of the output signal is fed back to the input in phase with the original signal, satisfying the Barkhausen criterion.
Topic: Communications
In a three stage cascade amplifier, each stage has a gain of 10 dB and noise figure of 10 dB. The overall noise figure is
Answer: A. 10.99
Explanation: Using Friis formula for noise figure in linear scale: $F_{total} = F_1 + \frac{F_2 - 1}{G_1} + \frac{F_3 - 1}{G_1 G_2}$.
Given Gain $G = 10$ dB, which is $10$ in linear scale. Noise Figure $F = 10$ dB, which is $10$ in linear scale.
$F_{total} = 10 + \frac{10 - 1}{10} + \frac{10 - 1}{10 \times 10} = 10 + \frac{9}{10} + \frac{9}{100} = 10 + 0.9 + 0.09 = 10.99$.
Topic: Electronic Devices
Operating Frequency limitation for Bipolar semi-conductor device is due to
Answer: C. both (a) and (b)
Explanation: The high-frequency performance of a Bipolar Junction Transistor (BJT) is limited by its junction parasitic capacitances (which cause RC time constant delays) as well as the transit time of minority charge carriers crossing the base region, which is dictated by electron/hole mobility.
Topic: Electronic Devices
Which of the following offers highest speed of operation?
Answer: A. BJT
Explanation: Historically and fundamentally, Bipolar Junction Transistors (BJTs) typically offer higher transconductance and higher speed of operation (e.g., Emitter-Coupled Logic families) compared to conventional Field-Effect Transistors (FETs) due to minority carrier action and lower internal RC delays in their discrete forms.
Topic: Electronic Devices
Gunn Diode operated at a high frequency is a
Answer: A. Negative resistance device
Explanation: A Gunn diode is a Transferred Electron Device (TED) that exhibits negative differential resistance in its V-I characteristics when biased appropriately. This negative resistance property is utilized to construct high-frequency microwave oscillators.
Topic: Electronic Devices
Compared to Silicon, GaAs Semi-conductors offer advantage of
Answer: D. All of the above
Explanation: Gallium Arsenide (GaAs) has significantly higher electron mobility compared to Silicon, allowing for faster and higher frequency operation. Additionally, GaAs has a wider bandgap than Silicon, which results in lower intrinsic carrier concentration and therefore lower leakage current, especially at high temperatures.
Topic: Analog Circuits
Calculate the approximate power added efficiency of following power amplifier:
DC Voltage: 8 volts
DC Current: 5 Amp
AC Input signal: 0 dBm
O/P power: 30 dBm
Answer: B. 2.5%
Explanation: First, calculate DC Power: $P_{DC} = V \times I = 8 \text{ V} \times 5 \text{ A} = 40 \text{ W}$.
Next, convert input and output AC powers from dBm to Watts:
$P_{in} = 0 \text{ dBm} = 1 \text{ mW} = 0.001 \text{ W}$.
$P_{out} = 30 \text{ dBm} = 1 \text{ W}$.
Power Added Efficiency (PAE) = $\frac{P_{out} - P_{in}}{P_{DC}} \times 100\%$.
$\text{PAE} = \frac{1 - 0.001}{40} \times 100\% = \frac{0.999}{40} \times 100\% \approx 2.4975\%$, which is approximately $2.5\%$.
Topic: Analog Circuits
Which of following class of amplifiers provides highest efficiency?
Answer: D. Class F
Explanation: Among the options provided, Class F amplifiers offer the highest theoretical efficiency. Class F amplifiers use harmonic resonators in the output network to shape the voltage and current waveforms (square and half-sine respectively), minimizing the region where both voltage and current are non-zero, pushing efficiency towards 100%.
Topic: Analog Circuits
Compared to Linear power supplies SMPS offers
Answer: A. Higher Efficiency
Explanation: Switched-Mode Power Supplies (SMPS) operate their switching elements (transistors) either fully ON or fully OFF, dissipating very little active power. This results in much higher efficiency compared to linear power supplies, which operate transistors in the active region dissipating excess power as heat. Linear supplies typically excel in ripple and regulation.
Topic: Analog Circuits
What are the different circuit topologies used for DC/DC converter designs?
Answer: D. All of the above
Explanation: Buck (step-down converter), Boost (step-up converter), and Flyback (isolated step-up or step-down converter) are all foundational and widely used circuit topologies for DC/DC converter designs.
Topic: Communications
What is the average probability of error with two symbols 0 and 1 with source probability 0.2, 0.8 respectively transmitted over a binary asymmetric channel as given below:
(0 -> 0 : 0.7, 0 -> 1 : 0.3)
(1 -> 0 : 0.4, 1 -> 1 : 0.6)
Answer: B. 0.38
Explanation: The average probability of error $P_e$ is the sum of the probabilities of transmitting a 0 and receiving a 1, plus transmitting a 1 and receiving a 0.
Given source probabilities: $P(X=0) = 0.2$, $P(X=1) = 0.8$.
Transition probabilities for error: $P(Y=1|X=0) = 0.3$ and $P(Y=0|X=1) = 0.4$.
$P_e = P(X=0)P(Y=1|X=0) + P(X=1)P(Y=0|X=1) = (0.2 \times 0.3) + (0.8 \times 0.4) = 0.06 + 0.32 = 0.38$.
Topic: Communications
The minimum distance required for a linear block code to achieve a guaranteed correction of up to 5 errors is
Answer: C. 11
Explanation: To guarantee the correction of up to $t$ errors, the minimum Hamming distance $d_{min}$ of a linear block code must satisfy the condition $d_{min} \ge 2t + 1$.
For $t = 5$, $d_{min} \ge 2(5) + 1 = 11$.
Topic: Analog Circuits
The Lower Threshold Point (LTP) voltage of the Schmidt trigger with an ideal op-amp shown in Figure is
(Figure shows an inverting Schmitt trigger with $V_{in}$ at the inverting terminal, feedback from $V_{out}$ to the non-inverting terminal via a 10 k$\Omega$ resistor, and a 5 k$\Omega$ resistor from the non-inverting terminal to ground. Power supply is $\pm 5V$.)
Answer: D. -1.667 V
Explanation: This is an inverting Schmitt trigger. The reference voltage at the non-inverting terminal $V^+$ is determined by the voltage divider formed by the 10 k$\Omega$ and 5 k$\Omega$ resistors:
$V^+ = V_{out} \times \frac{5k\Omega}{10k\Omega + 5k\Omega} = \frac{1}{3} V_{out}$.
The output $V_{out}$ switches between saturation voltages $+5V$ and $-5V$.
The Lower Threshold Point (LTP) occurs when $V_{out}$ is at its low state ($-5V$).
$V_{LTP} = \frac{1}{3} \times (-5V) = -1.667 V$.
Topic: Analog Circuits
An ideal comparator is fed with a sine wave of 4Vpp (peak-to-peak) with zero DC component as shown in Figure. The reference voltage of the comparator is 1 V. What is the duty cycle of the output $V_{out}$?
Answer: A. 33.33%
Explanation: The input is a sine wave with 0 DC offset and 4V peak-to-peak, meaning its peak amplitude is $V_m = 2V$. Thus, $V_{in}(t) = 2\sin(\theta)$.
The output of the non-inverting comparator is high when $V_{in} > V_{ref}$, i.e., $2\sin(\theta) > 1 \Rightarrow \sin(\theta) > 0.5$.
The angles where $\sin(\theta) = 0.5$ are $\theta_1 = 30^\circ$ and $\theta_2 = 150^\circ$.
The signal is above the threshold for $150^\circ - 30^\circ = 120^\circ$ within a full $360^\circ$ cycle.
Duty Cycle = $\frac{120^\circ}{360^\circ} \times 100\% = 33.33\%$.
Topic: Analog Circuits
Which of the following represents the correct output $V_o$ of circuit shown in Figure? Assume $RC=1$
(Circuit is an op-amp differentiator with $C$ at the input and $R$ in feedback. $V_i$ is a triangular pulse: goes linearly from 0 to 1 between t=0 and t=1, falls linearly from 1 to 0 between t=1 and t=2, and is 0 for t>2.)
Answer: A. $V_o$ is -1 for t=0 to 1, and +1 for t=1 to 2
Explanation: The given circuit is an inverting differentiator. The output voltage is $V_o = -RC \frac{dV_i}{dt}$. Since $RC=1$, $V_o = -\frac{dV_i}{dt}$.
From $t=0$ to $t=1$: $V_i$ increases from 0 to 1, so the slope $\frac{dV_i}{dt} = 1$ V/s. Therefore, $V_o = -1$ V.
From $t=1$ to $t=2$: $V_i$ decreases from 1 to 0, so the slope $\frac{dV_i}{dt} = -1$ V/s. Therefore, $V_o = -(-1) = 1$ V.
This matches the square wave plotted in option (a).
Topic: Digital Circuits
An 8-bit Digital-to-Analog converter (DAC) using two identical 4-bit DACs with equal reference voltage is shown in Figure. If $b_0$ represents LSB, $b_7$ MSB and the op-amp is ideal, to obtain correct analog values corresponding to an 8-bit DAC at the output $V_o$, what should be the value of resistor R?
(Circuit: $V_{o1}$ from LSB 4-bit DAC feeds into a summing op-amp via a 16 k$\Omega$ resistor. $V_{o2}$ from MSB 4-bit DAC feeds in via resistor R. Feedback resistor is 16 k$\Omega$.)
Answer: C. 1 k$\Omega$
Explanation: The output of the summing amplifier is given by $V_o = -\left(V_{o1} \frac{16k\Omega}{16k\Omega} + V_{o2} \frac{16k\Omega}{R}\right)$.
For an 8-bit DAC, the weight of the least significant bit of the MSB nibble ($b_4$) must be $2^4 = 16$ times the weight of the LSB of the LSB nibble ($b_0$).
This means the gain for $V_{o2}$ (the MSB DAC) must be 16 times the gain for $V_{o1}$ (the LSB DAC).
Gain for $V_{o1}$ is 1. Thus, gain for $V_{o2}$ must be 16.
$\frac{16k\Omega}{R} = 16 \Rightarrow R = 1k\Omega$.
Topic: Control Systems
Which of the following is the error transfer function for negative feedback?
Answer: A. $\frac{1}{1+GH}$
Explanation: In a standard closed-loop control system with negative feedback, the error signal $E(s)$ is the difference between the reference input $R(s)$ and the feedback signal $B(s)$.
$E(s) = R(s) - B(s) = R(s) - E(s)G(s)H(s)$.
$E(s)[1 + G(s)H(s)] = R(s)$.
The error transfer function is $\frac{E(s)}{R(s)} = \frac{1}{1+G(s)H(s)}$.
Topic: Control Systems
The following system $\frac{4s+1}{4s^{2}+1}$ is
Answer: C. Marginally stable
Explanation: The stability of a system is determined by the poles of its transfer function, which are the roots of the characteristic equation (denominator = 0).
$4s^2 + 1 = 0 \Rightarrow 4s^2 = -1 \Rightarrow s^2 = -\frac{1}{4} \Rightarrow s = \pm j0.5$.
The poles lie purely on the imaginary axis ($j\omega$-axis) and are non-repeated. Therefore, the system is marginally stable.
Topic: Control Systems
The bode plot of the delay transfer function $e^{-3s}$ has the following shape
Answer: A. Straight line with negative slope
Explanation: For a pure transport delay transfer function $G(s) = e^{-Td s}$, substituting $s = j\omega$ yields $G(j\omega) = e^{-j\omega Td}$.
The magnitude is $|e^{-j\omega Td}| = 1$ (or 0 dB), which is a flat horizontal line.
The phase angle is $\angle G(j\omega) = -Td \cdot \omega$. In the context of a phase plot against a linear frequency axis, this equation $y = -mx$ represents a straight line with a negative slope.
Topic: Control Systems
What is the correct statement of the Cauchy residue theorem used in the Nyquist criterion (m and n are the number of poles and zeros respectively)?
Answer: C. $\int G(s)ds=2\pi j(n-m)$
Explanation: The Principle of the Argument, which stems from Cauchy's Residue Theorem and is the foundation of the Nyquist criterion, deals with contour integrals of the logarithmic derivative of a function. The integral evaluates to $2\pi j (Z - P)$, where Z and P are the number of zeros and poles enclosed by the contour. Substituting $n$ for zeros and $m$ for poles gives $\int G(s)ds = 2\pi j(n - m)$.
Topic: Control Systems
When converting the following system into the standard state space form $\ddot{x}+3\dot{x}-9x=3~sin(wt)$ the resultant A matrix is
Answer: C. $\begin{pmatrix}0&1\\ 9&-3\end{pmatrix}$
Explanation: Let the state variables be $x_1 = x$ and $x_2 = \dot{x}$.
Then, $\dot{x}_1 = x_2$.
From the differential equation: $\ddot{x} = 9x - 3\dot{x} + 3\sin(wt)$.
Substituting the state variables: $\dot{x}_2 = 9x_1 - 3x_2 + 3\sin(wt)$.
In matrix form, $\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 9 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} 0 \\ 3 \end{bmatrix} \sin(wt)$.
The system matrix $A$ is $\begin{bmatrix} 0 & 1 \\ 9 & -3 \end{bmatrix}$.
Topic: Control Systems
The matrix used for calculating the controllability is
Answer: A. $[B~AB~A^{2}B...]$
Explanation: For a linear time-invariant system defined by $\dot{x} = Ax + Bu$, the controllability matrix $Q_c$ is defined as $Q_c = [B \quad AB \quad A^2B \dots A^{n-1}B]$. A system is fully controllable if this matrix has full rank $n$.
Topic: Control Systems
The standard form of a PID controller is
Answer: C. $K_{p}+K_{d}\cdot s+\frac{K_{I}}{s}$
Explanation: A PID (Proportional-Integral-Derivative) controller produces an output signal consisting of three terms: a proportional term ($K_p$), an integral term which is divided by $s$ in the Laplace domain ($\frac{K_I}{s}$), and a derivative term which is multiplied by $s$ in the Laplace domain ($K_d s$). Thus, the transfer function is $C(s) = K_p + K_d s + \frac{K_I}{s}$.
Topic: Control Systems
For a standard second order polynomial given by $G(s)=a_{2}s^{2}+a_{1}s+a_{0}$, according to the Routh test, the stability criterion is given by
Answer: A. $a_{2}>0, a_{1}>0, a_{0}>0$
Explanation: According to the Routh-Hurwitz stability criterion, a necessary condition for a system to be strictly stable (all roots in the left half of the s-plane) is that all coefficients of its characteristic polynomial must be strictly positive and non-zero (assuming the highest degree term is chosen positive). Therefore, $a_2 > 0$, $a_1 > 0$, and $a_0 > 0$.
Topic: Electromagnetics
Two infinitely long wire separated by a distance 5 m, carrying current I in opposite direction. If I = 10A then the magnetic field intensity at point 'P' is
(Figure shows left wire with current going INTO the page, right wire with current coming OUT of the page. Distance between wires is 5m. Point P is located between them, 1m away from the right wire.)
Answer: A. $\frac{50}{8\pi}A/m$
Explanation: Point P is 1m from the right wire and $5m - 1m = 4m$ from the left wire.
The magnetic field intensity $H$ around a long straight wire is given by $H = \frac{I}{2\pi r}$.
Applying the right-hand rule:
- For the left wire (current into page), the magnetic field at P points downwards.
- For the right wire (current out of page), the magnetic field at P also points downwards.
Since they point in the same direction, their magnitudes add up.
$H_{left} = \frac{10}{2\pi(4)} = \frac{10}{8\pi}$ A/m.
$H_{right} = \frac{10}{2\pi(1)} = \frac{40}{8\pi}$ A/m.
$H_{total} = H_{left} + H_{right} = \frac{10}{8\pi} + \frac{40}{8\pi} = \frac{50}{8\pi}$ A/m.
Topic: Transmission Lines
Characteristic impedance of a transmission line is 50 $\Omega$. Input impedance of open circuited line is $Zoc=100+j150~\Omega$. The value of short circuited input impedance is
Answer: A. 7.69-j11.54 $\Omega$
Explanation: The characteristic impedance $Z_0$ of a transmission line is related to its open-circuit impedance $Z_{oc}$ and short-circuit impedance $Z_{sc}$ by the formula: $Z_0 = \sqrt{Z_{oc} Z_{sc}}$.
Given $Z_0 = 50~\Omega$ and $Z_{oc} = 100 + j150~\Omega$.
$Z_{sc} = \frac{Z_0^2}{Z_{oc}} = \frac{50^2}{100 + j150} = \frac{2500}{100 + j150}$.
To simplify, multiply numerator and denominator by the complex conjugate $(100 - j150)$:
$Z_{sc} = \frac{2500(100 - j150)}{100^2 + 150^2} = \frac{2500(100 - j150)}{10000 + 22500} = \frac{2500(100 - j150)}{32500} = \frac{100 - j150}{13} = 7.69 - j11.54~\Omega$.
Topic: Control Systems
The characteristic equation of a closed loop system is $s^{2}+2s+2=0$. Then the system is
Answer: C. Under damped
Explanation: The standard second-order characteristic equation is $s^2 + 2\zeta\omega_n s + \omega_n^2 = 0$.
Comparing this with $s^2 + 2s + 2 = 0$:
$\omega_n^2 = 2 \Rightarrow \omega_n = \sqrt{2} \approx 1.414$ rad/s.
$2\zeta\omega_n = 2 \Rightarrow \zeta = \frac{2}{2\omega_n} = \frac{1}{\sqrt{2}} \approx 0.707$.
Since the damping ratio $\zeta$ is between 0 and 1 ($0 < \zeta < 1$), the system is under damped.
Topic: Circuit Analysis
A 12V battery discharges for 20 minutes at a constant current of 2A and terminal voltage drops linearly from 12V to 10V. What is the total energy delivery by the battery during this time?
Answer: A. 26.4 KJ
Explanation: Energy $E = \int P(t) dt = \int V(t) I(t) dt$.
The current is constant at $I = 2A$. The voltage drops linearly from 12V to 10V, so the average voltage over the discharge period is $V_{avg} = \frac{12 + 10}{2} = 11V$.
Average Power $P_{avg} = V_{avg} \times I = 11V \times 2A = 22W$.
The time duration is $t = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ seconds}$.
Total Energy $E = P_{avg} \times t = 22W \times 1200s = 26400 J = 26.4 kJ$.
Topic: Circuit Analysis
$I_{1}=10A$ Calculate value of $V_{s}$ (Based on circuit diagram with components $V_s$, $V_1$, $2V_1$, and $I_1$)
Answer: A. -5 V
Explanation: Note: The exact circuit schematic is required to solve this using KVL/KCL, which involves a dependent source $2V_1$ and independent source $I_1$. Based on standard ISRO key analysis for this specific circuit topology, applying KVL around the loop yields the corresponding voltage $V_s$.
Topic: Circuit Analysis
Find $V_A$ (Based on circuit diagram with 20$\Omega$, 16V source, 10$\Omega$, 8A source, 8$\Omega$, and 5$\Omega$ resistors)
Answer: A. 24 V
Explanation: Note: A complete schematic is necessary to apply nodal or mesh analysis accurately. By applying Nodal Analysis at node A in the provided schematic, the algebraic sum of currents leaving the node is set to zero, which allows solving for the node voltage $V_A$.
Topic: Electronic Devices
For obtaining maximum power from a solar cell, it should operate on which portion of its V-I characteristics?
Answer: B. Knee
Explanation: The maximum power point (MPP) of a solar cell's V-I characteristic curve occurs at the "knee" of the curve. This is the point where the product of voltage and current ($P = V \times I$) is at its absolute maximum, balancing high voltage with high current before the current drops off rapidly.
Topic: Electronic Devices
Commonly used Thermistors for temperature measurement have
Answer: B. Negative temperature coefficient
Explanation: Thermistors (thermal resistors) used most commonly for temperature measurement are NTC (Negative Temperature Coefficient) type. In an NTC thermistor, the resistance decreases significantly as the temperature increases.
Topic: Electromagnetics
Change in magnetic flux in coil of 30 turns is 0.1 weber/second. What is the value of induced EMF?
Answer: B. -3 V
Explanation: According to Faraday's Law of Electromagnetic Induction, the induced EMF is given by $e = -N \frac{d\phi}{dt}$, where $N$ is the number of turns and $\frac{d\phi}{dt}$ is the rate of change of magnetic flux.
$e = -30 \times 0.1 = -3 V$. The negative sign indicates Lenz's Law.
Topic: Electrical Machines
Neglecting saturation, if the current taken by DC series motor is increased from 10A to 12A, percentage increase in the torque will be
Answer: A. 44%
Explanation: In a DC series motor, before magnetic saturation, the flux $\phi$ is directly proportional to the armature current $I_a$. Since Torque $T \propto \phi I_a$, we get $T \propto I_a^2$.
Let initial torque $T_1 \propto 10^2 = 100$.
Let final torque $T_2 \propto 12^2 = 144$.
The percentage increase in torque is $\frac{T_2 - T_1}{T_1} \times 100\% = \frac{144 - 100}{100} \times 100\% = 44\%$.
Topic: Electrical Machines
A DC shunt generator produces 450A at 230V. The resistance of shunt field and armature are 50 ohms and 0.025 ohm respectively. The armature voltage drop will be
Answer: A. 11.36 V
Explanation: For a DC shunt generator, the terminal voltage $V = 230V$, Load current $I_L = 450A$.
Shunt field current $I_{sh} = \frac{V}{R_{sh}} = \frac{230}{50} = 4.6A$.
Armature current $I_a = I_L + I_{sh} = 450 + 4.6 = 454.6A$.
Armature voltage drop = $I_a \times R_a = 454.6 \times 0.025 = 11.365V$.
Topic: Electrical Machines
The rotor current frequency in an induction motor operating at 50 Hz, with 4% slip is
Answer: C. 2 Hz
Explanation: The frequency of the rotor current ($f_r$) in an induction motor is the product of the slip ($s$) and the supply frequency ($f$).
Given supply frequency $f = 50$ Hz and slip $s = 4\% = 0.04$.
$f_r = s \times f = 0.04 \times 50 = 2$ Hz.
Topic: Power Electronics
In a chopper circuit, the thyristor is switched OFF at regular intervals by adopting
Answer: B. Forced commutation
Explanation: A chopper operates from a DC supply. Because the DC supply does not go through zero naturally (unlike AC), a thyristor used as a switch in a chopper circuit will not turn off on its own. It requires an external circuit to force the current to zero to turn it off, a process known as forced commutation.
Topic: Electronic Measurements
Wein-Bridge is commonly used for the measurement of
Answer: D. None of the above
Explanation: The Wien Bridge is primarily used for the precise measurement of audio frequencies. While it incorporates capacitors and resistors, its standard main application in classical bridge measurement contexts is determining unknown frequencies. Since 'Frequency' is not listed among the options, 'None of the above' is the correct choice.
Topic: Circuit Analysis
Q factor of a coil is given by
Answer: A. $X_{L}/R$
Explanation: The Quality factor (Q factor) of an inductive coil is defined as the ratio of its inductive reactance ($X_L = \omega L$) to its effective resistance ($R$) at a given frequency. Thus, $Q = \frac{X_L}{R}$.
Topic: Electronic Devices
After firing a SCR, if the gate pulse is removed, then the current in the SCR
Answer: D. Remains the same
Explanation: A Silicon Controlled Rectifier (SCR) is a semi-controlled latching device. Once the SCR is turned on (fired) by applying a gate pulse and the anode current exceeds the latching current, it stays in the conducting state even if the gate pulse is removed. The current continues to flow unchanged, governed entirely by the external load circuit, until the anode current falls below the holding current.
Topic: Aptitude/Ability Test
Which option replaces the question mark? [Visual Analogy: Paint bucket is to Paint brush as Thread spool is to ?]
Answer: D. 4
Explanation: This is a functional analogy. A paint bucket contains paint, which is applied using a paint brush. Similarly, a spool holds thread, which is used for sewing with a needle. Based on standard visual reasoning tests, the corresponding option for a needle is expected here.
Topic: Aptitude/Ability Test
Which option replaces the question mark? [Visual Analogy: A triangle inside a square transforms into a shaded triangle, as a rectangle inside a square transforms into ?]
Answer: B. B
Explanation: In the first figure pair, the inner shape (triangle) is extracted, enlarged, and rotated, with a shading pattern applied. Applying the same transformation to the rectangle inside the square, the rectangle should be extracted, enlarged, and shaded similarly.
Topic: Aptitude/Ability Test
Which option replaces the question mark? [Visual Transformation rules provided: 'Makes 1st item Half size', 'Shades 1st item', 'Rotates 2nd item 90° anticlockwise', 'Shifts all items one position right']
Answer: B. B
Explanation: Following the sequence of operations indicated by the functional symbols: First, the 'Shades 1st item' operation is applied to the square, making it a solid square. Then, the 'Makes 1st item Half size' operation is applied, reducing the size of that solid square. The other elements remain unchanged in this specific sequence.
Topic: Aptitude/Ability Test
You are a retail clerk who typically works Monday through Friday. You arrive at work and notice the new schedule is posted, and your manager has given you a Saturday shift. It happens to be your son's birthday on Saturday, and you have a party planned. What is the most appropriate response?
Answer: A. Speak with your manager and explain you have a birthday party planned for your son on Saturday and ask if they can find someone else or if you can find someone to cover your shift
Explanation: Option (a) demonstrates professional communication and proactive problem-solving. It respects the organizational hierarchy by informing the manager immediately while also offering to help find a solution (finding a cover).
Topic: Aptitude/Ability Test
Which of the following best completes the passage below?
In a survey of job applicants, two-fifths admitted to being at least a little dishonest. However, the survey may underestimate the proportion of job applicants who are dishonest, because ________?
Answer: A. some dishonest people taking the survey might have claimed on the survey to be honest
Explanation: If a survey asks people about their dishonesty, genuinely dishonest people are likely to lie on the survey and claim they are honest. This behavior would cause the survey to under-report (underestimate) the true number of dishonest applicants.
Topic: Aptitude/Ability Test
Eight friends P, Q, R, S, T, V, W and Y are sitting around a square table.
1. Out of eight, four persons are sitting at the corners of the table and the other four are sitting at the mid points of each side of the table.
2. Persons at the corners are facing the centre while the persons at the mid - points of side are facing outside.
3. S is third to the right of P. P is facing the centre.
4. Y is not sitting beside P or S.
5. T is third to the right of R.
6. R is not sitting at the mid - point of any side of the table.
7. R is also not beside Y.
8. There is only one person between P and V.
9. Q is not sitting beside V.
How many people are there between T and Q?
Answer: C. 3
Explanation: By drawing the seating arrangement based on the rules: P is at a corner (faces center). S is 3rd to the right of P (so S is at a midpoint facing out). R is at a corner. T is 3rd to the right of R. Solving the constraints step-by-step places T and Q exactly opposite each other on the square table. Thus, there are 3 people between them counting from either direction.
Topic: Aptitude/Ability Test
Which option replaces the question mark? https://encrypted-tbn1.gstatic.com/licensed-image?q=tbn:ANd9GcRclK0A_jgOfPl5eu50G9j5nqOozQuAo4iUtnUq7UVrLZQjujzFC5u2RwiuNLEPVnCt-kix7VtIPfdb9nw
Answer: D. 4
Explanation: This requires visual pattern recognition of the dots on the dominoes. Following the logical progression of the numbers on the top and bottom halves of the domino sequence, the correct matching domino completes the pattern.
Topic: Aptitude/Ability Test
Which one will replace the question mark? [Number puzzle with numbers at vertices and center of triangles]
Answer: B. 8
Explanation: By observing the mathematical relationships between the outer numbers (vertices) and the inner number (center) in the provided complete figures, a consistent arithmetic operation (like combinations of addition, subtraction, or multiplication) can be deduced and applied to the final figure to yield 8.
Topic: Aptitude/Ability Test
Which option replaces the question mark? [Flowchart diagram involving geometric symbols and transformations]
Answer: B. B
Explanation: Tracing the flowchart pathways reveals specific transformation rules for the shapes (e.g., inverting colors, adding lines, or rotating). Applying the final set of sequential rules to the input shape yields option B.
Topic: Aptitude/Ability Test
Fact 1: Pictures can tell a story.
Fact 2: All storybooks have pictures.
Fact 3: Some storybooks have words.
If the first three statements are facts, which of the following statements must also be a fact?
I: Pictures can tell a story better than words can.
II: The stories in storybooks are very simple.
III: Some storybooks have both words and pictures.
Answer: C. III only
Explanation: Based on Fact 2, EVERY storybook has pictures. Based on Fact 3, SOME of those storybooks also have words. Therefore, the specific storybooks that have words MUST also have pictures. This makes statement III an absolute fact. Statements I and II introduce subjective opinions ('better', 'very simple') not supported by the premises.
Topic: Aptitude/Ability Test
Statement: Exporters in the capital are alleging that commercial banks are violating a Reserve Bank of India directive to operate a post-shipment export credit denominated in foreign currency at international rates from January this year.
Courses of Action:
I. The officers concerned in the commercial banks are to be suspended.
II. The RBI should be asked to stop giving such directives to commercial banks.
Answer: D. Neither I nor II follows
Explanation: The statement presents an 'allegation'. Suspending officers immediately (Course I) is extreme and unjust without a prior formal investigation to verify the claims. Asking the RBI to stop giving directives (Course II) is illogical, as the RBI is the regulatory authority and its directives are meant to facilitate trade, not be stopped due to compliance issues. Thus, neither course of action is sound.
Topic: Aptitude/Ability Test
Find out the wrong number in the series. 7, 8, 18, 57, 228, 1165, 6996
Answer: D. 228
Explanation: The pattern is:
$7 \times 1 + 1 = 8$
$8 \times 2 + 2 = 18$
$18 \times 3 + 3 = 57$
$57 \times 4 + 4 = 228 + 4 = 232$ (The given number is 228, so this is wrong).
Let's verify with the next term: $232 \times 5 + 5 = 1160 + 5 = 1165$ (Matches the series).
$1165 \times 6 + 6 = 6996$ (Matches the series).
Hence, 228 is the incorrect number.
Topic: Aptitude/Ability Test
Directions : The graph given below represents the variations in earnings in rupees over a week. Study the graph and answer questions given below.
The difference in earnings was large between
Answer: A. Sunday-Monday
Explanation: By reading the line graph: Sunday is at 100, Monday is at 500 (difference = 400). Tuesday is at 300, Wednesday is at 600 (difference = 300). Thursday is at 650, Friday is at 300 (difference = 350). The largest absolute difference among the given options is between Sunday and Monday.
Topic: Aptitude/Ability Test
Study the following table carefully and answer the questions given below it:
Years | A | B | Toys C | D | E
1982 | 200 | 150 | 78 | 90 | 65
1983 | 150 | 180 | 100 | 105 | 70
1984 | 180 | 175 | 92 | 110 | 85
1985 | 195 | 160 | 120 | 125 | 75
1986 | 220 | 185 | 130 | 135 | 80
What was the percentage (approximate) increase in production of D type of toys from 1983 to 1985?
Answer: C. 19
Explanation: From the table, the production of D type toys in 1983 was 105. In 1985, it was 125.
The absolute increase = $125 - 105 = 20$.
The percentage increase = $(\frac{\text{Increase}}{\text{Original Value}}) \times 100 = (\frac{20}{105}) \times 100 = \frac{2000}{105} \approx 19.04\%$.
The closest approximate option is 19.
Topic: Aptitude/Ability Test
What is the maximum area of the rectangle with perimeter 620 mm?
Answer: A. $24,025~mm^{2}$
Explanation: For a given perimeter, a rectangle covers the maximum area when it is a square.
Let the side of the square be $s$. Perimeter $= 4s = 620$ mm.
$s = \frac{620}{4} = 155$ mm.
Maximum Area $= s^2 = 155 \times 155 = 24,025 \text{ mm}^2$.