Attempt the ISRO Electronics & Communication (ECE) 2006 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions.
Topic: Network Theory
The value of C which gives the critical damping in the given circuit is (R = 1 ohm, L = 2H)
Answer: C. 8F
Explanation: In a series RLC circuit, for critical damping, the damping factor is equal to the resonant frequency. R/(2L) = 1/$\sqrt{LC}$. Solving for C gives C = 4L/(R²). Substituting R = 1 and L = 2 gives C = 8F.
Topic: Network Theory
A series RLC circuit resonates at 3 MHz and has 3-dB bandwidth of 10 kHz. The Q of the circuit at resonance
Answer: C. 300
Explanation: The quality factor (Q) of a resonant circuit is defined as the ratio of resonant frequency to the bandwidth. Q = f_r / B.W. = (3 * 10^6) / (10 * 10^3) = 300.
Topic: Network Theory
The value of resistance R shown in the given figure
Answer: A. 3.5 ohm
Explanation: Applying KCL at the node above the 7-ohm resistor, let the voltage be V. (V - 50)/6 + V/7 + 4 = 0. Solving this gives V = 14 V. Since the current through R is 4 A, R = V / 4 = 14 / 4 = 3.5 ohm.
Topic: Network Theory
At 3-dB frequencies, current in the series RLC circuit is equal to current at resonance multiplied by
Answer: B. 1/$\sqrt{2}$
Explanation: At the 3-dB cutoff frequencies, the power dissipated is half of the maximum power at resonance. Therefore, the current is 1/$\sqrt{2}$ times the current at resonance.
Topic: Network Theory
A series RLC circuit resonates at 1000 kHz. At frequency of 995 kHz, the circuit impedance is
Answer: D. Capacitive
Explanation: In a series RLC circuit, below the resonant frequency (f < f_0), capacitive reactance (1/wC) is greater than inductive reactance (wL). Thus, the overall impedance is capacitive.
Topic: Analog Circuits
If each stage had gain of 10 dB and noise figure of 10 dB, then the overall noise figure of two-stage cascade amplifier will be
Answer: D. 10.9
Explanation: First, convert dB to absolute ratios: Gain G1 = 10, Noise Figure F1 = 10, F2 = 10. Using Friis formula: F_overall = F1 + (F2 - 1) / G1 = 10 + (10 - 1) / 10 = 10 + 0.9 = 10.9.
Topic: Digital Circuits
In sigma delta ADC, high bit accuracy is achieved by
Answer: A. Over sampling and noise shaping
Explanation: Sigma-delta ADCs achieve high resolution through oversampling (sampling much faster than the Nyquist rate) combined with noise shaping, which pushes quantization noise into higher frequencies that are then filtered out.
Topic: Signals and Systems
Let delta(t) denote the delta function. The value of the integral from -a to a of delta(t)*cos(3t/2) dt is
Answer: A. 1
Explanation: Using the sifting property of the Dirac delta function, integral from -a to a of f(t)delta(t) dt = f(0) as long as the integration interval contains t=0. Here, f(t) = cos(3t/2). Evaluating at t=0 gives cos(0) = 1.
Topic: Signals and Systems
Consider the compound system shown in below figure. Its output is equal to the input with a delay of two units. If the transfer function of the first system is given by H1(Z) = (Z - 0.5)/(Z - 0.8), then the transfer function of the second system H2(Z) is
Answer: B. H2(Z) = (Z^-2 - 0.8Z^-3)/(1 - 0.5Z^-1)
Explanation: The overall transfer function is a delay of two units, so H(Z) = Z^-2. Since H(Z) = H1(Z) * H2(Z), H2(Z) = Z^-2 / H1(Z) = Z^-2 * (Z - 0.8)/(Z - 0.5). Multiplying numerator and denominator by Z^-1 yields (Z^-2 - 0.8Z^-3)/(1 - 0.5Z^-1).
Topic: Signals and Systems
The z-transform of the signal x(n) = {1 for n=-1; 2 for n=0; -1 for n=1; 1 for n=2; 0 otherwise}
Answer: A. z + 2 - z^-1 + z^-2
Explanation: By definition, X(z) = sum_{n=-infinity}^{infinity} x(n)z^{-n}. Plugging in the values: X(z) = 1*z^{-(-1)} + 2*z^{0} + (-1)*z^{-1} + 1*z^{-2} = z + 2 - z^-1 + z^-2.
Topic: Network Theory
For the circuit shown in the given figure, the voltage V_AB is
Answer: A. 6 V
Explanation: Applying KCL at node B with node A as reference (0V): (V_B - 10)/5 + V_B/5 + (V_B + 50)/10 = 0. Solving for V_B: 2(V_B - 10) + 2V_B + V_B + 50 = 0 => 5V_B + 30 = 0 => V_B = -6 V. Thus V_BA = -6V, so V_AB = 6V.
Topic: Network Theory
The equivalent capacitance across 'ab' will be (a combination of five 0.1 uF capacitors)
Answer: B. 0.1 uF
Explanation: The capacitors are arranged in a bridge structure. Because all capacitors are identical (0.1 uF), the bridge is balanced. The capacitor in the middle branch can be removed, reducing the circuit to two parallel branches each containing two series capacitors, yielding an equivalent capacitance of 0.1 uF.
Topic: Control Systems
The transfer function, T(s) = s / (s + a) is that of a
Answer: C. High-pass filter
Explanation: Substituting s = jw, the magnitude is |T(jw)| = w / $\sqrt{w² + a²}$. At w = 0, gain is 0. At w approaching infinity, gain approaches 1. This characteristic represents a high-pass filter.
Topic: Network Theory
A particular current is made up of two components: a 10 A dc and a sinusoidal current of peak value of 1.414 A. The average value of the resultant current is
Answer: C. 10 A
Explanation: The resultant current is i(t) = 10 + 1.414*sin(wt). The average value of a pure sinusoidal wave over a full cycle is zero. Therefore, the average value of the resultant current is simply the DC component, which is 10 A.
Topic: Communication Systems
By doubling the sampling frequency
Answer: A. Quantisation noise decreases by 3 dB
Explanation: Quantization noise power is inversely proportional to the sampling frequency when appropriately filtered. Doubling the sampling frequency reduces the noise by a factor of 2, which corresponds to a 3 dB decrease: 10*log10(1/2) = -3 dB.
Topic: Analog Circuits
The output voltage (V_0) of the circuit shown in the given figure is (V1=13V, input resistor 1k, two 6.3V zener diodes connected back-to-back)
Answer: C. 6.9 V
Explanation: In a back-to-back zener diode configuration with an input of 13V, one zener diode will be forward-biased (acting as a normal diode with approximately 0.6V drop) and the other will be reverse-biased in its breakdown region (acting as a 6.3V voltage regulator). The total output voltage is the sum of these voltage drops: V_0 = 6.3V + 0.6V = 6.9V.
Topic: Digital Circuits
Assuming that only the X and Y logic inputs are available and their complements X_bar and Y_bar are not available, what is the minimum number of two-input NAND gates required to implement X XOR Y?
Answer: C. 4
Explanation: To implement an XOR gate using only two-input NAND gates without inverted inputs readily available, a minimum of 4 NAND gates are required. The configuration uses one NAND gate for (X NAND Y), and then uses its output as inputs to two subsequent NAND gates along with X and Y respectively, feeding into a final fourth NAND gate.
Topic: Digital Circuits
In the given network of AND and OR gates f can be written as: (cascade of AND and OR gates)
Answer: D. X0X1X3......Xn-1 + X2 + X3 + X5....Xn-1 + ..... + Xn-2 + Xn-1 + Xn
Explanation: By tracing the logic network stage by stage, the Boolean function is evaluated accumulating the products and sums depending on the arrangement of AND and OR gates. Option (d) matches the accumulated output sequence described in the standard logic circuit problem this is referencing.
Topic: Digital Circuits
A pulse train with a frequency of 1 MHz is counted using a modulo 1024 ripple-counter built with J-K flip-flops. For proper operation of the counter the maximum permissible propagation delay per flip-flop stage is
Answer: A. 100 n sec
Explanation: For a ripple counter with N stages, the maximum operating frequency f_max = 1 / (N * t_pd). A modulo-1024 counter requires 10 flip-flops (since 2^10 = 1024), so N = 10. Thus, t_pd = 1 / (N * f_max) = 1 / (10 * 10^6) = 10^-7 seconds = 100 nsec.
Topic: Digital Circuits
The A/D converter used in a digital voltmeter, could be (1) successive approximation type (2) Flash converter type (3) Dual slope converter type. The correct sequence in the increasing order of their conversion times is
Answer: B. 2, 1, 3
Explanation: Flash converters are the fastest ADCs (shortest conversion time). Successive approximation ADCs are slower than flash but faster than dual-slope. Dual-slope integrating ADCs are the slowest but offer high resolution and noise rejection, hence widely used in digital voltmeters. Thus, increasing order of conversion time is Flash (2), SAR (1), Dual-slope (3).
Topic: Digital Circuits
The resolution of D/A converter is approximately 0.4% of its full-scale range. It is
Answer: A. An 8-bit converter
Explanation: The resolution of a D/A converter as a percentage is given by (1 / (2^n - 1)) * 100% or approximately (1 / 2^n) * 100%. For an 8-bit converter, 1 / 2^8 = 1 / 256 = 0.0039, which is approximately 0.39% or 0.4%.
Topic: Microprocessors
In a microprocessor, the resistor [register] which holds the address of the next instruction to be fetched is
Answer: B. Program counter
Explanation: The Program Counter (PC) is a dedicated register in a microprocessor that always contains the memory address of the next instruction to be fetched and executed.
Topic: Microprocessors
In microcomputer, WAIT states are used to
Answer: D. Interface slow peripherals to the processor
Explanation: WAIT states are inserted into the microprocessor's machine cycles to stall the CPU temporarily. This allows slower memory or peripheral I/O devices extra time to place data on the bus or read data from it.
Topic: Digital Circuits
Which of the following statements are correct? \n1. A flip-flop is used to store 1 bit of information. \n2. Race-around condition occurs in a J-K flip-flops when both the inputs are 1. \n3. Master-slave configuration is used in flip-flops to store 2 bits of information. \n4. A transparent latch consists of a D-type flip-flop
Answer: C. 1, 2 and 4
Explanation: Statement 1 is correct (a flip-flop stores 1 bit). Statement 2 is correct (toggling continuously during the high clock level causes race-around). Statement 4 is considered correct in standard multiple-choice contexts (a D-latch is transparent). Statement 3 is incorrect; a master-slave configuration still stores only 1 bit, but prevents the race-around condition.
Topic: Digital Circuits
How many 1's are present in the binary representation of 3 x 512 + 7 x 64 + 5 x 8 + 3?
Answer: B. 9
Explanation: Express the expression in powers of 2: \n3 x 512 = 3 x 2^9 = (2+1)2^9 = 2^10 + 2^9 (2 ones). \n7 x 64 = 7 x 2^6 = (4+2+1)2^6 = 2^8 + 2^7 + 2^6 (3 ones). \n5 x 8 = 5 x 2^3 = (4+1)2^3 = 2^5 + 2^3 (2 ones). \n3 = 2^1 + 2^0 (2 ones). \nTotal ones = 2 + 3 + 2 + 2 = 9.
Topic: Digital Circuits
For emitter-coupled logic, the switching speed is very high because
Answer: B. The transistors are not saturated when conducting
Explanation: Emitter-Coupled Logic (ECL) is the fastest logic family because its transistors are biased to operate in the active region, rather than saturation. This eliminates storage time delays that occur when bringing a transistor out of deep saturation.
Topic: Digital Circuits
The output of the circuit shown below is (1 MHz Oscillator -> Divider /10^6 -> Flip Flop -> Schmitt trigger -> Output)
Answer: B. A pulse train of duration 2 sec
Explanation: The 1 MHz oscillator output is divided by 10^6, resulting in a 1 Hz frequency. The flip-flop acts as a toggle (divide-by-2 circuit), changing state on the clock edges, which yields an output frequency of 0.5 Hz. The time period (duration of one full cycle) is T = 1 / f = 1 / 0.5 = 2 seconds.
Topic: Digital Circuits
Gray code for number 7 is
Answer: D. 0100
Explanation: The binary representation for 7 is 0111. To convert to Gray code: keep the MSB (0). Next bit = 0 XOR 1 = 1. Next bit = 1 XOR 1 = 0. Next bit = 1 XOR 1 = 0. Thus, the Gray code is 0100.
Topic: Digital Circuits
10 bit A/D converters, the quantization error is given by (in percent)
Answer: C. 0.1
Explanation: For an n-bit ADC, the maximum percentage resolution (which dictates quantization step) is roughly (1 / 2^n) * 100%. For n=10, 2^10 = 1024. Therefore, the resolution/error is approximately (1 / 1024) * 100% ≈ 0.1%.
Topic: Digital Circuits
For the switch circuit, taking open as 0 and closed as 1, the expression for the circuit in Y (A in series with parallel combination of (B in series with C) and D)
Answer: C. A(BC + D)
Explanation: Switches in series correspond to an AND operation (multiplication), and switches in parallel correspond to an OR operation (addition). Switches B and C are in series (B * C), and they are in parallel with D (BC + D). This entire block is in series with A, yielding A * (BC + D).
Topic: Digital Circuits
The Boolean expression for the shaded area in the Venn diagram is (Venn diagram with sets X, Y, Z where the intersection of X and Y is shaded, as well as the region of Z excluding X and Y)
Answer: D. X_bar*Y_bar*Z + X*Y
Explanation: Based on the provided Venn diagram, the shaded regions represent the intersection of sets X and Y (which is X*Y) and the region of Z that does not intersect with either X or Y (which is X_bar*Y_bar*Z). The logical OR of these two disjoint regions gives the expression X_bar*Y_bar*Z + X*Y.
Topic: Microprocessors / Computer Organization
If the memory chip size is 256*1 bits, then the number of chips required to make up 1 K bytes of memory is
Answer: A. 32
Explanation: 1 K byte of memory = 1024 bytes = 1024 * 8 bits. The size of one memory chip is 256 bits (256 * 1). The number of chips required is total capacity divided by single chip capacity: (1024 * 8) / 256 = 4 * 8 = 32 chips.
Topic: Digital Circuits
Given the decimal number -19, an eight bit two's complement representation is given by
Answer: B. 11101101
Explanation: The 8-bit binary representation of +19 is 00010011. To find the two's complement for -19, first take the one's complement by inverting all bits (11101100), then add 1 to the LSB: 11101100 + 1 = 11101101.
Topic: Digital Circuits
The function shown in the figure when simplified will yield a result with (A 4-variable K-map showing an irregular checkerboard pattern with 8 ones)
Answer: C. 7 terms
Explanation: Looking at the K-map provided in the paper, the placement of the 1s does not allow for heavy grouping (most terms cannot be combined into large blocks). Based on standard simplification for the given configuration, the optimal sum-of-products yields a 7-term expression.
Topic: Digital Circuits
The logic circuit given below converts a binary code Y1 Y2 Y3 into (Circuit connects Y1 to X1, Y1 XOR Y2 to X2, Y2 XOR Y3 to X3)
Answer: B. Gray code
Explanation: The circuit describes the standard binary-to-Gray code conversion algorithm. The most significant bit is kept the same (X1 = Y1), and each subsequent Gray code bit is generated by XORing adjacent binary bits (X2 = Y1 XOR Y2; X3 = Y2 XOR Y3).
Topic: Digital Circuits
A 4-bit synchronous counter uses flip-flops with propagation delay time of 25 ns each. The maximum possible time required for change of state will be
Answer: A. 25 ns
Explanation: In a synchronous counter, the clock pulse is applied simultaneously to all flip-flops. Therefore, the state of all flip-flops changes concurrently, and the maximum time required for the counter state to change is simply the propagation delay of a single flip-flop, which is 25 ns.
Topic: Electromagnetics
An electromagnetic wave incident on a perfect conductor is
Answer: A. Entirely reflected
Explanation: A perfect conductor has infinite conductivity. The skin depth is zero, meaning electromagnetic waves cannot penetrate into the conductor. Thus, the incident wave is entirely reflected.
Topic: Electromagnetics
The characteristic impedance of a lossless transmission line is given by
Answer: D. Z = $\sqrt{L/C}$
Explanation: For a transmission line, the characteristic impedance Z0 = $\sqrt{(R + jwL}$ / (G + jwC)). For a lossless line, the resistance (R) and conductance (G) are zero. Thus, Z0 simplifies to $\sqrt{jwL / jwC}$ = $\sqrt{L / C}$.
Topic: Electromagnetics
A lossless line of 50 ohms is terminated in a load of 100 ohms resistive. The VSWR is
Answer: B. 2:1
Explanation: The reflection coefficient (Gamma) is (ZL - Z0) / (ZL + Z0) = (100 - 50) / (100 + 50) = 50 / 150 = 1/3. The Voltage Standing Wave Ratio (VSWR) is (1 + |Gamma|) / (1 - |Gamma|) = (1 + 1/3) / (1 - 1/3) = (4/3) / (2/3) = 2, which is conventionally written as a 2:1 ratio.
Topic: Electromagnetics
Which of the following does not exist in waveguides
Answer: D. TEM waves
Explanation: Transverse Electromagnetic (TEM) waves require two separate conductors (an inner and an outer conductor, like in a coaxial cable) to propagate, as they need a return path for current at DC frequencies. A standard single-conductor hollow waveguide cannot support TEM waves.
Topic: Communication Systems
Two carriers of 2 GHz and 4 GHz respectively are frequency modulated by a signal of 10 kHz, such that bandwidth of the FM signal in the two cases are same. The peak deviation in the two cases are in the ratio of
Answer: D. 1:1
Explanation: According to Carson's rule, the bandwidth of an FM signal is BW = 2(Delta_f + fm), where Delta_f is the peak frequency deviation and fm is the modulating frequency. Since both the BW and fm (10 kHz) are the same for both signals, their peak deviations (Delta_f) must also be the same. The ratio is 1:1.
Topic: Communication Systems
The bandwidth required for QPSK modulated channel is
Answer: D. Half of the BW of BPSK
Explanation: Quadrature Phase Shift Keying (QPSK) transmits 2 bits per symbol, while Binary Phase Shift Keying (BPSK) transmits 1 bit per symbol. For the same total bit rate, the symbol rate in QPSK is halved, meaning the required transmission bandwidth is half that of BPSK.
Topic: Microwave Engineering
Magic T is
Answer: A. Four port junction
Explanation: A Magic T, also known as a hybrid tee, is a combination of an H-plane tee and an E-plane tee in a waveguide setup. It features four ports: one H-plane port, one E-plane port, and two collinear ports.
Topic: Microwave Engineering
Duplexer is made of
Answer: D. Both receive filter and transmit filter
Explanation: A duplexer is a device that allows a transmitter and a receiver to share the same antenna while operating at different frequencies. It generally consists of both a transmit filter and a receive filter to ensure proper isolation.
Topic: Electromagnetics
The gain G of an antenna of effective area A is given by
Answer: C. G = 4*pi*A/lambda²
Explanation: The relationship between the maximum gain (G) of an antenna and its effective aperture area (A) is given by the formula G = (4 * pi * A) / lambda², where lambda is the operating wavelength.
Topic: Electromagnetics
If the short circuit and open circuit impedance of line are 5 and 20 respectively the characteristic impedance is given by
Answer: B. 10 Ω
Explanation: The characteristic impedance of a transmission line can be found using the geometric mean of its open-circuit and short-circuit impedances: $Z_{0}=\sqrt{Z_{OC}\cdot Z_{SC}}$. Substituting the given values, $Z_{0}=\sqrt{20\times5}=\sqrt{100}=10~\Omega$.
Topic: Electromagnetics
The input impedance of short circuited line of length l where $\frac{\lambda}{4}<l<\frac{\lambda}{2}$ is
Answer: A. Capacitive
Explanation: For a short-circuited lossless transmission line, the input impedance is $Z_{in}=jZ_{0}\tan(\beta l)$. Since $\beta=2\pi/\lambda$, the given length range $\lambda/4<l<\lambda/2$ corresponds to $\pi/2<\beta l<\pi$. In this second quadrant, the tangent function is negative, making the input impedance purely capacitive.
Topic: Communication Systems
Maximum coding gain in
Answer: C. Turbo codes
Explanation: Among the given error-correcting codes, Turbo codes (along with Low-Density Parity-Check codes) provide the maximum coding gain as they can achieve performance very close to the theoretical Shannon limit.
Topic: Analog Circuits
Noise figure of an amplifier depends on
Answer: D. None of the above
Explanation: The Noise Figure (NF) is a measure of degradation of the Signal-to-Noise Ratio (SNR) caused by components in a signal chain. It is an intrinsic property of the amplifier determined by its internal components and operating temperature, not by the input/output signal power or the channel bandwidth.
Topic: Communication Systems
BCH code belongs to
Answer: A. Block codes
Explanation: BCH (Bose-Chaudhuri-Hocquenghem) codes are a prominent and powerful class of cyclic error-correcting block codes.
Topic: Communication Systems
When a carrier is phase modulated, with an integrated modulating signal, the resultant is
Answer: B. Frequency modulated signal
Explanation: Frequency modulation (FM) and phase modulation (PM) are closely related. If the message signal is integrated before being applied to a phase modulator, the instantaneous frequency (which is the derivative of the phase) becomes proportional to the original message signal, resulting in Frequency Modulation.
Topic: Communication Systems
A satellite orbiting in 600 km orbit transmits 5 GHz frequency. The Doppler shift observed at the ground station, when the satellite is over head of the station is
Answer: A. Zero
Explanation: Doppler shift is proportional to the relative radial velocity between the transmitter and the receiver. When the satellite is exactly overhead, its motion is purely tangential to the ground station, meaning the radial velocity is zero. Consequently, the Doppler shift observed at that instant is zero (Note: The provided source key incorrectly listed 'Maximum', but physics dictates it is zero).
Topic: Communication Systems
A communication channel disturbed by additive white Gaussian noise has a bandwidth of 4 kHz and SNR of 15. The highest transmission rate that such a channel can support (in k-bits/sec) is
Answer: A. 16
Explanation: According to Shannon's channel capacity theorem, $C=B\log_{2}(1+SNR)$. Substituting the values, $C=4\log_{2}(1+15)=4\log_{2}(16)=4\times4=16$ kbps.
Topic: Microwave Engineering
A dual directional coupler is connected in a microwave reflectometer measurement setup. The reading of the power meter in the forward direction is 100 mw and in the reverse direction 4 mw. The VSWR is
Answer: C. 1.5
Explanation: The magnitude of the reflection coefficient is $|\rho|=\sqrt{P_{r}/P_{i}}=\sqrt{4/100}=2/10=0.2$. The Voltage Standing Wave Ratio is given by $VSWR=\frac{1+|\rho|}{1-|\rho|}=\frac{1+0.2}{1-0.2}=\frac{1.2}{0.8}=1.5$.
Topic: Analog Circuits
Linear amplifier with a gain of 30 dB is fed with 1.0 µW power, the output power of the amplifier
Answer: B. 0 dBm
Explanation: First, convert the input power to dBm: $1.0~\mu\text{W}=10^{-6}\text{W}=-30\text{dBm}$. The output power is the input power plus the gain in dB: $P_{out}=-30\text{dBm}+30\text{dB}=0\text{dBm}$.
Topic: Communication Systems
10 Watts RF power is transmitted with a circular polarized antenna having gain of 10 dB. A receiving antenna has vertical polarization. The path loss is 100 dB. The receiving signal is
Answer: A. - 83 dBW
Explanation: Transmit power $P_{t}=10\text{W}=10\text{dBW}$. Transmit gain $G_{t}=10\text{dB}$. The path loss is $100\text{dB}$. Because there is a mismatch between a circularly polarized transmitting antenna and a linearly (vertically) polarized receiving antenna, there is a polarization loss of 3 dB. Received power $P_{r}=P_{t}+G_{t}-L_{path}-L_{pol}=10+10-100-3=-83\text{dBW}$.
Topic: Engineering Mathematics
$\vec{p}=2i-3j$, $\vec{Q}=-3i+4j-2k$ and $\vec{R}$ are in equilibrium, if $\vec{R}$ is
Answer: B. $i-j+2k$
Explanation: For the vectors to be in equilibrium, their sum must be zero: $\vec{p}+\vec{Q}+\vec{R}=0$. Therefore, $\vec{R}=-(\vec{p}+\vec{Q})=-((2-3)i+(-3+4)j+(0-2)k)=-(-i+j-2k)=i-j+2k$.
Topic: Engineering Mathematics
A rigid body is rotating with constant angular velocity $\omega$ about a fixed axis, if v is the velocity of a point of the body, then curl $v=$
Answer: C. $2\omega$
Explanation: The velocity vector of a point on the rigid body is given by $\vec{v}=\vec{\omega}\times\vec{r}$. Taking the curl of the velocity field yields $\nabla\times\vec{v}=\nabla\times(\vec{\omega}\times\vec{r})$. Since the angular velocity $\vec{\omega}$ is constant, this resolves to $2\vec{\omega}$, making the magnitude $2\omega$.
Topic: Engineering Mathematics
Laplace transform of $\sin^{3} 2t$ is
Answer: C. $\frac{48}{(s^{2}+4)(s^{2}+36)}$
Explanation: Using the trigonometric identity $\sin^{3}\theta=\frac{1}{4}(3\sin\theta-\sin 3\theta)$, we have $\sin^{3}(2t)=\frac{1}{4}(3\sin 2t-\sin 6t)$. Taking the Laplace transform: $L\{\sin^{3}(2t)\}=\frac{1}{4}\left(\frac{3\times2}{s^{2}+4}-\frac{6}{s^{2}+36}\right)$. Simplifying this gives $\frac{6}{4}\left(\frac{s^{2}+36-(s^{2}+4)}{(s^{2}+4)(s^{2}+36)}\right)=\frac{6}{4}\times\frac{32}{(s^{2}+4)(s^{2}+36)}=\frac{48}{(s^{2}+4)(s^{2}+36)}$.
Topic: Engineering Mathematics
The value of the determinant $\begin{vmatrix}\cos~\theta & 0 & \sin~\theta \\ 0 & 1 & 0 \\ -\sin~\theta & 0 & \cos~\theta\end{vmatrix}$ is
Answer: C. 1
Explanation: Expanding the determinant along the second row gives $1\times(\cos\theta\cdot\cos\theta-\sin\theta\cdot(-\sin\theta))$. This simplifies to $\cos^{2}\theta+\sin^{2}\theta$, which is equal to 1 based on standard trigonometric identities.
Topic: Engineering Mathematics
Solution of $(D^{2}+4)y=\sin^{2}x$ is
Answer: D. $y=A~\cos2x+B~\sin2x+\frac{1}{8}-\frac{x}{8}\sin~2x$
Explanation: The particular solution is given by $\frac{1}{D^{2}+4}[\frac{1-\cos 2x}{2}]$. This evaluates to $\frac{1}{2}[\frac{1}{D^{2}+4}\cdot e^{0x}-\frac{1}{D^{2}+4}\cos 2x] = \frac{1}{8}-\frac{x}{4}\frac{\sin 2x}{2} = \frac{1}{8}-\frac{x}{8}\sin 2x$. The general solution is the sum of the complementary function $A \cos 2x + B \sin 2x$ and the particular solution.
Topic: Engineering Mathematics
The value of k for which the lines $2x+y-1=0$, $4x+3y-3=0$ and $3x+ky-2=0$, are concurrent is
Answer: C. 2
Explanation: For lines to be concurrent, the determinant of their coefficients must be zero. Evaluating the determinant of the $3\times3$ matrix of coefficients gives $2(-6+3k)-1(-8+9)-1(4k-9)=0$. This simplifies to $2k-4=0$, which yields $k=2$.
Topic: Engineering Mathematics
A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is
Answer: D. $\frac{2}{9}$
Explanation: The probability for both balls being red is calculated by multiplying the probability of picking a red ball first by the probability of picking a second red ball: $\frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}$.
Topic: Engineering Mathematics
$x^{3}+x~\sin~x$ is
Answer: A. Constant function
Explanation: $f(x)=x^{3}+x~\sin~x$. Evaluating $f(-x)$ shows it is neither an even nor an odd function. Furthermore, a periodic + nonperiodic function is always nonperiodic. The source text concludes it as a 'constant function' based on eliminating the provided alternatives, although mathematically it represents a general non-periodic function.
Topic: Engineering Mathematics
$\int\frac{dx}{x\sqrt{x^{2}-a^{2}}}$ is
Answer: D. $\frac{1}{a}\sec^{-1}(\frac{x}{a})$
Explanation: By standard integration formulas, the integral of $\frac{1}{x\sqrt{x^{2}-a^{2}}}$ with respect to $x$ is equal to $\frac{1}{a}\sec^{-1}(\frac{x}{a})$.
Topic: Engineering Mathematics
Eigen values of $\begin{bmatrix}-5&2\\ 2&-2\end{bmatrix}$ are
Answer: A. -6, -1
Explanation: The characteristic equation is given by $\det(A-\lambda I)=0$. This becomes $(-5-\lambda)(-2-\lambda)-4=0$, which simplifies to $\lambda^{2}+7\lambda+6=0$. Factoring this gives $(\lambda+6)(\lambda+1)=0$, leading to the eigenvalues $\lambda = -1, -6$.
Topic: Network Theory
The approximate equivalent resistance at the points $X_{1}$ and $X_{2}$ in the circuit shown in below (A ladder network with series 10 $\Omega$ resistors and parallel 40 $\Omega$ resistors)
Answer: B. 40 $\Omega$
Explanation: Evaluating the resistive ladder network starting from the far right end, the series and parallel combinations simplify step-by-step. The equivalent resistance across the input terminals $X_{1}$ and $X_{2}$ is 40 $\Omega$.
Topic: Network Theory
An inductor supplied with 50 V ac with a frequency of 10 kHz passes a current of 7.96 mA. The value of inductor is
Answer: C. 100 mH
Explanation: Using the relation $V=IX_{L}$, we get $50 = 7.96\times10^{-3} \times 2\pi \times 10^{4} \times L$. Solving for L gives $L = 99.97\text{ mH}$, which is approximately 100 mH.
Topic: Electromagnetics
In a capacitor, the electric charge is stored in
Answer: B. Metal plates
Explanation: In a capacitor, the electric charge itself accumulates on the conductive metal plates, while the energy is stored within the electric field in the dielectric medium.
Topic: Analog Circuits
Oscillator requires
Answer: C. Positive feedback
Explanation: An oscillator requires positive feedback to satisfy the Barkhausen criterion, ensuring the loop gain is unity and the phase shift is zero (or an integer multiple of $360^\circ$) to sustain continuous oscillations.
Topic: Electrical Machines
Which loss in a transformer varies significantly with load?
Answer: C. Copper loss
Explanation: Copper losses ($I^{2}R$ losses) depend directly on the square of the current flowing through the transformer windings, making them highly dependent on the load.
Topic: Network Theory
The resistance of a parallel circuit consisting of two resistors is 12 $\Omega$. One of the resistance wires breaks and the effective resistance becomes 18 $\Omega$. The resistance of the broken wire is
Answer: C. 36 $\Omega$
Explanation: For two resistors in parallel, $\frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{12}$. If one breaks (say $R_{2}$ is left), the effective resistance is 18 $\Omega$, so $R_{2} = 18 \Omega$. Substituting this back gives $\frac{1}{R_{1}} = \frac{1}{12} - \frac{1}{18} = \frac{1}{36}$. Thus, $R_{1} = 36 \Omega$.
Topic: Network Theory
Time constant of a series R-L circuit equals
Answer: A. L/R second
Explanation: The time constant ($\tau$) for a series R-L circuit is defined as the ratio of inductance to resistance, $\tau = \frac{L}{R}$ seconds.
Topic: Network Theory
A dc voltage V is applied at time $t=0$ to a series RC circuit. The steady state current is
Answer: D. Zero
Explanation: For an RC circuit, the current response to a DC step is $i(t)=\frac{V}{R}e^{-t/RC}$. In the steady state ($t \to \infty$), the capacitor is fully charged and acts as an open circuit, making the current zero.
Topic: Network Theory
The given circuit represents a (T-network with series capacitors 2C and a shunt inductor L)
Answer: B. High pass filter
Explanation: The series capacitors block low-frequency signals and pass high frequencies, while the shunt inductor provides a low-impedance path to ground for low frequencies. This configuration operates as a high pass filter.
Topic: Analog Circuits
The very low frequency gain of the low pass filter shown in the given figure is (Op-amp with input resistor 1 k$\Omega$ and feedback loop of 100 k$\Omega$ parallel with 0.01 $\mu$F capacitor)
Answer: D. 40 dB
Explanation: At very low frequencies, the feedback capacitor acts as an open circuit. The gain magnitude of this inverting amplifier is $A_{f} = |-\frac{R_{f}}{R_{1}}| = |\frac{100\text{k}}{1\text{k}}| = 100$. Converting to decibels: $20 \log_{10}(100) = 40 \text{ dB}$.
Topic: Network Theory
The time-constant of the network shown in the figure is (Two parallel branches, each with a resistor R and capacitor C arranged to form an equivalent circuit)
Answer: A. CR
Explanation: The circuit simplifies to an equivalent resistance $R_{eq}=\frac{R}{2}$ and equivalent capacitance $C_{eq}=2C$. The time constant is the product of these equivalent values: $R_{eq}C_{eq} = \frac{R}{2} \cdot 2C = RC$.
Topic: Network Theory
For the circuit shown below the current I flowing through the circuit will be (A 2V source across three 3 $\Omega$ resistors configured in parallel)
Answer: C. 2 A
Explanation: The bridge-like structural layout of the resistors forms three parallel branches. The equivalent resistance is $R_{eq} = 3 || 3 || 3 = 1 \Omega$. The total current drawn from the 2 V source is $I = \frac{V}{R_{eq}} = \frac{2}{1} = 2 \text{ A}$.
Topic: Network Theory
When L is doubled and C is halved, the resonance frequency of series tuned circuit becomes
Answer: D. Unchanged
Explanation: The resonance frequency of a series tuned circuit is $f_{r}=\frac{1}{2\pi\sqrt{LC}}$. If L becomes $2L$ and C becomes $\frac{C}{2}$, the product $LC$ remains unchanged, thus the resonance frequency stays unchanged.
Topic: Network Theory
In a series resonant circuit, with the increase in L
Answer: D. All of these
Explanation: With an increase in inductance L: (1) Resonant frequency $f_{r} = \frac{1}{2\pi\sqrt{LC}}$ decreases. (2) Bandwidth $\frac{R}{L}$ decreases. (3) Quality factor $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ increases. Therefore, all the given statements are correct.