A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.
According to the question
100R + 120 R = 2200
2200 R = 2200
R = 10%
Hence required rate % = 10%
Amit rows a boat 9 kilometres in 2 hours downstream and returns upstream in 6 hours. The speed of the boat (in kmph) is:
Downstream rate = 9/2 = 4.5 kmph
Upstream rate = 9/6= 1.5 kmph
The speed of the boat = (4.5 – 1.5) kmph = 3 kmph
The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.
They abandoned their comrades ______the wolves.
Hence the correct answer is option D.
In the following question, out of the four alternatives, select the word opposite in meaning to the given word.
Turgid
Bloated = inflated
Humble = plain, simple
Puffy = inflated
Tumescent = swollen
Hence, humble is the correct answer.
Direction: Study the following information carefully and answer the given questions:
The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week :
The pie chart 2 shows the percentage distribution of total number of Apples (Wet) sold in different days of a week.
Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?
Directions: In each of the questions below, some statements are given followed by some conclusions. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You have to decide which of the following conclusions logically follows from the given statements. Give answer.
Statements:
Some jacket is shirt.
Some shirts are trouser.
No shoes are tshirt.
All trousers are shoes.
All pants are tshirt.
Conclusions:
I. All shirts being tshirt is a possibility.
II. Some shoes are trouser.
III. Some jackets are tshirt.
IV. Some shirts being tshirt is a possibility.
The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.
P: July 1969 was to see a transformed Indira Gandhi.
Q: Quite a few people contributed with ideas.
R: She sounded the bugle through her historic ‘Note on Economic Policy and Programme’ that was circulated among delegates at Bangalore on July 9, 1969.
S: But the pivot was P.N.Haksar who gave shape, structure and substance with the help of some of his colleagues in the Prime Minister’s Secretariat.
P is clearly the first sentence as it introduces the main person of the paragraph, i.e. Indira Gandhi. The pronoun ‘she’ in R refers to Indira Gandhi mention in P. Thus, R forms the second statement. Now, in the sentence R, ‘delegates’ is mentioned which says about the people, who are given in the sentence Q. So, sentence Q must follow R.
Thus, the sequence after rearrangement is PRQS and option A is the correct answer.
Direction: Which of the following is the MOST SIMILAR in meaning to the given word?
Remnant
‘Remnant’ means ‘remainder’ which is same as ‘residue’.
Direction: Read the information carefully and give the answer of the following questions:
Total number of children = 2000
If twoninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?
57.4% (If 2/9 th of children play football are female , then male children = 1  2/9 = 7/9th of children play football;
Let's take z% of male football children equal to cricket children , then
⇒ z% of cricket children = 7/9 th of football;
⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);
⇒ z% of 23 = 7/9 th of 17
⇒ z = 7× 17 × 100 = 57.4%
Shortcut :
Required percentage = (7/9)×17×100/ 23 = 57.4%
Direction: Study the following information carefully and answer the given questions.
Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.
If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?
From the given conditions, we can conclude
After arranging in alphabetical order,
Hence, only Q's position will remain unchanged.
The figure shows the circuit of a gate in the resistor transistor logic (RTL) Then the circuit represent which gate?
Hence gate is OR Gate.
For the given p+ n junction diode the minority carrier life time in the n region is Υ_{P} = 1.1 μsec. If the excess minority charges present in the n region at any time t is given as Q_{P}(t) then Q_{P}(t) at t = 1.5 μsec is _____ μC?
I_{0} = 2A
Diode in FB has diffusion capacitance, as current increases the charge inside increase. Hence voltage increase.
Now
Here
Y = life time of holes in Nregion
= 1.1μsec
Multiplying equation (1) by C gives
And t = 1.5 \musec
Hence
For the given network, the power given to the load is ____ kW? [Assume R =1 Ω and ideal diode]
V = 100. sinωt
Using mirror symmetry
The resistive network between AB can be simplified as.
Further simplifying
A current source with an internal resistance of R_{S}, supplied power to load R_{L}. The plot of power delivered as a function of load R_{L} is?
Hence option B is correct.
where (x) denotes greatest integer
For given system
Angle of arrival is?
Poles: s = 1 ± j
ϕ’ = 180  tan11 = 135°
ϕA = 180° + ϕ
Where ϕ = SP  SZ
= 180 + 135  90
= 225°
∴ ϕA = 180° + ϕ = 405°
= 45°
Hence ϕA = ±45°
A system is shown in figure as follows
Where, h_{k} (n) = δ [n – k.(1/2)^{k}]
Then value of overall impulse response of above system h(n) at h = 2 is ______?
Therefore h(n) can be drawn as
∴ at n = 2h(n) = 1
A square wave with frequency = 5 kHz
This signal is passed through an ideal lowpass filter with 0 dB of passband gain and 10 kHz of cutoff frequency. The filtered signal is subsequently buried additively into a zeromean noise processed with onesided power spectral density (PSD) of 20 nW Hz^{–1}. Upto a frequency of 2.5 MHz. The PSD of noise is zero beyond 2.5 MHz. The signal to noise ratio of the output is _______dB.
Now, Noise power = Area
= One sided PSD × Frequency range
The output of filter is a
Hence impulse response of matched filter h(t) = x(T – t) is
∴ output y(t) = x(t) * h(t)
∴ Hence option D.
For a MOD 30 ripple counter, the maximum operating frequency is 8 MHz. The propagation delay of each flip flop must be _________nsec?
∴ Propagation delay pe flip flops
For a successive approximation type of ADC shown.
If V_{a} = 8V and resolution of D/A = 0.25 volt
Then the error voltage between V_{a} and SAR output at 5^{th} clock is_______volt?
V_{a} = 8v
V_{DAC }= Resolution × Decimal equivalent of binary o/p of SAR
∴at 5th clock max o/p = 7.75 volt
∴ error = 8 – 7.75 = 0.25 volt.
For the transfer function G(jω) = 6 + jω. The corresponding Nyquist plot for positive frequency has the form
Only imaginary port increases with ω.
Hence option D is correct.
Consider a random process as z(t) = x(t) + y(t)
Where x(t) and y(t) are two random processes with zero mean and they are individually WSS. The PSD of x(t) is S_{x}(f) and y(t) is S_{y}(f). Then PSD of z(t) is?
R_{z}(T) = E[z(t).z(t − T)]
Here z(t) = x(t) + y(t)
∴ R_{z}(T) = E[(x(t) + y(t))⋅(x(t − T) + y(t − T))]
= E[x(t)⋅x(t − T)] + E[y(t)⋅x(t − T)] + E[x(t)⋅y(t − T)] + E[y(t)⋅y(t − T)]
∴ R_{z}(T) = R_{x}(T) + R_{y}(T) + R_{xy}(T) + R_{yx}(T)
We know
R_{z}(T) ⟶ PSD S_{z(f)}
∴ S_{2}(f) = S_{x}(f) + S_{y}(f) + S_{yα}(f) + S_{xy}(f)
Hence option D is correct.
A digital circuit is designed in a way that it accept 3bit number and generate it’s square. The circuit is to be designed using ROM with minimum hardware. Then the size of ROM is _______?
It can be seen that
D_{1} = 0 and D_{0} = z
Hence ROM size = 23 × 4 = 8 × 4
The design is
Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the timedivision multiplexed signal will be (in kbps)____
256 levels = 8 bits.
So one sample is represented by 8 bit. So total = 8×8×10^{3} bits/sec
For 8 voice signal = 8×8×8×10^{3} bits/sec
= 512 kbps
When a relatively small d.c. voltage is applied across a thin slice of Gallium Arsenide with thickness of order of few tenths of a micrometer and the voltage gradient across the slice is in excess of about 3300 V/cm, then the negative resistance will manifest itself. Which of the following devices operate based on the above phenomenon?
A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N_{0}/2 = 0.5 × 10^{−20}ω/Hz then the signal to noise ratio at the input of the receiver is
In a superheterodyne AM receiver, the image channel selectivity is determined by :
Consider a medium such that μ=μ_{0},ε = 81ε_{0},σ = 20S/m. The frequency at which the conduction current density is 10 times the displacement current density in magnitude is ____(GHz)
on putting all these values.
= 36 GHz
A 78 Ω lossless planar line was designed but did not meet a requirement. The fraction of the widths of strip (in %)should be added or removed to get the characteristic impedance of 75Ω is ___
i.e. the width must be increased by 4%.
A 16K × 8 ROM IC is interfaced to 8085 microprocessor. What is the address space?
16K × 8 ⇒ 2^{4 }× 2^{10 }× 8 = 2^{14 }× 8
i.e. 14 address input lines and 8 data lines
With 14 address means there are 14 different address lines out of 16 address lines
Starting address should be 0000 H
Hence starting address will be
0000 0000 0000 0000 = 0000 H
And end address
0011 1111 1111 1111 = 3FFF H
Hence address space is 0000 H – 3FFF H
A 16 bit computer can address 1M word and has 128K words installed from 64K 1 chips. Calculate the no. of chips required and no. of bits of address bus that allows to select the memory chips
As it is a 64K memory chips hence 16 bits are needed to address them
i.e. 64K = 2^{16}
As we want 128K × 16 from 64K × 1 chips
Now as we have two rows of 16 chips each, we will need 1 bit to difference one row from another. So A_{16} will be used to select the memory chips
Maximum number of Self Dual possible for 4variable function.
A minimized Boolean expression is a combination of
The prime implicant which contains at least one 1 which cannot be covered by any other prime implicant is called an essential prime implicant (EPI).
The prime implicant whose each 1 is covered by at least one EPI is called Redundant prime implicant (RPI).
The prime implicant which is neither EPI nor RPI is called selective prime implicant (SPI).
If then which of the following holds
Partially differentiating with respect to y,
Partially differentiating with respect to x,
Hence,
Calculate the value of
If dt/dx = sin(t + x) + cos(t + x) then find out the value of t at x = π/3 correct upto 2 decimal places. It is given that t = π/2 at x = 0
A poisson variate satisfies P(x = 1) = 0.5 P(x = 2). Then find the value of P(x = 4) is
Find the general solution of the equation
This is Cauchy’s Homogeneous Equation,
Therefore,
A discrete time signal with input as x[n] has impulse response h[n]=kδ[n] where k is constant. The output y[n] of the system is given as:
For a discrete time signal with input as x[n] has impulse response h[n] =kδ[n], the output y[n] of the system is given as
y[n]=x[n]*h[n]
y[n]=x[n]*kδ[n]
y[n]=k(x*δ)[n]
y[n]=kx[n]
Taking Fourier transform on both sides
Y(ω)=X(ω) x H(ω)
Therefore H(ω)= Y(ω)/X(ω)
This is known as transfer function of the system.
h[n] = F^{1}[H(ω)]
Find the complete solution of the state equation given below
Given the initial condition as
Consider the state equation be,
X˙= AX+BU
We know that,
State transition matrix,
For the given system, determine the overall response Y(s) of the system corresponding to both inputs.
When multiple inputs are present in a linear system, output is computed by treating each input independently. The overall response of the system is the obtained by adding the outputs corresponding to each input.
Assuming X_{2}(s) = 0,
The block diagram can be reduced to,
Reducing the inner feedback loop,
The block diagram is reduced to
Thus,
The block diagram is reduced to,
Thus, Y_{X2}(s) = output due to X_{1}(s) acting alone
Hence, the response of the system,
An integral controller is used for speed control with a setpoint of 120 rpm within a range of 100 to 150 rpm. The controller output is 22% initially. The constant KI for controller output is 0.15%/sec for unit percentage error. If the speed goes to 105 rpm, then the controller output after two seconds for a constant error e_{p} is (in %).
Where r = set point value and b = measured value
The controller output for constant error will be
A junction diode is fabricated in which the p and n regions are doped equally with 5 × 10^{16} atoms/cm^{3}. Assume n_{i} = 1.5 × 10^{10}/cm^{3}. If the crosssectional area of the junction is 20 μm^{2}, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 1014 F/cm, V_{0} (barrier potential) = 0.78V)
We have the formula to find out Charge stored on either side on a junction given as follows,
We don’t have the value of W with us, it needs to be calculated. This can be easily calculated using the following formula,
is the applied Voltage and V_{0} is the barrier potential.
here V = 0 as no external bias is applied.
therefore,V_{B} = V_{0} = 0.78V
Then the value of W is 2.01 x 10^{6} cm.
So the value of QJ is found out to be 1.61 x 10^{15} C
A thin Ptype silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to
Recombination rate, R = B(n_{n0 }+ n_{n})(P_{n0} + P_{n})Electron and hole concentrations respectively under thermal equilibrium n_{n0} & P_{n0} Excess elements and hole concentrations respectively. Thus, option B is the correct answer.
Two ideal pn junction have exactly the same electrical and physical parameters except for the band gap of the semiconductor materials. The first has a bandgap energy of 0.525 eV and a forwardbias current of 10 mA with V_{a} = 0.255 V. The second pn junction diode is to be designed such that the diode current I = 10μA at a forwardbias voltage of V_{a} = 0.32 V. The bandgap energy of second diode would be_____ (eV)
Consider the given figure and calculate the feedback factor of the wave generator if R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF. Also calculate the one cycle period?
The given circuit is that of a square wave generator.
Given: R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF.
To calculate feedback factor:
The period T is given by,
T = 0.01 x 0.405
T = 4.05 x 10^{3}s
T = 4.05ms
Calculate the hysteresis width (in Volts) for an inverting Schmitt trigger with feedback fraction 0.5. Assume the supply voltage to be 10V.
For a inverting Schmitt trigger.
Given: β=0.5, Vcc = 10V
Hysteresis Voltage V_{H} = V_{UT}  V_{LT}
V_{UT} = +β x Vcc
V_{LT }= β x Vcc
V_{H} = 2 x β x Vcc
V_{H} = 2 x 0.5 x 10
V_{H} = 10V
The combination of resistors is shown in figure below. If R_{1} is operated at T and R_{2} is operated at temperature (T+t) ,then the equivalent noise temperature if the circuit is
The circuit can be redrawn as
Decide the stability of the system whose characteristic equation is given by s^{5} + 2s^{4} + 5s^{3} + 10s^{2} + 4s + 8 = 0
s^{3} row elements are all zero, thus the even polynomial above is given by:
f(s) = 2s^{4} + 10s^{2} + 8
f′(s) = 8s^{3} + 20s
There is no sign change in the first column, thus no zeros of the polynomial are in
the RHP, with 4 poles on the imaginary axis and one in LHP. Hence the system response is converging oscillations, hence stable.
Consider the electronic PID controller shown in figure below
If R_{1} = R_{2} = R and C_{1} = C_{2} = C_{2} = C, then find K_{p}, K_{d} and K_{i}
Two point charges of 10 µC at point M and 20 µC at point N exerts a force of 4.8 N and 10 N respectively on a point charge of 40 µC at point O. Determine the electric field intensity at point O.
E = F/Q
The electric field intensity due to two point charges Q_{1} at point M and Q_{2} at point N is the sum of the forces on the test charge Qt at point O, caused by the point charges Q_{1} and Q_{2} acting alone. Thus,
An opamp is used in a feedback configuration having open loop forward gain value of 50 as shown. The feedback gain is β in the circuit, the feedback circuit is given below. For which of the combination of R_{1} and R_{2} respectively would the configuration work as an oscillator?
Here, we can see from the figure that the feedback is connected in a negative feedback configuration. Then, overall gain is given by,
Where, A_{f} is the open loop forward gain. β is the gain provided by the feedback component. According to Barkhausen Criteria, in this situation the given arrangement would work as an oscillator only if A_{f}∗β would be ′−1′
Thus, here β should be 0.02. Among the given opamp configurations for feedback element, given that gain for opamp as inverting amplifier İs given as
Only C provides the required gain β.
Find the h parameters of the following circuit
For parallel connection Y parameters gets added Y = Y_{1} + Y_{2}
The expression for unit tangent ∅ = e^{2x} casyz at the or igin at the or igin in the direction of the tangent to curve x = a sin t, y = a cos t,z = 2α sin t, at t = T/4
Root locus of 2s(s+4)+k(s+8) is a circle. The coordinates of the centre of the circle?
Characteristic Equation is
Now for the centre of the root locus we have to find the breakaway point of the plot For breakaway point,
On solving for s, s_{1} = −1.1255 and s_{2} = 12.74
So, the centre will be at and the coordinates are (7, 0)
In the following table, x is a discrete random variable and p(x) is the probability density. The difference between the variance & the standard deviation is
Difference of variance & standard deviation = 3.06  1.749 = 1.311
Find the state transition diagram for the logic circuit shown below
Hence option C is correct
Consider the following circuit consisting of three Dflip flops.
If all the flipflops were reset to 0 at power on, then the total number of different output states represented by the counter ABC is equal to
So it requires 4 states
Consider a silicon PN junction at T = 300K with current density 3.6 10^{11} A/cm^{2}. If the photocurrent density is 10 mA/cm^{2}. Then find the opencircuit voltage (in V) of a solar cell when the solar intensity is increased by a factor of
Assume thermal voltage V_{T} = 26 mV
Current density J_{S} = 3.6 × 10^{−11}A/cm^{2}since the solar intensity is increased by 10 times
J_{L} = 10 × 10mA/cm^{2} = 100mA/cm^{2}The opencircuit voltage of solar cell is
DTFT of x(n) is X(ω). Hence x(n) = (0.5)^{n}u(n). Then the time signal corresponding to X^{2}(ω) is
A varying magnetic flux linking a coil is φ=1/3 λt^{3}. If the value of λ = −1Wb/s^{2}, then at what time the induced EMF is 9V?
Determine length of a cavity resonator(rectangular) if it has dimensions a = 3cm, b = 2cm, and is excited by TE_{101 }mode at 30GHz frequency.
Given: f = 30GHz = 30 × 109Hz
We have TE_{101} mode, which means: m = 1,n = 0 and p = 1 Also, given 1 = 3cm = 0.03m and b = 2cm = 0.02m.
Taking square on both sides, we get:
Now, we have a = 0.03m. So, on putting the values and solving for 'd' we get:
(1/d)2 = 38888.89 ⇒ d = 5.07 x 10^{3} m
If directive gain of a halfwave dipole is 2.64 W/W, and ohmic losses is 8Ω, then determine its power gain(in dB).
R_{rad} = 73Ω
Now, antenna efficiency is given as:
ƞ_{r} = (R_{rad}) / (R_{rad} + R_{loss}) = 73 / (73 + 8) = 0.901.
Also, we know that efficiency of antenna can also be written as:
ƞ_{r} = G_{p}/ G_{d} = Power gain / directive gain
Now, we have, ƞ_{r} = 0.901 and directive gain Gd = 2.64
So, power gain G_{p} can be given as:
0.901 = G_{p}/ (2.64)
So, G_{p} = 2.37
G_{p}(in db) can be give as : G_{p}(dB)= 10log_{10} G_{p} = 10log(2.37)= 3.74 dB.
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