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Choose the word or phrase which is nearest in meaning to the key word:
Pilfer
The following question comprises two words that have a certain relationship between them followed by four pairs of words. Select the pair that has same relationship as the original pair of words
Fox : Cunning
Fox is cunning animal similarly
Ants are Industrious.
The second word shows the quality of the first.
Replace the phrase printed in bold to make it grammatically correct ?
A twentyfirst century economy "cannot be held" hostage by power cuts nor travel on nineteenth century roads.
Improve the sentence with suitable options by replacing the underlined word.
He "lay" on the grass enjoying the sunshine.
Fill in the blanks :
At times, we are all ______ to be mistaken.
At times, we are all likely to be mistaken.
There are two solutions of wine and water, the concentration of wine being 0.4 and 0.6 respectively. If five liters of the first solution be mixed with fifteen liters of the second, find the concentration of wine in the resultant solution.
Resultant concentration of wine
Rajiv bought some apples at the rate of 25 for Rs. 400 and an equal number at the rate of 30 for Rs. 270. He then sold all at the rate of 10 for Rs. 180. Find his profit/loss percentage
Let Rajiv bought x apples of each kind.
Total cost price of Rajiv =
Total selling price of Rajiv =
SP > CP , Hence he makes profit;
Profit percentage =
Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is alteast one graduate among them ?
P (at least one graduate) = 1 – P (no graduate)
2 men and 3 boys can do a piece of work in 10 days, while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?
Let work done by 1 man in 1 day = x units
Let work done by 1 boy in 1 day = y units
2 men and 1 boy can finish work
No. of days taken by 2 men and 1 boy = = 12.5 days
In an exam, 60% of the candidates passed in Maths and 70% candidates passed in English and 10% of the candidates failed in both the subjects, 300 candidates passed in both the subjects. Find the total number of candidates appeared in the exam, if they took test in only two subjects that is Math and English.
Let total number of students appear be X.
X = 750
In the figure, the value of resistor R is (I 5) ohms, where I is the current in amperes. The value of current I is
150 – IR = 0
150 – I (I +5) = 0
150 – I^{2}  5I = 0
I^{2} +5I – 150 = 0
I^{2} +15I – 10I  150 = 0
I (I +15) – 10 (I +15) = 0
(I +15) (I10) = 0
I = 10A & I = 15A
The velocity of propagation of electromagnetic wave in an underground cable with relative permittivity of 9 is
The function shown in the figure can be represented as
Consider an LTI system with impulse response h(t) = e^{5t }u(t). Determine the output of the system for the input x(t) = 5 u(t)
y(t) = h(t) * x(t)
= e^{5t} u(t) * 5u(t)
y(t) = (1  e^{5t}) u(t)
= e^{0} × 0
= 0
A rectangular pulse x(t) is shown in the figure. Determine the value of
According to Parswell’s theorem
If and f(t) is as t → ∞ . Determine the value of K
⇒K = 2
In the signal flow graph shown in the given figure, the value of C/R ratio is
There are three path
What is the openloop transfer function for an unity feedback system having root locus shown in the following figure ?
The digital circuit shown in the figure works as a
For D flipflop
Q_{n 1} = D
Here, D = X Q_{n}
So Q_{n 1} = X Q_{n}
So it becomes T flipflop
The simplified form of Boolean function F(A,B,C) implemented by the given 3 × 8 decoder is
The contents of the accumulator in an 8085 microprocessor is altered after the execution of the instruction
Compare operation does not affect the accumulator and ORAA does not change the content of accumulator.
Assuming that all the diodes are ideal in figure, the current in the diode D_{2} is
Assume both the diodes are ON
So, I_{1} = 1A
But current can not flow from n to p side in a diode so D_{1} will be off and equivalent circuit becomes
Therefore,
Determine the upper and lower threshold voltage of the schmitt trigger shown below :
The transistor in the given circuit, should always be in saturation region. Take V_{CE} (sat) = 0V and V_{BE} = 0.7V. the minimum value of R_{C} is
A differential amplifier has input at non inverting terminal is 1050 μV and at inverting terminal 950 μV. What is the error in the differential output if CMRR is 1000.
V_{0} = A_{d}V_{d} A_{cm}V_{c}
Which one of the following type of negative feedback increases the input resistance and decreases the output resistance of an amplifier ?
The spectral density and auto correlation function of white noise are respectively.
Auto corelation function and PSD are fourier transform pair
Here, PSD = N_{0}
For a random variable x having the probability density function is shown in the figure, what are the variance of random variable.
= 4/3
An amplitude modulated signal is given as, S(t) = 100 Cos 2pf_{c}t 50 Cos 2pf_{m}t.Cos 2pf_{c}t, where f_{c } f_{m}. and f_{c} is the carrier frequency and f_{m} is the modulating signal frequency. The power efficiency is.
S(t) = 100 Cos 2pf_{c}t 50 Cos 2pf_{m}t.Cos 2pf_{c}t
= 100 [1 0.5 Cos 2pf_{m}t] Cos 2pf_{c}t
Here modulation index, μ = 0.5
= 11.11%
Signal x(t) = 4Sinc^{2} 100t. Sin 500pt .At what sampling frequency should this signal be sampled to avoid aliasing ?
So, f_{m} = f_{1} f_{2}
= 350 Hz
So, f_{s} = 2f_{m}
= 2 × 350
= 700 Hz
The solution of differential equation dy = (1y) dx is
dy = (1 – y) dx
y = 1 ce^{x}
The rank of the matrix A is 2, when K is
= 2 {3k}  3{2k} 4{2´15  3´16}
= 0 {for any value of k}
Given If C is a counterclockwise path in the zplane such that z  i = 1, the value of is
Poles are at z = 1, 2
There is no any pole inside the circle, so integration will be zero
In the circuit shown in figure, determine the maximum power absorbed by the load resistance R_{L}
R_{th} = 4 ohms
For V_{th}
V_{2} – V_{3} V_{2} – V_{1} 12 = 0
But V_{1} = 2V
So, 2 V_{2} – V_{3 }– V_{1} 12 = 0
2 V_{2} – V_{3 }– 2 12 = 0
2 V_{2} – V_{3 }= 10…………....2
From equation 1 and 2
V_{3} – V_{2 }= 12………………..1
V_{3} 2V_{2 }= 10…………….2
V_{2} = 2 Volt
So V_{3} = 14 Volt
So V_{th} = 14V
In the circuit shown below, the switch is closed at t = 0. What is the initial value of the current through the capacitor C.
At t = 0^{}, the equivalent circuit is shown below :
At t = 0^{ }, the equivalent circuit
So, i_{c} = 4 – 3.2 = 0.8A
Determine the zparameter for the network shown below.
Since it is symmetrical ladder network
At which frequency (in rad/sec) the thevenins equivalent impedance looking at terminals AB becomes purely resistive?
From the circuit it is clear, that Z_{AB} becomes real when circuit is in resonance and resonance frequency of this circuit is given as
= 1000 rad/sec
Consider if y(t) = Ae^{t} for then value of A is
The 4point DFT of sequence x[n] = [1,2,2,1] is
The output y(t) of a continuoustime system for the input x(t) is given by . Which one of the following is correct ?
It is an integrator that generally an unstable system because integrator produces an unbounded output and for causal
Let at t = 2
It means y(2) is depends on x(4) that means present output is depends on future input, so it is noncausal
For the NMOS and PMOS shown in the figure are matched and = 1mA/V^{2} and V_{ n} = V_{ p} = 1 V. Determine V_{0}
For the circuit, it is clear that since Q_{N} and Q_{P} are perfectly matched and are operating at V_{GS} equal values of i.e. are 2.5V, the symmetrical which dictates that V_{0} = 0V
Determine the region of operation of BJT in the given circuit.
Since, V_{EB} = 0.7
Here V_{B} = 0, So, V_{E} = 0.7 V
So, V_{C} = 10 4.6 × 1
=  5.4 V
Since, V_{EC} = 0.7 5.4 = 6.1V
That is greater than 0.2V So BJT is in active region.
In the silicon BJT circuit shown below. Assume the emitter area of transistor Q_{2} is two time of area of Q_{1}. The value of I_{o} { Let b of the transistors are very large}
= 2mA
Since β is very large,
So, I_{ref} = I_{c1}
i.e. I_{c1 }= 2mA
Also the emitter current area of transistor Q_{2} is times of Q_{1}
So, I_{c2} = 2 I_{c1}
= 2 × 2 mA
= 4 mA
So, I_{0} = +4 mA
Determine the current i_{o} in the given circuit.
Circuit can be redrawn
Consider a bar of silicon in which a hole concentration profile describe by P(x) = P_{0}.e^{x/Lp}. Find the current at x = 0, given P_{0} = 10^{16}/cm^{3}, L_{p} = 1 mm and crosssectional area of the bar is 100 mm^{2} and D_{p} = 12 cm^{2}/sec.
= 192 A/cm^{2}
So the current I_{p},
I_{p} = J_{p} × A
= 192 × 10^{8}
= 192 mA
A system with the open loop transfer function is connected in a negative feedback gain of unity. What is the value of K for which the system is unstable.
The characteristic equation
S (S^{3} 8S^{2} 24S 32) K = 0
Routh – Hurwitz array
S^{4} 1 24 K
S^{3} 8 32 O
S^{2} 20 K
S' (20×328K) 0
S^{0} K
So for stable system,
K > 0
and 20×32  8K > 0
K < 80
So for unstable system K > 80
i.e. K = 90
In the system shown below x(t) = e^{2t} u(t). Determine the output y(t) in steady state.
Here, x(t) = e^{2t} u(t)
The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for
System is stable in the region 0.1 to 5 on the left side of 10 as number of enrichment there is zero
0.1K < 1 ⇒ K < 10
5K > 1 ⇒ K > 0.2
0.2 < K < 10
Also 1 > 10 K
K < 0.1
So for the stable system 0.2 < K < 10 and K < 0.1
The sensitivity of transfer function with respect to the parameter K is 0.5, Determine the value of K
8K^{2} 10K 3 = 4K
8K^{2} 14K 3 = 0
4K (2K 3) 1 (2K 3) = 0
(2K 3) (4K 1) = 0
K = 1.5 and K = 0.25
Given two binary FSK signals i = 1, 2.The minimum frequency separation of these two signals such that they are orthogonal is
For the two signals to be orthogonal
A block code is transmitted in NRZ inventoneone format with voltage levels
6V → ‘1’
6V → ‘0’
Three code words are sent : 10010, 11110, 01001. Over these three transmitted code words the average voltage is
There are total 15 digits so, In which 8 are 1’s and 7 are 0’s
So average value = = 0.4V
A signal is given by,
γ(t) = s(t) n(t)
where s(t) = 20 Cos2π4000t 40 Cos2π2000t and the noise n(t) is white with power spectral density η_{0} = 1 W/Hz. The total received signal is put through a band pass filter with pass band between 100 to 110 Hz. The value of SNR at the output of filter is
Signal power
= 200 800
= 1000
and noise power, P_{n} = n_{0} × (BW)
= 1 × 10
= 10W
So, SNR = 1000/10 = 100
In dB = 10log100 = 20dB
The base band signal m(t) is recovered from the DSBSc signal s(t) = m(t). Cos2πf_{c}t, by multiplying s(t) by the locally generated carrier Cos (2πf_{c}t φ). The product is passed through a LPF which rejects the double frequency signal. If the base band signal is band limited to 4KHz, what is the minimum value of f_{c} for which m(t) can be recovered by filtering ?
In order to recover m(t) from s(t) by filter method it is necessary that the lowest frequency contained in the first term of s(t) must be greater than the highest frequency contained in the second term that is
So minimum value of f_{c} = 4 KHz
A plane wave in free space is incident normally on a large block of material . which occupies z > 0. If the incident electric field is = 10 Cos (ωt z) V/m, the reflected magnetic field is :
A plane wave with is incident normally on a thick plane conductor lying in the XY plane. Its conductivity is 6 × 10^{6} s/m and surface impedance is 5 × 10^{4}
?. The skin depth of conductor is
The volume of the paralleopipe whose sides are given by is
Volume of paralleopipe
2 {1} 2 {1 3} 0 = 2 4 = 2
A number in 4 bit 2 5 complement representation is 1101. This number when stored using 8bits will beA number in 4 bit 2’5 complement representation is 1101. This number when stored using 8bits will be
The number in 4 bit 1101
So original number, 0011
Now the original number in 8bit,
00000011
So, in 2s complement, 11111101
Four variable Kmap is shown in the figure. The essential prime implicants are
An 8085 microprocessor is interfaced to a 8KB RAM as shown below. The address range of the RAM is
Range of Address
The RAM will be active when is 0
when, A_{13} A_{14} A_{15} = 011 as decoder is active low.
So the range of RAM address is from COOOH to DFFFH
If z = x – iy and = p iq, then is
= P iq
Z = P^{3} (iq)^{3} 3P (iq) (P iq)
x – iy = P^{3} – 3Pq^{2} i (3P^{2}q – q^{3})
X = P^{3} – 3Pq^{2}
The Newton – Raphson method is used to find the roots of the equation f(x) = x  cosπx If the initial guess for the root is 0.5, then the value of x after the first iteration is
f(x) = X  Cosπx
f(0.5) = 0.5 – Cos 0.5π
= 0.5
= 0.38
The eign value of the matrix are,
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