Firstranker's choice
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General Aptitude
- Newspapers are a constant source of delight and recreation for me. The only (what bother's) trouble is
that I read too (a lot/ large) many of them. - 343 = 73
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1331 = 113
4913 = 173
All numbers given are cube of prime numbers so 133 = 2197 satisfy the missing number. - The passengers were angry with the airline staff about the delay.
- Time taken by X to mow the lawn = 2 hrs.
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.: Work done by X in 1hr = 1/2
Similarly,
Work done by Y in 1 hr = ¼--- Content provided by FirstRanker.com ---
Work done by x + Y in 1 hr =1/2+1/4=3/4
Total time taken by X & Y together = 4/3 hours--- Content provided by FirstRanker.com ---
= 4/3 ×60 minutes
= 80 Minutes - I am not sure if the bus that has been booked will be able to accommodate (occupy) all the students.
- Given that X = {1, 2, 3}
Y = {2, 3, 4}--- Content provided by FirstRanker.com ---
Z = {1/2, 1/2, 2/3, 2/3, 3/4, 3/2, 3/4, 4/2, 4/3}
Minimum value in z = 1/4
Maximum value in z = 3/2--- Content provided by FirstRanker.com ---
Product = 3/8 - Let number of boys participated = 4x
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Number of girls participated = 3x
Total number of students participated = 7x
Total passed candidates = 80/100 x7x = 28/5 X--- Content provided by FirstRanker.com ---
Girls candidate who passed = 90/100 x3x = 27/10 X
Boys candidate who passed = Total passed candidate – Girls candidate who passed--- Content provided by FirstRanker.com ---
= 28/5 X- 27/10 X
= 29/10 X
= 29x/10x4x ×100= 72.5% - The correct statement can be concluded from Venn diagram or using the Syllogism.
- For all digits of a number which lie between 100 and 1000 are even,
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Unit and tens digits can be filled from the set {0, 2, 4, 6, 8}
But hundred's digit does not include 0 as it will not remain a number which lie between 100 and 1000
.: Hundreds digit set is {2, 4, 6, 8}--- Content provided by FirstRanker.com ---
Total integer be = 5 X 5 X 4
Total integer = 100 numbers - Given that
Ganga > Rekha, Lakshmi
Lakshmi > Sana--- Content provided by FirstRanker.com ---
Mita > Ganga
.: Mita > Ganga > Rekha, Lakshmi > Sana--- Content provided by FirstRanker.com ---
.. 2 and statement 4 are correct
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Firstranker's choice
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- Given that
Electrical Engineering--- Content provided by FirstRanker.com ---
Mean square of random process = E(x²) =
Mean is given zero ? E (x) = 0--- Content provided by FirstRanker.com ---
We know that E(x²) – [E(x)]² = variance
Variance = KT/C
Standard deviation = v variance = v(KT/C) - Applying R.H criteria for stability
?(S) = S4 + 3S³ + 3S² + S + K = 0--- Content provided by FirstRanker.com ---
S4 1 3 K S3 3 1 0 S2 8/3 K 0 S1 (8/3 -3K)/(8/3) 0 --- Content provided by FirstRanker.com ---
S0 K
For stability, first column should be greater than zero--- Content provided by FirstRanker.com ---
(8/3 -3K)/(8/3) >0 and k>0
.:. 0 < K < 8/9 - L-1[1/((S+1)2 + 1)2] = L-1[1/(S+3)]2 = L-1[1/(S2 + 2S+1)] = L-1[-1/((S+1)2 + 2]
= L-1[1/(S+1)2] + L-1[1/(S+1)]2
H (t) = e-t+2te-t - We know that
Voltage Regulation = (VNL-VFL)/VNL ×100--- Content provided by FirstRanker.com ---
Given that VFL = 95V
VNL = 100 V
% VR = (100-95)/100 ×100=5% - We know that P = VI cos f, as load and voltage are same
I cos f = constant--- Content provided by FirstRanker.com ---
I1 cos f1 = I2 cos f2
I1 = 200A
Cos f1 = 1--- Content provided by FirstRanker.com ---
Cos f2 = 0.5
I2 = I1 cos f1/COS f2 = 400 A - We know that
Er2
R--- Content provided by FirstRanker.com ---
Figure (i)
Er1 = 2
Figure (ii)--- Content provided by FirstRanker.com ---
C1
R
ln(R/r) C1 = 4peo/ln(R/r)
Total portion cover 2p--- Content provided by FirstRanker.com ---
p/2 portion covers = p/2 length for Er1
3p/2 and length for er2
Both are connected in parallel--- Content provided by FirstRanker.com ---
C2 = Cr1 + Cr2
Given C2 = 2C1
[3+e2] = 2(4peo)/ln(R/r)--- Content provided by FirstRanker.com ---
er2 = 10
By reducing the rms value of supply voltage at rated frequency, magnetizing current changes which changes the magnetizing reactance- H(s) = 10/(s(s²+s+100v2))
For finding steady state value, we will apply final value theorem--- Content provided by FirstRanker.com ---
lim y(t) = lim sy(s)
t?8 s-0
y(8) = lim s x 10/(s(s²+s+100v2))--- Content provided by FirstRanker.com ---
s-0
y(8) = 10/100v2 - G(s) = pe-0.25s/s
Nyquist plot cut the negative real Axis at w = phase cross over frequency
f = -90° -0.25?×180/p--- Content provided by FirstRanker.com ---
_G(j?)|?=@pc=-180°
f?=@pc = -90° -90°= - 180°--- Content provided by FirstRanker.com ---
?@pc = 2p
?@pc = 2p
x =|G(jw)|--- Content provided by FirstRanker.com ---
x = p/(2p) = 1/2
Then, the co-ordinates becomes (-0.5, j0). - Given Zin = 10O, Zo/p = 100O
For CCCS
Series connection is output--- Content provided by FirstRanker.com ---
Zo/p = Zo/p (1+?ß) = 100 (1 + 9)
= 100 ?O - We know that,
For 6-pulse converter harmonic present in AC current are 6K ± 1
General expression NK ± 1--- Content provided by FirstRanker.com ---
For 6 pulse n = 6
Lowest order harmonic = 5--- Content provided by FirstRanker.com ---
[k = 0, 1, 2, 3]
Lower harmonic frequency = 5 × 50 = 250 Hz --- Content provided by FirstRanker.com ---
Applying nodal analysis at point 1 whose voltage is assumed as V1.
(V1-20)/2 = -1; V1= 18
(V1-5I)/3= 2; V1 -5I =6
Solving 1 and 2--- Content provided by FirstRanker.com ---
18 -5I =6; I=2.4- Wave equation ?2u/?x2 + ?2u/?y2 = 1/c2 ?2u/?t2
Laplace equation ?2U = ?2u/?x2 + ?2u/?y2 = 0
Poission equation ?2U = f--- Content provided by FirstRanker.com ---
Heat equation ?u/?t -a(?2u/?x2 + ?2u/?y2 + ?2u/?z2) = 0 - ?(z2-1)/(z+2) dz = 0 for|z|=1
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For , the singularity z = -2 lies outside the |Z| < 1
? By Cauchy's integral theorem ?(z2-1)/(z+2) dz = 0 for|z|= 1 - Given that
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y = 2x3 + 3y2 + 4z
[grad f.dr = ?--- Content provided by FirstRanker.com ---
dr = dxi + dyj+dzk
grad f = ?f/?x i + ?f/?y j + ?f/?z k
= 6x2i + 6yj + 4zk--- Content provided by FirstRanker.com ---
? grad f.dr = ?6x2dx + ?6ydy + ?4zdz
Applying the limits--- Content provided by FirstRanker.com ---
? grad f.dr = ?2-3 6x2dx+ ?2-3 6ydy+ ?2-1 4dz
= [2x3]2-3 +[3y2]2-3 +[4z]2-1
=70+81-12 = 139 - Net reactance of generator
X = 0.25/5 = 0.05p.u.--- Content provided by FirstRanker.com ---
Pre - fault voltage
Isc = 1/X = 20
Short Circuit MVA = Isc Base MVA--- Content provided by FirstRanker.com ---
= 20 × 5 = 100 MVA - For NMOS transistor to be in saturation the condition will be
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VGS > Vth
And VDS > VGS - VTh - Isec = 5 x 20 = 100 A
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V = Isec R = 100 × 0.01 = 1V
VA output of CT = VIsec = 100 × 1 = 100 VA - Y12 = - (y12) = - j20
Series admittance of each line = Y12/2 = -j20/2 = -j10
Series reactance of each line = 1/-j10 = j0.1p.u. - M=
101
110--- Content provided by FirstRanker.com ---
Determinant of M = |M|
|M| = 1(0 - 1) - 1(0 - 1) + 1(1 - 0)
|M| = 2--- Content provided by FirstRanker.com ---
|M| ? 0
.. Rank of M = number of columns--- Content provided by FirstRanker.com ---
P (M) = 3 - H (t) = 1 + e-at u (t)
'1' is a constant and two sided so the impulse response cannot be causal as for causal it should satisfy--- Content provided by FirstRanker.com ---
h (t) = 0 t<0
Which it is not satisfying due to presence of constant--- Content provided by FirstRanker.com ---
:: It is not causal - H(s) = (a1s² + b1s + C1)/(a2s² + b2s+ C2)
a1 = b1 = 0--- Content provided by FirstRanker.com ---
H(s) = C1/(a2s² + b2s+C2)
At s = 0--- Content provided by FirstRanker.com ---
H (0) = constant
At s = 8
H(S)? S--- Content provided by FirstRanker.com ---
? It is a low par filter - Waveform for output voltage of single phase full bridge PWM inverter
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vo = S(4Vdc/np) sin(npd/2) sin(np/2)
n=6k±1
Vo1rms = fundamental rms output voltage--- Content provided by FirstRanker.com ---
Vo1 = (2v2/p)Vdc sind
Given, Vo1 = 0.754 Vdc--- Content provided by FirstRanker.com ---
0.75 Vdc = (2v2/p)Vdc sind
d = sin-1(0.75p/(2v2)) = 56.44
Pulse width = 2d = 112.88 - For series R - L circuit, I (t) expression is
i(t) = (Vm/v(R²+XL2)) sin(?t-f) + Ae-t/t--- Content provided by FirstRanker.com ---
f = tan-1(XL/R)
DC offset = A = -VMsin(-f)/Z
For Maximum value of DC offset A--- Content provided by FirstRanker.com ---
? - f = - 90
?-tan-1(377*10*10-3/3.77) = -90
? = -45° - M is a 2 x 2 Matrix with Eigen value 4 and 9 If M has ?1, ?2 _____ Eigen values
?n ? ?1n, ?2n
M² ? 42, 92--- Content provided by FirstRanker.com ---
M² has Eigen values as 16 and 81 - Vs = 400 KV
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l = 300 km
L1 = 1 mH / km / phase
C1 = 0.01 µF / km / phase--- Content provided by FirstRanker.com ---
ß' = 1/v(L1C1)
= 1/v (1×10-3 ×0.01×10-6)--- Content provided by FirstRanker.com ---
= 3.16×105 km/s
ßl = (2pfl)/V
= (2p×50×300)/(3.16×105) = 0.29--- Content provided by FirstRanker.com ---
A=1- (ßl)2/2 =1- (0.29)2/2 = 0.955
VR = Vs/A = 400/0.955 = 418.85KV - According to Millman's Theorem, the equivalent circuit of the given circuit is
Eeq = V = (E1/R1+E2/R2 +E3/R3+ E4/R4) / (1/R1+1/R2 +1/R3+1/R4)
Eeq = (200/50+160/40 +100/25+ 80/20) / (1/50+1/40 +1/25+1/20) = 0V--- Content provided by FirstRanker.com ---
So, the current I flowing is 0 A - For synchronous motor
Eg = V1 - IZ--- Content provided by FirstRanker.com ---
V1 = 220/v3 V (Phase)
Z = (0.25 + j 2.5)O--- Content provided by FirstRanker.com ---
I = 10?-36.86 A
Eg = 220/v3 -(0.25+ j2.5)×10?-36.86
Eg = 141.658 ?-8.7 V (phase)--- Content provided by FirstRanker.com ---
Eg = 245.36 V (line) - lim(z³ + z² +8)/(z+2) = 2pj× lim(z+2)³/(z+2) = 2pj× lim(-8+4+8)=8pj
- V (t) = - 170 sin (377t-p/6)
I (t) = 8 cos (377t+p/3)
V(t) = - 170 sin (377t-p/6) = 170 cos(377t+p/3)--- Content provided by FirstRanker.com ---
P = Vrms Irms cosp
P = (170/v2)*(8/v2)cos 30--- Content provided by FirstRanker.com ---
P = 588.89 watts - Given R1 = 5.39O, R2 = 5.72O, X1 = X2 = 8.22O
for frequency ? 10 Hz--- Content provided by FirstRanker.com ---
X1 = X2 = 8.22x(10/50) = 1.644O
Starting phase current at 10 Hz--- Content provided by FirstRanker.com ---
Ipn = Vpn/v(R1+R2)²+(X1+X2)²
= 100/v(5.39+5.72)² (1.644+1.644)²
Ipn = 8.63A--- Content provided by FirstRanker.com ---
Starting line current = IL = v31ph
IL = v3×8.63 = 14.95A - Given data L = 50mH, C = 0.05 µF
Critical resistance to avoid current shopping will be given as
R=1/2v (L/C)--- Content provided by FirstRanker.com ---
R=1/2v ((50×10-3)/(0.05×10-6))
R = 500O
XL1=0.2 pu
XL2=0.4 pu--- Content provided by FirstRanker.com ---
V=1?0°
Xa=0.25 pu
Xeq = 0.25 +0.2 + (0.4/2) = 0.65 PU--- Content provided by FirstRanker.com ---
P = VPU IPU COSf
0.8 = 1 x IPU × 0.8--- Content provided by FirstRanker.com ---
IPU = 1 PU [as 0.8 pf lagging]
V = Ieq +IL1(jXL1)
1?0° = Ieq (1 ? - 36.86× j0.65) =1.484?20.51Pu--- Content provided by FirstRanker.com ---
d = 20.51 degrees--- Content provided by FirstRanker.com ---
Vo/V2 = (R1+R2)/R1 = (100+10)/10 =11; V2 =50mv
Vo/V1 = -R2/R1 = -100/10 = -10; V1 = 10mv
Vout = 11*50 + (-10)*10 = 450 mv
Output = XY + XY = X + Y
The above expression is for XOR gate- Discharging of capacitor equation
Vc (t) = Voe-t/t--- Content provided by FirstRanker.com ---
Where T = RC = (103) (10-7) = 10-4
Vo = 100V
Vc(t) = 100 e-104t--- Content provided by FirstRanker.com ---
Vc(t) = 1V
1 = 100 e-104t--- Content provided by FirstRanker.com ---
T = 0.46 msec - f(t) = a0 + ?n=1 an cos nt +?n=1 bn sinnt.
an = 2/T ?x(t)cosn?t d(?t)--- Content provided by FirstRanker.com ---
?=1
T=2p--- Content provided by FirstRanker.com ---
?Asint cost dt
= p/2 sint -A/2 cost
a1 = 0--- Content provided by FirstRanker.com ---
bn = 2/T ?x(t)sin n?t d(?t)
= 2/2p ?Asint sint dt
2/2p ?Asin2t dt--- Content provided by FirstRanker.com ---
=Ap -1/2 cos 2t
b1 = A/2 - A=2xi +3yj+4zk, U = x² + y² + z²
UA = (2x³ + 2xy² + 2xz²)i +(3x²y + 3y³ +3yz²)j +(4x²z+4y²z+4z³)k
div(UA) = (?/?x)(2x³ +2xy² + 2xz²)i +(?/?y)(3x²y + 3y³ +3yz²)j +(?/?z)(4x²z+4y²z+4z³)k--- Content provided by FirstRanker.com ---
div (UA) = (6x² + 2y² + 2z2) + (3x² + 9y² + 3z²) + (4x² + 4y² + 12z²)
at (1, 1, 1) ? x = 1, y = 1, z = 1--- Content provided by FirstRanker.com ---
div (UA) = 45 - PMMC Instrument
Ifs = 10 mA--- Content provided by FirstRanker.com ---
Rm = 100O
100 = Ifs (Rm + Rse)--- Content provided by FirstRanker.com ---
100 = 10 × 10-3 (10 + Rsc)
Rse = 10000-10 = 9990O - y' = -2ßX2
--- Content provided by FirstRanker.com ---
y = [1 0]
X1 + (a/a) u
X2
We know--- Content provided by FirstRanker.com ---
Y = CX + Du
Comparing the above equation with the given problem--- Content provided by FirstRanker.com ---
A=
B=
C = (1 0)
Characteristic equation is--- Content provided by FirstRanker.com ---
|SI-A | = 0
| |= 0
s² + 2?ns + ?n² = 0--- Content provided by FirstRanker.com ---
Comparing 1 and 2
?n² = a--- Content provided by FirstRanker.com ---
?n = va
2?n = 2ß
? = ß/?n = ß/va - VDD-2V
OV out
W = 10 µm--- Content provided by FirstRanker.com ---
L = 1 µm
15 x 10-6 ==(1/2)(Mn Cox) (W/L) (VGS - Vt)² = (1/2)(100x10-6)x(10)x(Vout -0.5)²--- Content provided by FirstRanker.com ---
(Vout -0.5)² = 0.01
Vout = 0.6V = 600mV - From the given Bode plot,
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T(S) = Transfer function = K/(s(s+1)(1+s/20))
It has three poles and no zero--- Content provided by FirstRanker.com ---
So, statement 1 is false
?T(s) = - 90-tan-1? - tan-1(?/20)
?T(jw) |w ? 8 = - 270°--- Content provided by FirstRanker.com ---
So, statement 2 is true - Load supplied previously before adding extra load
--- Content provided by FirstRanker.com ---
12 KW at pf of 0.6
SLoad = 12 + j16
Now, Let P be extra load added--- Content provided by FirstRanker.com ---
(Qs unity p.f)
SLoad = 12 + P + j16--- Content provided by FirstRanker.com ---
Rated KVA | Srated | = 25
v(12 + P)² + 16²= 25
25² = (12+ P)² + 16²--- Content provided by FirstRanker.com ---
P = 7.5,-31.2
So, 7.20 KW is extra load which is added - M-1 M = I
[V1 V2] =
U1TV1 U1TV2 =
U2TV1 U2TV2--- Content provided by FirstRanker.com ---
UTV1 = 1; UTV2 = 0
UTV1 = 0; UTV2 = 1
Statement 1 and 2 are both correct - Vsr Isr cos f = Volo
For single phase fully - controlled converter
lo = Isr = 10A--- Content provided by FirstRanker.com ---
cos f = V0/Vsr = 180/230 = 0.78 - Given that
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Switch frequency, fs = 250Hz
Load resistance RL = 24O
Supply voltage Vs = 48V--- Content provided by FirstRanker.com ---
TON = 1 msec
T = 1/fs = 4ms--- Content provided by FirstRanker.com ---
a = TON/T = 0.25
Load power = Vo²/R = (aVs)²/R = (0.25×48)²/24 = 6 watts - Po = 120w, Vs = 24V, Vo = 48V
--- Content provided by FirstRanker.com ---
Vo = Vs/(1-a)
1-a = Vs/Vo = 24/48 = 1/2--- Content provided by FirstRanker.com ---
a = 0.5 [Duty cycle]
Po = Volo = 120; Io = Po/Vo = 120/48 = 2.54A
Vsls = Volo--- Content provided by FirstRanker.com ---
Is = VoIo/Vs = 120/24 =5A
At boundary of continuous & discontinuous--- Content provided by FirstRanker.com ---
?IL/2 = IS
?IL = 2×5
?IL = (aVs)/fLc--- Content provided by FirstRanker.com ---
Lc = (aVs)/(?IL f) = (0.5×24)/(50×10³×10) = 24 µH - No load
--- Content provided by FirstRanker.com ---
INL = 3A
If = 220/Rf = 220/220 = 1A
Ia = IL - If = 2A--- Content provided by FirstRanker.com ---
Back cmf = Ebn = V - laRa = 220-2 × 0.5 = 219V
Full load--- Content provided by FirstRanker.com ---
IFL = 25A
If = 1A
Ia = IFL - If = 24A--- Content provided by FirstRanker.com ---
Nf = 1500 rpm
EbF = V - laRa = 220 - 24 × 0.5 = 208 V--- Content provided by FirstRanker.com ---
We know E a speed (N)
EbF/Ebn = Nf/NN
(NN = speed at no load)--- Content provided by FirstRanker.com ---
208/219 = 1500/NN
NN = 1579.33 rpm - Acline current I(l)= Isrms/(v3/2 )=(100/v3)/1/2=81.65A
- PQ
RS 00 01 11 10
00 0 1 1 0
01 1 1 1 1--- Content provided by FirstRanker.com ---
11 1 1 1 1
10 0 0 0 0 PQ
RS 00 01 11 10
00 0 1 1 0
01 1 1 1 1--- Content provided by FirstRanker.com ---
11 1 1 1 1
10 0 0 0 0
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This download link is referred from the post: GATE Previous Last 10 Years 2010-2020 Question Papers With Solutions And Answer Keys