Statement: All bags are cakes. All lamps are cakes.
I. Some lamps are bags.
II. No lamp is bag.
Deduce which of the above conclusion logically follows statements:
Conclusion I and II are complementary. So either conclusion I or conclusion II follows the statement.
Man does not live by __________ alone.
Man does not live by bread alone
Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything __________ to the requirements of the exam.
Extraneous – irrelevant or unrelated to the subject being dealt with
Select the pair that best expresses a relationship similar to that expressed in the pair
LIGHT : BLIND
If = log(a + b), then:
RHS : Log a + Log b = Log ab
For LHS : Log a/b + Log b/a
= Log ((a/b) *(b/a))
=Log (ab/ab) =1
LHS = RHS
In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and average came down to 45 marks. Total number of candidates who took the exam was:
Let number of candidates be n
n = 600
A solid 4cm cube of wood is coated with red paint on all the six sides. Then the cube is cut into smaller 1cm cubes. How many of these 1cm cubes have no colour on any side?
Number of cubes with 3 faces red = 8
Number of cubes with 2 faces red = 24
Number of cubes with 1 face red = 24
Number of cubes with no face red = 64-(24+24+8) = 8
One of the warmest winters on record has put consumers in the mood to spend money. Spending is to be the strongest in thirteen years. During the month of February, sales of existing single family homes hit an annual record rate of 4.75 million. This paragraph best supports the statement that:
The age of father is 4 times more than the age of his son Amit. After 8 years, he would be 3 times older than Amit. After further 8 years, how many times will he be older than Amit?
Let Amit’s age be n years
Age of his father = (4*n) = 4n years
After 8 years:
Amit’s age = n + 8
Father’s age =
3(n + 8) = 4n + 8
3n + 24 = 4n + 8
16 = n
n = 16 years
After further 8 years:
Amit’s age = 24 years
Father’s age = 4n +16 = 80 years
So father is 2.5 times older than Amit.
Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes. Find t?
A is kept open for all t minutes and fills one-third the tank. Or, A should be able to fill the entire tank in '3t' minutes.
A and B together can fill the tank completely in t minutes. A alone can fill it in 3t minutes.
A and B together can fill 1/t of the tank in a minute. A alone can fill 1/3t of the tank in a minute. So, in a minute, B can fill 1/t - 1/3t=2/3t. Or, B takes 3t/2 minutes to fill an entire tank.
To fill one-third the tank, B will take t/2 minutes. B is kept open for t -10 minutes.
t/2 = t - 10,
∴ t = 20 minutes.
A takes 60 minutes to fill the entire tank, B takes 30 minutes to fill the entire tank. A is kept open for all 20 minutes. B is kept open for 10 minutes.
C, which is kept open for 8 minutes also fills one-third the tank. Or, C alone can fill the tank in 24 minutes.
The value of ‘V’ from the given below circuits is_____________ volts.
By redrawing the given circuit
Current flowing through 30Ω is zero. So circuits becomes
Current flowing through 10Ω resistor is
For eliminating Harmonics from the emf generated the coil span must be.
, Full pitch
Determine the value of
∴ Integral becomes
= sin(x/2) * δ(x-(pie/2)) * (dx/2)
∴ (1/2) *sin(pie/4) =1/(2√2)
In HVDC converter stations equipment using thyristor it is necessary to use a large number of thyristors in series because
The holding current of a SCR is 18mA. Its latching current will be
Ratio between latching current to holding current is 2-3 times.
In a 132 kV system, the phase to ground capacitance is 0.01 µF and the inductance is 4H. Calculate the voltage appearing across the electrodes of C.B. If a magnetizing current at 5 amp is interrupted. Calculate the critical resistance to be connected, to eliminate restriking.
A phase lead compensating network has its transfer function The maximum phase lead occurs at a frequency of:
Comparing with the standard phase lead compensating network,
So, maximum phase lead occurs at frequency
= 50 rad /sec
The time taken for the output to settle within 2% of step input for the control system represented by is given by
Comparing the above transfer function with the standard second order transfer function.
Settling time for 2% tolerance band =
= = 1.6 sec
When 30V dc is applied to the vertical deflection plates of CRO, the bright spot moves 1cm away from the center if 30V (rms) is applied then the picture on screen
A compound dc generator is delivering full load current at a terminal voltage of 230V. It’s
series field winding get’s short circuit. If it’s terminal voltage becomes.
1. More than 230V it is over compounded
2. More than 230V it is differential compound
3. Less than 230V it is over compounded
4. Less than 230V it is differential compounded
5. Less than 230V it is level compounded
The correct statement’s are?
An (star connected) Alternator is connected to Delta-star transformer with start neutral earthed. On the star side of transformer a single line to ground fault occurs. This is
If the winding is delta connected, it does not associated with 0 sequence component current, but only +ve, -ve sequence components exist. Also only +ve & -ve sequence components exist only in line to line fault
Snubber circuit is used to limit the rate of
The unit of mobility is:
Mobility= drift velocity/Electric field
In the circuit shown in figure, a Silicon transistor with VBE = 0.7 V and βdc = 100 is used. The Q-point is established at
DC equivalent of the given circuit
Step (1): KVL for B-E loop
0 - IB 220 k - VBE - IE (2. 2k) + 20 V =0
=> IB (220k) + (1 + β) IB 2.2k = 19.3 V
Step (2): KVL for C-E loop
A power supply has a full load voltage of 24V. What is its no-load voltage for 5% regulation.
Voltage - Regulation,
Assume that the op-amp as the circuits shown below is ideal. Find the current passing through the 2kΩ resistor in the circuit.
But the current entering into the op-amp terminals is zero. So current through 12k is zero.
The diode in the circuit given below has VON = 0.7 V but is ideal otherwise. The current (in mA) in the 4kΩ resistor is______ mA
Correct answer is 0.6 Given circuit is
Here, we have
So bridge is balanced, and hence, no current will flow through diode. The equivalent circuit is shown below.
Current through 4kΩ resistor is
The system of linear equations:
3x + 4y = 6
2x + 3y = 5
Since rank of matrix A as well as auguemented matrix is 2, therefore it has unique solution
A fair dice is tossed two times. The probability that the second results in a value that is higher than the first toss is:
In the first toss, results can be 1, 2, 3, 4, 5
For 1, the second toss results can be 2, 3, 4, 5, 6
For 2, the second toss results can be 3, 4, 5, 6
For 3, the second toss results can be 4, 5, 6
For 4, the second toss results can be 5, 6
For 5, the second toss results can be 6
So the required probability
where P is a vector is equal to :
Gauss divergence theorem will explain the given expression.
Identify the instruction that affects only Carry Flag.
The initial of MOD-16 down counter is 0110. After 37 clock pulses, the state of the counter will be:
After 37 clock pulses, the state of MOD-16 DOWN counter will be five states below the present state.
12 MHz clock frequency is applied to a cascaded counter of modulus-3 counter, modulus-4 counter, and modulus-5 counter. Determine the output frequency.
Modulus of cascaded counter,
So, frequency of the output signal
= 200 KHz
In the circuit shown, the power supplied by the voltage source is:
There is no current flowing through voltage source that's why the power delivered by voltage source is zero.
Total energy stored on network in kjoules at steady state is _______.
At steady state inductor behaves as short circuit and capacitor behaves as open circuits.
By drawing the circuit at steady state
If the current changes from 3A to 5A in 2s and the emf is 10V. Calculate the inductance.
We know that:
Substituting the values from the question we get L=10H.
The output y(t) of a continuous – time system S for the input x(t) is given by Which one of the following is correct?
For a continuous time system,
Therefore s(t) = u(t)
Which is linear and time-invariant
Consider the following block diagram
Overall gain C(S)/R(s) of the above system is
"What is the inverse lap lace transform of
Determine the value of K for a unity feedback system with G(s) = to have a peak overshoot of 50%
4K = 21.56
= K = 5.39
A there phase alternator is wound with a 60 degree phase-spread armature windings and develops 300 kVA. If the armature is reconnected utilizing all the coils for single phase operation with a phase spread of 180 degrees, the new rating of the machine is
The flux density distribution for 50 Hz cylindrical rotor alternator is B(q) = sin q+ 0.3sin3q+ 0.2sin5q where is measured from neutral axis,alternator has pole pitch 40 cm and length 35 cm, stator coil span is four-fifth of the pole pitch. Find the emf generated in one turn coil?
Peak flux density
If the current changes from 5A to 3A in 2 seconds and the inductance is 10H, calculate the emf.
We know that:
Substituting the values from the question, we get emf=10V.
Find the type of feedback topology for the given circuit
The power factor of pure resistive circuit is
For the purely resistive circuit, the power factor is 1 (perfect), because the reactive power equals zero.
A 400V / 100V, 10kVA two winding transformer is to be used as an autotransformer to supply a 400V circuit from 500V source when tested as two winding transformer at rated load .85 P.F. lag it’s efficiency is 97%. Determine efficiency as autotransformer at rated load and 0.85 power factor
2 wdg TF : 10KVA, 400/100KV
Efficiency : 97%
Load PF : 0.85 Lag
Output Power = 10KVA × 0.85 = 8500W
Efficiency = Output / (Output + Losses) = 97%
Losses = 262.88W
Auto TF : Vh = 500V ; Ih=100A & VL= 400V ; IL=125A. P= (50KVA)
Output Power = 50KVA × 0.85 = 42500W
Losses (as same as 2Wdg) = 262.88W
Efficiency = Output / (Output + Losses) = 42500/( 42500 + 262.88)
= 99.38 %
A solid conductor of 4 cm radius is converted into four solid conductors of equal radius without changing net current rating. The four solid conductors are used to form a stranded conductor as follows.
GMR. of stranded conductor is____________ cm____________.
An amplifier using BJT has two identical stages each having a lower cut-off (3dB) frequency of 64Hz due to coupling capacitor. The emitter bypass capacitor also provides a lower cut-off (3dB) frequency due to emitter degeneration alone of 64Hz. The lower 3dB frequency of the overall amplifier is nearly
So overall lower cut-off frequency,
= 100 Hz
A 240V, 50 Hz supply feeds a highly inductive load of 50 ohm resistance through a half controlled thyristor bridge. When the firing angle the load power is:
A pulse train with a frequency of 1MHz is counted using a mod-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:
Here mod – 1024
So to count 1024 number of flip-flops requires
n = 10
So propagation delay per flip flop
All flip-flops are triggered with negative edge.
The above counter shown in figure (1) is Mod 12 down counter. Then the Gate name is
It starts with 1111 and after reaching 0011 the OR gate output becomes low. So, 'PR’ inputs get enabled.
If the input to the gate is 0011 then it preset's all the flip flops. The truth table for the gate is as follows.
From the above truth table we conclude that the given Gate is 'OR' Gate.
Minimum number of 2 input NOR gates needed to implement half sub tractor is ____________ ,
If a fault occurs near an impedance relay, the V/I ratio will be
If the fault is away from the relay, then the impedance seen by the relay is more.
A d.c. to d.c. chopper operates from a 48 V battery source into a resistive load
of 24 Ω. The frequency of the chopper is set to 250Hz. When chopper on-time is 1 ms the load power is
When converting 7,000 nA to microamperes, the result is
1A = 1000(10-3)A.
1mA = 1000(10-6)A.
1uA = 1000(10-9)A.
1nA = 1000(10-12)A.
So, than the solution is: 7000nA = 7uA.
The diodes in the circuit shown below have Vγ = 0.7 V. Then output voltage Vo is __________
Apply open circuit test for the diodes and voltages across diodes are as
D1 —> 10 V
D2 —> 5V
D3 -A -5V
So D1 becomes ON first