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Assertion (A): A half wavelength line can be used as a 1 : 1 transformer. Reason (R): The input impedance of a half wavelength line is equal to load impedance.  
A.  Both A and R are correct and R is correct explanation of A [Correct Answer] 
B.  Both A and R are correct but R is not correct explanation of A [Wrong Answer] 
C.  A is correct but R is wrong [Wrong Answer] 
D.  A is wrong but R is correct [Wrong Answer] 
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Explanation:
Answer : A Discuss it below : !! OOPS Login [Click here] is required to post your answer/result 
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TRUE
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increase
decrease
be unaffected
depend on value of β
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Full wave rectified sinusoid
Exponentially increasing sinusoid
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⇒ In a ratio detector
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100°
50°
5°
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⇒ If sound intensity is 8 W/m^{2}, the dB level of this sound is about
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129 dB
1290 dB
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in the offstate
in the ON state
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at the point of breakdown
⇒ In a digital measuring device the input electrical signal is in the frequency range of dc to ac (50) Hz. It must be sampled at a rate of
f times/sec
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0.25 f times/sec
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1.672 x 10^{26} kg and 9.107 x 10^{31} kg respectively
9.107 x 10^{26} kg and 9.107 x 10^{31} kg respectively
9.107 x 10^{29} kg and 9.107 x 10^{31} kg respectively
9.107 x 10^{30} kg and 9.107 x 10^{31} kg respectively
⇒ Which of the following is incorrect?
(8)_{16} = (8)_{8}
(5)_{16} = (5)_{8}
(8)_{2} = (2)_{10}
(2)_{16} = (2)_{10}
⇒ In order to rectify sinusoidal signals of millivolt range (< 0.6 V)
bridge rectifier using diodes can be employed
full wave diode rectifier can be used
a diode is to be inserted in the feedback loop of an opamp
a diode is to be inserted in the I/P of an opamp
⇒ The output of a half adder is
SUM
CARRY
SUM and CARRY
none of the above
⇒ The power density spectrum of noise voltage of resistor R can be expressed as
2 kTR
1 kTR
0.5 kTR
0.25 kTR
⇒ A transistor amplifier has pole
s_{1}
s_{1} + s_{2} + s_{3} + s_{4}
s_{1} + s_{4}
s_{4}
⇒ In figure the total inductance of the circui
L_{1} + L_{2}
L_{1} + L_{2} + M
L_{1} + L_{2} + 2M
L_{1} + L_{2} + M
⇒ A constant k low pass filter has f_{c} = 1000 Hz. At f = 500 Hz, the phase shift is
p
zero
more than p
less than p
⇒ The open wired circuit in the given figure works
EX  NOR gate
AND gate
XOR gate
NOR gate
⇒ For the transport lag G(jω) = e^{jωT}, the phase angle
is constant
varies linearly with frequency
varies linearly with low frequencies only
values as per square law
⇒ The mean free path of conduction electrons in copper is about 4 x 10^{8} m. For a copper block, find the electric field which can give, on an average, 1 eV energy to a conduction electron
2.62 x 10^{7} V/m
2.64 x 10^{7} V/m
2.5 x 10^{7} V/m
2.58 x 10^{7} V/m
⇒ For a series RLC circuit bandwidth is ω_{2}  ω_{1} = R/2L.
TRUE
FALSE