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Assertion (A): Silicon is less sensitive to changes in temperature than germanium. Reason (R): Cut in voltage in silicon is less than that in germanium. | |
A. | Both A and R are true and R is correct explanation of A [Wrong Answer] |
B. | Both A and R are true but R is not correct explanation of A [Wrong Answer] |
C. | A is true but R is false [Correct Answer] |
D. | A is false but R is true [Wrong Answer] |
View Answer
Explanation:-
Answer : C Discuss it below : !! OOPS Login [Click here] is required to post your answer/result |
Also Read Similar Questions Below :
⇒ In figure, the current I_{410 ∠05 ∠90°5 ∠270°10 ∠90°}
⇒ A difference amplifier using op-amp has closed loop gain = 50. If input is 2 V to each of inverting and non-inverting terminals, output is 5 mA. Then CMRR =
20000
2000
200
20
⇒ In the given figure the E and H lines in a coaxial c
The radial lines are E lines and circular lines are H lines
Radial lines are H lines and circular lines are E lines
The directions of E and H lines are wrong
Radial lines may be E and H lines depending an direction of current
⇒ Bandwidth for practical PCM system is (where n is no. of channels, N length of PCM codes)
2n N f_{m} Hz
4n f_{m} Hz
2n f_{m}
(N + 1)f_{m}
⇒ To couple a coaxial line to a parallel wire line, it is best to use a
balun
slotted line
quarter wave transformer
directional coupler
⇒ 7483 is a TTL binary adder. It can add
two 4-bit binary numbers
four 6-bit binary numbers
four 8-bit binary numbers
any number of 4-bit numbers
⇒ The Boolean function realized by the logic circuit show
F = Σm(0, 1, 3, 5, 9, 10, 14)
F = Σm(2, 3, 5, 7, 8, 12, 13)
F = Σm(1, 2, 4, 5, 11, 14, 15)
F = Σm(2, 3, 5, 7, 8, 9, 12)
⇒ The Boolean expression for the circuit of the given fi
A {F + (B + C) (D + E)}
A [F + (B + C) (DE)]
A + F + (B + C) (D + E)]
A [F + (BC) (DE)]
⇒ Which of these has poor overload capacity?
PMMC instrument
Hot wire instrument
Dynamometer instrument
Both (a) and (c)
⇒ The Depth of penetration of EM wave in medium having conductivity σ at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be
6.25 cm
12.50 cm
50 cm
100 cm
⇒ Radiation resistance of an antenna is a
dc resistance
ac resistance
either (a) or (b)
none of the above
⇒ A shift register can be used for all of the following EXCEPT
ring counter
A/D conversion
series to parallel conversion
generation of delay
⇒ A device has two resistances 300 Ω and 200 Ω respectively. The noise voltages due to these two resistance calculated separately are 5.38 μV and 4.38 μV respectively. When both of them are used together, the noise voltage will be
9.76 V
1 μV
23.56 μV
6.94 μV
⇒ In the given figure R_{C} = R_{L} = 1 kΩ, then V_{05 V2.5 V1 V0 V}
⇒ A video transmission system transmits 625 picture frames per second. Each frame consist of a 400 x 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is
16 Mbps
100 Mbps
600 Mbps
6.4 Gbps
⇒ The transfer function of a system is given by H(s) = 1/s^{2} (s - 2). The impulse response of the system is
(t^{2*} e^{-2t}) U(t)
(t^{*} e^{2t}) U(t)
(te^{-2t}) U(t)
(te^{2t}) U(t)
⇒ A half adder can be used only for adding
1 s
2 s
4 s
8 s
⇒ A 555 timer can be used as
an astable multivibrator
monostable multivibrator
frequency Divider only
any of the above
⇒ For holding the drain current constant in spite of large changes in JFET parameters
current source bias is used
voltage divider bias is used
reverse bias is used
no bias is used
⇒ An AM signal and a narrowband FM signal with identical carriers, modulating signals and modulation index of 0.1 are added together. The resultant signal can be closely approximated by
Broadband FM
SSB with carrier
DSB-SC
SSB without carrier
⇒ For resonant long wire antenna 'n' half wavelength long in free space, if 'n' is increased then inclination of the major lobe with wire axis will __________ .
increases
decreases
remains same
none
⇒ For a NPN bipolar transistor, what is the main stream of current in the base region?
Drift of holes
Diffusion of holes
Drift of electrons
Diffusion of electrons
⇒ An npn transistor has a Beta cutoff frequency f_{β} of 1 MHz, and a common emitter short circuit low frequency current gain β_{0} of 200. It unity gain frequency f_{T} and the alpha cut off frequency f_{a2} respectively are
200 MHz, 201 MHz
200 MHz, 199 MHz
199 MHz, 200 MHz
201 MHz, 200 MHz
⇒ A 1 ms pulse can be stretched to 1 sec pulse by using
an astable multivibrator
a monostable multivibrator
a bistable multivibrator
a Schmitt trigger circuit
⇒ In a CE amplifier the Q point is very close to saturation point on the dc load line. This results in
positive clipping of collector current
negative clipping of collector current
both positive and negative clipping of collector current
either positive or negative clipping of collector current
⇒ The spectral component of noise associated with k^{th} frequency is given, in the limit as Δf → 0 by n_{k}(t) written as n_{k}(t) = C_{k} cos (2pk Δft + θ_{k1, 2 only1, 3 only1, 2, 32, 3 only}
⇒ An MTI compares a set of received echos with those received during the previous sweep.
TRUE
FALSE
⇒ A 4 bit down counter starts counting from 1111 irrespective of modulus.
TRUE
FALSE
⇒ Refer to the NAND and NOR latches shown in the figure. The inputs (P_{1}, P_{2}) for both the latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs (Q_{1}, Q_{2})
NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0)
NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (1, 0)
NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (0, 0)
NAND: first (1, 0) then (1, 1) NOR: first (0, 1) then (0, 1)
⇒ A better SNR can be obtained in a PCM by
increasing sampling rate above Nyquist rate
decreasing the code bits
reducing channel bandwidth
increasing Quantization levels
⇒ In figure, the current I_{410 ∠05 ∠90°5 ∠270°10 ∠90°}
⇒ A difference amplifier using op-amp has closed loop gain = 50. If input is 2 V to each of inverting and non-inverting terminals, output is 5 mA. Then CMRR =
20000
2000
200
20
⇒ In the given figure the E and H lines in a coaxial c
The radial lines are E lines and circular lines are H lines
Radial lines are H lines and circular lines are E lines
The directions of E and H lines are wrong
Radial lines may be E and H lines depending an direction of current
⇒ Bandwidth for practical PCM system is (where n is no. of channels, N length of PCM codes)
2n N f_{m} Hz
4n f_{m} Hz
2n f_{m}
(N + 1)f_{m}
⇒ To couple a coaxial line to a parallel wire line, it is best to use a
balun
slotted line
quarter wave transformer
directional coupler
⇒ 7483 is a TTL binary adder. It can add
two 4-bit binary numbers
four 6-bit binary numbers
four 8-bit binary numbers
any number of 4-bit numbers
⇒ The Boolean function realized by the logic circuit show
F = Σm(0, 1, 3, 5, 9, 10, 14)
F = Σm(2, 3, 5, 7, 8, 12, 13)
F = Σm(1, 2, 4, 5, 11, 14, 15)
F = Σm(2, 3, 5, 7, 8, 9, 12)
⇒ The Boolean expression for the circuit of the given fi
A {F + (B + C) (D + E)}
A [F + (B + C) (DE)]
A + F + (B + C) (D + E)]
A [F + (BC) (DE)]
⇒ Which of these has poor overload capacity?
PMMC instrument
Hot wire instrument
Dynamometer instrument
Both (a) and (c)
⇒ The Depth of penetration of EM wave in medium having conductivity σ at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be
6.25 cm
12.50 cm
50 cm
100 cm
⇒ Radiation resistance of an antenna is a
dc resistance
ac resistance
either (a) or (b)
none of the above
⇒ A shift register can be used for all of the following EXCEPT
ring counter
A/D conversion
series to parallel conversion
generation of delay
⇒ A device has two resistances 300 Ω and 200 Ω respectively. The noise voltages due to these two resistance calculated separately are 5.38 μV and 4.38 μV respectively. When both of them are used together, the noise voltage will be
9.76 V
1 μV
23.56 μV
6.94 μV
⇒ In the given figure R_{C} = R_{L} = 1 kΩ, then V_{05 V2.5 V1 V0 V}
⇒ A video transmission system transmits 625 picture frames per second. Each frame consist of a 400 x 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is
16 Mbps
100 Mbps
600 Mbps
6.4 Gbps
⇒ The transfer function of a system is given by H(s) = 1/s^{2} (s - 2). The impulse response of the system is
(t^{2*} e^{-2t}) U(t)
(t^{*} e^{2t}) U(t)
(te^{-2t}) U(t)
(te^{2t}) U(t)
⇒ A half adder can be used only for adding
1 s
2 s
4 s
8 s
⇒ A 555 timer can be used as
an astable multivibrator
monostable multivibrator
frequency Divider only
any of the above
⇒ For holding the drain current constant in spite of large changes in JFET parameters
current source bias is used
voltage divider bias is used
reverse bias is used
no bias is used
⇒ An AM signal and a narrowband FM signal with identical carriers, modulating signals and modulation index of 0.1 are added together. The resultant signal can be closely approximated by
Broadband FM
SSB with carrier
DSB-SC
SSB without carrier
⇒ For resonant long wire antenna 'n' half wavelength long in free space, if 'n' is increased then inclination of the major lobe with wire axis will __________ .
increases
decreases
remains same
none
⇒ For a NPN bipolar transistor, what is the main stream of current in the base region?
Drift of holes
Diffusion of holes
Drift of electrons
Diffusion of electrons
⇒ An npn transistor has a Beta cutoff frequency f_{β} of 1 MHz, and a common emitter short circuit low frequency current gain β_{0} of 200. It unity gain frequency f_{T} and the alpha cut off frequency f_{a2} respectively are
200 MHz, 201 MHz
200 MHz, 199 MHz
199 MHz, 200 MHz
201 MHz, 200 MHz
⇒ A 1 ms pulse can be stretched to 1 sec pulse by using
an astable multivibrator
a monostable multivibrator
a bistable multivibrator
a Schmitt trigger circuit
⇒ In a CE amplifier the Q point is very close to saturation point on the dc load line. This results in
positive clipping of collector current
negative clipping of collector current
both positive and negative clipping of collector current
either positive or negative clipping of collector current
⇒ The spectral component of noise associated with k^{th} frequency is given, in the limit as Δf → 0 by n_{k}(t) written as n_{k}(t) = C_{k} cos (2pk Δft + θ_{k1, 2 only1, 3 only1, 2, 32, 3 only}
⇒ An MTI compares a set of received echos with those received during the previous sweep.
TRUE
FALSE
⇒ A 4 bit down counter starts counting from 1111 irrespective of modulus.
TRUE
FALSE
⇒ Refer to the NAND and NOR latches shown in the figure. The inputs (P_{1}, P_{2}) for both the latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs (Q_{1}, Q_{2})
NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0)
NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (1, 0)
NAND: first (1, 0) then (1, 0) NOR: first (1, 0) then (0, 0)
NAND: first (1, 0) then (1, 1) NOR: first (0, 1) then (0, 1)
⇒ A better SNR can be obtained in a PCM by
increasing sampling rate above Nyquist rate
decreasing the code bits
reducing channel bandwidth
increasing Quantization levels