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Above a certain temperature, the specific heat of a metal becomes constant. This temperature is called | |
A. | Debye temperature [Correct Answer] |
B. | Curie temperature [Wrong Answer] |
C. | Neel temperature [Wrong Answer] |
D. | Transition temperature [Wrong Answer] |
View Answer
Explanation:-
Answer : A Discuss it below : !! OOPS Login [Click here] is required to post your answer/result |
Also Read Similar Questions Below :
⇒ In the given figure A = 1, B = 1. B is now changed to a sequence 101010................The outputs X and Y wil
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X = 1010.......and Y = 0101........
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⇒ The input impedance of short-circuited line of length l where l < λ/4
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X = 1010.......and Y = 0101........
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⇒ The input impedance of short-circuited line of length l where l < λ/4
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capacitive
resistive
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-81.56
73
10254,1
42
⇒ Consider the z-transform X(z) = 5z^{2} + 4z^{-1} + 3; 0 < |z| < ∞ The inverse transform x[n] is
5δ[n + 2] + 3δ[n] + 4δ[n - 1]
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⇒ Two D flip-flop as shown below are to be connected as a synchronous counter that goes through the following Q_{1}Q_{0} sequ
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⇒ Which DOS command improves memory usage?
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MOVE
MEM IM
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⇒ A . 0 =
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⇒ The circuit given below
counter
digital to analog converter
two-bit series-to-parallel converter
analog-to digital converter
⇒ The phase angle characteristic in the given figure is
all pass function
transport lag
minimum phase function
none of the above
⇒ In a transistor CE mode, V_{CC} = +30 V. If the transistor is in cut off region, V_{CE} =
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A 6 H, 1
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00 H, 0
00 H, 1
⇒ Astable multivibrator has
one stable state
two stable state
two quasi-stable states
one quasi stable state
⇒ An ideal rectifier should have
100% efficiency, zero ripple factor, zero harmonic factor and unity displacement factor
100% efficiency, zero tipple factor, zero harmonic factor and zero displacement factor
100% efficiency, zero ripple factor, unity harmonic factor and unity displacement factor
100% efficiency, unity ripple factor, zero harmonic factor and unity displacement factor
⇒ If operator 'a' = 1 ∠120° then (1 - a^{2}) =
1 ∠ - 30°
1 ∠ 30°
1 ∠ - 60°
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⇒ Graphic equaliser system in stereophonic systems generally have
3 or 4 bands
1 band
more than 8 bands
large number of bands
⇒ At a given probability of error, binary coherent FSK is interior to binary coherent PSK by
6 dB
3 dB
2 dB
0 dB
⇒ The number of valence electrons in germanium atom is
4
2
1
zero
⇒ The transition of electron from energy level W_{1} to W_{2} is associated with emission or absorption of electro magnetic radiation of frequency f such that
hf = |W_{1} - W_{2}|
|W_{1} - W_{2}|f = h
hf = 0.5|W_{1} - W_{2}|
(hf)^{2} = |W_{1} - W_{2}|
⇒ When a p-n Junction is forward biased
the current flow is due to electrons only
the majority carriers in both p and n materials are driven toward the junction.
the majority carriers in both p and n materials are away from the junction.
both (a) and (c).
⇒ For a symmetrical network
h_{11} = h_{22}
h_{12} = h_{21}
h_{11} h_{22} - h_{12} h_{21} = 0
h_{11} h_{22} - h_{12} h_{21} = 1
⇒ In force current analogy, displacement x is analogous to
charge
magnetic flux linkage
electrostatic energy
voltage
⇒ In the given figure shows a 10dB directional coupler. If power input at port 1 is 100 mW, power output at port
10 mW
15 mW
20 mW
25 mW
⇒ The sum of two or more arbitrary sinusoidal is
always periodic
periodic under certain conditions
never periodic
periodic only if all the sinusoidals are identical in frequency and phase
⇒ If operator 'a' = 1 ∠120° then (1 + a + a^{2}) =
-a
-a^{2}
1
0
⇒ Johnson noise is
white for all Practical application
always white
depends on temperature
never white
⇒ If 4 in binary system is 100 then 8 will be
10
100
111
1000
⇒ Two sinusoidal voltage sources v_{1} = 50 sin (100t) and v_{1} = 50 sin (100t + p) are connected in parallel and fed an inductance X_{L} = 2Ω. The cuirent through X_{L} is
25 A
about 17.5 A
zero
50 A
⇒ A signal is x + f(t) where x is constant and f(t) is a power signal with zero mean value. The power of the signal is
x^{2}
x^{2} + f^{2}(t)
more than x^{2} + f^{2}(t)
less than x^{2} + f^{2}(t)
⇒ Refractive index of glass is 1.5. Find the wavelength of a beam of light with a frequency of 10^{14} Hz in glass. Assume velocity of light is 3 x 10^{8} m/sec in vacuum.
4 μm
3 μm
2 μm
1 μm
⇒ In figure V_{CE} (sat) = 0.1 V. Then I_{C} (sa
10 mA
20 mA
30 mA
40 mA