ENGINEERING PHYSICS KTU 2016 PREVIOUS QUESTION PAPER WITH ANSWERS FOR S1,S2 FIRST YEAR STUDENTS

PART B


ktu previous questions with answers,ktu solved questions,ENGINEERING PHYSICS KTU 2016 PREVIOUS QUESTION PAPER  WITH ANSWERS FOR S1,S2 FIRST YEAR STUDENTS

Answer any 10 questions. Each question carries 4 marks

13.Compare an electrical and mechanical oscillator.
ANS:
Electrical oscillator
An oscillator can be defined as an electronic circuit which converts dc power from the supply to ac power in the load. It is circuit which generates an output signal without any external input signal.
Mechanical Oscillator
In mechanical oscillator there is only a transformation from K.E to P.E and vice versa. Total Energy of a harmonic oscillator with no dissipative forces remains constant.
14. A transverse wave on a stretched string is described by y(x,t)=4.0sin(25t+0.016x+π/3) where x and y are in cm and t is in second.
Obtain the (i) Speed (ii) Amplitude (iii) Frequency and(iv)Initial phase at the origin.
ANS:
(i) Instantaneous velocity is,
ENGINEERING PHYSICS KTU 2016 PREVIOUS QUESTION PAPER  WITH ANSWERS FOR S1,S2 FIRST YEAR STUDENTS
(ii) amplitude is 4
(iii) Frequency,
ω=2πf
f=ω/2π
where, ω=25;
f=25/2π;
=3.98Hz
(iv) Initial phase at origin is, π/3
15. With Newton’s rings arrangement, nth dark ring formed by light of wavelength 6000 A° coincides with the (n+1)th dark ring for light of wavelength 4500A°. If the radius of curvature of the convex surface is 90 cm, find the diameter of the nth ring for light wavelength 6000 A°.
ANS:
We know that,
for dark ring
2t=nλ; nth dark ring of first wave coincide with (n+1)th dark ring;
So, 1=(n+1)λ2
n x 6000=(n+1) x 4500 by solving we get,
n=3
We know that,
rn2=Rnλ
rn=√Rnλ
r=√(90×10-2x3x6000x10-10)
r=1.272mm.
16. A plane transmission grating has 6000 lines/cm. Find the angular separation between two wavelengths 500nm and 510nm in the 3rd order.
ANS:
We know that;
SinΘ=Nnλ
N be the number of lines in one metre of the grating
Θ1=Sin-1(6000x100x3x500x10-9)
Θ1=64.16o
Θ2=Sin-1(6000x100x3x510x10-9)
Θ2=66.64o
angular separation= Θ– Θ1
=2.48o
17. The refractive index of a calcite is 1.658 for ordinary ray and it is 1.486 for extraordinary ray. A slice having thickness 0.9×10-4cm is cut from the crystal. For what wavelength the slice will acts as a (i) Quarter wave plate. (ii) Half wave plate.
ANS:
(i) For Quarter Wave Plate, thickness
 qwp
(ii) For Half wave plate, thickness
18. Distinguish between Type I and Type II superconductors with examples.
ANS:

Type I SuperconductorsType II Superconductors
1. Soft superconductors are those which can tolerate impurities without affecting the superconducting properties.1. Hard superconductors are those which cannot tolerate impurities, i.e., the impurity affects the superconducting property.
2. They have low critical field.2. They have high critical field.
3. Show complete Meissner effect.

3. Hard super conductors trap magnetic flux and hence Meissner effect is not complete.
4. The current flows through the surface only.4. It is found that current flows throughout the material.
5. Eg. Tin, Aluminum5. Eg. Tantalum, Neobium
19. Calculate the De Broglie wavelength of electron whose kinetic energy is 10Kev.
ANS:
Kinetic Energy KE=1/2(mv2)
λ=h/p
2Em=P2
P=√(2Em)
λ=h/√(2Em)
1eV=1.6×10-19J ; mass of e=9.1×10-31kg ; h=6.626×10-34 m2kg/s
Substituting the value we get,
λ=1.22nm.
20. Distinguish between Macrostate and Microstate of a system.
ANS:
Macrostate and Microstate
The state of a physical system characterized by its bulk parameters like pressure, volume, temperature, mass etc. is called Macro states.
Microstate  of a system is characterized by the value of coordinates of position and momentum for all particle on a system at any instant.
Microstate changes with time without affecting the Macrostates. Thus an enormously large number of microstates  correspond to each macrostate of the  system.
21. The volume of a hall is 3000 m. It has a total absorption of 100m2 sabine. If the hall is filled with audience who add another 80m2 sabine, then find the difference in reverberation time.
ANS:
We know Sabine’s formula;
T = 0.16 V / A
Here,
V- volume
A- area of absorption
reverberation time T1= 0.16×3000/100
=4.8 Sec
T2=0.16×3000/180
=2.67 Sec
difference in reverberation time,
T2 – T1
=2.13 Sec
22. What is NDT? and how ultrasonic wave used for NDT?
ANS:
Nondestructive testing is a method of finding defects in an object without harming the object.  NDT is used to look for internal changes or signs of wear on airplanes. Discovering defects will increase the safety of the passengers. The railroad industry also uses nondestructive testing to examine railway rails for signs of damage. Sound with high frequencies, or ultrasound, is one method used in NDT. Basically, ultrasonic waves are emitted from a transducer into an object and the returning waves are analyzed. If an impurity or a crack is present, the sound will bounce off of them and be seen in the returned signal. In order to create ultrasonic waves, a transducer contains a thin disk made of a crystalline material with piezoelectric properties, such as quartz. When electricity is applied to piezoelectric materials, they begin to vibrate, using the electrical energy to create movement. Remember that waves travel in every direction from the source. To keep the waves from going backwards into the transducer and interfering with its reception of returning waves, an absorptive material is layered behind the crystal. Thus, the ultrasound waves only travel outward.
23. What is the difference between spontaneous emission and stimulated emission?
ANS:
spontaneous emission
The emission of photons by the natural de-excitation transition of atoms, molecules, ions etc. is known as spontaneous emission process. This is an uncontrolled and natural  phenomenon in which each excited atom emit photons independently in random direction, phase and polarization.
stimulated emission
The process of de-excitation transition with the emission of radiation on finding an identical radiation as stimuli to trigger the emission process and is known stimulated emission.
24. What is LED? Give its working principle?
ANS:
 LED operating on the principle of injection of luminescence. Basically LED is a semiconductor pn junction device capable of emission of electromagnetic radiation under forward-biased condition. LED emits light under forward biased condition. When the external voltage exceeds the barrier potential, recombination between electrons and holes take place. The minority carrier concentration on both sides of the junction are considerably increased under this condition, which increases their rate of recombination. Thus electromagnetic radiation emitted.

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