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CLASS XII
Physics
Sample Question Paper 2018-19
Time allowed: 3 hrs Max Marks: 70
General Instructions:
1. All questions are compulsory. There are 27 questions in all. 2. This question paper has four sections: Section A, Section B, Section C and Sec-
tion D. 3. Section A contains five questions of one mark each, Section B contains seven
questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains three questions of five marks each.
4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of three marks
and three questions of five marks weightage. You have to attempt only one of
the choices in such questions. 5. You may use the following values of physical constants wherever necessary.
c = 3 × 108 m/s h = 6.63 × 10–34 Js
e = 1.6 × 10–19 C μo = 4π × 10–7 T m A–1
ε0 = 8.854 × 10–12 C2 N-1 m-2
o
1
4= 9 × 109 N m2 C–2
me = 9.1 × 10–31 kg mass of neutron = 1.675 × 10–27 kg
mass of proton = 1.673 × 10–27 kg Avogadro’s number = 6.023 × 1023 per gram mole
Boltzmann constant = 1.38 × 10–23 JK–1
SECTION – A (Question numbers 1 to 5 carries 1 mark each)
1. State the SI unit of the electric polarization vector P.
2. Define temperature coefficient of resistivity.
3. Name the electromagnetic waves that are widely used as a diagnostic tool in medicine.
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OR
Name the current which can flow even in the absence of electric charge.
4. A ray of light is incident on a medium with the angle of incidence ‘i’ and is refracted into a second medium with the angle of refraction ‘r’. The graph of sin i versus sin r is as shown. Find the ratio of the velocity of light in the first medium to the velocity of light in the second medium.
5. Two particles have equal momenta. What is the ratio of their de-Broglie wave-lengths?
OR
Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. What is the energy of a photon in the light beam?
SECTION – B (Question numbers 6 to 12 carries 2 marks each)
6. A network of resistors is connected to a 16 V battery with an internal resistance of 1 Ω as shown in the following figure. Compute the equivalent resistance of the network.
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OR
A 9 V battery is connected in series with a resistor. The terminal voltage is found to be 8 V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?
7. The diagram below shows a potentiometer set up. On touching the jockey near to the end X of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end Y of the potentiometer, the galvanom-eter pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it
leads to such a one-sided deflection.
OR
Following circuit was set up in a meter bridge experiment to determine the value X of unknown resistance.
(a) Write the formula to be used for finding X from the observations.
(b) If the resistance R is increased, what will happen to balancing length?
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8. The figure shows two sinusoidal curves representing oscillating supply voltage and current in an AC circuit.
Draw a phasor diagram to represent the current and supply voltage appropri-
ately as phasors. State the phase difference between the two quantities.
9. Compare the following
(i) Wavelengths of the incident solar radiation absorbed by the earth’s surface and the radiation re-radiated by the earth.
(ii) Tanning effect produced on the skin by UV radiation incident directly on the skin and that coming through the glass window.
10. A narrow slit is illuminated by a parallel beam of monochromatic light of wave-length λ equals to 6000 Å and the angular width of the central maxima in the resulting diffraction pattern is measured. When the slit is next illuminated by
the light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’.
11. What are the universal gates? How can AND gate be realized using an appro-priate combination of NOR gates?
12. A TV transmission tower antenna is at the height of 20 m. How much range can it cover if the receiving antenna is at the height of 25 m?
SECTION – C (Question numbers 13 to 24 carries 3 marks each)
13. A particle, having a charge +5 μC, is initially at rest at the point x = 30 cm on the x-axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (a) Q =+15μC and (b) Q = -15μC.
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14. (a) An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The ar-rows P1 and P2 show the directions of its electric dipole moment in the two cases.
Identify for each case, whether the dipole is in stable or unstable equilibrium.
Justify each answer.
(b) Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density.
Will the dipole be in equilibrium at these two positions? Justify your answer.
15. Two material bars A and B of equal area of cross-section, are connected in
series to a DC supply. A is made of usual resistance wire and B of an n-type semiconductor.
(a) In which bar is drift speed of free electrons greater?
(b) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected?
Justify each answer.
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16. Derive an expression for the velocity vC of positive ions passing undeflected through a region where crossed and uniform electric field E, and magnetic field B are simultaneously present.
Draw and justify the trajectory of identical positive ions whose velocity has a magnitude less than vC.
OR
A particle of mass m and charge q is in motion at speed v parallel to a long straight conductor carrying current I as shown below.
Find the magnitude and direction of the electric field required so that the par-
ticle goes undeflected.
17. A sinusoidal voltage of peak value 10 V is applied to a series LCR circuit in which resistance, capacitance and inductance have values of 10 Ω, 1μF and 1H respectively. Find (i) the peak voltage across the inductor at resonance (ii) quality factor of the circuit.
18. (a) What is the principle of the transformer?
(b) Explain how laminating the core of a transformer helps to reduce eddy current losses in it
(c) Why the primary and secondary coils of a transformer are preferably wound on the same core
OR
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
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19. Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope in the normal adjustment position. Write two draw-backs of refracting type telescopes.
OR
(a) Define resolving power of a telescope. Write the factors on which it de-pends.
(b) A telescope resolves whereas a microscope magnifies. Justify the state-ment.
20. A jar of height h is filled with a transparent liquid of refractive index μ. At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when it is placed on the top surface symmetrically about the centre, the dot is invisible.
21. (a) In the photoelectric effect, do all the electrons that absorb a photon come out as photoelectrons irrespective of their location? Explain.
(b) A source of light, of frequency greater than the threshold frequency, is placed at a distance ‘d’ from the cathode of a photocell. The stopping potential is found to be V. If the distance of the light source is reduced to d/n (where
n>1), explain the changes that are likely to be observed in the (i) photoelectric current and (ii) stopping potential.
22. A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them?
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23. Binding energy per nucleon versus mass number curve is as shown. 1 2
1 2
A AA
Z Z ZS, W, X,
and 3
3
A
ZY are four nuclei indicated on the curve
Based on the graph:
(a) Arrange X, W and S in the increasing order of stability.
(b) Write the relation between the relevant A and Z values for the following nuclear reaction.
S → X + W
(c) Explain why binding energy for heavy nuclei is low.
OR
How are protons, which are positively charged, held together inside a nucleus? Explain the variation of potential energy of a pair of nucleons as a function of their separation. State the significance of negative potential energy in this re-gion?
24. A sinusoidal carrier wave of amplitude AC and angular frequency ωC is modulated in accordance with a sinusoidal information signal of amplitude Am
and angular frequency ωm. Show that the amplitude modulated signal contains three frequencies centred around ωC. Draw the frequency spectrum of the re-sulting modulated signal.
SECTION – D (Question numbers 25 to 27 carries 5 marks each)
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25. (a)Write the expression for the equivalent magnetic moment of a planer cur-rent loop of area A, having N turns and carrying a current I. Use the expression to find the magnetic dipole moment of a revolving electron.
(b) A circular loop of radius r, having N turns and carrying current I, is kept in
the XY plane. It is then subjected to a uniform magnetic field B→
= Bxi + By j + Bzk.
Obtain expression for the magnetic potential energy of the coil-magnetic field system.
OR
(a) A long solenoid with air core has n turns per unit length and carries a current I. Using Ampere’s circuital law, derive an expression for the magnetic
field B at an interior point on its axis. Write an expression for magnetic intensity H in the interior of the solenoid.
(b) A (small) bar of material, having magnetic susceptibility χ, is now put along the axis and near the centre, of the solenoid which is carrying a d.c. current through its coils. After some time, the bar is taken out and suspended freely with an unspun thread. Will the bar orient itself in magnetic meridian if (i) χ < 0 (ii) χ ˃ 1000?
Justify your answer in each case.
26. (a) There are two sets of apparatus of Young’s double slit experiment. In set
A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two set ups?
(b) Deduce the expression for the resultant intensity in both the above-mentioned set ups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength λ.
OR
(a) The two polaroids, in a given set up, are kept ‘crossed’ with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated.
Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of first and the third polar-oid. Draw a graph showing the dependence of I on ϴ.
(b) When an unpolarized light is incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarization, of re-flected and refracted light, under this condition?
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27. (a) Draw the circuit diagram to determine the characteristics of a p- n-p tran-sistor in common emitter configuration.
b) Explain, using I-V characteristics, how the collector current changes with the base current. How can (i) output resistance and (ii) current amplification factor be determined from the I-V characteristics?
OR
(a) Why are photodiodes preferably operated under reverse bias when the cur-rent in the forward bias is known to be more than that in reverse bias?
(b) The two optoelectronic devices: - Photodiode and solar cell, have the same working principle but differ in terms of their process of operation. Explain the difference between the two devices in terms of (i) biasing, (ii) junction area and (iii) I-V characteristics.
***
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SOLUTIONS
1. A Dielectric develops a net dipole moment in the presence of an external field whether it is polar or non-polar. The dipole moment per unit volume is called Polarization and is denoted by P. For
linear isotropic dielectrics,
P =XeE
where Xe is a constant characteristic of the dielectric and is known as the elec-tric susceptibility of the dielectric medium, so the SI unit of the electric polarization vector P is coulombs per square meter (Cm-2).
2. The resistivity of materials depends on the temperature at which they are operated. Different materials exhibit different dependences on temperatures. Over a limited range of small temperatures, the resistivity of a metallic con-ductor is approximately given by,
RT = R0 [1 + α(T– T0)], where RT is the resistivity at a temperature T and R0 is the same at a reference temperature T0, α is called the temperature co-efficient of resistivity.
3. The electromagnetic waves that are widely used as a diagnostic tool in medi-cine are X-Rays because it penetrates much into the muscle layers and is used to obtain images of bones which are not much penetrated by these rays.
OR
The current which can flow even in the absence of electric charge is the Dis-placement current because it flows in a region where there is a time-varying electric field. It can be seen by the Ampere-Maxwell’s law which is,
∮ B→
. d l→
= μ0iC + μ0ϵ0
dϕE
dt
Where,
iD = ϵ0
dϕE
dt
Is the displacement current.
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4. As we know that,
velocityoflightinmedium1
velocityoflightinmedium2=
sini
sinr,
Also, as per the graph given above, λ sini
sinr= tanθ
So, v1
v2= tanθ
5. Given: -
Two particles have equal momenta ‘p’
Formula: -
the de Broglie wavelength associated with a given particle is,
λ =h
p,
Where λ is the de-Broglie wavelength of the particle, h is the Planck’s constant and p is the momentum.
So, we can write the ratio as
λ1
λ2=
hp1
hp2
= 1
(∵p1 = p2 = p)
Conclusion: -
The ratio of de-Broglie wavelengths is 1.
OR
Given: -
Frequency of Monochromatic light, v = 6.0x 1014 Hz
Formula: -
We know that the energy of a photon is given by the equation,
E = h v
Where, E is the energy of the photon,
h is the Planck’s constant and,
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v is the frequency.
So, substituting the values we get,
E = 6.626 × 10-34 × 6 × 1014= 39.756 × 10-20 J.
Conclusion: -
The energy of the light is, E = 3.9756 × 10-19 J.
6.
The equivalent resistance between AB is 4×4
4+4= 2Ω,
between CD is 12×6
12+6= 4Ω,
So, the net resistance of the circuit is 2+1+4+1=8 Ω
OR
The equivalent resistance of the circuit is R + r Ω.
The current in the circuit as per ohm’s law is,
9
R + rA
So, we can write, 9
R + r= 5 … . (1)
Also, the terminal voltage across the resistor is 8 V, so,
R × 5 = 8,
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R =8
5Ω
Substituting the value of R in (1),
⇒9
85
+ r= 5
⇒9
5=
8
5+ r
⇒ r =1
5Ω
So, the internal resistance of the cell is 1
5Ω.
7.
As the potentiometer wire has a uniform cross section throughout, the potential
difference between X and any point at a distance l from A till point Y is,
V (l)= k l
Where k is the potential drop per unit length, As the positive terminal of E1 is connected to X and on touching the jockey near to the end X of the potenti-ometer wire, the galvanometer pointer deflects to left and on touching the jockey near to end Y of the potentiometer, the galvanometer pointer again
deflects to left but now by a larger amount, this indicates that the cell E1 is of larger emf than E which indicates that the galvanometer will never show zero deflection. For a galvanometer to show zero deflection somewhere on XY the emf of E > E1
OR
a) Formula: -
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The meter bridge uses the concept of a balanced Wheatstone bridge to find the value of the unknown resistor so that the formula can be written as,
RResistancebox
lAB=
x
1 − lAB
Where,
X is the unknown resistance,
R is the resistance of the resistance box,
LAB is the length of wire where till the null point is found.
It is written as to fulfill the condition for the balanced bridge.
b) The above formula can be re-written as
R − lABR = lABx
lAB =R
R + x
lAB =1
1 +xR
The above expression indicates that if the value of R is increased, then the value of lAB will increase.
Conclusions: -
The value of x can be found as,
x = R ×1 − lAB
lAB
The value of R increases then the value of lAB also increases.
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8. The figure clearly indicates that voltage is lagging the current by a phase dif-ference say Φ so that we can write as,
V(t)=V0 sinωt
Moreover,,
I(t)=I0 sin(ωt+ϕ)
At t=T
8, I=I0
I0 = I0sin (ωT
8+ ϕ),
Also, at t=T
4, V=V0
⇒ V0 = V0sin (ωT
4)
⇒ sin (ωT
4) = 1
⇒ωT
4=
π
2
⇒ ω =2π
T
Substituting the value of angular frequency in the current relation we get,
⇒ sin (2π
T×
T
8+ ϕ) = 1
⇒π
4+ ϕ =
π
2
⇒ ϕ =π
4
The corresponding phasors can be drawn as,
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9. i) Wavelengths of the incident solar radiation absorbed by the earth’s surface are much smaller than that of the radiation re-radiated by the earth as the energy received from the sun has a very broad spectrum containing the visible rays, the UV rays, the infrared rays having higher frequencies and thus smaller wavelengths.
ii) Tanning effect produced on the skin by UV radiation incident directly on the skin is much more than when it is coming through glass window because when the radiation is incident on the glass, some part it gets reflected and some gets transmitted so the intensity of the UV radiation reduces.
10. Given: -
The wavelength of incident radiation 6000 Å
The decrease in the wavelength is of 30%
Formula: -
We know that,
d sin θ = λ (First diffraction)
where d is the slit width,
θ is the angle of incidence,
λ is the wavelength of the incident light.
In the case of the small angle, sinθ≈θ,
So, d θ =λ
Then we can say half angular width will be,
θ =λ
d
Then, Full- angular width will be,
w = 2θ =2λ
d… (1)
For the second wavelength,
w′ =2λ′
d… (2)
So, by dividing the two equations, we get,
λ′
λ=
w′
w
λ′ =w′
w× λ
Substituting the values, we get,
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λ'=0.7×6000=4200A0
Conclusion: -
The wavelength of light will be 4200 Å.
11. Universal Gates are called so because by using these gates other basic gates like OR, AND and NOT are realized, and these gates are inexpensive to manu-facture and simple to construct large circuits.
The AND gate can be realized by NOR gates as follows,
12. Given: -
The height of the transmitting antenna, 20 m.,
The height of the receiving antenna, 25 m.,
Formula: -
The maximum line-of-sight distance dM between the two antennas having heights hT and hR above the earth is given by,
DM = (2 R hT)1/2 + (2 R hR)1/2
where hR is the height of receiving antenna, where hT is the height of the trans-mitting antenna and R is the radius of the earth (6400 Km approx.).
So, substituting the values, we get,
DM = (2 × 6400000 × 20)1/2 + (2 × 6400000 × 25)1/2
≈33888 m
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DM = 33.8 Km
Conclusion:
The Range of the Transmission is, DM = 33.8 Km.
13. Given
Charge Q=+15 μC
Charge q =+ 5 μC
The distance between resting position of charge q and Q=30 cm
Displacement of charge q due to charge Q= 15 cm
Formula used:
The kinetic energy= change in potential energy =UF-Ui
Uf is final potential energy, Ui is initial potential energy
Potential energy U =k×Q×q
r2
Where,
K is the constant its value=9 × 109 Nm2/c2
Q, q are the charges
R is the distance between the charges
(a) Finding the initial potential energy when q is at res
Putting the values of Q=15 μC, q=5 μC and r =30 cm in the equation(2), we
get
Ui =9 × 109 × 5 × 10−6C × 15C × 10−6
(0.3m)= 2.25J
Putting the value of Q=15 μC, q=5 μC and r=45 cm in the equation(2), we get final potential energy
Uf =9 × 109 × 15 × 10−6C × 5 × 10−6C
(0.45m)= 1.5J
The kinetic energy = Ui - Uf = 2.25 – 1.5 =0.75 J
(b) Finding the initial potential energy when q is at res
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Putting the values of Q=15 μ, q=5 μC and r=30 cm in the equation(2), we get
Ui =9 × 109 × 5 × 10−6C × (−15) × 10−6C
(0.3m)= −2.25J
Putting the value of Q=15 μC, q=5 μC and r=45 cm in the equation (2), we get final potential energy
Uf =9 × 109 × 15 × 10−6C × (−15) × 10−6C
(0.15m)= −4.5J
The kinetic energy = Ui - Uf = -2.25 – (-4.5) =2.25 J
Conclusion:
The final kinetic energy of the charge q(5 μC) when Q is 15 μC is 0.75 J.
The final kinetic energy of the charge q(5 μC) when Q is -15 μC is 2.25 J
14. a) As for a negatively charged infinite plane sheet having a uniform surface charge density the value of electric field is
E =σ
ϵ0
Which is independent of the distance from the sheet, so the dipole will be in stable equilibrium whether it is kept first to the left and then to the right.
b) The value of the electric field for an infinitely long straight wire having uni-form negative linear charge density λ is given as,
E =λ
2πϵ0r
Where r is the distance from the line charge.
As the value of electric field near the vicinity of the wire depends upon r the dipole will not be in stable equilibrium whether its direction is towards the wire or away from it.
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Note:
The electric field due an infinite line charge is derived using the Gauss’s law,
Flux through the Gaussian surfaces shown in the diagram aside
= flux through the curved cylindrical part of the surface
= E × 2 πrl
The surface includes charge equal to λ l. Gauss’s law then gives
E × 2πrl =λl
ϵ0
E =λ
2πϵ0r
15. a) As we know that the drift velocity depends on the current as,
vd =I
neA
Where n is the no. of charge carriers per unit volume, e is the charge of an electron, I is the current flowing and A is the area of cross section of the re-sistance.
As we know that,
nmetals>nsemiconductors
so, for the same value of the current and same area of cross-section, the drift velocity of electrons in metal resistance is smaller than in the n-type semicon-ductor.
b) If the same constant current continues to flow for a long time, the voltage drop across A will increase as the resistance of metal increases with increase in temperature but will decrease in B as the resistance of a semiconductor decreases with increase in temperature.
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16. Formula: -
We know that a charge q moving with velocity v in the presence of both electric and magnetic fields experiences a force (Lorentz’s Force) given by, i.e.,
F→
= q (E→
+ v→
× B→
) … (1)
Where E is the Electric field
B is the magnetic field
V is the velocity of the charged particle and,
q is the charge on the particle.
Let us consider that the velocity of the particle is perpendicular to both electric
and magnetic fields and they are mutually perpendicular also, depicted in the figure.
We have,
E→
= Ej,
B→
= Bk, and,
v→
= vi
Substituting the values in the equation (1), we get,
A = πr2F→
= q(E − vB)j
Therefore,
F→
= q(E − vB)j
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Thus, electric and magnetic forces are in opposite directions. The charge will move in the fields undeflected only when the magnitudes of the two forces are equal. Then, the total force on the charge is zero, and this happens when,
qE=qvB,
vC =E
B
If the velocity v <vC, then,
v<vC
v <E
B
E-vB<0
As
F→
= q(E − vB)j
So, the force is in the positive y-direction so there will be an acceleration in the positive y-direction.
Conclusions: -
The expression for the velocity vC of positive ions passing undeflected through a region where crossed and uniform electric field E and magnetic field B are simultaneously present is,
vC =E
B
In the second part there will be an acceleration in the positive y direction
OR
Formula: -
When a charge q moves with a velocity v in the vicinity of both electric and magnetic fields, it experiences a force which is called the Lorentz’s Force and mathematically is,
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F→
= q (E→
+ v→
× B→
)
the magnetic field due to the wire above the x-z plane is,
B→
= Bkˆ
So,
F→
B = qvB(i × k) = −qvBj
In order to balance this force, we need a force due to electric field as,
F→
E + F→
B = 0
⇒ F→
E = −(−qvBj) = qvBj
Also, we know that,
F→
E = qE→
So,
⇒ F→
E = qE→
= qvBj
⇒ E→
= vBj
⇒ E = v ×μ0
2π×
I
r[∵ B =
μ0
2π×
I
r]
Conclusion: -
The magnitude of the electric field is,
E =vμ0I
2πr
Moreover, the direction is in the positive y-axis.
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17.
Given:
Voltage, V=10 v
Resistance, R= 10 Ω
Inductive impedance=1 H
Capacitor =1 uF
Fomula used:
1. Angular velocity, w0=1
√LC … (1)
Where,
L is the inductive impedance
C is capacitor impedance.
2. Peak current,Ipeak =Vpeak
R … (2)
Where,
Vpeak is the peak current
R is the resistance
3. Voltage across inductor, Vinductor=Ipeak × w0L … (3)
Where,
Ipeak is the peak currents
w0 is the angular frequency
L is the inductive impedance
4.The quality factor of the circuit is given as,
Q =1
R× √
L
C… (4)
Where,
Q is the quality factor,
R is the resistance,
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C is the capacitance, and L is the inductance.
Putting the value in equation 1, we get
ω0 =1
√LC=
1
√1 × 1 × 10−6= 103rads−1
The net current in circuit only depends upon the resistance as the circuit is in resonance,
Putting the value in equation 2
Ipeak =Vpeak
R=
10
10= 1A
Now the rms voltage drop across the inductor is given as,
Putting the value in equation 3, we get
Vinductor = Ipeak × XL = Ipeak × ω0L
V = 1 × 103 × 1 = 1000V.
Also, the quality factor Q (which is ratio of reactance to resistance) of the given series LCR circuit given by equation 4 is as,
Q =1
10√
1
10−6
Q =1
10−2= 102
So, Q = 100.
Conclusion: -
The peak voltage across the inductor at resonance is 1000 V.
The quality factor of the circuit is, Q = 100.
18. a) Transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as shown in the diagram 1 or on separate
28 | P a g e
limbs of the core as shown in the diagram 2. One of the coils called the primary coil has NP turns. The other coil is called the secondary coil; it has NP turns.
b) Lamination is done to reduce the eddy current loss by increasing the re-
sistance of the core. The core is made up of thin sheets of steel, therby having relatively high resistance, each lamination being insulated from others by a thin layer of varnish. The planes of these sheets are placed perpendicular to the direction of the current that would be set up by the induced emf. The planes of these sheets are arranged parallel to the magnetic fields so that they can cut across the eddy current paths. Thus, the laminated sheet will have an eddy current that circulates within it but the sum of individual eddy currents of all such laminations are very less compared to that of using single solid iron core.
c) The primary and secondary coils of a transformer are preferably wound on
the same core so as to reduce the flux losses and thereby increasing the effi-ciency of the transformer.
OR
Formula: -
The total energy in the system can be written as,
Etotal(t) =L[i(t)]2
2+
[q(t)]2
2C
Where L is value of inductance, C is the value of capacitance, i(t ) and q(t), are the current and charge depending upon the time respectively. Calculations: -
We can write,
Etotal(t) =L(i0sinω0t)2
2+
(q0cosω0t)2
2C
Etotal(t) =L
2i02sin2ω0t +
q02
2Ccos2ω0t
Also, we know that,
29 | P a g e
i0=ω0 q0
And,
ω0 =1
√LC
Which gives,
⇒ q0=i0 √(LC)
Using the above result, we can write,
Etotal(t) =Li0
2
2(sin2ω0t + cos2ω0t) =
Li02
2
So, the total energy is,
Etotal(t) =Li0
2
2
Which is independent of time and so is a constant.
Conclusions: -
The total energy of the system at any given time is,
Etotal(t) =Li0
2
2
which is a constant so the total energy remains constant all the time.
19. Ray diagram of the image formation in a refracting type astronomical telescope in the normal adjustment position,
The two main drawbacks with an astronomical telescope are: -
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1. Refracting telescopes of practical applications are made of very large and thick lenses, such big lenses are very heavy and therefore, difficult to make and support by their edges.
2. Manufacturing such large sized lenses is difficult and expensive which form images that are free from any kind of chromatic aberration and distortions.
OR
a) Resolving power of a telescope is defined as the ability of a telescope to distinguish two close images as being separate. An example of resolving power is how well a telescope can show two stars as being separate stars.
It depends upon the area of the objective and on the diameter of the objective.
b) A microscope produces magnified images of very small near objects. On the other hand, a telescope produces image of far objects nearer to our eye. Ob-jects which are not resolved at far distance, can be resolved by telescope.
20. Given: -
Height of the jar is h,
Transparent liquid of refractive index μ,
There is a dot at the center of bottom surface of the jar.
The problem involves the phenomenon of Total Internal Reflection,
Formula: -
From the required condition for TIR we know that,
Sin i≥μ,
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Where,
i is the incident angle,
μ is the refractive index of the medium
So, from geometry we can write as,
tan i =R
h
Where, h is the height of the jar,
R is the radius of the disc,
i is the incident angle,
So,
sin i =R
√R2 + h2
Using the above expressions, we can write, R
√R2 + h2≥ μ
⇒R2≥μ2 R2+μ2 h2
⇒R2 (1-μ2 )≥μ2 h2
⇒ R2 ≥μ2h2
(1 − μ2)
⇒ R ≥μh
√1 − μ2
Conclusion: -
The minimum radius of the Disc is,
Rmin =μh
√1 − μ2
21. a) It is not necessary that if the energy supplied to an electron is more than the work function it will come out. It comes out of the metal atom only when
it absorbs a single photon with energy greater than or equal to the work func-tion. The electron after receiving energy may loose its energy due collisions with other metal atoms. So most of them get scattered into the metal. Only a few electrons near the surface may come out of the surface of the metal for whom the incident energy is greater than the minimum threshold energy to loose an electron (work function) of the metal. So, it is not necessary that all
32 | P a g e
the electrons that absorb a photon come out as photoelectrons irrespective of their location.
b) When the light source is brought near the photocell, the intensity of the light increases, which doesn’t effects the stopping potential but the photo-current increases, the following diagram justifies the above statements,
In the plot, I denotes the intensity of the light.
22. Given: -
The wavelength of the monochromatic radiation, λ=975 A0
Ionization energy for the hydrogen atom= 13.6 eV
Formula used: -
The energy of monochromatic radiation of wavelength 975 A0 is given by the formula,
E =hc
λ
Where, E is the energy of the incident photon,
h is the Planck’s constant,
λ is the wavelength of the incident radiation and,
c is the speed of light.
So, after substituting the given values we get,
E =6.63 × 10−34 × 3 × 108
975 × 10−10 × 1.6 × 10−9eV
33 | P a g e
E = 12.75 eV
Also, by using the formula for energy of electron of hydrogen atom in nth shell, we can write,
ΔE = 13.6 × Z [1
nf2 −
1
ni2] eV
Where, nf is the final principal quantum number and ni is the initial quantum number and Z is the atomic mass (for hydrogen Z=1),
12.75 = 13.6 [1
12−
1
n2]
1
n2= 1 −
12.75
13.6=
1
16
n2 = 16 ⇒ n = 4
Thus, the number of lines possible in the resultant spectrum is 6
The transition from n=4 to n=3 corresponds to the emission of minimum en-
ergy thus has maximum wavelength.
Conclusion: -
The number of lines possible in the resultant spectrum is 6.
The transition from n=4 to n=3 corresponds to the emission of minimum en-ergy thus has maximum wavelength.
23. a) The increasing order of stability is X > W > S.
b) S → X + W
the relation between the relevant A and Z values are,
A = A2 + A1
Z = Z2 + Z1 ,
c) Binding energy is the minimum energy required to separate every particle from a system of particles. A nucleon inside a fairly large nucleus is under the influence of only some of its neighbors, which come within the range of the nuclear force as seen from the potential energy v/s separation diagram of two nucleons.
34 | P a g e
When the distance between two nucleons increases more than few femtometers, the nuclear force acting between them rapidly reduces to zero. This small range of distance is the reason of saturation of forces in a medium or a large-sized nucleus. The nuclear force will have no influence on the binding energy of the nucleon under consideration and the coulomb forces will increase the repulsion between the two.
If Atomic number increases by the addition of nucleons, they will not change the binding energy of a nucleon inside, since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. Thus, the binding energy for heavy nuclei are
generally found to be low.
OR
Protons, which are positively charged, held together inside a nucleus by the virtue of a special kind of force called the nuclear force which overcomes the repulsive electric force between the positively charged particles. The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force dominates over the Coulomb repulsive force between protons inside the nucleus
The variation of potential energy of a pair of nucleons as a function of their separation is as shown in the diagram.
When the distance between two nucleons increases more than 2 femtometers,
the nuclear force acting between them rapidly reduces to zero. This small range of distance is the reason of saturation of forces in a medium or a large-sized
nucleus. This leads in steadiness in the binding energy per nucleon.
The potential energy is at its minimum at a distance of about 0.8 femtometers between the nucleons which indicates that the force is attractive for distances larger than 0.8 fm. Nuclear force becomes repulsive if the distance becomes less than 0.8 fm. The nuclear force acting in neutron-neutron, proton-neutron and proton-proton interaction is approximately the same as it does not depend on the electric charge of the nucleon.
35 | P a g e
24. The carrier wave is,
c(t)=Ac sinωc t
The message signal is,
m(t)=Am sinωm t
So the modulated signal cm(t) can be written as,
cm (t)=(Ac+Am sinωm t)sinωc t
cm(t) = Ac (1 +Am
Acsinωmt) sinωct
we can write,
cm (t)=Ac sinωc t+μAc sinωc tsinωm t
Here,
μ =Am
Ac
Is called the modulation index, Using the trigonometric relation,
sinAsinB =1
2(cos(A − B) − cos(A + B)),
we can write,
cm(t) = Acsinωct +μAc
2cos(ωc − ωm)t −
μAc
2cos(ωc + ωm)t
Here ωc-ωmand ωc+ωmare respectively called the lower side and upperside fre-quencies. The modulated signal now consists of thecarrier waveof frequency ωc plus two sinusoidal waves each with a frequency slightly different from, known as side bands. The frequency spectrum of the amplitude modulated signal is shown below,
25. a) The required expression is,
M = n I A,
36 | P a g e
Where m is the magnetic dipole moment, n is the no. of turns, I is the current and A is the cross-section area of the loop.
the magnetic dipole moment of a revolving electron is found as,
The electron of charge (–e) (e = + 1.6 × 10-19 C) performs uniform circular
motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I, where
I =e
T
and T is the time period of revolution. Let r be the orbitalradius of the electron, and v the orbital speed. Then,
T =2πr
v
Which gives,
I =ev
2πr
So the magnetic dipole moment can be written as,
μl = I × πr2 =ev
2πr× πr2 =
evr
2
We can also write,
μl =e(mevr)
2me
Where me is the mass of the electron and l is angular momentums so,
μl =e
2mel
b) The magnetic dipole moment of the loop is given as,
m→
= ±N × πr2 × Ik
(± is used the direction of current is not given whether it is clockwise or anti-clockwise)
37 | P a g e
So, the magnetic potential energy is given as,
E = m→
. B→
Substituting the values,
E = ±Nπr2Ik. (Bxi + By j + Bzk)
E=±Nπr2 IBz
Conclusion: -
The magnetic potential energy of the system is E = Nπr2IBz
OR
a) A long wire wound in a close-packed helical structure carrying a current ‘I’
is called a solenoid. For practical applications, the length of the solenoid is taken much greater than its diameter. The magnetic field in the interior space of the solenoid can be found as the vector summation of all vector fields pro-duced by the different individual turns which constitute the overall structure of the solenoid.
Magnetic field B is nearly uniform and parallel to the axis of the solenoid at
interior points near its center and external field near the center is very small.
To find the net magnetic field intensity, assume a rectangular Amperian loop abcd. Along cd the field is zero. Alongside crosswise sections bc and ad, the field component is zero. Thus, these two sections make no contribution. Let
the field along ab be B. Thus, the appropriate length of the Amperian loop is, L = h.
Let n be the number of turns per unit length, then the total number of turns is nh.
The enclosed current is, Ie = I (n h), where I is the current in the solenoid. From Ampere’s circuital law,
BL = μ0Ie,
38 | P a g e
B h = μ0(Inh) ,
B = μ0 n I
The direction of the field is given by the right-hand rule.
The magnetic intensity H, inside a solenoid with air inside is given as,
H→
=B→
μ0− M
→
Where, M→
is the magnetization of the material and for air M→
is zero, so,
H =B
μ0
H =μ0nI
μ0
H=nI
b) i)for χ<0, the material is diamagnetic.
The materials which are diamagnetic in nature show weak, negative sensitivity towards magnetic fields. These materials are weakly repelled by magnetic fields and does not preserve the magnetic properties when the external field is removed. The bar will not orient itself in magnetic meridian.
ii) for χ>1000, the material is ferromagnetic.
The materials which are ferromagnetic in nature show huge, positive sensitivity towards magnetic fields. These materials are strongly attracted by magnetic fields and preserve the magnetic properties when the external field is removed. Yes the bar will orient itself in magnetic meridian.
26. a) In Set A, as the phase difference between the two waves emanating from the slits does not change with time so alternate bright and dark bands would be seen on the screen as per Young’s double slit experiment as two slits are acting as coherent sources whereas in Set B, the Phase difference changes rapidly with time, the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent, when this happens, the intensities on the screen will add up and only bright light will be seen.
b) Given: -
The waves emanating from the two slits have the same amplitude A and same wavelength λ
39 | P a g e
Derivation: -
Let the displacement produced by wave emanating from slit 1 be y1 and from slit 2 be y2
So, we can write,
y1=Acosωt
And,
y2=Acos(ωt+ϕ)
Where ϕ is the phase difference between the two sources, and A is the ampli-
tude,
the resultant displacement will be given by,
y = y1 + y2
so, after substituting the values we get,
y = Acosωt + Acos(ωt + ϕ)using the trigonometric identity,
cosA + cosB = 2cosA + B
2cos
A − B
2
We get,
y = A (2cosϕ
2cos (ωt +
ϕ
2))
So, the amplitude of the resultant displacement is,
A′ = 2Acosϕ
2
As we know that the intensity is directly proportional to the square of the am-plitude, so,
I0∝A2
And
I∝A'2
So, we can write,
I
I0=
A′2
A2
Which gives,
I = 4I0cos2ϕ
2
Conclusions: -
The resultant intensity is given as,
40 | P a g e
I = 4I0cos2ϕ
2
For set A,
As ϕ is constant, the value of I remains constant for a particular band on the
screen, which leads to formation of alternate bright and dark bands.
For set B,
As ϕ is changing rapidly with time, the value of I changes rapidly with time on
the screen. The value of I changes so rapidly from 0 to 4I0 and again back to zero so rapidly, so it seems to the human eye that the intensity is constant as I = 4I0 and it seems that constructive inference is occurring.
OR
a) Given: -
The intensity of the incident light I0
Derivation: -
After passing the first polaroid (P1) the intensity be I1
We know that I1 reduces to half
After passing the third polaroid (P3) the intensity be I2
We can write by malus’ law as,
I2 =I0
2cos2θ
Also the angle between the pass axis of P3 and P2 is 90 - ϴ,
So the intensity of light transmitted through P2 is,
I3 = I2cos290 − θ I0
2cos2θsin2θ
I0
8(2cos θsin θ)
2
I3 =I0
8sin22θ
41 | P a g e
a) When unpolarized light is incident on a plane glass surface, the reflected light is polarized with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. Thus, when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarized wave. The angle of incidence in this case is called Brewster’s angle and is denoted by iB. We can see that iB is related to the refractive index of the denser medium.
Since we have,
iB + r =π
2
from Snell’s law we get,
μ =siniB
sinr=
siniB
sin (π2 − iB)
μ =siniB
cosiB= taniB
So, finally,
μ=taniB
The above expression is called the Brewster’s Law.
The state of the above light is partially polarized because When such light is viewed through a rotating analyzer, one sees a maximum and a minimum of intensity but not complete darkness.
27. a) The circuit diagram to determine the characteristics of a p-n-p transistor is drawn below: -
42 | P a g e
b) The output I-V characteristics obtained by observing the variation of IC (col-
lector current) as VCE is varied keeping IB (base current) constant are shown below. If VBE is raised by a small amount, the hole current from the emitter region raises. Also, the electron current from the base region raises. This leads to increase in both IB and IC proportionately.
If IB increases, IC also increases. The plot of IC versus VCE for different fixed
values of IB gives one output characteristic. So, there will be different output characteristics corresponding to different values of IB as shown,
43 | P a g e
i) Determination of Output Resistance: -
Output resistance (r0) is defined as the ratio of change in
collector-emitter voltage (ΔVCE) to the change in collector current (ΔIC) at a constant base current IB. So, it can be written as,
r0 = (ΔVCE
ΔIC)
IB
i) Determination of Current Amplification Factor: -
Current Amplification factor (β) is defined as the ratio of the change in collector
current (IC) to the change in base current (IB) at a constant collector-emitter voltage (VCE) when the transistor is in active state. So, mathematically,
βAC = (ΔIC
ΔIB)
VCE
βAC also known as small signal current gain and its value is found generally very large. The ratio of IC and IB gives the DC β of the transistor. Hence,
βDC =IC
IB
Since IC increases with IB almost linearly and IC = 0 when IB = 0, the values of both βDC and βAC are approximately equal.
OR
a) Photodiodes are preferably operated under reverse bias when the current in the forward bias is known to be more than that in reverse bias because the photodiode converts incident light to electric current efficiently in reverse bias condition than in forward bias. This is due to the expansion of the depletion
44 | P a g e
region of the diode. Photons after absorption generate electron hole pairs. The pairs generated in and around the region only contribute the electric current. In the forward bias condition, there is a strong electric field there to separate the two different charge carriers, so the pairs outside the region rapidly re-combine and vanish also the width of depletion region decreases as there is an increase in the applied voltage so the effective number of photons contributing to the electric current decreases. But in reversed bias condition, the width of depletion region increases as there is an increase in the applied voltage, so, a large number of incident photons get converted into electric current, conse-quently the efficiency increases.
b) i) On the basis of biasing
A photodiode converts incident light to electric current in reverse bias condition whereas a solar cell is basically a diode which generates emf when solar radi-ation falls on the p-n junction. It works on the same principle (photovoltaic effect) as the photodiode, except that no external bias is applied.
ii) On the basis of junction area
photodiode is a simple p-n junction diode which is operated in reversed bias condition whereas a solar cell is a p-n junction diode in which the junction area is kept much larger for solar radiation to be incident and is not biased.
iii) On the basis of I-V characteristics
The I – V characteristics of a photodiode is drawn same as of a p-n junction diode in reverse bias condition as shown below
45 | P a g e
The I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes as shown in the diagram.
This is because a solar cell does not draw current but supplies the same to the load.
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