CBSE 12th Physics 2017 Guess Paper By 4ono 12th Physics 2017 Guess Paper By 4ono.com ... This is...
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CBSE 12th Physics 2017 Guess Paper
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CBSE 12th Physics 2017 Guess Paper By 4ono.com
TIME - 3HR. | QUESTIONS - 26
THE MARKS ARE MENTIONED ON EACH QUESTION _________________________________________________________________________
SECTION โ A
Q.1. Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? 1 mark
Ans. Since, the magnetic field induction outside the toroid is zero.
Q.2. A plot of magnetic flux (๐) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of
self-inductance? 1 mark
Ans. Since ๐ = LI โ L = ๐
1 = slope of ๐ โ I graph โด slope of inductor A = slope of
inductor B. Hence the inductor A has larger value of self-inductance.
Q.3. Show graphically, the variation of the de-Broglie wavelength (ฮป) with the potential (V) through which an electron is accelerated from rest. 1 mark
Ans.
Q.4. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 ฮฉ as shown in the figure. Find the value of the current in circuit. 1 mark
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Ans. Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by
= 200 -10 = 190v.
Hence, the current in the circuit is given by ๐ผ =ฮต
๐ .
๐ผ =190
38= 5๐ด
Q.5. Define electric dipole moment. Write its S.I. unit. 1 mark
Ans. Electric dipole moment: Dipole moment ( ) is a measure of strength of electric dipole. It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and the distance between them. Si unit of dipole moment is coulomb-meter (cm).
SECTION - B
Q.6. How does the resistivity of a conductor depend upon temperature electrical conductivity? 2 marks
Ans. (i) The resistivity of a conductor increases with increase with increase in temperature
โด ๐๐ = ๐0[1 + ๐ผ(๐ โ ๐0)]
(ii) The resistivity of a conductor is the reciprocal of electrical conductivity
โด ๐ =1
๐
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Q.7. In the meter bridge experiment, balance point was observed at J with
AJ = l. 2 mark
(i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected? 2 mark
Ans. (i) the value of R and X were doubled and then interchanged. Hence the new position of balance point will 100 โ ๐ผ.
(ii) ๐ด๐ฝ = ๐.
From the principle of Wheat Stones Bridge,
๐
๐=
๐
100 โ ๐
๐ = ๐ (100 โ ๐
๐)
Hence, the galvanometer and tell are interchanged, the condition for a balance bridge is still satisfied. Therefore, the galvanometer will not show any deflection.
Q. 8. State Kirchhoff's rules. Explain briefly how these rules are justified. 2 marks
(a) Kirchhoffโs First law:
Ans. Junction rule: The algebraic sum of all the emf meeting at a point in an electrical circuit is always zero.Let the currents be ๐ผ1, ๐ผ2, ๐ผ3 ๐๐๐ ๐ผ4
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Convention: Current towards the junction is always positive, while current away from the junction is negative
๐ผ3 + (โ๐ผ1) + (โ๐ผ2) + (โ๐ผ) = 0
Kirchhoffโs Second law: Loop rule
In a closed loop, the algebraic sum of the emfโs is equal to the algebraic sum of the products of the resistance and current flowing through them.
For closed part BACB, ๐ธ1 โ ๐ธ2 = ๐ผ1๐ 1 + ๐ผ2๐ 2 โ ๐ผ3๐ 3
For closed part CADC, ๐ธ2 = ๐ผ3๐ 3 + ๐ผ4๐ 4 + ๐ผ5๐ 5
Demonstration:
Wheatstone Bridge: - The Wheatstone Bridge is an arrangement pf four resistance as shown in the following figure.
๐ 1, ๐ 2, ๐ 3 ๐๐๐ ๐ 4 are the four resistances.
Galvanometer(a) has a current ๐ผ๐flowing through it at
balanced condition ๐ผ๐ = 0
Applying junction rule at B,
โด ๐ผ2 = ๐ผ4
Applying junction rule at D,
โด ๐ผ1 = ๐ผ3
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Applying loop rule to closed loop ADBA,
โ๐ 1 + ๐ + ๐ผ1๐ 2 = 0
โด ๐ผ1๐ผ2
=๐ 2
๐ 1โฆ(๐)
Applying loop rule to closed loop CBDC,
๐ 1๐ 1 + ๐ โ ๐ 1๐ 1 = ๐ [๐ผ3 = ๐ผ1, ๐ผ4 = ๐ผ2]
โด ๐ 2
๐ 1=
๐ 4
๐ 3โฆ(๐๐)
From Eq. (i) and (ii)
๐ 2
๐ 1=
๐ 4
๐ 3
This is the required balanced condition of wheatstone bridge.
Q.9. Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions ๐พ๐ and (๐พ๐ > ๐พ๐). On what factors does the (i) slope and (ii) intercept of the lines depend? 2 mark
Ans. The graph showing the variation of stopping potential (๐0) with the frequency of incident radiation (๐ฃ0) for two different photosensitive materials having work functions W1 and W2(W1>W2) is shown in fig.
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(i) Slope of the line = โV
โ๐ฃ =
โ
๐
[โด ๐โV = h โ๐ฃ] โด Slope of the line depends on the Planckโs constant h and the electronic charge ๐.
(ii) Intercept of graph A on the potential axis
=work function(W)
๐= โ
โ๐ฃ0
๐
โด Intercept of the line depends upon Planckโs constant h, threshold frequency (๐0) and the electronic charge (e).
Q.10. Using Rutherfordโs model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron? 2 mark
Ans. From Rutherfordโs model of the atom, the magnitude of this force is
๐น =1
4 ๐๐0 .2๐. (๐๐)
๐2
For hydrogen atom, Let, ๐น๐ โCentripetal force required to keep a revolving electron in orbit. Fe-Electrostatic force of attraction between the revolving electron and the nucleus. Then, for a dynamically stable orbit in hydrogen atom, where Z = 1
๐น๐ = ๐น๐
๐๐ฃ2
๐=
(๐)(๐)
4๐๐0๐2 โฆ (๐)
๐ =๐
4๐๐0๐๐ฃ2 โฆ (๐๐)
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K.E. of electron in the orbit,
K =1
2๐๐ฃ2,
Form equation (๐),
K =๐2
8๐๐0๐
Potential energy of electron in
Orbit, U = (๐)(๐)
4๐๐0๐=
โ๐2
4๐๐0๐
โด Total energy of electron in hydrogen atom
E = k + U = ๐2
8๐๐0๐โ
๐2
4๐๐0๐; E = โ
๐2
8๐๐0๐
Here, negative sign indicates that the revolving electron is bound to the positive nucleus.
OR
Using Bohrโs postulates of the atomic model, derive the expression for radius of
nth electron orbit, thus obtaining the expression for Bohrโs radius.
Ans. Form de-Broglie hypothesis, wavelength associated with electron
๐ =โ
๐๐ฃ
๐๐ฃ =โ
๐
Substituting this value in ๐๐ฃ๐ = ๐h
2๐,
we get, โ
๐ ๐ = ๐
h
2๐
2๐ ๐ = ๐๐
๐. ๐., circurmference (๐ = 2๐๐) of nth permitted orbit for the electron can contains exactly ๐ wavelength of de-Broglie wavelength associated with electron in that orbit.
SECTION - C
Q.11. A convex lens made up of glass of refractive index 1ยท5 is dipped, in turn, in (i) a medium of refractive index 1ยท65, (ii) a medium of refractive index 1ยท33.
(a) Will it behave as a converging or a diverging lens in the two cases?
(b) How will its focal length change in the two media? 3 marks
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Ans. Here,
๐๐ = 1.5๐
Let ๐๐๐๐ be the focal length of the lens in air, Then, 1
๐๐๐๐( ๐๐
๐ โ 1) (1
๐ 1โ
1
๐ 2)
or (1
๐ 1โ
1
๐ 2) =
1
๐น๐๐๐( ๐๐๐ โ1)
=1
๐๐๐๐(1.5โ1)
Or (1
๐ 1โ
1
๐ 2) =
1
๐น๐๐๐ โฆ. (i)
(i) When lens is dipped in medium A Here, ( ๐๐
๐ = 1.65)
Let ๐น๐ด be the focal length of the lens, when dipped in medium A. Then,
1
๐น๐ด( ๐๐
๐ โ 1) (1
๐ 1โ
1
๐ 2)
= ( ๐๐
๐
๐๐๐ โ 1)(
1
๐ 1โ
1
๐ 2)
using the equation (i), we have
1
๐๐ด= (
1.5
1.65โ 1)ร
1
๐๐๐๐= โ
1
5.5๐๐๐๐
As the sign of ๐๐ดis opposite to that of ๐๐๐๐ the lens will behave as a diverging lens.
(ii) When lens is dipped in medium B:
Here, ๐๐๐ = 1.33
Let ๐น๐ตbe the focal length of the lens, when dipped in medium B. Then,
1
๐น๐ต( ๐๐ โ 1๐ต ) (
1
๐ 1โ
1
๐ 2) = (
๐๐๐
๐๐๐ โ 1)(
1
๐ 1โ
1
๐ 2)
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Using the equation (i), we have
1
๐น๐ต= (
1.5
1.33โ 1)ร
2
๐๐๐๐=
0.34
1.33๐๐๐๐
Or ๐๐ต = 3.91 ๐๐๐๐
As the sign of ๐๐ตis same as that of ๐๐๐๐ the lens will behave as a converging lens.
Q.12. (a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm? 3 marks
(b) Which transition corresponds to emission of radiation of maximum wavelength?
Ans. (a) For element A
Ground state energy, ๐ธ1 = โ2๐๐
Exicted state energy, ๐ธ2 = 0 ๐๐
Energy of photon emitted, E = ๐ธ2 โ ๐ธ1
= 0 โ (โ2) = 2๐๐
โด Wavelength of photon emitted,
๐ =โ๐
๐ธ=
6.626ร10โ34ร3ร108
2ร1.6ร10โ19=
19.878ร108
3.2
6.211ร10โ7๐ = ๐๐๐. ๐๐ง๐ฆ
For element B
๐ธ๐ = โ4.5 ๐ ๐, ๐ธ2 = 0๐ ๐
๐ธ = 0 โ (โ4.5) = 4.5 ๐ ๐
โด ๐ =6.626ร10โ34ร3ร108
4.5ร1.6ร10โ19
19.878ร10โ7
7.2= 2.760ร10โ7 = ๐๐๐ ๐ง๐ฆ
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For element C
๐ธ๐ = โ4.5 ๐ ๐, ๐ธ2 = โ2๐ ๐
๐ธ = โ2 โ (โ4.5) = โ2 + 4.5 = 2.5 ๐๐
โด ๐ =6.626ร10โ34ร3ร108
2.5ร1.6ร10โ19
19.878ร10โ7
4= 4.969ร10โ7๐ = ๐๐๐. ๐ ๐ง๐ฆ
For element D
๐ธ๐ = โ10 ๐ ๐, ๐ธ2 = โ2๐ ๐
๐ธ = โ2 โ (โ10) = 8 ๐๐
โด ๐ =6.626ร10โ34ร3ร108
8ร1.6ร10โ19
=19.878ร10โ7
12.8= 1.552ร10โ7๐ = ๐๐๐. ๐ ๐ง๐ฆ
โด Element B has a proton of wavelength ๐๐๐ ๐ง๐ฆ
(b) Element A has radiation of maximum wavelength 621nm
Q.13. An air solenoid of length 0.3m, area of cross section is 1.2 x ๐๐โ๐๐๐and has 2500 turns. Around its central section, a coil of 350 turns is wound. The solenoid and the coil are electrically insulated from each other. Calculate the emf induced in the coil if the initial current of 3A in the solenoid is reversed in 0.25s. 3 marks
Ans. N1 = 2500
N2 = 350
A = 1.2 ร10โ3๐2
l = 0.3m.
dl = 3-(-3) = 3+3 = 6A
dt = 0.25s
Since Mutual inductance,
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๐ = ๐0๐1๐2๐ด
๐=
4๐ร10โ7ร2500ร350ร1.2ร10โ3
0.3
=4ร3.14ร1.05ร10โ4
0.3= 43.96ร10โ4 = 4.39ร10โ3H
Induced ๐๐๐|E| = MdI
dt
=4.39ร10โ3ร6
0.25= 105.36ร10โ3 = 0.10536 V.
Q.14. Draw a labelled ray diagram of a refracting telescope. Define its magnifying power write the expression for it. Write two important limitations of a refracting telescope over reflecting type
telescope. 3 marks
Ans. Refracting telescope:
Magnifying power- The magnifying power is in the ratio of the angle โ subtended at the eye by the final image to the angle ๐ฝ which the object subtends at the lens or the eye.
๐ โ๐ฝ
โโ
โ
๐๐.๐0โ
=๐0๐๐
Limitations of refracting telescope over the reflecting type telescope โ
(i) Refracting telescope suffers from chromatic aberration as it uses large sized lenses.
(ii) The requirements of big lenses tend to be very heavy and therefore difficult to make and support by their edges.
Q.15. In a Geiger-Marsden experiment, calculate the distance of closet approach to the nucleus of ๐ = ๐๐, when an ๐ถ-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closet approach be affected when the kinetic energy of the ๐ถ-particle is doubled? 3 marks
Ans. Z=80, KE=8MeV.
Potential energy =๐พ๐๐2
๐0=
1
2 ๐1๐ฃ0
2
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๐0 =๐พ๐๐2
(๐2๐0
2
2)
=9 ร 109 ร 80 ร (1.6ร 1019)2
8 ร 1.6 ร 10โ13
[โต 1๐๐๐ = 1.6ร10โ13๐ฝ]
=18 ร 18 ร 109 ร 1.6 ร 1010
8 ร 106= 128.8 ๐๐
Since ๐0๐ผ1
๐ธ๐
So, when kinetic energy is doubled the distance of closet to halved. If, the kinetic energy is doubled the distance of ๐ผ- particles is halved.
OR
The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition form an energy level -0.85 eV to -3.4eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?
Ans. ๐ธ๐ = โ13.6
๐2 ๐๐ฃ. Here ground state energy for ๐ = 1, ๐ธ1 = โ13.6 ๐๐
Now electron transits form ๐ธ๐ = โ0.85๐๐ ๐ก๐ ๐ธ๐ = โ3.4๐๐
โ0.85 =โ13.6
๐๐2
๐๐2 =
13.6
0.85= 16
Thus,
๐๐ = 4
Again, โ3.4 = 13.6/๐๐2
๐๐2 =
13.6
3.4= 4
๐๐ = 2
Thus electron makes transition from n = 4 to n = 2. Hence, it is blamer series. Now
๐ = 1.0974 ร 107
1
๐= ๐ (
1
22โ
1
๐2)
1
๐= 1.0974ร 107
(1
22โ
1
42) =
1.09ร107ร12
4ร16
1
2
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1
๐= 0.2057 ร 107
๐ = 4.861 ร 107
๐ = 4861
intensity of light after passing through second polarizer ๐2 is given by ๐ผ = ๐ผ0 ๐๐๐ 2๐ .
Q.16. An illuminated object and a screen are placed 90cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object. 3 marks
Ans.
m =๐ฃ
๐ขโ 2 =
๐ฃ
๐ขโ ๐ฃ = 2๐ข โฆ . . (๐๐)
Putting the value of v in (i), we get
u + 2u = 90 โ ๐ข =90
30= 30
โด ๐ฃ = 2ร30 = 60
Using lens formula, we get
1
๐=
1
๐ฃโ
1
๐ขโ
1
๐=
1
60โ
1
โ30
โ1
๐=
1
60+
1
30โ
1
๐=
1 + 2
60=
3
60
โ ๐ =60
3โด ๐ = 20 cm.
Q. 17. (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture ๐ร๐๐โ๐๐.The distance between the slit and the screen is 1.5.m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases. 3marks
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Ans.(a). If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of two slits.
(b). Given that: Wavelength of the light beam
๐1 = 590๐๐ = 5.9ร10โ7๐
Wavelength of another light beam,
๐2 = 596๐๐ = 5.96ร10โ7๐
Distance of the slits from the screen = D = 1.5m
Distance of the two slits = a = 2ร10โ4๐
For the first secondary maxima
sin๐ =3๐1
2๐=
๐ฅ1
๐ท
OR
๐ฅ1 =3๐1๐ท
2๐๐๐๐ ๐ฅ1 =
3๐2๐ท
2๐
โด Spacing between the positions of first secondary maxima of two sodium lines
๐ฅ1 โ ๐ฅ2 =3๐ท
2๐(๐1 โ ๐2) = 6.75 ร10โ5๐.
Q.18. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity. 3 marks
Ans. (a)
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(b) Magnifying power: The linear magnification (๐0) due to the objective is
๐0 =๐ดโฒ๐ตโฒ
๐ด๐ต=
โโฒ
โโฆโฆ(๐)
Also tan๐ฝ =โ
๐0=
โโฒ
๐ฟ
โด โโฒ
โ=
๐ฟ
๐0โฆโฆ . . (ii)
From (i) and (ii) we have
๐0 =๐ฟ
๐0โฆโฆ(๐๐๐)
Where hโ is the size of the first image, the object size being h and ๐0 being the focal length of the objective and L be the distance between the second focal point of the objective and first focal point of the eye piece (focal length ๐๐) is called the tube length of compound microscope.
When the final image is formed at the near point, then the angular magnification (๐๐) of the eye piece is
๐๐ = (1 +๐
๐๐)โฆโฆ . (๐๐ฃ)
โด Total magnification of compound microscope is
๐๐ = ๐0. ๐๐
๐ =1
๐0+ (1 +
๐ท
๐๐)
OR
(a) State Huygensโs principle. Using this principle draw a diagram to show how a
plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snellโs law of
refraction. (b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons: (i) Is the frequency of reflected and reflected light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave?
Ans.(a) Huygensโs Principle: It is based on the assumptions:
(i) Each point on the primary wave front acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does.
(ii) The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant
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Refraction on the basis of wave theory:
(i) Consider any point Q on the incident wave front. (ii) Suppose when disturbance from point P on incident wave front reaches point
pโ on the refracting surface XY. (iii) Since, PโAโ represents the refracted wave front, the time taken by light to
travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time taken by light to go from Q to Qโ will be
๐ก =๐๐พ
๐+
๐พ๐โฒ
๐ฃ โฆ . . (๐)
In right angled โ๐ด๐๐พ,< ๐๐ด๐พ = ๐
โด ๐๐พ = ๐ด๐พ sin ๐ ---(ii)
in right- =angled โ๐โฒ๐โฒ๐พ,< ๐โฒ๐โฒ๐พ = ๐ ๐๐๐ ๐พ๐โฒ = ๐พ๐โฒ sin ๐. ---(iii)
Substituting (ii) and (iii) in equation (i)
We get
๐ก =๐ด๐พ sin ๐
๐+
๐พ๐โฒ sin ๐
๐ฃ
or,
๐ก =๐ด๐พ sin ๐
๐+
(๐ด๐โฒ โ ๐ด๐พ) sin ๐
๐ฃ(KPโ = APโ โ AK)
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or,
๐ก =๐ด๐โฒ
๐sin ๐ + ๐ด๐พ (
sin ๐
๐โ
sin ๐
๐ฃ)โฆ . (๐๐ฃ)
The rays from different point on the incident wave front will take the same time to reach the corresponding points on the refracted wave from i.e., t given by equation
(iv) is independent of AK. It will happen so,
If
sin ๐
๐โ
sin ๐
๐ฃ= 0
sin ๐
sin ๐=
๐
๐ฃ= ๐ =
sin ๐
sin ๐
This is the Snellโs law for refraction of light.
(b)(i). The frequency of refracted light remains same as the frequency of incident light frequency only depends on the source of light.
(ii) Since, the frequency remains same, hence there is no reduction in energy.
Q.19. State the law of radioactive decay. Plot a graph showing the number (N) of undecided nuclei as a function of time (t) for a given radioactive sample having half life ๐ปยฝ. Depict in the plot the number of undecided nuclei at (i) t = 3 ๐ปยฝ and (ii) t = 5 ๐ปยฝ. 3 mark Ans. Radioactive decay Law: The number of atoms disintegrated per second at any instant is directly Proportional to the number of radioactive atoms actually present at that time. The following graph showing the number (N) of Undecided nuclei as a function of time (t) for a given radioactive sample having half-life. And undecided nuclei at (i) t = 2 ๐1/2 (ii) t = 4 ๐1/2 included also,
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Q.20. Three light rays red (R), green (G) and blue (B) are incident on a right angled prism 'abc' at face 'ab'. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. Out of the three which color ray will emerge out of face 'ac' ? Justify your answer. Trace the path of these rays after passing through face 'ab'. 3 marks
Ans. The red light ray (R) will emerge out of face ac. The path of green (G) and blue (B) light rays will be as,
Q.21. (a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
(b) write the basis features of photon picture of electromagnetic radiation on which Einsteinโs photoelectric equation is based. 3 marks
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Ans. (a) Wave nature of radiation cannot explain the following:
(i) The instantaneous ejection of photoelectrons. (ii) The existence of threshold frequency for a metal surface (iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of
light and depends upon its frequency. Thus, the photoelectric effect cannot be explained on the basis of wave
nature of light.
(b) Photon picture of electromagnetic radiation on which Einsteinโs photoelectric equation is based on particle nature of light. Its basic features are:
(i) In interaction with matter, radiation behaves as if it is made up of particles called photons.
(ii) Each photon has energy E = hv and momentum ๐ =โ๐ฃ
๐ and speed C, the speed of
light. (iii) All photons of light a particular frequency v, or wavelength ๐ have the same energy
๐ธ = โ๐ฃ =โ๐ฃ
๐ and momentum ๐ =
โ๐ฃ
๐=
โ
๐whatever the intensity of radiation may be.
(iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus photon energy is independent of intensity of radiation.
Photons are electrically neutral and are not deflected by electric and magnetic fields.
Q.22. You are given three lenses ๐ณ๐, ๐ณ๐ and ๐ณ๐each of focal length 20 cm. An object is kept at 40 cm in front of ๐ณ๐, as shown. The final real image is formed at the focus โIโ of ๐ณ๐. Find the separations between ๐ณ๐, ๐ณ๐ ๐๐๐ ๐ณ๐. 3 marks
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Ans.
1
๐ฃ1โ
1
๐ข=
1
๐1
1
๐ฃ1=
1
๐1+
1
๐ข1
1
20+
1
โ40=
1 โ 2 โ 7
โ40
1
40
๐ฃ1 = 40๐๐.
Here, image by ๐ฟ3 is formed at focus. So the object should lie at infinity for ๐ฟ3. Hence. ๐ฟ2 will produce image at infinity. So we can conclude that object for ๐ฟ2 should be at its focus.
But, we have seen above that image by ๐ฟ1 is formed at 40 right of ๐ฟ1which is at 20 cm left of ๐ฟ2 focus of ๐ฟ2.
So ๐1= distance between ๐ฟ1 and ๐ฟ2 = (40 + 20) cm = 60 cm
Again distance between ๐ฟ2 and ๐ฟ3 does not matter as the image by ๐ฟ2 is formed at infinity so ๐2 can take any value.
SECTION - D
Q.23. With the help of a suitable ray diagram, derive the mirror formula for a concave mirror. 4 marks
Ans. Mirror Formula for Concave Mirror:
Let, AB be the length of an object placed beyond C in front of a concave mirror. The image AโBโ is real, inverted and between C and F.
Applying sign conventions, we have
Object distance PB = โ๐ข image distance PBโ = โ๐ฃ focal length PF = โ๐ and radius of curvature PC = โ2๐
In similar โs ABC and AโฒBโฒF
๐ด๐ต
๐ดโฒ๐ตโฒ=
๐ต๐ถ
๐ตโฒ๐ถ โฆ (๐)
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and similar โs RSF and AโฒBโฒF
๐ ๐
๐ดโฒ๐ตโฒ=
๐๐น
๐ตโฒ๐น
โต ๐ ๐ = ๐ด๐ต
๐ด๐ต
๐ดโฒ๐ตโฒ=
๐๐น
๐ตโฒ๐น โฆ (๐๐)
From eq. (i) and (ii), we have
๐ต๐ถ
๐ตโฒ๐ถ=
๐๐น
๐ตโฒ๐น
Since, the aperture of the concave mirror is small so the point S and P coincides.
โด๐ต๐ถ
๐ตโฒ๐ถ=
๐๐น
๐ตโฒ๐น
๐๐ต โ ๐๐ถ
๐๐ถ โ ๐๐ตโฒ=
๐๐น
๐๐ตโฒ โ ๐๐น
โ๐ข + 2๐
โ2๐ + ๐ฃ=
โ๐
โ๐ฃ + ๐
๐ข๐ฃ โ ๐ข๐ โ 2๐ฃ๐ + 2๐2 = 2๐2 โ ๐๐ฃ
โ ๐ข๐ฃ = ๐ข๐ + ๐ฃ๐
Dividing both side by ๐ข๐ฃ๐, we get
๐ข๐ฃ
๐ข๐ฃ๐=
๐ข๐
๐ข๐ฃ๐+
๐ฃ๐
๐ข๐ฃ๐
โด1
๐=
1
๐ฃ+
1
๐ข
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SECTION โ E
Q.24. (a) for a ray of light travelling from a denser medium of refractive index ๐๐to a rarer medium of refractive index ๐๐, prove that
๐๐
๐๐= sin ๐๐ ,
where ๐๐is the critical angle of incidence for the media.
(b) Explain with the help of a diagram. how the above principle is used for transmission of video signals using optical fibers? 5 marks
Ans.(a) Relation between refractive index and critical angle: Let O be a point object in the denser medium of refractive index (๐1). A ray incident along OA1 deviates away from normal and is refracted along A1 B1in the rarer medium of refractive index (๐2). It increases with increase in the angle of incidence: For particular value of i = C, the critical angle, the incident ray OA2 is refracted at < r = 90ยฐ and goes along A2 B2.
Applying Snellโs Law at A2
n1 ๐ ๐๐ ic = n2 sin 900 โn1 ๐ ๐๐ ic = n2ร1
โด ๐ ๐๐ ic =n2
n1 ๐๐
n2
n1= ๐ ๐๐ ic
(b) Optical Fiber: Optical fiber make use the phenomenon of total internal reflection. Optical fibers consist of many long high quality composite glass or quartz fibers. Each fiber consists of a core and cladding. The refractive index of the material of the core is higher than that of the cladding.
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Propagation of light through an optical fiber: When light is incident at one end of the fiber at a small angle, it suffers several total internal reflection at the glass boundary because the angle of incidence is greater than the critical angle. The intensity of the out coming beam is almost equal to that of the incident beam.
OR
(a) What is plane polarized light? Two polaroidโs are placed at ๐๐๐ to each other
and the transmitted intensity is zero. What happens when one more polaroid is placed between these two bisecting the angle between them? how will the intensity of transmitted light vary on further rotting the polaroid?
(b) If a light beam shows no intensity variation when transmitted. Through a polaroid which is rotated, does it mean that the light is un โ polarized? explain briefly.
Ans.(a) Plane polarized light: When polarized, light is passed through a tourmaline crystal cut with its face parallel to its crystallographic axis AB. Only those vibrations of light pass through the crystal, which are parallel to AB, all other vibrations are absorbed. The emerged light from the Crystal is said to be plane polarized light. If E is the amplitude of electric field component emanating-from 1st polaroid, then from 2nd polaroid at 45ยฐ.The amplitude of
electric field component is E1 = E cos 45ยฐ = E ร1
โ2ร
E
โ2
Again amplitude of electric field component coming from 3rd polaroid at 45ยฐ to 2nd polaroid would be
E2 = E1 cos 45ยฐ =
E
โ2.1
โ2=
E
2
= Half of E
As Intensity ๐ผE2
โด Intensity transmitted from three polaroidโs will be 1
4th of the intensity
transmitted from the first polaroid. (b) No, The light which is made up of electric field components Ex, Ey with 90ยฐ phase difference but equal amplitudes. The tip of electric vector executes uniform circular motion at the frequency of the light itself.
When such light is passed through a polaroid, which is rotated, the transmitted average intensity remains constant.
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Q.25. An a.c. source generating a voltage ๐ = ๐๐ ๐ฌ๐ข๐ง๐๐ is connected capacitor of capacitance C. find the expression for the current, i , through it. Plot a graph of v and i versus t to show that the current is ๐ /๐ ahead of the voltage. A resistor of 200 ๐ and a capacitor of ๐๐. ๐ ๐๐ญ Fare connected in series to a 220 V, 50 Hz a.c. source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? if yes, resolve the paradox. 5 marks
Ans. A.C source containing capacitor: Let a source of alternating ๐๐๐. ๐ = ๐๐ sin๐๐ก be connected to a capacitor of capacitance C only
๐ = ๐๐ sin๐๐ก โฆ (๐)
At every instant, the potential V is given by
๐ =๐
๐ถ โ ๐๐
sin๐๐ก =๐
๐ โด ๐ = ๐ถ ๐๐ sin๐๐ก.
If I is instantaneous value of current in the circuit at instant t, then
๐ =๐๐
๐๐ก=
๐
๐๐ก (๐ถ ๐๐ sin๐๐ก)
๐ = ๐ ๐๐(cos๐๐ก). ๐ =
1๐1๐๐
sin (๐๐ก +๐
2)
The current will be maximum. When (sin๐๐ก +๐
2) = 1
โด ๐๐ =๐๐1
๐๐
ร 1 = ๐๐1
๐๐
๐ = ๐๐ sin (๐๐ก +๐
2)
Therefore, alternating current I lead the alternating voltage by a phase angle of ๐
2.
Numerical: Here R = 200 ฮฉ
๐ถ = 15.0 ๐๐น = 15 ร 10โ6๐น, ๐๐๐๐ = 220 ๐,
๐ = 50 ๐ป๐ง
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๐ผ๐๐๐ =?
๐๐ =1
๐๐=
1
2๐๐๐
=1
2 ร227
ร 50 ร 15 ร 10โ6
=7 ร 106
33000
= 212.12 โ 212 ฮฉ
โด ๐ = โ๐ 2 + โ๐๐2 = โ(200)2 + (212)2
= โ40000 + 44944
= โ84944
= 291.45 ฮฉ
โด ๐ผ๐๐๐ =๐๐๐๐
๐=
220
291.45
= 0.75 ๐ด
โด ๐๐ = ๐๐๐๐ ๐ = 0.75ร200
= 125 ๐
๐๐ = ๐ผ๐๐๐ . ๐๐
= 0.75 ร 212 = 159 ๐
โด ๐๐ + ๐๐ถ = 150 + 159 = 309
โด ๐๐ + ๐๐ถ > ๐
This is because these voltages are not in same phase and they cannot be added like ordinary numbers.
โด ๐ = โ๐๐ 2 + ๐๐ถ
2
= โ(150)2 + (159)2
= โ47781
= 218.18 ๐.
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Q.26. State Faradayโs law of electromagnetic induction. Figure shows a rectangular conductor PORS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x=0 to x=b and is zero for x>b. Assume that only the arm PQ possesses resistance r, when the arm PQ is pulled outward from x=0 with constant speed v, absinthe expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 x 2b. 5 marks
Ans. Part I: Faradayโs law of induction: It states that the emf induced in a coil of N turns is directly related to the rate of change of flux through it.
โด ๐ = โ๐๐โ ๐ต
๐๐ก
Where โ ๐ต is the flux linked with one turn of the coil? If the circuit is closed, a
current I =๐
๐ is set up in it.
Part II: refer to following fig (a). the arm PQ of the rectangular conductor is moved from x = 0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and zero the situation when the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as joule heat. Sketch the variation of these quantities with distance.
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Let us first consider the forward motion from ๐ฅ = 0 to x = 2b the flux โ ๐ linked with the circuit SPQR is
โ ๐ = ๐ต๐๐ฅ 0 โค ๐ฅ < ๐
= Blb b โค ๐ฅ < 2๐
The induced emf is,
๐ = โ๐โ ๐
๐๐ก= โ๐ต๐๐ฃ ; 0 โค ๐ฅ < ๐ = 0 ; ๐ โค< 2๐
When the induced emf is non-zero, the current I is (in magnitude)
I =๐ต๐๐ฃ
๐
The force required to keep the arm PQ in constant motion is IlB. Its direction is to the left. In magnitude
F =๐ต2๐2๐ฃ
๐= 0 ; 0 โค ๐ฅ < ๐ = 0 ; ๐ โค ๐ฅ < 2๐
The joule heating loss is
๐๐ = I2๐
=๐ต2๐2๐ฃ2
๐ 0 โค ๐ฅ < ๐
= 0 b โค ๐ฅ < 2
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OR
Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils?
How is the transformer used in large scale transmission and distribution of electrical energy over long distances?
Ans. Step up transformer: principle: It is a device which converts low voltage. A.C. into high voltage A. C. It is based upon the principle of mutual induction. When alternating current passed through a coil, an induced e. m. f. is set up in the neighboring coil.
Construction: A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as primary (p) is connected to A.C. supply. The other coil known as secondary (s) is connected to the load.
Working: When an alternating current is passed through the primary, the magnetic flux through the iron core changes which does two things. It produces e. m. f in the primary and an induced e. m. f is also set up in the secondary, if we assume that the resistance of primary is negligible, the back e. m. f will be equal to the voltage applied to the primary.
โด ๐๐ = โ๐๐
๐โ
๐๐ก
and ๐๐ = โ๐๐
๐โ
๐๐ก
Where ๐๐and ๐๐ are number of terms in the primary and secondary respectively. V๐
and V๐ are their respective voltages.
๐๐ ๐๐
=N๐
N๐
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This ratio N๐
N๐ called the turms ratio.
In a step โ up transformer: N๐ > N๐
So ๐s > V๐
In a step โ down transformer: N๐ < N๐
So, V๐ > V๐
Large scale transmission: The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-upโso that reduced and power loss I2R is cut down. It is then transmitted over long distances to an area sub- station near the consumers. There the voltage is stepped down. It is further stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes.
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