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Old January 31st, 2014, 12:39 PM
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I am looking to get Question Papers for Class 12th Physics of CBSE Board. So can you provide the Question Papers for Physics?

As you want to collect 12th class of CBSE Board Question Papers for Physics , so I have its paper and share with you.

1. Two identical charged particles moving with same speed enter a region of uniform magnetic field. If one of these
enters normal to the field direction and the other enters along a direction at 300 with the field, what would be the
ratio of their angular frequencies?
2. Why does a metallic piece become very hot when it is surrounded by a coil carrying high frequency alternating
current?
3. How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers?
4. Name the characteristics of electromagnetic waves that
(i) increases
(ii) remains constant
in the electromagnetic spectrum as one moves from radiowave region towards ultravoilet region.
5. How would the angular separation of interference fringes in young’s double slit experiment change when the
distance of separation between the slits and the screen is doubled?
6. Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its,

Complete paper is in given below attachment:

CBSE 12 physics question paper





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Last edited by Aakashd; May 29th, 2019 at 10:38 AM.
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  #2  
Old February 3rd, 2014, 06:37 PM
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Join Date: Dec 2011
Default Re: 12th class of CBSE Board Question Papers for Physics

As you want to get Question Papers for Class 12th Physics of CBSE Board, so here I am providing the following Question Papers:

CBSE Class 12th Physics Question Paper 2011
1. What physical quantity is the same for X-ray of wavelength 10-10 m red light of wavelength 6300Å and radiowaves of wavelength 500m?
Solution:
The speed in vacuum c = 3  108m/s is the physical quantity which is constant for X-rays, red light and radiowaves.



2. A plane electromagnetic wave travels in vacuum along z direction. What can you say about the direction of its electric and magnetic field vector? If the frequency of the wave is 30 MHz. What is its wavelength?
Solution:
The electric field and the field lie perpendicular to each other and are also perpendicular to the z direction. Hence it is in x-y plane.

The frequency of the wave  = 30MHz

The velocity of the wave c = 3 x108 m/s

... The wavelength of the wave  =
=
= 10 m.



3. A radio can tune to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Solution:
The velocity of the band 'c' =3 x 108 m/s

for 12 MHz and 7.5 MHz
The wavelength corresponding to 7.5 Hz = =
= 40 m
The wavelength corresponding to 12 MHz =
= 25 m

... The corresponding wavelength band = 40 to 25 m



4. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Solution:
The charged particle oscillation results in high frequency oscillating electric field. Consequently an oscillating magnetic field is set up of the same frequency. These oscillating electric and magnetic fields constitute electromagnetic waves. The frequency of the electromagnetic waves in the same frequency of varying electric and magnetic field which is the same as the oscillation of the charged particle. Hence the frequency of the electromagnetic waves is 109 Hz which is the frequency of the charged particle.



5. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is Bo = 510 nT.What is the amplitude of the electric field part of the wave.
Solution:
Magnetic field B = E0/c = 510 nT
Therefore electric field E0 = B c
= 510 nT 3 108
= 153 N/C



6. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is  = 50.0 MHz. Determine B0, , k and . (b) Find expressions for E and B.
Solution:
Amplitude of an electromagnetic wave E0 = 120 N/C
Frequency  = 50 MHz

(a) Amplitude of magnetic field B = E0/c
= 120 N/C / 3 108
= 400 nT

Angular velocity  = 2 = 2 3.14 50 MHz
= 3.14 108 rad/s

Constant k = /c = 3.14 108 / 3 108
= 1.05 rad/m

Wavelength  = c/ = 3 108 / 50106
= 6.0 m
(b) E = E0 sin[2(x/) - (t/T)] }j -UNIT VECTOR
E = { (120 N/C) sin [(1.05 rad/m)x - (3.14 108 rad/s)t] }j
B = B0 sin[2(x/) - (t/T)] }k-UNIT VECTOR
B = { (400 nT) sin [(1.05 rad/m)x - (3.14 108 rad/s)t] }k



7. The terminology for different parts of the electromagnetic spectrum is given in the text. Use the formula E = hfor energy of a quantum of radiation (photon) and obtain the photon energy in units of eV for different parts of the em spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electro magnetic radiation?
Solution:
Photon energy (for  = 1m)
= = eV = 1.24 x 10-6 eV
Photon energy for other wavelength for em spectrum can be obtained by multiplying appropriate powers of ten. Energy of a photon that a source produces indicates the spacing of the relevant energy levels of the source. For example, = 10-12 m corresponds to photon energy = 1.24 x 106 eV = 1.24 MeV. This indicates that nuclear energy levels (transition between which cause -ray emission) are typically spaced by 1MeV or so. Similarly, a visible wavelength  = 5 x 10-7 m corresponds to photon energy = 2.5 eV. This implies that energy levels (transitions between which give visible radiations) are typically spaced by a few eV.



8. In a plane em wave, the electric field oscillates sinusoidally at a frequency of 2.0  1010 Hz and amplitude 48 Vm-1
a. What is the wavelength of a wave?
b. What is the amplitude of the oscillating magnetic field?
c. Show that the average energy density of the field equals the average energy density of the field, [c = 3 x108 ms-1].
Solution:
a. The velocity c = 3 x 108 ms -1
The frequency n = 2.0 x 1010 Hz
... The wavelength  = = = 0.015 = 1.5 x 10-2m

b. The amplitude of electric field Eo = 48 Vm-1
The velocity c = 3 x 108 ms-1
The amplitude of magnetic field = = = 1 .6 x 10-7 T

c. Energy density in field, UE = ()E2

Energy density in field, UB = B2

Using = c x and c = 1/o), UE = UB



9. Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 106 rad/s)t]} i
(a) What is the direction of propagation?
(b) What is the wavelength ?
(c) What is the frequency ?
(d) What is the amplitude of the magnetic field part of wave?
(e) Write an expression for the magnetic field part of wave.
Solution:
(a) -j vector
(b)  = = = 3.5 m
(c)  = 2
2 = 5.4 10-6 rad/s
= (5.4 10-6)/2 = 86 MHz
(d) Amplitude of magnetic field B = E0/c
= 3.1 / 3 108
= 100 nT
(e) {(100 nT) cos [(1.8 rad/m) y + (5.4  106 rad/s)t] } i



10. About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Solution:
Intensity of the radiation I =
(a) If the bulb is at 1m distance, area = 4r2
= 4 1 1
= 12.56
Therefore I = = 0.4 W/m2

(b) If the bulb is at 10m distance, area = 4r2
= 4 10 10
= 1256
Therefore I = = 0.004 W/m2



11. Use the formula m= 0.29cmK to obtain the characteristic temperature ranges for different parts of the em spectrum. What do the number that you obtain tell you?
Solution:
A body at temperature T produces a continuous spectrum of wavelength. For a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck's law, e.g the relation m = 0.29 cm k/T. For m = 10 -6m, T = 2900K. Temperatures for other wavelength can be found. These numbers tell us the temperature ranges required for obtaining radiations in different parts of the em spectrum. Thus to obtain visible radiations, say = 5 x 10-7 m the source should have a temperature of about 6000K. Note, a lower temperature will also produce this wavelength but not with maximum intensity.




12. Magnetic field lines can never emanate from a point nor end on a point. Yet the field lines, outside a bar magnet do seem to start from the North pole and end on the south pole. Does the second fact contradict the first? Explain.
Solution:
There is no contradiction. Field lines inside the bar magnet go away from S and towards N. The net flux over any surface fully enclosing N or S must be identically zero.



13. Given below are some famous numbers associated with electromagnetic radiation in different contexts in physics. State the part of the em spectrum to which each belongs.
(i) 21 cm (wavelength emitted by atomic hydrogen in inter stellar space).
(ii) 10.57 MHz (frequency of radiation arising from two close energy levels in hydrogen, known as Lamb shift).
(iii) 2.7 K (temperature associated with the isotopic radiation filling all space - thought to be a relic of the 'bigbang' origin of the universe).
(iv) 5890 Å - 5896 Å (double lines of sodium).
(v) 14.4 KeV (energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy).
Solution:
(i) Radio (short wavelength end).

(ii) Radio (short wavelength end).

(iii) Microwave.

(iv) Visible (yellow).

(v) X ray or soft -ray region.



14. Answer the following questions:
(a) Long distance radio broadcasts use short wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission why?
(c) Optical and radio telescopes are built on the ground but X ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe 'nuclear winter', with a devastating effect on life on earth. What might be the basis of this prediction?
Solution:
(a) Ionosphere reflects waves in these bands.

(b) Television signals are not properly reflected by the ionosphere. Therefore, reflection is effected by satellites.

(c) Atmosphere absorbs X rays, while visible and radiowaves can penetrate it.

(d) It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth's surface and causing damage to life.

(e) The temperature of the earth would be lower because the Green house effect of the atmosphere would be absent.

(f) The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a 'winter'.
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