|Paper:||XII Physics CBSE Board Paper 2013 Delhi Set 3|
|Total marks of the paper:||70|
|Total time of the paper:||3 hrs|
1. All questions are compulsory.
2. There are 29 questions in total. Question Nos. 1 to 8 are very short answer type questions and carry one mark each.
3. Question Nos. 9 to 16 carry two marks each, Question Nos. 17 to 25 carry three marks each and Question Nos. 27 to 29 carry five marks each.
4. There is no overall choice. However, an internal choice has been provided in one question of two marks, one questions of three marks and all three questions of five marks each. You have to attempt only one of the choices in such questions.
5. Question No. 26 is value based questions carry four marks.
6. Use of calculators is not permitted. However, you may use log tables if necessary.
7. You may use the following values of physical constants wherever necessary:
c = 3 108 m/s
h = 6.63 10-34Js
e = 1.6 10-19C
= 4 10-7 TmA1
= 9 109 Nm2C-2
me = 9.1 10-31 kg
Mass of the Neutron = 1.675 10-27 kg
Mass of the Proton = 1.673 10-27 kg
The equipotential surfaces due to a single isolated charge are concentric spherical surfaces. As the distance from the charge increases the electric field strength will decrease and the distance between the spherical surfaces will increase.
The relation between the angle of incidence i , angle of prism A and angle of minimum deviation is:
Angle of minimum deviation ==2i-A
Electric flux through the plates of the capacitor =
As q is constant after the capacitor is fully charged, so will also be constant.
So displacement current Id= =0
Conduction current=Ic=C =0 as V is constant.
Ic= Id when the capacitor will be fully charged.
Curve 1 and 3 and curves 2and 4 corresponds to different materials.
For a given frequency of incident radiation the stopping potential is independent of intensity of light. As the pair of curves: 1and3 and 2and 4 have different stopping potential so those materials are different.
(i) Heat waves can be polarized as it is transverse wave and propagates in perpendicular direction to the direction of vibration / disturbance in individual particle
(ii) Sound waves cannot be polarized as it is longitudinal wave. In longitudinal wave the disturbance/oscillation in individual particle take place in the direction of propagation of wave.
Value of current in the circuit:
Applying Kirchoff's loop rule:
Al and Ca.
I max=1.0 A
Shunt resistance=Rs=? so that the ammeter can measure up to 5.0 A.
For this the voltage difference across the ammeter when allowing max current =voltage difference across the (shunt+ammeter) allowing max 5.0 A current
Solving the above expression: Rs = 0.15
The combined resistance of the ammeter and the shunt==0.12
(a) Necessary conditions for total internal reflection to occur are:
(i) The incident ray on the interface should travel in optically denser medium.
(ii) The angle of incidence should be greater than the critical angle for the given pair of optical media.
(b) ,where, a and b are the rarer and denser media respectively. C is the critical angle for the given pair of optical media.
(a)Magnetic field will be along Z- axis and electric field along Y- axis.
The sketch of the em wave indicating the direction of electric and magnetic field will be:
(b) E=c x B
where E is magnitude of electric field and B is magnitude of magnetic field, c= speed of light.
(a) X represents Intermediate Frequency (IF) stage while Y represents an Amplifier.
(b) At IF stage, the carrier frequency is changed to a lower frequency and in this process, the modulated signal is detected. While the function of amplifier is to amplify the detected signal which not be strong enough to be made use of and hence is required.
Lenz's law states that the polarity of induced emf is such that it opposes the change in magnetic flux which caused it to produce.
Yes, there will be an emf induced due to changing magnitude of earth's magnetic field.
Photodiode is a special type of photo-detector. The general principle is electronic excitation from the valence band to conduction band by photons. If an optical photon of frequency is incident on a semiconductor, such that its energy is greater than the band gap of the semiconductor, it will excite an electron from the valence band to the conduction band, leaving a vacancy or hole in the valence band. Thus, an electron-hole pair will be created. These are called 'photo generated charge carriers' which increases the conductivity of the semiconductor.
It can detect optical signals :
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. The magnitude of the emf generated across a length dr of the rod as it moves at right angles
to the magnetic field is given by
If voltage applied have increased by 100 V:
Charge stored will be=
(a)This can be explained by the Q value concept of nuclear reaction:
mn is mass of neutron, mass of proton , mass of electron antineutrino, mass of electron.If these mass defect is calculated and the Q value is calculated from the above equation.That value will be 12.86 MeV
(b) Photon picture of electromagnetic radiation on which Einstein's photoelectric equation is based on particle nature of light. Its basic features are:
i. When interacting with matter radiation behaves like particles called photon.
ii. Each photon has energy E= h and momentum p= h/c and c= speed of light.
iii. All photons of light of a particular frequency ,wave length have same energy E=h=hc/ and momentum p= h/c what ever is the intensity of radiation 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. So photon energy is independent of incident intensity.
v. Photons are electrically neutral and are not deflected by electric and magnetic fields.
vi. Photon particle collision is completely elastic in nature.
Current amplitude is sharper for R2 than R1.
Significance of Q factor:
(i) For finding Kinetic energy of electrons:
(b) the negative binding energy corresponds to the fact :stability.
(a) Because the car acts like electric shield.We know that the electric field inside the closed conductor is zero.
(b) Awareness and humanity or concern.
(c) Gratitude and obligation.
(d) I was struck in severe thunder storm once in an isolated place. I insisted to go out of the car and enjoy the rain. My parents advised not to go out of the car otherwise I may get thunderstruck.
A proton is a positively charged particle. So it will execute a circular trajectory.
Deuteron is uncharged particle. So it will not be deflected in the magnetic field.
Differences between myopia and hyper metropia :
For hyper metropia:
(b) i. Frequency of reflected and refracted light will be the same.
ii. No the energy remains the same. As the speed decreases wavelength also decreases so the frequency remains same.Energy of light wave=hx frequency h=Planck's constant
(a) Kirchhoff's First Law ? Junction Rule
The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.
Let the currents be
Current towards the junction ? positive
Current away from the junction ? negative
The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.
This is the required balanced condition of Wheatstone Bridge.
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