Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

Class 12 : Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

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PHYSICS (Theory)

General Instructions

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
μ0 = 4π × 10–7 T m A–1
ε0 = 8.854 × 10–12 C2 N–1 m–2
Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
Mass of electron = 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

Q1. A point charge +Q is placed at point O as shown in the figure. Is the potential difference VA – VB positive, negative or zero ?

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(1)

Ans: Positive

Q2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ?

(1)

Ans: Electric flux remains unaffected

Q3. Write the underlying principle of a moving coil galvanometer.

(1)

Ans: A current carrying coil, in the presence of magnetic field, experiences a torque, which produces proportionate deflection.
[Alternatively

(deflection) θ α Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics (Torque)]

Q4. Why are microwaves considered suitable for radar systems used in aircraft navigation ? 

(1)

Ans:Due to their short wavelengths, (they are suitable for radar system used in aircraft navigation).

Q5. Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ?

(1)

Ans: Quality factor Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

[Alternatively]

Quality factor Q = Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics , Alternatively, It gives the sharpness of the  resonance circuit.] It has no unit.

Q6. Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.

(2)

Ans:

(i) The loss of strength of a signal while propagating through a medium.

(1)

(ii) The process of retrieval of information, from the carrier wave, at the receiver.

(1)

Q7.  Plot a graph showing variation of de-Broglie wavelength λ versus 1/√V,  where V is accelerating potential for two particles A and B carrying same charge but of masses m1, m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why ?

(2)

Ans:

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½ + ½)

 

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

As the charge of two particles is same , therefore

Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

Hence, particle with lower mass ( m2) will have greater slop.

(½)

Q8. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.

OR

Calculate the energy in fusion reaction :

Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

(2) 

Ans:Binding energy of nucleus with mass number 240,

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics = 240 × 7.6 MeV

(½)

Binding energy of two fragments
= 2 × 120 × 8.5 MeV

(½)

Energy released = 240 (8.5 – 7.6) MeV
= 240 × 0.9
= 216 MeV

(½)

OR

Total Binding energy of Initial System

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics = (2.23 + 2.23) MeV
                                = 4.46 MeV

(½)

Binding energy of Final System i.e Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

= 7.73 MeV

(½)

Hence energy released = 7.73 MeV- 4.46 MeV
= 3.27 MeV

(1)

Q9. Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

(2)

Ans:

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

Q10. State Brewster’s law.The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.

(2)

Ans: When unpolarised light is incident on the surface separating two media, the reflected light gets (completely) polarized only when the reflected light and refracted light become perpendicular to each other.
[ Alternatively
If the student draws the diagram, as shown, and writes Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics as the polarizing angle, award this 1 mark.
If the student just writes Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics award half mark olny.]  Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(1)

The refractive index of denser medium, with respect to rarer medium, is given by  Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(½)

Since Refractive index (μ) of a transparent medium is different for different colours, hence Brewster angle is different for different colours.

(½)

 

Q11. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

(3)

Ans;

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Net Electric Field at point P = Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

dE = Electric field due to a small element having charge dq

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Let λ = Linear charge density

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

At large distance i.e. x>>a

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

This is the Electric field due to a point charge at distance x.

(NOTE: Award two marks for this question, if a student attempts this question but does not give the complete answer)

Q12. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.

(3)

Ans; The three characteristic features which can’t be explained by wave theory are:

i. Kinetic energy of emitted electrons are found to be independent of intensity of incident light.

(1)

ii. Below a certain frequency (threshold) there is no photo-emission.

(1)


iii. Spontaneous emission of photo-electrons.

(1)

Q13. (a) Write the expression for the magnetic force acting on a charged particle moving with velocity n in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer.

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(3)

Ans; Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Justification: Direction of force experienced by the particle will be according to the Fleming’s Left hand rule / (any other alternative correct rule.)

Q14. (i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux inkage with the other coil ?

(1)

Ans.(i) Magnetic flux, linked with the secondary coil due to the unit current flowing in the primary coil, Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
[Alternatively
Induced emf associated with the secondary coil, for a unit rate of change of current in the primary coil Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics . ]
[Also accept the Definition of Mutual Induction, as per the
Hindi translation of the question]
[i.e. the phenomenon of production of induced emf in one coil
due to change in current in neighbouring coil

(ii) Change of flux linkage

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Q15.Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric
medium of er = 4.

 

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
(i) Calculate capacitance of each capacitor if equivalent capacitance of the
combination is 4 μF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y.

(3)

Ans;

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

ii) Total charge Q = CV
= 4μF × 15 V= 60μC

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

(Also accept any other correct alternative method)

Q.16Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same irection, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.

(3)

 

Ans.

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

As shown in Figure, the direction of force on conductor b is attractive
[Alternatively:

[Alternatively:
Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  at a point on wire 2, is along - Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  on wire 2, due to the , is along -  Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
towards wire1. Hence the force is attractive.

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Magnetic field, due to current in conductor a,

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
The magnitude of force on a length L of conductor b,

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
One ampere is that steady current which, when maintained in each of the two very long, straight, parallel conductors, placed one meter apart in vacuum, would produce on each of these conductors a force equal to Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  newton per meter of their length.

Q17. How are em waves produced by oscillating charges ?
Draw a sketch of linearly polarized em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.

OR

Write Maxwell’s generalization of Ampere’s Circuital Law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics


where  Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics is the electric flux produced during charging of the capacitor plates.

(3)

Ans.

;A charge oscillating with some frequency, produces an oscillating electric field in space, which in turn produces an oscillating magnetic field perpendicular to the electric field, this process goes on repeating , producing em waves in space perpendicular to both the fields.

Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Directions of and are perpendicular to each other and also perpendicular to direction of propagation of em waves.
                                                                       OR
Ampere’s circuital law is given by as . Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
But for a circuit containing capacitor, during its charging / discharging the current within the plates of the capacitor varies, (producing displacement current Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  Therefore, the above equation, as generalized by Maxwell, is given as  Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics
During the process of charging of capacitor, electric flux Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics  between the plates of capacitor changes with time, which produces the current within the plates of capacitor. This current, being proportional to Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics , we have Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics

Q18. (a) Explain any two factors which justify the need of modulating a low frequency signal.

(b) Write two advantages of frequency modulation over amplitude modulation.

(3)

Ans; a) A low frequency signal is modulated for the following purposes:
(i) It reduces the wavelength of transmitted signal, and the minimum height of antenna for effective communication is Past YEar Question paper,CBSE Question paper,CBSE Class 12,Class 12 Physics .Therefore
height of antenna becomes practically achievable.

(ii) Power radiated into the space by an antenna is inversely proportional to λ2 . Therefore, the power radiated into the space increases and signal can travel larger distance.
(Give full credit of this part for any other correct answer)
b)
(i) High efficiency
(ii) Less noise
(iii) Maximum use of transmitted power (any two)

Q19. (i) Write the functions of three segments of a transistor.
(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.

(3)

Ans; (i) Emitter : Supplies the large number of majority charge carriers for the flow of current through the transistor.
Base : Controls the movement of charge carriers coming from emitter region
Collector: Collects a major portion of the majority carriers supplied by the emitter.
(NOTE: Also accept the following explanation of these parts of the transistor as asked in Hindi translation)
Emitter: Heavily doped and of moderate size.
Base: Central region, thin and lightly doped.
Collector: Moderately doped and large sized.

(ii)

Input characteristics are obtained by recording the values of base curren t , for different values of at constant
Output characteristics are obtained by recording the values of  for different values of at constant

[Alternatively
Also accept input/output characteristic curves for this part of the question.]

Q20. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also.
(b) Using mirror formula, explain why does a convex mirror always produce a
virtual image

(3)

Ans; a) Given R = -20 cm, and magnification m= -2
Focal length of the mirror 

Magnification (m) =

Using mirror formula

b)

Using sign convention, for convex mirror, we have

From the formula

∵ f is positive and u is negative,

is always positive, hence image is always virtual.

Q21. (i) State Bohr’s quantization condition for defining stationary orbits. How does de Broglie hypothesis explain the stationary orbits ?
(ii) Find the relation between the three wavelengths λ1, λ2 and λ3 from the energy level diagram shown below.

(3)

Ans; (i) Only those orbits are stable for which the angular momentum, of revolving electron, is an integral multiple of

[Alternatively

i.e. angular momentum of orbiting electron is quantised.]

According to de Broglie hypothesis

Linear momentum

And for circular orbit L =rnp where ‘rn ’ is the radius of quantized orbits.

Circumference of permitted orbits are integral multiples of the wavelength λ

Q23. Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eye-piece. Write its two important advantages over a refracting telescope.

(3)

Ans;

(i) Large gathering power
(ii) Large magnifying power
(iii) No chromatic aberration
(iv) Spherical aberration is also removed
(v) Easy mechanical support
(vi) Large resolving power
(Any Two)

Q23. Meeta’s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.Answer the following questions based on above information :
(i) What were the values displayed by Meeta and her father ?
(ii) What answer did Meeta’s father give ?
(iii) What are the tiny lights in traffic signals called and how do these operate ?

(4)

Ans.(i) Values displayed by Meeta:
Inquisitive/ Keen Observer/ Scientific temperament/ (Any other value.)
Values displayed by Father:
Encouraging/ Supportive /(Any other value)
(ii) Meeta’s father explained that the traffic light is made up of tiny bulbs
called light emitting diodes (LED)
(Also accept other relevant answers)
(iii)Light emitting diode
These diodes (LED’s) operate under forward bias, due to which the majority charge carriers are sent from these majority zones to minority zones. Hence recombination occur near the junction boundary, which  eleases energy in the form of photons of light.

Q24. (i) An a.c. source of voltage V = V0 sin wt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition
called ?
(ii) In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1/P2.

OR

(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220 V and 1100 W. Calculate
(a) number of turns in secondary
(b) current in primary
(c) voltage across secondary
(d) current in secondary

(e) power in secondary

(5)

Ans:

From Figure

From Figure

Condition for current and voltage are in phase :

Circuit is called Resonant circuit.
(ii) Power factor   

Power factor when capacitor C of Reactance is put in
series in the circuit

(i) Conversion of ac of low voltage into ac of high voltage & vice versa

Mutual induction: When alternating voltage is applied to primary windings, emf is induced in the secondary windings.

(Any one of the above diagram)
Energy losses:
a. Leakage of magnetic flux
b. Eddy currents
c. Hysterisis loss
d. Copper loss

(Any two)

(ii)
Np= 100
Transformation ratio= 100
a) Number of turns in secondary coil

Ns= 100 x 100 = 10000
b) Input Power = Input voltage x current in primary
1100 = 220 xlp
  = 5A

e) Power in secondary = Power in Primary
=1100 W

Q25. (i) In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position ‘x’ on the screen.
(ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.

OR

(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light ? What is its cause ?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass ? Give relevant calculations.

  Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

(5)

Ans.

(i)

From figure
Path difference

For x, d << D

For constructive interference

For destructive interference

Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

n = 0, 1, 2 .....

(ii)
(a) The Interference pattern has number of equally spaced bright and dark bands, while in the diffraction pattern the width of the centralm maximum is twice the width of other maxima.
(b) In Interference all bright fringes are of equal intensity, whereas in the diffraction pattern the intensity falls as order of maxima increases.
(c) In Interference pattern, maxima occurs at an angle , where a is the slit width, whereas in diffraction pattern, at the same angle λ/a', first minimum occurs. (Here ‘a’ is the size of the slit)
(Any other distinguishing feature)

From figure which implies r1=r2

(ii) The phenomenon of splitting of white light into its constituent
colours.
Cause: Refractive index of the material is different for different colours According to the equation, where A is the angle of prism, different colours will deviate through different amount.

Past Year Question Paper(Delhi All SET) - 2016, Physics, Class 12 Class 12 Notes | EduRev

For total internal reflection,
 

Hence, the minimum value of refractive index must be 

Q26.(i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend ?
(iii) Why alloys like constantan and manganin are used for making standard resistors ?
                                                           OR
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 W. Calculate the potential gradient along the wire and balance length AO ( = l).

(5)

Ans. i) Average velocity acquired by the electrons in the conductor in the presence of external electric field.
[Alternatively:

We have E =  where V is potential difference across the length   of
the conductor


 

Also, R = -----------(ii)
Comparing (i) and (ii)


Resistivity of the material of a conductor depends on the relaxation time, i.e., temperature and the number density of electrons.
iii) Because constantan and manganin show very weak dependence of resistivity on temperature

OR

i) When constant current flows through a conductor of uniform area of cross section, the potential difference, across a length l of the wire, is directly proportional to that length of the wire.
(Provided current and area are constant]

ii) Current flowing in the potentiometer wire

∴ Potential difference across the two ends of the wire


Hence potential gradient  
Current flowing in the circuit containing experimental cell,

Hence, potential difference across length AO of the wire

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