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MARKING SCHEME
SET 42/1 (SPL)
Q. No. Expected Answer / Value Points Marks Total Marks
- Net dipole moment per unit volume developed in the dielectric medium in presence of an external electric field is called electric polarization. SI unit : coulomb/m^2
- i. When a charged particle is moving parallel/anti parallel to the magnetic field. ii. When a charged particle is moving perpendicular to the magnetic field.
- Two factors on which drift velocity of charge carriers depends; Applied electric field and temperature of the conductor. (or any other correct 2 factors.)
Alternatively : = (
- A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receiver.
- i. There is no chromatic aberration. ii. Spherical aberration is reduced. iii. Easy mechanical support. (Any one /any other one correct reason.) 1 1
[ Note: Award this one mark, if a student writes answer directly without calculation.]
a) Limiting factor – break down field of air. b) c = as dielectric strength E = =
(a) Identification of factor ½ (b) Determination of charge on the capacitor 1 ½ 1
c =
q= 9 x 10-^2 C = 90mC
i. For circuit I, Xc > XL Reason: Current is leading the voltage in phase. ii. For circuit II, Xc < XL Reason: Current is lagging the voltage in phase.
a) =
b)
For a convex mirror, and for an object on left side i.e. , Using the mirror formula =
(as and )
< Image is between Pole and focus of the mirror, Linear magnification m= for Hence, image is diminished
OR
Identification of circuits ½ + ½ Reasons ½ + ½
(a) Mathematical expressions for electric and magnetic field components ½ + ½ (b) Drawing of the sketch of the plane electromagnetic wave 1
Showing, that the virtual image produced by convex mirror is always diminished and between focus & pole 2
junction , holes diffuse from p n, and electrons from n p. This movement of charge carriers leaves behind ionised acceptors on the p-side and donors on the n- side of the junction. This space charge region on either side of the junction, together, is known as depletion region.
i. Analogue signals are in the form of continuos, time varying voltage or current. Digital signals are those which can take only discrete stepwise values i.e. only two levels of voltage / current. ii.
Electric field at a point distant 10cm from each wire, = + E = [ ]
= 9 x 10^9 x ( 2 ) = 9 x 5 x 2 x 10 N/C = 900 N/C (towards the wire having negative linear charge density.) [ Note : Give full credit, if a student doesn’t draw diagram.]
for electron (^) ½
a) Distinction between digital & analogue signals 1 b) Logic symbol and truth table ½ + ½
Finding the magnitude and direction of the electric field. 2
Calculation of ratio of kinetic energy of electron and energy of photon. 2
K.E. =
For photon, energy(E) =
66 x 10-^4
= q ( ) Since, charged particle is moving along + x axis and experiences the force along – y axis, therefore magnetic field should be along + z axis. (using right hand thumb rule/ fleming’s left hand rule/ any relevant rule.) i.e. should be along. F = qv BSin 2 x 10-^5 = 2 x 10-^9 x 10^5 B(as angle = )
= 0.1 T
i. For series combination of the given capacitors
ii. For parallel combination of the given capacitors
a) As energy stored U , for same voltage source. Energy stored in parallel combination of capacitors will be more, b) Charge acquired, Q for same Voltage source, Hence charge acquired will be more in parallel combination. [ Note : Also give full credit, for alternative methods.]
Linear width of principal maxima( w )
= 4.8 x 10-^7 m = 480nm
Calculation of wavelength ( ) of light used 1 ½ Calculation of linear width of second dark fringe ½ Variation of Intensity of fringes. 1
Finding the direction and magnitude of magnetic field. 2
Calculation of energy stored and charge acquired in given cases. 1½+1½
=
=
=
= 2.0V
OR
In loop abcda
- 20 I 1 + 40 – 40 I 3 = 0 I 1 +2 I 3 = 2 ----------(1) As I 3 = I 1 + I 2 I 1 +2 I 1 +2 I 2 = 2 3 I 1 + 2I 2 = 2 ----------(2) In loop adefa 40 I 3 – 40 – 80 +20 I 2 = 0 2 I 3 + I 2 = 6 -----------(3) Substituting I 3 = I 1 + I 2 2 I 1 + 2I 2 +I 2 = 6 2 I 1 + 3I 2 =6 -----------(4) solving equation 2 and 4 I 2 = A
And I 1 = A
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. This law is in conformity with the principle of conservation of energy, consider a situation where induced current is in the direction opposite to that given by lenz’s law. Then in this situation kinetic energy will continuously increase without expanding any energy and hence one can construct a perpetual motion machine by a suitable arrangement, which violates the conservation of energy. In irregular loop induced current will be in the sense. pqrsp, as induced current will oppose the increase in magnetic flux. In other loop, induced current will be in the sense abcda, as the induced current will oppose the decrease in magnetic flux.
Statement of Lenz’s law 1 Description of a simple example 1 Prediction of direction of induced current ½ + ½
Finding the value of I 1 and I 2 3
a) It is because, light which is produced from an ordinary source (like a sodium lamp) is unpolarised, when an unpolarised light wave is passed through a Polaroid sheet, it gets linearly polarised with the electric vector oscillating along the direction perpendicular to the aligned molecules. b) Ray of light parallel to the pricipal axis, gets deviated away from the normal at the first surface and towards the normal at the second surface since deviation at two surfaces is in same sense, net deviation from both the surfaces make the ray diverge away from the principal axis. Alternatively: It is because the relative refractive index of lens becomes less than 1 and becomes – ve. Hence, lens behave like a diverging lens. [ Note : Also accept the explanation through diagram.] c) Due to short focal length of eye piece, angular magnification of eye piece increases and magnification of objective is large when is slighty greater than , since the microscope is used for viewing very close object therefore is small and hence , should be small.
V =
I =
Instantaneous power ‘p’ supplied by the source P = VI =. = [ Average power over a complete cycle P = (because the average of time dependent term is zero.) = = At resonance V and I are in same phase i.e. P = , which is maximum
i. Concerned/ Caring(or any other one relevant value) ii. Because, one can get electric shock, as water conducts electricity, Yes, relatively safer, because distilled water is a bad conductor of electricity. iii. Any example displaying similar value in real life situation.
Explanation of parts a, b and c 1 +1+
Deduction of the expression for the average power 2 Showing the average power dissipation is maximum at resonance 1
Answer of the parts i, ii and iii 1+1+
a) Derivation of the expression for torque 3 b) Mentioning the conditions for (i) & (ii) 1+
Three physical quantities are Magnetic declination, angle of dip (inclination) and horizontal component of Earth’s field. The angle between the true geographic north and north shown by compass needle is called magnetic declination. Dip is the angle that the total magnetic field of earth makes with the surface of Earth. Component of total magnetic field of earth along the surface of earth is called horizontal component. = , = And tanI = where and are horizintal and vertical component of the earth’s field and I is the inclination at a place. b)
The field lines gets concentrated inside the material and field inside is enhanced.
The field lines are repelled or expelled and the field inside the material is reduced.
a)
In the quadrilateral AQNR --------------------(1) From the triangle QNR
- -----------------(2) Comparing these two equations
- =A -----------------------------------(3) The total devation is the sum of derivations at the two faces. =
Ray diagram showing the passage of ray 1 Obtaining the relation for refractive index 3 Nature of graph. 1
That is = i+ -------------------(4) At the minimum derivation Dm, the refracted ray inside the prism becomes parallel to its base. = Dm, i = e which implies = Equation (3) becomes
2r = A or r =
In the same way equation (4) becomes
Dm = 2i-A, => i=
The refractive index of the prism is
OR
a) Wave front is the surface of constant phase Alternatively: locus of points, which orscillate in phase. i.
To produce the plane wave at a later time t, draw spheres of radius vt from each point on the plane wave front. By drawing the common tangent to all these spheres a plane wave at a later time t is obtained. (where v is the speed of waves in the medium)
a) Definition of wave front 1 Showing (i) and (ii) using Huygen’s construction 1+ b) Verification of Snell’s law. 2
b) Bohr’s third postulate , gives the Rydberg Formula
= R
For first member of spectral line in lyman series,
= 129nm OR
a)
i. Alternatively
ii.
Alternatively:
b) Binding energy per nucleon of the fussed muclei is more than the binding energy per nucleon of the lighter nuclei i.e. final system is more tightly bound than the initial system therefor energy is released. c) Let A be the mass number and m be the average mass of a nucleon, Nuclear density =
=
Substituting R =
Which is constant and independent of Mass Number(A).
a) One example each for (i) and (ii) 1+ b) Explanation of release of energy 1 c) Showing that nuclear density is independent of mass number 1
