Instructions:
1. SectionA Q. no. 18 carry 1 mark each.
2. SectionB Q. no. 918 carry 2 marks each. 3. SectionC Q. no. 1927 carry 3 marks each. 4. SectionD Q. no. 2830 carry 5 marks each.
SectionA
1. How
will the magnetic field intensity at the centre of circular coil of
radius R is changed if current through coil doubled and radius of coil
is halved?
Ans. As i.e Bo will be four times.
2. What is the power dissipated in a.c. circuit in which voltage and current are given by V = 230sin(ωt+π/2) and I = 10sinwt?
Ans. zero as φ = π/2
3. State Einstein’s Photoelectric equation.
Ans. E = h = h_{0} + K_{max}
4. A point charge ‘q’ is placed at 0 is V_{P} – V_{Q} positive or negative when (i) q > 0 (ii) q < 0.
Ans.
q > 0 V_{P} – V_{Q} = +ve
q < 0 V_{P} – V_{Q} =  ve
5. What is induced current?
Ans. Current setup due to change in magnetic flux is known as induced current.
6. The Horizontal component is times the vertical component of earth’s magnetic field at a place. What is the angle of dip at that place?
Ans. BH = Bv ⇒ B cosθ = B sinθ
⇒ tanθ =1/ ⇒ θ = 30^{o}
7. A wire of resistivity ρ is streched to twice its length. What will be its new resistivity?
Ans. No Effect.
8. Compare the radii of two nuclei of mass number 1 & 27?
Ans. R_{1}:R_{2} = 1:3
SECTIONB
9. Draw
the graph showing the variation of inductive reactance and capacitive
reactance with frequency of applied a.c. voltage source.
Ans. X_{L} = ωL and X_{C} = 1/WC
10. Write two factors justifying the need of modulation for transmitting the signal.
Ans. (i) Due to very large size of antenna.
(ii) To avoid mixing of signal with other signals, modulation is required.
11. Two
charges –q and +q are located at A(0,0,a) and B(0,0,+a) respectively .
How much work is done in moving a test charge from P(7,0,0) to Q
(3,0,0)?
Ans. As P & Q lies on bisector of dipole
∴ W_{PQ }= q(V_{Q } V_{P}) = 0
12. Name
two elements one having positive susceptibility and other having
negative susceptibility. What does negative susceptibility signify?
Ans. +X → Al
X → Bi
ve
susceptibility (x) indicates that diagrammatic materials gets .
Weakly magnetized in direction opposite to the direction of magnetic
field.
13. Calculate the number of photons emitted per second by transmitter of 10 KW power; radio waves of frequency 6×10^{5} Hz.
Ans.
14.
Draw a sketch of a plane e.m. wave propagating along x axis. Depict
clearly the directions of electric and magnetic field varying
sinusoidally.
Ans.
15. Show that energy stored in an inductor L, when a current I is established through it is 1/2LI^{2}.
Ans. Total work done
16. The value of ground state energy of Hatom is 13.6ev.
(i) What does the negative sign signify
(ii) How much energy is required to take an electron from the ground state to the first excited state.
Ans. (i) 13.6ev energy is required to make electrons free
(ii) E_{1} to E_{2} =E_{2} – E_{1} = 10.2ev
17. Explain the term chain reaction what are the functions of moderator & control rods in nuclear reactor.
Ans. When a nuclear fission of _{92}U^{238} is initiated with _{0}n^{1} it releases energy and three neutron particle. Which will carry the nuclear reaction which is termed as chain reaction.
Moderator – It slows down the fast moving secondary neutrons.
Control rods It captures slow neutrons to control chain reaction.
18. Three
indentical cells each of emf 4v and internal resistance r are
connected in series to a 6Ω resistor. If the current flowing in the
circuit is 1.5A. What will be internal resistance of each cell?
Ans. E = 12 V, r’ = 3r, R = 6?, I = 1.5A
R = ?
⇒ 9 + 4.5r’ = 12
SECTIONC
20. Derive an expression for the energy stored in a parallel plate capacitor.
Ans.
At any time charge on plates of capacitor is +q and –q respectively Then q = CV (1)
If
additional charge ‘dq’ is supplied by an external source then work
done will be dw = (dq)V = (dq)q/c (2)
Total work done to store max charge ± Q on each plate is given as
Which is stored as electrostatic potential energy.
i.e.,
21. A
proton and an alpha particle having the same K.E. are in turn allowed
to pass through a uniform magnetic field perpendicular to their
direction of motion Compare the radii of paths of proton & α
particle.
Ans.
23.
State the principle of potentiometer. Draw a circuit diagram used to
compare the emf of two primary cells. Write the formula used?
Ans.
Principle:  If P.D. is applied b /w the ends of a resistance wire of
uniform cross section then potential drops along the two ends of wire
is directly proportional length
i.e Vα L
24. Three point charges of +2μc; and 3μc
are Kept at the vertices A, B and C respectively of an equilateral
triangle of side 20cm. what should be the sign and magnitude of the
charge to be placed at the midpoint M of side BC. So that charge at a
remains in equilibrium.
Ans. For equilibrium of charge at A
Which indicate nature of charge at M must be
26.
Define self inductance . Derive an expression for self induction of a
long solenoid of length L, cross section area A, having N number of
turns.
Ans.
Self inductance is numerically equal to the induced emf in coil it
self when the rate of change of current is unity in the coil itself.
Derivation
Let N be total number of turns
Φ = N(B A) = M (µ_{0}NI/L)πr^{2} Φ = LI = (µ_{0}N^{2}I/L) πr^{2} ⇒ L = (µ_{0}N^{2}I/L) πr^{2}
27. Deduce the Bohr’s quantization on the basis of debroglie concept.
Ans. or an electron in n^{th} order,
SECTIOND
28.
(a) Explain with the help of suitable nuclear reaction in each of the
following case how neutron proton ratio changes during α decay & β
decay
(b) Define actively of a radioactive sample give its S.I unit.
Ans. (a) α – decay , _{92}U^{238} → _{90}Th^{234} + _{2}He^{4} + Q
(b) Activity – Activity is the total decay rate of sample of one or more radionuclide.
Unit  Curie
29. Describe a method to find out the resistivity of a material in laboratory along with necessary circuit diagram.
Ans. Experimentally we can determine the resistance of a material (of wire) by using meter bridge principle as given.
According to W.S.B principle
If l be length of wire of resistance XΩ & A be Area of cross section then
30. Deduce an expression for electric potential due to dipole at a point in general.
Ans. For an ideal dipole r_{1}= AP ≈ CP = r + oc
r_{1 }= r + acosθ
Similarly r_{2} =BP=racosθ
=
=
For r >> a

Tuesday, September 16, 2014
Class  XII Physics Sample Paper 3
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XII Physics Sample Paper
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