Download CBSE 2008 Physics Solved Paper All India XII

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Answers to 2008 CBSE Physics Paper
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Physics (Theory)−
−2008
Code No. 55/3
Time Allowed: 3 hours]
1.
[Maximum Marks: 70
What is the nuclear radius of 125Te, if that of 27Al is 3.6 fermi?
R = R0 A1/3
Ans.
R Al AAl1/3
271/3
=
=
(
)
RTe ATe1/3
561/3
1251/3
RTe = (
) × R Al
271/3
2.
A diverging lens of focal length ‘F’ is out into two identical parts each
forming a plano-concave lens, what is the focal length of each part?
1 n2
2
= ( − 1)(− )
f
n1
R
n2
1
1
=
(
−
1)(
−
)
R
f ' n1
2f = f '
3.
What is the geometrical shape of the wavefront when a plane wave passes
through a convex lens.
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4.
What is the stopping potential applied to photocell, if the maximum kinetic
energy of electrons emitted is 5 eV.
KE = eV0
5eV = eV0
V0 = 5volts
5.
Draw an equipotential surface for a uniform electric field.
7. Identify the part of the electromagnetic spectrum to which the following
wavelengths belong:(i) 10-1m (ii) 10-12 m
1) 10-1 m corresponds to microwaves ii) gamma rays
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8. When current in a coil changes with time, how is the back emf induced in the coil
related to it?
ε =−
9.
dφ
dt
State the law of radioactive decay. If N0 is the number of radioactive nuclei in the
sample at some initial time, t0, find out the relation to determine the number N
present at a subsequent time. Draw a plot of N as a function of time.
OR
Draw a plot of the binding energy per nucleon as a function of mass number for a
large number of nuclei.Explain the energy release in the process of nuclear fission from
the above plot. Write a typical nuclear reaction in which a large amount of energy is
released in the process of nuclear fission.
The law of radioactive decay states that the rate of decay is proportional to the number of
dN
nuclei present.
∝N
dt
dN
∝N
dt
dN
= −λ N
dt
dN
= −λdt
N
Integrating
N
t
ln
N
= −λ (t − t 0 )
N0
dN
∫N N = −t∫ λdt
0
0
N = N 0e − λ (t −t0 )
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OR
Elements with high mass number have low values of binding energy. This implies that
they are less stable. By undergoing nuclear fission they can have lower values of A and
hence higher values of binding energy which means they are more stable.
235
89
1
U 92
+ n 10 → Ba 144
56 + Kr36 + 3n 0
10.Why do we need carrier waves of very high frequency in the modulation of signals? A
carrier wave of peak voltage 20V is used to transmit a message signal. What should be
the peak voltage of the modulating signal, in order to have a modulation index of 80%.
As size of antenna required is comparable to wavelength of waves, high frequency
waves will require smaller antenna. High frequencies are transmitted better as power
radiated increases with increasing frequency.
Modulation index=
message or modulating signal amplitude
carrier amplitude
80
x
=
100 20
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11. Using Gauss’s law derive an expression for the electric field intensity at any point
near a uniformly charged thin wire of charge/length λ C/m.
Derivation is given in NCERT text book. Choose a cylinder as a gaussian surface and
find flux through it, equate it to charge enclosed.
12. The following graph shows the variation of terminal potential difference V,
across, combination of three cells, in series to a resistor, versus the current.
(i) calculate the emf of each cell.
(ii) For what current i, will the power dissipation of the circuit be maximum?
The internal resistance of each cell is r, then the 3 cells placed in series have a total
resistance of 3r.
V = E − ir , for current equal to zero, V = E. So emf of each cell is 2V.
Power dissipation is maximum for R= 3r, i =
3E
3E
E
=
=
R + 3r 3r + 3r 2r
We need to find r.
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V = E − ir
3 = 6 − 1(3r ) From graph we saw that for V=3, i = 1A and E is fixed
3r = 3
r = 1Ω
Power dissipation maximum for i =
E
2
=
= 1A
2r 2 × 1
13. What is meant by the transverse nature of electromagnetic waves? Draw a
diagram showing the propagation of an electromagnetic wave along the xdirection, indicating clearly the directions of the oscillating electric and magnetic
fields associated with is
Transverse nature of em waves means that the electric field and magnetic field are
perpendicular to each other and to the direction of propagation of the waves. The
vector product E × B point in the direction of propogation.
14.
In the figure given below, light rays of blue, green, red – wavelengths are incident
on an isosceles right-angled prism. Explain with reason, which ray of light will be
transmitted through the face AC. The refractive index of the prism for red, green,
blue light are 1.39, 1.424, 1.476 respectively.
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Angle of incidence is 450. Critical angles corresponding to these 3 values of refractive
index is
sin i
1
=
sin 90 1.39
critical angle for 1.39 is 46 deg rees
critical angle for 1.476 is42 deg rees
As angle of incidence is greater than critical angle, the light will emerge through AC.
This is true for red.
15.
Two wires X, Y have the same resistivity, but their cross sectional areas are in
the ratio 2 : 3 and lengths in the ratio 1 : 2 . They are first connected in series
and then in parallel to a d.c. source. Find out the ratio of the drift speeds of the
electrons in the two wires for the two cases.
R1 = ρ
l
l1
; R2 = ρ 2
A1
A2
A
A
R1
l
l
1 2 1
= 1 × 2 = 1× 2 = × =
R2 A1 l 2
l 2 A1 2 3 3
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i1 =
V
V
=
R1 + R2 4x
i2 =
V
V
4
=
=
V
R1R2
3x / 4 3x
R1 + R2
i1
V
3x
3
=
×
=
i2 4x 4V 16
i1 = neA1v1
i2 = neA2v 2
i1
Av
2 v1
= 1 1 =
i2 A2v 2 3 v 2
3
2 v1
=
16 3 v 2
v1
9
=
v 2 32
16.
Define the term: magnetic dipole moment of a current loop. Write the
expression for the magnetic moment when an electron revolves at a speed ‘v’
around an orbit of radius ‘ r’ in hydrogen atom.
Magnetic moment m = IA where I is the current and A is the area of the loop.
Time period and hence frequency of electrons is
2π r
v
v
f =
2π r
T =
v
)
2π r
We need to find out current, or charge/time = ch arg e / time =
1sec ond
e(
m = [e (
17.
v
evr
)][π r 2 ] =
Am2
2π r
2
In a single slit diffraction experiment, the width of the slit is made double the
original width. How does this affect the size and intensity of the central
diffraction band? Draw a plot of the intensity distribution.
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Central diffraction band =
2λ
a
If ‘a’ is doubled, central band =
2λ λ
= so size reduces to half
2a a
Intensity increases 4 times.
18.
A jet plane is travelling west at 450 ms-1. If the horizontal component of
earths magnetic field at that place is 4 × 10-4 T and the angle of dip is 300, find
the emf induced between the ends of wings having a span of 30 m.
Induced emf = Blv
Horizontal component of earth's field = Bcos300 = 4 × 10−4 T
B=
4 × 10−4 T
cos300
4 × 10 −4 T
sin 30
cos300
T) (tan 300 )(30m )(450m / s )
Vertical component of earth's field = Bsin300 =
Induced emf = Bcosθ lv = (4 × 10 −4
19.
What does the term LOS communication mean ? Name the types of waves
that are used for this communication. With the help of a suitable figure, of
communication systems that use space wave mode propagation.
Line of sight communication. It refers to direct reception of waves. The space waves
travel in a straight line from the transmitting antenna to the receiving antenna.
Space waves are used for LOS. Television , microwaves and satellite communication
use space waves.
Diagram given in NCERT text book.
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20.
Draw a plot showing the variation of power of a lens, with the wavelength of the
incident light. A converging lens of refractive index 1.5 and of focal length 15
cm in air, has the same radii of curvature for both sides. If it is immersed in a
liquid of refractive index 1.7, find the focal length of the lens in the liquid.
n glass
1
1
1
=(
− 1)( −
)
f air
nair
R1 R 2
n glass
1
1
1
=(
− 1)( −
)
f liq
nliq
R1 R 2
Taking ratio of both equations
n
( glass − 1)
f liq
n air
=
n
f air
( glass − 1)
nliq
Substitute and find the answer
21.
Draw V-I characteristics of a zener diode. Explain with the help of a circuit
diagram how a zener diode can be used as a voltage regulator.
OR
With the help of a circuit diagram, explain the working principle of full wave
rectifier. Draw the input and output waveforms.
Diagram and waveforms are given in NCERT textbook
.
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Across the zener diode voltage remains constant even though the current through it
changes. Example if the input voltage decreases, the current through the series
resistor in the circuit decreases and through the zener diode also decreases. The
voltage across the series resistor will decrease but the voltage across the zener
diode will not change. Thus the zener diode acts as a voltage regulator.
22.
A resistance R = 4Ω is connected to one of the gaps in a meter bridge, which
uses a wire of length 1 m. An unknown resistance X > 4Ω is connected in the
other gap as shown in the figure. The balance point is noticed at T from the
positive end of the battery. On interchanging R and X, it is found that the
balance point further shifts by 20 cm (away from end A) . Neglecting the end
correction, calculate the value of unknown resistance X used.
4
l
=
X (100 − l )
X
l + 20
=
4 100 − (l + 20)
Mutiple the two equations and find l
OR
Show that Bohr’s second postulate, the electron revolves around the nucleus only in
certain fixed orbits without radiating energy can be explained on the basis of the
Broglie hypothesis of wave nature of electron.
According to Bohr’s second postulate the electrons revolves in orbits for which the
h
angular momentum is an integral multiple of
. The angular momentum is
2π
nh
quantised and L =
.
2π
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For electron in an orbit, its circumference contains an integral number of deBroglie
wavelengths.
2π r = nl = n (
h
); l is the deBroglie wavelength of electron
mv
nh
2π
nh
L=
2π
mvr =
24.
If a particle of charge q is moving with velocity v along the x-axis and the
magnetic field B is acting along the y-axis, use the expression
the direction of the force F acting on it.
to find
A beam of proton passes undeflected with a horizontal velocity v, through a
region of electric and magnetic fields, mutually perpendicular to each other and normal
to the direction of the beam. If the magnitudes of the electric and fields are 50 kV/m, 100
mT respectively, calculate
(i) velocity v of the beam (ii) force with which it strikes a target on the screen,
if the proton beam current is equal to 0.80 mA..
v = viˆ; B = Bjˆ
F = qvBkˆ
As proton passes undeflected, qE=qvB
v=
E 50 × 1000
=
m /s
B 100 × 10 −3
Force is equal to dp/dt. This is force exerted by 1 proton.
F =
dp d (mv )
=
dt
dt
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Using current we can find the number of protons hitting the screen.
I=q/t
Q = (0.80 x10-3A)(1 second) = 0.80 x10-3C
Number of protons hitting the screen = Q/charge of a proton
Total force = (mvQ)/ charge of a proton.
Substitute all values, mass of proton, vel of proton(solved above), charge of proton, and
given in question.
25.
Distinguish between isotopes and isobars. Give one example for each of the
species. A radioactive isotope has a half-life of 5 years. How long will it take the activity
to reduce to 3.125%?
Isotopes have same atomic no. Isobars have same mass no.
Isotopes of hydrogen are deuteron and triton. Isobars are argon and calcium.
3.125% =
3.125
1
= 0.03125 = ( )5
100
2
5 half lifes will be required. Hence time required is 5x(half life or isotope) = 25 years
26.
Distinguish between paramagnetic and diamagnetic substances. A magnetizing
field of 1500 A/m produces a flux of 2.4 × 10-5 weber in a bar of iron of cross
sectional area 0.5 cm2. Calculate the permeability and susceptibility of the iron
bar used.
Properties of para and dia given in book.
H = 1500 A / m
B = µH
Φ = B .A;
Φ 2.4 × 10 −5 Wb
=
= 0.48T
A
0.5 × 10 −4 m 2
B
0.48
=
= 0.00032 = 3.2 × 10 −4
µ=
H 1500
B=
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µr = 1 + χ
Susceptibility µr =
µ 3.2 × 10−4
=
= 255
µ0 4π × 10−7
χ = 254
27.
Two signals A, B as given below, are applied as input to (i) AND (ii) NOR and
(iii) NAND gates. Draw the output wave – form in each case.
Using the truth table for each of these gates combine the 2 inputs to get the output.
28.
Draw a labeled ray diagram of a compound microscope and write an expression
for its magnifying power.
The focal lengths of the objective and eye-lens of a compound microscope are
2cm, 6.25 cm respectively. The distance between the lenses is 15 cm. (i) How far from
the objective lens, will the object be kept, so as to obtain the final image at the near point
of the eye ? (ii) Also calculate its magnifying power.
OR
Draw a labeled ray diagram of an astronomical telescope, in the normal
adjustment position and write the expression for its magnifying power. An astronomical
telescope uses an objective lens of focal length 15 m and eye-lens of focal length 1 cm.
What is the angular magnification of the telescope? If this telescope is used to view
Moon, what is the diameter of the image of Moon formed by the objective lens?
(Diameter of Moon = 3.5 × 106 m and Radius of lunar orbit = 3.8 × 108 m)
Ray diagrams can be checked from your NCERT textbook
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i )For objective lens
1
f 0bjective
=
1 1
−
v u
f = 2cm; u = u ; u and v are both unknown so first use eyepiece
For eyepiece
1
=
f eye
1 1
−
v u
f = 6.25cm ; u = u;v = −25cm
1 1 1
1
1
= − =
−
u v f
−25 6.25
u = −5cm
1
f 0bjective
=
1 1
−
v u
f = 2cm; u = ? cm;v = (15 − 5) = 10cm
1
1 1
=
−
2 10 u
u = −2.5cm
ii ) f 0 = 2cm ; f e = 6.25cm ; L = 15cm; D = 25cm
m=
L
D
(1 + )
f0
fe
Substitute values and calculate magnifying power.
m=
OR
f 0 15cm
=
= 15
fe
1cm
Diameter of moon
3.5 × 106
=
= 0.92 × 10 −2
radius of Lunarorbit 3.8 × 108
Size of image = 15 × 0.92 × 10 −2 m
29.
State the condition for resonance to occur in a series LCR a.c. circuit and derive
an expression for the resonant frequency. Draw a plot showing the variation of the peak
current (im) with frequency of the a.c. source used. Define the quality factor, Q of the
circuit.
Calculate the (i) impedance, (ii) wattless current of the given a.c. circuit.
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Draw a labeled circuit arrangement showing the windings of primary and
secondary coil in a transformer. Explain the underlying principle and working of a step
up transformer. Write any two major sources of energy loss in this device. How much
current is drawn by the primary coil of a transformer that step down 220 V to 22 V to
operate device with an impedance of 220 ohm?
Ans:Refer to the book for the derivation:
ωL =
1
ωC
Re sonant frequency=
im =
Q=
1
LC
V
R + ( X L − X C )2
2
ωL
R
can also be given by Q =
1
ωCR
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40 = iX C
30 = iR
i = 2A
X c = 20ohms
R = 15ohms
Z = R 2 + X 2c = 152 + 202
2=
Vrms
Z
Wattless component of voltage is across the reactive components , here the capacitor.
XC
R
X
sin θ = C
Z
V sin θ is the wattless component of voltage.
tan θ =
If the circuit has only inductive and capacitive
components then the current is called wattless current.
OR
Refer to the book for description of transformer and losses in transformers.
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Vs N s i p
=
=
V p N p is
Vs = i s R ;
22 = is (220)
is =
1
10
i
22
= p
1
220
10
1
ip =
A = 0.01A
100
30.
(a) Derive an expression for the energy stored in a parallel plate capacitor C,
charged to a potential difference V.
(b) Obtain the equivalent capacitance of the network given below. For a supply
of 300 V, determine the charge and voltage across C4.
Explain the principle on which Van-de-Graaff generator operates. Draw a labeled
schematic sketch and write briefly its working.
.
A Va-de-Graaff type generator is capable of building up potential difference of
15× 10 V. The dielectric strength of the gas surrounding the electrode is 5 × 107V m-1.
What is the minimum radius of the spherical shell required?
6
You can look up your NCERT book for the Van-de-Graaff generator and derivation of
energy stored in a capacitor.
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C2 and C3 are in series so their combination is 100pF. This is in parallel with C1. This
combination is 300pF. This 300pF is in series with C4. So total is 120pF.
Q = CV = 120 × 10 −12 × 300C = 36 × 10 −9C = 36nC
This charge is present in C4. Voltage across C4 is V =
Q
36 × 10 −9
=
= 180V
C 200 × 10 −12
OR
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V = Ed
15 × 106 = (5 × 107 )d
d=
15 × 106
= 0.3m
5 × 107
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