Physics (Code-A) Sample Question Paper for Class XII
Transcript of Physics (Code-A) Sample Question Paper for Class XII
(1)
Physics (Code-A) Sample Question Paper for Class XII
Time : 3 Hours Max. Marks : 70
General Instructions :
(i) All questions are compulsory.
(ii) There is no overall choice. However an internal choice has been provided in one question of 2 marks,
one question of 3 marks and all three questions of 5 marks each. You have to attempt only one of the
choices in such questions.
(iii) Questions 1 to 8 are very short answer type and carry 1 mark each.
(iv) Questions 9 to 16 are short answer type and carry 2 marks each.
(v) Questions 17 to 25 are short answer type and carry 3 marks each.
(vi) Question 26 is a value based question and carry 4 marks
(vii) Questions 27 to 29 are long answer type and carry 5 marks each.
(viii) Use of calculator is not permitted.
(ix) You may use the following values of physical constants wherever necessary :
c = 3 × 108
ms–1
h = 6.626 × 10–34
Js
e = 1.602 × 10–19
C
�0
= 4� × 10–7
T mA–1
9 2 2
0
1
9 10 Nm C
4
� ���
Mass of neutron mn
= 1.675 × 10–27
kg
Boltzmann’s constant k = 1.381 × 10–23
JK–1
Avogadro’s number NA
= 6.022 × 1023
/mol–1
SECTION-A
Very Short Answer Type Questions :
1. A wire of resistivity � is stretched to three times its length. What will be its new resistivity? [1]
2. What oscillates in electromagnetic waves? Are these waves transverse or longitudinal? [1]
3. State the essential condition for diffraction of light to occur. [1]
4. The electric current flowing in a wire in direction B to A is decreasing. What is the direction of induced current
in the metallic loop kept above wire as seen from top? [1]
A B
5. What are the two possible extreme values of the magnifying power of a simple microscope made of convex
lens of focal length 5.0 cm? [1]
6. If the maximum kinetic energy of a photoelectron is 3 eV, what is its stopping potential? [1]
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Sample Question Paper for Class XII Physics (Code-A)
7. The frequency of the alternating current applied to a series circuit containing resistance, inductance and
capacitance is doubled. What happens to R, XL
and XC
? [1]
8. An electric dipole of dipole moment 20 × 10–6
cm is enclosed in a closed surface. What is net flux coming
out of the surface? [1]
SECTION-B
Short Answer Type Questions :
9. Calculate the capacitance of the unknown capacitor C if the equivalent capacitance between P and Q is
40 �F. [2]
Q
10 F�
80 F�
C
P
10. Using Gauss law, derive an expression for electric field intensity due to a uniformly charged spherical shell at
a point : [2]
(a) Outside and
(b) Inside the shell
11. Define self inductance and give its SI unit. Derive an expression for self inductance of a long air cored solenoid
of length l, radius r and turns per unit length n. [2]
12. State Ampere’s circuital law. Use it to derive an expression for the magnetic field along the axis of a current
carrying toroidal solenoid of N number of turns having radius r. [2]
13. What do you mean by power of a lens? What is its unit? What is the expression for power of two thin lenses
in contact with each other? [2]
14. Draw a neat and labelled diagram of image formed by a compound microscope. Write an expression for its
magnifying power. [2]
15. On the basis of energy band diagrams, distinguish between [2]
(i) A metal
(ii) An insulator
(iii) A semiconductor.
16. A coil has an inductance of 1 H. [2]
(a) At what frequency will it have a reactance of 3142 �?
(b) What should be the capacitance of a capacitor which has the same reactance at that frequency?
OR
A 15.0 �F capacitor is connected to 110 V, 50 Hz source. Find the capacitive reactance and the rms current.
(3)
Physics (Code-A) Sample Question Paper for Class XII
SECTION-C
Short Answer Type Questions :
17. Using Biot-Savart’s law, derive an expression for the magnetic field intensity at the centre of a current carrying
circular coil. Also find the magnitude of field due to a semicircular coil of radius r and carrying current I. [3]
18. Define capacitance. Derive an expression for the capacitance of a parallel plate capacitor. How will the
capacitance of the capacitor be affected if a conducting slab of some thickness is partially filled in the space
between two plates. [3]
19. Define the term electron mobility. Explain how electron mobility changes for a good conductor when : [3]
(a) The temperature of the conductor is decreased, keeping potential difference constant.
(b) Applied potential difference is doubled keeping temperature constant.
20. In certain cases when an object and screen are separated by a distance d, two positions x of the converging
lens relative to the object will give an image on the screen. Show that these two values are
1/2
4
1 1
2
d f
x
d
⎡ ⎤⎛ ⎞� ⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
.
Under what conditions will no image be formed? [3]
21. What is electron emission? Discuss various forms of electron emission in brief. Establish Einstein’s photoelectric
equation? [3]
22. A series LCR circuit is connected to an AC source. Using the phasor diagram, derive the expression for the
impedance of the circuit. Plot a graph to show the variation of current with frequency of source, explaining the
nature of its variation. [3]
23. Mention three modes of propagation used in communication system. Explain with the help of a neat and labelled
diagram how long distance communication can be achieved by ionosphere reflection of radio waves. [3]
24. Obtain the binding energy of the nuclei 56
26
Fe in units of MeV from the following data : [3]
1.007825 up
m �
1.008665 un
m �
� �56
26
Fe 55.934939 um � . Also calculate binding energy/nucleon.
25. With the help of an example, explain how the neutron to proton ratio changes during -decay. Also briefly explain
what are nuclear forces. Give their important properties. [3]
OR
A doubly ionised lithium atom is hydrogen like with atomic number 3. Find
(a) The wavelength of radiation required to excite the electron in Li++
from the first to the third Bohr’s orbit.
(b) How many spectral lines are observed in the emission spectrum of the above excited system?
26. Mr. Feynmann and his students went to north pole on a picnic. They found beautiful coloured curtain like
structures are hanging from the polar sky. What is the reason behind this? [4]
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Sample Question Paper for Class XII Physics (Code-A)
SECTION-D
Long Answer Type Questions :
27. (a) In Young’s double slit experiment, derive the condition for (i) Constructive interference and (ii) Destructive
interference at a point on the screen. Also derive expression of fringe width. [3]
(b) Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength
400 nm. [2]
OR
State Huygens Principle. Using the geometrical construction of secondary wavelets, explain the refraction of a
plane wavefront incident at a plane surface. Hence verify the Snell’s law. Illustrate with the help of a diagram the
action of (i) Convex lens and (ii) concave mirror on a plane wavefront incident on it.
28. (a) Explain briefly with the help of a circuit diagram, how V-I characteristics of a p-n junction diode are obtained
in
(i) Forward bias
(ii) Reverse bias
Draw the shapes of curves obtained. [3]
(b) A photodiode is fabricated from a semiconductor with a band gap of 2.8 eV. Can it detect a wavelength of
6000 nm? Justify. [2]
OR
(a) With the help of a circuit diagram, explain working of a transistor as an amplifier.
(b) Calculate the base current, collector current and collector to emitter voltage for the circuit given
RC = 2 k�
RB = 300 k�
C
B
VCC
= 9 V
E � = 50
29. (a) Prove that for a short magnetic dipole axial
equatorial
2
B
B � . [3]
(b) The susceptibility of magnesium at 300 K is 1.2 × 10–5
. At what temperature will the susceptibility increase
to 1.8 × 10–5
? [2]
OR
(a) State Biot-Savart’s Law. Use it to obtain the magnetic field at an axial point, distance x from the centre of a
circular coil of radius a, carrying a current I.
(b) A galvanometer having 30 divisions has a current sensitivity of 20 �A/div. It has a resistance of 25 �. How will
you convert it into an ammeter of range 0 - 1 A?
� � �
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Chemistry (Code-A) Sample Question Paper for Class XII
Time : 3 Hours Max. Marks : 70
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consists of four sections A, B, C and D.
Section A contains 8 questions of 1 mark each.
Section B contains 10 questions of 2 marks each.
Section C has 9 questions of 3 marks each, whereas Section D contains 3 questions of 5 marks each.
(iii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks,
one question of 3 marks and all three questions of 5 marks weightage. A student has to attempt only one of the
alternatives in such questions.
(iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.
SECTION-A
Very Short Answer Type Questions :
1. Give IUPAC name of the compound: [1]
CH — C — CH — C — CH3 2 3
—
F
—
C H2 5
—
C H2 5
—
Br
2. Consider the reaction R ��� P. The change in the concentration of R with time is shown in the following
plot. [1]
Time
[R]t
(i) Predict the order of the reaction.
(ii) Write the expression for half life of this reaction.
3. Predict the shape of BrF3
on the basis of VSEPR theory. [1]
4. What are the dispersed phase and dispersion medium in milk? [1]
5. A compound contains two type of atoms A and B. It crystalises in a cubic lattice with atom A at the corners
of the cube and atom B at the face centres. What is the simplest formula of the compound? [1]
6. An ionic compound AB is 50% dissociated in aqueous solution. Determine the value of Van’t Holf factor for
this compound. [1]
7. An ore sample of galena (PbS) is contaminated with zinc blende (ZnS). Name one chemical reagent which
can be used to concentrate selectively by froth floatation method. [1]
8. Write balanced equation for the complete hydrolysis of XeF4
. [1]
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Sample Question Paper for Class XII Chemistry (Code-A)
SECTION-B
Short Answer Type Questions :
9. Write the structure of 4-Methylpent-3-en-2-one. Will it respond positive to iodoform test? If so, Why? [2]
10. Define osmotic pressure. How does it change with [2]
(a) Temperature
(b) Atmospheric pressure
11. Write the equations of the reactions involved in the extraction of copper from its sulphide ore (Cu2
S). [2]
12. Name the two components of starch. How do they differ from each other structurally? [2]
13. Give the equations of the reactions involved when glucose is treated with [2]
(a) HI
(b) (CH3
CO)2
O
14. Describe the mechanism of the formation of ethene from ethanol in the presence of concentrated sulphuric
acid. [2]
15. Give one example each of [2]
(a) Hofmann bromamide reaction
(b) Gabriel phthalimide reaction
16. Distinguish chemically between [2]
(a) Butan-1-ol and 2-Methyl propan-2-ol
(b) Phenol and benzyl alcohol
17. Account for the following: [2]
(a) Aniline does not undergo Friedel Crafts alkylation
(b) Aniline undergoes bromination even in the absence of halogen carrier
OR
(a) Unlike alkylation, acylation of amines stops after first step
(b) Tertiary amines do not undergo acylation
18. Explain the type of deviation shown by a mixture of acetone and chloroform. [2]
SECTION-C
Short Answer Type Questions :
19. An element X with atomic mass 60 g/mole has a density 6.23 g/cm3
. If the edge length of unit cell is 400 pm,
identify the type of cubic unit cell. Calculate the radius of an atom of this element. (NA
= 6 × 1023
mole–1
) [3]
20. Write the names of the monomers of these polymers and classify them as addition or condensation polymers. [3]
(a) Terylene
(b) Teflon
(c) Natural rubber
(7)
Chemistry (Code-A) Sample Question Paper for Class XII
21. (a) Name the isomerism exhibited by the following pair of coordination compounds: [3]
[Co(NH3
)5
Br]SO4
& [Co(NH3
)5
SO4
]Br. Give one chemical test to distinguish between these two
compounds.
(b) Using valence bond theory compare the structures of [FeF6
]4–
and [Fe(CN)6
]4–
(Atomic number of Fe = 26)
22. Give three differences each between [3]
(a) Physisorption and chemisorption
(b) Lyophilic sol and lyophobic sol
23. Account for the following : [3]
(a) ICl is more reactive than l2
(b) NO2
dimerises to form N2
O4
(c) H3
PO3
is a reducing agent but not H3
PO4
24. Give the structures of the following compounds : [3]
(a) XeOF4
(b) PF5
(c) H2
S2
O8
OR
Give the products and balance the following equations:
(a) Ca3
P2
+ H2
O ���
(b) XeF6
+ NaF ���
(c) Cu + H2
SO4
(conc.) ���
25. Activation energy of a reaction at 300 K is 55 kJ/mole. When the same reaction is carried out in the presence
of a catalyst, activation energy is lowered to 45 kJ/mole at 300 K. Determine the extent to which the rate of
the reaction is increased. [3]
26. Account for the following: [3]
(a) Haloalkanes react with KCN to give alkyl cyanide as a major product and not alkyl isocyanide.
(b) In chloro benzene, Cl is electron withdrawing group but it undergoes electrophilic substitution at ortho and
para position.
(c) Allyl chloride is more reactive towards nucleophilic substitution than n-propyl chloride.
27. Why are vitamin A and C essential to us? Also write their chemical names. [3]
SECTION-D
Long Answer Type Questions :
28. (a) Carry out the following conversions: [3]
(i) Ethanal to butane-1, 3-diol
(ii) Acetone to tert. butyl alcohol
(iii) Benzoic acid to m-nitro benzoic acid
(8)
Sample Question Paper for Class XII Chemistry (Code-A)
(b) Give one example each of [2]
(i) Clemmensen reduction
(ii) Cannizzaro reaction
OR
(a) Carry out the following conversions: [3]
(i) Ethanal to butan-1-ol
(ii) Benzaldehyde to 3-phenyl propanol
(iii) Ethyl benzene to benzene
(b) Give one example each of [2]
(i) Aldol condensation
(ii) HVZ reaction
29. (a) At what pH of HCl solution will the standard hydrogen electrode have a potential – 0.118 V at 298 K? [2]
(b) Write the equations of the reactions involved at each electrode in a H2
– O2
fuel cell. [2]
(c) Why does the potential of a mercury cell remain constant? [1]
OR
(a) Calculate the equilibrium constant and work done by the cell [2]
Ni + Cu2+
��� Ni2+
+ Cu, given 2 2
o o
Ni /Ni Cu /Cu
E 0.25 V; E 0.34 V� �� �
(b) Write the equations of the reactions involved at each electrode during the electrolysis of [2]
(i) CuSO4
(aq) using inert electrode
(ii) Dilute solution of sulphuric acid
(c) How many coulombs of electricity are needed for the conversion of 9 grams of Al from molten AlCl3
?
(Atomic mass of Al = 27) [1]
30. Police usually disperse the indisciplined mob by using tear gas shell. One of the person in the mob advised
the people either to use water wetted cloth on eyes or to avoid smoke.
(i) Write the chemical formula of tear gas. [1]
(ii) Write the value involved as advised by the person present in the mob. [2]
(iii) Write IUPAC name of tear gas. [2]
OR
Arrange the following in order of property indicated:
(i) F2
, Cl2
, Br2
, I2
– Increasing bond dissociation enthalpy [1]
(ii) HF, HCl, HBr, HI – Increasing acid strength [1]
(iii) NH3
, PH3
, AsH3
, SbH3
– Increasing base strength [1]
(iv) H2
O, H2
S, H2
Se, H2
Te – Increasing boiling point [1]
(v) HF, HCl, HBr, HI – Increasing boiling point [1]
� � �
(9)
Mathematics (Code-A) Sample Question Paper for Class XII
Time : 3 Hours Max. Marks : 100
General Instructions :
(i) All questions are compulsory.
(ii) The question paper consist of 29 questions divided into three sections A, B and C. Section A comprises of
10 questions of one mark each, section B comprises of 12 questions of four marks each and
section C comprises of 7 questions of six marks each.
(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of
the question.
(iv) There will be no overall choice. However, internal choice has been provided in Four Questions of 4 Marks
each and Two Questions of 6 Marks each. You have to attempt only one of the alternatives in all such
questions.
(v) Use of calculator is not permitted. You may ask for logarithmic tables, if required.
SECTION-A
Very Short Answer Type Questions :
1. Using principal values, write the value of � �1 1 11 3
sin cos tan 3
2 2
⎛ ⎞ � � ⎜ ⎟
⎝ ⎠
[1]
2. Let � be a binary operation on N given by HCF( , )a b a b� � for all a, b � N. Find 12 27.� [1]
3. If
2 4 1
3 2 1
0 0 2
� � , write cofactor of the element a31
. [1]
4. Find value of x for which the matrix
1 2
3 1
x
x
�⎡ ⎤
⎢ ⎥⎣ ⎦
is singular. [1]
5. Evaluate � �� �� �1 1 1x x x dx � �∫ . [1]
6. If ˆ ˆ ˆˆ 2a i j k� � and ˆ ˆ ˆ ˆ,b i j k� � find | |a b�
��
[1]
7. Write the degree of differential equation [1]
22
2
0
dy d y
y
dx dx
⎛ ⎞⎛ ⎞ � � �⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
8. If
2 1 4 0 9 2
,
3 2 1 2 14 x
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
�⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
then find value of x. [1]
(10)
Sample Question Paper for Class XII Mathematics (Code-A)
9. Write the vector equation of a line given by [1]
1 1
2 3 4
x y z � �
10. Find distance of point (1, 2, 3) from the plane 2x + 3y – 4z = 0. [1]
SECTION-B
Short Answer Type Questions :
11. If x = 2(� – sin�) and y = 2(1 + cos�). Find
dy
dx
at .
2
�� � . [4]
12. Find the value of ‘a’ for which the function f defined as [4]
3
( 1), if 0
( ) tan sin
, if 0
a x x
f x x x
x
x
� �⎧
⎪� ⎨ �⎪⎩
is continuous at x = 0
13. Evaluate
1
2
sin
1
x x
dx
x
∫ [4]
OR
Evaluate
sin( )
sin
a x
dx
x
∫
14. Show that 1 1 1
3 8 84
sin sin cos
5 17 85
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
[4]
15. Let
3
: {–1}
2
f R R
⎧ ⎫ � ⎨ ⎬
⎩ ⎭
be defined as
2 3
( ) .
3 2
x
f x
x
��
Find the function g(x) such that fog = gof = IR
. Also
find domain of g(x) [4]
OR
Show that f : R � R defined as f(x) = 2x + 7 is one-one onto function. Also find its inverse
16. Solve the differential equation [4]
sec sec tan
dy
y x x x
dx
� �
17. If , ,a b c
�� �
are three vectors such that | | 5, | | 8a b� ��
�
and | | 3c ��
and ˆ ˆ8 6 .a b c i j� � � �
�� �
Find the value of
.a b b c c a� � � � �� �
� � � �
[4]
(11)
Mathematics (Code-A) Sample Question Paper for Class XII
18. Evaluate
1
11
x
dx
e �∫ [4]
OR
Evaluate
0
(| cos | | sin |)x x dx
�
�∫
19. Find the point on the curve y = x3
– 6x + 3 at which equation of tangent is 3x + y – 1 = 0. [4]
OR
Using differentials approximate 25.2
20. Using matrix method, solve the system of equations [4]
x + 2y + z = 1
x + y – z = 2
2x + 3y + z = 1
21. Write the equation of line in vector form which is perpendicular to the two lines
1 1 3
2 2 4
x y z � � and
1
1
4 3
x y
z
� � � , and passes through point (2, 1, 2). [4]
22. How many times a fair coin must be tossed, so that the probability of getting atleast one tail is more than
90%? [4]
SECTION-C
Long Answer Type Questions :
23. Show that of all the rectangles inscribed in a given fixed circle, the square has maximum area. [6]
24. Write the equation of plane passing through points A(1, 2, 1), B(2, 3, 1) and C(2, 1, 0). Also find its distance
from point (1, 1, 1). [6]
25. Using integration, find the area of bounded region between the curve y = x2
and y = –|x| + 2. [6]
OR
Find the area of region
� �2
( , ) : 0 1, 0 1, 0 2x y y x y x x� � � � � � � �
26. A factory makes nuts and bolts. A nut takes 1.5 hrs of machine time and 3 hrs of worker’s time in its making
while a bolt take 3 hrs of machine time and 1 hr of worker’s time. In a day, the factory has the availability of
not more than 42 hrs of machine time and 24 hrs of worker’s time. If the profit on a nut is Rs. 20 and on a
bolt is Rs. 10, find the number of nuts and bolts the factory should manufacture to earn maximum profit. Make
it as L.P.P and solve it graphically. [6]
(12)
Sample Question Paper for Class XII Mathematics (Code-A)
27. Using elementary transformations, find the inverse of matrix [6]
1 3 2
3 0 1
2 1 0
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎣ ⎦
28. Find the particular solution of the differential equation log 3 4
dy
x y
dx
⎛ ⎞ � ⎜ ⎟
⎝ ⎠
given that y = 0 when x = 0. [6]
29. In an examination of a school having 60 students in XII A and 40 students in XII B, 15 and 20 students failed
in XII A and XII B, respectively. A student is called by the principal at random, find the probability that he is
from XII A if he passed in the examination. [6]
OR
Let bag A contains 2 red and 3 white balls, another bag B contains 2 white and 3 black balls and another
bag C contains 2 black and 3 red balls. A bag and a ball out of it are selected at random. What is the
probability that selected ball is white assuming selections of bag A, B and C are equally likely?
� � �
(1)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
SECTION-A
1. No change, resistivity is material dependent.
2. Electric and magnetic fields. Transverse waves.
3. Diffraction of light occurs when size of obstacle (or aperture) is comparable to the wavelength of light used.
4. By Lenz law it is in clockwise direction.
5. When image is formed at distinct vision,
25
1 1 6
5
D
M
f
� � � � �
When image is formed at infinity,
25
5
5
D
M
f
� � �
6. KEmax
= eVSP
3 eV
3 volt.SP
V
e
� �
7. R - Remains unchanged
XL
- Becomes doubled
XC
- Becomes halved
8. Since dipole consists of equal and opposite charge hence net charge and thus net flux is zero.
SECTION-B
9. 80 �F and 10 �F are in parallel
Therefore, equivalent is 80 + 10 = 90 �F
This is in series with C
Hence, equivalent
90
40
90
C
C
��
72 FC⇒ � �
SOLUTIONS
(2)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
10. (a) Outside the shell,
enc
0
q
E ds� ��∫
��
�
2
0
4
q
E r⇒ � ��
Rr R>
2
0
4
q
E
r
⇒ �� �
(b) Inside the shell, Take Gaussian surface
enc
0
q
E ds� ��∫
��
�
as enc
0q �
r
R
0E⇒ �
11. The property of an electric circuit by virtue of which it opposes any change of flux or current in it by inducing a
current in itself is called self induction. It is numerically equal to the flux linked when unit current flows through it.
��� i
Li� �
L
i
��
SI unit is henry.
Consider, air cored solenoid
I I
BAN� �
� �� �� �2
0
ni r nl� � � �
2 2
0
L n r l
i
�� � � � �
12. Ampere’s circuital law : The magnetic circulation around any closed curve is equal to �0
times the electric
current threading the curve.
Amperian curve
0 enclosed
B dl I� � �∫
� �
�
Magnetic field due to a toroid
r
Current
in each
turn is I
0 enclosed
B dl I� � �∫
� �
�
0
2B r NI� � � �
0
2
NI
B
r
���
(3)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
13. Power of lens is a measure of its ability to converge or diverge a light beam falling on it. Power of lens is defined
as the tangent of the angle by which the lens converges (or diverges) a beam of light falling on it at a unit distance
from its optical centre.
tan
h
f
� �
when h = 1,
1
tan
f
� �
Power of a lens
1
tanP
f
� � � f
hF
�
Unit of power of lens is dioptre (D) or m–1
.
When lenses are in contact
eq 1 2
P P P� �
14.
Magnification :
(a) For strained eye, 1
o e
L D
M
f f
⎡ ⎤� �⎢ ⎥
⎣ ⎦
(b) For relaxed eye,
o e
LD
M
f f
�
15. (i) Metal : A solid is a conductor if its valence band overlaps its conduction band.
VB
CB
(4)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
(ii) Insulator : A solid is an insulator if the valence band and the conduction band do not overlap and are separated
by an energy gap between 3 eV and 6 eV.
VB
CB
Eg (~3 to 6 eV)
(iii) Semiconductor : A solid is a semiconductor if the energy gap is much smaller as compared to that in case
of insulators. Band energy gap is in the range (0.1 - 1.0 eV)
VB
CB
Eg (~0.1 to 1.0 eV)
16. (a) Here, L = 1 H, XL
= 3142 �
XL
= 2�fL
f = 500 Hz
(b)
1
0.10 F
2
C
X C
fC
� ⇒ � ��
OR
15 FC � �
1
212.3
2
C
X
fC
� � ��
rms
rms
0.52 A
C
V
I
X
� �
SECTION-C
17.
0
3
,
4
I dl r
dB B
r
� ���
��
� �
directed outwards
0
2
4
I dl
dB
r
���
r
dl
I
0 0
2
4 2
Idl I
B dB
r r
� �� � ��∫ ∫
For half coil
0
4
I
B
r
��
(5)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
18. Capacitance of a conductor is defined as the ratio of charge on it to its potential
Numerically,
Q
C
V
� .
Numerically, capacitance of a conductor is equal to the amount of charge required to raise potential through unity.
0
E
��
V
E
d
�
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
E
P
Q –Q
0
d
V
��
0
AQ
C
V d
�� �
On insertion of a conducting slab, capacitance of the conductor increases as 0
A
C
d t
��
, t is thickness of the
conducting slab inserted between the plates.
19. Mobility is defined as the magnitude of drift velocity per unit electric field.
,
d
e d e
V eE
V
E m
� � � !
e
e
e
m
!⇒ � �
(a) When temperature of the conductor decreases, e
! decreases and consequently e
� decreases.
(b)e
� doesn’t depend upon potential difference and thus it remains unchanged.
20. For a lens
1 1 1
f v u
�
Since the object is real, u = –x
Thus,
vx
f
v x
��
Further, as the distance between the object and the screen is d,
� �v d x�
Thus,
� �� �d x x
f
d x x
� �
2
0x dx fd⇒ � �
2
4
2
d d fd
x
⇒ �
For the image to be formed, x should be real,
i.e. 1 4 0
f
d
�
4d f⇒ �
When d < 4f, no image is formed.
(6)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
21. The emission of free electron from metal surface is called electron emission. It can take place through any of
following physical processes.
(i) Thermionic Emission : The release of electron from metal as a result of its temperature, i.e. by heating.
(ii) Field emission : It is a kind of electron emission in which a very strong electric field pulls the electron out of
metal surface.
(iii) Photoelectric Emission : It is that kind of electron emission in which light of suitable frequency ejects the
electrons from a metal surface.
Photoelectric effect equation max 0
KE h h� " " .
22. Phasor diagram,
#
VR
I0
E0
VC
V – L
VC
VL
0
sinE E t#�
L R C
0 0
,
R L L
V I R V I X� � and 0C C
V I X�
� �22
0 R L C
E V V V� �
� �22
0 0 L C
E I R X X� �
Impedance � �22
L C
Z R X X� �
0
I is max, when L C
# � #
1
2
2
fL
fC
� ��
( )I0 max
I
#r
#
1
2
f
LC
��
(7)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
23. Modes of propagation
(i) Ground wave propagation : Travel along curved earth’s surface from transmitter to the receiver.
(ii) Sky wave propagation : Radio waves travel skywards and if its frequency is below certain critical frequency
(typically 30 MHz). It is returned to the earth by ionosphere.
(iii) Space wave : In space wave propagation, radio waves travel in a straight line from transmitting antenna to the
receiving antenna.
Sky wave : The sky wave below critical frequency travels from the transmitting antenna to receiving antenna
via ionosphere. The ionosphere consists of layers of air molecules which have become positively charged by
removal of electrons by sun’s ultraviolet radiations. On striking the earth, the sky wave bounces back to the
ionosphere where it is again gradually refracted and returned earthwards as if by reflection. This continues
until it is completely attenuated.
24. In 56
26
Fe nucleus, there are 26 protons and 30 neutrons
Mass defect � �56
26
26 30 Fep n
m m m� �
= 0.528461 u
Binding energy = (mass defect)c2
$ 0.528461 931.5 MeV 492.26 MeV� �
�Binding Energy 492.26 MeV
8.76 MeV
Nucleon 56
� �
(8)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
25. Consider the -decay of 238
92
U
238 234 4
92 90 2
U Th He��� �
Neutron-proton ratio before -decay =
238 92
1.587
92
�
Neutron-proton ratio after -decay =
234 90
1.6
90
�
Thus, neutron-proton ratio increases during -decay.
Nuclear Force : Force between nucleons is called nuclear forces. It is one of the four fundamental forces in
nature.
Properties :
(i) It is an attractive force.
(ii) It is independent of the interacting nucleons.
(iii) It is a short range force.
(iv) It is a non-central force.
OR
(a)� �
2
2
13.6
eVn
Z
E
n
�
For Li , 3Z
�� �
2
122.4
eVn
E
n
�
For transition from n = 1 to n = 3
3 1
hc
E E E� � �%
114.2 Å% �
(b) No. of spectral lines = 3
26. During solar flare, a large number of electrons and protons are ejected from the sun, some of them get trapped in
the earth’s magnetic field and move in helical paths along the field lines. The magnetic field lines come closer to
each other near the magnetic pole. Hence, density of charges increases near the poles. These particles collide
with atoms and molecules of the atmosphere. Excited oxygen atom emit green light and erected nitrogen atoms
emit pink light.
(9)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
SECTION-D
27. (a) Consider point P on the screen, path difference
2 1
x S P S P� �
2
2 2
2
2
d
S P y D
⎛ ⎞� � �⎜ ⎟
⎝ ⎠
2
2 2
1
2
d
S P y D
⎛ ⎞� �⎜ ⎟
⎝ ⎠Q
P
S
D
y
d
S1
S2
2 2
2 1
2S P S P yd �
� �2 1 2 1
2
2
yd
S P S P S P S P D
D
� & &
yd
x
D
⇒ � �
For constructive Interference,
x n%� �
yd
n
D
⇒ % �
� �0,1,....
n D
y n
d
%⇒ � �
For destructive interference
� �2 1
2
x n
%� � �� � � �2 1
0,1,....
2
n D
y n
d
� %⇒ � �
Hence, fringe width
D
d
%�
(b) Fresnel distance,
2
F
a
Z �%
40 mF
Z �
OR
Huygens Principle : Every point on wavefront may be considered as a source that produces secondary wavelets.
These wavelets propagate in the forward direction with a speed equal to speed of wave motion. The surface which
touches these wavelets at any later instant is the position of new wavefront, called secondary wavefront.
1 2
AD CE
v v
�
sin
AD
i
CD
� from �CAD
sin
CE
r
CD
� from �CED
1
2
sin
sin
vi AD CD
r CD CE v
� � �
(10)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
As 1 2
2 1
,
vc
v
v
�� �� �
2
1
sin
sin
i
r
���
sin i� = constant (Snell’s law)
Convex Lens
Refracted wavefrontIncident wavefrontRefracted
wavefront
Incident
wavefront
28. (a) V-I characteristic
VF(V)
VR (V)
IF (mA)
IR ( A)�
Minority
CarriersForward Bias
Reverse
Breakdown
Breakdown
Majority
Carriers
V (Knee Voltage)K
80
60
40
20
–80 –60 –40 –20
–1
–2
Forward Bias
P
n
V
Reverse Bias
P
n
V
(b)
1242 eV nm
0.207 eV
6000 nm
g
hc
E � � �%
Since g
E E' , the photodiode cannot detect wavelength of 6000 nm.
(11)
Physics (Code-A) Solutions of Sample Question Paper for Class XII
OR
(a) Transistor as a common - emitter amplifier.
Circuit Diagram.
ac inputac output
VBB
BE
BIB
CIC
VCE
RL
VCC
I
E
Operation :
(i) With no signal input
CE CC C L
V V I R� �
(ii) With signal applied to the emitter base circuit for positive half cycle, increases the forward bias resulting
in an increase in collector current. During negative half cycle, the input signal opposes the forward bias
of the input circuit, thereby reducing the emitter and consequently the collector current.
dc current gain C
B
I
I
� �
(b)3
9 V
30 A
300 10
CC BE CC
B
B B
V V V
I
R R
� & � � �� �
(as BE CC
V V'' )
� �50 30 A 1.5 mAC B
I I� � � � �
� �� �3 3
9 V 1.5 10 A 2 10 6 VCE CC C C
V V I R
� � � � � �
29. (a) Magnetic field at a point on Axial line : End on position
� �0
2
4
S
m
B
r a
�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �
�
� �0
2
4
N
m
B
r a
�⎛ ⎞� ⎜ ⎟�⎝ ⎠
�
2a
r
–m +m
S N
� �0
2
2 2
4
4
P S N
m ra
B B B
r a
�� � � ��
� � �
Since r >> a
0
3
2
4
P
M
B
r
�⎛ ⎞⎛ ⎞� � ⎜ ⎟⎜ ⎟� ⎝ ⎠⎝ ⎠
�
Magnetic field at a point on equatorial line
� �0
2 2
4
N
m
B
r a
�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �
�
� �0
2 2
4
S
m
B
r a
�⎛ ⎞� ⎜ ⎟�⎝ ⎠ �
�
S
B
P
B
N
B
S N
P
(12)
Solutions of Sample Question Paper for Class XII Physics (Code-A)
� �0
3/2
2 2
2
4
P N S
am
B B B
r a
�⎛ ⎞� � � ⎜ ⎟�⎝ ⎠ �
� � �
r >> a
� �0
3
4
P
M
B
r
�⎛ ⎞� ⎜ ⎟�⎝ ⎠
�
axial equatorial
2B B�� �
(b) As m
C
T
( �
'
'
m
m
T
T
(⇒ �
(
' 200 KT⇒ �
OR
(a) Biot Savart’s law,
dB
�
the magnetic field at point P
0
2
sin
4
Idl
dB
r
� ��� r
dl
�
P
I
⇒net
2 sindB dB� ��
⇒
0
net 2
2 sin
4
Idl
dB
r
�� ��
�
⇒
0
net 2
2
4
Idl a
dB
r r
���
dB
dB
x
r
��
a
x
I⇒
0
net 3
2
4
Ia
dB dl
r
���∫ ∫
⇒ � �
2
0
net 3/2
2 2
Ia
B
a x
���
(b)
420 A
30 div 6 10 A
div
g
I
�⎛ ⎞� � � �⎜ ⎟
⎝ ⎠
Shunt required 0.015
1
g
G
S
I
I
� � �
G
S
Ammeter
� � �
(13)
Chemistry (Code-A) Solutions of Sample Question Paper for Class XII
SECTION-A
1. 3-Bromo-5-fluoro-3, 5-dimethylheptane
2. (i) It is a zero order reaction.
(ii)
0
1/2
[R]
t
2K
�
3. Out of the five electron pairs around the central Br-atom, the two lone pairs are at equatorial positions to minimise
the repulsive interactions. Hence, BrF3
has T-shaped structure.
F Br
F
F
4. Dispersed phase – milk fats
Dispersion medium – water
5. Number of A-atoms per unit cell
1
8(corners) 1
8
� � �
Number of B-atoms per unit cell
1
6 3
2
� � �
Hence, the formula of the compound is AB3
.
6. ABaq.
���⇀�↽����
A–
(aq.) + B+
(aq.)
Initially 1 0 0
At equilibrium 0.5 0.5 0.5
van’t Hoff factor 1 1.5� � �
7. NaCN is used as a depressant for ZnS and prevents it from coming with froth.
8. 6XeF4
+ 12H2
O ��� 4Xe + 2XeO3
+ 24HF + 3O2
SECTION-B
9. The structural formula of 4-methylpent-3-en-2-one is
H C — C — CH C — CH3 3
O
—
CH3
Due to the presence of H C — C — 3
O
group, it will respond positive to iodoform test.
SOLUTIONS
(14)
Solutions of Sample Question Paper for Class XII Chemistry (Code-A)
10.Osmotic pressure may be defined as the excess pressure which must be applied to the solution side to just
prevent the osmosis. Osmotic pressure is directly proportional to temperature as well as pressure.
11. 2Cu2
S + 3O2
��� 2Cu2
O + 2SO2
Cu2
S + 2Cu2
O ��� 6Cu + SO2
12. The two components of starch are amylose and amylopectin. Amylose is a linear polymer of -D-(+)-Glucose
while amylopectin is heavily branched polymer of -D(+)-Glucose.
13. (a) H—C — (CHOH) — CH — OH CH CH CH CH CH CH4 2 3 2 2 2 2 3
�����
O
Reduction
HI/P
n-Hexane
(b)(CHOH) + 5(CH CO) O (CHOCOCH ) + 5CH COOH
4 3 2 3 4 3���
——
CH OH2
Glucose
CHO
——
CH OCOCH2 3
Pentacetyl glucose
CHO
14.H SO H+ :OSO H
2 4 3���
H C — CH — O – H + H H C —CH — O — H3 2 3 2
+����
—
+
H
Slow
H C — CH + H O3 2 2
+
H — C — C — H + :OSO H H C = CH + H SO3 2 2 2 4
���
——
H
H
—
H
)Fast
15. (a) Hofmann’s Bromamide reaction:
H C — C — NH + Br + 4KOH H C – NH + 2KBr + K CO + 2H O3 2 2 3 2 2 3 2
���Heat
O
(b) Gabriel phthalimide reaction:
—
CO
CO
—
NH ����KOH(alc.)
–H O2
—
CO
CO
—
NK ����C H –I2 5
–KI
—
CO
CO
—
N — CH CH2 3
H /H O
+
2
—
COOH
COOH
+ H C — CH — NH3 2 2
(15)
Chemistry (Code-A) Solutions of Sample Question Paper for Class XII
16. (a) Distinction can be made by Lucas reagent. Treat both the solutions separately with Lucas reagent which is
a mixture of HCl(g) and anhy. ZnCl2
. The compound that shows turbidity instantly is 2-methylpropan-2-ol
while the solution that shows turbidity after 5 minutes of heating, is Butan-1-ol.
(b) Distinction can be made by litmus test. Add a few drops of blue litmus separately to the solutions of both
the compounds. The solutions which changes the colour of the blue litmus to red is that of phenol while
the other is benzyl alcohol.
17. (a) Aniline does not undergo Friedel Crafts reaction. It being a Lewis base, coordinates with anhy. AlCl3
.
—
H N AlCl2 3
���
The amino group now, is not in a position to activate the benzene ring towards electrophilic substitution,
that is alkylation or acylation. Also, AlCl3
does not remain free to generate the carbonation from alkyl or
acyl halide.
(b) The –NH
2
group over benzene ring makes it so electron-rich site that Br2
molecule itself gets polarized
under its influence. This helps in the generation of the electrophile Br+
, even in the absence of a halogen
carrier.
OR
(a) During acylation aniline gives CH – C – NH3
O
, which is a resonance stabilized compound.
(b) 3°-amine does not contain replaceable hydrogen on N-atom.
18. The solution of acetone and chloroform will show negative deviation from their ideal behaviour.
Cl–C–H..........O=C
Cl
Cl
�(+) �(–)CH
3
CH3
There is formation of hydrogen bond between acetone and chloroform molecules, so the escaping tendency of
both components is lowered.
SECTION-C
19. We know that, 3
A
z M
,
N a
�� ��
here ‘a’ is edge length
23 8 3
6.23 6.023 10 (4.00 10 )
z 4.002 4
60
� � � �� � � �
� The cubic unit cell is ‘face-centered cubic’.
(16)
Solutions of Sample Question Paper for Class XII Chemistry (Code-A)
Now, in fcc lattice,
a
r
2 2
�
8
4 10
2 2
��
= 1.414 ×10–8
cm
20. (a) Terylene is a condensation polymer and its monomers are Ethylene glycol (HOCH2
CH2
OH) and terephthalic
acid HOOC COOH
(b) Teflon is an addition polymer and its monomer is tetrafluoroethylene (CF2
= CF2
)
(c) Its monomer unit is isoprene i.e., CH –C=CH–CH3 3
CH3
21. (a) These two compounds are ionization isomers to each other and on treating the aq. solutions of the two with
AgNO3
(aq), [CO(NH3
)5
(SO4
)]Br gives yellow ppts of AgBr while the other compound does not.
(b) The electronic configuration of Fe(Z = 26) is [Ar], 3d6
, 4s2
while that of Fe(II) is [Ar], 3d6
since F–
is a
weak field ligand that does not cause pairing up of electrons, sp3
d2
hybridization takes place.
[FeF ] :6
4–
3d 4s 4p 4d
sp d3 2
hybridization
The structure of the complex is octahedral and it is a high-spin complex. On the other hand, CN–
is a strong
field ligand that causes greater crystal field splitting and hence, pairing up of electrons.
[Fe(CN) ] :6
4–
3d 4s 4p
d sp2 3
This way, d2
sp3
hybridization takes place and the octahedral structure is diamagnetic.
22. (a)
(i) The adsorbate molecules
are held to the surface of
adsorbent by weak van
der Waals forces.
(ii) It is not specific in nature.
(iii)It forms multimolecular
layers.
(i) Adsorbate molecules are
held to the surface of
adsorbent by strong
chemical forces (chemical
bonds).
(ii) It is highly specific.
(iii) It forms monomolecular
layers.
Physisorption Chemisorption
(17)
Chemistry (Code-A) Solutions of Sample Question Paper for Class XII
(b)
(i) When dispersion medium
likes dispersed phase,
the sol is called lyophilic.
(ii) They are quite stable and
are not precipitated easily.
(iii) They are reversible in
nature.
Lyophilic sol Lyophobic sol
(i) When dispersion medium
dislikes dispersed phase,
the sol is called lyophobic.
(ii) They are easily precipitated
by addition of a small
amount of a suitable
electrolyte.
(iii) They are reversible in
nature.
23. (a) ICl is polar due to electronegativity difference between I and Cl. But I2
is nonpolar covalent compound.
So stability of polar bond is less than covalent bond and ICl is more reactive than I2
.
(b)
N 1 odd electron (incomplete octet of N)
O
N — N
O
O
O
(Complete octet of N)
O
O
At lower temperature NO2
is converted into N2
O4
due to incomplete octet of N in NO2
.
(c) In H3
PO3
, oxidation number of P is +3 and in H3
PO4
, P present in higher oxidation number +5, so H3
PO4
is not further oxidised.
24. (a)
F
FF
F
Xe
O
sp d
3 2
, Square pyramidal
(b)
F
F
FP
F
F
Triangular bipyramidal
(c)
S
O
H
O
S
OO
O O
O O
H
sp
3
-Hybridised Sulphur
(18)
Solutions of Sample Question Paper for Class XII Chemistry (Code-A)
OR
(a) Ca P + 6H O 3Ca(OH) + 2PH3 2 2 2 3
���Cal. Phosphate Phosphine
(b) XeF + NaF Na [XeF ] [Sodium heptafluoroxenate (vi)]6 7
��� + –
(c) Cu + 2H2
SO4
(conc.) ��� CuSO4
+ SO2
+ 2H2
O
25. KP
= Ae–EP
/RT
; P � Presence of catalyst
KA
= Ae–EA
/RT
; A � Absence of catalyst
P A(E E )/RT E/RTP
A
K
e e
K
�� �
P A A
10
A
K E E
Ine log 10
K RT 2.303RT
� �
P P A
A
K (E E ) (45 55)
Antillog Antilog
K 2.303RT 2.303R300K
⎡ ⎤ ⎡ ⎤� �⎢ ⎥⎢ ⎥
⎣ ⎦ ⎣ ⎦
P
A
K 10
Antilog 1.003
K 2.303 8.314 300
� �� �
P AK K�
26. (a) KCN is ionic compound and generate C
–
N ion in solution, the attack takes place mainly through carbon atom
due to C — C bond strength is higher than C – N bond strength.
(b) Resonating structure of chlorobenzene
CI CI
)CI
)CI
)CI
—
— — — — — — —
Lone pair of Cl is conjugated with double bonds of ring. So the electron density of ortho and para
positions increases by resonance and it undergoes electrophilic substitution at ortho and para positions
while Cl is deactivating group.
(c)CH = CH – CH – Cl CH = CH – CH + Cl
2 2 2 2���
Allyl Chloride Allyl (more stable carbocation
by resonance)
Nu)
CH – CH – CH – Cl CH – CH – CH + Cl3 2 2 3 2 2
���n-Propylchloride 1° carbocation
Nu)
Intermediate allyl carbocation is more stable than 1° carbocation. So the allyl chloride is more reactive towards
nucleophilic substitution.
(19)
Chemistry (Code-A) Solutions of Sample Question Paper for Class XII
27. Deficiency diseases by vitamin – A � Xerophthalmia, Night blindness
C � Scurvy (Bleeding gums)
Vitamin-A � Retinol
Vitamin-C � Ascorbic acid
SECTION-D
28. (a) (i) CH — CH + CH — CH CH — CH — CH — CH CH — CH — CH 3 2 3 2 3 2
������ ���
O
H
O
Aldol
(Condensations)
Ba(OH)2
OH O
LiAlH4
OH
CH — OH2
Butan-1-3-diol
(ii) CH — C — CH + CH — Mg — I CH CH — C — OH 3 3 3 3 3
�� ����
O
C
CH3
CH3
OMgI
–Mg(OH)I
H O2
CH3
CH3
Acetone Methyl
Mag. Iodide
Adduct
tert-butyl alcohol
(iii)
COOH
��������
( HNO +H SO )3 2 4
COOH
NO2
Benzoic Acid Meta ntiro benzoic acid
(Conc.)
(b) (i) Clemmensen reduction
CH — CHO + 4[H] CH — CH + H O3 3 3 2
����Zn–Hg
�Ethanal Ethane
(ii) Cannizzaro reaction
2HCHO + NaOH CH OH + HCOONa���3
Formaldehyde Methanol
�
OR
(a) (i)
CH — CH CH — CH CH — CH — Cl CH — CH — CH — CH3 3 3 3 2 3 2 2 3
������� ����� ���
Ethanal
Clemmenson
reductionEthane
Mono
Chlorination
Wurtz
Reaction
O
Zn–Hg (Conc. HCl) Cl /U.V.2
Na
Cl /U.V.2
CH — CH — CH — CH CH — CH — CH = CH CH — CH — CH — CH3 2 2 2 3 2 2 3 2 3
*����� *�����
Cl
(CH ) COK3 3
(i) B H ,THF2 6
(ii) H O /OH2 2
–
Butan-1-ol
OH
(20)
Solutions of Sample Question Paper for Class XII Chemistry (Code-A)
(ii)
CHO
Benzaldehyde
���[O]
KMnO4
COOH
���NaOH
CaO
��������Cl – CH – CH = CH2 2
AlCl3
CH – CH = CH2 2
����B H , THF2 6
H O2 2
CH – CH2 2
HO – CH2
3-Phenyl Propanol
(iii)
CH – CH2 3
Ethyl Benzene
������Baeyer’s reagent
KMnO4
COOH
Benzoid Acid
������Soda Lime
NaOH + CaO
Decarboxylation
Benzene
(b) (i) Aldol condensation
CH — CH + CH — CHO CH — CH — CH — CHO3 2 3 2
����
O
Ba(OH)2
OH
Ethanal Aldol
OH
(ii) HVZ reaction
CH — C — OH CH — C — OH CH — C — OH 3 2
��� ���
O
�Br /P
2
Br
O
Br /P2
�
Br
Br
O
Br /P2
� CBr — C — OH2
O
29. (a) H+
+ e–
� H2
(g)
Ecell
= 2
pH0.0591
– log
1 H�
�⎡ ⎤
⎣ ⎦
or –0.118 =
0.0591 1
– log
1 H�
�⎡ ⎤
⎣ ⎦
� [H+
] = 10–2
mol/L
� pH = 2
(b)At anode H (g) 2H + 2e
At cathode 2H + 2e +1/2 O (I) H O(l)
Net reaction
2
2 2
��
��
+ –
+ –
H (g) +1/2 O (g) H O(l)2 2 2
��
(21)
Chemistry (Code-A) Solutions of Sample Question Paper for Class XII
(c) Anode Zn–Hg
Cathode Carbon Paste
��
Mercury cell
Overall reaction,
E = 1.35 V0
Zn(Hg) + HgO(s) ZnO(s) + Hg(l)��
In the final equation of reaction, no solvent or ion is involved therefore, the emf does not drop easily.
OR
(a) E°cell
= E°cathode
– E°anode
= 0.34 – (–0.25)
= 0.59 V
� �G° = –2 × 96500 × 0.59 J
= –113870 J
= –113.9 kJ
Now, �G° = –2.303 RT log KC
� logKC
=
–113900
–2.303 8.314 298� � = 19.96
� KC
= antilog (19.96)
� KC
� 1020
(b) (i) At cathode
Cu+2
(aq) + 2e–
�� Cu(s)
At anode
OH (aq) OH + e�� –
4OH �� 2H2
O + O2
(ii) 2H SO 2H + 2HSO2 4 4
+
At anode
2HSO H S O + 2e4 2 2 8�� –
H S O + H O 2H SO + ½ O2 2 8 2 2 4 2
��
At cathode
2H + 2e H
) – �� +2
Net reaction
H O H +1/2 O2 2 2
����Electric
Current
(22)
Solutions of Sample Question Paper for Class XII Chemistry (Code-A)
(c) W = ZQ
W W 9
Q 96500 96500
Z E 9
� � � � �
Q 96500 coulomb�
30. (i) Cl – C – NO2
Cl
Cl
.
Which is also called as tear gas. When it reaches in eye it irritates gland and brings tears.
(ii) Social responsibility and social justice.
(iii) 1, 1, 1 Trichloro-1-nitromethane
OR
(i) I2
< F2
< Br2
< Cl2
: Increasing bond dissociation energy
(ii) HF < HCl < HBr < HI : Increasing acid strength
(iii) SbH3
< AsH3
< PH3
< NH3
: Increasing base strength
(iv) H2
S < H2
Se < H2
Te < H2
O : Increasing boiling point
(v) HCl < HBr < HI < HF : Increasing boiling point
� � �
(23)
Mathematics (Code-A) Solutions of Sample Question Paper for Class XII
SECTION-A
1. � �1 1 11 3
sin ,cos ,tan 3
2 6 2 6 3
⎛ ⎞� � �⎛ ⎞ � � �⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
0
6 6 3
� � �� �
2. 12 × 27 = HCF(12, 27) = 3
3. Cofactor of a31
= (–1)3 + 1
[(4 × 1) – (2 × 1)] = 2
4. For singular matrix
1 2
0
3 1
x
x
��
$ (x + 1) – 2(3 – x) = 0
$ x + 1 – 6 + 2x = 0
5
3
x⇒ �
5. � �� �1 1 ( 1)x x x dx � �∫
3
2
( 1)( 1) ( 1)
3
x
x x dx x dx x C� � � � �∫ ∫
6.ˆ ˆ ˆ
3 2 2a b i j k� � � �
�
| | 9 4 4 17a b� � � � ��
�
7. Power of
2
2
1
d y
dx
� � degree of given differential equation
8. x = 4
9.ˆ ˆ ˆ ˆ ˆ( ) (2 3 4 )r i j i j k� � � % � �
�
10.
2 2 2
2(1) 3(2) 4(3) 8 12 4
29 292 3 4
� � �� �
SOLUTIONS
(24)
Solutions of Sample Question Paper for Class XII Mathematics (Code-A)
SECTION-B
11. � �2 2cos 2sin
dy d
d d
� � � � �� �
� �2 2sin 2 2cos
dx d
d d
� � � � �� �
2sin
2 2cos
dy
dyd
dxdx
d
��� � �
�
Now,
( /2)
dy
dx ���
⎛ ⎞
⎜ ⎟
⎝ ⎠
2(1) 2
1
2 2(0) 2
� � �
12. For continuity left hand limit at x = 0 should be equal to right hand limit at x = 0.
0
lim (0 1)
x
a a
�� � �
=
3
0
sin
sin
coslim
x
x
x
x
x
��
2
2
0 0
sin
sin (1 cos ) sin 2 1 12
lim lim . . .
4 cos 2cos
2
x x
x
x x x
xx x xx x
� �� �
⎡ ⎤⎛ ⎞
⎜ ⎟⎢ ⎥ ⎝ ⎠⎢ ⎥� � �⎢ ⎥
⎢ ⎥⎣ ⎦
1
2
a⇒ �
13. Put sin–1
x = �
2
and sin
1
dx
d x
x
� � � �
sin d⇒ � � �∫
Apply integration by parts, we get
cos cos d� � � � �∫
1 2
sin cos sin 1C x x x C
⇒ � � � � � � �
OR
sin( )
sin
a x
dx
x
∫
sin cos cos sin
sin sin
a x a x
dx
x x
⎛ ⎞� ⎜ ⎟
⎝ ⎠∫
sin cot cosa xdx adx� ∫ ∫
= sina�log|sinx| – xcosa + C
(25)
Mathematics (Code-A) Solutions of Sample Question Paper for Class XII
14.
1 1 13 8 84
sin sin cos
5 17 85
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Now,1 1
3 3
sin tan
5 4
⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
$ 1 18 8
sin tan
17 15
⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
$ 1 184 13
cos tan
85 84
⎛ ⎞ ⎛ ⎞�⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Now we have to prove,
1 1 13 8 13
tan tan tan
4 15 84
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ �⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
L.H.S.
1
3 8
4 15tan
3 8
1 ·
4 15
⎛ ⎞⎜ ⎟
� ⎜ ⎟
⎜ ⎟�⎜ ⎟
⎝ ⎠
=
1
45 32
60tan
60 24
60
⎛ ⎞
⎜ ⎟
⎜ ⎟�⎜ ⎟⎜ ⎟
⎝ ⎠
=
113
tan
84
⎛ ⎞
⎜ ⎟
⎝ ⎠
R.H.S. proved.
15. Given fog = gof = IR
$ g(x) is inverse of f(x)
2 3
3 – 2
x
y
x
�⇒ �
$ 3y – 2xy = 2x + 3 $ 2x(1+y) = 3(y – 1)
$
3 1
2 1
y
x
y
⎛ ⎞�⎜ ⎟�⎝ ⎠
13 1
( ) ( )
2 1
x
g x f x
x
⎛ ⎞⇒ � �
⎜ ⎟�⎝ ⎠
Domain of g(x), x + 1 , 0
1x⇒ ,
Domain of � �( ) 1g x R�
(26)
Solutions of Sample Question Paper for Class XII Mathematics (Code-A)
OR
At x = x1
, f(x1
) = 2x
1
+ 7
and at x = x2
, f(x2
) = 2x
2
+ 7
Let f(x1
) = f(x2
)
$ 2x
1
+ 7 = 2x
2
+ 7
$ x
1
= x2
is the only solution
Hence f(x) is an one-one function, also range of f is (–-, -)
Hence f is onto
Finding inverse of f(x),
y = 2x + 7
7
2
y
x
⇒ �
17
( )
2
x
f x
⇒ �
16. sec sec tan
dy
y x x x
dx
� �
Comparing with ( ) ( )
dy
yP x Q x
dx
⎛ ⎞� �⎜ ⎟⎝ ⎠
sec ln(sec tan )
I.F (sec tan )
xdxx x
e e x x
�∫� � � �
2 2
(sec tan ) (sec tan )y x x x x dx⇒ � � ∫
(sec tan )y x x x C⇒ � � �
17.ˆ ˆ
8 6a b c i j� � � ��
� �
Squaring both sides
2 2ˆ ˆ( ) (8 6 )a b c i j� � � �
�� �
$ ˆ ˆ ˆ ˆˆ( )( ) (8 6 )(8 6 )a b c a b c i j i j� � � � � � �� �
� � �
$ 2 2 2
| | | | | | 2( ) 64 36a b c a b b c c a� � � � � � � � � �� � �
� � � � � �
$ 25 64 9 2( ) 100a b b c c a� � � � � � � � �� �
� � � �
$ 2( ) 2a b b c c a� � � � � �� �
� � � �
$1a b b c c a� � � � � �
� �� � � �
18.
1
11
x
dx
I
e
��∫ … (i)
Using property ( ) ( )
b b
a a
f x dx f a b x dx� � ∫ ∫
1 1
1 11 1
x
x x
dx e dx
I
e e
⇒ � �� �∫ ∫ …(ii)
(27)
Mathematics (Code-A) Solutions of Sample Question Paper for Class XII
Adding (i) and (ii), we get
1
1
1
2
1
x
x
e
I dx
e
⎛ ⎞�� ⎜ ⎟⎜ ⎟�⎝ ⎠
∫
1
1
2I dx
⇒ � ∫
1
1
2I x ⇒ �
1 ( 1) 2
1
2 2
I
⇒ � � �
OR
0
| cos | | sin |x x dx
�
�∫
/2
0 /2 0
cos cos sinxdx xdx xdx
� � �
�
� �∫ ∫ ∫
/2
0 /2 0
(sin ) ( sin ) ( cos )x x x
� � ��� �
$ 1 + (–(–1)) + (1 – (–1)
$ 1 + 1 + 2 = 4
19.2
3 6
dy
x
dx
� = Slope of line
3x
2
– 6 = – 3
$ 3x
2
= 3 $ x = ±1
for x = 1 y = –2
and x = –1 y = 8
Point (1, –2) satisfies the line but (–1, 8) doesn’t, so required point (1, –2)
OR
Let ( )f x x�
Let x = 25 and �x = 0.2 then
�y = (x + �x)1/2
– x1/2
�y = (25.2)1/2
– (25)1/2
$ 25.2 25 5y y� � � � � �
Now approximating �y
dy
y x
dx
⎛ ⎞� � �⎜ ⎟
⎝ ⎠
1
(0.2)
2 x
�(as )y x�
1
0.2 0.02
2 5
� � ��
25.2 5 0.02 5.02⇒ � � �
(28)
Solutions of Sample Question Paper for Class XII Mathematics (Code-A)
20. x + 2y + z = 1
x + y – z = 2
2x + 3y + z = 1
1 2 1 1
1 1 –1 2
2 3 1 1
x
y
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥�⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
AX = B, X = A
–1
B
Now,
| A |
= 1(1 + 3) – 2 (1 + 2) + 1(3 – 2) = 4 – 6 + 1 = –1
1
4 1 –3 –4 –1 3
adj.
1 –3 –1 2 = 3 1 –2
1 1 –1 –1 –1 1
A
A
A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� � ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
1
–4 –1 3 1 3
= 3 1 –2 2 3
–1 –1 1 1 2
x
x y A B
z
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥� � �⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
� x = –3, y = 3, z = –2
21. Vector direction of lines perpendicular to both given lines is
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ2 2 4 (2 12) (2 16) (6 8) 10 14 2
4 3 1
i j k
i j k i j k� � � �
Line passing through (2, 1, 2) and parallel to ˆ ˆ ˆ
( 10 14 2 )i j k � in vector form is
ˆ ˆ ˆ ˆ ˆ ˆ(2 2 ) ( 10 14 2 )r i j k i j k� � � � % �
�
22. Let the coin tossed n times
Probability of getting at least one tail = 1 – (Probability of getting no tail)
1
1
2
n
⎛ ⎞� ⎜ ⎟
⎝ ⎠
Given that
1
1 0.9
2
n
⇒ .
1
0.1
2
n
⇒ .
1
2
0.1
n
⇒ .
2 10
n
⇒ .
4n⇒ .
So coin must be tossed 4 times.
(29)
Mathematics (Code-A) Solutions of Sample Question Paper for Class XII
SECTION-C
23. Let ABCD be a rectangle inscribed in the circle of given radius r.
AB = x and BD = y
Area = x × y
Also, x2
+ y2
= 4r
2
For maximum area,
2
2
0, 0
dA d A
dx dx
� '
� �1
2 22
4 0
dA d
x r x
dx dx
⎡ ⎤
� �⎢ ⎥
⎢ ⎥⎣ ⎦
A B
C D
y
x
r
r
2 3
2 2 4
(8 4 )
0
2 4
r x x
r x x
⇒ �
2x r⇒ � , also
2
2
0
d A
dx
' for 2 x r� . And for 2 , 2x r y r� �
Hence for maximum area ABCD must be square
24. Let the plane be Ax + By + Cz + 1 = 0
Passing through (1, 2, 1), (2, 3, 1) and (2, 1, 0), we get
A + 2B + C + 1 = 0 …(i)
2A + 3B + C + 1 = 0 …(ii)
2A + B + 1 = 0 …(iii)
Given A = –1, B = 1 and C = –2
$ Required plane –x + y – 2z + 1 = 0
$�x – y + 2z = 1
Distance from point (1, 1, 1) is
2 2
1 1 2 1 1
61 1 4
d
� � �� �
25. Required area = area of ABCD
1
2
0
2 ((– 2) )x x dx� � ∫
1
2 3
0
2 2
2 3
x x
x
⎛ ⎞/ � ⎜ ⎟⎜ ⎟
⎝ ⎠
1 1
2 2 0
2 3
⎛ ⎞/ � ⎜ ⎟
⎝ ⎠
(–2, 0)
(2, 0)
(1, 1)(–1, 1)
y
x
=
+ 2
y x = – + 2
A
B
C
0
5
2 2
6
⎛ ⎞/ ⎜ ⎟
⎝ ⎠
12 5 7
2 units
6 3
⎛ ⎞/ �⎜ ⎟
⎝ ⎠
(30)
Solutions of Sample Question Paper for Class XII Mathematics (Code-A)
OR
Area of shaded region required
1 2
2
0 1
( 1) ( 1)x dx x dx� � �∫ ∫
1 2
3 2
0 1
3 2
x x
x x
⎛ ⎞ ⎛ ⎞
� � � �⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
P(0, 1)
Q
(1, 2)
(0, 2)
x = 2
x = 1
1 1 23
1 2 2 1 units
3 2 6
� � � � �
26. Let number of nuts be ‘x’ and bolts be ‘y’
� L.P.P. is
Maximise z = 20x + 10y
such that 1.5x + 3y � 42
and 3x + y � 24
B(4, 12)A(0, 14)
C (8, 0)O
3 + = 24x y1.5 +3 = 42x y
x
y
x . 0, y . 0
Vertices of feasible region are A(0, 14), B(4, 12), C(8, 0)
Z(A) = Rs. 140, Z(B) = Rs. 200, Z(C) = Rs. 160
For maximum profit
Number of nuts = 4
Number of bolts = 12
27. Given matrix can be written as
1 3 2 1 0 0
3 0 1 0 1 0
2 1 0 0 0 1
A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥�⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
Applying C2
� C2
+ 3C
1
, C3
��C3
– 2C
1
1 0 0 1 3 2
3 9 5 0 1 0
2 7 4 0 0 1
A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥⇒ �
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
2 2 3
1 0 0 1 1 2
2 3 1 5 0 1 0
2 1 4 0 2 1
C C C A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� � ⇒ �⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
3 3 2
1 0 0 1 1 3
5 3 1 0 0 1 5
2 1 1 0 2 9
C C C A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� ⇒ � ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
2 2 3 3 3
1 0 0 1 2 3
, – 3 1 0 0 4 5
2 0 1 0 7 9
C C C C C A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� � � ⇒ � ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
(31)
Mathematics (Code-A) Solutions of Sample Question Paper for Class XII
1 1 3 2 2
1 0 0 5 2 3
2 , 3 1 0 10 4 5
0 0 1 18 7 9
C C C C C A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� � � ⇒ �⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
1 1 2
1 0 0 1 2 3
3 0 1 0 2 4 5
0 0 1 3 7 9
C C C A
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥� ⇒ � ⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
1
1 2 3
2 4 5
3 7 9
A
⎡ ⎤
⎢ ⎥� � ⎢ ⎥
⎢ ⎥⎣ ⎦
28.
3 4x ydy
e
dx
�
4 3y x
e dy e dx⇒ �
Integrating both sides
4 3y x
e dy e dx�∫ ∫
4 3
4 3
y x
e e
C⇒ � �
4 3
4 3
y x
e e
C⇒ � � Putting x = 0, y = 0 we get
1
12
c �
4 3
1
0
4 3 12
y x
e e � �
4 3
3 – 4 1 0
y x
e e⇒ � �
29. Let Selected student of section A
Eventof student that student is passed
Student of section B
A
E
B
///
( )
( ) ( )
E
P P A
A A
P
E EE
P P A P P B
A B
⎛ ⎞ �⎜ ⎟
⎛ ⎞ ⎝ ⎠�⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠ � � �⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
45 30 3 2
0.75, 0.5, ( ) , ( )
60 40 5 5
E E
P P P A P B
A B
⎛ ⎞ ⎛ ⎞� � � � � �⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3
0.75
2.25 95
3 2 3.25 13
0.75 0.5
5 5
A
P
E
�⎛ ⎞ � � �⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠ � � �⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
(32)
Solutions of Sample Question Paper for Class XII Mathematics (Code-A)
OR
Bag A Bag B Bag C
2R + 3W 2W + 3B 2B + 3R
P(A) = Probability of selecting bag A =
1
3
P(B) = Probability of selecting bag B =
1
3
P(C) = Probability of selecting bag C =
1
3
P(W) = Probability of selecting white ball.
( ) ( ). ( ). ( ).
W W W
P W P A P P B P P C P
A B C
⎛ ⎞ ⎛ ⎞ ⎛ ⎞� � �⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
1 3 1 2 1 1 2 3 2 1
0
3 5 3 5 3 5 15 15 3
�� � � � � � � � � �
� � �