Subject: English
1.
Solve 4 unseen reading
comprehension passages (2 of each kind).
2.
Learn and write the difficult
words-meanings, Q-A, summary of the novel- Chapters1-4.
3.
Section B Writing (Write 1 each)
Notice, Advertisements(2 kinds),
Posters,Reports writing, Factual Description, process writing, Letter writing 4
types(1 each), Articles -2, Speeches -2, Debates-2,
4.
Write and learn the vocabulary
words of lessons- Lost spring, Deep
Water, The Enemy, Should Wizard Hit money
Subject: Physics
Prepare
a note on following topics and learnt it:
Chapter–2:
Electrostatic Potential and Capacitance
·
Electric potential
·
Potential difference.
·
Electric potential due to a
point charge.
·
A dipole and system of
charges.
·
Equipotential surfaces.
·
Electrical potential energy
of a system of two point charges and of
·
Electric dipole in an
electrostatic field.
·
Conductors and insulators,
free charges and bound charges inside a conductor.
·
Dielectrics and electric polarisation
·
Capacitors and capacitance.
·
Combination of capacitors in
series and in parallel.
·
Capacitance of a parallel
plate capacitor with and without dielectric medium between the plates.
·
Energy stored in a
capacitor.
·
ALSO SOLVE THE NCERT
QUESTION
Chapter–3:
Current Electricity
·
Electric current
·
Flow of electric charges in
a metallic conductor
·
Drift velocity, mobility and
their relation with electric current
·
Ohm's law
·
Electrical resistance
·
V-I characteristics (linear
and nonlinear),
·
Electrical energy and power
·
Electrical resistivity and
conductivity
·
Carbon resistors, colour
code for carbon resistors
·
Series and parallel
combinations of resistors
·
Temperature dependence of
resistance
·
Internal resistance of a
cell
·
Potential difference and emf
of a cell
·
Combination of cells in
series and in parallel
·
Kirchhoff's laws and simple
applications
·
Wheatstone bridge
·
Metre bridge
·
Potentiometer - principle
and its applications to measure potential difference and for comparing EMF of
two cells.
·
Measurement
of internal resistance of a cell.
·
ALSO SOLVE THE NCERT
QUESTION
Subject: Chemistry
UNIT (1) –
SolidState Practice Paper-1 02.05.2016
Q1 (i) Why is Frenkel defect not found in pure
alkali metal halides.
(ii)
What happens to the structure of CsCl when it is heated to about 760 k?
(iii) Fe3O4
is ferrimagnetic at room temperature and becomes paramagnetic at 850 K.Why?
(iv)
When atoms are placed at the corners of all 12 edges, how many atoms are
present per unit cell? 4
Q2 The composition of a sample of wustite is Fe0.93O1.00.
What percentage of iron is present in the form of
Fe(III)?
4
Q3 Calculate the packing efficiciency in cubic close
packing arrangement.
4
Q4 If the radius of octahedral void is r& radius of atoms in close packing is R,
drive the relation between
r
&R.
4
Q5 Niobium crystallizes in body centered cubic
structure. If density is 8.55 gm cm-3. Calculate atomic
radius
of niobium . Atomic mass of niobium is 93 u.
4
Q6 Explain the following term with suitable example
(a) Schottky defect (b)Frenkel defect
(c) Interstitials (d) F-centres
4
Q7 What is a semiconductor? Describe the two main
types of semiconductors & contrast their conduction
mechanism.
4
Q8 (a) In
terms of band theory, what is difference between
(i) Conductor and an insulator (ii) Conductor and semiconductor
2
(b) If NaCl is doped with 10-3
mol % of SrCl2 , what is the
concentration of cation vacancy? 2
Q9 (a) What type of stoichiometric defect is shown
by :
2
(i) ZnS (ii) AgBr
(b)
Classify the following as p-type or n-type semiconductor: 2
(i) B doped with Si (ii) Si doped with B
Q10 (i) Why is glass considered a super cooled
liquid?
(ii)
Name the parameters that characterize a unit cell.
(iii) What
is the two dimensional coordination number of a molecule in square close-packed
layer ?
(iv)
Give the significance of a lattice point.
4
Answers→
Q1 Because
ions can not get into interstitial sites due to their larger size.(b) CsCl structure transforms into
NaCl
structure.(c) due to randomization of spins at high temp.(d) at corners 8 x 1/8
= 1 atom(ans).
Q2 Let atoms
of Fe+3 = x , atoms of Fe+2
= 0.93—x . For electrical neutrality,
Total positive charge = Total negative charge
2(0.93—x ) + 3x = 2,
then x = 0.14( Fe+3atoms), %Fe+3 = 0.14x100/0.93 =
15.05
% 0f Fe+2 = 100—15.05 =
84.95.
Q3 Packing
efficiency in ccp or fcc = 74.06 %
derivation in book
Q4 r / R = 0.414
(ans).
Q5 Z = 2, M
= 93, d = 8.55 gm/cm3, N0 = 6.022 x 1023, d =
Z x M / N0 x 10—30 x a3
Calculate a3 = 3.61 x 107 , a = 330.5 pm.
Q6 (a)
Schottky—it arises if some of the atoms or ions are missing from their normal
lattice sites. Equal
number
of cations and anions are missing. Density is lowered. Occurs in compounds
having high C.N
&Cations and anions are of similar size.eg Alkali metal halides (B)
Frenkel—it arises if some oftheatoms
or ions are missing from their
normal positions and occupy an interstitialsite between lattice points.Density
remains same. Occurs in compounds having low C.N. Cations are smaller
than anions. Ag halidesAgBr contains both
defects. (c) Vacant sites in the
structure of crystal lattice. (d) Electrons trapped in anion
vacancies arerefered as F—centres.They
are responsible for colour. Nonstoichiometric KCl is violet in colour.
Q7 Those
substances which are insulator at lower temperature but conduct electricity at
higher temperature
are
called semiconductors.Thermal energy available at room temperature is
sufficient to excite electrons
from
highest occupied band to the next permitted band.
Intrinsic—Electrical conductivity is due to ther effect. No extra
impurity is added.(Ge& Si at room temp).
Extrinsic—Electrical conductivity is due to impurity effect. Extra
substance is added. They are of two
types,
n—type and p—type.
n-type—Electrical conductivity is due to electrons. eg group 13 elements
doped with group 14 elements.
p-type—Electrical conductivity is due to electron deficient bond or
electron vacancy or holes.eg group 14
elements
doped with group 13 elements.
Q8 (a) In
conductors energy gap is very-very small less than 50 kj/mole(metals).
In insulators energy is large more than 300 kj/mole(rubber).
In semiconductors energy gap is 30—300 kj/mole.
(b)
One cation of Sr+2 would create one vacancy. Conc of cation vacancy
= 10—3mole % = 10—3/ 100 =
10—5mole.
No of Sr+2 ions in 10—5mole = 6.022 x 1023x 10—5= 6.022 x 1018 ions = cation
vacancies.
Q9 (a) ZnS—Frenkel or dislocation. AgBr—Frenkel and
schottky both. (b) i—n-type, ii—p-type.
Q10 (a)
Because it has the property to flow
slowly.
(b)
Three dimensions and three angles.
(c) Four.
(d)
lattice points are occupied by constituent particles.
Practice Paper- 2
Q1 (a)Why
is the window glass of old buildings thick at the bottom?
3
(b)What happens when CaCl2 is introduced to the AgCl crystal?
(c)What happens when CsCl is heated to 760 K?
Q2 (a)Give
significance of a lattice point.
(b)What type of defect can arise when a solid is heated? Which physical
property is affected by it
and in what way?
(c)What is meant by co-ordination number?
3
Q3 Analysis
shows that Nickel Oxide has oxide Ni0.98O1.00. What
fraction of Ni exists as Ni+3& Ni+2 ions? 3
Q4 KF has
NaCl structure, What is the distance between K+ ions and F-
ions, if density is 2.48 gm/cm3. 3
Q5
(a)Explain intrinsic and extrinsic semiconductors with suitable
examples. 3
(b)Classify the following as n-type or p-type semiconductors
(i) B doped with Si (ii) Si doped with B
Q6 Explain
the following with suitable examples 3
Frenkel defect, F-centres, Ferrimagnetism
Q7 Derive
the relation between r and R for tetrahedral void. 3
Q8 An
element has BCC structure with a edge length 288 pm. The density of element is
7.2 gm/cm3.
How
many atoms are present in 208 gm of this element?
3
Q9
Calculate the efficiency of packing in case of a metal crystal for face
centred cubic.
3
Q10 A metal
crystallises into two cubic phases FCC & BCC whose edge lengths are 3.5
& 3.0 A0 respectively.
Calculate the ratio of densities of FCC & BCC.
3
ANSWERS
Q1 (a) Glass
has the property to flow slowly (b) vacancies are created(schottky defect).
(c) it
changes to NaCl structure. C.N changes from 8 to 6
Q2 (a)
particles are present on lattice points (b) schottky defects, density decreases
(c) definition
Q3 NCERT page No-31, Q -1.16 Ni+2 = 96 %, Ni+3 = 4 %
Q4 Z = 4, M = 39 + 19 = 58, d = 2.48 , NA
= 6.022 x 1023, calculate a (5.375 x 10—10cm )
distance between K+ and F—ions =
5.375 x 10—10cm / 2 = 2.688 x 10—10 cm.
Q5 extrinsic
contains impurities & intrinsic no impurities (i) n-type (ii) p-type
Q6
definitions in book
Q7 derivation r / R = 2.25 in book
Q8 Z =
2, M = ?, NA = 6.022 x 1023,
d = 7.2, calculate M(51.8),
No of
atoms = 6.022 x 1023 x 208/51.8
= 24.16 x 1023ANS
Q9
efficiency FCC = 74 % derivation in book
Q10 d(fcc) = 4 x M / NA x (3.5)3, d(bcc)
= 2 x M / NA(3)3
d(fcc) / d(bcc) = 4 x 33 / 2
x (3.5)3 = 1.26 ans
Practice paper-3
Q1 (a) Give
significance of a lattice point
(b)
What type of defect can arise when a solid is heated? Which physical property
is
affected by it & in what way?
(c)
What is meant by co-ordination number?
3
Q2 Analysis
shows that Nickel oxide has the formula
Ni0.98O1.00 . What fraction of Nickel
exists as Ni+3& Ni+2 ions?
3
Q3 (a)
Explain extrinsic and intrinsic semiconductors with examples.
(b)
Classify the followings as p-type or n-type of semiconductors
(i) B doped with Si (b) Si doped
with B
Q4 Derive
the relation between r & R in tetrahedral void.
3
Q5 An element
has BCC structure with a edge length of 288 pm. The density of element is
7.2gm/cm3. How many atoms are present in 208 gm of this
element? 3
Q6 State
Henry,S law for solubility of a gas in liquid. Explain the
significance of Henry,S
Constant(KH) at the same temperature, H2 is more
soluble than He in water.Which of
them
will have higher value of KH and why?
1+1+1/2+1/2
Q7 Define
the followings;
(i) Raoult,S law
(ii) Azeotropes (iii)
Colligative properties
3
Q8 What do
you understand by relative lowering in V.P? Show that it is a colligative
Property. How will determine the molar mass by relative lowering in V.P? 3
Q9 How
many ml of 0.1 M HCl are required to react completely with 1gm mixture of
Na2CO3 and NaHCO3 containing equimolar
amounts of both. 3
Q10 1.00 g
of a non-electrolye solute dissolved in 50 g of benzene lowered the freezing
point of benzene by 0.40 K. Find the molar mass
of the solute.( Kf for
benzene = 5.12 K kg mol-1)
ANSWERS
Q1 (a)
particles are presents on lattice points (b) schottky defect, density decreases
(c) definition
Q2 NCERT page 31 Q 1.16 Ni+2
=96% Ni+3= 4%
Q3 (a)
extrinsic contains impurities , intrinsic no impurities (b) (i) n-type (ii)
p-type
Q4 derive
r/R = 0.225
Q5 Z = 2,
M =?, Na = 6.022 X 1023, No
of atoms = Na x 208 / M , answer = 24.16 x 1023.
Q6
Statement, solubility α 1/ KH,
H2 has lower value of KH
Q7
definitions in book
Q8
Derivation in book
Q9 NCERT page 60 Q-2.6, ans = 158.7 ml
Q10 
UNIT(2)
—SOLUTION Practice paper-1
Q 1 (i)
Which is more
concentrated, 1Molar aqueous solution
or 1 molal aqueous
solution & why?
(ii) Will the
elevation in boiling
point be same
if 0.1 mole NaCl or 0.1 mole of
sugar is
dissolved in 1 L of
water?
Equimolar
solutions of NaCl&
glucose are not
isotonic . Why?
(iv)
Distinguish between boiling
point & normal boiling
point of a
liquid . 1+1+1+1
Q 2
State Raoult,s law
for solution containing
nonvolatile solute in volatile
solvent.Derive a
Mathematical expression for
the law.
1+3
Q 3
What are ideal
and nonideal solutions?
Give reasons for
their formation . Give one example
in each
case. 1+1+1+1
Q 4
What isVant ,Hoff factor ? What
possible values can it
have if the
solute molecules undergo
(i)
Association (ii) dissociation
in the solution .Prove that
osmotic pressure is
a colligative
Property.
1+1+1+1
Q 5
Define the following
terms :
1+1+1+1
(i) Mole fraction (ii)
Azeotropes (iii) Henry,s law (iv)
Molal depression constant
Q 6 How many
mL of 0.1 M HCl
are required to
react completely with 1
gm mixture of
Na2CO3 and
NaHCO3 containing equimolar
amount of both ?
4
Q 7 Calculate the
depression in freezing
point of water
when 10 gm of CH3CH2CHClCOOH
is added
to 250 gm of
water .Ka= 1.4 x 10-3, Kf = 1.86 K Kg mole-1. 4
Q8 Henry ,s law
constant for CO2 in
water is 1.67 x 108
Pa at
298 K. Calculate the quantity
of
CO2 in
500 ml of soda
water when packed
under 2.5 Atm CO2 pressure
at 298 K. 4
Q9 18 gm
of glucose, is dissolved
in 1 kg of
water in a saucepan.
At what temperature
will
Water boil
at 1.013 bar? Kb
for water is 0.52 K kg mol -1
4
Q10
What volume of 95
wt % sulphuric acid
(density 1.85 gm/cm3
) and
what mass of
water
must be taken
to pre pare 100 cm3
0f 15 wt % solution
sulphuric acid (density 1.10 gm/cm3)
2+2
Answers→
Q1 (i) 1
Molar because it contains both solvent & solute.(ii) No. More in case of
0.1M NaCl because it
dissociates into ions.(iii) NaCl dissociates to give ions & osmotic
pressure is directly proportional to
number of particles.(iv) b.p is when v.p of liquid becomes equal to
surrounding pressure.Normalb.p
is
the temp at which v.p is 1 atm.
Q6 mass
of mixture = 1gm , let Na2CO3 = x gm, mass of NaHCO3 = 1—x , moles of Na2CO3
= x / 106.
Moles of NaHCO3 = 1—x / 84.
equivimolar mixture means x/106 =
1—x / 84. x = 0.557
Mass of Na2CO3 = 0.557gm, mass of NaHCO3
= 1—0.557 = 0.443gm.
Na2CO3 +
2 HCl → 2 NaCl+
H2O + CO2
106gm 2x36.5=73 106 gm Na2CO3need = 73gm HCl
0.557gm will need = 73 x
0.557 / 106 =0.384gm HCl
Similarly—
NaHCO3 + HCl
→ NaCl+ CO2 + H2O
84gm 36.5gm 84gm NaHCO3 need = 36.5gm HCl
0.443 will need = 36.5x
0.443 / 84 = 0.192gm HCl
Total HCl used = 0.384 + 0.192 = 0.576gm
MHCl= moles of HCl / vol in L
0.576/ 36.5
0.1 = —
-------- , V = 0.157
Litre(ans)
V
Q7 . Molar mass of CH3CH2CHClCOOH
= 122.5,
moles of CH3CH2CHClCOOH= 10/122.5 = 0.0816, m =
0.0816/ 0.250kg = 0.3264
α = √ Ka/C = √ 1.4 x 10—3/ 0.3264 = 0.065, i—1 i—1 i—1
α = --------, 0.065 = -------- =
-------, i = 1.065
m—1 2—1 1
∆Tf = i x Kf x m = 1.065 x 1.86 x 0.3264 = 0.650
(ANS)
Q8
Q9 Moles of glucose = 18
g/ 180 g mol–1 = 0.1 mol
Number
of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg-1
For
water, change in boiling point
ÄTb = Kb
× m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K
Since water boils at 373.15
K at 1.013 bar pressure,therefore, the boiling point of solution will be 373.15
+ 0.052 = 373.202K.
Q10 M of 95
% H2SO4 = moles x 1000 / V of solution in L
Moles of H2SO4 = 95/98, volume of solution = m/d = 100/1.85
= 54.05ml.
M = 95 x 1000/ 98 x 54.05 = 17.93 M
Similarly Molarity of 15% H2SO4 can be calculated
as— moles = 15/98, V = 100/1.10 =
90.91ml
M = 15 x 1000 / 98 x 90.91 = 1.68 M
Apply molarity equation — M1V1 =
M2V2
95 % 15 %
17.93 x V1 = 1.68 x 100 , V1 = 9.37 ml (ans)
Mass of
100 ml 15 % H2SO4to
prepared = 1.10 x 100 = 110 gm
Mass of
9.4 ml H2SO4 of 95 % = 9.4 x 1.85 = 17.4 gm, Mass of H2O =
110—17.4 = 92.6 gm(ans)
Practice paper-2
Q1 (a)Why is elevation in boiling point of
water different in the following cases; 3
(i) 0.1 M NaCl solution
(ii) 0.1 M sugar solution
(b)Which is more concentrated 1 M or
1 m solution & why?
(c)Two liquids A & B boil at 1450C
1900C respectively. Which of them will have higher V.P at 800C?
Q2 State Henry,S law for
solubility of a gas in liquid.Explain the significance of Henry,S
constant(KH).
At the same temperature, H2
is more soluble than He in water. Which of them will have higher value of
KH&
Why?
3
Q3 Define the terms: (a) Raoult,Slaw (b) Azeotropes (c)
Colligative properties 3
Q4 What do you understand by relative
lowering in V.P? Show that it is a colligative property. How will you
determine
the molar mass of by relative lowering in V.P ?
3
Q5 What
are ideal and non-ideal solutions? Discuss the positive and negative deviations
from ideal
behavior with the help of graphs.
3
Q6 Calculate (a) Molality (b) Molarity and (c) Mol fraction of KI if
density of 20 % (mass/mass) aqueous
KI is 1.202 gm /ml. (K = 39, I = 127).
3
Q7 Vapour pressure of pure liquids A & B
are 450 and 700 mm Hg respectively at 350 K. Find the
composition
of liquid mixture if total VP is 600 mm Hg. Also find the composition in vapour
phase. 3
Q8 How
many mL of 0.1
M HCl are required
to react completely
with 1 gm mixture
of
Na2CO3 and
NaHCO3 containing equimolar
amount of both ?
3
Q9 Two elements A & B form compounds
having formula AB2 and AB4. When dissolved in 20 gm
benzene
1 gm of AB2 lowers the
freezing point by 2.3 K whereas 1 gm of AB4 lowers it by 1.3 K. Kf
for benzene
is
5.1 K kg mol—1. Calculate atomic masses of A & B.
3
Q10 Calculate the depression in freezing point
of water when 10 gm of CH3CH2CHClCOOH is added to
250 gm of water.(Ka = 1.4
x 10—3, Kf = 1.86
= K kg mol—1 . C = 12, H = 1,
Cl =35.5, O =16) 3
Answers-
1
(a) 0.1M NaCl due to more number of ions (b) 1Molar
because it contains solvent & solute both(c) liq A
2
statement,
Solubility α 1 / KH,
H2 has lower value
of KH
3
definitions, Q4- derivation, Q5- book,
4
book
5
book
6
NCERT page 37( 2.5)
m =1.5, M = 1.44, XKI =
0.0263
(a) mass of solvent = 100—20 = 80gm = 0.08
kg, molar mass of KI = 166, moles of
KI= 20/166= 0.120
m = 0.120/0.08 = 1.5
(b)
vol of solution = m/d = 100/1.202 = 83.2ml= 0.0832 L, M = 0.120/ 0.0832 =
1.44,
(c) moles of H2O = 80/18= 4.44,
XKI = 0.120/0.120 +4.44 = 0.0263
7
NCERT PAGE 47(2.8) XA = 0.40, XB = 0.60, vapour pressure XA = 0.30, XB = 0.70
PA0 = 450, PB0
= 700 , Total = 600mm of Hg, Raoults law(PTOTAL) = PA0 + PB0 = PA0
XA + PB0XB
(PTOTAL) = PA0 XA + PB0(1—XA)
, 600 = 450 x XA + 700(1—XA), XA = 0.40 and XB = 0.60
PA = PA0 XA = 450 x 0.40 =
180 mm of Hg, PB = PB0XB
= 700 X 0.60 = 420 mm of Hg
XA in
vapour phase = partial pressure of A / Total vapourpressure(PA + PB)
= 180/ 180 + 420 = 0.30
XB in vapour phase =
1—0.30 = 0.70
8
NCERT page 60 (2.6) ans—158.7ml ,
mass of mixture = 1gm , let Na2CO3
= x gm, mass of NaHCO3 =
1—x , moles of Na2CO3 = x / 106.
Moles of NaHCO3 = 1—x / 84. equivimolar mixture means x/106 = 1—x / 84. x = 0.557
Mass of Na2CO3 = 0.557gm, mass of NaHCO3
= 1—0.557 = 0.443gm.
Na2CO3
+ 2 HCl →
2 NaCl+ H2O +
CO2
106gm 2x36.5=73 106 gm Na2CO3need = 73gm HCl
0.557gm will need = 73 x
0.557 / 106 =0.384gm HCl
Similarly—
NaHCO3 + HCl → NaCl+
CO2 + H2O
84gm 36.5gm 84gm NaHCO3 need = 36.5gm HCl
0.443 will need =
36.5x 0.443 / 84 = 0.192gm HCl
Total HCl used = 0.384 + 0.192 = 0.576gm
MHCl= moles of HCl / vol in L
0.576/ 36.5
0.1 = — -------- , v =
0.157 Litre(answer)
V in L
9
Kf x WB x 1000 Kf
x WB x 1000
NCERT page 60(2.21), For AB2, ∆Tf = ----------------------
, For AB4 , ∆Tf
= -------------------
MB x WA
MB x WA
5.1 x 1 x 1000 5.1 x 1 x
1000
2.3
= --------------------- ,
1.3 = -------------------
MB x 20 MB x 20
MB = 110.86 MB
= 196.15
AB2 = A + 2B = 110.86, AB4 = A +
4B =
196.15, solve equations and
find A & B values
A = 25.58,
B = 42.64 (ans)
10
NCERT page 61(2.32) Molar mass of CH3CH2CHClCOOH
= 122.5,
moles of CH3CH2CHClCOOH= 10/122.5 = 0.0816, m =
0.0816/ 0.250kg = 0.3264
α = √ Ka/C =
√ 1.4 x 10—3/ 0.3264 = 0.065, i—1 i—1 i—1
α = --------, 0.065 = --------
= -------, i = 1.065
m—1 2—1 1
∆Tf = i x Kf
x m = 1.065 x 1.86 x 0.3264 = 0.650 (ANS)
HINDI
ग्रीष्मकालीन अवकाश कार्य
कक्षा १२ वीं
Ø कीसी एक विषय पर औपचारिक पत्र लिखीए |
Ø कीसी एक विषय पर अनौपचारिक पत्र लिखीए |
Ø कीसी भी दो वर्त्तमान समस्याओ पर निबंध लिखीए |
Ø जनसंचार माध्यम से १० प्रश्न उत्तर लिखीए |
Ø किसी एक विषय पर फिचर लेखन लिखिए |
Ø कक्षा में दिए गये पूर्व प्रश्नपत्र हल करे |
Ø पाठ्यक्रम पुनरावृत्ति करे |
BIOLOGY
1. Completion
of practical record.
2. As per Theory chapters 5, 8, and 13 write
comments with neat labeled diagram on Pedigree analysis, amoebiosis, ascariasis
and adaptive features of an aquatic plant (Hydrilla/lotus) an aquatic animal
(Marine/fresh water fish), Any One
desert plant and one desert animal in practical record.
3. Preparation
of chapter 1 to 5 for the board examination thoughrly with the help of NCERT
text book. Write point wise notes. No
details required
4. Diagram
practice and schematic representation from Chapter 1 to 6.
6. Solve Two
CBSE sample paper’s questions of chapter 1 to 5.
7. Write
abbreviations from biology text-book(approx 100 abbreviations)
8. Write 25
common microbes with their scientific name and their diseases.
CLASS-XII (MATHEMATICS)
SUMMER VACATION HOME WORK AND ASSIGNMENT
1.
Chapter-2: Inverse
trigonometric functions
2. Chapter-3: Matrices
3. Chapter-4: Determinants
4. Chapter-5: Continuity And Differentiation
5. Chapter-6: Application of differentiation (Exercise 6.1,
Exercise6.2, Exercise 6.3 and Exercise 6.5)
Note: Solve all the
questions of exercises with examples
Subject: Physics
Prepare
a note on following topics and learnt it:
Chapter–2:
Electrostatic Potential and Capacitance
·
Electric potential
·
Potential difference.
·
Electric potential due to a
point charge.
·
A dipole and system of
charges.
·
Equipotential surfaces.
·
Electrical potential energy
of a system of two point charges and of
·
Electric dipole in an
electrostatic field.
·
Conductors and insulators,
free charges and bound charges inside a conductor.
·
Dielectrics and electric polarisation
·
Capacitors and capacitance.
·
Combination of capacitors in
series and in parallel.
·
Capacitance of a parallel
plate capacitor with and without dielectric medium between the plates.
·
Energy stored in a
capacitor.
·
ALSO SOLVE THE NCERT
QUESTION
Chapter–3:
Current Electricity
·
Electric current
·
Flow of electric charges in
a metallic conductor
·
Drift velocity, mobility and
their relation with electric current
·
Ohm's law
·
Electrical resistance
·
V-I characteristics (linear
and nonlinear),
·
Electrical energy and power
·
Electrical resistivity and
conductivity
·
Carbon resistors, colour
code for carbon resistors
·
Series and parallel
combinations of resistors
·
Temperature dependence of
resistance
·
Internal resistance of a
cell
·
Potential difference and emf
of a cell
·
Combination of cells in
series and in parallel
·
Kirchhoff's laws and simple
applications
·
Wheatstone bridge
·
Metre bridge
·
Potentiometer - principle
and its applications to measure potential difference and for comparing EMF of
two cells.
·
Measurement
of internal resistance of a cell.
·
ALSO SOLVE THE NCERT
QUESTION
Subject: Chemistry
UNIT (1) –
SolidState Practice Paper-1 02.05.2016
Q1 (i) Why is Frenkel defect not found in pure
alkali metal halides.
(ii)
What happens to the structure of CsCl when it is heated to about 760 k?
(iii) Fe3O4
is ferrimagnetic at room temperature and becomes paramagnetic at 850 K.Why?
(iv)
When atoms are placed at the corners of all 12 edges, how many atoms are
present per unit cell? 4
Q2 The composition of a sample of wustite is Fe0.93O1.00.
What percentage of iron is present in the form of
Fe(III)?
4
Q3 Calculate the packing efficiciency in cubic close
packing arrangement.
4
Q4 If the radius of octahedral void is r& radius of atoms in close packing is R,
drive the relation between
r
&R.
4
Q5 Niobium crystallizes in body centered cubic
structure. If density is 8.55 gm cm-3. Calculate atomic
radius
of niobium . Atomic mass of niobium is 93 u.
4
Q6 Explain the following term with suitable example
(a) Schottky defect (b)Frenkel defect
(c) Interstitials (d) F-centres
4
Q7 What is a semiconductor? Describe the two main
types of semiconductors & contrast their conduction
mechanism.
4
Q8 (a) In
terms of band theory, what is difference between
(i) Conductor and an insulator (ii) Conductor and semiconductor
2
(b) If NaCl is doped with 10-3
mol % of SrCl2 , what is the
concentration of cation vacancy? 2
Q9 (a) What type of stoichiometric defect is shown
by :
2
(i) ZnS (ii) AgBr
(b)
Classify the following as p-type or n-type semiconductor: 2
(i) B doped with Si (ii) Si doped with B
Q10 (i) Why is glass considered a super cooled
liquid?
(ii)
Name the parameters that characterize a unit cell.
(iii) What
is the two dimensional coordination number of a molecule in square close-packed
layer ?
(iv)
Give the significance of a lattice point.
4
Answers→
Q1 Because
ions can not get into interstitial sites due to their larger size.(b) CsCl structure transforms into
NaCl
structure.(c) due to randomization of spins at high temp.(d) at corners 8 x 1/8
= 1 atom(ans).
Q2 Let atoms
of Fe+3 = x , atoms of Fe+2
= 0.93—x . For electrical neutrality,
Total positive charge = Total negative charge
2(0.93—x ) + 3x = 2,
then x = 0.14( Fe+3atoms), %Fe+3 = 0.14x100/0.93 =
15.05
% 0f Fe+2 = 100—15.05 =
84.95.
Q3 Packing
efficiency in ccp or fcc = 74.06 %
derivation in book
Q4 r / R = 0.414
(ans).
Q5 Z = 2, M
= 93, d = 8.55 gm/cm3, N0 = 6.022 x 1023, d =
Z x M / N0 x 10—30 x a3
Calculate a3 = 3.61 x 107 , a = 330.5 pm.
Q6 (a)
Schottky—it arises if some of the atoms or ions are missing from their normal
lattice sites. Equal
number
of cations and anions are missing. Density is lowered. Occurs in compounds
having high C.N
&Cations and anions are of similar size.eg Alkali metal halides (B)
Frenkel—it arises if some oftheatoms
or ions are missing from their
normal positions and occupy an interstitialsite between lattice points.Density
remains same. Occurs in compounds having low C.N. Cations are smaller
than anions. Ag halidesAgBr contains both
defects. (c) Vacant sites in the
structure of crystal lattice. (d) Electrons trapped in anion
vacancies arerefered as F—centres.They
are responsible for colour. Nonstoichiometric KCl is violet in colour.
Q7 Those
substances which are insulator at lower temperature but conduct electricity at
higher temperature
are
called semiconductors.Thermal energy available at room temperature is
sufficient to excite electrons
from
highest occupied band to the next permitted band.
Intrinsic—Electrical conductivity is due to ther effect. No extra
impurity is added.(Ge& Si at room temp).
Extrinsic—Electrical conductivity is due to impurity effect. Extra
substance is added. They are of two
types,
n—type and p—type.
n-type—Electrical conductivity is due to electrons. eg group 13 elements
doped with group 14 elements.
p-type—Electrical conductivity is due to electron deficient bond or
electron vacancy or holes.eg group 14
elements
doped with group 13 elements.
Q8 (a) In
conductors energy gap is very-very small less than 50 kj/mole(metals).
In insulators energy is large more than 300 kj/mole(rubber).
In semiconductors energy gap is 30—300 kj/mole.
(b)
One cation of Sr+2 would create one vacancy. Conc of cation vacancy
= 10—3mole % = 10—3/ 100 =
10—5mole.
No of Sr+2 ions in 10—5mole = 6.022 x 1023x 10—5= 6.022 x 1018 ions = cation
vacancies.
Q9 (a) ZnS—Frenkel or dislocation. AgBr—Frenkel and
schottky both. (b) i—n-type, ii—p-type.
Q10 (a)
Because it has the property to flow
slowly.
(b)
Three dimensions and three angles.
(c) Four.
(d)
lattice points are occupied by constituent particles.
Practice Paper- 2
Q1 (a)Why
is the window glass of old buildings thick at the bottom?
3
(b)What happens when CaCl2 is introduced to the AgCl crystal?
(c)What happens when CsCl is heated to 760 K?
Q2 (a)Give
significance of a lattice point.
(b)What type of defect can arise when a solid is heated? Which physical
property is affected by it
and in what way?
(c)What is meant by co-ordination number?
3
Q3 Analysis
shows that Nickel Oxide has oxide Ni0.98O1.00. What
fraction of Ni exists as Ni+3& Ni+2 ions? 3
Q4 KF has
NaCl structure, What is the distance between K+ ions and F-
ions, if density is 2.48 gm/cm3. 3
Q5
(a)Explain intrinsic and extrinsic semiconductors with suitable
examples. 3
(b)Classify the following as n-type or p-type semiconductors
(i) B doped with Si (ii) Si doped with B
Q6 Explain
the following with suitable examples 3
Frenkel defect, F-centres, Ferrimagnetism
Q7 Derive
the relation between r and R for tetrahedral void. 3
Q8 An
element has BCC structure with a edge length 288 pm. The density of element is
7.2 gm/cm3.
How
many atoms are present in 208 gm of this element?
3
Q9
Calculate the efficiency of packing in case of a metal crystal for face
centred cubic.
3
Q10 A metal
crystallises into two cubic phases FCC & BCC whose edge lengths are 3.5
& 3.0 A0 respectively.
Calculate the ratio of densities of FCC & BCC.
3
ANSWERS
Q1 (a) Glass
has the property to flow slowly (b) vacancies are created(schottky defect).
(c) it
changes to NaCl structure. C.N changes from 8 to 6
Q2 (a)
particles are present on lattice points (b) schottky defects, density decreases
(c) definition
Q3 NCERT page No-31, Q -1.16 Ni+2 = 96 %, Ni+3 = 4 %
Q4 Z = 4, M = 39 + 19 = 58, d = 2.48 , NA
= 6.022 x 1023, calculate a (5.375 x 10—10cm )
distance between K+ and F—ions =
5.375 x 10—10cm / 2 = 2.688 x 10—10 cm.
Q5 extrinsic
contains impurities & intrinsic no impurities (i) n-type (ii) p-type
Q6
definitions in book
Q7 derivation r / R = 2.25 in book
Q8 Z =
2, M = ?, NA = 6.022 x 1023,
d = 7.2, calculate M(51.8),
No of
atoms = 6.022 x 1023 x 208/51.8
= 24.16 x 1023ANS
Q9
efficiency FCC = 74 % derivation in book
Q10 d(fcc) = 4 x M / NA x (3.5)3, d(bcc)
= 2 x M / NA(3)3
d(fcc) / d(bcc) = 4 x 33 / 2
x (3.5)3 = 1.26 ans
Practice paper-3
Q1 (a) Give
significance of a lattice point
(b)
What type of defect can arise when a solid is heated? Which physical property
is
affected by it & in what way?
(c)
What is meant by co-ordination number?
3
Q2 Analysis
shows that Nickel oxide has the formula
Ni0.98O1.00 . What fraction of Nickel
exists as Ni+3& Ni+2 ions?
3
Q3 (a)
Explain extrinsic and intrinsic semiconductors with examples.
(b)
Classify the followings as p-type or n-type of semiconductors
(i) B doped with Si (b) Si doped
with B
Q4 Derive
the relation between r & R in tetrahedral void.
3
Q5 An element
has BCC structure with a edge length of 288 pm. The density of element is
7.2gm/cm3. How many atoms are present in 208 gm of this
element? 3
Q6 State
Henry,S law for solubility of a gas in liquid. Explain the
significance of Henry,S
Constant(KH) at the same temperature, H2 is more
soluble than He in water.Which of
them
will have higher value of KH and why?
1+1+1/2+1/2
Q7 Define
the followings;
(i) Raoult,S law
(ii) Azeotropes (iii)
Colligative properties
3
Q8 What do
you understand by relative lowering in V.P? Show that it is a colligative
Property. How will determine the molar mass by relative lowering in V.P? 3
Q9 How
many ml of 0.1 M HCl are required to react completely with 1gm mixture of
Na2CO3 and NaHCO3 containing equimolar
amounts of both. 3
Q10 1.00 g
of a non-electrolye solute dissolved in 50 g of benzene lowered the freezing
point of benzene by 0.40 K. Find the molar mass
of the solute.( Kf for
benzene = 5.12 K kg mol-1)
ANSWERS
Q1 (a)
particles are presents on lattice points (b) schottky defect, density decreases
(c) definition
Q2 NCERT page 31 Q 1.16 Ni+2
=96% Ni+3= 4%
Q3 (a)
extrinsic contains impurities , intrinsic no impurities (b) (i) n-type (ii)
p-type
Q4 derive
r/R = 0.225
Q5 Z = 2,
M =?, Na = 6.022 X 1023, No
of atoms = Na x 208 / M , answer = 24.16 x 1023.
Q6
Statement, solubility α 1/ KH,
H2 has lower value of KH
Q7
definitions in book
Q8
Derivation in book
Q9 NCERT page 60 Q-2.6, ans = 158.7 ml
Q10
UNIT(2)
—SOLUTION Practice paper-1
Q 1 (i)
Which is more
concentrated, 1Molar aqueous solution
or 1 molal aqueous
solution & why?
(ii) Will the
elevation in boiling
point be same
if 0.1 mole NaCl or 0.1 mole of
sugar is
dissolved in 1 L of
water?
Equimolar
solutions of NaCl&
glucose are not
isotonic . Why?
(iv)
Distinguish between boiling
point & normal boiling
point of a
liquid . 1+1+1+1
Q 2
State Raoult,s law
for solution containing
nonvolatile solute in volatile
solvent.Derive a
Mathematical expression for
the law.
1+3
Q 3
What are ideal
and nonideal solutions?
Give reasons for
their formation . Give one example
in each
case. 1+1+1+1
Q 4
What isVant ,Hoff factor ? What
possible values can it
have if the
solute molecules undergo
(i)
Association (ii) dissociation
in the solution .Prove that
osmotic pressure is
a colligative
Property.
1+1+1+1
Q 5
Define the following
terms :
1+1+1+1
(i) Mole fraction (ii)
Azeotropes (iii) Henry,s law (iv)
Molal depression constant
Q 6 How many
mL of 0.1 M HCl
are required to
react completely with 1
gm mixture of
Na2CO3 and
NaHCO3 containing equimolar
amount of both ?
4
Q 7 Calculate the
depression in freezing
point of water
when 10 gm of CH3CH2CHClCOOH
is added
to 250 gm of
water .Ka= 1.4 x 10-3, Kf = 1.86 K Kg mole-1. 4
Q8 Henry ,s law
constant for CO2 in
water is 1.67 x 108
Pa at
298 K. Calculate the quantity
of
CO2 in
500 ml of soda
water when packed
under 2.5 Atm CO2 pressure
at 298 K. 4
Q9 18 gm
of glucose, is dissolved
in 1 kg of
water in a saucepan.
At what temperature
will
Water boil
at 1.013 bar? Kb
for water is 0.52 K kg mol -1
4
Q10
What volume of 95
wt % sulphuric acid
(density 1.85 gm/cm3
) and
what mass of
water
must be taken
to pre pare 100 cm3
0f 15 wt % solution
sulphuric acid (density 1.10 gm/cm3)
2+2
Answers→
Q1 (i) 1
Molar because it contains both solvent & solute.(ii) No. More in case of
0.1M NaCl because it
dissociates into ions.(iii) NaCl dissociates to give ions & osmotic
pressure is directly proportional to
number of particles.(iv) b.p is when v.p of liquid becomes equal to
surrounding pressure.Normalb.p
is
the temp at which v.p is 1 atm.
Q6 mass
of mixture = 1gm , let Na2CO3 = x gm, mass of NaHCO3 = 1—x , moles of Na2CO3
= x / 106.
Moles of NaHCO3 = 1—x / 84.
equivimolar mixture means x/106 =
1—x / 84. x = 0.557
Mass of Na2CO3 = 0.557gm, mass of NaHCO3
= 1—0.557 = 0.443gm.
Na2CO3 +
2 HCl → 2 NaCl+
H2O + CO2
106gm 2x36.5=73 106 gm Na2CO3need = 73gm HCl
0.557gm will need = 73 x
0.557 / 106 =0.384gm HCl
Similarly—
NaHCO3 + HCl
→ NaCl+ CO2 + H2O
84gm 36.5gm 84gm NaHCO3 need = 36.5gm HCl
0.443 will need = 36.5x
0.443 / 84 = 0.192gm HCl
Total HCl used = 0.384 + 0.192 = 0.576gm
MHCl= moles of HCl / vol in L
0.576/ 36.5
0.1 = —
-------- , V = 0.157
Litre(ans)
V
Q7 . Molar mass of CH3CH2CHClCOOH
= 122.5,
moles of CH3CH2CHClCOOH= 10/122.5 = 0.0816, m =
0.0816/ 0.250kg = 0.3264
α = √ Ka/C = √ 1.4 x 10—3/ 0.3264 = 0.065, i—1 i—1 i—1
α = --------, 0.065 = -------- =
-------, i = 1.065
m—1 2—1 1
∆Tf = i x Kf x m = 1.065 x 1.86 x 0.3264 = 0.650
(ANS)
Q8
Q9 Moles of glucose = 18
g/ 180 g mol–1 = 0.1 mol
Number
of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg-1
For
water, change in boiling point
ÄTb = Kb
× m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K
Since water boils at 373.15
K at 1.013 bar pressure,therefore, the boiling point of solution will be 373.15
+ 0.052 = 373.202K.
Q10 M of 95
% H2SO4 = moles x 1000 / V of solution in L
Moles of H2SO4 = 95/98, volume of solution = m/d = 100/1.85
= 54.05ml.
M = 95 x 1000/ 98 x 54.05 = 17.93 M
Similarly Molarity of 15% H2SO4 can be calculated
as— moles = 15/98, V = 100/1.10 =
90.91ml
M = 15 x 1000 / 98 x 90.91 = 1.68 M
Apply molarity equation — M1V1 =
M2V2
95 % 15 %
17.93 x V1 = 1.68 x 100 , V1 = 9.37 ml (ans)
Mass of
100 ml 15 % H2SO4to
prepared = 1.10 x 100 = 110 gm
Mass of
9.4 ml H2SO4 of 95 % = 9.4 x 1.85 = 17.4 gm, Mass of H2O =
110—17.4 = 92.6 gm(ans)
Practice paper-2
Q1 (a)Why is elevation in boiling point of
water different in the following cases; 3
(i) 0.1 M NaCl solution
(ii) 0.1 M sugar solution
(b)Which is more concentrated 1 M or
1 m solution & why?
(c)Two liquids A & B boil at 1450C
1900C respectively. Which of them will have higher V.P at 800C?
Q2 State Henry,S law for
solubility of a gas in liquid.Explain the significance of Henry,S
constant(KH).
At the same temperature, H2
is more soluble than He in water. Which of them will have higher value of
KH&
Why?
3
Q3 Define the terms: (a) Raoult,Slaw (b) Azeotropes (c)
Colligative properties 3
Q4 What do you understand by relative
lowering in V.P? Show that it is a colligative property. How will you
determine
the molar mass of by relative lowering in V.P ?
3
Q5 What
are ideal and non-ideal solutions? Discuss the positive and negative deviations
from ideal
behavior with the help of graphs.
3
Q6 Calculate (a) Molality (b) Molarity and (c) Mol fraction of KI if
density of 20 % (mass/mass) aqueous
KI is 1.202 gm /ml. (K = 39, I = 127).
3
Q7 Vapour pressure of pure liquids A & B
are 450 and 700 mm Hg respectively at 350 K. Find the
composition
of liquid mixture if total VP is 600 mm Hg. Also find the composition in vapour
phase. 3
Q8 How
many mL of 0.1
M HCl are required
to react completely
with 1 gm mixture
of
Na2CO3 and
NaHCO3 containing equimolar
amount of both ?
3
Q9 Two elements A & B form compounds
having formula AB2 and AB4. When dissolved in 20 gm
benzene
1 gm of AB2 lowers the
freezing point by 2.3 K whereas 1 gm of AB4 lowers it by 1.3 K. Kf
for benzene
is
5.1 K kg mol—1. Calculate atomic masses of A & B.
3
Q10 Calculate the depression in freezing point
of water when 10 gm of CH3CH2CHClCOOH is added to
250 gm of water.(Ka = 1.4
x 10—3, Kf = 1.86
= K kg mol—1 . C = 12, H = 1,
Cl =35.5, O =16) 3
Answers-
1
|
(a) 0.1M NaCl due to more number of ions (b) 1Molar
because it contains solvent & solute both(c) liq A
|
2
|
statement,
Solubility α 1 / KH,
H2 has lower value
of KH
|
3
|
definitions, Q4- derivation, Q5- book,
|
4
|
book
|
5
|
book
|
6
|
NCERT page 37( 2.5)
m =1.5, M = 1.44, XKI =
0.0263
(a) mass of solvent = 100—20 = 80gm = 0.08
kg, molar mass of KI = 166, moles of
KI= 20/166= 0.120
m = 0.120/0.08 = 1.5
(b)
vol of solution = m/d = 100/1.202 = 83.2ml= 0.0832 L, M = 0.120/ 0.0832 =
1.44,
(c) moles of H2O = 80/18= 4.44,
XKI = 0.120/0.120 +4.44 = 0.0263
|
7
|
NCERT PAGE 47(2.8) XA = 0.40, XB = 0.60, vapour pressure XA = 0.30, XB = 0.70
PA0 = 450, PB0
= 700 , Total = 600mm of Hg, Raoults law(PTOTAL) = PA0 + PB0 = PA0
XA + PB0XB
(PTOTAL) = PA0 XA + PB0(1—XA)
, 600 = 450 x XA + 700(1—XA), XA = 0.40 and XB = 0.60
PA = PA0 XA = 450 x 0.40 =
180 mm of Hg, PB = PB0XB
= 700 X 0.60 = 420 mm of Hg
XA in
vapour phase = partial pressure of A / Total vapourpressure(PA + PB)
= 180/ 180 + 420 = 0.30
XB in vapour phase =
1—0.30 = 0.70
|
8
|
NCERT page 60 (2.6) ans—158.7ml ,
mass of mixture = 1gm , let Na2CO3
= x gm, mass of NaHCO3 =
1—x , moles of Na2CO3 = x / 106.
Moles of NaHCO3 = 1—x / 84. equivimolar mixture means x/106 = 1—x / 84. x = 0.557
Mass of Na2CO3 = 0.557gm, mass of NaHCO3
= 1—0.557 = 0.443gm.
Na2CO3
+ 2 HCl →
2 NaCl+ H2O +
CO2
106gm 2x36.5=73 106 gm Na2CO3need = 73gm HCl
0.557gm will need = 73 x
0.557 / 106 =0.384gm HCl
Similarly—
NaHCO3 + HCl → NaCl+
CO2 + H2O
84gm 36.5gm 84gm NaHCO3 need = 36.5gm HCl
0.443 will need =
36.5x 0.443 / 84 = 0.192gm HCl
Total HCl used = 0.384 + 0.192 = 0.576gm
MHCl= moles of HCl / vol in L
0.576/ 36.5
0.1 = — -------- , v =
0.157 Litre(answer)
V in L
|
9
|
Kf x WB x 1000 Kf
x WB x 1000
NCERT page 60(2.21), For AB2, ∆Tf = ----------------------
, For AB4 , ∆Tf
= -------------------
MB x WA
MB x WA
5.1 x 1 x 1000 5.1 x 1 x
1000
2.3
= --------------------- ,
1.3 = -------------------
MB x 20 MB x 20
MB = 110.86 MB
= 196.15
AB2 = A + 2B = 110.86, AB4 = A +
4B =
196.15, solve equations and
find A & B values
A = 25.58,
B = 42.64 (ans)
|
10
|
NCERT page 61(2.32) Molar mass of CH3CH2CHClCOOH
= 122.5,
moles of CH3CH2CHClCOOH= 10/122.5 = 0.0816, m =
0.0816/ 0.250kg = 0.3264
α = √ Ka/C =
√ 1.4 x 10—3/ 0.3264 = 0.065, i—1 i—1 i—1
α = --------, 0.065 = --------
= -------, i = 1.065
m—1 2—1 1
∆Tf = i x Kf
x m = 1.065 x 1.86 x 0.3264 = 0.650 (ANS)
HINDI |
ग्रीष्मकालीन अवकाश कार्य
कक्षा १२ वीं
Ø कीसी एक विषय पर औपचारिक पत्र लिखीए |
Ø कीसी एक विषय पर अनौपचारिक पत्र लिखीए |
Ø कीसी भी दो वर्त्तमान समस्याओ पर निबंध लिखीए |
Ø जनसंचार माध्यम से १० प्रश्न उत्तर लिखीए |
Ø किसी एक विषय पर फिचर लेखन लिखिए |
Ø कक्षा में दिए गये पूर्व प्रश्नपत्र हल करे |
Ø पाठ्यक्रम पुनरावृत्ति करे |
1. Completion
of practical record.
2. As per Theory chapters 5, 8, and 13 write
comments with neat labeled diagram on Pedigree analysis, amoebiosis, ascariasis
and adaptive features of an aquatic plant (Hydrilla/lotus) an aquatic animal
(Marine/fresh water fish), Any One
desert plant and one desert animal in practical record.
3. Preparation
of chapter 1 to 5 for the board examination thoughrly with the help of NCERT
text book. Write point wise notes. No
details required
4. Diagram
practice and schematic representation from Chapter 1 to 6.
6. Solve Two
CBSE sample paper’s questions of chapter 1 to 5.
7. Write
abbreviations from biology text-book(approx 100 abbreviations)
8. Write 25
common microbes with their scientific name and their diseases.
CLASS-XII (MATHEMATICS)
SUMMER VACATION HOME WORK AND ASSIGNMENT
1.
Chapter-2: Inverse
trigonometric functions
2. Chapter-3: Matrices
3. Chapter-4: Determinants
4. Chapter-5: Continuity And Differentiation
5. Chapter-6: Application of differentiation (Exercise 6.1,
Exercise6.2, Exercise 6.3 and Exercise 6.5)
Note: Solve all the
questions of exercises with examples
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