STATE BOARD - Deccan Chronicle

Transcription

STATE BOARD - Deccan Chronicle
Answers for December 20th
STD – X (STATE BOARD)
ONE MARKS
1.
Interferon
2.
BCG
3.
a. Pepo
4.
c. 180 L
5.
c. Bio – plastic
6.
10-10 m
7.
Decomposition reaction
8.
Froth flotation
9.
660° C
10.
Ethyne
11.
b. zero
12.
c. current
13.
b. non-conventional
14.
b. Magnetic field
15.
b.Reflection
TWO MARKS
16.
Fill in the blanks.
a. Intra specific variation.
b. Inter generic variation.
17.
Spot the error and correct.
No, the symptoms of the above diseases are not correct. Marasmus and Kwashiorkar
are both protein deficiency defects of children. But
The symptoms of Marasmus are: loses weight and suffers severe diarrhoea and it will
appear as though bones are covered by the skin.
The symptoms Kwashiorkar are: enlarged belly with swelling in the face and feet.
18.
Functional Areas of Human brain
Answers for December 20th
19.
STD – X (STATE BOARD)
To raise the questions.
a. Have our daughter cells been formed at the end of cell division?
b. Do the daughter cells contain half the number of chromosomes than the parent cell.
20.
Reason and answer
a) A is right R is wrong
Reason: Motor areas of cerebrum are sites of command but Sensory areas are the sites
where the sensory functions of the various sense organs are received through the sensory
nerves.
21.
Structure of pollen grain
Exine
Tube nucleus
22.
Match:
a. Autotrophs – Chlorophyll – Hibiscus
b. Parasites
– Haustoria
– Cuscuta
c. Saprophytes – Mycorrhiza – Monotropa
23.
Mulberry → Sparrow → Caterpillar → kite.
Kite
Tertiary
1KJ
T4 (Forth tropical level)1kJ
Sparrow
Secondary consumer
T3 (Third tropical level)10kJ
Cater pillar
Primary consumer
T2 (Second tropical) 100kJ
10KJ
Mulberry
Producer
T1 (First tropical level)1000kJ
100KJ
1000KJ
Answers for December 20th
27.
STD – X (STATE BOARD)
To raise question.
a. Is the drinking water is contaminated with human feces and animal wastes?
b. Are they drinking the contaminated water?
28.
Spot the error and correct the mistakes.
a. Nitrosomonas nitri can be used to treat sewage.
Correction: Nitrosomonas europaea can be used to treat sewage.
b. The coal has been found to be a good choice among all the alternative fuel option.
Correction: The hydrogen has been found to be a good choice among the entire alternative
fuel option.
29.
Reason and assertion.
b) A is Wrong R is correct
Reason: Denmark is called the country of “Winds”. Because More than 25% of their
electricity needs are generated through a vast network of wind mills.
30.
To interpret what happens in the given situation.
a. Suspension: It is a heterogeneous mixture of small insoluble particles in a solvent
in a suspension, the particles of solid stay in clusters that are large enough to be seen.
b. True solution: It is a homogeneous mixture that contains small solute particles that
dissolved through out the solvent.
31.
Raise the question.
a. What forms the dispersed medium in the blood?
b. What compounds of the blood from the dispersed phase?
c. How can you confirm that a solution is colloidal?
Answers for December 20th
32.
STD – X (STATE BOARD)
Problems.
Number of particles
= 3.0115x1023
Avogadro number
= 6.023x1023
Number of particles
Number of moles
=
Avogadro number
=
3.0115 × 10 23
6.023 × 10 23
=
3.0115
6.023
= 0.5 mole
Number of moles
33.
= 0.5 mole
Problem.
[H+] = 1.0x10-9m
PH = - log10 [H+]
= - log10 (1.0x10-9)
= - (log101.0 + log10 10-9)
= - (0+ (-9) log10 10) (log10 1.0=0)
= - (0-9 x log1010) (log1010=1)
= - (-9x1)
= - (-9)
=9
PH = 9
PH >7
Therefore the given solution is basic.
34.
Find out the odd one.
a. Vinegar
Reason: Vinegar is acidic in nature and others are basic in nature.
b. H2So4
Reason: H2So4 is an example for dibasic acid and others are example for monobasic
acid.
Answers for December 20th
35.
STD – X (STATE BOARD)
To spot the error and correct the mistake.
a. Gold does not occur in the combined form. It is react with air (or) water.
Correction: Gold does not occur in the combined form. It does not react with air (or)
water.
b. Coating the surface of ion with other metal prevents it from rusting. If it is coated
with thick layer of zinc it is called cathodic protection.
Correction: Coating the surface of ion with other metal prevents it from rusting. If it is
`coated with thin layer of zinc it is called galvanization.
36.
Match the following:
a. galena
– Pbs
b. Fluorspar
– CaF2
c. Cryolyte
– Na3AlF 6
d. Zinc blende – Zns
37.
Fill in the blanks.
a. Little affinity
b. 4.5%
38.
Match the following:
a. Current
– Q/T
b. Ohm’s law
– Vα I
c. Potential difference – W/Q
d. Joule’s law
–I2RT
Answers for December 20th
39.
STD – X (STATE BOARD)
Assertion and Reason.
d) Both A and R are correct
Reason: Two equal and opposite forces where lines of action coincide constitute a
Couple.
40.
Schematic diagram of electric circuit
41.
B1, B2=Brushes S1, S2=splitrings
Answers for December 20th
42.
STD – X (STATE BOARD)
Problem
The mass of bullet, m1 = 15 g = 0.015 kg
Mass of the pistol, m2 = 2 kg
Initial velocity of the bullet, u1 = 0
Initial velocity of the pistol, u2 = 0
Final velocity of the bullet, v1 = + 100 m s-1
Recoil velocity of the pistol, = v
Total momentum of the pistol and bullet
Before fire,
m 1 u1 + m 2 u2
= (0.015 × 0 + 2 × 0) kg m s-1
= 0 kg m s-1
Total momentum of the pistol and bullet after fire,
= (0.015 × 100 + 2 × v)
= (1.5 + 2v) kg m s-1
According to the law of conservation of momentum,
Total momentum after fire = total momentum before fire
1.5 + 2v = 0
2v = -1.5
v = -0.75 m s-1
Negative sign indicates that the direction in which the pistol would recoil is
opposite
to that of the bullet, that is, right to left.
43.
To interpret what happens in the given situation.
a. The earth attracts an apple. Why?
Due to the gravitational force apple is attracted.
b. Does the apple also attract the earth?
Yes, the apple does attracts the earth.
44.
Find out the odd one.
a. Retina-Reason: Retina is one of the part in eye but others are related to eye defects.
b. Black-Reason : Black doesn’t include VIBGYOR
Answers for December 20th
45.
STD – X (STATE BOARD)
Fill in the blanks.
a. Permanent magnet.
b. Ammeter.
FIVE MARKS
GROUP – A
46.
a. Stem cells:
One of the most fascinating branches in applied embryology is stem cell culture. The stem
cells are the most unspecialized mass of cells. They are derived from animals and plants.
They have two important characteristic features. They are:
1. Unspecialized cells which have the potentiality of growing and multiplying into
enormous number of same type of cells by repeated mitosis.
2.They can be introduced to become any other type of tissues with specific functions i.e.,
they can be induced to become a cardiac muscle, beta cells of pancreas (which produce
insulin), special neurons in brain etc.,
b. Explain the two types of stem cells.
There are two kinds of stem cells:
1. Embryonic Stem Cells: The embryonic stem cells can be derived from early embryo
which is developed by “invitro fertilization” (fertilisation made artificially in the
laboratory).
After fertilization the zygote develops into a hollow blastula by cell division. The inner
mass of undifferentiated cells are isolated and they are considered as embryonic stem cells.
2. Adult or Somatic Stem Cells: The body of higher animals and human beings have
many well differentiated tissues like epithelial, connective, muscular, vascular,
supporting, nervous and reproductive tissues. In these tissues, there are some
undifferentiated cells and
are considered as the adult or somatic stem cells. They can grow, multiply and can be
differentiated into same type of tissues into which they are implanted. The mechanism of
adult or somatic stem cell culture is similar to that of embryonic stem cell culture. The
somatic stem cells are derived from sources such as bone marrow, embryos, amniotic fluid
and umbilical cord.
Answers for December 20th
47.
STD – X (STATE BOARD)
Tuberculosis:
It is an airborne disease affecting the lungs and also parts of our body such as bones, joints,
lymph glands, alimentary tract, liver, kidney, etc,.
Causative agent: Mycobacterium tuberculosis, a rod shaped bacterium cause’s
tuberculosis
Symptoms: i) The affected parts develop lesions in the form of small nodules called
tubercles from which the disease gets its name.
ii) Persistant cough
iii) Loss of body weight
Transmission:
Tuberculosis is transmitted through air. Large number of bacteria
leave the patients through the droplets of sputum expelled by the patients while eating,
sneezing, talking, laughing and so on by the patients. The droplets may remain suspended
in the air for a long time. The dust arising from the sputum may also contain viable germs.
The waxy cell wall of the tuberculosis bacillus prevents it from drying up and so it can
remain viable outside the body for a long period. The germs suspended in the air may be
inhaled by a healthy person.
Prevention:
i) Keeping oneself healthy and avoiding insanitary conditions, overcrowding and poor
ventilation.
ii) Sunlight and fresh air are important agents, as they act as natural disinfectants readily
destroying the germs.
iii) Isolation of the patients and frequent sterilization of articles used by them are also
important.
iv) Incineration (burning) of the droplets, the sputum from the patients to prevent its
occurrence in the air.
v) Immunization with BCG vaccine is an effective measure to prevent this disease.
Answers for December 20th
STD – X (STATE BOARD)
GROUP – B
48.
a) The process of fruit development.
* Fertilization is the process by which the fruit is developed.
* Fruits are the product of Fertilization.
* In higher plants double fertilization takes place.
* The ovary will became fruit after fertilization.
* It has two parts namely pericarp (fruit wall) and seeds.
b) Draw a neat diagram and label the “process of fertilization”.
Pollen grain
Style
Pollen tube
Embryo sac
Ovule
c)Vegetative propagation and their types: Vegetative propagation is the ability of plants
to reproduce by bringing forth new plants from existing vegetative structure without
sexual reproduction.
Fragmentation: In multicellular organisms with simple body organization, simple
reproductive methods have been noticed. In Spirogyra algae, the plant body breaks up into
smaller fragments. Each fragment grows into a new individual.
Sprogyra
Nucleus
Septum
Spiral Chloroplast
Answers for December 20th
STD – X (STATE BOARD)
Budding: In Hydra, a bud develops as an outgrowth due to repeated cell division at one
specific site. These buds develop into tiny individuals and, when fully mature, get detached
from the parent body to become new independent individuals. Similarly, buds produced in
the notches along the leaf margin of Bryophyllum fall on the soil and develop into new
plants (in Tamil katti pottal kutti podum).
Bryophyllum
Shoot
Notch
Bud
49.
a) Classify the following substances-wood, paper, plastic and grasses.
Bio-degradable
Non-bio degradable
Wood-paper
Plastics
Producers
Grasses –It is also producer
b) Classify the following.
Goat – Herbivores - Primary consumer
Deer – Herbivores - Primary consumer
Lion – Carnivores – Tertiary consumer
O2 - Abiotic factor
Goat, deer • Consume the green plants. So they are called herbivore – primary consumer.
Lion • Carnivore. They consume the deer and goat. So they are called tertiary consumer.
O2 • It is an inorganic substance and comes under abiotic factor of the ecosystem.
c) Ecosystem.
A community of organisms that interact with one another and with the environment is
called an ecosystem.
Answers for December 20th
STD – X (STATE BOARD)
GROUP – C
50.
Avogadro’s law
Equal volumes of all gases under the same condition of temperature and pressure contin
the equal number of molecules.
To establish the relationship between vapour density and relative molecular mass of a gas :
(i) Relative Molecualr mass :
It is defined as the ratio of the mass of 1 molecule of the gas or vapour to the mass of
1 atom of hydrogen.
Relative molecular mass of a gas =
Massof 1molecule of the gas or vapour
Mass of 1 atom of hydrogen
(ii) Vapour density : (V.D).
It is defined as the ratio of the mass of a certain volume of the gas or vapour to the
mass of the same volume of hydrogen at the same temperature and pressure.
V.D =
Massof 1 volume of the gas or vapour
Mass of 1 volume of hydrogen
Applying Avagadro's law,
V.D.
=
Massof 1 molecule of the gas or vapour
Mass of 1 molecule of hydrogen
Since hydrogen is diatomic,
V.D.
=
Mass of 1 volume of the gas or vapour
2 × Mass of 1 atom of hydrogen
Mutiplying bothsides by 2, we get
2 x V.D =
Mass of 1 molecule of the gas or vapour
Mass of 1 atom of hydrogen
2 xV.D. = Relative molecular mass of a gas (or) vapour
2 x vapour density
= Relative molecular mass.
Answers for December 20th
51.
STD – X (STATE BOARD)
Chemical properties of carbon and its compounds.
i) Carbon and its compounds burn in oxygen to give carbon- di- oxide along with heat and
light.
(Ex)
C + O2 • CO2 + heat + light
CH4 + 2O2 • CO2 + 2H2O + heat + light
C2H5OH + 3O2 •2CO2 + 3H2O + heat + light
ii) Carbon compounds can be easily oxidized using suitable oxidizing agent alkaline
potassium permanganate to form carboxylic acids.
iii) Unsaturated carbon compounds undergo addition reactions with hydrogen in the
presence of palladium or nickel catalyst.
(Ex) Addition of hydrogen
CH2= CH2 ⎯⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯⎯→ CH3 - CH3
Ethene
Ni - Catalyst
Ethane
iv) Carbon compounds undergo substitution reaction in the presence of either sunlight or
any other reagents.
(Ex) Methane undergoes substitution reaction to form different types of products
v) Carbon compounds such as alcohol react with sodium to liberate hydrogen gas.
(Ex) 2CH3CH2 OH + 2Na • 2CH3CH2ONa+H2
Answers for December 20th
STD – X (STATE BOARD)
GROUP – D
52.
Refraction of light through a prism
PE - Incident ray
EF - Refracted ray
FS - Emergent ray
A - Angle of the Prism
•D - Angle of deviation
•i - Angle of incident
•r - Angle of refraction
•e - Angle of emergence
53.
Primary cell
The cells from which the electric energy is derived by irreversible chemical reaction are
called primary cells. The primary cell is capable of giving an emf, when its constituents,
two electrodes and a suitable electrolyte, are assembled together. The main primary cells
are Daniel cell and Leclanche cell. These cells cannot be recharged. Lechlanche cell is
discussed here.
The construction and working of a Lechalanche cell.
A Leclanche cell consists of a glass vessel which is filled with ammonium chloride solution.
Ammonium chloride solution is acting as electrolyte. In it there stands a zinc rod and
porous pot containing a carbon rod which is packed round with a mixture of manganese
Answers for December 20th
STD – X (STATE BOARD)
dioxide and powdered carbon. Therefore the carbon rod forms the positive pole and the
zinc rod the negative pole.
Ammonium chloride, splits into ammonium and chloride ions. The chloride ions migrate
to the zinc rod and deposit their negative charge at the zinc rod. Hence zinc becomes
negatively charged and the reaction takes place in which zinc is converted to zinc chloride.
The ammonium ions migrate to the carbon rod and make it positively charged. When the
car-bon rod and zinc rod are connected by a wire, the current flows from carbon to zinc
through the wire. The e.m.f of the cell is about 1.5V.