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Tuesday, September 15, 2015
Monday, September 14, 2015
answer key of mock test Science
398CI7G
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Marking Scheme
SUMMATIVE ASSESSMENT – I (2014-15)
Science (Class–IX)
General
Instructions:
1. The Marking Scheme provides general guidelines to
reduce subjectivity and maintain uniformity. The answers given in the marking
scheme are the best suggested answers.
2. Marking be done as per the instructions provided
in the marking scheme. (It should not be done according to one’s own
interpretation or any other consideration).
3. Alternative methods be accepted. Proportional
marks be awarded.
4. If a question is attempted twice and the candidate
has not crossed any answer, only first attempt be evaluated and ‘EXTRA’ be
written with the second attempt.
5. In case where no answers are given or answers are found wrong in
this Marking Scheme, correct answers may be found and used for valuation
purpose.
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SECTION-A
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1
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Columnar
epithelial tissue.
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1
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2
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Distance
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1
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3
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Weight of the stone 5 20 N
Mass of the stone
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1
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4
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(a) The solvent will evaporate and a residue will be left
behind.
(b) (i) Because it
does not have a uniform composition throughout
(ii) No new compound is formed in a mixture
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2
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5
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Striated squamous epithelium –
arranged in many layers to prevent wear and tear.
Squamous epithelium – extremely
thin, flat, single layered.
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2
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6
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According to Newton’s law of Inertia of motion the body remains
in motion after applying the brakes.
Safety belt helps to keep the grip and body do not get enough
jerk and get saved from getting smashed against the windscreen of the car.
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2
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7
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Property of solid, liquid and
gas
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3
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8
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Explanation of reason :
Large latent heat of vaporization of water helps to cool the area.
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3
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9
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It is the amount of solute
present in given amount (mass or volume) of solution or mass or volume of
solvent.
Ways of expressing
concentration :
(a) Mass by Mass
percentage 5
(b) Mass by volume
percentage 5
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3
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10
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3
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11
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Refer fig. 6.7, page 73, NCERT
Structure-made up of four types of elements :
Sieve tubes-tubular cells with perforated walls
Phloem parenchyma
Phloem fibres-are dead
Companion cells
Function – transports food from leaves to other parts
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3
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12
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(a) 0 to 10 sec
(b) Displacement =
125 m, average velocity = 4.17 m/s
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3
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13
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Yes, value of g50 at centre of earth. G has the same value at the
centre of
the earth as every where else as
it is a universal constant.
Weight of object at the centre of
the earth is 0.
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3
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14
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The reason is that the
acceleration is inversely proportional to the mass of the body. The mass of stone being very small as
compared to the mass of the earth the
acceleration produced in the stone is large. But the mass of the earth being very large the acceleration
produced in the earth is so small that the earth does not move towards the
stone.
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3
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15
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Meaning of uniform velocity, eg
Meaning of variable velocity, eg
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3
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16
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Acceleration
= (v-u)/t
M
= F/a
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3
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17
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Expected
Answer / Value Points of Test item – 39
(i) Air, water and soil
(ii) Reproduction, growth, susceptibility
to diseases
(iii) Family bonding, applying knowledge
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3
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18
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One
variety can grow in different climatic conditions.
Explain
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3
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19
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(a) (i) gas (ii) solid
(b) (i) 258 C (ii) 278
C (iii) 78 C
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5
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20
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5
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21
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(a) (i) in such
habitat, protection against water loss is essential
(ii) protection against water loss, mechanical
injury
(b) Cells are elongated, flattened
Closely
packed. No intercellular spaces and form a continuous layer
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5
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22
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(a) 2gh5v22u2
23103h5022(20)2
h520m
Displacement 50
Total distance 52320540 m
(b) G5
constant value, G56.7310211 Nm2kg22
g 5
changes from place to place, g on earth59.8
ms22
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5
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23
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(a) A
(b) B
(c) A
(d) B
(e) B
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5
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24
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Criteria
of crop selection are
(i) different nutrient requirement.
(ii) availability of moisture and irrigation facility
Advantages of
intercropping
Advantages of crop
rotation
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5
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/SECTION - B
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25
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(c) Iodine solution
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1
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26
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(c) Potato
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1
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27
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(c)
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1
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28
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(d)
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1
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29
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(a)
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1
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30
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(c)
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1
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31
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(c)
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1
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32
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(a) sugar
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1
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33
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(a)
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1
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34
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Thin
paste of starch is added to hot water with constant stirring.
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2
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35
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Melting
point of ice 00C 273K
Boiling
point of water 1000C 373 K
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2
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36
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Mass
of dry raisins (w1) =
3g
Mass
of soaked raisins (w2) =
4.8g
Mass
of water absorbed = 4.8-3 = 1.8g
w1
3
= 60%
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2
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