Latest Pattern Sample Paper {Science} Term-I Examination (SA–I)

Transcription

Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Latest Pattern Sample Paper {Science}
Term-I Examination (SA–I)
Time: 3 Hours
Max. Marks: 90
General Instructions:
(i) The question paper comprises of 36 questions.
(ii) All questions are compulsory.
(iii) There is no overall choice. However, internal choice has been provided in all the six
questions of five marks category.
Only one option in such questions is to be attempted.
(iv) Questions 1 to 12 are 1 mark questions. These are to be answered in one word or
in one sentence.
(v) Questions 13 to 18 are 2 marks questions. These are to be answered in about 30
words each.
(vi) Questions 19 to 30 are 3 marks questions. These are to be answered in about 50
words each.
(vii) Questions 31 to 36 are 5 marks questions. These are to be answered in about 70
words each.
1.
How are kelvin scale and celsius scale related to each other?
Solution: Kelvin scale (°K) = Celsius scale (°C) + 273.
2.
Name the term used for a functional segment of DNA.
Solution: The functional segment of DNA is gene.
3.
A force of 10 N produces an acceleration of 2.5 ms–2, when applied on an object. What
is the mass of the object?
Solution: Mass (m) =
4.
F
a
10N
2.5ms
2
4kg
A teacher has provided maize grain, nut grain, wheat grain and potato to a student
and asked him to select a plant material, which would give blue-black colour with
iodine solution. Then what would he select?
Solution: He would select potato as it has starch.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721
1
5.
Name the process in which a solid changes directly into gaseous form on
heating without going in liquid state?
Solution: Sublimation.
6.
What do you mean by a resultant force?
Solution: When two or more forces act on a body simultaneously, then the single
force which produces the same effect as produced by all the forces acting together is
known as the resultant force.
7.
By which process recovery of salt from salt solution in water can be done?
Solution: The recovery of salt from salt solution in water can be done by evaporation.
8.
What is plasma?
Solution: Plasma is the fourth state of matter. It consists of super energetic particles.
9.
What is the colour of pure ammonium chloride?
Solution: The colour of pure NH4 Cl is white.
10. Which law is associated with this activity?
Solution: Newton's first law of motion is associated with it.
11. Substance 'X' was added to a test-tube containing water and grounded arhar dal to
test the presence of metanil yellow. The colour of solution changed to pink.
Identify 'X'.
Solution: 'X’ is hydrochloric acid (HCl).
12. A mixture of iron filings and sulphur is heated. What colour change in the mixture will
appear?
Solution: The initial colour is yellow due to sulphur powder whereas on heating, iron
sulphide is formed, which is black in colour.
2
13. When sodium sulphate and barium chloride a solutions are mixed in a test-tube. What
do you observe? Give chemical reaction also.
Solution: White precipitate of barium sulphate is formed.
Na2SO4 aq + BaCl2 aq
BaSO4 s + 2NaCl aq
14. A group of students recorded the following readings while performing the
experiment to calculate the percentage of water absorbed by raisins.
Mass of dry raisins = 2.0 g
Mass of raisins after absorbing water =3.0 g
Calculate the percentage of water absorbed by raisins.
Solution: Percentage of water absorbed by raisins
3.0 2.0
100
2
1
100 50%
2
15. Which elements do the following compounds contain?
(a) Sugar
(b) Common salt
Solution: (a) Sugar is Cl2H22O11. It contains carbon (C), hydrogen (H) and oxygen (O)
elements.
(b) Common salt is NaCl. It contains sodium (Na) and chlorine (Cl) elements.
16. While preparing a temporary mount of onion peel cells or human cheek cells, a
cover slip is put on the mounted material on a slide very gently and pressed down.
Give reasons.
Solution: A cover slip is put on a mounted material gently because
(i) To avoid entry of air bubbles.
(ii) To allow the entry of glycerin into the slide.
17. Differentiate the following activities on the basis of voluntary and involuntary
muscles.
(a) Jumping of frog
(b) Pumping of the heart
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(c) Writing with hand
(d) Movement of chocolate in your intestine
Solution: (a) V
(b)IV
(c)V
(d)IV
18. Make a pile of similar carom coins on a table. Then to remove the lower coin
without touching the other coins with your fingers, you may give a sharp
horizontal hit at the bottom of the pile using sticker.
(a) What will happen if the hit is strong enough?
(b) What will happen to the remaining coins once the lower coin is removed?
Solution: (a) In case the hit is strong enough, the bottom coin moves quickly so that
any horizontal force between it and the above will not move the rest of the pile in
horizontal direction.
(b) Once the lower coin is removed, the inertia of the remaining coins makes them fall
vertically on the table.
19. (a) 4g of solute are dissolved in 36g of water. What is the mass percentage of the
solution?
(b) How can we make a saturated solution unsaturated?
Solution: (a) Mass percentage
=
Mass of solute m 100
Mass of solute m
Mass of solvent M
Here m = 4g, M = 36g Then, mass percentage
4
400
100
10%
4 36
40
(b) On heating or adding more solvent, saturated solution will become unsaturated.
4
20. What is green manuring? Name some commonly used green manure crops.
Solution: Green manuring is a practice of cultivation of quickly growing crop and its
ploughing into soil to improve the physical nature and fertility of soil.
Green manuring provides organic matter and additional nitrogen to the soil.
Commonly used green manure crops are
(i) Sun hemp
(ii) Dhaincha
(iii) Guar
(iv) Lentil (Masoor)
(v) Cowpea (Lobhia)
21. Compare the force of attraction between iron, rubber band and chalk.
Solution: Take an iron nail, a rubber band and a piece of chalk. Try to break them by
hammering or cutting. It is observed that it is most difficult to break an iron
nail but easiest to break a piece of chalk. This shows that force of attraction between
the particles is maximum in an iron nail but minimum in case of a
piece of chalk and between these two, in case of rubber band.
22. Explain what happens, when a beam of light is passed through a colloidal
solution?
Solution: When a beam of light is passed through a colloidal solution, its path
becomes visible. This is due to the scattering of light by colloidal particles. This is
known as Tyndall effect.
This is shown in the diagram below
23. What is cell theory and who proposed this theory? What modification Virchow made
in this theory?
Solution: Cell theory was presented by two biologists, Schleiden and Schwann.
According to this
(i) cell is the basic unit of structure.
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(ii) cell is the basic unit of function.
Rudolf Virchow further modified the cell theory by stating that all cells arise from
pre-existing cells.
24. What will happen if cells are not organized in tissue?
Solution: Organisms whether It is unicellular or multicellular need to perform a lots
of functions like respiration, digestion, locomotion, etc. Cells that are present in group
and specialize in one particular function forms tissues. Some tissue helps in growth,
while others in locomotion and some in body movement.
So, if tissues are not present in bodies of living organism then these kinds of highly
organized and specialized process will become disorganized. There will be no
coordination in the functioning of cells and body.
25. (a) If a potted plant is covered with a glass jar, water vapours appear on the wall of
the glass jar. Explain.
(b) What is the function of cardiac muscle fibre?
(c) Name the cells of bone and cartilage.
Solution: (a) Water vapour released through the stomatal aperture during
transpiration, accumulate on the wall of glass jar.
(b) These muscles are found only in heart. They help in beating of heart and work 24
hours a day till death.
(c) Bone—Osteocytes
Cartilage—Chondrocytes.
26. Starting from a stationary position, Anil paddles his bicycle to attain a velocity of 10
ms–1 in 25 s. Then, he applies brakes such that he again comes to rest after next 50 s.
Calculate the acceleration of the bicycle in both cases. Also find the total distance
covered by Anil.
6
Solution: In first case, initial velocity u1 = 0, time t1 = 25 s and final velocity, v1= 10
ms–1
acceleration, a1 =
v1 u1
t1
10 0
25
0.4ms
2
and distance covered in time t1 is
s1 u1t 1
1 2
a1 t 1
2
= 0 2.5
1
2
0.4
25
2
= 0 + 125
= 125 m
After applying the brakes in second case,
Initial velocity, u2 = 10ms –1
Time, t2 =50s
Final velocity,
v2 = 0
acceleration,
a2
v 2 u2
t2
0 10
50
0.2ms
2
and distance covered in time t2 is
s2 u2t 2
1 2
1
a2t 2 10 50
2
2
0.2
50
2
= 500– 250 = 250 m
So, total distance covered by Anil =s = s1 +s2
=125+ 250 = 375m
27. Describe an activity to demonstrate Newton's third law of motion.
Solution: Describe an activity to demonstrate Newton's third law of motion.
Take a rubber balloon and inflate it. Attach it to a straw and connect it between two
rigid supports as shown in the figure. Prick the balloon with a pin. The balloon moves
in opposite direction of escaped air. This shows that the force with which air is
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pushed out, is exerted in opposite direction by air on the balloon.
28. Glass and china wares are wrapped in straw or paper before packing. Briefly
explain why?
Solution: Glass and china wares are wrapped in straw or paper before packing. If
during loading or unloading in transit, packing gets some jerk, then straw or paper,
being compressible increases the time of jerk. As a result for given change in
momentum, the force of jerk will be less and chances of breakage of glass and china
wares will be reduced.
29. Shikha buys a few grams of precious stones in Delhi. She takes them to
equator and is surprised to see that their weight is reduced. She thinks
that she has been cheated. She goes back to the seller and claims the
exchange amount. The shopkeeper explains her how weight differs due to
value of g. Satisfied Shikha goes back.
(a) List few values of Shikha.
(b) What are the values of shopkeeper?
(c) How does g vary with distance?
Solution: (a) Shikha jumps to conclusion without thinking twice.
(b) Shopkeeper is logical, patient and scientific
(c) g
1
, at equators, g is minimum and at poles, g is maximum.
R2
30. Two objects of masses m1 and m2, when separated by a distance d, exerts a force F on
each other. What happens when
(a) value of mass of first is doubled?
(b) masses of both objects are doubled?
(c) masses are brought so closer that distance between them becomes d/2?
(d) the space between the two objects has no air, i.e., it is complete vacuum?
8
Solution: We know that initial value of gravitational force between two objects,
F=
Gm1m2
d2
(a) If the mass of first object is doubled, then,
F’
2
G 2ml m2
d2
Gm1m2
d2
2F
Hence, the force is doubled.
(b) If the masses of both objects are doubled, then
F'
G 2m1 2m2
d2
4Gm1m2
d2
4F
Hence, the force is quadrupled.
(c) If distance between the objects is reduced to one
half of its original value, then
F'
Gm 1m2
d /2
2
4Gm1m2
d2
4F
hence, the force is quadrupled.
(d) There is no effect on force when air medium is removed.
31. (a) What is vermicompost?
(b) Name two crops used for preparing green manure.
(c) What are the characteristics of storage structure for grains?
(c) Name two nitrogen fixing bacteria.
(d) Which type of crop is generally grown between two cereal crops to
restore soil fertility?
OR
Figure shows the two crop fields [plots A and B] have been treated by manures and
chemical fertilizers respectively, keeping other environmental factors same.
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Observe the graph and answer the following questions:
(a) Why does plot B show sudden increase and then gradual decrease in yield?
(b) Why is the highest peak in plot A graph slightly delayed?
(c) What is the reason for the different pattern of the two graphs?
Solution: (a) The process of preparation of compost by using earthworms, which
hasten the process of decomposition of plant and animal refuse is called
vermi-compost.
(b) Sun hemp
(b) Dhaincha
(c) Storage structures for grains should be air-tight, easy to clean, moisture proof and
rodent proof.
(d) Rhizobium and Nitrosomonas
(e) Leguminous crop.
OR
(a) With the addition of chemical fertilizers there is sudden increase in yield due to
release of nutrients N, P, K, etc in high quantity. The gradual decline in the graph may
be due to continuous use and high quantity of chemicals which kills microbes useful
for replenishing the organic matter in the soil. This decreases the soil fertility.
(b) Manures supply small quantities of nutrients to the soil slowly as it contains large
amounts of organic matter.
It enriches soil with nutrients; thereby increasing soil fertility continuously.
(c) The differences in the two graphs indicate that use of manure is beneficial for long
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duration in cropping as the yield tends to remain high when
the quantity of manure increases.
In case of plot B the chemical fertilizers may cause various problems when used
continuously for long time. Loss of microbial activity reduces decomposition of
organic matter and as a result, soil fertility is lost that affects the yield.
32. (a) Why do solids have fixed shape and fixed volume?
(b) Why is air dense at the sea level?
(c) On melting of ice, there is decrease in volume instead of increase. Why?
(d) What is the binding force between molecules if a substance is a gas under
ordinary conditions of temperature and pressure?
(e) Why are average kinetic energies of hydrogen, carbon dioxide and ethane the
same at the same temperature?
OR
(a) When we light an incense stick (agarbatti) in a corner of our room, why
does its fragrance spread in the whole room quickly?
(b) Liquid A has higher vapour pressure than liquid B. Which liquid out of A
and B will have lower boiling point?
(c) Name the five states of matter, which the scientists are now talking of?
(d) What do you understand by plasma?
(e) Name the Indian physicist, who worked for a fifth state of matter.
Solution: (a) In solids, the particles are closely packed and their positions are fixed
due to strong forces of attraction existing between them. So, solids have fixed shape.
Since, the space between the particles are also fixed, the solids have fixed volume.
(b) Air at sea level is compressed by the mass of air above it. Hence, air is more dense
at sea level than at an altitude.
(c) There are empty spaces in the packing of water molecules in ice. On heating,
different strings of hydrogen bonded water molecules break and water
molecules come closer to each other. Hence, the volume decreases on melting of ice.
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(d) The binding force between the molecules of a gas is van der waals' force.
(e) Average kinetic energy of any gas is directly proportional to its absolute
temperature. Since, all the gases are at the same temperature) their average
kinetic energies will also be the same.
OR
(a) The fragrance of 'agarbatti' spreads all around due to diffusion of its smoke into
the air. The particles of gas produced on burning of incense stick move rapidly in all
the directions. They mix with air and reach the whole room very quickly.
(b) The boiling point of a liquid is the temperature at which its vapour pressure
becomes equal to the atmospheric pressure. The vapour pressure of liquid A is higher
than liquid B. So, vapour pressure of A will become equal to atmospheric pressure at a
lower temperature than liquid B. Consequently boiling point of liquid A is lower than
that of liquid B.
(c) The five states of matter are solid, liquid, gas, plasma and Bose-Einstein
condensate.
(d) Plasma is the state, which consists of super energetic and super excited particles
in the form of ionized gases. The non-sign bulbs and the fluorescent tube consist of
plasma. The plasma glows with a special colour depending on the nature of the gas,
when electric energy flows through it. The sun and stars glow due to the presence of
plasma in them at very high temperature.
(e) In 1920, Indian physicist Satyendra Nath Bose did some calculations for the fifth
state of matter.
33. How will you separate a mixture containing kerosene and petrol (difference in their
boiling points is more than 25 °C), which are miscible with each other?
OR
Benzoic acid is used as a food preservative. The graph below shows the heating curve
for benzoic acid. Study the graph and answer the following questions
(i) At what time does benzoic acid begin to
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(a) melt?
(b) boil?
(ii) What is the melting point of benzoic acid?
(iii) What happens to the temperature while benzoic acid melts?
(iv) What is the physical state of benzoic acid during the time interval of 35-45 min?
Solution: A mixture of kerosene and petrol (b.p. differ by more than 25 oC) can be
separated by the process of simple distillation.
Method In a distillation flask, a mixture of kerosene and petrol is taken as shown in
figure. The mixture is heated slowly and the temperature is noted with the help of
thermometer. Petrol (b.p. =70 oC to 120 oC) vaporizes first and the temperature
becomes constant for sometime (till all petrol evaporates from the mixture). Vapours
of petrol are condensed and collected in another container, while the kerosene
remains in the distillation flask. Again the temperature starts rising and the heating is
stopped and both the components are collected separately.
This is shown in the diagram below:
13
OR
(i) (a) Benzoic acid begin to melt at 20 min.
(b) Benzoic acid begin to boil at 52 min.
(ii) The melting point of benzoic acid is 120° C
(iii) The temperature remains constant at 120°C until all the benzoic acid has melted.
(iv) The physical state of benzoic acid is liquid during the tune interval of 35-45 min.
34. An athlete completes one round of a circular track of diameter 200 m in 40 s.
What will be the distance covered and the displacement at the end of 2 min 20s?
OR
Deduce the following equations of motion
(i) s = ut+ (1/2)at2
(ii) v2 =u2+2as
Solution: (i) Diameter of circular track = 200 m
Radius of circular track =
200
= 100 m
2
Circumference of circular track
2 r 2
22
4400
100
m
7
7
Total time given = 2 min 20 s
= 2×60+20 = 140s
According to question, athlete takes 40 s to complete one round.
Distance covered in 40 s =
4400
m
7
Distance covered in one round is equal to the circumference of the track).
Distance covered m 1 s =
4400
7 40
or distance covered in 140 s =
4400
140
7 40
= 2200 m
= 2.2 km
14
(ii) As per the question, athlete completes one round in 40 s, i.e, after 40 s, athlete
comes back to its initial position or after 40 s his displacement is
zero. Similarly after 120s (40 ×3), his displacement is zero.
Displacement after 2 min 20 s or 140 s
= Displacement after 20 s.
( After 120 s (2 min), displacement is zero.)
In 20 s, athlete will cover half the circular track from A to B (in 40 s one complete
round is taken, so in 20 s half round will be taken).
Displacement after 20 s = AB|
(Initial position is A and after half round, final position is B)
= Diameter of circular track
= 200 m
OR
(i) Consider a body which starts with initial velocity u and due to uniform
acceleration a, its final velocity becomes after time t. Then, its average velocity is
given by
Average velocity
Initial velocity
Final velocity
2
=
u v
2
The distance covered by the body in time t is given by Distance,
s = average velocity × time
or s
u v
t
2
s
u
u at
2
t
15
2ut at 2
1 2
s
ors ut
at
2
2
(ii) We know that,
1 2
at
2
s ut
Also, a =
....................(1)
u v
t
u v
s u
a
t
v u
a
1 v a
a
2
a
uv u2
or s
a
2
v 2 u2 2uv
2a
or 2as 2uv 2u2 v2 u2 uv
or
v2 – u2 = 2as
35. Two objects of masses 100 g and 200 g are moving along the same line and
direction with velocities of 2 ms–1 and 1ms–1, respectively. They collide and
after the collision, the first object moves at a velocity of 1.67 ms –1. Determine the
velocity of the second object.
OR
(i) Explain why is it difficult to walk on sand?
(ii) Why is the recoil of a heavy gun, on firing, not so strong as that of a light gun using
the same cartridge?
(iii) A constant force acts on an object of 5 kg for a period of 2 s. It increases the
velocity of an object from 3 m/s to 7 m/s. Find the magnitude of the applied force.
Now if the force was applied for a period of 5 s, what would be the final velocity of the
object?
Solution: Before collision,
m1 100g
100
0.1kg  1kg 1000g
1000
16
m2 200g
200
0.2kg
1000
v1 = 2 ms–2 and v2 = 1 ms–1
After collision, m1 =100 g = 0.1 kg
m2 =200g = 0.2kg
v1' = 1.67ms–1
Let velocity of second object after collision be v '2 ms–1. Since, there is no external force
on the system.
So, total momentum before collision = total momentum after collision
m1v1 + m2v2 = m1 v1' +m2 v '2
0.1×2 + 0.2 × 1 = 0.1 × 1.67 + 0.2 v '2
0.2+0.2 = 0.167 + 0.2 v '2
0.2 v '2 = 0.4 – 0.167 = 0.233
v '2 =
0.233
1.165ms
0.2
1
OR
(i) While walking on sand, the sand gets pressed down, impacting less reaction force
on the person.
(ii) The mass of the gun is higher. So, it recoils velocity is less.
(iii) Given, m = 5
kg, t1 = 2s,
u = 3 m/s, v = 7 m/s, F=?
t 2 5s, v2 ?
F= m
= 5
v u
t1
7 3
2
10N
v 2 u at 2 3
10 5
13m / s
5
17
36. Name the protective tissue of animal body. State the types of this tissue.
OR
List the characteristics of cork. How is it formed? Mention its role.
Solution: Epithelial tissue is the protective tissue of animal body.
Depending upon the shape and function of the cells, they are of following types
(i) Squamous epithelium Simple squamous epithelium consists of extremely thin and
flat cells forming delicate lining, e.g,, the oesophagus and lining of mouth.
(ii) Cuboidal epithelium It consists of cube-like cells with rounded nuclei and forms
the lining of kidney tubules and duct of salivary glands, where it provides mechanical
support.
(iii) Columnar epithelium It consists of pillar-like cells having elongated nuclei. It is
found in inner lining of intestine where absorption and secretion occur.
(iv) Ciliated epithelium The columnar epithelium tissue also has cilia, which are
hair-like projections on the outer surface of epithelial cells.
(v) Glandular epithelium The columnar epithelium is often modified to form glands,
which secrete chemicals.
OR
(i) It is the outer protective tissue of older stem and roots.
(ii) It is formed by secondary lateral meristem called cork cambium,
(iii) The mature cork become dead and filled with tannin, resin and ah-,
(iv) The cells are arranged compactly without intercellular spaces.
(v) The cells become several layers thick, which are impermeable due to deposition of
suberin in their wall.
Formation of Cork
As plant grow older, the outer protective tissue undergoes certain changes. A strip of
secondary meristem replaces the epidermis of stem. Cells on the outside are cut-off
from this layer. This form several layers thick cork or bark of the no overlap.
18
Role of Cork
(i) It prevents loss of water by evaporation.
(ii) It protects plant from the invasion of parasites and other harmful
microorganisms,
(iii) It is used for manufacture of insulation boards, sport goods, shock absorber, etc.
19

Latest Pattern Sample Paper {Science} Term-I Examination (SA–I)

Transcription

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