Class X_Delhi_Science_Set-3
General Instructions:
(i)
The question paper comprises of two Sections, A and B. You are to attempt both the
sections.
(ii)
All questions are compulsory.
(iii) All question of Section-A and all questions of Section-B are to be attempted separately.
(iv)
Question numbers 1 to 3 in Section-A are one mark questions. These are to be answered
in one word or in one sentence.
(v)
Question numbers 4 to 7 in Sections-A are two marks questions. These are to be
answered in about 30 words each.
(vi)
Question number 8 to 19 in Section-A are three marks questions.
These are to be answered in about 50 words each.
(vii) Question numbers 20 to 24 in Section-A are five marks questions. These are to be
answered in about 70 words each.
(viii) Question numbers 25 to 42 in Section-B are multiple choice questions based on practical
skills. Each question is a one mark question. You are to select one most appropriate
response out of the four provided to you.
SECTION A
1.
Write two reasons responsible for late discovery of noble gases?
1
Ans.
Two reasons for late discoveries of noble gases are:(a)
They are very inert.
(b)
They are present in extremely low concentrations in our environment.
2.
―Cell division is a type of reproduction in unicellular organism.‖ Justify.
Ans.
In unicellular organisms all the important functions are carried out by the single cell
because they lack all cellular organelles. The two types of reproduction i.e. sexual and
asexual need the involvement of many organelles and biomolecules. Hence, these modes
of reproduction is absent in prokaryotes. Unicellular organisms are the simple organisms.
Therefore, cell division acts as the mode of reproduction and results in the production of
daughter cells.
3.
List any two measures that you suggest for better management of water resources.
Ans.
Two measures for better management of water resources are:
(1)
Dams
(2)
Water harvesting
1
1
1
Class X_Delhi_Science_Set-3
4.
(a)
(b)
Ans.
(a)
(b)
5.
Trace the path of sperms from where they are produced in human body to the
exterior.
Write the functions of secretions of prostate gland and seminal vesicles in
humans.
2
The formation of germ-cells or sperms takes place in the testes. The sperms
formed are delivered through the vas deferens which unites with a tube coming
from the urinary bladder. Along the path of the vas deferens, prostate glands and
the seminal vesicles add their secretions so that the sperms are now in a fluid
which makes their transport easier and this fluid also provides nutrition to the
sperms.
Seminal vesicles and prostate glands secrete fluid which
(i)
provide nutrients to the sperm in the form of fructose, calcium, and some
enzymes.
(ii)
act as lubricant and enhance sperm motility.
A ray of light falls normally on the surface of a transparent glass slab. Draw a ray
diagram to show its path and also mark angle of incidence and angle of emergence.
2
Ans.
As the incident light is normal to the surface thus the incidence angle will be zero which
will correspond to zero degree refracted angle and zero degree emergent angle.
6.
Ans.
List and explain any two advantages associated with water harvesting at community
level.
2
Two advantages associated with water harvesting at community level are:
(a)
It will reduce the overexploitation of water resources.
(b)
It also helps in mitigation of droughts and floods.
2
Class X_Delhi_Science_Set-3
7.
We often observe domestic waste decomposing in the bylanes of residential colonies.
Suggest way to make people realise that the improper disposal of waste is harmful to the
environment.
2
Ans.
Some of the ways to make people realize that the improper disposal of waste is harmful
to the environment are:
(a)
Improper disposal of waste will serve as a breeding ground for mosquitoes and
will create favourable conditions for the spread of various diseases.
(b)
Improper disposal of waste will release harmful gases in the environment. It will
make the environment unclean and unhygienic for normal living of the organisms.
(c)
The waste will flow to water bodies along with the rain water and become a threat
to aquatic organisms.
8.
What happens when:
(a)
ethanol is burnt in air,
(b)
ethanol is heated with excess conc. H2SO4 at 443 K,
(c)
a piece of sodium is dropped into ethanol ?
Ans.
(a)
3
When Ethanol burns in air, we get CO2 and H2O
C2 H5OH 3O2  2CO2  3H 2O  l 
Ethanol
Oxygen
Carbon dioxide
(b)
When ethanol is heated with excess conc. H2SO4 at 443K then ethene is formed.
conc. H2SO4
C2 H5OH 
CH2  CH 2  H 2O
443K
(c)
When a piece of sodium is dropped into ethanol we get sodium ethoxide and
hydrogen gas.
2C2 H5OH 2Na  2C2H5ONa  H 2
Ethene
Ethanol
Ethanol
Sodium
Sodium ethoxide
Hydrogen
9.
Why homologous series of carbon compounds are so called? Write chemical formula of
two consecutive members of a homologous series and state the part of these compounds
that determines their (i) physical properties, and (ii) chemical properties.
3
Ans.
Homologous series of carbon compounds are so called because in such a series of
compounds, the same functional group dictates the properties of the carbon compound
regardless of the length of the carbon chain. The two consecutive members of a
homologous series are CH3OH and C2H5OH (belong to alcohol)
3
Class X_Delhi_Science_Set-3
10.
The elements Li, Na and K, each having one valence electron, are in period 2, 3 and 4
respectively of modern periodic table.
3
(i)
In which group of the periodic table should they be?
(ii)
Which one of them is least reactive?
(iii)
Which one of them has the largest atomic radius? Give reason to justify your
answer in each case.
Ans.
Electronic configuration of Li, Na and K can be written as:
Li → 2, 1
Na → 2, 8, 1
K → 2, 8, 8, 1
(a)
As all of them have only on electron in their valence shell they belong to Group 1.
(b)
Li is the least reactive among them due to its smallest size as it has only 2 shells.
(c)
K has the largest atomic radius among the three because it has maximum number
of shells.
Higher the number of shells, larger is the atomic radius.
11.
Given below are some elements of the modern periodic table:
3
Be,
Fe,
Si,
K,
Ca
4
9
14
19
20
(i)
Select the element that has one electron in the outermost shell and write its
electronic configuration.
(ii)
Select two elements that belong to the same group. Give reason for your answer.
(iii)
Select two elements that belong to the same period. Which one of the two has
bigger atomic size?
Ans.
(i)
(ii)
19K
has one electron in the outermost shell and its electronic configuration is
2, 8, 8, 1
4Be and 20Ca belongs to same group i.e. Group-2. Electronic configuration:
4Be – 2, 2
20Ca- 2, 8, 8, 2
The number of electrons in the outermost shell of 4Be and 20Ca is same hence they
belong to the same shell.
4
Class X_Delhi_Science_Set-3
(iii)
and 4Be belongs to the same period, Period 2. Electronic configuration:
9F – 2, 7
4Be – 2, 2
4Be has bigger atomic size then 9F because the atomic radius decreases as we move from
left to right due to increase in nuclear charge which tends to pull the electrons closer to
the nucleus and hence size of F reduces.
12.
9F
Write two examples each of sexually transmitted diseases caused by (i) virus, (ii)
bacteria. Explain how the transmission of such diseases be prevented?
3
Ans.
Sexually transmitted diseases by virus- Genital Herpes by herpes simplex virus and
AIDS by HIV.
Sexually transmitted diseases by bacteria- Gonorrhoea by Nisseria gonorrhoeae and
Syphilis by Trepanoma pallidum
Prevention of transmission of STD‘s
 Having sex with infected or any unknown person should be avoided.
 Sharing of needles, syringes etc. must be prohibited.
 The surgical and dental instruments should be sterilized properly before use.
 Avoid blood transfusion from infected person. Blood should be tested before
transfusion.
 Adequate medical treatment should be provided to the pregnant woman to protect the
child from getting infected.
13.
(a)
(b)
Ans.
(a)
(b)
Name the following:
3
(i)
thread like non-reproductive structures present in Rhizopus.
(ii)
‗blobs‘ that develop at the tips of the non-reproductive threads in
Rhizopus.
Explain the structure and the function of the structures released from the ‗blobs‘
in Rhizopus.
(i)
The thread like non reproductive structures in rhizopus are known as
hyphae
(ii)
Spores develop at the tips of the non reproductive threads in rhizopus.
In fungi Rhizopus, some specialized hyphae give rise to a globular structure
known as sporangia, which contains spores. The sporangia burst to release
spores. Each of these spores germinates to produce a new individual. Since the
spores disperse through air, they can land on various sites. These sites may be
favorable or unfavorable. During unfavourable conditions, these spores are
protected by their thick walls until they come in contact with some favourable
conditions.
5
Class X_Delhi_Science_Set-3
14.
―The sex of a newborn child is a matter of chance and none of the parents may be
considered responsible for it.‖ Justify this statement with the help of flow chart showing
determination of sex of a newborn.
3
Ans.
In human beings, the females have two X chromosomes and the males have one X and
one Y chromosome. Therefore, the females are XX and the males are XY.
At the time of mating, large number of sperms is ejaculated from the male reproductive
organ (penis), into the female reproductive organ i.e. vagina. They travel towards the
fallopian tubes, where only one sperm meets with the egg.
The process of fusion of the sperm and ovum is called fertilization. The sperm has either
X or Y chromosome and egg has only X chromosome. So if sperm carrying Y
chromosome fuses with egg the newly born child will be male and if sperm carrying X
chromosome fuses with the egg the newly born child will be female.
There is equal chance of fusion of either X or Y chromosome with the egg so we can say
that the sex of new born child is a matter of chance and none of the parent is responsible
for it.
Sex determination in human is shown below:
6
Class X_Delhi_Science_Set-3
15.
Ans.
What are fossils? State their importance in the study of evolution with the help of a
suitable example.
3
Fossils are the remains of the organism that once existed on earth.
They provide evidences of evolution by revealing the characteristics of the past
organisms and the changes that have occurred in these organisms to give rise to the
present organisms. Let us explain the importance of fossils in deciding evolutionary
history with the help of the following example:
Around 100 million years ago, some invertebrates died and were buried in the soil in that
area. More sediment accumulated on top of it turning it into sedimentary rock.
At the same place, millions of years later, some dinosaurs died and their bodies were
buried on top of the sedimentary rock. The mud containing dinosaurs also turned into a
rock.
Then, millions of years later, some horse-like creatures died in that area and got fossilized
in rocks above the dinosaur fossils.
Sometime later, due to soil erosion or floods in that area, the rocks containing horse-like
fossils are exposed.
If that area is excavated deeper, then the dinosaur and invertebrates fossils can also be
found. Thus, by digging that area, scientists can easily predict that horse-like animals
evolved later than the dinosaurs and the invertebrates.
Thus, the above example suggests that the fossils found closer to the surface of the earth
are more recent ones than the fossils present in deeper layers.
16.
Ans.
An object of height 4 cm is kept at a distance of 30 cm from a concave lens. Use lens
formula to determine the image distance, nature and size of the image formed if focal
length of the lens is 15 cm.
3
We have height of object, h1= 4 cm, focal length of lens, f = -15 cm, and object distance,
u = -30 cm
1 1 1
1
1
1
1 1
1
Using lens formula, we have    

 

v u f
v (30) (15)
v 30
15
1
1 1
30
     v    10 cm
v
30 15
3
v
 1  1
Magnification, M   10       0.33
u
 30  3
h
v h
1
4
Again, Magnification, M   2  2   h 2   1.33 cm
u h1
4 3
3
Thus the image will be formed in front of the lens at a distance of 10 cm from the lens,
virtual and erect of size 1.33 cm.
7
Class X_Delhi_Science_Set-3
17.
Mention the types of mirrors used as (i) rear view mirrors, (ii) shaving mirrors. List two
reasons to justify your answers in each case.
3
Ans.
(i)
(ii)
Rear view mirrors: Convex mirrors
Reason: (a) Image is formed virtual and erect always.
(b) Image formed is always diminished in size, so it covers large view area.
Shaving mirrors: Concave mirrors
Reason: (a) Image formed is larger in size so as to see clearly details of the object.
(b) Image formed is virtual and erect when object is placed close to the mirror.
18.
State the difference in colours of the sun observed during sunrise/sunset and noon. Give
explanation for each.
3
Ans.
Differences in colours of the sun at sunrise/sunset and at noon are as follows:
(i)
Sunrise/ sunset: Reddish in colour
Reason: Sunlight travels longer distance at this time of the day, so the short
wavelength colours get scattered away and the sunlight is left with only longer
wavelength lights which is reddish or orange. So the sun appears reddish in
colour.
(ii)
Noon: white in colour
Reason: Sunrays travel shorter distance at noon and contains all the wavelengths
of light which combine to form white colour. This makes the colour of sun white.
19.
(a)
(b)
Ans.
Ecosystem is a self sustaining system where the biotic and abiotic organisms of various
communities interact with each other.
(a)
The two components of the ecosystem are – Biotic and Abiotic. Biotic system
consists of all the living organisms of particular area like humans, animals etc and
the nonliving or abiotic component consists of air, mineral soil, water and
sunlight.
(b)
Ponds are the example of natural ecosystem whereas the aquarium is an example
of artificial ecosystem. Ponds do not need to be cleaned but aquarium needs to be
cleaned. This is because aquarium does not contain soil and decomposing bacteria
which helps in degrading complex organic substance into simple inorganic
substance. But pond has this facility. Therefore, it does not need to be cleaned.
20.
(a)
(b)
What is an ecosystem? List its two main components.
3
We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly.
Explain.
Draw a sectional view of human female reproductive system and label the
following parts :
(i)
Where the development of egg occurs.
(ii)
Where fertilization takes place.
Describe the changes the uterus undergoes:
(i)
to receive the zygote.
(ii)
if zygote is not formed.
5
8
Class X_Delhi_Science_Set-3
Ans.
(a)
Human female reproductive system consists of a pair of ovaries, a pair of
oviducts, uterus, and vagina.
(i)
(ii)
(b)
(i)
(ii)
The development of egg occurs in the ovary (labeled in the fig shown
above)
Fertilization takes place in the fallopian tubes (labeled in the fig shown
above))
The uterus prepares itself every month to receive a fertilized egg/ zygote.
The inner uterus lining (endometrium) becomes thick and is supplied with
blood to nourish the embryo.
If the egg is not fertilized, then the uterus lining is not required. Hence, it
breaks down and is released in the form of blood and mucous through the
vagina. This process lasts for 2-8 days. This cycle occurs every month and
is known as menstruation
21.
(a)
(b)
Identify A, B and C in the given diagram and write their functions.
Mention the role of gamete and zygote in sexually reproducing organisms.
5
Ans.
(a)
A - Stigma, Function - The stigma is a sticky surface where the pollen lands and
later germinates
B - Pollen tube, Function- It carries the pollen to the egg cell for fertilization
C - Egg cell, Function- It fuses with the male gamete and leads to the formation of
zygote.
9
Class X_Delhi_Science_Set-3
22.
(b)
Role of gametes- The gametes play an important role in the sexually reproducing
organisms as they carry the entire genetic information of the organism. These
gametes upon fusion result in the formation of zygote, which develops into a new
individual. Any deformation in the gametes will lead to the deformity in the
newly formed offspring.
Role of zygote- Zygote is the diploid cell formed by the fusion of male and
female gametes during fertilization in sexual reproduction. Zygote is the first
stage in the development process of an organism and it contains all the genetic
information of both the parents, essential for the growth of the new organism.
(a)
(b)
Define the term ‗isomers‘.
5
Draw two possible isomers of the compound with molecular formula C3H6O and
write their names.
Give the electron dot structures of the above two compounds.
(c)
Ans.
(a)
(b)
Molecules having same molecular formula but different structural formula are
known as isomers.
Two possible isomers of the compound with molecular formula C3H6O:
O
O
||
||
CH3 — C — CH3 and CH3 — CH 2 — C — H
Acetone
(c)
Electron dot structure:
O
||
CH3 — C — CH3 and
Propanal
O
||
CH3 — CH 2 — C — H
10
Class X_Delhi_Science_Set-3
23.
(a)
(b)
Ans.
(a)
A person cannot read newspaper placed nearer than 50 cm from his eyes. Name
the defect of vision he is suffering from. Draw a ray diagram to illustrate this
defect. List its two possible causes. Draw a ray diagram to show how this defect
may be corrected using a lens of appropriate focal length.
5
We see advertisements for eye donation on television or in newspapers. Write the
importance of such advertisements.
The person is suffering from Hypermetropia i.e. far-sightedness.
Hypermetropia is caused due to:
(i) The focal length of the eye lens is too long
(ii) Decrease in the length of the eyeball
This defect can be corrected using convex lens of appropriate focal length.
(b)
Eye donation:
Our eye can live even after our death, so by donating our eyes, we can give vision
to any blind person and make him see. By giving such advertisement in
newspaper we can aware more people for this noble cause and raise the number of
donations for blind people.
11
Class X_Delhi_Science_Set-3
24.
Define the term absolute refractive index. The absolute refractive index of diamond is
2.42. What is the meaning of this statement? Refractive indices of media A, B C and D
are given below:
5
Media
Refractive index
A
1.33
B
1.44
C
1.52
D
1.65
In which of these four media is the speed of light (i) minimum and (ii) maximum?
Find the refractive index of medium C with respect to medium B.
Ans.
Absolute refractive index: The refractive index of any medium with respect to the
vacuum is called absolute refractive index of that medium.
The absolute refractive index of diamond is 2.42. This means the speed of light in
diamond is 2.42 times slower than that in the vacuum.
c
n   2.42
v
Where, n = absolute refractive index of diamond
c = speed of light in vacuum
v = speed of light in medium (i.e. diamond)
From the above expression we can see that the speed of light varies inversely with respect
to the refractive index. Thus,
(i)
Minimum speed of light in: medium A
(ii)
Maximum speed of light in: medium D
Refractive index of medium C with respect to medium B is
n C 1.52

 1.06
n B 1.44
12