General Instructions:
- All questions would be compulsory. However, an internal choice of approximately 33% would be provided. 50% marks are to be allotted to competency-based questions.
- Section A would have 16 simple/complex MCQs and 04 Assertion-Reasoning type questions carrying 1 mark each.
- Section B would have 6 Short Answer (SA) type questions carrying 02 marks each.
- Section C would have 7 Short Answer (SA) type questions carrying 03 marks each.
- Section D would have 3 Long Answer (LA) type questions carrying 05 marks each.
- Section E would have 3 source based/case based/passage based/integrated units of assessment (04 marks each) with sub-parts of the values of 1/2/3 marks.
SECTION-A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions.
1.
Identify '\(x\)', '\(y\)', and '\(z\)' in the following balanced reaction:
$$ x Al(s) + y O_2(g) \rightarrow z Al_2O_3(s) $$
(a) 4, 3, 2
(b) 2, 1, 1
(c) 4, 2, 2
(d) 2, 3, 2
Ans: (a) 4, 3, 2
The balanced reaction is: \( 4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s) \)
2.
4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide. Consider the following table:
| Substance | pH |
|---|---|
| Lemon | 2.3 |
| Battery acid | \(x\) |
| Sea water | 8.5 |
| Apple | 3.1 |
The value of \(x\) in above table is:
(a) 0
(b) 1.3
(c) 2.5
(d) 1.9
Ans: (a) 0
The value of \(x\) is 0. The value of pH of battery acid is zero. pH may be defined as a number by which negative power of 10 has to be raised in order to express the concentration of hydrogen ion of solution.
3.
Magnesium ribbon is rubbed with sand paper before making it to burn. The reason of rubbing the ribbon is to:
(a) remove moisture condensed over the surface of ribbon.
(b) generate heat due to exothermic reaction.
(c) remove magnesium oxide formed over the surface of magnesium.
(d) mix silicon from sand paper (silicon dioxide) with magnesium for lowering ignition temperature of the ribbon.
Ans: (c) remove magnesium oxide formed over the surface of magnesium.
When magnesium is exposed to air, a layer of oxide is formed on its surface and it gets corroded. So, as to remove the layer of oxide formed (MgO), magnesium ribbon is rubbed.
4.
A student traces the path of a ray of light through a glass prism for different angles of incidence. He analyzes each diagram and draws the following conclusion:
I. On entering prism, the light ray bends towards its base.
II. Light ray suffers refraction at the point of incidence and point of emergence while passing through the prism.
III. Emergent ray bends at certain angle to the direction of the incident ray.
IV. While emerging from the prism, the light ray bends towards the vertex of the prism.
Out of the above inferences, the correct ones are:
I. On entering prism, the light ray bends towards its base.
II. Light ray suffers refraction at the point of incidence and point of emergence while passing through the prism.
III. Emergent ray bends at certain angle to the direction of the incident ray.
IV. While emerging from the prism, the light ray bends towards the vertex of the prism.
Out of the above inferences, the correct ones are:
(a) I, II and III
(b) I, III and IV
(c) II, III and IV
(d) I and IV
Ans: (a) I, II and III
Incident ray and the emergent ray on passing through the prism follows the different path. On refraction of a ray of light through the prism, the emergent ray always bends at an angle with the incident ray called angle of deviation. Light ray suffers refraction at the point of incidence and point of emergence while passing through the prism.
5.
Which of the following structures is involved in gaseous exchange in woody stem of a plant?
(a) Stomata
(b) Guard cell
(c) Lenticel
(d) Epidermis
Ans: (c) Lenticel
In woody stems, there is a special organ called lenticels which helps in the respiratory exchange of gases. Lenticels are the tissue which consists of large inter-cellular spaces in the periderm layer. The tissues are porous in nature and functions as a pore which helps in direct exchange of gases in the woody stems.
6.
A feature of reproduction that is common to Amoeba, Spirogyra and yeast is that
(a) They reproduce asexually
(b) They are all unicellular
(c) They reproduce only sexually
(d) They are all multicellular
Ans: (a) they reproduce asexually
Amoeba, Spirogyra, and Yeast all reproduce by the asexual method. Amoeba reproduces by binary fission, Spirogyra by fragmentation, and Yeast by the budding method. All are unicellular organisms.
7.
Ethane (\(C_2H_6\)) on complete combustion gave \(CO_2\) and water. It shows that the results are in accordance with the law of conservation of mass. Then, the coefficient of oxygen is equal to
(a) 7/2
(b) 3/2
(c) 5/2
(d) 9/2
Ans: (a) 7/2
Balanced chemical equation w.r.t. law of conservation of mass.
\( C_2H_6 + \frac{7}{2}O_2 \longrightarrow 2CO_2 + 3H_2O \)
The coefficient of \(C_2H_6\) is 1, \(\frac{7}{2}\) for \(O_2\), 2 for \(CO_2\) and 3 for \(H_2O\).
\( C_2H_6 + \frac{7}{2}O_2 \longrightarrow 2CO_2 + 3H_2O \)
The coefficient of \(C_2H_6\) is 1, \(\frac{7}{2}\) for \(O_2\), 2 for \(CO_2\) and 3 for \(H_2O\).
8.
When white light passes through the achromatic combination of prisms, then what is observed?
(a) Deviation
(b) Dispersion
(c) Both deviation and dispersion
(d) Atmospheric refraction
Ans: (a) Deviation
When white light passes through a prism It disperses into band of seven colours. But when two prism are combined in such a way that sum of angular dispersions of crown glass prism and flint glass prism is zero then such a combination is achromatic combination of prisms. When white light passes through achromatic combination of prisms, internally dispersed components of white light from crown glass prism refract and meet together at the outer surface edge of flint glass prism, so the refracted light from the achromatic combination of prisms become white light again with deviation only without any dispersion.
9.
Exposure of silver chloride to sunlight for a long duration turns grey due to
Which among the following statement(s) is(are) true?
(i) the formation of silver by decomposition of silver chloride.
(ii) sublimation of silver chloride.
(iii) decomposition of chlorine gas from silver chloride.
(iv) oxidation of silver chloride.
Which among the following statement(s) is(are) true?
(i) the formation of silver by decomposition of silver chloride.
(ii) sublimation of silver chloride.
(iii) decomposition of chlorine gas from silver chloride.
(iv) oxidation of silver chloride.
(a) Only 1
(b) 1 and 3
(c) 2 and 3
(d) Only 4
Ans: (a) Only 1
Silver chloride decomposes to silver in presence of sunlight hence turns grey.
10.
Magnesium reacts with hot water and steam both. Human body stores energy in form of:
(a) Glucose
(b) Insulin
(c) Glycogen
(d) Fructose
Ans: (c) glycogen
Carbohydrates, such as sugar and starch, are readily broken down into glucose, the body's principal energy source. Glucose can be used immediately as fuel, or can be sent to the liver and muscles and stored as glycogen.
11.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-
(a) Plane
(b) Concave
(c) Convex
(d) Either plane or convex
Ans: (d) Either plane or convex
No matter whatever is the position of object, a convex mirror always forms a virtual, erect and diminished image of the object placed in front of it, whereas a plane mirror always forms a virtual, erect and of the same size image as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.
12.
What must be preserved in an ecosystem, if the system needs to be maintained?
(a) Producers and carnivores
(b) Producers and decomposers
(c) Carnivores and decomposers
(d) Herbivores and carnivores
Ans: (b) producers and decomposers
The most important characteristics of any ecosystem are energy flow and cycling of materials. Producers and decomposers are indispensable for any ecosystem. Producers trap solar energy and convert it into usable form as carbohydrates which are passed on to successive levels through food chains and food webs. The other important characteristic of the ecosystem is cycling of materials. This is called as biogeochemical cycles, in which decomposers are most important because they will release the mineral nutrients back to the environment.
13.
Posture and balance of the body is controlled by
(a) cerebrum
(b) cerebellum
(c) medulla
(d) pons
Ans: (b) cerebellum
Cerebellum is the region which plays a major role in coordination of voluntary motor movement, balance and equilibrium and muscle tone. Cerebellar damage causes disorder in coordination, speed, posture and motor learning.
14.
A student determines the focal length of a device \(X\), by focusing the image of a far off object on the screen positioned as shown in figure. The device \(X\) is a
[Diagram showing rays reflecting off a mirror onto a screen]
(a) Convex lens
(b) Concave lens
(c) Convex mirror
(d) Concave mirror
Ans: (d) Concave mirror
A concave mirror alone can form real image of distant object on the screen held in position, as shown in figure.
15.
Which among the following statements is incorrect for magnesium metal?
(a) It burns in oxygen with a dazzling white flame.
(b) It reacts with cold water to form magnesium oxide and evolves hydrogen gas.
(c) It reacts with hot water to form magnesium hydroxide and evolves hydrogen gas.
(d) It reacts with steam to form magnesium hydroxide and evolves hydrogen gas.
Ans: (b) It reacts with cold water to form magnesium oxide and evolves hydrogen gas.
Magnesium does not react with cold water to give magnesium hydroxide and hydrogen.
\( Mg(s) + H_2O(l) \longrightarrow Mg(OH)_2 + H_2(g) \uparrow \)
\( Mg(s) + H_2O(l) \longrightarrow Mg(OH)_2 + H_2(g) \uparrow \)
16.
Mineral acids are stronger acids than carboxylic acids because
(i) mineral acids are completely ionized.
(ii) carboxylic acids are completely ionized
(iii) mineral acids are partially ionized
(iv) carboxylic acids are partially ionized
(i) mineral acids are completely ionized.
(ii) carboxylic acids are completely ionized
(iii) mineral acids are partially ionized
(iv) carboxylic acids are partially ionized
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (iii) and (iv)
Ans: (a) (i) and (iv)
Mineral acids are stronger acids than carboxylic acids because mineral acids are completely ionised whereas carboxylic acids are partially ionised.
Question no. 17 to 20 are Assertion-Reasoning based questions.
17.
Assertion : Photosynthesis is considered as an endothermic reaction.
Reason : Energy gets released in the process of photosynthesis.
Reason : Energy gets released in the process of photosynthesis.
(a) Both Assertion and Reason are True and Reason is the correct explanation of the Assertion.
(b) Both Assertion and Reason are True but Reason is not the Correct explanation of the Assertion.
(c) Assertion is True but the Reason is False.
(d) Both Assertion and Reason are False.
Ans: (c) Assertion (A) is true but reason (R) is false.
Photosynthesis is considered as an endothermic reaction because energy in the form of sunlight is absorbed by the green plants.
18.
Assertion : Our body maintains blood sugar level.
Reason : Pancreas secretes insulin which helps to regulate blood sugar levels in the body.
Reason : Pancreas secretes insulin which helps to regulate blood sugar levels in the body.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Pancreas secretes insulin which helps to regulate blood sugar levels in the body. If the sugar level in blood rises, they are detected by the cells of the pancreas which respond by producing more insulin. As the blood sugar level falls, insulin secretion is reduced.
19.
Assertion : Artificial kidney is a device used to remove nitrogenous waste products from the blood through dialysis.
Reason : Reabsorption does not occur in artificial kidney.
Reason : Reabsorption does not occur in artificial kidney.
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Ans: (c) Assertion is true but Reason is false.
Kidney failure can be managed by artificial kidney. It is a device used to remove nitrogenous waste products from the blood through dialysis. Artificial kidney is different from natural kidney as the process of reabsorption does not occur in artificial kidney.
20.
Assertion : The product of resistivity and conductivity of a conductor depends on the material of the conductor.
Reason : Because each of resistivity and conductivity depends on the material of the conductor.
Reason : Because each of resistivity and conductivity depends on the material of the conductor.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Ans: (c) Assertion (A) is true but reason (R) is false.
\( \text{Conductivity} = \frac{1}{\text{Resistivity}} \)
\( \text{Conductivity} \times \text{resistivity} = 1 \)
\( \text{Conductivity} \times \text{resistivity} = 1 \)
SECTION-B
Question no. 21 to 26 are very short answer questions.
21.
If you keep the potted plant horizontally for 2-3 days, what type of movements would be shown by the shoot and root after two or three days. Why?
Ans:
If we keep the potted plant horizontally for 2-3 days, shoots may grow upwards and away from the earth while roots always grow downwards. It happens because shoots are negatively geotropic and positively phototropic while roots are positively geotropic.
22.
What are the rules of inheritance?
Ans:
On the basis of his experiments, Mendel established some rules which are called the rules of inheritance. They are:
(i) Law of Dominance,
(ii) Law of Segregation and
(iii) Law of Independent Assortment.
(i) Law of Dominance,
(ii) Law of Segregation and
(iii) Law of Independent Assortment.
23.
A. What prevents the metals such as magnesium, aluminium, zinc and lead from oxidation at ordinary temperature?
Ans:
At ordinary temperature, the surface of metals such as magnesium, aluminium, zinc and lead, etc., are covered with a thin layer of oxide. The protective oxide layer prevents the metals from further oxidation.
or
B. Explain why sodium hydroxide solution cannot be kept in aluminium containers? Write equation for the reaction that may take for the same.
Ans:
Aluminium is an amphoteric metal. It reacts with NaOH to form \(NaAlO_2\). So, NaOH cannot be stored in an aluminium container.
\( 2Al + 2NaOH \rightarrow 2NaAlO_2 + H_2(g) \)
\( 2Al + 2NaOH \rightarrow 2NaAlO_2 + H_2(g) \)
24.
What is meant by pollination? Name and differentiate between the two modes of pollination in flowering plants.
Ans:
Pollination: Is the process of transfer of pollen grains from the anther to the stigma of flower. If this transfer of pollen occurs in the same flower or flowers of same plant, it is referred to as self-pollination whereas if the pollen is transferred from one flower to another of same species, it is known as cross-pollination.
25.
A. State two positions in which a concave mirror produces a magnified image of a given object. List two differences between the two images.
Ans:
(i) When the object is placed in front of the mirror:
(a) between the pole and focus.
(b) between the focus and centre of curvature.
(ii) In case (a) the image is virtual and erect.
(iii) In case (b) the image is real and inverted.
(a) between the pole and focus.
(b) between the focus and centre of curvature.
(ii) In case (a) the image is virtual and erect.
(iii) In case (b) the image is real and inverted.
or
B. What is the difference between virtual images produced by concave, plane and convex mirror?
Ans:
Virtual image produced by concave mirror is magnified, that produced by plane mirror is of the same size and the virtual image produced by convex mirror is diminished.
26.
In the cartoon below, a rabbit is shown eating grass and later a fox is seen hunting the rabbit. In the next frame, after the fox dies, mushrooms and earthworms are feeding on its body. What role do the rabbit and fox play in the food chain?
Ans:
(i) The rabbit is a primary consumer (herbivore) because it eats plants (grass).
(ii) The fox is a secondary consumer (carnivore) because it preys on the rabbit.
(ii) The fox is a secondary consumer (carnivore) because it preys on the rabbit.
SECTION-C
Question no. 27 to 33 are short answer questions.
27.
Aman creates a compact device that uses an organic compound (\(C_3H_8O\)) reacting with sodium metal to produce hydrogen gas. This hydrogen powers a fuel cell, providing a clean and immediate energy source. Deduce the possible structure of the compound. Write the balanced chemical equation of the reaction.
Ans:
\( 2CH_3CH_2CH_2OH + 2Na \longrightarrow 2CH_3CH_2CH_2O^-Na^+ + H_2(g) \)
(Propanol) -> (Sodium propoxide)
The possible structure of compound is propanol.
$$ H-C(H)(H)-C(H)(H)-C(H)(H)-O-H $$
(Propanol) -> (Sodium propoxide)
The possible structure of compound is propanol.
$$ H-C(H)(H)-C(H)(H)-C(H)(H)-O-H $$
28.
A. Our government launches campaigns to provide information about AIDS prevention, testing and treatment by putting posters, conducting radio shows and using other agencies of advertisements. To which category of diseases AIDS belongs? Name and explain. What is its causative organism? Also give two more examples of such diseases.
Ans:
AIDS belongs to STDs (Sexually Transmitted Diseases). The diseases which are spread by sexual contact with an infected person are called sexually transmitted disease.
Its causative organism is a virus — HIV.
Some examples are:
(i) Gonorrhoea caused by bacteria
(ii) Syphilis caused by bacteria
(iii) Warts
Its causative organism is a virus — HIV.
Some examples are:
(i) Gonorrhoea caused by bacteria
(ii) Syphilis caused by bacteria
(iii) Warts
or
B. Distinguish between pollination and fertilisation. Mention the site and the product of fertilisation in a flower.
Ans:
(i) The transfer of pollen grains from anther of a stamen to the stigma of a carpel is called pollination whereas fertilisation is the process when the male gamete present in pollen grain joins the female gamete present in ovule.
(ii) Pollination is an external mechanism whereas fertilisation is an internal mechanism which takes place inside the flower.
Site of fertilisation in flower is ovary.
Product of fertilisation in flower is zygote.
(ii) Pollination is an external mechanism whereas fertilisation is an internal mechanism which takes place inside the flower.
Site of fertilisation in flower is ovary.
Product of fertilisation in flower is zygote.
29.
Explain the following chemical changes, giving one example in each case:
(i) Displacement or substitution,
(ii) Dissociation,
(iii) Isomerisation reaction.
(i) Displacement or substitution,
(ii) Dissociation,
(iii) Isomerisation reaction.
Ans:
(i) Displacement or substitution reaction: The reaction in which an atom or a group of atoms in the molecule is replaced by another atom or a group of atoms is called displacement or substitution reaction. For example, zinc displaces copper from its sulphate solution.
\( CuSO_4 + Zn \longrightarrow ZnSO_4 + Cu \)
(ii) Dissociation reaction: When a substance breaks up into positive and negative ions in water, it is called dissociation reaction. For example, acetic acid in water dissociates into \(CH_3COO^-\) and \(H^+\) ions.
\( CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \)
(iii) Isomerisation reaction: When a compound changes into another compound by simple rearrangement of atoms, it is called an isomerisation reaction. For example,
\( NH_4CNO \xrightarrow{\text{heat}} NH_2CONH_2 \) (Ammonium cyanate to Urea)
\( CuSO_4 + Zn \longrightarrow ZnSO_4 + Cu \)
(ii) Dissociation reaction: When a substance breaks up into positive and negative ions in water, it is called dissociation reaction. For example, acetic acid in water dissociates into \(CH_3COO^-\) and \(H^+\) ions.
\( CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \)
(iii) Isomerisation reaction: When a compound changes into another compound by simple rearrangement of atoms, it is called an isomerisation reaction. For example,
\( NH_4CNO \xrightarrow{\text{heat}} NH_2CONH_2 \) (Ammonium cyanate to Urea)
30.
Why does a ray of light passing through the centre of curvature of a concave mirror after reflection, is reflected back along the same path?
Ans:
It is because the incident ray falls on the mirror along the normal to the reflecting surface. Hence the angle of incidence is zero and according to law of reflection, angle of incidence is always equal to angle of reflection. Therefore the reflected ray back along the same path.
31.
(i) A compound lens is made of two lenses in contact having powers \(+12.5 D\) and \(-2.5 D\). Find the focal length and power of the combination.
(ii) The magnification produced by a mirror is \(+1\). What does this mean?
(ii) The magnification produced by a mirror is \(+1\). What does this mean?
Ans:
(i) \( P = P_1 + P_2 = 12.5 + (-2.5) = 10 D \)
\( f = \frac{1}{P} = \frac{1}{10} = 0.1 \text{ m} \)
(ii) '+' sign means image is virtual and erect. 1 means it is of the same size.
\( f = \frac{1}{P} = \frac{1}{10} = 0.1 \text{ m} \)
(ii) '+' sign means image is virtual and erect. 1 means it is of the same size.
32.
In the given circuit, find:
(i) Total resistance of the network of resistors
(ii) Current through ammeter \(A\)
[Circuit description: A 6V battery. Two parallel branches. Branch 1 has \(4\Omega\) and \(2\Omega\) in series. Branch 2 has \(3\Omega\) and \(3\Omega\) in series.]
(i) Total resistance of the network of resistors
(ii) Current through ammeter \(A\)
[Circuit description: A 6V battery. Two parallel branches. Branch 1 has \(4\Omega\) and \(2\Omega\) in series. Branch 2 has \(3\Omega\) and \(3\Omega\) in series.]
Ans:
(i) In the given circuit diagram \(4 \Omega\) and \(2 \Omega\) resistances are connected in series combination and \(3 \Omega\) and \(3 \Omega\) resistance are also connected in the series combination.
\( R_1 = 4 + 2 = 6 \Omega \)
\( R_2 = 3 + 3 = 6 \Omega \)
Now the equivalent resistance of circuit
\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \)
\( R_{eq} = \frac{6}{2} = 3 \Omega \)
(ii) According to ohm's law,
\( V = IR_{eq} \)
\( I = \frac{V}{R_{eq}} = \frac{6}{3} = 2 A \)
\( R_1 = 4 + 2 = 6 \Omega \)
\( R_2 = 3 + 3 = 6 \Omega \)
Now the equivalent resistance of circuit
\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \)
\( R_{eq} = \frac{6}{2} = 3 \Omega \)
(ii) According to ohm's law,
\( V = IR_{eq} \)
\( I = \frac{V}{R_{eq}} = \frac{6}{3} = 2 A \)
33.
(i) How many eggs are produced every month by either of the ovaries in a human female? Where does fertilization take place in the female reproductive system?
(ii) What happens in case the eggs released by the ovary are not fertilized?
(ii) What happens in case the eggs released by the ovary are not fertilized?
Ans:
(i) One egg is produced every month by one of the ovaries. Fertilization takes place in the fallopian tube.
(ii) In case the egg released by the ovary is not fertilized, it lives for about one day. Since the uterus prepares itself every month to receive a fertilized egg its lining becomes thick and spongy and since it is not required anymore, this lining slowly breaks and comes out through the vagina as blood and mucous. This is known as menstruation.
(ii) In case the egg released by the ovary is not fertilized, it lives for about one day. Since the uterus prepares itself every month to receive a fertilized egg its lining becomes thick and spongy and since it is not required anymore, this lining slowly breaks and comes out through the vagina as blood and mucous. This is known as menstruation.
SECTION-D
Question no. 34 to 36 are Long answer questions.
34.
Discuss the physical properties of non-metals.
Discuss the exceptions in the properties of metals and non-metals.
Discuss the exceptions in the properties of metals and non-metals.
Ans:
Physical properties of non-metals are following:
(i) Non-metals are brittle: They break into pieces when hammered or stretched, i.e., they are not malleable.
(ii) Non-metals are non-ductile: These cannot be drawn into wires.
(iii) Non-metals are bad conductors of heat and electricity: Non-metals do not have free electrons. There is one exception graphite, which is a good conductor of heat and electricity.
(iv) Non-metals are non-lustrous and cannot be polished: The exceptions are graphite and iodine which are lustrous.
(v) Non-metals are generally soft: Except diamond (allotropic form of carbon) which is the hardest substance known, non-metals are soft.
(vi) Non-metals generally have low melting and boiling points: Except graphite which has high melting point, non-metals have weak intra-molecular force.
(vii) Non-metals have low densities: Most non-metals are light.
Exceptions in the properties of metals and non-metals:
(i) All metals except mercury exist as solids at room temperature. Metals have high melting points but gallium and caesium have very low melting points. These two metals will melt if they are kept on palm.
(ii) Iodine is a non-metal but it is lustrous.
(iii) Carbon is a non-metal that can exist in different forms. Each form is called an allotrope. Diamond, an allotrope of carbon, is the hardest natural substance known and has a very high melting and boiling points. Graphite, another allotrope of carbon, is a conductor of electricity.
(iv) Alkali metals (lithium, sodium, potassium) are so soft that they can be cut with a knife. They have low densities and low melting points.
(i) Non-metals are brittle: They break into pieces when hammered or stretched, i.e., they are not malleable.
(ii) Non-metals are non-ductile: These cannot be drawn into wires.
(iii) Non-metals are bad conductors of heat and electricity: Non-metals do not have free electrons. There is one exception graphite, which is a good conductor of heat and electricity.
(iv) Non-metals are non-lustrous and cannot be polished: The exceptions are graphite and iodine which are lustrous.
(v) Non-metals are generally soft: Except diamond (allotropic form of carbon) which is the hardest substance known, non-metals are soft.
(vi) Non-metals generally have low melting and boiling points: Except graphite which has high melting point, non-metals have weak intra-molecular force.
(vii) Non-metals have low densities: Most non-metals are light.
Exceptions in the properties of metals and non-metals:
(i) All metals except mercury exist as solids at room temperature. Metals have high melting points but gallium and caesium have very low melting points. These two metals will melt if they are kept on palm.
(ii) Iodine is a non-metal but it is lustrous.
(iii) Carbon is a non-metal that can exist in different forms. Each form is called an allotrope. Diamond, an allotrope of carbon, is the hardest natural substance known and has a very high melting and boiling points. Graphite, another allotrope of carbon, is a conductor of electricity.
(iv) Alkali metals (lithium, sodium, potassium) are so soft that they can be cut with a knife. They have low densities and low melting points.
35.
Suggest three contraceptive methods to control the size of human population. Mention two factors that determine the size of population.
OR
How do the following organisms reproduce by asexual methods?
(i) Euglena (ii) Spirogyra (iii) Ginger (iv) Chrysanthemum (v) Strawberry (vi) Mango
OR
How do the following organisms reproduce by asexual methods?
(i) Euglena (ii) Spirogyra (iii) Ginger (iv) Chrysanthemum (v) Strawberry (vi) Mango
Ans:
(i) Three contraceptive methods to control the size of human population:
(a) One category is the use of a mechanical barrier, e.g., condoms.
(b) Another category of contraceptive acts by changing the hormonal balance of the body so that eggs are not released and fertilization doesn't occur, e.g., oral pills.
(c) Use of intrauterine device like copper-T to prevent pregnancy.
(ii) Two factors that determine the size of population:
(a) Rate of birth;
(b) Rate of death.
OR
(i) Euglena — Binary fission.
(ii) Spirogyra — Fragmentation.
(iii) Ginger — Natural vegetative propagation by stems.
(iv) Chrysanthemum — Artificial vegetative propagation by cutting.
(v) Strawberry — Artificial vegetative propagation by layering.
(vi) Mango — Artificial vegetative propagation by grafting.
(a) One category is the use of a mechanical barrier, e.g., condoms.
(b) Another category of contraceptive acts by changing the hormonal balance of the body so that eggs are not released and fertilization doesn't occur, e.g., oral pills.
(c) Use of intrauterine device like copper-T to prevent pregnancy.
(ii) Two factors that determine the size of population:
(a) Rate of birth;
(b) Rate of death.
OR
(i) Euglena — Binary fission.
(ii) Spirogyra — Fragmentation.
(iii) Ginger — Natural vegetative propagation by stems.
(iv) Chrysanthemum — Artificial vegetative propagation by cutting.
(v) Strawberry — Artificial vegetative propagation by layering.
(vi) Mango — Artificial vegetative propagation by grafting.
36.
A household uses the following electric appliances:
(i) refrigerator of rating 400 W for 10 hours each day.
(ii) two electric fans of rating 80 W each for 6 hours daily.
(iii) six electric tubes of rating 18 W each for 6 hours daily.
Calculate the electricity bill for the household for month of June, if cost of electrical energy is ₹3.00 per unit.
OR
The values of current \(I\) flowing in a given resistor for the corresponding values of potential difference \(V\) across the resistor are given below:
\(I\) (ampere): 0.5, 1.0, 2.0, 3.0, 4.0
\(V\) (volt): 1.6, 3.4, 6.7, 10.2, 13.2
Plot a graph between \(V\) and \(I\) and calculate the resistance of the resistor.
(i) refrigerator of rating 400 W for 10 hours each day.
(ii) two electric fans of rating 80 W each for 6 hours daily.
(iii) six electric tubes of rating 18 W each for 6 hours daily.
Calculate the electricity bill for the household for month of June, if cost of electrical energy is ₹3.00 per unit.
OR
The values of current \(I\) flowing in a given resistor for the corresponding values of potential difference \(V\) across the resistor are given below:
\(I\) (ampere): 0.5, 1.0, 2.0, 3.0, 4.0
\(V\) (volt): 1.6, 3.4, 6.7, 10.2, 13.2
Plot a graph between \(V\) and \(I\) and calculate the resistance of the resistor.
Ans:
Energy consumed per day by refrigerator = \( 0.4 \text{ kW} \times 10 \text{ h} = 4 \text{ kWh} \)
Energy consumed per day by fans = \( 2 \times 0.08 \text{ kW} \times 6 = 0.96 \text{ kWh} \)
Energy consumed by lights = \( 6 \times 0.018 \text{ kW} \times 6 \text{ h} = 0.648 \text{ kWh} \)
Total energy consumed per day = \( 4 + 0.96 + 0.648 = 5.608 \text{ kWh} \)
Energy consumed in 30 days = \( 30 \times 5.608 = 168.24 \text{ kWh} \)
Cost of 168.24 units @ ₹3.00 = \( 168.24 \times 3 = \text{₹}504.72 \)
OR
From the graph (plotting V vs I), we can take values of \(V\) and \(I\).
\( \Delta V = (6.7 - 3.4) = 3.3 \text{ Volt} \)
\( \Delta I = (2.0 - 1.0) = 1.0 \text{ A} \)
\( R = \frac{V}{I} = \frac{3.3}{1.0} = 3.3 \Omega \)
So, Resistance = \( 3.3 \Omega \) (ohm)
Energy consumed per day by fans = \( 2 \times 0.08 \text{ kW} \times 6 = 0.96 \text{ kWh} \)
Energy consumed by lights = \( 6 \times 0.018 \text{ kW} \times 6 \text{ h} = 0.648 \text{ kWh} \)
Total energy consumed per day = \( 4 + 0.96 + 0.648 = 5.608 \text{ kWh} \)
Energy consumed in 30 days = \( 30 \times 5.608 = 168.24 \text{ kWh} \)
Cost of 168.24 units @ ₹3.00 = \( 168.24 \times 3 = \text{₹}504.72 \)
OR
From the graph (plotting V vs I), we can take values of \(V\) and \(I\).
\( \Delta V = (6.7 - 3.4) = 3.3 \text{ Volt} \)
\( \Delta I = (2.0 - 1.0) = 1.0 \text{ A} \)
\( R = \frac{V}{I} = \frac{3.3}{1.0} = 3.3 \Omega \)
So, Resistance = \( 3.3 \Omega \) (ohm)
SECTION-E
Question no. 37 to 39 are case-based/data-based questions.
37.
After coming from playground, Tanu feels very hungry. But still some more time was required by her mother to cook food. While waiting on dining table Tanu was playing with her spoon. All of sudden she observed two different orientations of her face when she looked her face from both sides of spoon. She was confused why the orientation of her face changed in two cases.
(i) Which type of image is formed on the both surface of spoon?
(ii) As tanu move concave surface of spoon towards her face, again she find that there comes a point (provided the spoon is big enough) where her image flips from inverted to upright. State the condition under which it happens? Is this image real or virtual?
(iii) The given ray diagram depict the correct explanation of the image formed by one surface of the spoon. Name the surface which can form the image as depicted in given ray diagram?
OR
(iv) Tanu was trying to form image using a concave mirror. She got an inverted and real image of same size of the object. Given figure shows four possible positions of the image formed. Figure out the correct position and justify it.
(i) Which type of image is formed on the both surface of spoon?
(ii) As tanu move concave surface of spoon towards her face, again she find that there comes a point (provided the spoon is big enough) where her image flips from inverted to upright. State the condition under which it happens? Is this image real or virtual?
(iii) The given ray diagram depict the correct explanation of the image formed by one surface of the spoon. Name the surface which can form the image as depicted in given ray diagram?
OR
(iv) Tanu was trying to form image using a concave mirror. She got an inverted and real image of same size of the object. Given figure shows four possible positions of the image formed. Figure out the correct position and justify it.
Ans:
(i) Erect image in convex surface and inverted image in concave surface.
(ii) When object is place between focus and pole of the mirror, it forms virtual and erect image. So as we move concave surface of spoon towards our face as soon as our face comes between principal focus and pole we observe erect and virtual image of our face.
(iii) Convex surface of the spoon will form the image as depicted in the given ray diagram.
(iv) Image A is the correct image as concave mirror forms real and inverted image of the same size of the object when object is placed at centre of curvature of the mirror.
(ii) When object is place between focus and pole of the mirror, it forms virtual and erect image. So as we move concave surface of spoon towards our face as soon as our face comes between principal focus and pole we observe erect and virtual image of our face.
(iii) Convex surface of the spoon will form the image as depicted in the given ray diagram.
(iv) Image A is the correct image as concave mirror forms real and inverted image of the same size of the object when object is placed at centre of curvature of the mirror.
38.
Acids, bases and salts are three main categories of chemical compounds. These have certain definite properties which distinguish one class from the other. The acids are sour in taste while bases are bitter in taste... (content on indicators)...
(i) Give two examples each of natural and artificial indicators.
(ii) An aqueous solution turns red litmus solution blue. Excess addition of which solution would reverse the change-ammonium hydroxide solution or hydrochloric acid?
(iii) What will be the change in colour when a few drops of phenolphthalein is added to a solution having pH 8.5.
OR
(iv) What is universal indicator?
(i) Give two examples each of natural and artificial indicators.
(ii) An aqueous solution turns red litmus solution blue. Excess addition of which solution would reverse the change-ammonium hydroxide solution or hydrochloric acid?
(iii) What will be the change in colour when a few drops of phenolphthalein is added to a solution having pH 8.5.
OR
(iv) What is universal indicator?
Ans:
(i) Natural Indicators: Turmeric and red cabbage, Artificial Indicators: Methyl red and methyl orange.
(ii) Hydrochloric acid because adding excess acid to the base would turn blue litmus solution red.
(iii) It changes into pink.
(iv) Universal indicator is a mixture of dyes that changes colour gradually over a range of pH and is used in testing for acids and alkalis.
(ii) Hydrochloric acid because adding excess acid to the base would turn blue litmus solution red.
(iii) It changes into pink.
(iv) Universal indicator is a mixture of dyes that changes colour gradually over a range of pH and is used in testing for acids and alkalis.
39.
Questions are based on the two table given below. Study these tables related to blood pressure level and answer the question that follow:
Table-A: Hypertension Guidelines (Normal: 120/80, High Stage 1: 130-139 / 80-90, etc.)
Table-B: BP of Patients X and Y at Morning, Afternoon, Evening.
(i) In the table B, at which time patent-Y have ideal normal blood pressure?
(ii) Identify the patient, which have hypertension stage-1 blood pressure?
(iii) Which Diet is the best for high blood pressure patient?
OR
(iv) What is the ideal blood pressure measurement of a human?
Table-A: Hypertension Guidelines (Normal: 120/80, High Stage 1: 130-139 / 80-90, etc.)
Table-B: BP of Patients X and Y at Morning, Afternoon, Evening.
(i) In the table B, at which time patent-Y have ideal normal blood pressure?
(ii) Identify the patient, which have hypertension stage-1 blood pressure?
(iii) Which Diet is the best for high blood pressure patient?
OR
(iv) What is the ideal blood pressure measurement of a human?
Ans:
(i) Afternoon (80-120)
(ii) Patient X (82-132) Evening
(iii) Grain and fruits
(iv) 80-120 mm Hg
(ii) Patient X (82-132) Evening
(iii) Grain and fruits
(iv) 80-120 mm Hg
[Diagram of Anatropous Ovule]
[Diagram of hromosomal aberrations]
[Diagram of Transcription Unit]
[Sketch of Brain Ventricles]
