CBSE Sample Papers for Class 10 Science Set 9 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 9 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 9 with Solutions

Time : 3 Hr.
Max. Marks : 80

General Instructions:

This question paper consists of 39 questions in 5 sections.
Alt questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
Section A consists of 20 objective type questions carrying 1 mark each.
Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should in the range of 30 to 50 words.
Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should in the range of 50 to 80 words.
Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.

SECTION – A (20 Marks)
(Select and write one most appropriate option out of the four options given for each of the questions 1-20)

Question 1.
If we eat sweets, it is suggested that we should brush our teeth. What quality of tooth paste would you suggest?
(a) Basic
(b) Acidic
(c) Neutral
(d) Sweet (1)
Answer:
(a) Basic

Explanation: If we eat sweets it is suggested that we brush our teeth as the bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth after eating. Using toothpastes which are generally basic for cleaning the teeth can neutralise the excess acid and prevent tooth decay.

Question 2.
Identify the balanced equation for the chemical reaction taking place when dilute hydrochloric acid is added to copper oxide.

Answer:

Explanation: When dilute hydrochloric acid is added to copper oxide, it gives green coloured copper chloride and water. The balanced equation for the chemical reaction taking place is given below:
CuO + 2HCl(dil) → CuCl₂ + H₂O

Question 3.
The diagram below represents:

(a) dispLacement reaction
(b) electrolysis reaction
(c) photodecomposition reaction
(d) thermal decomposition reaction (1)
Answer:
(d) thermal decomposition reaction

Explanation: When mercury (II) oxide is heated strongly, it decomposes to give the elements mercury and oxygen.

Question 4.
Which of the following is correct about structure marked as ‘X’ in the figure ?

(a) It is a bag like structure.
(b) Digests the food.
(c) It has digestive enzymes.
(d) All of these. (1)
Answer:
(d) All of these.

Explanation: Food vacuole is the structure ‘X’ which is a bag like structure inside the cell of certain protozoan such as Amoeba in which the ingested food is digested. When the food is completely encircled by Amoeba and the tips of encircling pseudopodia touch each other, the membrane at that point dissolves and the food is encaptured into the cell like a bag called food vacuole.

Related Theory
Food gets digested inside the food vacuole by digestive enzymes. The digested food diffuses into the cytoplasm and is utilized by the cell- Assimilation. The undigested food remains in the food vacuole and is thrown out of the body- Egestion.

Question 5.
Two gases ‘G’ having suffocating odour are obtained when a green solid ‘S’ is heated, along with a residue ‘R’. These gases are major air pollutants. When the vapours of the gases are collected and dissolved in water, the solution turns blue litmus to red. The colour of the residue becomes red.

What would be S, R and G? (1)

Answer:
(c) S: FeSO₂; R :Fe₂O₃; G: SO₂, SO₃

Explanation:

Caution
In order to solve such question. student may use exclusive method, by crossing out the incorrect options first For e.g., option (b) is wrong as H₂O is not a suffocating gas.

Question 6.
Observe the cross shown:

Which one of the following are new combinations?
(I) Tall with wrinked seeds
(II) Tall with round seeds.
(III) Short with wrinkled seeds.
(IV) Short with round seeds.
Select the correct option:
(a) (I) and (III)
(b) (I) and (IV)
(c) (II) and (III)
(d) (II) and (IV) (1)
Answer:
(b) (I) and (IV)

Explanation: Mendel’s dihybrid cross is shown here:

with wrinkled seeds (Q) = 3 Short plants with round seeds (U) = 3 Short plants with wrinkled seeds (S) = 1 From this cross we find out two new combinations are formed i.e. short plants with round seeds and tall plants with wrinkled seeds.

Question 7.
If you want to see an enlarged image of your face, which of the following should be used:

(a) (i)
(b) (ii)
(c) Both (i) and (ii)
(d) First (i) then (ii) (1)
Answer:
(a) (i)

Explanation: A convex mirror always forms a virtual but diminished image of an object placed anywhere in front of the mirror. A concave mirror forms a virtual and magnified image of an object when it is placed between its pole and focus. Therefore, we will use a concave mirror to see an enlarged image of our face.

Caution
Students usually get confused between the two types of lenses and the nature of image formed by them. The ray diagram for the image formed by concave mirror when object is placed between pole and focus is drawn below:

Ray diagram for image formed by a convex mirror is drawn as given:

Question 8.
If the person looks at different objects through the hot air over fire, the objects appear to him to be
(a) shaking highly
(b) moving slightly
(c) moving highly
(d) stable with no movement. (1)
Answer:
(b) moving slightly

Explanation: The air just above the fire becomes hotter and this hotter air is optically rarer but the colder air further up is optically denser. So when we see the objects by the light coming from them through hot and cold air layers having different optical densities, then refraction of light takes place randomly due to which the objects appear to be moving slightly.

Question 9.
In the following diagram, a ray of light is incident on a convex mirror.

Which of these following option shows the path of this ray, after reflection? (1)

Answer:

Explanation: When a ray of light appears to pass through the focus of a convex mirror, the reflected ray will be parallel to the principal axis.

Question 10.
Two wires, one of copper and other of manganin, have equal lengths and equal resistances. Which wire is thicker? Given that resistivity of copper is lower than that of manganin.
(a) Copper
(b) Manganin
(c) Both have equal thickness
(d) Can not say (1)
Answer:
(b) Manganin

Explanation: The resistance R of a wire is given by R = p\(\frac { l }{ A }\) where l is the resistivity of the
material of the wire, l is the length and A area of cross section of the wire. As resistivity of copper (p_c) is lower than that of manganin (p_m) and both wires have equal length and equal resistances,

where A_c and A_m are the area of cross-section of copper and manganin wires respectively.
Since, p_c < p_m A_c < A_m.
Therefore, wire made of manganin will be thicker.

Related Theory
Kilowatt hour is the commercial unit of electric energy and is defined as the energy consumed when 1 kW is used for 1 hour.
1 kWh = 1 kW x 1 hour
= 1000 watt x 3600 second = 3.6 x 10₆ joule

Question 11.
Two bulbs of 60 W and 40 W are connected in parallel. The current through the 60 W bulb is 0.6 A.

The current through the 40 W bulb will be:
(a) 0.4 A
(b) 0.6 A
(c) 0.8 A
(d) 1A (1)
Answer:
(a) 0.4 A

Explanation: When devices are connected in parallel, the potential difference across them is equal but the currents are in inverse ratios of their resistances. Let V be the potential difference, I current a cross 60 W bulb and I₂ across 40 W bulb.
P = VI
V = \(\frac { P }{ l }\)
Therefore, V = \(\frac { 6 }{ 0.6 }\) = 100 V.
The current I through 40 W bulb is:
I = \(\frac { P }{ V }\) = \(\frac { 40 }{ 100 }\) A = 0.4A

Question 12.
What does the divergence of magnetic field lines near the ends of a current carrying straight solenoid indicate?
(a) Relative strength of magnetic field in that region,
(b) Relative strength of magnetic field in another region
(c) Relative strength of force in another region
(d) Both (b) and (c) (1)
Answer:
(a) Relative strength of magnetic field in that region.

Explanation: The relative closeness of the magnetic field lines indicates the relative strength of the magnetic field in that region. As the magnetic field lines are diverging near the ends of a current carrying straight solenoid it indicates a decrease in the strength of the magnetic field near the ends of the solenoid.

Related Theory
Characteristics of magnetic lines of force:

The magnetic lines of force indicate the direction in which a N-pole would move if placed at that point.
The relative strength of the magnetic field is shown by the degree of closeness of the field lines.
No two lines of force intersect each other, for if they did, it would mean that there would be two directions of magnetic field at the point of intersection.
The direction of magnetic field at any point is found by drawing a tangent at that point.

Question 13.
Match column I with Column II

	Column I (Salt)		Column II (Use of Salt)
(A)	Plaster of paris	(i)	Removes permanent hardness of water
(B)	Washing soda	(ii)	Antacid
(C)	Bleaching powder	(iii)	Makes surfac-es smooth
(D)	Baking soda	(v)	Decolourisa-tion

(a) (A) – (i), (B) – (Hi), (C) – (ii), (D) – (iv)
(b) (A) – (iii), (B) – (i), (C) – (iv), (D) – (ii)
(c) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii)
(d) (A) – (i), (B) – (ii), (C) – (iii), (D) – (iv) (1)
Answer:
(b) (A) – (iii), (B) – (i), (C) – (iv), (D) – (ii)

	Column I (Salt)		Column II (Use of Salt)
(A)	Plaster of paris	(iii)	Decolourisation
(B)	Washing soda	(i)	Makes surfaces smooth
(C)	Bleaching powder	(iv)	Antacid
(D)	Baking soda	(ii)	Removes permanent hardness of water

Question 14.
The total resistance between the points A and B is:

(a) 10 Ohm
(b) 7.67 Ohm
(c) 5 Ohm
(d) 0.205 Ohm
Answer:
(b) 7.67 Ohm

Explanation: The resistances 1 Ohm and 2 Ohm are in parallel. Their equivalent resistance is
\(\frac{1}{R_p}\) = \(\frac { 1 }{ 1 }\) + \(\frac { 1 }{ 2 }\) = \(\frac { 3 }{ 2 }\)
R₂ = \(\frac { 2 }{ 3 }\) Ω = 0.67 Ω
This combination is connected in series with the 4 Ohm and 3 Ohm resistances. Therefore, equivalent resistance between A and B is (4 + 0.67 + 3) ohm or 7.67 ohm.

Question 15.
Two students performed the experiment on series and parallel combination of two given resistances R₁ and R₂ and plotted the graph (a) and (b) as shown:

Select the correct statement:
(a) Graph drawn by both students A and B are correct.
(b) Graph drawn by student A is correct, but graph drawn by student B is not correct.
(c) Graph drawn by student A is not correct, but graph drawn by student B is correct.
(d) Graph drawn by both students A and B is not correct. (1)
Answer:
(a) Graph drawn by both students A and B are correct

Explanation: When resistances are connected in series, their equivalent resistance is more than when they are connected in parallel.

As Resistance = \(\frac { V }{ I }\) = slope of V – I graph = Reciprocal of slope of I-V graph, the slope for resistances in series is greater than slope of resistance in parallel as correctly drawn by student A. Similarly, slope of resistance in series is less than the slope of resistance in parallel in the graph drawn by student B. But as R = \(\frac { V }{ I }\) or = \(\frac { 1 }{ R }\) = \(\frac { I }{ V }\) , graphs drawn by both students are correct as in both graphs, resistance in series > resistance in parallel.

Question 16.
Accumulation of harmful chemicals in our bodies can be avoided. How this can be achieved?
(a) Using pesticides in crops
(b) Using insecticides in crops
(c) Both (a) and (b)
(d) Avoiding use of pesticides in crops. (1)
Answer:
(d) Avoiding use of pesticides in crops.

Explanation: Accumulation of harmful chemicals in our bodies can be avoided by not using harmful and toxic pesticides in farms as these are non-biodegradable and keep accumulating at each trophic level. Biological magnification refers to the process where toxic substances move up the food chain and become more concentrated at each trophic level.

Related Theory
As human beings are at the top of the trophic level in any food chain, maximum concentration of these harmful chemicals get accumulated in our bodies.

Caution
Students usually get confused and do not know how biological magnification is related to food chain.
The pesticides and chemicals or industrial effluents are washed down into the soil or water bodies from where they are absorbed by the plants, aquatic plants or animals. This is how they enter our food chain and being non-biodegradable, their concentration increases at each trophic level.

Q. no 17 to 20 are Assertion – Reasoning based questions.
These consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true and R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

Question 17.
Assertion (A) : Bags of chips are flushed with oxygen gas.
Reason (R) : It prevents chips from getting rancid. (1)
Answer:
(d) A is false but R is true.

Explanation: Chips bags are flushed with inert gas such as nitrogen gas to prevent oxidation of fried food items or in other words, to prevent chips from getting rancid. But oxygen is not an inert gas and in fact will help in rapid oxidation of fried food items such as chips.

Related Theory
The condition produced by oxidation of fats and oils in foods marked by unpleasant smell and taste is called rancidity.
Prevention of Rancidity:

By adding anti-oxidants to foods containing fats and oils: Usually substances which prevent oxidation (antioxidants) are added to foods containing fats and oil. The two common anti-oxidants are BHA (Butylated Hydroxy-Anisole) and BHT (Butylated Hydroxy-Toluene).
Keeping food in air tight containers helps to slow down oxidation.
By keeping food in a refrigerator.
By storing foods away from light

Question 18.
Assertion (A) : Bile is essential for fat digestion.
Reason (R) : Fats cannot be digested without emulsification. (1)
Answer:
(a) Both A and R are true and R is correct explanation of A.

Explanation: Bile contains bile salts, which are essential for digestion. Fats are present in the intestines in the form of larger globules which makes it difficult for enzymes to act on them. Bile salts break them into smaller globules and this is called emulsification.

Question 19.
Assertion (A): Food chains generally consist of only three to four steps.
Reason (R) : Autotrophs capture the solar energy and convert it into chemical energy. (1)
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Explanation: While it is true that autotrophs capture the solar energy and convert it into chemical energy, an average of 10% of the food eaten by an organism is turned into its own body and made available for the next level of consumers. Since so little energy is available for the next level of consumers, food chains generally consist of only three to four steps.

Question 20.
Assertion (A) : Components of an electrical circuit can be connected in series or only.
Reason (R) : The two simplest of these are called series and parallel and occur frequently. (1)
Answer:
(c) A is true but R is false

Explanation: Components of an electrical circuit or electronic circuit can be connected in series, parallel, or series-parallel. The two simplest of these are called series and parallel and occur frequently. Components connected in series are connected along a single conductive path, so the same current flows through all of the components but voltage is dropped across each of the resistances.

In a series circuit, the sum of the voltages consumed by each individual resistance is equal to the source voltage. Components connected in parallel are connected along multiple paths so that the current can split up; the same voltage is applied to each component.

SECTION – B (12 Marks)
(Q. no. 21 to 26 are very short answer questions.)

Question 21.
Draw a diagram to show how three resistors Rlt R2 and R3 should be connected so that the total resistance of the circuit is minimum.
OR
Determine the following quantities for circuit shown below:
(A) The total current from the power supply.
(B) The voltage drop across each resistor.

Answer:
The equivalent resistance of three resistances R₁, R₂ and R₃ will be minimum when they are connected in parallel as shown:

OR
(A) To determine the total current drawn from the power supply, we need to calculate the total resistance in the circuit. As the resistances are connected in series, the equivalent resistance is given by the sum of the individual resistances.
R_s = R₁ + R₂ + R₃ = 20 + 30 + 50 = 100 Q.
Current drawn, I = \(\frac { V }{ R }\)
I = \(\frac { 125 }{ 100 }\) = 1.25A

(B) The voltage drop across each resistance is given by Ohm’s law V = IR.
If V₁, V₂ and V₃ are the voltage drop across R₁, R₂ and R₃, then V₁ = IR₁
= 1.25 x 20V = 25V;
V₂ = IR₂ = 1.25 x 30 = 37.5V;
V₃ = IR₃ = 1.25 x 50= 62.5V

Question 22.
What is a homologous series of carbon compounds? What is the difference in the molecular mass of any two adjacent homologues? (2)
Answer:
Homologous series is a series of carbon compounds in which the hydrogen in a carbon chain is replaced by the same functional group. For example, the compounds CH₄, C₂H₆, C₃H₈, C₄H₁₀ (all alkanes having general formula C_nH_2n+2) belong to the same homologous series.

Difference in molecular mass of any two adjacent homologues is 14 amu since any two adjacent homologues differ by CH₂ in their molecular formula.

Related Theory

All members of a homologous series can be represented by the same general formula. General formula for alkanes is C_nH_2n+2, where n is the number of carbon atoms.
They have similar chemical properties.
All the compounds belonging to the same homologous series have similar chemical properties.
The members of a homologous series show a gradual change in their physical properties with increase in molecular mass.

Question 23.
List in tabular form differences between binary fission and multiple fission.
OR
What are the advantages of growing plants by vegetative propagation? (2)
Answer:

Binary fission	Multiple fission
Two daughter cells are formed form the splitting of the parent’s cell or nucleus.	Many daughter are formed from the splitting of the parent’s cell or nucleus.
It occurs during favorable conditions.Parent cell divides only once.	It occurs during unfavorable conditions. Parent cell divides repeatedly.
Both the nucleus and cytoplasm divide simultaneously.	First the nucleus divides and then it is surrounded by cytoplasm.
Example: Amoeba,Bacteria, Euglena, etc.	Example : Plasmodium, Sporozoans. Algae, etc.

Advantages of growing plants by vegetative propagation are:
(1) Plants raised by this method can bear flowers and fruits earlier than those produced from seeds.

(2) It is cheaper, easier and more rapid method of propagation in plants as compared to growing plants from their seeds.

(3) The traits or characters of the parent plant are preserved. That is, all plants produced are genetically similar enough to the parent plant to have all its characteristics.

(4) Better quality of the plants can be maintained. Those plants which do not produce viable seeds or seeds with prolonged period of dormancy, such as banana, orange, rose, jasmine etc, can be propagated by this method.

Question 24.
What are the methods used by plants to get rid of excretory products? (2)
Answer:
Plants use the following methods to get rid of their excretory products:
(1) The plants produce carbon dioxide as a waste product during respiration and oxygen as a waste product during photosynthesis. The plants get rid of gaseous waste products through stomata in their leaves and lenticels in stems.

(2) Many waste products are stored in vacuoles of the cells. Plant cells have comparatively large vacuoles. Some waste products are stored in the leaves. They are removed as the Leaves fall off

(3) Some waste products such as resins and gums are stored, especially in non-functional old xylem. Some waste products such as tannins, resins, gums are stored in bark, thereby removed as barks are peeled off

(4) Plants excrete some waste products through roots into the soil around them. Plants also get rid of excess water through transpiration.

Question 25.
Shalini was having problem in eye vision so she went to a doctor. A doctor has prescribed the corrective lens of power +1.5D. Find the focal length of the lens. What would you conclude about the prescribed lens-diverging or converging? (2)
Answer:
It is given that power of the corrective lens P = + 1.5 D.
Power of a lens is the reciprocal ofits focal length (in metres).
P = \(\frac{1}{f(\text { in } m)}\)
+ 1.5 = \(\frac { 1 }{ f }\)
f = \(\frac { 1 }{ 1.5 }\) m
= \(\frac { 10 }{ 15 }\) m = 0.667 m = 66.67 cm
The focal length of the lens = 66.67 cm
As the focal length is positive, the prescribed lens will be converging lens (convex lens).

Question 26.
Write a chemical equation when dilute sulphuric acid reacts with zinc granules. State the type of the reaction. Name the gas evolved. How will you test for the gas? (2)
Answer:
The reaction between dilute sulphuric acid with zinc granules is given below:
H₂SO₄ + Zn → ZnSO₄ + H_2(g)
The type of reaction is displacement reaction as zinc metal displaces hydrogen from the acid (H₂SO₄) as hydrogen gas (H₂) and forms a salt zinc sulphate.
The gas evolved is hydrogen gas which can be tested by passing it through soap solution and then bringing burning candle near the soap bubbles filled with gas. Hydrogen gas will burn with pop sound.

SECTION – C (21 Marks)
(Q.no. 27 to 33 are short answer questions.)

Question 27.
Identify the type of each of the following reactions stating reason for your answers: (3)

Answer:
(A) This is a thermal decomposition reaction as crystals of iron sulphate upon heating decompose to form three products, namely, iron oxide, sulphur dioxide and sulphur trioxide.
Since a single reactant gives two or more than two products, it is a decomposition reaction.

(B) This is a double displacement reaction in which a black precipitate of copper sulphide is formed along with sulphuric acid solution when hydrogen sulphide gas is passed through copper sulphate solution. As exchange of ions takes place between the reactants (copper sulphate and hydrogen sulphide), it is a double displacement reaction.

(C) This is a displacement reaction as magnesium metal, being more reactive than copper metal, displaces copper from copper sulphate solution and forms magnesium sulphate and copper metal.

Quesiton 28.
(A) Under what condition does a farmer need to treat the soil in his field with quick lime or slaked lime or chalk?
(B) You are given two solutions A and B and their pH is 6 and 8 respectively. Answer the following:
(i) Which of the two solutions have more hydrogen ion concentration?
(ii) Which is acidic and which is basic?
OR
(A) Equal lengths of magnesium ribbons , are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH₃COOH) is added to test tube B. In which test tube will the fizzing occur more vigorously and why?
(B) On what does the strength of an acid or base depend? (3)
Answer:
(A) If the soil that is on the field is too acidic then the farmer should treat the soil with quicklime or slaked lime or chalk as they are all alkaline substances and adding them to the soil will neutralize the acidic nature of soil. For healthy growth of plants, the soil should be neither alkaline nor highly acidic. (B) pH of solution A = 6; pH of solution B = 8
(i) Solution A will have higher hydrogen ion concentration. The pH of any solution is inversely proportional to the hydrogen ion concentration. This means that the solution that has lower pH number will have the higher hydrogen ion concentration.
(ii) Solution A is acidic and solution B is basic because H⁺ ion concentration is higher in acidic solutions. Moreover, all substances having pH less than 7 are acidic and those having pH greater than 7 are basic.

(A) In both cases, hydrogen gas is evolved. When metal reacts with acid it forms salt and hydrogen gas.
Metal + Acid → Salt + Hydrogen gas
Fizzing will occur more vigorously in test tube A containing hydrochloric acid. This is because hydrochloric acid is a stronger acid than acetic acid and reaction between magnesium ribbon and hydrochloric acid is faster in test tube A than the reaction between Magnesium and acetic acid in test tube B.
The reaction between HCl and Mg is:
Mg + 2HCl → MgCl₂ + H₂

(B) The strength of an acid or base depends upon the number of H⁺ ions and OH^– ions produced, respectively. Acids producing more H+ ions are said to be strong acids and acids that produce less H⁺ ions are said to be weak acids.

Question 29.
A squirrel was roaming around in a garden. Suddenly, a snake arrived, and it ran away

(A) Describe how a squirrel uses its hormonal system to react to a dangerous situation.
(B) How does the hormone action get regulated by the feedback mechanism? (3)
Answer:
(A) The hormone adrenaline is released into a squirrel’s blood when it detects danger, increasing heart rate and blood flow to tissues. As a result, its cells and tissues receive energy more quickly, allowing it to flee dangerous situations.

(B) Through a regulatory mechanism known as a feedback mechanism, the presence or absence of a specific hormone can control its subsequent formation.

Question 30.
(A) Identify P, Q, Rand S in the given diagram.

(B) Explain the process of fertilization in a flower. What happens to the (i) ovary and (ii) ovule after fertilization? (3)
Answer:
(A) P represents the pollen grain.
Q represents the male germ cell.
R represents the embryo sac.
S represents the female germ cell.

(B) Fertilisation occurs when the male gamete present in a pollen grain fuses with the female gamete or egg present in ovule. When the pollen grain falls on the stigma of the carpel, it bursts open and grows a pollen tube downward through the style towards the female gamete in the ovary. Here, zygote is formed which later grows into embryo. Zygote divides repeatedly to form an embryo within the ovule. Ovule develops a thick coat and gradually forms seed. Ovary grows rapidly and ripens to form a fruit. The petals, sepals, stamens, style and stigma may shrivel and fall off

Question 31.
In human beings, the statistical probability of getting either a male or a female child is 50:50. Explain with the help of diagram. (3)
Answer:
Human beings possess 23 pairs ofchromosomes. Out of these, the 23rd pair is known as sex chromosomes (XX in females and XY in males).

The statistical probability of getting either a male or female child is 50 : 50 as the sex of the child in determined by what chromosome he/ she inherits from father. A child who inherits a X chromosome from the father would be a girl (XX) while a child who inherits a Y chromosome from the father would be a boy (XY).

Question 32.
Draw a neat diagram to show the refraction of light through a glass prism.
On the diagram clearly show:
(A) Angle of incidence
(B) Angle of refraction
(C) Angle of emergence
(D) Angle of deviation (3)
Answer:
Diagram showing the refraction of light through
a glass prism is drawn below:
(A) Angle of incidence – ∠i
(B) Angle of refrction – ∠r
(C) Angle of emergence – ∠e
(D) Angle of deviation – ∠D

Question 33.
Will the levels of this magnification be different at different levels of the ecosystem? Explain with the help of a food chain comprising of three trophic levels. (3)
Answer:
Several pesticides and harmful chemicals enter our body through food chain and remain there without being decomposed. The accumulation of these harmful chemicals increases towards the higher side of the food chain that is the level or concentration of the harmful chemicals increases

with increasing trophic level. So, it is different at different levels of the ecosystem. There are different ways in which poisonous chemicals enter different food chain at producer level.

For example, conc. of DDT in:

If water bodies were found to contain 0.02 ppm (parts per million) of DDT (use of DDT is now banned). The phytoplankton and zooplankton which consume this water were found to contain 5 ppm concentration of DDT in their body. Fish feeding on such plankton had 240 ppm DDT in their body tissue. The level of DDT concentration reached to 1600 ppm in birds feeding on these fish. Humans which perhaps occupy the highest point in the food chain will face highest accumulation.

SECTION – D (15 Marks)
(Q.no. 34 to 36 are Long answer questions.)

Question 34.
Analyze the following reactivity series given below and answer the questions that follow:

(A) Which one of the two metals is more reactive: copper or silver?

(B) What will happen if a strip of zinc is immersed in a solution of copper sulphate?

(D) Which one of the metals out of Mg, Ag, Zn and Cu would be displaced from the solution of its salts by other three metals?

(E) What will you observe when some silver pieces are put into green coloured ferrous sulphate solution?
OR

(A) When a metal X is treated with cold water, it gives a basic salt Y with molecular formula XOH (molecular mass = 40) and liberates a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.

(B) What are ionic compounds? Why do ionic compounds not conduct electricity in the solid state? (5)
Answer:
(A) Copper is more reactive than silver as it is placed above silver in the reactivity series of metals.

(B) When a strip of zinc metal is put in copper sulphate solution, then the blue colour of copper sulphate solution fades gradually and red brown coating of copper is deposited on zinc strip.
Zn + CuSO₄ → ZnSO₄ + Cu
Zinc is more reactive than copper so it will displace copper from its salt solution so displacement reaction takes place.

(C) If copper is kept open in air, it slowly loses its shining brown surface and gains a green coating as it forms basic copper carbonate, CuCO₃. Copper reacts with moist carbon dioxide in the air and forms green colour basic copper carbonate.

(D) Out of Mg, Ag, Zn and Cu, Ag will be displaced from the solution of its salts by other three metals. Least reactive metal can be displaced from its solution by other three metals. Silver is least reactive out of given four metals.

(E) No reaction takes place when some silver pieces are put into green coloured ferrous sulphate solution as silver is less reactive than iron and hence cannot displace iron from ferrous sulphate solution.

(A) Given, molecular formula of Y = XOH and molecular mass of Y = 40
So, atomic mass of metal X would be,
M + 16 + 1 = 40
M = 40 – 17
= 23
Metal with atomic mass 23 is sodium (Na) and reaction of it with cold water will form the base sodium hydroxide (NaOH) and liberates H₂ gas which easily catches fire.
2Na + 2H₂O(Cold) → 2NaOH + 2H₂
X = Sodium (Na)
Y = Sodium hydroxide (NaOH)
Z = Hydrogen gas (H₂)

(B) Ionic compounds are the compounds formed by the combination of oppositely charged ions formed by transfer of electrons from one atom to another. Ionic compounds do not conduct electricity in solid state as they have a rigid structure as ions are not free to move due to strong electrostatic forces of attraction between them.

Question 35.
(A) Draw the diagram of alimentary canal of a human and label the following parts:
(i) Mouth
(ii) Oesophagus
(iii) Stomach
(iv) Intestine

(B) How is small intestine designed to absorb digested food?

(A) Drawthe diagram of human respiratory system and label the following parts:
(i) Larynx
(ii) Trachea
(iii) Lungs
(iv) Bronchi
(B) How are lungs designed in human beings to maximize the area of exchange of gases? (5)
Answer:

(B) The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption. The villi are richly supplied with blood vessels which take the absorbed food to each and every cell of the body, where it is utilized for obtaining energy, building up new tissues and repair of old tissues.

(A) Human respiratory system:

(B) Human beings have network of respiratory tubes in their lungs. Trachea divided into bronchi. Each bronchus divides in lungs to form bronchioles and bronchioles finally terminate into balloon-like structure which are called alveoli. The walls of alveoli contain an extensive network of blood-vessels and they provide a vast surface area for exchange of gases.

Question 36.
Shikha took a small aluminium rod AB and suspended it horizontally by means of two connecting wires from a stand. Then she placed a strong horse shoe magnet in such a way that the rod is between the two poles with the field directed upwards.
(A) What happens:
(i) when current is passed in the rod from B to A?
(ii) when direction of current is reversed?
(iii) poles of the magnet are interchanged?
(B) When will the displacement of the rod largest? (5)
Answer:
(A) (i) As the field is directed upwards, the aluminium rod is placed in such a way between the horse shoe magnet that the north pole of the magnet is vertically below and south pole vertically above the rod.
When the current is passed in the rod from B to A, we will observe that the rod is displaced towards the left. The displacement of the rod is due to the force acting on the current carrying conductor (aluminium rod) when it is placed in a magnetic field.

(ii) The direction of force acting on the rod will be reversed on reversing the direction of current. The rod will now be displaced towards the right as direction of current (given by central or middle finger) is reversed.

(iii) On reversing the direction of poles of the magnet, the direction of magnetic field is reversed due to which direction of force exerted on the rod will be reversed. If current is flowing from end B to A of the rod, the rod will be displaced towards right.

(B) Displacement of the rod will be largest when the direction of current is at right angles to the direction of the magnetic field.

SECTION – E (12 Marks)
(Q.no. 37 to 39 are case – based/data -based questions with 2 to 3 short sub – parts. Internal choice is provided in one of these sub-parts.)

Question 37.
Blood types are determined by the presence or absence of certain antigens – substances that can trigger an immune response if they are foreign to the body. Since some antigens can trigger a patient’s immune system to attack the transfused blood, safe blood transfusions depend on careful blood typing and cross matching.

Everyone has an ABO blood type (A. B, AB, or O) and an Rh factor (positive or negative). Just like eye or hair colour, our blood type is inherited from our parents.

The table shows what blood group a child is likely to have depending upon the blood group inherited from his parents:
Father’s Blood Type

Combinations of Mother’s Blood and Father’s Blood type The genotype and phenotype of various blood groups is shown below:

(A) A child has blood group A. The mother also has blood group A and the maternal grandmother is also A and a brother is O. What are the possible blood types of father?
(B) What do you mean by law of dominance? Explain.
OR
(B) What does “genotype” refer to?
Answer:
(A) The mother in question is blood type A. Her genotype is either I^AI^A or I^AI⁰. The maternal grandmother is also blood type A and a brother is blood type O tells us that the maternal grandmother of the child has genotype I^AI⁰, since she is type A but donated an O allele to one of her children.

The son is blood type A. The father can have any of the four possible blood types, type A, type AB, type B, or type O.

Related Theory

If both the parents have blood group B, the child can have blood group B or O.
If at least one parent has blood group B or AB, the blood group of the child will be A, B or AB.
If one parent has O and the other has B, the child can have blood group B or O.
If both parents have AB, the child will have A, B or AB.

(B) Mendel’s Law of Dominance states that recessive alleles will always be masked by dominant alleles. Therefore, a cross between a homozygous dominant and a homozygous recessive will always express the dominant phenotype, while still having a heterozygous genotype. Each character is controlled by a pair of dissimilar factors. Only one of the characters expresses. The one which expresses in the F1 generation is called Dominant.

(B) Genotype is the description of genes present in an organism. It is always a pair of letters such as TT, Tt or tt (where T and t are the different forms of the same gene). Phenotype is the characteristic or trait which is visible in an organism is called its phenotype. Being tall or dwarf are phenotypes of a plant as these traits are visible.

Question 38.
The table given below shows six organic compounds A, B, C. D, E and F having different molecular formula

Organic	Molecular
Compound	Formula
A	C₇H₁₆
B	C₈h₁₆
C	C₄h₆
D	C₆H₁₀
E	C₅H₁₀
F	C₉H₂₀

(A) Which compounds belong to same homologous series?
(B) Which is the member of the same homologous series as E?
(C) A and F are saturated hydrocarbons while all others are unsaturated hydrocarbons. Justify.

(C) What type of compound is B and F?
Answer:
(A) A and F belong to same homologous series of alkenes. C and D belong to same homologous series of alkynes.

(B) B is an alkene having general formula C_nH_2n, the homologous series to which E belongs.

(C) A and F have general formula C_nH_2n+2 i.e., of alkanes which are saturated hydrocarbons. C and D belong to a homologous series having general formula CnH_2n-2. B and E are alkenes.

Caution
Students usually get confused with the general formulas of alkene, alkene and alkynes. General formula for:

alkane – C_nH_2n+2
alkene – C_nH_2n.
alkyne – C_nH_2n-2

Question 39.
There are two basic types of optical microscopes: simple microscopes and compound microscopes. A simple microscope uses the optical power of single lens or group of lenses for magnification. A compound microscope uses a system of lenses (one set enlarges the image produced by another) to achieve much enlarges higher magnification of an object. The vast majority of modem research microscopes are compound microscopes while some cheaper commercial digital microscopes are simple single lens microscopes. Compound microscopes can be further divided into a variety of other types of microscopes which differ in their optical configurations, cost, and intended purposes.

(A) Based on the diagram shown, what is the nature of image formed by the objective lens and the eyepiece lens respectively?

(B) If the focal lengths of the eyepiece and objective lenses are in the ratio 10 : 1, find the ratio of power of eyepiece and objective lens.

(C) A student found that that the magnification of the objective lens of a compound microscope ranges from 5 to 100. Suppose the magnification of the objective is 10, where will be the image formed by the objective lens if an object is placed at 2 cm in front of it?
Answer:
(A) The objective forms a real image of the object whereas the eyepiece forms a virtual image of the object.

(B) Power of a lens is the reciprocal of its focal length. As the ratio of the focal lengths of the eyepiece and objective lenses is 10 : 1, the ratio of power of eyepiece and objective lens would be 1 : 10

(C) Magnification produced by a lens is the ratio of size of image(h’) to the size of object(h). It can also be written as
m = \(\frac { h’ }{ h }\) = \(\frac { V }{ U }\)
where u and v are the object distance and image distance respectively.

Related Theory
Magnification is always positive in case of a concave lens as it always produces a virtual and erect image. Magnification is positive in case of a convex lens when the image formed is virtual and negative when image formed is real.

(C) As the magnification of the objective is 10 and objective forms a real image of the object, m = -10.
the Since object distance, u = -2 cm, using
formula m = \(\frac { V }{ U }\)
v = mu = (-10) (-2) = 20.
The image formed is real and at a distance of 20 cm.