Category: CBSE Sample Papers

Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Science Set 4 for Practice

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper. All questions are compulsory.
(ii) Section-A – question no. 1 to 20 – all questions and parts there of are of one mark each.
These questions contain multiple choice questions (MCQs), very short answer questions and assertion – reason type questions. Answers to these should be given in one word or one sentence.
(iii) Section-B – question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should in the range of 30 to 50 words.
(iv) Section-C – question no. 27 to 33 are short answer type questions, carrying 3 marks each.
Answers to these questions should in the range of 50 to 80 words. :
(v) Section—D – question no. – 34 to 36 are long answer type questions carrying 5 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vi) There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
(vii) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section-A

1. What is the radius of curvature of a plane mirror?

2. Why is red colour selected for danger signal lights?

3. What is the cause of dispersion of light by prism.
OR
Why does the sky appear dark instead of blue to an astronaut?

4. Name the components of blood which transport:
(i) Food, carbon dioxide and nitrogeneous wastes
(ii) Oxygen

5. Write the relationship between heat energy produced in a conductor when potential difference V is applied across its terminals and a current, I flows through it for the time ‘F.

6. A beam of light is incident through the holes on side A and emerges out of the hole on the other face of the box as shown in the figure. What type of lens could be inside the box?

7. Why do stars twinkle?

8. A lens forms an image at Y of an object at X. Where does the lens form an image of an object at P?

9. What indication do we get by appearance of dwarf plant in F2 generation?
OR
What is sex of the baby that inherits Y-chromosome from the father?

10. If a women is using copper-T, will it help in protecting against sexually transmitted diseases?
OR
Why is temperature of scrotal sac 2°C less than the body temperature?

11. Name the intermediate and the end products of glucose breakdown on aerobic respiration.

In the experiment “Chlorophyll is necessary for photosynthesis’, why does the green part of variegated leaf turns blue black after putting in iodine solution?

12. Out of Li and K, which one have stronger metallic character and why?
OR
The formula of magnesium oxide is MgO. State the formula of barium nitrate and barium sulphate, if barium belongs to the same group.

13. X + YSO₄ → XSO₄ + Y
Y + XSO₄ → No reaction
Out of two elements, ‘X’ and ‘Y’ which is more reactive and why?

Assertion (A) and Reason (R)
For question numbers 14,15 and 16, two statements are given- one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (i), (iii) and (iv) as given below:

(i) Both A and R are true, and R is correct explanation of the assertion.
(ii) Both A and R are true, but R is not the correct explanation of the assertion.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

14. A. Copper sulphate solution turns blue litmus red.
R. Copper sulphate is salt of strong acid H₂SO₄, weak base Cu(OH)₂, therefore, acidic in nature.
Answer:
(i)

15. A. The angle of incidence and reflection are equal.
R. The angle between incidence ray and normal is called angle of incidence.
Answer:
(ii)

16.
R. It is a combination reaction because CO combines with H₂ to form CH₃OH i.e. two substances combine to form a single compound.
OR
1. The reaction, MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂ is an example of redox reaction.
2. In this reaction, HCl is reduced to Cl₂ whereas MnO₂ is oxidized to MnCl.
Answer:
(ii) OR (iii)

Answer Q. No 17 – 20 contain five sub-parts each. You are expected to answer any four sub­parts in these questions.

17. Read the following and answer any four questions from 17 (i) to 17 (v) (4 × 1 = 4)

The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on.

Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.

(i) Which of the following is not a common sign of sexual maturation in boys and girls.
(a) Pubic hair
(b) Menstrual cycle in females
(c) Broad shoulders in females
(d) Wider chest in males
Answer:
(c) Broad shoulders in females

(ii) Which contraceptive method changes the hormonal balance of the body?
(a) Condoms
(b) Orals pills
(c) IUCD
(d) None of these
Answer:
(b) Orals pills

(iii) Sex ratio of females/1000 males in different states are as follows:

States	2013-15	2012-14
Delhi	869	876
Haryana	831	866
Uttar Pradesh	879	869
Punjab	889	870

Name the state which improves the ratio to maximum extent in 2013-15.
(a) Delhi
(b) Haryana
(c) Uttar Pradesh
(d) Punjab
Answer:
(d) Punjab

(iv) According to the tabel in above question, which state declines this ratio to maximum extent in 2013-15.
(a) Delhi
(b) Haryana
(c) Uttar Pradesh
(d) Punjab
Answer:
(b) Haryana

(v) What is the major cause of less females than males in India?
(a) Males live longer than females
(b) Naturally, male child birth rate is higher than female child
(c) Female foeticide
(d) None of these
Answer:
(c) Female foeticide

18. Read the following and answer any four questions from 18 (i) to 18 (v) (4 × 1 = 4)
The data for four elements is given below:

Element	Carbon	Nitrogen	Fluorine	Neon
Symbol	C	N	F	Ne
Structure	Macromolecular	Simple molecules N7	Simple molecules F,	Single atoms Ne
Boiling point/°C	4200	-196	-188	-246

(i) Name the element that is solid at room temperature.
(a) Carbon
(b) Nitrogen
(c) Fluorine
(d) Neon
Answer:
(a) Carbon

(ii) Why is the boiling point of carbon the highest?
(a) Carbon atoms linked through strong ionic bonds to form molecules.
(b) Carbon atoms are close together with strong covalent bonds.
(c) Carbon atoms are small in size thus have more stronger attractions among them.
(d) Carbon atoms are not linked to each other.
Answer:
(b) Carbon atoms are close together with strong covalent bonds.

(iii) Which of the following element is inert?
(a) Carbon
(b) Nitrogen
(c) Fluorine
(d) Neon
Answer:
(d) Neon

(iv) Carbon has many allotropes with macromolecular structure. Which of the following is not such kind of allotrope of carbon?
(a) Diamond
(b) Fullerene
(c) Graphite
(d) Coke
Answer:
(d) Coke

(v) Which of the following is the most reactive element?
(a) Carbon
(b) Nitrogen
(c) Fluorine
(d) Neon
Answer:
(c) Fluorine

19. Read the following and answer any four questions from 19 (i) to 19 (v) (4 × 1 = 4)

(i) What is the range of resistivity in metals which are conductor of electricity?
(a) 10^-8 to 10^-7 Ωm
(b) 10^-7 to 10^-6 Ωm
(c) 10 ⁸ to 10^-6 Qm
(d) 10 ⁸ to 10 ⁴ Qm
Answer:
(c) 10 ⁸ to 10^-6 Qm

(ii) Why are alloys used in heating devices like electric iron, toasters, immersion rods, etc.?
(1) Alloys have high resistivity.
(2) Alloys do not oxidise or bum at high temperature.
(3) Alloys have high conductivity thus allow high current through it
(a) III only
(b) I only
(c) II and III
(d) I and II
Answer:
(d) I and II

(iii) Which of the following is used in transmission wires?
(a) Cu
(b) Al
(c) Zn
(d) Fe
Answer:
(d) Fe

(iv) Which is the best conducting metal?
(a) Cu
(b) Ag
(c)  Au
(d) Hg
Answer:
(d) Hg

(v) Based on resistivity, which of the following is better to use as insulation to protect us from electric current.
(a) Glass
(b) Hard rubber
(c) Nichrome
(d) Diamond
Answer:
(c) Nichrome

20. Read the following and answer any four questions from 20 (i) to 20 (v)  (4 × 1 = 4)

When a single parent is involved in the formation of a new individual, without the fusion of gametes, it is called asexual reproduction. It involves repeated mitotic divisions.

Sexual reproduction is a process of combining DNA from two different individuals during reproduction. This is brought about by the fusion of the male and female gametes.

(i) The type of reproduction taking place is

(a) Budding
(b) Fragmentation
(c) Regeneration
(d) Fission
Answer:
(c) Regeneration

(ii) Choose the right option

A		B	C	D	E
(a)	Fallopian tube	Oviduct	Uterus	Cervix	Vagina
(b)	Oviduct	Vas deferens	Ovary	Vagina	Cervix
(c)	Ovary	Oviduct	Uterus	Cervix	Vagina
(d)	Ovary	Fallopian tube	Uterus	Vagina	Cervix

Answer:
(c)

(iii) Where does fertilisation occur in human females?
(a) Uterus
(b) Cervix
(c) Oviduct
(d) None of these
Answer:
(d) None of these

(iv) The simple animals like Planaria can be cut into a number of pieces and each piece grows into a complex organism. What is the process known as?
(a) Budding
(b) Fragmentation
(c) Spore formation
(d) Regeneration
Answer:
(d) Regeneration

(v) Spirogyra reproduces by
(a) Fission
(b) Regeneration
(c) Fragmentation
(d) Budding
Answer:
(c) Fragmentation

Section-B

21. Identify the type of magnetic field represented by the magnetic field lines given below and name the type of conductors which can produce them.

22. 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 (amperes)	0.5	1.0	2.0	3.0	4.0
V (volts)	1.6	3.4	6.7	10.2	13.2

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
3.314 Ω

23. You have three test tubes. One of them contains distilled water and the other two contain an acidic and a basic solution, respectively. If you are given only red litmus paper; how will you identify the contents of each test tube?
OR
Equal lengths of magnesium ribbon 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?

24. A solution of potassium chloride when mixed with silver nitrate solution, an insoluble white substance is formed. Write the chemical reaction involved and also mention the type of reaction.

25. A ray of light incident on rectangular glass slab emerges parallel to itself. Draw a labelled diagram to justify this statement.
OR
Define absolute refractive index. Mention its unit. Can the value of absolute refractive index be smaller than 1 ? Justify your answer.

26. How does the metallic character of elements changes along a period of the periodic table from left to right and why?

Section – C

27. Show the path of the reflected ray, after reflection in each case.

(i) Draw a diagram of human alimentary canal and label on it:
Oesophagus, Gall bladder, Liver and Pancreas

(ii) Explain the statement, “Bile does not contain any enzyme, but it is essential for digestion.”
OR
How respiration does takes place in plants?

29. Based on the group valency of elements, state the formula of the following, giving justification for each.
(i) Oxides of Group 1 elements.
(ii) Halides of the elements of Group 13.
(iii) Compounds formed when an element of group 2 combines with an element of Group 16.

30. A magnetic compass needle is placed in the plane of paper near point A as shown in the figure. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the “A deflection of the compass? Under what conditions is the deflection maximum and why?

31. (i) What is the nature of food coming from stomach?
(ii) Mention any two structural modifications in small intestine which helps in absorption of food.

32. An electron enters a magnetic field, at right angle to the field direction.
(i) State the rule to find the direction of force acting on the electron.

(ii) What will be the direction of force acting on the electron, in the above case?

33. (i) Two conductors A & B of resistance 5 ohms and 10 ohms respectively can be arranged in parallel and later on in series. In each arrangement, the total voltage applied is 120V. In which arrangement will the voltage across A and B be the same and in which case will the current flowing through A and B be the same?
(ii) Calculate the total resistance of each arrangement.
Answer:
(ii) (a) 3.33Ω
(b) 15Ω

Section-D

34. The structural formula of five compounds are given below.

(i) Which two compounds belong to the same homologous series?
(ii) Which compound belongs to the same homologous series as ethanol?
(iii) Which compound on hydrogenation produces E?
(iv) Which compound when dissolved in water turns blue litmus red?
(v) Name the homologous series for compound B.

35. (i) Identity A, B, C and D in the given diagram and write their names.

(ii) What is pollination? Explain its significance.
(iii) Explain the process of fertilisation in flowers. Name the parts of the flower that develop after fertilisation into
(a) seed
(b) fruit
OR
(i) Draw the diagram of female reproductive system and match and mark the part (s):
(a) Where block is created surgically to prevent fertilisation.
(b) Where CuT is inserted.
(c) Inside which condom can be placed.
(ii) Why do more and more people prefer to use condoms? What is the principle behind use of condoms?

36. (i) Name the organisms which belong to first and third trophic level in food chain comprising of the following:
Insects, birds, frog and grass.
(ii) What will happen if we kill all the organisms in one trophic level?
OR
(i) Give reason to justify the following:
(a) The existence of decomposers is essential in a biosphere.
(b) Flow of energy in a food chain is unidirectional.
(ii) Why is improper disposal of waste a curse to environment?

Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. This question paper contains two parts, A and B.
2. Both Part A and Part B have internal choices.

Part-A:
1. It consists of two sections, I and II.
2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.
3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.

Part-B:
1. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

Section 1 has 16 questions of 1 mark each. Internal choice is provided in 5 questions.

Question 1.
Find the zeroes of the of the polynomials p(x) = 4x² – 12x + 9
Solution :

Question 2.
In the given figure, DE || BC. Find \(\frac{x}{y}\) DE || BC

Solution :

Question 3.
If 2k + 1, 6, 3& + 1 are in AP, then find the value of k.
OR
If the sum of n terms of an AP is 2n² + 5n, then find the 2^nd term
Solution :
For an AP a, b, c; 2b = a + c ⇒ (2k + 1) + (3k + 1) = 2×6
5k+2=12
⇒ 5k = 10
⇒ k = 2

Question 4.
If x = a, y = b is the solution of the pair of equations x-y-2 and x + y = 4, then find the values of a and b.
Solution :

Question 5.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle (see figure). Find the length of the chord CD parallel to XY and at a distance 8 cm from A.

Solution :
OD² = OM² + DM²
⇒ 5² = 3² + x²
x² = 25 – 9 = 16
x = 4
CD =2 x = 2 × 4 = 8 cm

Question 6.
Find the radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm.
Solution :
Edge of the cube = 4.2 cm
∴ Radius of the largest right circular cone \(\frac{1}{2}\) (Edge of the Square) \(=\frac{4.2}{2}=2.1 \mathrm{~cm}\)

Question 7.
The circumference of a circle exceeds its diameter by 180 cm. Find its radius
Solution :

Question 8.
A bag contains 40 balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is \(\frac{11}{20} \) and that of blue ball is \(\frac{1}{5} \) , then find the number of black balls.
OR
Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Solution :

Let the number of black balls = x.

Total number of balls = Total possible outcomes = 40

⇒ Number of black balls =10
OR
Number of possible outcomes = 4 as possible outcomes are HH, HT, TH, TT. Favourable outcomes for getting at least one tail are HT, TH, TT No. of favourable outcomes = 3
∴ P(getting at least one tail) =\(\frac{3}{4}\)

Question 9.
If Cos θ \(\frac{7}{8}\) then find the value of \(\frac{(1+\cos \theta)(1-\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}\)
Solution :

Question 10.
Write whether the rational number \(\frac{7}{75}\) will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
OR
Write the HCF of the smallest composite number and the smallest prime number.
Solution :
\(\frac{7}{75}=0.0933333 \ldots \ldots=0.09 \overline{3}\)
So, it is a non-terminating repeating decimal expansion.
OR
The smallest composite number is 4 and the smallest prime number is 2.
The prime factorisation of 4 = 2 x 2 = 2² and the prime factorisation of 2 = 2¹
Now, the HCF of 2 and 4 is the product of smallest power of each common prime factor in . the numbers.
HCF (2, 4) = 2¹ = 2

Question 11.
Find the value(s) of k, if the quadratic equation 3x² – k √3x + 4 = 0 has equal roots.
OR
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.
Solution :

Question 12.
In the given figure, ABCD is a rectangle. Find the values of x and y.

Solution :

Question 13.
If ax² + bx + c = 0 has equal roots, find the value of
Solution :
For equal roots D = 0
i.e., b² – 4ac =0
⇒ b² = 4ac
\(c=\frac{b^{2}}{4 a}\)

Question 14.
In the given figure, PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of APCD.

OR
In the given figure, the quadrilateral PQRS circumscribes a circle with centre O. If ∠POQ = 115°, then find ∠ROS.

Solution :
Since PA and PB are tangents from same external point P.
PA = PB = 15 cm
Now, Perimeter of APCD = PC + CD + DP = PC + CQ + QD + DP
= PC + CA + DB + DP = PA + PB = 15 cm +15 cm = 30 cm
OR
Solution :
Since  ∠POQ = ∠ROS   (Vertically opposite angles)
⇒ ∠ROS = 115°

Question 15.
To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn such that ∠BAX is an acute angle and then at equal distances points are marked on the ray. Find the minimum number of these points.
Solution :
Since 5+ 7 = 12
So, the number of points marked on ray AX = 12.

Question 16.
Find the value of (sin 30° + cos 60°).
Solution :
sin 30° + cos 60° \(=\frac{1}{2}+\frac{1}{2}=\)

Section-II

Question 17.
Case Study Based-1
Medicinal Garden
A medicinal garden is a garden in which different kinds of medicinal plants, like Aloe Vera, Mint, Lemon Balm, etc. are planted with the goal of serving the need of general health maintenance. Observe the following diagram.

Refer to IJKL
(a) The mid-point of the segment joining the points 1(6, 6) and J(6, 18) is
(i) (7,9)
(ii) \(\left(12, \frac{11}{2}\right) \)
(iii) (6, 12)
(iv) (12,24)

Refer to EFGH
(b) The distance between points H(10, 6) and F(14, 18) is
(i) \(8 \sqrt{5}\)unit
(ii) \(4 \sqrt{10}\) unit
(iii) 18 unit
(iv) 24 unit

Refer to ABCD

(c) The coordinates of the points A and B are (22,6) and (22,18) respectively. The x-coordinate of a point R on the line segment AB such that \(\frac{A R}{A B}=\frac{3}{5}\) is…………….
(i) 18
(ii) 24
(iii) 22
(iv) 31

Refer to MQ

(d) The ratio in which the points (20, k) divides the line segment joining the points M(4, 2) and Q(24, 2) is
(i) 4 : 1
(ii) 16 : 15
(iii) 8 : 21
(iv) 10 : 17

Refer to MH and HP

How much longer is HP than MH given that coordinates of H(10, 6), M(4, 2) and P(19,2)
(i) \((\sqrt{95}-2 \sqrt{3}) \text { unit }\)
(ii) \((\sqrt{97}-2 \sqrt{13}) \text { unit }\)
(iii) \((\sqrt{61}-4 \sqrt{5}) \text { unit }\)
(iv) None of there.
Solution:

Question 18.
Case Study Based-2
A Frame House
A frame-house is a house constructed from a wooden skeleton, typically covered with timber board. The concept of similar triangles is used to construct it. Look at the following picture:

Refer to House (i)
(a) The front view of house (7) is shown below in which point P on AB is joined with point Q

If PQ || BC, AP = x m  PB = 10 m. AQ = (x – 2) m, QC = 6 m, then the value of a is
(i) 3m
(ii) 4m
(iii) 5m
(iv) 8 m

(b) The side vies of house (i) is shown below in which point F on AC is joined with point G on DE.

If ACED is a trapezium with AD || CE, F and G are points on non-parallel sides AC and AF
DE respectively such that FG is parallel to AD, then =\(\frac{\mathrm{AF}}{\mathrm{FC}}=\)
(i) \(\frac{\mathrm{DG}}{\mathrm{GE}}\)
(ii) \(\frac{\mathrm{AD}}{\mathrm{CE}}\)
(iii) \(\frac{\mathrm{AF}}{\mathrm{GE}}\)
(iv) \(\frac{\mathrm{DG}}{\mathrm{FC}}\)

(c) The front view of house (ii) is shown below in which point S on PQ is joined with point T on PR.

\(\frac{\mathrm{PS}}{\mathrm{QS}}=\frac{\mathrm{PT}}{\mathrm{TR}} \text { and } \angle \mathrm{PST}=70^{\circ}, \angle \mathrm{QPR}=50^{\circ}\) then the angle ∠QRP =
(i) 70°
(ii) 50°
(iii) 80°
(iv) 60°

(d) Again consider the front view of house (ii). If S and T are points on side PQ and PR respectively such that
ST || QR and PS : SQ = 3 : 1. Also TP = 6.6 m, then PR is

(i) 6.9 m
(ii) 8.8 m
(iii) 10.5m
(iv) 9.4 m

(e) Sneha has also a frame house whose front view is shown below

If MN || AB, BC = 7.5 m, AM = 4 m and MC = 2 m, then length of BN is
(i) 5 m
(ii) 4 m
(iii) 8 m
(iv) 9 m
Solution :




Question 19.
Case Study Based-3

Rainbow is an arch of colours that is visible in the sky, caused by the refraction and dispersion of the sun’s light after rain or other water droplets in the atmosphere. The colours of the rainbow are generally said to be red, orange, yellow, green, blue, indigo and violet.

Each colour of rainbow makes a parabola. We know that for any quadratic polynomial ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas.

(a) A rainbow is represented by the quadratic polynomial x² + (a + 1 )x + b whose zeroes are 2 and-3. Then
(i) a = -7, b   = -1
(ii)  a = 5, b =-1
(iii)  a – 2, b = – 6
(iv)   a – 0, b = – 6

(b) The polynomial x² – 2x – (7p + 3) represents a rainbow.   If -4 is zero of it, then the value of p is
(i) 1
(ii)  2
(iii) 3
(iv)  4

(c) The graph of a rainbow y=f(x) is shown below.

The number of zeroes of f(x) is
(i) 0
(ii) 1
(iii) 2
(iv) 3

(d) If graph of a rainbow does not intersect the x-axis but intersects y-axis in one point, then number of zeroes of the polynomial is equal to
(i) 0
(ii) 1
(iii) 0 OR 1
(iv) none of these

(e) The representation of a rainbow is a quadratic polynomial. The sum and the product of its zeroes are 3 and -2 respectively. The polynomial is 1
(i) k(x² – 2x – 3), for some real k.
(ii) k(x² – 5x – 9), for some real k.
(iii) k(x² – 3x – 2), for some real k.
(iv) k(x² – 8x + 2), for some real k.
Solution :
(a) x² + (a + 1)x + b =(2)² + (a + 1)2 + b = 0 and (-3)² + (a + l)(-3) + b = 0
⇒ 4 + 2a + 2 + 6=0 and 9 — 3a — 3 + 6 = 0
⇒ 2a + b=-6      … (i)
and  -3a + b=-6   …(ii)
Solving (i) and (ii), we get a = 0 and b = -6
So, option (iv) is correct answer.

(b) p(-4) = 0 ⇒ (-4)² – 2(-4) – (7p + 3) = 0
⇒ 16 + 8 – 7p-3 = 0    ⇒ 21 – 7p = 0
⇒ p = 3
So, option (iii) is correct answer.

(c) ∵ Graph f(x) intersects x-axis at two different points.
∴  Number of zeroes of f(x) = 2.
So, option (iii) is correct answer.

(d) We know that the number of zeroes of a polynomial is equal to number of points of intersection of the graph of polynomial with x-axis.
Since the graph of rainbow does not intersect the x-axis, so it has no zeroes.
So, option (i) is correct answer.

(e) Let the required polynomial be f(x).
Then f(x) = k(x² – 3x – 2) for some real k.
So, option (iii) is correct answer.

Question 20.
Case Study Based-4

Cost of Living Index	140-150	150-160	160-170	170-180	180-190	190-200	Total
Number of weeks	5	10	20	9	6	2	52

(a) The mid-value (class-mark) of 160-170 is ……..
(i) 140
(ii) 145
(iii) 155
(iv) 165

(b) The approximate mean weekly cost-of-living index is 1
(i) 166.4
(ii) 184.5
(iii) 190
(iv) 201.8

(c) The sum of lower and upper limits of modal class is
(i) 290
(ii) 310
(iii) 330
(iv) 350

(d) Mode is the value of the variable which has
(i) maximum frequency
(ii) minimum frequency
(iii) mean frequency
(iv) middle most frequency

(e) The median class of above data is
(i) 150-160
(ii) 160-170
(iii) 170-180
(iv) 190-200

Solution :
(a) Class-mark of 160-170 = \(160-170=\frac{160+170}{2}=\frac{330}{2}=165\)
So, option (iv) is correct answer.

Cost of Living Index	No. of Weeks (fi)	Mid-point (xi)	fixi
140-150	5	145	725
150-160	10	155	1550
160-170	20	165	3300
170-180	9	175	1575
180-190	6	185	1110
190-200	2	195	390
Total	n = 52		8650

\(\text { Mean }=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{8650}{52}=166.4(\text { approx. })\)
So, option (i) is correct answer.

(c) Maximum frequency is 20
∴ Modal class = 160-170
Lower limit of modal class = 160
Upper limit of modal class =170
Sum of lower and upper limits = 160 + 170 = 330
So, option (iii) is correct answer.

(d) (i) Maximum frequency

Cost of Living Index	f	cf
140-150	5	5
150-160	10	15
160-170	20	35
170-180	9	44
180-190	6	50
190-200	2	52

n= 52 \(\Rightarrow \frac{n}{2}=26\)
Median class is 160-170.
So, option (ii) is correct answer.

Part-B
Section-III

Question 21.
If two positive integers p and q are written as p = a²b² and q = a³b; a, b are prime numbers, then verify: LCM (p, q) x HCF (p, q) =pq.
Answer:
LCM (p, q) =a³b³ and HCF (p, q) = a²b
LCM (p, q) x HCF (p, q) =a⁵b⁴ = (a²b³) (a³ b) = pq

Question 22.
Draw a line segment of length 6 cm. Using compass and ruler, find a point P on it which divides it in the ratio 3:4.

Solution :
Steps of construction:
1. Draw a line segment AB = 6 cm.
2. Draw an acute angle <BAX.
3. Along AX take 7 points, such that
AA₁ — A₁ A₂ — A₂ A₃ — A₃A₄ — A₄A₅ – A₅A₆ — A₆A₇
4. Join BA₇
5. Through A₃ draw A₃P A₇B which meets AB at P.
6. AP: PB = 3 : 4 and P is the required point.

Question 23.
If 7 sin² A + 3 cos² A = 4, show that tan \(A=\frac{1}{\sqrt{3}}\)
OR
Prove that \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\)
Solution :

Question 24.
In the given figure, XY and XT’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XT’ at B. Prove that ∠AOB = 90°.

Solution :

Question 25.
The coordinates of the points P and Q are respectively (4,—3) and (—1, 7). Find the x-coordinate (abscissa) of a point R on the line segment PQ such that \(\frac{P R}{P Q}=\frac{3}{5}\)
OR
Find the ratio in which the point (—3, k) divides the line segment joining the points (—5, —4) and (—2, 3). Also find the value of k.
Solution :

Question 26.
Find k, if the sum of the zeroes of the polynomial x² – (k + 6) x + 2 (2k – 1) is half of their product.
Answer:

Section-IV

Question 27.
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being of the same subject.
Solution :

Question 28.
In the given figure, ABPC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution :

Question 29.
Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares
of the other two sides.
OR
P is the mid-point of side BC of AABC, Q is the mid-point of AP, BQ when produced meets AC at L. Prove that AL = \(\frac{1}{3} \) AC.
Solution :

Question 30.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency k.

Daily pocket allowance (in ₹)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	3	6	9	13	k	5	4

Solution :

Question 31.
Using quadratic formula, solve the following equation for X:
abx²+(b²—ac)x—bc=O
OR
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Solution :

Let the two natural numbers be x and y such that x >y
According to the question,
Difference of numbers,
x – y =  5    ⇒  x = 5 +y ……………(1)
Difference of their reciprocals,
\(\frac{1}{y}-\frac{1}{x}=\frac{1}{10}\)

= (y – 5)(y + 10) = 0  ∴ y = 5  or y = – 10
y is a natural number. ∴ y = 5
Putting the value of)’ in (i), we get
x= 5+5=10
Thus, the required numbers are 10 and 5.

Question 32.
Find the angle of depression from the top of 12m high tower of an object lying at a point 12 m away from the base of the tower.
Solution :
Let AB be the tower of 12 m height and B its base. Let C be a point A 12 m away from base B of tower AB where an object situated.

Question 33.
If the median of the distribution given below is 28.5, find the values of x and y.

Class interval	0-10	10-20	20-30	30-40	40-50	50-60	Total
Frequency	5	X	20	15	y	5	60

Solution :
Here, median = 28.5, n = 60

Class interval	Frequency (f)	Cumulative frequency (cf)
0-10	5	5
10-20	X	5 + x
20-30	20	25 + x
30-10	15	40+ x
40-50	y	40 + x + y
50-60	5	45+x + y
Total	∑ fi = 60

Section – V

Question 34.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
OR
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Solution :

Thus, height of the other tower = 10 m and the distance between two towers = BD = 10√3 m.

Question 35.
A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block.
Solution :

Question 36.
A person can row 8 km upstream and 24 km downstream in 4 hours. He can row 12 km downstream and 12 km upstream in 4 hours. Find the speed of the person in still water and also the speed of the current. 5
Solution :
Let the person’s speed of rowing in still water be, x km/h and speed of the current y km/h.
∴ Speed of boat in downstream = (x + y) km/h
and speed of boat in upstream = (x -y) km/h
Since \(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\)

Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Science Set 3 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper. All questions are compulsory.
(ii) Section-A – question no. 1 to 20 – all questions and parts there of are of one mark each.
These questions contain multiple choice questions (MCQs), very short answer questions and assertion – reason type questions. Answers to these should be given in one word or one sentence.
(iii) Section-B – question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should in the range of 30 to 50 words.
(iv) Section-C – question no. 27 to 33 are short answer type questions, carrying 3 marks each.
Answers to these questions should in the range of 50 to 80 words. :
(v) Section—D – question no. – 34 to 36 are long answer type questions carrying 5 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vi) There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
(vii) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

Question 1.
Name a cyclic unsaturated carbon compound.
Answer:
Cyclopentene / Cyclohexene-formula or structure (or any other).

Question 2.
The change in magnetic field lines in a coil is the cause of induced electric current in it. Name the underlying phenomenon.
Answer:
Electromagnetic Induction

Question 3.
Why do we store silver chloride in dark coloured bottles?
Answer:
It is because it undergoes photochemical decomposition reaction in the presence of sunlight.

Question 4.
How many male gametes are produced by pollen grains?
Answer:
Two

Question 5.
Two beakers with chemicals are shown below. Name the beaker which will show exothermic reaction and the one which will be endothermic in nature.

Answer:
Exothermic : Beaker A
Endothermic : Beaker B

Question 6.
What will happen, if a solution of sodium hydrogen carbonate is heated? Give the equation of the reaction involved.
OR
How chloride of lime differs chemically from calcium chloride?
Answer:
Sodium carbonate, carbon dioxide and water will be formed.
\(2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O} \text { ( } \mathrm{l}\)

Question 11.
What will happen to a plant if its xylem is removed?
OR
Why is it advisable to breathe through nose?
Answer:
If xylem is removed from a plant, transport of water and nutrients will stop and plant will die.
OR
There are fine hair and mucus gland in the inner lining of nose which can filter the incoming air from germs and dust. Moreover, the air attains the optimum temperature before reaching the lungs.

Question 12.
An element ‘X’ on reacting with oxygen forms an oxide X₂O. This oxide dissolves in water and turns blue litmus red. State whether ‘X’ is metal or non-metal.
Answer:
Non-metal

Question 13.
What is the speed of blue light travelling in vacuum?
Answer:
The speed of blue light is same as that of light i.e. 3 x 10⁸ m s^–1.
Assertion (A) and Reason (R)
For question numbers 14,15 and 16, two statements are given- one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (i), (iii) and (iv) as given below:

(i) Both A and R are true, and R is correct explanation of the assertion.
(ii) Both A and R are true, but R is not the correct explanation of the assertion.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Question 14.
A. The element carbon forms the basic structural framework of more compounds than any other element.
R. The carbon-carbon bond is ionic.
Answer:
(iii)

Question 15.
A. Pollination is different from fertilisation.
R. The process of transfer of pollens from anther to stigma is called pollination which takes place in plants.
Answer:
(iv)

Question 16.
A. A 200 W bulb glows with more brightness than 100 W bulb.
R. 100 W bulb has more resistance than 200 W bulb.
OR
A. When two resistances of 4Ω are connected in series, total resistance is 8Ω.
B. When two resistances of 4Ω are connected in parallel total resistance is 2Ω.
Answer:
(i) OR (ii)

Answer Q. No 17 – 20 contain five sub-parts each. You are expected to answer any four sub­parts in these questions.

Question 17.
Read the following and answer any four questions from 17 (i) to 17 (v) (4 × 1 = 4)

The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual

chemicals are not biodegradable, they get accumulated progressively at each trophic level. The maximum concentration of these chemicals gets accumulated in our bodies and greatly affects the health of our mind and body.

(i) Why is the maximum concentration of pesticides found in human beings?
(a) Humans are at the top level in any food chain
(b) Major part of crops cultivated using pesticides are consumed by humans
(c) Pesticides are stored in different organs of humans
(d) None of these
Answer:
(a) Humans are at the top level in any food chain

(ii) Which method could be applied to reduce our intake of pesticides through food to some extent.
(a) Washing crops thoroughly
(b) Organic fanning
(c) Use of bio-pesticides
(d) All of these
Answer:
(d) All of these

(iii) Various steps in a food chain represent:
(a) Food web
(b) Trophic level
(c) Ecosystem
(d) Biomagnification
Answer:
(b) Trophic level

(iv) With regard to various food chains operating in an ecosystem, man is a:
(a) Consumer
(b) Producer
(c) Producer and consumer
(d) Producer and decomposer
Answer:
(a) Consumer

(v) Food web is constituted by
(a) relationship between the organisms and the environment.
(b) relationship between plants and animals.
(c) various interlinked food chains in an ecosystem.
(d) relationship between animals and environment
Answer:
(c) various interlinked food chains in an ecosystem.

Question 19.
Read the following and answer any four questions from 19 (i) to 19 (v) (4 × 1 = 4)

Electrical resistivity and its inverse, electrical conductivity are fundamental properties of a material that quantifies how strongly it resists or conducts electric current. A low resistivity indicates a material that readily allows electric current to flow.

Resistance is defined as the ratio of potential difference across a conductor to the current passing through it. Resistance is the property of a conductor to resist the flow of charges through it.

Both resistance and resistivity describe how difficult it is to G make electrical current flow through a material, but unlike w resistance, resistivity is an intrinsic property. This means that g all pure copper wires irrespective of their shape and size, have the same resistivity, but a long, thin copper wire has a much larger resistance than a thick, short copper wire. Every material has its own characteristic resistivity. For example, rubber has a far larger resistivity than copper. The resistivity of metals also varies with temperature.

The resistance of almost all alloys increases with increase in temperature but the rate of change of resistance is less than that of metals. In fact, the resistance of certain alloys such as Manganin, Eureka and Constantan show practically no change in resistance to a considerable range of temperature.                        ‘

(i) What happens to the resistivity of a wire if it is stretched?
(a) It will increase
(b) It will decrease
(c) First increases then decreases
(d) Remains the same
Answer:
(d) Remains the same

(ii) Though silver is a good conductor, why it is not an ideal choice for transmission of electricity?
1. It is expensive
2. It oxidizes and tarnishes when it comes in contact with air
3. Its resistivity decreases with increase in temperature
(a) Only 1
(b) Only 2
(c) Both 1 and 2
(d) Only 3
Answer:
(c) Both 1 and 2

(iii) The area of cross section of a wire becomes half when its length is stretched to double. How the resistance of the wire is affected in the new condition?
(a) Resistance of the wire remains unchanged
(b) Resistance of the wire decreases to half
(c) Resistance of the wire increases to double
(d) Resistance of the wire increases four times
Answer:
(d) Volume of the wire remains the same before and after stretching. Assuming wire to be cylindrical in shows them  volume = Area of base x height/length
Due to stretching if length is increased, area will decrease.
R = ρl/A
New length = 2l
New area = A/2
Resistance increases four times.

(iv) If there are two wires, W₁ and W₂ of same material and same length but have radius r and 2r respectively, then which wire will have more resistance?
(a) Wire W₁ has more resistance as compare to W₂
(b) Wire W₂ has more resistance as compare to W₁
(c) Both the wires have same resistance
(d) Wire W₂ has three times resistance than wire W₁
Answer:
(a) Wire W₁ has more resistance as compare to W₂

(v) Ajay made the following conclusions regarding resistance of metals/alloy. Which one would you disagree with?
(a) The resistivity of a metal decreases with temperature
(b) A thick copper wire has less resistance than a thin copper wire of the same length
(c) Alloys are often used in the manufacture of standard resistors
(d) Low resistivity means the material easily allow the flow of electric current
Answer:
(a) The resistivity of a metal decreases with temperature

Question 20.
Read the following and answer any four questions from 20 (i) to 20 (v) (4 × 1 = 4)
The following graph shows the volume of hydrogen collected in four different experiments when dil. hydrochloric acid was reacted with zinc, copper, magnesium and iron metal.

(i) Name the graph that produces hydrogen gas faster.
(a) Graph 1
(b) Graph 2
(c) Graph 3
(d) Graph 4
Answer:
(d) Graph 4

(ii) Name the metal used in experiment that gave the result for graph-4.
(a) Mg
(b) Zn
(c) Fe
(d) Cu
Answer:
(a) Mg

(iii) Ways to speed up the slow reaction are
1. heating the reactants
2. cooling the reactants
3. using more diluted acid
4. using more concentrated acid
(a) 1 only
(b) 1 and 4
(c) 1 and 3
(d) 2 and 4
Answer:
(b) 1 and 4

(iv) The amount of zinc metal used for the above reaction is same but the concentration of acid is changed, what would be the effect on the rate of reaction?
(a) no change
(b) the speed will increase
(c) the speed will decrease
(d) the information is incomplete
Answer:
(d) the information is incomplete

(v) Which of the following metal liberates hydrogen gas on reaction with dilute HNO₃.
(a) Zn
(b) Cu
(c) Mg
(d) Fe
Answer:
(c) Mg

Section – B

Question 21.
The period 3 oxides are given below.
[Na₂O, MgO] [Al₂O₃]……………………. [SO₂, Cl₂O]
A                               B                                           C
State 2 characters of these oxides grouped as A & C
Answer:
Oxide A: Na₂O, MgO are basic in nature and ionic Oxide C : SO₂, Cl₂O are acidic in nature and covalent

Question 22.
Name two human organs that perform dual function and explain their functions to justify their dual nature.
Answer:
Two human organs with dual nature are pancreas and gonads. Pancreas secretes insulin hormone and digestive enzymes like trypsin ( pancreatic enzyme) similarly human gonads like testis and ovaries secretes the gametes like sperms and ovum respectively along with the sex hormones.

Question 23.
Why are small number of surviving tigers a cause of worry from genetics point of view?
Answer:
(i) Tigers may get extinct in near future due to natural calamities, lack of available food and due to their small number.
(ii) Cross breeding may be less possible. Variations will not occur, which is essential for the survival.
(iii) Adaptation due to changing environment, like cutting of forests, global warming becomes difficult.
(iv) Protection from enemies is easy if they are more in number.

Question 24.
An object is placed at a distance of 15 cm from a concave lens of focal length 30 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.
Answer:

The image will be formed at 10 cm from ‘O’ on the same side as the object. It will be virtual, erect and diminished.

Question 25.
Write chemical equation for the reaction when
(i) Steam acts on red hot iron.
(ii) Zinc is added to iron(II) sulphate solution.
OR
State two reasons for the following facts:
(i) Sulphur is a non-metal.
(ii) Magnesium is a metal.
One of the reason must be supported with a chemical equation.
Answer:
(i) \(3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) \underset{\text { steam }}{\longrightarrow} \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g)\)
(ii) Zn(s) + FeSO₄(aq) → ZnSO₄(aq) + Fe(s)
OR
(i) Sulphur is a poor conductor of electricity.
It reacts with oxygen to form SO₂ which forms acidic solution in water.
S + O₂ → S0_2
SO2 + H₂O →H₂SO₃

(ii) Magnesium is a metal because it is a good conductor of electricity.
It forms basic oxide with oxygen
\(2 \mathrm{Mg}+\mathrm{O}_{2} \stackrel{\text { Burning }}{\longrightarrow} 2 \mathrm{MgO}\)
\(\mathrm{MgO}+\underset{(\mathrm{Hot})}{\mathrm{H}_{2} \mathrm{O}} \rightarrow \underset{(\text { Base })}{\mathrm{Mg}(\mathrm{OH})_{2}}\)

Question 26.
Study the following circuit and answer the questions that follows:

(i) How much current is flowing through
(a) 10 Q and
(b) 15 Q resistor?
(ii) What is the ammeter reading?
OR
Find the current drawn from the battery by the network of four resistors shown in the figure.

Answer:
(i) Current through

Question 27.
1 g of copper powder was taken in a China dish and heated. What change takes place on heating? When hydrogen gas is passed over this heated substance, a visible change is seen in it. Give the chemical equations of reactions, the name and the color of the products formed in each case.
Answer:
A black colour is formed on the surface

Original/brown colour is restored

Question 28.
List the important products of the Chlor-alkali process. Write one important use of each.
OR
How is washing soda prepared from sodium carbonate? Give its chemical equation. State the type of this salt. Name the type of hardness of water which can be removed by it?
Answer:
Products: Hydrogen, Chlorine, Sodium hydroxide Uses:
Hydrogen: In the production of margarine/ammonia/as a fuel
Chlorine: Water treatment / swimming pools / production of PVC / Disinfectants / CFCs / Pesticides
Sodium hydroxide: For degreasing metal surface / in making soaps and detergents / paper making / artificial fibres.
OR
1. By recrystallisation of sodium carbonate
2. Na₂CO₃ + 10H₂O → Na₂CO₃.10H₂O
3. Basic Salt
4. Permanent hardness

Question 29.
Study the following cross showing self pollination in F₁, fill in the blank and answer the question that follow:

What are the combinations of character in the F₂ progeny? What are their ratios?
Answer:
F₁ progeny is Rr Yy-Round, Yellow
Combinations of character in the F₂ progeny are:

• Round yellow — 9 (Both dominant traits)
• Round green — 3 (One recessive, one dominant)
• Wrinkled yellow — 3 (One dominant, one recessive)
• Wrinkled green — 1 (both recessive traits)

The ratio is 9 : 3 : 3 : 1

Question 30.
Define the term pollination. Differentiate between self pollination and cross pollination. What is the significance of pollination?
Answer:
Pollination is the transfer of pollen from another to stigma.

Self Pollination	Cross Pollination
Transfer of pollen in the same flower.	Transfer of pollen from one flower to another.

• Pollination leads to fertilization resulting in the formation of zygote.

Question 31.
Why is Tyndall effect shown by colloidal particles? State four instances of observing the Tyndall effect.
Answer:
Because of the scattering of light.
Instances:

• When a fine beam of light enters a smoke-filled dark room through a small hole.
• When sunlight passes through a canopy of dense forest in foggy/ misty conditions.
• Blue colour of sky.
• Red colour of the sun during sunrise or sunset.

Question 32.
A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.

Answer:
V α I or Potential difference is directly proportional to current

Note: If circuit diagram is correct but labelling of ammeter and voltmeter are incorrect, deduct 1 mark.

Question 33.
(i) Write the mathematical expression for Joule’s law of heating.
(ii) Compute the heat generated while transferring 96000 coulombs of charge in two hours through a potential difference of 40 V.
Answer:
(i) H = I²Ri
(ii) H = V.I.t = V.Q
Given:     V = 40 volts, Q = 96000 C
H = 40 V x 96000 C
= 3.84 x 10⁶ J

Section – D

Question 34.
The position of certain elements in the Modem Periodic Table are shown below.

Using the above table answer the following questions giving reasons in each case:
(i) Which element will form only covalent compounds?
(ii) Which element is a non-metal with valency 2?
(iii) Which element is a metal with valency 2?
(iv) Out of H, C and F which has largest atomic size?
(v) To which family does H, C and F belong?
OR
Define atomic size. Give its unit of measurement. In the modem periodic table what trend is observed in the atomic radius in a group and a period and why is it so?
Answer:
(i) E, it has 4 valence electrons.
(ii) B, it needs only 2 electrons to attain stable configuration.
(iii) D, it loses two electrons to attain stable configuration.
(iv) F, it has the largest size since size increases down the group.
(v) Noble gases, outermost shell is complete.
OR

• Atomic size is the distance between the centre of the nucleus and the outermost shell of an isolated atom.
• Picometer /pm
• Trends in Atomic radius
  In a group: increases down the group; due to addition of a new shell.
  In a period: atomic radius decreases from left to right; due to increase in pulling power of nucleus / due to addition of electrons in the same shell.

Question 35.
(i) What is the law of dominance of traits? Explain with an example.
(ii) Why are the traits acquired during the life time of an individual not inherited? Explain.
Answer:
(i) Law of dominance of traits: In a cross between a pair of contrasting characters, only one parental character will be expressed in F₁ generation which is called dominant trait and the other is called a recessive trait.
For example – in pea plants.

All plants in F₁ generation were tall proving that the gene for tallness is dominant over the gene for dwarfness/short, which is not able to express itself in the presence of dominant trait.

(ii) Traits acquired by an organism during its lifetime are known as aquired traits. These traits are not inherited because they occur in somatic cells only/do not cause any change in the DNA of the germ cells.

Question 36.
Draw a ray diagram in each of the following cases to show the formation of image, when the object is placed:
(i) between optical centre and principal focus of a convex lens.
(ii) anywhere in front of a concave lens.
(iii) at 2F of convex lens.
State the signs and values of magnifications in the above mentioned cases (i) and (ii).
OR
An object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.
(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?
(ii) Find the size of the image.
(iii) Draw a ray diagram to show the formation of image in this case.
Answer:

Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Science Set 2 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper. All questions are compulsory.
(ii) Section-A – question no. 1 to 20 – all questions and parts there of are of one mark each.
These questions contain multiple choice questions (MCQs), very short answer questions and assertion – reason type questions. Answers to these should be given in one word or one sentence.
(iii) Section-B – question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should in the range of 30 to 50 words.
(iv) Section-C – question no. 27 to 33 are short answer type questions, carrying 3 marks each.
Answers to these questions should in the range of 50 to 80 words. :
(v) Section—D – question no. – 34 to 36 are long answer type questions carrying 5 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vi) There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
(vii) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section-A

Question 1.
List any two observations when ferrous sulphate is heated in a dry test tube?
OR
Identify the products formed when 1 mL of dil. hydrochloric acid is added to 1g of sodium metal?
Answer:

• Initial light green colour changes to reddish-brown colour
• Colourless gas is evolved
• Gas with choking smell is evolved (Any two)
  OR
• Sodium chloride and hydrogen gas

Question 2.
Write the chemical name and chemical formula of the salt used to remove the permanent hardness of water.
Answer:
Sodium carbonate decahydrate, Na₂CO₃.10H₂O

Question 3.
Which of the following is not observed in a homologous series? Give reason for your choice,
(a) Change in chemical properties
(b) Difference in -CH₂ and 14u molecular mass
(c) Gradation in physical properties
(d) Same functional group
Answer:
(a) It does not occur due to the presence of the same functional group.

Question 4.
Why does the sun appear white at noon?
Answer:
The light is least scattered at noon.

Question 5.
Both a spherical mirror and a thin spherical lens have a focal length of (-)15 cm. What type of mirror and lens are these?
Answer:
Both are concave.
Alternative answer that should be given credit: Plano-concave lens.

Question 6.
The image formed by a concave mirror is observed to be real, inverted and larger than the object. Where is the object placed?
OR
Name the part of a lens through which a ray of light passes without suffering any deviation.
Answer:
Between the principal focus and the centre of curvature.
OR
Optical centre.

Question 7.
In the arrangement shown in figure there are two coils wound on a nonconducting cylindrical rod. Initially the key is not inserted in the circuit. Later the key is inserted and then removed shortly after.

What are the two observations that can be noted from the galvanometer reading?
Answer:
There are momentary galvanometer deflections that die out shortly; the deflections are in opposite directions.

Question 8.
Draw the magnetic field lines around a straight current carrying conductor.
Answer:
The field consists of concentric circles centred on the wire

Question 9.
Two unequal resistances are connected in parallel. If you are not provided with any other parameters (eg. numerical values of I and R), what can be said about the voltage drop across the two resistors?
OR
Some work is done to move a charge Q from infinity to a point A in space. The potential of the point A is given as V. What is the work done to move this charge from infinity in terms of Q and V?
Answer:
Voltage-drop is same across both OR W = QV

Question 10.
Veins are thin walled and have valves. Justify.
Answer:
Veins have thin walls because the blood there is no longer under pressure and they have valves to ensure blood flow in one direction.

Question 11.
How is the wall of small intestine adapted for performing the function of absorption of food?
OR
Out of a goat and a tiger, which one will have a longer small intestine? Justify your answer.
Answer:
The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption.
OR
Goat because herbivores eating grass need a longer small intestine to allow the cellulose to be digested.

Question 12.
Explain how ozone being a deadly poison can still perform an essential function for our environment.
OR
Give reason why a food chain cannot have more than four trophic levels.
Answer:
Ozone layer protects us from harmful effects of UV radiation.
OR
The loss of energy at each step is so great that very little usable energy remains after four trophic levels.

Question 13.
State the role of pancreas in digestion of food.
Assertion (A) and Reason (R)
For question numbers 14,15 and 16, two statements are given- one labeled Assertion (A) and the other labeled Reason (R).

Select the correct answer to these questions from the codes (i), (ii), (iii) and (iv) as given below:
(i) Both A and R are true, and R is correct explanation of the assertion.
(ii) Both A and R are true, but R is not the correct explanation of the assertion.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:
The pancreas secretes digestive juice which contains enzymes like trypsin for digesting proteins and lipase for breakdown of emulsified fats.

Question 14.
A. After white washing the walls, a shiny white finish on walls is obtained after two to three days. R. Calcium oxide reacts with carbon dioxide to form calcium hydrogen carbonate which gives a shiny white finish.
Answer:
(iii)

Question 15.
A. Food chain is responsible for the entry of harmful chemicals in our bodies. R. The length and complexity of food chains vary greatly.
OR
A. Greater number of individuals are present in lower trophic levels.
R. The flow of energy is unidirectional.
Answer:
(ii) OR (ii)

Question 16.
A. A geneticist crossed a pea plant having violet flowers with a pea plant with white flowers, he got all violet flowers in first generation. R. White colour gene is not passed on to next generation.
Answer:
(i)

Answer Q. No 17 – 20 contain five sub-parts each. You are expected to answer any four sub¬parts in these questions.

Question 17.
Read the following and answer any four questions from 17 (i) to 17 (v) (4 x 1 = 4)
All living cells require energy for various activities. This energy is available by the breakdown of simple carbohydrates either using oxygen or without using oxygen. 4 x 1 = 4
(i) Energy in the case of higher plants and animals is obtained by
(a) Breathing
(b) Tissue respiration
(c) Organ respiration
(d) Digestion of food
Answer:
(b) Tissue respiration

(ii) The graph below represents the blood lactic acid concentration of an athlete during a race of 400 m and shows a peak at point D. Respiration in athletics: The blood of an athlete was tested before, during and after a 400 m race:

Lactic acid production has occurred in the athlete while running in the 400 m race. Which of the following processes explains this event?
(a) Aerobic respiration
(b) Anaerobic respiration
(c) Fermentation
(d) Breathing
Answer:
(b) Anaerobic respiration

(iii) Study the graph below that represents the amount of energy supplied with respect to the time while an athlete is running at full speed.

Choose the correct combination of plots and justification provided in the following table.

	Plot A	Plot B	Justification
(a)	Aerobic	Anaerobic	Amount of energy is low and inconsistent in aerobic and high in anaerobic
(b)	Aerobic	Anaerobic	Amount of energy is high and consistent in aerobic and low in anaerobic
(c)	Anaerobic	Aerobic	Amount of energy is high and consistent in aerobic and low in anaerobic
(d)	Anaerobic	Aerobic	Amount of energy is high and inconsistent in anaerobic and low in aerobic

Answer:
(b)

(iv) The characteristic processes observed in anaerobic respiration are
(1) presence of oxygen
(2) release of carbon dioxide
(3) release of energy
(4) release of lactic acid
(a) (1), (2) only
(b) (1), (2), (3) only
(c) (2), (3), (4) only
(d) (4) only
Answer:
(c) (2), (3), (4) only

(v) Study the table below and select the row that has the incorrect information.

		Aerobic	Anaerobic
(a)	Location	Cytoplasm	Mitochondria
(b)	End Product	CO2 and H2O	Ethanol and CO7
(c)	Amount of ATP	High	Low
(d)	Oxygen	Needed	Not needed

Answer:
(d)

Question 18.
Read the following and answer any four questions from 18 (i) to 18 (v). (4 x 1 = 4)
Answer:
Metallic Character
The ability of an atom to donate electrons and form positive ion (cation) is known as electropositivity or metallic character. Down the group, metallic character increases due to increase in atomic size and across the period, from left to right electropositivity decreases due to decrease in atomic size.

Non-Metallic Character
The ability of an atom to accept electrons to form a negative ion (anion) is called non-metallic character or electronegativity. The elements having high electronegativity have a higher S tendency to gain electrons and form anion. Down the group, electronegativity decreases due to increase in atomic size and across the period, from left to S right electronegativity increases due to decrease in atomic size.

(i) Which of the following correctly represents the decreasing order of metallic character of alkali metals plotted in the graph?

(a) Cs > Rb > Li > Na > K
(b) K > Rb > Li > Na > Cs
(c) Cs > Rb > K > Na > Li
(d) Cs > K > Rb > Na > Li
Answer:
(c) Cs > Rb > K > Na > Li

(ii) Hydrogen is placed along with alkali metals in the modem periodic table though it shows non-metallic character
(a) as hydrogen has one electron & readily loses electron to form negative ion
(b) as hydrogen can easily lose one electron like alkali metals to form positive ion
(c) as hydrogen can gain one electron easily like halogens to form negative ion
(d) as hydrogen shows the properties of non-metals
Answer:
(b) as hydrogen can easily lose one electron like alkali metals to form positive ion

(iii) Which of the following has highest electronegativity?
(a) F
(b) Cl
(c) Br
(d) I
Answer:
(a) F

(iv) Identify the reason for the gradual change in electronegativity in halogens down the group.
(a) Electronegativity increases down the group due to decrease in atomic size
(b) Electronegativity decreases down the group due to decrease in tendency to lose electrons
(c) Electronegativity decreases down the group due to increase in atomic radius/ tendency to gain electron decreases
(d) Electronegativity increases down the group due to increase in forces of attractions between nucleus & valence electrons
Answer:
(c) Electronegativity decreases down the group due to increase in atomic radius/ tendency to gain electron decreases

(v) Which of the following reason correctly justifies that “Fluorine (72pm) has smaller atomic radius than Lithium (152pm)”?
(a) F and Li are in the same group. Atomic size increases down the group
(b) F and Li are in the same period. Atomic size increases across the period due to increase in number of shells
(c) F and Li are in the same group. Atomic size decreases down the group
(d) F and Li are in the same period and across the period atomic size/radius decreases from left to right.
Answer:
(a) F and Li are in the same group. Atomic size increases down the group

Question 19.
Read the following and answer any four questions from 19 (1) to 19 (v). (4 x 1 = 4)
Sumati wanted to see the stars of the night sky. She knows that she needs a telescope to see those distant stars. She finds out that the telescopes, which are made of lenses, are called refracting telescopes and the ones which are made of mirrors are called reflecting telescopes.

So she decided to make a refracting telescope. She bought two lenses, Lj and L2. out of which Lj was bigger and L2 was smaller. The larger lens gathers and bends the light, while the smaller lens magnifies the image. Big, thick lenses are more powerful. So to see far away, she needed a big powerful lens.

Unfortunately, she realized that a big lens is very heavy. Heavy lenses are hard to make and difficult to hold in the right place. Also since the light is passing through the lens, the surface of the lens has to be extremely smooth. Any flaws in the lens will change the image. It would be like looking through a dirty window.

(i) Based on the diagram shown, what kind of lenses would Sumati need to make the telescope?
(a) Concave lenses
(b) Convex lenses
(c) Bifocal lenses
(d) Flat lenses

(ii) If the powers of the lenses L₁ and L₂ are in the ratio of 4:1, what would be the ratio of the focal length of L₁ and L₂?
(a) 4:1
(b) 1:4
(c) 2:1
(d) 1:1

(iii) What is the formula for magnification obtained with a lens?
(a) Ratio of height of image to height of object
(b) Double the focal length.
(c) Inverse of the radius of curvature.
(d) Inverse of the object distance.

(iv) Sumati did some preliminary experiment with the lenses and found out that the magnification of the eyepiece (L₂) is 3. If in her experiment with L₂ she found an image at 24 cm from the lens, at what distance did she put the object?
(a) 72 cm
(b) 12 cm
(c) 8 cm
(d) 6 cm

(v) Sumati bought not-so-thick lenses for the telescope and polished them. What advantages, if any, would she have with her choice of lenses?
(a) She will not have any advantage as even thicker lenses would give clearer images.
(b) Thicker lenses would have made the telescope easier to handle.
(c) Not-so-thick lenses would not make the telescope very heavy and also allow considerable amount of light to pass.
(d) Not-so-thick lenses will give her more magnification.
Answer:

Question 20.
Read the following and answer any four questions from 20 (i) to 20 (v). (4 x 1 = 4)
A solenoid is a long helical coil of wire through which a current is run in order to create a magnetic field. The magnetic field of the solenoid is the superposition of the fields due to the current through each coil.

It is nearly uniform inside the solenoid and close to zero outside and is similar to the field of a bar magnet having a north pole at one end and a south pole at the other depending upon the direction of current flow. The magnetic field produced in the solenoid is dependent on a few factors such as, the current in the coil, number of turns per unit length etc.

The following graph is obtained by a researcher while doing an experiment to see the variation of the magnetic field with respect to the current in the solenoid.

The unit of magnetic field as given in the graph attached is in milli-Tesla (mT) and the current is given in Ampere.

(i) What type of energy conversion is observed in a linear solenoid?
(a) Mechanical to Magnetic
(b) Electrical to Magnetic
(c) Electrical to Mechanical
(d) Magnetic to Mechanical
Answer:
(c) Electrical to Mechanical

(ii) What will happen if a soft iron bar is placed inside the solenoid?
(a) The bar will be electrocuted resulting in short-circuit.
(b) The bar will be magnetised as long as there is current in the circuit.
(c) The bar will be magnetised permanently.
(d) The bar will not be affected by any means.
Answer:
(b) The bar will be magnetised as long as there is current in the circuit.

(iii) The magnetic field lines produced inside the solenoid are similar to that of…
(a) a bar magnet
(b) a straight current carrying conductor
(c) a circular current carrying loop
(d) electromagnet of any shape
Answer:
(a) a bar magnet

(iv) After analysing the graph a student writes the following statements.
I. The magnetic field produced by the solenoid is inversely proportional to the current.
II. The magnetic field produced by the solenoid is directly proportional to the current.
III. The magnetic field produced by the solenoid is directly proportional to square of the current.
IV. The magnetic field produced by the solenoid is independent of the current.
Choose from the following which of the following would be the correct statement(s).
(a) Only IV
(b) I and III and IV
(c) I and II
(d) Only II
Answer:
(d) Only II

(v) From the graph deduce which of the following statements is correct.
(a) For a current of 0.8 A the magnetic field is 13 mT
(b) For larger currents, the magnetic field increases non-linearly.
(c) For a current of 0.8A the magnetic field is 1.3 mT
(d) There is not enough information to find the magnetic field corresponding to 0.8A current.
Answer:
(a) For a current of 0.8 A the magnetic field is 13 mT

Section-B

Question 21.
Bile juice does not have any digestive enzyme but still plays a significant role in the process of digestion. Justify the statement.
OR
In birds and mammals the left and right side of the heart are separated. Give reasons.
Answer:
Bile juice makes the acidic food coming from the stomach alkaline for the action of pancreatic enzymes. Bile salts break the large globules of fat in the intestine to smaller globules increasing the efficiency of enzyme action. This is similar to the emulsifying action of soaps on dirt.
OR
The separation keeps oxygenated and deoxygenated blood from mixing allowing a highly efficient supply of oxygen to the body. This is useful in animals that have high energy needs (birds and mammals) which constantly use energy to maintain their body temperature.

Question 22.
State the events occurring during the process of photosynthesis. Is it essential that these steps take place one after the other immediately?
Answer:

Absorption of light energy by chlorophyll.

• Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
• Reduction of carbon dioxide to carbohydrates.

These steps need not take place one after the other immediately. For example, desert plants take up carbon dioxide at night and prepare an intermediate which is acted upon by the energy absorbed by the chlorophyll during the day

Question 23.
Give a test that can be used to confirm the presence of carbon in a compound. With a valency of 4, how is carbon able to attain noble gas configuration in its compounds?
OR
The number of carbon compounds is more than those formed by all other elements put together. Justify the statement by giving two reasons.
Answer:
A bum compound in air/ oxygen; Gas evolved turns lime water milky
By sharing its four valence electrons with other elements.
OR

• Due to self linking ability of carbon/catenation
• Since carbon has a valency of four it can form bonds with four other atoms of carbon or atoms of some other mono-valent element.
• Due to small size of carbon it forms very strong and (or) stable bonds with other elements

Question 24.
The following observations were made by a student on treating four metals P, Q, R and S with the given salt solutions:

Sample	MgS04(aq)	Zn(NO3)2(aq)	CaS04(aq)	Na2S04(aq)
P	No reaction	Reaction occurs	Reaction occurs	No reaction
Q	Reaction occurs	Reaction occurs	Reaction occurs	Reaction occurs
R	No Reaction	Reaction occurs	No Reaction	No Reaction
S	No Reaction	No Reaction	No Reaction	No Reaction

Based on the above observations:
(i) Arrange the given samples in the increasing order of reactivity
(ii) Write the chemical formulae of products formed when Q reacts with CuS04 solution.
Answer:
(i) S > R > P > Q
(ii) Cu and QSO₄

Question 25.
A student observes the given phenomenon in the lab as a white light passes through a prism. Among many other colours, he observed the position of the two colours red and violet. What is the phenomenon called? What is the reason for the violet light to bend more than the red light?

Answer:

• The phenomenon is called dispersion.
• Speed of violet light inside the prism is slowest and that of red is highest. Hence, the deviation of violet light is maximum and that of red is minimum.

Question 26.
A student has two resistors- 2 Q and 3 Q. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose.

Answer:
The overall current needed = 9A. The voltage is 12V Hence by Ohm’s law V=IR,
R₂ = 2Q
The resistance for the entire circuit = \(=\frac{12}{9}=\frac{4}{3} \Omega=\mathrm{R}\)
R₁ and R₂ are in parallel.
Hence, R = \(R=\frac{\left(R_{1} R_{2}\right)}{\left(R_{1}+R_{2}\right)}=\frac{4 R_{2}}{\left(4+R_{2}\right)}=\frac{4}{3}\)

Section-C

Question 27.
After self-pollination in pea plants with round, yellow seeds, following types of seeds were obtained by Mendel:

Seeds	Number
Round, yellow	630
Round, green	216
Wrinkled, yellow	202
Wrinkled, green	64

Analyse the result and describe the mechanism of inheritance which explains these results.
OR
In humans, there is a 50% probability of the birth of a boy and 50 % probability that a girl will be born. Justify the statement on the basis of the mechanism of sex-determination in human beings.
Answer:
The ratio obtained is 9:3:3:1 in which parental as well as new combinations are observed. This indicates that progeny plants have not inherited a single whole gene set from each parent. Every germ cell takes one chromosome from the pair of maternal and paternal chromosomes. When two germ cells combine, segregation of one pair of characters is independent of other pair of characters.
OR
In human beings, the genes inherited from our parents decide whether we will be boys or girls. Women have a perfect pair of sex chromosomes (XX). But, men have a mismatched pair (XY).

All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of the children will be determined by what they inherit from their father. A child who inherits an X chromosome from her father will be a girl, and one who inherits a Y chromosome from him will be a boy.

Question 28.
Plastic cups were used to serve tea in trains in early days- these could be returned to the vendors, cleaned and reused. Later,Kulhads were used instead of plastic cups. Now, paper cups are used for serving tea. What are the reasons for the shift from Plastic to Kulhads and then finally to paper cups?
Answer:

• Use of Plastic cups raised the concern towards hygiene thus they were replaced by disposable plastic cups.
• Plastic cups are non-biodegradable and harm the environment. They were thus replaced by Kulhads.
• Making Kulhad made of clay on a large scale resulted in the loss of top fertile soil.
• Now, disposable paper cups are used because – the paper can be recycled, it is biodegradable and is eco-friendly material which does not cause environmental pollution.

Question 29.
Explain where and how urine is produced?
Answer:
Blood passes through filtration units in the kidney called nephron

• Passes through glomerulus in the Bowman’s capsule – Ultra filtration
• Filtrate initially has glucose, amino acids, water, salts and nitrogenous waste
• Reabsorption – Water (as per the need of the body), Glucose and amino acids are all reabsorbed
• Secretion of excess water, salts and urea (nitrogenous waste) which makes up the urine.

Question 30.
(i) Which of the following reactions is/ are an endothermic reaction(s) where decomposition also happens?
• Respiration
• Heating of lead nitrate
• Decomposition of organic matter
• Electrolysis of acidified water
(ii) Silver chloride when kept in the open turns grey. Illustrate this with a balanced chemical equation.
Answer:
(i) Heating of lead nitrate; and electrolysis of acidified water

(ii) \(2 \mathrm{AgCl}(\mathrm{s}) \stackrel{\text { Sunlight }}{\longrightarrow} 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g})\)
(No deduction for not mentioning state of reactants and products.)

Question 31.
The following table shows the position of five elements A, B, C, D and E in the modem periodic table.

Answer:
(i) D, As it is on the left side of the table in group 2.
(ii) C, as it is in the group 18/ Noble gas.
(iii) E, as we move from left to right across a period, atomic radius decreases.

Question 32.
(i) Explain the formation of calcium chloride with the help of electron dot structure. (At numbers: Ca = 20; Cl = 17)
(ii) Why do ionic compounds not conduct electricity in solid state but conduct electricity in molten and aqueous state?
Answer:

(ii) Ionic compounds do not conduct in solid state due to absence of free ions but they conduct electricity in molten and aqueous state due to presence of free ions.

Question 33.
Refractive index of water with respect to air is 1.33 and that of diamond is 2.42.
(i) In which medium does the light move faster, water or diamond?
(ii) What is the refractive index of diamond with respect to water?
Answer:
Refractive index = speed of light in vacuum / speed of light in medium.
Since the refractive index of diamond is more, hence the speed of light is lesser in diamond.
Let speed of light in water be v_w and in diamond be v_d.
Refractive index of diamond w.r.t water is say n = speed of light in water / speed of light in diamond
n = v_wv_d
Dividing both numerator and denominator by speed of light [c] we get
n = (v_w/c) (v_d/c)
= Inverse ratio of refractive index of water and diamond.
n = 2.42/1.33 = 1.82 (approx.)

Section-D

Question 34.
Match the following pH values 1,7, 10, 13 to the solutions given below:
• Milk of magnesia
• Gastric juices
• Brine
• Aqueous sodium hydroxide
Amit and Rita decided to bake a cake and added baking soda to the cake batter.

Explain with a balanced reaction, the role of the baking soda. Mention any other use of baking soda. 5
OR
(i) Four samples A, B, C and D change the colour of pH paper or solution to green, reddish- pink, blue and orange. Their pH was recorded as 7,2, 10.5 & 6 respectively. Which of the samples has the highest amount of hydrogen ion concentration? Arrange the four samples in the decreasing order of their pH.

(ii) Rahul found that the Plaster of Paris, which he stored in a container, has become very hard and lost its binding nature. What is the reason for this? Also, write a chemical equation to represent the reaction taking place.

(iii) Give any one use of Plaster of Paris other than for plastering or smoothening of walls.
Answer:

• Milk of magnesia – 10
• Gastric juices – 1
• Brine – 7
• Aqueous sodium hydroxide – 13

Baking soda undergoes thermal decomposition to form Na₂CO₃, CO₂ and H₂O; CO₂ makes the cake fluffy & soft
\(\mathrm{NaHCO}_{3} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Uses:
Used in fire extinguishers/ antacid to neutralize excess acid in stomach / to neutralize the effect of acid in insect sting.
OR
(i) (a) B
(b) C,A,D,B
(ii) Due to moisture in the atmosphere it converted into gypsum
\(\mathrm{CaSO}_{4} \cdot 1 / 2 \mathrm{H}_{2} \mathrm{O}+1 / 2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
(iii) Making toys/dolls or statues /fixing broken limbs/making decorative materials.

Question 35.
Trace the changes that take place in a flower from gamete formation to fruit formation.
Answer:
[Diagram drawn and annotated with the following points will also be considered]

• Stamen is the male reproductive part and it produces pollen grains.
• The ovary contains ovules and each ovule has an egg cell.
• The pollen needs to be transferred from the stamen to the stigma.
• If this transfer of pollen occurs in the same flower, it is referred to as self-pollination.

On the other hand, if the pollen is transferred from one flower to another, it is known as cross-pollination.
After the pollen lands on a suitable stigma, it has to reach the female germ-cells which are in the ovary. For this, a tube grows out of the pollen grain and travels through the style to reach the ovary/Figure

• The male germ-cell produced by pollen grain fuses with the female gamete present in the ovule.
• This fusion of the germ-cells or fertilisation gives the zygote.
• After fertilisation, the zygote divides several times to form an embryo within the ovule.
• The ovule develops a tough coat and is gradually converted into a seed. The ovary grows rapidly and ripens to form a fruit.
• Meanwhile, the petals, sepals, stamens, style and stigma may shrivel and fall off.

Question 36.
In the given circuit, A, B, C and D are four lamps connected with a battery of 60V.

Analyse the circuit to answer the following questions.
(i) What kind of combination are the lamps arranged in (series or parallel)?
(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?
(iii) Explain with proper calculations which lamp glows the brightest?
(iv) Find out the total resistance of the circuit.
OR
PQ is a current carrying conductor in the plane of the paper as shown in the figure below.
(i) Find the directions of the magnetic fields produced by it at points R and S?
(ii) Given r₁> r₂ where will the strength of the magnetic field be larger? Give reasons.
(iii) If the polarity of the battery connected to the wire is reversed, how would the direction of the magnetic field be changed?

(iv) Explain the rule that is used to find the direction of the magnetic field r₂ for a straight current carrying conductor.
Answer:
(i) The lamps are in parallel.

(ii) Advantages:
If one lamp is faulty, it will not affect the working of the other lamps.
They will also be using the full potential of the battery as they are connected in parallel.

(iii) The lamp with the highest power will glow the brightest.
P= VI
In this case, all the bulbs have the same voltage. But lamp C has the highest current. Hence, for lamp C, P = 5 x 60 Watt = 300 W. (the maximum).

(iv) The total current in the circuit = (3 + 4 + 5 + 3) A = 15A
The Voltage = 60V
V = IR and hence R = V/I
R= 60/15 Ω = 4Ω

OR

(i) The magnetic field lines produced is into the plane of the paper at R and out of it at S.

(ii) Field at S > Field at P Magnetic field strength for a straight current carrying conductor is inversely proportional to the distance from the wire.

(iii) The current will be going from top to bottom in the wire shown and the magnetic field lines are now in the clockwise direction on the plane which is perpendicular to the wire carrying current.

(iv) Right hand thumb rule. The thumb is aligned to the direction of the current and the direction in which the fingers are wrapped around the wire will give the direction of the magnetic field.

Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Maths Standard Set 2 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. This question paper contains two parts, A and B.
2. Both Part A and Part B have internal choices.

Part-A:
1. It consists of two sections, I and II.
2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.
3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.

Part-B:
1. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

Section 1 has 16 questions of 1 mark each. Internal choice is provided in 5 questions.

Question 1.
If xy = 180 and HCF(x, y) = 3, then find the LCM(x, y).
OR
The decimal representation of will terminate after how many decimal places?
Solution :
(LCM)(3)= 180
LCM = 60
OR
Four decimal places.

Question 2.
If the sum of the zeroes of the quadratic polynomial 3×2 – kx + 6 is 3, then find the value of k.
Solution :

Question 3.
For what value of k, the pair of linear equations 3x + y = 3 and 6.x + ky = 8 does not have a solution.
Solution :

Question 4.
If 3 chairs and 1 table costs ₹ 1500 and 6 chairs and 1 table costs ₹ 2400, form linear equations to represent this situation.
Solution :
Let the cost of 1 chair = ₹ x
And the cost of 1 table = ₹ y
3 x + y= 1500
6x + y = 2400

Question 5.
Which term of the AP 27, 24, 21, is zero?
OR
In an Arithmetic Progression, if d = -4, n = 7, an = 4, then find a.
Solution :
a_n = a + (n – 1) d
0= 27 + (n-1)(-3)
30= 3n
n= 10
10th term.
OR
a_n= a + (n — 1 )d
4 = a + 6 X (—4)
a = 28

Question 6.
For what values of k, the equation 9x² + 6kx + 4 = 0 has equal roots?
Solution :
9x² +6kx + 4 = 0
(6k)²-4 x 9 x 4 = 0
36k² = 144
⇒ k² = 4
k = ±2

Question 7.
Find the roots of the equation x² + 1x + 10 = 0.
OR
For what value(s) of ‘a’ quadratic equation 3ax² – 6x + 1 = 0 has no real roots?
Solution :
x² + 7x+ 10= 0
x² + 5x + 2x+ 10= 0
(x + 5)(x + 2) = 0
x = -5, x = -2
OR
3ax² – 6x + 1 = 0
(-6)²– 4(3a)(1) < 0
12a > 36
⇒ a > 3

Question 8.
If PQ = 28 cm, then find the perimeter of ΔPLM.

Answer:
PQ = PT

PL + LQ= PM + MT
PL + LN = PM + MN
Perimeter (ΔPLM) = PL + LM + PM
= PL + LN + MN + PM = 2 (PL + LN) = 2(PL + LQ)
= 2 x 28 = 56 cm

Question 9.
If two tangents inclined at 60° are drawn to a circle of radius 3 cm, then find length of each tangent.
OR
PQ is a tangent to a circle with centre O at point P. If ∠OPQ is an isosceles triangle, then find ∠OQP.
Answer:

Question 10.
In the AABC, D and E are points on side AB and AC respectively such that DE || BC. If
AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE.
Answer:

Question 11.
In the figure, if B₁, B₂, B₃,…. and A₁, A₂, A₃,….. have been marked at equal distances. In what ratio C divides AB?

Answer:
8:5

Question 12.
sin A + cos B =1, A = 30° and B is an acute angle, then find the value of B.
Answer:

Question 13.
If x = 2 sin² θ and y = 2 cos² θ + 1, then find x + y
Answer:
x + y = 2sin²
0 + 2 cos² 9 + 1
= 2(sin² 0 + cos² 0) + 1
= 3

Question 14.
In a circle of diameter 42 cm,if an arc subtends an angle of 60° at the centre where n = then what will be the length of arc?
Answer:

Question 15.
12 solid spheres of the same radii are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. Find the diameter of the each sphere.
Answer:

Question 16.
Find the probability of getting a doublet in a throw of a pair of dice.
OR
Find the probability of getting a black queen when a card is drawn at random from a well- shuffled pack of 52 cards.
Answer:

Section-II

Case Study based questions are compulsory. Attempt any four sub-parts of each question. Each sub-part carries 1 mark.

Case Study Based-1 Sun Room

Question 17.
The diagrams show the plans for a sun room. It will be built onto the wall of a house. The four walls of the sun room are square clear glass panels. The roof is made using four clear glass panels, trapezium in shape, all the same size

• one tinted glass panel, half a regular octagon in shape.

Refer to Top View
(a) Find the mid-point of the segment joining the points J (6, 17) and I (9, 16).
(i) \(\left(\frac{33}{2}, \frac{15}{2}\right)p\)
(ii) \(\left(\frac{3}{2}, \frac{1}{2}\right)\)
(iii) \(\left(\frac{15}{2}, \frac{33}{2}\right)\)
(iv) \(\left(\frac{1}{2}, \frac{3}{2}\right)\)
Solution:
(iii) \(\left(\frac{15}{2}, \frac{33}{2}\right)\)

Refer to Front View
(b) The distance of the point P from the y-axis is
(i) 4
(ii) 15
(iii) 19
(iv) 25
Solution:
(i) 4

Refer to Front View
(c) The distance between the points A and S is
(i) 4
(ii) 8
(iii) 16
(iv) 20
Solution:
(iii) 16

Refer to Front View
(d) Find the coordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally.
(i) (8.5, 2.0)
(ii) (2.0, 9.5)
(iii) (3.0, 7.5)
(iv) (2.0, 8.5)
Solution:
(iv) (2.0, 8.5)

Refer to Front View
(e) If a point (x,v) is equidistant from the Q(9. 8) and S( 17, 8), then 1
(i) x + y = 13
(ii) x – 13 = 0
(iii) x – 13 = 0
(iv) x – y – 13
Solution:
(ii) x – 13 = 0

Case Study Based-2

Question 18.
Scale Factor and Similarity
Scale Factor

Solution :
A scale drawing of an object is the same shape as the object but a different size.
The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio.

Similar Figures

The ratio of two corresponding sides in similar figures is called the scale factor.

If one shape can become another using resizing, then the shapes are similar.

Hence, two shapes are similar when one can become the other after a resize, flip, slide or turn.
(a) A model of a boat is made on the scale of 1 : 4. The model is 120 cm long. The full size of the boat has a width of 60 cm. What is the width of the scale model?
(i) 20 cm
(ii) 25 cm
(iii) 15 cm
(iv) 240 cm
Solution :
(iii) 15 cm

(b) What will effect the similarity of any two polygons?
(i) They are flipped horizontally.
(ii) They are dilated by a scale factor.
(iii) They are translated down.
(iv) They are not the mirror image of one another.
Solution :
(iv) They are not the mirror image of one another.

(c) If two similar triangles have a scale factor of a : b, which statement regarding the two triangles is true?
(i) The ratio of their perimeters is 3a : b
(ii) Their altitudes have a ratio a : b
(iii) Their medians have a ratio \(\frac{a}{2}: b\)
(iv) Their angle bisectors have a ratio a²: b²
Solution :
(ii) Their altitudes have a ratio a : b

(d) The shadow of a stick 5 m long is 2 m. At the same time, the shadow of a tree 12.5 m high is

(i) 3 m
(ii) 3.5 m
(iii) 4.5 m
(iv) 5 m
Solution :
(iv) 5 m

(e) Below you see a student’s mathematical model of a farmhouse roof with measurements. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edges of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT. and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m.

What is the length of EF, where EF is one of the horizontal edges of the block?
(i) 24 m
(ii) 3 m
(iii) 6 m
(iv) 10 m
Solution :
(iii) 6 m

Case Study Based-3

Question 19.
Applications of Parabolas-Highway Overpasses/Underpasses A highway underpass is parabolic in shape.

Answer:
Parabola: A parabola is the graph that results from p(x) — ax² + bx + c. Parabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.

(a) If the highway overpass is represented by x² – 2x – 8, then its zeroes are
(i) (2. -4)
(ii) (4, -2)
(iii) (-2, -2)
(iv) (-4, -4)
Answer:
(ii) (4, -2)

(b) The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial
(i) intersects x-axis
(ii) intersects y-axis
(iii) intersects y-axis or x-axis
(iv) None of the above
Answer:
(i) intersects x-axis

(c) Graph of a quadratic polynomial is a
(i) straight line.
(ii) Circle
(iii) Parabolla
(iv) ellipse
Answer:
(iii) Parabolla

(d) The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is 1
(i)  x² – 6x + 2
(ii) x² – 36
(iii) x² – 6
(iv) x² – 3
Answer:
(ii) x² – 36

(e) The number of zeroes that polynomial f(x) = (x – 2)² + 4 can have is:
(i) 1
(ii) 2
(iii) 0
(iv) 3
Answer:
(iii) 0

Case Study Based-4

Question 20.

(a) Estimate the mean time taken by a student to finish the race.
(i) 54
(ii) 63
(iii) 43
(iv) 50
Solution :
(iii) 43

(b) What will be the upper limit of the modal class?
(i) 20
(ii) 40
(iii) 60
(iv) 80
Solution :
(iii) 60

(c) The construction of cumulative frequency table is useful in detennining the
(i) mean
(ii) median
(iii) mode
(iv) All of the above
Solution :
(ii) median

(d) The sum of lower limits of median class and modal class is
Solution :
(i) 60
(ii) 100
(iii) 80
(iv) 140
Solution :
(iii) 80

(e) How many students finished the race within 1 minute?
(i) 18
(ii) 37
(iii) 31
(iv) 8
Solution :
(iii) 31

Part-B
Section-III

All questions are compulsory. In case of internal choices, attempt anyone.

Question 21.
3 bells ring at an interval of 4, 7 and 14 minutes. All three bell rang at 6 am, when the three balls will the ring together next?
Solution :
4 = 2 x 2
7 =7 x 1
14 = 2 x 7
LCM = 2 x 2 x 7 = 28
The three bells will ring together again at 6 : 28 am

Question 22.
Find the point on x-axis which is equidistant from the points (2, -2) and (-4, 2).
OR
P(-2, 5) and Q(3, 2) are two points. Find the coordinates of the point R on PQ such that PR = 2QR
Solution:

Question 23.
Find a quadratic polynomial whose zeroes are \(5-3 \sqrt{2} \text { and } 5+3 \sqrt{2}\)
Solution:

Question 24.
Draw a line segment AB of length 9 cm. With A and B as centres, draw circles of radius 5 cm and 3 cm respectively. Construct tangents to each circle from the centre of the other circle.
Solution:

Question 25.
If \(tan \mathrm{A}=\frac{3}{4}\), find the value of \(\frac{1}{\sin A}+\frac{1}{\cos A}\) and 3 cm respectively. Construct tangents to each circle from the centre of the other circle. If \(\sqrt{3} \sin \theta-\cos \theta=0 \text { and } 0^{\circ}<\theta<90^{\circ} \) find the value of θ.
Solution:

Question 26.
In the figure, quadrilateral ABCD is circumscribing a circle with centre O and AD ⊥ AB. If radius of incircle is 10 cm, then find the value of x.

Solution:

Question 27.
Prove that 2 – √3 is irrational, given that √3 is irrational.
Solution:

Question 28.
If one root of the quadratic equation 3x² + px + 4 = 0 is \(\frac{2}{3}\), then find the value of p and the other root of the equation.
OR
The roots α and β of the quadratic equation x² – 5x + 3(k – 1) = 0 are such that α – β = 1. Find the value k.
Solution :

Question 29.
In the figure, ABCD is a square of side 14 cm. Semicircles are drawn with each side of square as diameter. Find the area of the shaded region.

Solution:

Question 30.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, find the length of the corresponding side of the second triangle.
OR
In an equilateral triangle ABC, D is a point on side BC such that BD = -BC. Prove that 9 AD² = 7 AB².
Solution:

Question 31.
The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.

Class	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	12	Cl	12	15	b	6	6	4

Solution:

Class	Frequency	Cumulative Frequency
0-5	12	12
5-10	a	12 + a
10-15	12	24 +a
15-20	15	39 + a
20-25	b	39 + a + b
25-30	6	45 + a + b
30-35	6	51+0 + 6
35-40	4	55 + o + 6
Total	N = 70

Question 32.
If the angles of elevation of the top of the candle from two coins distant ‘a’ cm and ‘b’ cm (a > b) from its base and in the same straight line from it are 30° and 60°, then find the height of the candle.

Solution:

Let AB = candle
C and D are two coins

Question 33.
The mode of the following data is 67. Find the missing frequency x:
Solution:

Question 34.
The two palm trees are of equal heights and are standing opposite to each other on either side of the river, which is 80 m wide. From a point O between them on the river, the angles of elevation of the top of the trees are 60° and 30° respectively. Find the height of the trees and the distances of the point O from the trees.
OR
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60° respectively. Find the height of the tower, and also the horizontal distance between the building and the tower.
Solution:

Hence, height of the tower = h = 75 m
Distance between the building and the tower = 25 √3 = 43.25 m

Question 35.
Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm at the rate of 0.7 m/sec. By how much will the water rise in the tank in half an hour?
Solution:
For pipe, r = 1cm
Length of water flowing in 1 sec, h = 0.1 m = 70 cm
Cylindrical Tank, R = 40 cm, rise in water level = H
Volume of water flowing in 1 sec = n×h = a × 1 x 1 × 70 = 70a
Volume of water flowing in 60 sec = 70a: x 60
Volume of water flowing in 30 minutes = 70a × 60 ×30
Volume of water in Tank = ar²H = a × 40 × 40 × H
Volume of water in Tank = Volume of water flowing in 30 minutes
a × 40 × 40 × H = 70a x 60 x 30

Question 36.
A motorboat covers a distance of 16 km upstream and 24 km downstream in 6 hours. In the same time, it covers a distance of 12 km upstream and 36 km downstream. Find the speed of the boat in still water and that of the stream.
Solution:
Let speed of the boat in still water = x km/hr,
and Speed of the stream = y km/hr
Downstream speed = (x + y) km/hr
Upstream speed = (x – y) km/hr 24

Thus, speed of the boat in still water = 8 km/hr,
Speed of the stream = 4 km/hr

Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Maths Standard Set 3 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

1. This question paper contains two parts, A and B.
2. Both Part A and Part B have internal choices.

Part-A:
1. It consists of two sections, I and II.
2. Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.
3. Section II has 4 questions on case study. Each case study has 5 case-based sub-parts. An examinee is to attempt any 4 out of 5 sub-parts.

Part-B:
1. It consists of three sections III, IV and V.
2. In section III, Question Nos. 21 to 26 are Very Short Answer Type questions of 2 marks each.
3. In section IV, Question Nos. 27 to 33 are Short Answer Type questions of 3 marks each.
4. In section V, Question Nos. 34 to 36 are Long Answer Type questions of 5 marks each.
5. Internal choice is provided in 2 questions of 2 marks, 2 questions of 3 marks and 1 question of 5 marks.

Part – A
Section-I

Section I has 16 questions of 1 mark each. Internal choice is provided in 5 questions.

Question 1.
If one of the zeroes of the quadratic polynomial x¹ + 3x + k is 2, then find the value of k
Solution :
Since 2 is a zero of p(x) = x² + 3x + k, then p(x) =0
⇒(2)² + 3 x 2 + k = 0
⇒ 4 + 6 + k = 0
⇒ k = -10

Question 2.
Find the total number of factors of a prime number.
OR
Find the HCF and the LCM of 12, 21, 15.
Solution :
We know that prime number is a number which has exactly two factors, i.e., 1 and the number itself. So, the total number of factors of a prime number is 2.
OR
12 = 2 x 2 x 3 = 2² x 3¹
15 = 3 x 5 = 3¹ x 5¹ 21 = 3 x 7 = 3¹ x 7¹ HCF(12, 15,21)= 3
LCM(12, 15, 21) = 2² x 3¹ x 5¹ x 7¹ = 4 x 3 x 5 x 7 = 420

Question 3.
Find the value of k for which the system of equations x + y – 4 = 0 and 2x + ky = 3, has no solution.
Solution :
For the equations, x + y -4 = 0 and 2x + ky = 3
⇒ 2x + ky  –  3 = 0

Question 4.
Find the value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP.
OR
The first term of an AP is p and the common difference is q. Find its 10th term.
Solution :
2x (x + 10) and (3x + 2) are three consecutive terms of an AP.
Then, (x+10)-2x= (3x + 2)-(x+10)
⇒ x + 10 – 2x = 3x + 2 – x – 10
⇒ -x + 10 = 2x- 8
⇒ 3x = 18
⇒ x = 6
a_n = a + (n – 1 )d  ⇒ a_m =p + (10 – 1 )q =p + 9q
a₁₀ = p + 9q

Question 5.
In figure, ΔABC is circumscribing a circle. Find the length of BC.

Solution :
Line segments AB, BC and AC are tangents of the circle.
AP = AR = 4 cm     …(i)
[Tangents drawn from the same point to a circle are equal]
Similarly, BP = BQ = 3 cm ……………… (ii)
and   CQ = CR     ………….. (iii)
Since, AC =11 cm  ⇒ AR + CR = 11 cm
⇒ 4 cm + CR = 11 cm [from (i)]
⇒ CR = 7 cm ……………… (iv)
BC = BQ + CQ = 3 cm + 7 cm  [From (ii),(iii) and (iv)]
= 10 cm

Question 6.
If in the given figure, DE || BC, then find EC.

Solution :

Question 7.
Find the value of  \(\left(\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}\right)\)
Solution :
\(\sin ^{2} \theta+\frac{1}{1+\tan ^{2} \theta}=\sin ^{2} \theta+\frac{1}{\sec ^{2} \theta}=\sin ^{2} \theta+\cos ^{2} \theta=1\)

Question 8.
Find the value of (1 + tan² θ) (1 – sin θ) (1 + sin θ).
Solution :
(1 + tan² θ)(1 – sin θ)(1 + sin θ) = (1 + tan² θ)(1 – sin² θ)
\(=\sec ^{2} \theta \cos ^{2} \theta=\frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta=1\)

Question 9.
Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3:1. What is the ratio of their volumes? 1
Solution :
Let h₁ and h₂ be the heights of two cones and r₁ and r₂ by the radii

Question 10.
If ABC is an equilateral triangle of side 2a, then find the length of one of its altitudes.
OR
In the given figure, MN || QR. If PM = x cm, MQ = 10 cm, PN = (x – 2) cm, NR = 6 cm, then find the

Solution :
In ∠ADB and ∠ADC,
AB = AC [Sides of equilateral triangle ABC]
AD = AD [Common]
∠ADB = ∠ADC [90°, AD ⊥ BC]
ΔADB ≅ ΔADC
ABC is an equilateral triangle such that each side is 2a and AD ⊥ BC.

Question 11.
Base of an isosceles triangle is \(\frac{2}{3}\) times its congruent sides. Perimeter of the triangle is 32 cm.
Formulate this problem as a pair of equations.
Solution :
Let the congruent side of isosceles triangle be x cm and its base be y cm.
Then \(y=\frac{2}{3} x\)
2x – 3y = 0 ……………. (i)
Also,   x + x + y=32  ⇒ 2x + y = 32 ………….. (ii)

Question 12.
Check whether x(x + 2) – 3 = (x+ 4)x is a quadratic equation.
Solution :
Since  x(x + 2) – 3 = x(x + 4)
⇒ x² + 2x – 3 = x² + 4x
⇒ 2x + 3 = 0
This is linear equation not a quadratic equation.

Question 13.
Is x = -2 a solution of 3x² + 13x + 14 = 0?
OR
State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution :
Putting the value of x in the quadratic equation,
LHS = 3x² + 13x + 14 = 3 (-2)² + 13 (-2) + 14
= 12 – 26 + 14 = 0 = RHS
Hence, x = -2 is a solution.
OR
We have (x + 1) (x – 2) + x = 0
⇒ x²-x-2 + x= 0  ⇒ x²-2 = 0
D = b² – 4ac = 0 – 4(1) (-2) = 8 > 0
∴ Given equation has two distinct real roots.

Question 14.
Two concentric circles of radii a and b (a> b) are given. Find the length of the chord of the larger circle which touches the smaller circle.
OR
In the given figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC.

Solution :

OR
In the given figure, CP = CQ  [tangents drawn from an external point are equal]
So,  CP = CQ = 11 cm
Also,   BR = BQ   [tangents drawn from an external point are equal]
So,  BR = BQ = 4 cm
∴ Now, BC = CQ – BQ
= (11 -4) cm = 7 cm

Question 15.
Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3:4.
Solution :

Steps of construction:
1. Draw a line segment AB = 6
2. Draw any ray AX making an acute angle XAB with AB.
3. Along AX mark 7 (3 + 4) points A₁ A₂, A₃, A₄,……………, A₇ at equal distances such that
AA₁ = A₁A₂ = A₆A₇
4. Join A₇B
5. From A₃, draw A₃P parallel to A₇B (by making an angle equal to ∠AA₇B at A₃) to meet AB at point P.
Then  AP : PB = 3 : 4.

Question 16.
The radius of a circle is 5 cm. Find the circumference of the circle whose area is 49 times the area of given circle.
Solution :
The area of the given circle = πr² = π(5)² = 25π sq. cm
Area of the other circle = 49 × 25π
Let the radius of this circle be R Then
Then πR² = 49 x 25 x π  ⇒ R² = (7)² x (5)²
R = 7 x 5 = 35 cm
The required circumference = 2πR = 2 x \(\frac{22}{7}\) x 35 = 220 cm

Section-II

Case Study based questions are compulsory. Attempt any four sub-parts of each question. Each sub-part carries 1 mark.

Case Study Based-1

Question 17.
Treasure Island Shikha and Sanjana are playing a board game of

Treasure Island.
Shikha and Sanjana are playing a board game of Treasure Island.

Refer to Skull Rock and Cave of Death
(a) The mid-point of the segment joining A(3, 5) and C(5, 3) is……..
(i) (2,3)
(ii) (3,5)
(iii) (4,3)

Refer to Three Palms
(b) The distance of point D(6, 4) from origin is
(i) \(5 \sqrt{7}\)units
(ii) \(7 \sqrt{5}\) units
(iii) \(4 \sqrt{10}\) units
(iv) \(2 \sqrt{23}\) units

Refer to 4-Cross Cliffs and Three Palms
(c) The distance between the points B(2,3)and D(6, 4) is
(i) \(\sqrt{13} \text { units }\)units
(ii) 4 units
(iii) \(5 \sqrt{3}\)units
(iv) \(\sqrt{17}\)units

(d) The coordinate of the point which divides the join of B(2,3) and D(6,4) in the ratio 2 : 3 is
(i) \(\left(\frac{13}{4}, \frac{19}{4}\right)\)
(ii) \(\left(\frac{18}{5}, \frac{17}{5}\right)\)
(iii) \(\left(\frac{11}{8}, \frac{13}{8}\right)\)
(iv) \(\left(\frac{19}{3}, \frac{16}{3}\right)\)

(e) If \(P\left(\frac{x}{3}, 4\right)\) is the mid-point of the line segment joining the points C(5,3) and
A(3, 5), then
(i) 8
(ii) 10
(iii) 12
(iv) 16
Solution :

Case Study Based-2

Question 18.
What are you Smoking?
Given below is Air Quality Index of different localities of Delhi on 27th December 2019 by Times of India Newspaper on 28th. December 2019

The data recorded from above AQI (Air Quality Index) is given below:

AQI	310-320	320-330	330-340	340-350	350-360	360-370	370-380	380-390
Frequencies	2	4	3	4	5	7	5	4

(a) The sum of the lower limits of the median class and modal class is ……………
(i) 650
(ii) 660
(iii) 750
(iv) 710

(b) The modal class is …………
(i) 310-320
(ii) 330-340
(iii) 360-370
(iv) 380-390

(c) The upper limit of the median class is ………
(i) 330
(ii) 350
(iii) 360
(iv) 390

(d) The difference of the upper limit of the median class and the lower limit of the modal class is ……..
(i) 0
(ii) 1
(iii) 2
(iv) 3

(e) The mean AQI is ……..
(i) 335.8
(ii) 354.7
(iii) 360.4
(iv) 395.9
Solution :

Class	Frequency	Cumulative Frequency
310-320	2	2
320-330	4	6
330-340	3	9
340-350	4	13
350-360	5	18 ← Median Class
360-370	7	25 ← Model Class
370-380	5	30
380-390	4	34
	n= 34

We have \(n=34, \frac{n}{2}=\frac{34}{2}=17\)

∴ The sum of the lower limits of the median class and modal class = 350 + 360 = 710 So, option
(iv) is the correct answer.

(b) Here, the maximum frequency is 7 and the corresponding class is 360-370.
So, modal class is 360-370. So, option (iii) is the correct answer.

(c) The median class is 350-360 which has upper limit 360. So, option (iii) is the correct answer.

(d) The difference between upper limit of the median class and the lower limit of the modal class = 360 – 360 = 0. So, option (i) is the correct answer.

(e)

Class	Frequency (fi)	Class Mark (xi)	fi xi
310-320	2	315	630
320-330	4	325	1300
330-340	3	335	1005
340-350	4	345	1380
350-360	5	355	1775
360-370	7	365	2555
370-380	5	375	1875
380-390	4	385	1540
	∑fi = 34		∑fi  xi = 12060

\(\text { Mean }=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{12060}{34}=354.7(\text { approx. })\)
So, option (ii) is the correct answer.

Case Study Based-3

Question 19.
Skipping Rope
Skipping rope is a good exercise. It bums calories, makes bones strong and improves heart health.

During skipping, when rope goes up and down it makes the shape of parabolas (graphs of quadratic polynomials). Observe the following skipping pictures.

Refer Picture 1
(a) The graph of polynomial p(x) represented by Picture 1 is shown below. The number of zeroes of the polynomial is

(i) 0
(ii) 1
(iii) 2
(iv) 4

Refer Picture 2
(b) The graph of polynomial p(x) represented by Picture 2 is shown below. Which of the following has negative (-) sign?

(i) a
(ii) b
(iii) c
(iv) All of there

(c) If \(\frac{1}{4}\) and 1 are the sum and product of zeroes of a polynomial whose graph is represented by Picture (3), the quadratic polynomial is
(i) \(k\left(x^{2}-\frac{1}{4} x-1\right)\)
(ii) \(k\left(\frac{1}{4} x^{2}-x-1\right)\)
(iii) \(k\left(x^{2}+\frac{1}{4} x+1\right)\)
(iv) \(k\left(\frac{1}{4} x^{2}+x+1\right)\)

(d) Let the Picture (1) represent the quadratic polynomial f(x) = x² – 8x + k whose sum of the
squares of zeroes is 40, The value of k is
(i) 8
(ii) 10
(iii) 12
(iv) 20

(e) Let the Picture (3) represent the quadratic polynomial f(x) = x² + 7x + 10. Then its zeroes are
(i) -1, -5
(ii) -2, -5
(iii) 1, 5
(iv) 2, 5
Solution :
(a) Since the graph y p(x) cuts the x-axis at two different points, so the polynomial has two zeroes. So, option (iii) is the correct answer.

(b) The parabola)’ = ax² + bx + c open downwards. Therefore, a <O. The vertex \(\left(\frac{-b}{2 a}, \frac{-\mathrm{D}}{4 a}\right)\) of the parabola is on OX¹.
\( \quad \frac{-b}{2 a}<0 \quad \Rightarrow b<0\)
Parabola y =p(x) cuts y-axis at P(O, c) which lies on OY’. Therefore e <O.
Hence, O, b <O and c< O.
So, option (iv) is the correct answer.

(c) We have sum \(\frac{1}{4}\) and product
\(f(x)=k\left(x^{2}-\frac{1}{4} x-1\right)\)
So, option (iii) is the correct answer.

(d)

(e) We have
x² + 7x + 10 =x² + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 2)(x + 5)
Now, when x + 2 = 0 or x + 5 = 0, i.e., whenx = -2 andx = -5
Therefore, the zeroes of x² + lx + 10 are -2 and -5

Question 20.

Fair Play
Garima has two children, Tapan and Maya. Every Sunday is a game night in the family. Tonight Garima has planned for a game with three cubes, one purple and two yellow. She placed the three cubes in a bag and called for her children.
Garima: Do you want to play a game of probability?
Maya: What is probability?
Garima: Let me ask you something before I answer you. Can you predict what is in this bag?
Tapan: I cannot guess that!
Maya: I am 100% sure it is a toy!
Garima: I am glad you think that Maya. Just now you used the concept of probability.
Whether an event can happen or not, can’t be predicted with total certainty. But we can always predict how- likely or unlikely it is for an event to happen.
And for predicting that, we use a concept called probability.
\(\text { Probability (an event to happen) }=\frac{\text { Number of ways event can happen }}{\text { Total number of ways all events can happen }}\)
Placing the bag of cubes in the centre, Garima explained the rules of the game to the children. Garima: Without looking, the first player will pick out a cube from the bag and then the second player will also pick out one cube without looking. If the two cubes picked out were the same colour, then the first person will win the game. If the boxes picked out are of two differently coloured cubes, then the second player will be the winner.

(a) In the first round, Maya pulled out a cube, which was yellow. What is the probability that Tapan will win the game?
(i) \(\frac{1}{2}\)
(ii) \(\frac{1}{3}\)
(iii) \(\frac{2}{3}\)
(iv) 0

(b) in the second round, Tapan started by picking out a purple cube. What is the probability for Tapan to win the round?
(i) 1
(ii) \(\frac{1}{3}\)
(iii) 0
(iv) \(\frac{2}{3}\)

(c) In the third round, Maya pulled out a cube. The probability that the pulled out cube is not of yellow colour is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)

(d) In the fourth round, Tapan pulled out a cube. The probability that the pulled out cube is either purple or yellow is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)

(e) In the last round, Maya pulled out a cube. The probability that the pulled out cube is of green colour is
(i) 1
(ii) \(\frac{1}{2}\)
(iii) \(\frac{1}{3}\)
(iv) 0
Solution :
(a) After taking out one yellow cube, the bag is left with 1 yellow cube whose probability of pulling out \(\frac{1}{2}\) So, option (i) is the correct answer.

(b) After taking out one purple cube, the bag has no purple cube. So, the probability for Tapan to win the round is
\(\frac{0}{3}\) = 0
So, option (iii) is the correct answer.

(c) The number of yellow cube is 1. So, the probability of pulling out a cube not of yellow colour is \(\frac{1}{3}\)
So, option (iii) is the correct answer.

(d) The number of purple and yellow cubes =1 + 2 = 3.
∴ The required probability =\(\frac{3}{3}\) = 1
So, option (i) is the correct answer.

(e) The number of green cube = 0
.’. The required probability = \(\frac{0}{3}\) = 0
So, option (iv) is the correct answer.

Part-B
Section-III

All questions are compulsory. In case of internal choices, attempt any one.

Question 21.
Write a rational number between \(\sqrt{2} \text { and } \sqrt{3}\)
Solution :
We have  √2 =1.4142135…. and √3 = 1.7320508…
Since every terminating decimal or repeating decimal represents a rational number.
So, 1.666666… is a rational number between \(\sqrt{2} \text { and } \sqrt{3}\)
Also 1.5, 1.6, 1.7 are rational numbers between \(\sqrt{2} \text { and } \sqrt{3}\)

Question 22.
Find the ratio in which the point (-3, p) divides the line segment joining the points (-5,-4) and (-2,3).
Hence find the value of p.
OR
If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Solution :

Question 23.
Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x)=ax²+bx+c,a≠0,c≠0.
Solution :
Let the zeroes of the polynornial f(x) = ax² + bx + c be α and β.
Then \(\alpha+\beta=\frac{-b}{a} \text { and } \alpha \beta=\frac{c}{a}\)
Now, the zeroes of the required polynomial are reciprocals of a and p.
∴ The required quadratic polynomial is given by

Question 24.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.
Solution :

Steps of construction:
1. Draw a line segment AB of length 7 cm.
2. With A as centre, draw a circle of radius 3 cm.
3. With B as centre, draw a circle of radius 2 cm.
4. Draw the perpendicular bisector of AB. Let P be the mid-point of segment AB.
5. With P as centre and radius PA draw a circle which intersects the circle with centre A at M and the circle with centre B at R and S.
6. Join BM and BN. Also join AR and AS. Then, BM, BN, AR and AS are required tangents.

Question 25.
The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in figure. If AC = 1.5 m and CD = 3 m, then find
(a) tan θ
(b) sec  θ + cosec θ

If sin θ + cosθ = √3, then prove that tan θ + cot θ = 1.
Solution :

Question 26.
In the given figure, two tangents TP and TQ are drawn to a circle with centre O from an external point T

Prove that: ∠PTQ = ∠OPQ.
Solution :
We know that, tangents drawn from same external point are equal.

Section-IV

Question 27.
Find HCF and LCM of 404 and 96 and verify that HCF x LCM = product of the two given numbers.
Solution :
404 = 2 x 2 x 101 =2² x 101
96 = 2 x 2 x 2 x 2 x 2 x 3 = 2⁵ x3
∴ HCF of 404 and 96 = 2² = 4
LCM of 404 and 96 = 101 x 2⁵ x 3 = 9696 HCF x LCM
= 4 x 9696 = 38784
Also   404 x 96 = 38784
Hence HCF x LCM = Product of 404 and 96.

Question 28.
If the roots of the equation (a – b)x² + (b – c )x + (c – a) = 0 are equal, prove that 2a = b + c.
OR
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution :
For real and equal roots, D = 0 ⇒ (b – c)² – 4(a – b)(c -a) = 0
⇒ b² + c² – 2be – 4ac + 46c + 4a² – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2 be – 4ac = 0
⇒ (-2a)² + (b)¹ + (c)² + 2(-2 a)b + 2 (b)(c) + 2c(-2a) = 0
⇒ ( – 2a + b + c)² = 0  ⇒ – 2a + b + c = 0
⇒ 2 a = b + c
OR
Let the two consecutive odd numbers be x and x +2
x² + (x + 2)² = 394                                ⇒ x² + x² + 4 + 4x = 394
⇒ 2x² + 4x + 4 = 394                            ⇒ 2x² + 4x – 390 = 0
⇒  x² + 2x – 195 = 0                               ⇒ x² + 15x – 13x – 195 = 0
⇒ x (x + 15) – 13 (x + 15) = 0                ⇒ (x- 13) (x+ 15) = 0
x – 13 = 0                                                  ⇒ or x + 15 = 0
⇒  x= 13                                                   ⇒ or x = – 15 (neglected)

When the first number x = 13, then the second number x + 2 = 13 + 2= 15.

Question 29.
ABCD is a square of side 4 cm. At each comer of the square, a A quarter circle of radius 1 cm, and at the centre, a circle of radius 1 cm, are drawn, as shown in the given figure. Find the area of the shaded region.

Solution :
Area of the shaded portion
= Area of given square ABCD
– 4 x area of each quarter circle
– area of the circle at the centre.

Question 30.
In the given figure, \(\angle \mathrm{D}=\angle \mathrm{E} \text { and } \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) prove that ΔBAC is an isosceles triangle.

OR
Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

[ ∴ AB = BC = CD = AD, sides of a rhombus]
Hence, the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Question 31.
In the given figure, PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

Solution :
If three or more line segments are perpendiculars to one line, then they are parallel to each other.

Question 32.
The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Solution:
Let AB be the tower and angle of elevation from point C = 30°

Question 33.
The median of the following data is 525. Find the values ofx andy, if total frequency is 100.

Class	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800	800-900	900-1000
Frequency	2	5	X	12	17	20	y	9	7	4

Solution:

Class	Frequency	cf
0-100	2	2
100-200	5	1
200-300	X	7 + x
300-400	12	19+x
400-500	17	36+x
500-600	20	56 +x
600-700	y	56 + x +y
700-800	9	65 + x + y
800-900	7	72 +x+y
900-1000	4	16+ x + y

Section – V

Question 34.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 6 m. At a point on the plane, the angle of elevation of the bottom and top of the flag­staff are 30° and 45° respectively. Find the height of the tower. (Take 73 = 1.73)
OR
A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height 5 m. From a point on the ground the angles of elevation of the top and bottom of the flag-staff are 60° and 30° respectively. Find the height of the tower and the distance of the point from base of the tower.
Solution :

Or

Let height of tower (TR) be x m, distance (RP) of a point from the base of tower be y m, height of the flag-staff (QT) be 5 m. Then in the ΔTRP,

Question 35.
A toy is in the form of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

Solution :
Let r be the radius and h the height of the cylindrical part of the toy.
Then, r = 5 cm and h = 13 cm.
Let, r₁ be the radius of the conical part, h₁ its height and 1 its slant height.

Question 36.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Solution :
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
∴ Speed of the boat going upstream = (x-y) km/h and
speed of the boat going downstream = (x + y) km/h



On solving (v) and (vi), we get x = 8, y = 3
Hence, speed of the boat in still water = 8 km/h and speed of the stream = 3 km/h.

Students can access the CBSE Sample Papers for Class 10 Science with Solutions and marking scheme Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 10 Science Set 1 with Solutions

Time: 3 Hours
Maximum Marks: 80

General Instructions:

(i) The question paper comprises four sections A, B, C and D. There are 36 questions in the question paper. All questions are compulsory.
(ii) Section-A – question no. 1 to 20 – all questions and parts there of are of one mark each.
These questions contain multiple choice questions (MCQs), very short answer questions and assertion – reason type questions. Answers to these should be given in one word or one sentence.
(iii) Section-B – question no. 21 to 26 are short answer type questions, carrying 2 marks each. Answers to these questions should in the range of 30 to 50 words.
(iv) Section-C – question no. 27 to 33 are short answer type questions, carrying 3 marks each.
Answers to these questions should in the range of 50 to 80 words. :
(v) Section—D – question no. – 34 to 36 are long answer type questions carrying 5 marks each. Answer to these questions should be in the range of 80 to 120 words.
(vi) There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
(vii) Wherever necessary, neat and properly labelled diagrams should be drawn.

Section – A

Question 1.
A women gave birth to a girl. Name the chromosome from sperm responsible for the offspring.
Answer:
Sperm with X chromosome from male is responsible for the birth of female and Y chromosome would be responsible for the male offspring. So for the given case, X chromosome from sperm is responsible for the offspring.

Question 2.
Name the gases which have replaced CFCs.
Answer:

• HFCs (Hydrofluorocarbons)
• Perfluorocarbons (PFCs) have replaced CFCs.

Question 3.
When calcium oxide is added to water, it completely dissolves in water without forming bubbles. What product is formed in this reaction?
Answer:
Ca(OH)₂

Question 4.
A plant gets rid of excess water through transpiration. What is the method used by plants to get rid of solid waste products?
Answer:
Shedding of yellow leaves

Question 5.
In peas, a pure tall plant (TT) is crossed with a short plant (tt). What is the ratio of pure tall plants to short plants in F2?
Answer:
3 : 1

Question 6.
State the direction of magnetic field in the following case.
Answer:

Perpendicular to the plane of paper in the outward direction by using Fleming’s left hand rule.

Question 7.
Why covalent compounds have low melting and boiling points?
OR
What is catenation?
Answer:
As the bond is formed by sharing of electrons between two atoms. Intermolecular forces are small between the covalent compounds. These bonds break easily.
OR
Carbon has the unique ability to form bonds with the other atoms of carbon which gives rise to large molecules. This property of self linking is called catenation.

Question 8.
What is the formula of oxide and hydride of Group I elements?
OR
What are isotopes?
Answer:
Oxide formula → R₂O; Hydride formula → RH.
‘R’ represents Group I element. F-31
OR
Isotopes are the atoms of same element having same atomic number but different mass number.

Question 9.
Why are the rings of cartilage present in air tube—Trachea?
OR
In what form is food energy stored in plants and animals?
Answer:
Rings of cartilage prevents the trachea from collapsing.
OR
Plants—Starch; Animals—Glycogen.

Question 10.
List two functions performed by the testis in human beings.
OR
What happens when a mature spirogyra filament attains considerable length?
Answer:

• To produce sperms
• To produce male sex hormone/testosterone

OR
Its filament breaks up into smaller fragments or pieces, and each fragment grows into a new filament/individual.

Question 11.
Name two human traits that show variations.
OR
How does the creation of variations in a species promote survival?
Answer:
Colours of eyes and shape of external ears.
OR
Variations increases the adaptability of an organism to its changing environmental conditions.

Question 12.
What is silvering of mirror?
Answer:
Silvering of mirror means coating the surface of mirror with a thin layer of silver, aluminium or some other shiny, opaque material.

Question 13.
Write the two raw materials for making food, used by living organisms of first tropic level.
Answer:
CO₂ and Water

Assertion (A) and Reason (R)
For question numbers 14,15 and 16, two statements are given- one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (i), (ii), (iii) and (iv) as given below:

(i) Both A and R are true, and R is correct explanation of the assertion.
(ii) Both A and R are true, but R is not the correct explanation of the assertion.
(iii) A is true, but R is false.
(iv) A is false, but R is true.

Question 14.
A. Magnetic field lines have both direction and magnitude.
R. The field lines, emerge from the north pole and merge at south pole
OR
A. Compass is a small magnet and gives direction of magnetic field lines.
R. It gets deflected when brought near a bar magnet.
Answer:
(ii) OR (i)

Question 15.
A. Copper does not displace hydrogen from dilute acids.
R. Copper is less reactive than hydrogen.
Answer:
(i)

Question 16.
A. When two resistances of 4 Ω are connected in parallel total resistance is 2 Ω.
R. When two resistances of 4 Ω are connected in series total resistance is 8 Ω.
Answer:
(ii)

Answer Q. No 17 – 20 contain five sub-parts each. You are expected to answer any four sub­parts in these questions.

Question 17.
Read the following and answer any four questions from 17 (i) to 17 (v) (4 × 1 = 4)
The table below shows the results of different experiment conducted on two acids X and Y of equal concentration:

	X	Y
Conductivity of Solution	400 S m1	7.0 S m1
Temperature change when excess acid reacted with 100 mL of 1.0 M NaOH solution	6.9°C	6.7°C
Reaction with Mg	Hydrogen gas is produced rapidly with rapid effervescence	Hydrogen gas is produced slowly with gradual effervescence

(i) Which of the following statement is correct?
(a) X is stronger acid because it has higher conductivity.
(b) Y is stronger acid because hydrogen is produced slowly when reacts with Mg.
(c) X is weaker acid because temperature change is higher when reacts with NaOH.
(d) Y is weaker acid because is has higher conductivity.
Answer:
(a) X is stronger acid because it has higher conductivity.

(ii) Select the correct relation from the following:
(a) (H+) x < (H+)y
(b) (pH) x < (pH)Y
(c) (H+) x = (H+)Y
(d) (PH) y > 7
Answer:
(b) (pH) x < (pH)Y

(iii) What will be colour of universal indicator in ‘X’?
(a) Blue
(b) Orange
(c) Green
(d) Red
Answer:
(d) Red

(iv) How many moles of H2 gas will be liberated by ‘X’ and ‘Y’ during reaction with 1 mole of Mg?
(a) 1/2
(b) 1/4
(c) 2
(d) 1
Answer:
(d) 1

(v) The gas formed when X and Y reacts with a metal carbonate is:
(a) Hydrogen
(b) Hydrogen Sulphide
(c) Oxygen
(d) Carbon dioxide
Answer:
(d) Carbon dioxide

Question 18.
Read the following and answer any four questions from 18 (i) to 18 (v) (4 × 1 = 4)
Answer:
Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter.

Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

(i) Graphs between electric current and potential difference across two conductors A and B are shown in the figure. Which of the following conductor has more resistance?

(a) B
(b) A
(c) Both have equal resistance
(d) None of these
Answer:
(b) A

(ii) Seven identical lamps of resistance 220 Ω each are connected to a 220 V line as shown in figure. Then reading of the ammeter will be:

(a) \(\frac{1}{10} \mathrm{~A}\)
(b) \(\frac{2}{5} \mathrm{~A}\)
(c) \(\frac{3}{10} \mathrm{~A}\)
(d) None of there
Answer:
(d) None of there

(iii) For metallic conductor voltage uses current graph is shown at two different temperatures T₁ and T₂. From the graph it follows:

(a) T₁ = T₂
(b) T₁ > T₂
(c) T₁ < T₂
(d) None of above
Answer:
(c) T₁ < T₂

(iv) If the resistance of wire A is four times resistance of wire B then the ratio of cross sectional areas of wires is
(a) 1:2
(b) 1:4
(c) 1:8
(d) 1:6
Answer:
(b) 1:4

(v) Ohm’s law is valid only when
(a) Temperature increases
(b) Temperature decreases
(c) Graph between V and I is a straight line
(d) Pressure remains constant.
Answer:
(c) Graph between V and I is a straight line

Question 19.
Read the following and answer any four questions from 19 (i) to 19 (v) (4 × 1 = 4)
Answer:
A student took HC1 in a conical flask and was placed on a white tile with cross mark. When seen through the flask the mark was visible. On adding some sodium thiosulphate in the flask the cross marked disappeared when seen through the flask. The student observed some reaction as

(i) Name the type of reaction seen in the setup.
(a) Redox reaction
(b) Precipitation reaction
(c) Displacement reaction
(d) Double displacement reaction
Answer:
(b) Precipitation reaction

(ii) Why did the cross mark disappear?
(a) The insoluble precipitate that is formed in the flask does not allow the light to pass through it to see the cross mark.
(b) The resultant solution is dark coloured which does not allow the light to pass through it.
(c) The resultant solution has high viscosity due to which it refracts the light to other angle.
(d) The solution so formed has ability to absorb black colour hence black coloured cross is not seen.
Answer:
(a) The insoluble precipitate that is formed in the flask does not allow the light to pass through it to see the cross mark.

(iii) Give the chemical formula of the gas formed in the reaction.
(a) SO₂
(b) SO₃
(c) O₂
(d) Cl₂
Answer:
(a) SO₂

(iv) Name the resultant insoluble substance formed in the flask.
(a) Sodium sulphide
(b) Sodium sulphate
(c) Sodium sulphite
(d) Sulphur
Answer:
(d) Sulphur

(v) Which of the following reaction is of similar type as observed in the given set up.

Answer:
(a)

Question 20.
Read the following and answer any four questions from 20 (i) to 20 (v) (4 × 1 = 4)
Answer:
Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow, without doing any calculations:

S. No.	Object distance u (cm)	Image distance v (cm)
1	-90	+ 18
2	-60	+ 20
3	-30	+ 30
4	-20	+ 60
5	-18	+ 90
6	-10	+ 100

(i) What is the focal length of the convex lens? Give reason in support of your answer.
(a) -10
(b) +10
(c) -15
(d) +15

(ii) Write the serial number of that observation which is not correct.
(a) S.No. 1
(b) S.No. 4
(c) S.No. 3
(d) S.No. 6

(iii) The approximate value of magnification in case of S.No. 4 is
(a) -2
(b) -3
(c) +3
(d) +2

(iv) The image formed in case of S.No. 2 is
(a) virtual and enlarged
(b) virtual and diminished
(c) real and diminished
(d) real and enlarged

(v) If a convex lens is used to focus sunlight on a paper, where the paper should be placed so that it catches fire.
(a) At centre of curvature
(b) At principal focus.
(c) At optical centre of lens
(d) At 25 cm away from lens
Answer:
(i) (d) From S.No.3 we can say that the radius of curvature of the lens is 30 cm because when an object is placed at the centre of curvature of a convex lens its image is formed on the other side of the lens at the same distance from the lens. And, we also know that focal length is half of the radius of curvature. Thus, focal length of the lens is + 15 cm.

(ii) (d) S.No.6 is not correct as the object distance is between focus and pole, so for such cases the image formed is always virtual but in this case it is written that a real image is formed as the image distance is positive.

(iii) \(\text { (b) } m=\frac{v}{u}=\frac{+60}{-20}=-3\)

(iv) (c)

(v) (b)

Section-B

Question 21.
A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 18 cm at a distance of 12 cm from it. Use lens formula to determine the position, size and nature of image formed.

The image formed is on the same side as the object, at a distance of 36 cm from the optical centre. The -ve sign shows that the image is virtual.
\(m=\frac{h_{i}}{h_{o}}=\frac{v}{u}=\frac{-36}{-12}=3\)
h₀ – 5 cm, then h_i = 3 x 5 = 15 cm The image is three times larger than the size of the object.

Question 22.
How is bleaching powder prepared? Give its chemical equation and write its two uses.
OR
A white coloured powder is used by doctors for supporting fractured bones.
(i) Write chemical name and formula of the powder
(ii) When this white powder is mixed with water a hard solid mass is obtained. Write balanced chemical equation for the change.
Answer:
Preparation: On passing chlorine gas through a test tube containing dry slaked lime Ca(OH)₂, bleaching powder is prepared.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O
Uses:
(i) It is used as. bleaching agent in textile industry and paper factories.
(ii) It is used as an oxidising agent in many industries.
OR
(i) Calcium sulphate hemihydrate, \(\mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}\)
(ii) \(\mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}+\frac{3}{2} \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Question 23.
(i) Choose the amphoteric oxides from the following oxides.
Na₂O, ZnO, Al₂O₃, CO₂, H₂O.

(ii) Why is it that non-metals do not displace hydrogen from dilute acids?
Answer:
(i) Al₂O₃, H₂O and ZnO are amphoteric oxides.
(ii) Non-metals cannot loose electrons so that H⁺ ions from acids become hydrogen gas and thus, cannot be displaced.

Question 24.
Find the current drawn from the battery by the network of four resistors shown in the figure.
Answer:

Equivalent resistance of the given network is

Question 25.
An element ‘E’ has following electronic configuration:

(i) To which group of the periodic table does element ‘E’ belong?
(ii) To which period of the periodic table does element ‘E’ belong?
(iii) State the number of valence electrons present in element ‘E’
(iv) State the valency of the element ‘E’
Answer:
(i) ‘E’ belongs to group 16
(ii) It belongs to 3rd period
(iii) It has 6 valence electrons
(iv) Its valency is equal to 2

Question 26.
Why is respiration considered as exothermic reaction? Explain.
OR
(i) Hydrogen being a highly inflammable gas and oxygen being a supporter of combustion, yet water which is a compound made up of hydrogen and oxygen is used to extinguish fire. Why?

(ii) What change in colour is observed when silver chloride is left exposed to sunlight? What type of chemical reaction is this?
Answer:
In respiration, glucose gets oxidized to form carbon dioxide, water and heat is evolved.
As heat energy is released during respiration it is regarded as an exothermic reaction.
OR
(i) It is because properties of compound (H₂O) are different from properties of its constituting elements, i.e., H₂ and O₂.
(ii) Silver turns black, the reaction is decomposition.

Section – C

Question 27.
Leaves of healthy potted plants were coated with vaseline to block the stomata. Will this plant remain healthy for long? State 3 reasons for your answer.
Answer:
No, this plant will not stay healthy for long. The plant will start dying because:

• Stomata gets blocked and no exchange of gases takes place.
• No intake of CO₂ for photosynthesis and O₂ for respiration.
• No transpiration and hence water transportation will be affected.

Question 28.
Describe briefly the three ways in which individual with a particular trait may increase in population.
Answer:

• Variation helps in survival of individuals.
• Sexual reproduction results in variation.
• Adaptation and natural selection.

Question 29.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel conducted a monohybrid cross with pea plants, and he observed that one of the contrasting characters disappears in F₁ generation. This character reappears in F₂ generation (obtained by selfing F₁) in just 25% of the progeny.

Mendel conclude that the character which expresses itself in F₁ is the dominant character while the other one which is not able to express in F₁ individuals is recessive character. This recessive character is able to express only in its pure form i.e. in 25% of F₂ individuals.

Question 30.
Study the given flow chart and answer the below given questions.

(i)
(a) Which form of the Sun’s energy is trapped by the producer?
(b) Into which energy form is the Sun’s energy converted when it is trapped by the producer?

(ii)
(a) Calculate X in the flow chart.
(b) Calculate Y on the flow chart.
(c) Label the above box on flow chart.
Answer:
(i)
(a) Light (or solar) energy
(b) Chemical energy

(ii)
(a) 1200 units
(b) 48 units
(c) Decomposers like bacteria or fungi.

Question 31.
Samples of four metals A,B, C and D were taken and added to the following solutions one by one. The results obtained have been tabulated as follows.

Metal	Iron (II) sulphate	Copper (II) sulphate	Zinc sulphate	Silver nitrate
A	No reaction	Displacement	—	—
B	Displacement	—	No reaction	—
C	No reaction	No reaction	No reaction	Displacement
D	No reaction	No reaction	No reaction	No reaction

Use the table above to answer the following questions about metals A, B, C and D.
(i) Which is the most reactive metal?
(ii) What would you observe if B is added to a solution of copper (II) sulphate?
(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.
Answer:
(i) B is most reactive.
(ii) Blue colour of copper sulphate solution disappears and reddish brown copper metal is deposited on the metal B.
(iii) B > A > C > D is the order of reactivity.

Question 32.
Foetus derives its nutrition from the mother.
(i) Identify the tissue used for above purpose. Explain its structure.
(ii) Explain how wastes generated by developing embryo are removed.
(iii) How does the birth of child take place?
Answer:
(i) Placenta is a special tissue connection between embryo and uterine wall. It acts as an endocrine gland. It possesses villi that increases the surface area for absorption of nutrients. Facilitates passage of nutrition and oxygen to embryo from mother through blood.

(ii) Waste substances produced by embryo are removed through placenta into mother’s blood.

(iii) Birth of child takes place after the gestation period (9 months inside the womb of mother). When the pain in the uterine walls is at highest then the birth of child takes place by the birth canal.

Question 33.
(i) Show the formation of Na₂O by transfer of electrons between the combining atoms.
(ii) Why are ionic compounds usually hard?
(iii) How is it that ionic compounds in the solid state do not conduct electricity, but they do so when in molten state?
OR
(i) Which product is formed in the chemical reaction between a small strip of magnesium and nitric acid?
(ii) What happens when a pellet of sodium is dropped in water?
(iii) Identify the ‘X’ in the given reaction: P₄(s) + SO₂(g) → X
Answer:

(ii) Ionic compounds are hard due to strong force of attraction between oppositely charged i0ns.

(iii) In solid-state, ions are not free to move, whereas in molten state ions are free to move. Therefore, they conduct electricity in a molten state.
OR
(i) Mg(NO₃)₂ and 2H₂
(ii) It catches fire and forms sodium hydroxide
(iii) P₄O₁₀(s)

Section-D

Question 34.
(i) Write the functions of the following parts in human female reproductive system:
(a) Ovary
(b) Oviduct
(c) Uterus

(ii) Describe the structure and function of placenta.
OR
What is binary fission in organisms? With the help of suitable diagrams, describe the mode of reproduction in amoeba.
Answer:
(i) (a) Ovary –
• Production of female hormone
• Production of female gamete

(b) Oviduct –
• Transfer of female gamete from the ovary
• Site of fertilization

(c) Uterus –
• Implantation of the zygote
• Nourishment of the developing embryo/placenta formation.

(ii) Structure of Placenta: It is a disc-like structure embedded in the uterine wall, connected to the embryo. It has villi on the embryo’s side of the tissue and on the mother side it has blood spaces, which surround the villi.

(iii) (a) Circuit diagram

V	I = V/R	V/I
0.8 V	0.8/2.67 = 0.30	8/3
1.2 V	1.2/2.67 = 0.45	8/3
1.6 V	1.6/2.67 = 0.60	8/3

Conclusion: ‘R’ is constant at given temperature. VI graph shows straight line.

Question 36.
(i) A thin converging lens forms a
(a) Real magnified image.
(b) Virtual magnified image of an object place in front of it.
Write the positions of the object in each case.
(ii) Draw a labelled ray diagram to show the image formation in each case.
(iii) How will the following be affected on cutting this lens into two halves along the principal axis?
(a) Focal length
(b) Intensity of image formed by half lens.
OR
Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow, without doing any calculations:

S.No.	Object distance u (cm)	Image distance v (cm)
1	-90	+ 18
2	-60	+ 20
3	-30	+ 30
4	-20	+ 60
5	– 18	+ 90
6	– 10	+ 100

(i) What is the focal length of the convex lens? Give reason in support of your answer.
(ii) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?
(iii) Take an appropriate scale to draw a ray diagram for the observation at S. No. 4 and find the approximate value of magnification.
Answer:
(i) (a) When the object is placed between ‘F’ and ‘2F’
(b) When the object is placed between pole ‘P’ and ‘F’

(ii)

(iii) No, focal length will remain the same because radius of curvature remains the same. Therefore, intensity of the image will decrease.
OR

(i)  From S.No.3 we can say that the radius of curvature of the lens is 30 cm because when an object is placed at the centre of curvature of a convex lens its image is formed on the other side of the lens at the same distance from the lens. And, we also know that focal length is half of the radius of curvature. Thus, focal length of the lens is + 15 cm.

(ii) S.No.6 is not correct as the object distance is between focus and pole, so for such cases the image formed is always virtual but in this case it is written that a real image is formed as the image distance is positive which is not possible.

(ii) Approximate value of magnification for object distance – 20 cm and image distance + 60 cm is 3.

Solved CBSE Sample Papers for Class 11 2021-2022 Pdf with Solutions: Solving Pre Board CBSE Sample Papers for Class 11 with Solutions Answers Pdf Download Term 1 & Term 2 will give students an idea about the question paper pattern, frequently asked questions, and on which topics to emphasize more in a subject. CBSE Sample Papers of Class 11th Term 1 & Term 2 will boost their confidence level, and they will feel optimistic about the board exams.

CBSE Sample Papers Class 11 2022 with Solutions Term 1 & Term 2

• CBSE Sample Papers for Class 11 Maths
• CBSE Sample Papers for Class 11 Applied Mathematics
• CBSE Sample Papers for Class 11 Physics
• CBSE Sample Papers for Class 11 Chemistry
• CBSE Sample Papers for Class 11 Biology
• CBSE Sample Papers for Class 11 English Core
• CBSE Sample Papers for Class 11 Hindi
• CBSE Sample Papers for Class 11 Geography
• CBSE Sample Papers for Class 11 History
• CBSE Sample Papers for Class 11 Political Science
• CBSE Sample Papers for Class 11 Economics
• CBSE Sample Papers for Class 11 Accountancy
• CBSE Sample Papers for Class 11 Business Studies
• CBSE Sample Papers for Class 11 Computer Science

We hope these Solved Sample Paper of Class 11 CBSE 2021 2022 Pdf with Solutions Term 1 & Term 2 will help in self-evaluation. Stay tuned for further updates on CBSE Class 11th Sample Papers Term 1 & Term 2 for their exam preparation.

Solved Sample Paper of Social Science SST for Class 10 CBSE with Answers 2021-2022 Term 1 & Term 2: Solving Pre Board CBSE Sample Papers for Class 10 Social Science with Solutions Answers 2021-2022 Pdf Download Term 1 & Term 2 will give students an idea about the question paper pattern, frequently asked questions, and on which topics to emphasize more in a subject. CBSE SST Sample Paper Class 10 2022 with Solutions Term 1 & Term 2 will boost their confidence level, and they will feel optimistic about the board exams.

CBSE SST Sample Paper Class 10 2021-2022 with Solutions | SST Sample Paper Class 10 Term 1 & Term 2

Download Social Science Class 10 Sample Paper and practice for your 2021-22 board exams. Sample Paper of Social Science Class 10 is prepared according to the CBSE Examination pattern and syllabus with each answer according to CBSE marking scheme. According to the new CBSE Exam Pattern, MCQ Questions for Class 10 Social Science Carries 20 Marks. Click on the link below to access the Solved Sample Paper of SST Class 10.

SST Sample Papers Class 10 with Solutions 2021-22 Term 2

• CBSE Sample Papers for Class 10 Social Science Term 2 Set 1
• CBSE Sample Papers for Class 10 Social Science Term 2 Set 2
• CBSE Sample Papers for Class 10 Social Science Term 2 Set 3
• CBSE Sample Papers for Class 10 Social Science Term 2 Set 4

SST Sample Papers Class 10 with Solutions (Old Pattern)

• SST Sample Paper Class 10 2020 with Solutions Set 1 (New 2020-21)
• Class 10 SST Sample Paper Set 2 (New 2020-21)
• Social Science Class 10 Sample Paper Set 3 (New 2020-21)
• Sample Paper of Social Science Class 10 Set 4 for Practice (New 2020-21)
• Sample Paper SST Class 10 Set 5 for Practice (New 2020-21)

CBSE Class 10 Social Science Sample Question Paper Design 2022

Section	VSA* (2 Marks)	SA (3 Marks)	LA (5 Marks)	Case-based (4 Marks)	Map (3 Marks)
A	5Q	–	–	–	–
B	–	3Q (1 Internal Choice)	–	–	–
C	–	–	2Q (2 Internal Choices)	–	–
D	–	–	–	2Q	–
E	–	–	–	–	1Q (1 Internal Choice)
Total	10M	9M	10M	8M	3M

Social Science SST Sample Paper Class 10 Question Paper Design 2020-21

Typology of Questions	Total marks	% Weightage
1. Remembering and Understanding: Exhibiting memory of previously learned material by recalling facts, terms, basic concepts, and answers; Demonstrating an understanding of facts and ideas by organizing, translating, interpreting, giving descriptions, and stating main ideas.	28	35%
2. Applying: Solving problems to new situations by applying acquired knowledge, facts, techniques, and rules in a different way.	15	18.75%
3. Formulating, Analysing, Evaluating, and Creating: Examining and breaking information into parts by identifying motives or causes; making inferences and finding evidence to support generalizations; Presenting and defending opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria; Compiling information together in a different way by combining elements in a new pattern or proposing alternative solutions.	32	40%
4. Map Skill	5	6.25%
Total	80	100

CBSE Sample Paper of Class 10 Social Science

• CBSE SST Sample Paper Class 10 Paper 1
• SST Class 10 Sample Paper Paper 2
• Sample Paper Class 10 SST Paper 3
• Sample Paper SST Class 10 Paper 4
• Social Science Sample Paper Class 10 Paper 5
• Sample Paper of Social Science Class 10 Paper 6
• Sample Paper Social Science Class 10 Paper 7

We hope these Sample Papers of Social Science for Class 10 CBSE with Answers 2021-2022 Term 1 & Term 2 will help in self-evaluation. Stay tuned for further updates on NCERT CBSE SST Sample Paper Class 10 2022 with Solutions Term 1 & Term 2 for their exam preparation.

Students can access the CBSE Sample Papers for Class 11 English Education with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 11 English Term 2 Set 1 with Solutions

Time : 2 Hours
Maximum Marks : 40

General Instructions :

• The Question Paper contains THREE sections-READING, WRITING & GRAMMAR and LITERATURE.
• Attempt questions based on specific instructions for each part.

Section – A
Reading (13 Marks)

Question 1.
Read the passage given below :

We live in an age of wonders and miracles. It has been called the ‘Age of Science’, and different aspects of our life that changed in the preceding centuries have been attributed to science. This is completely true, but it is only one side of the coin. The flip side is that as we have advanced more in the field of technology, values such as empathy and concern for our fellow human beings are gradually being eroded.

Take the example of the Internet. On one hand, access to information and knowledge at the click of a button is a veritable boon to everyone (especially students) and this has made our lives much simpler. On the other hand, it has severely limited actual contact with teachers, friends and elders. Thus, the learning is incomplete because it lacks personal advice that a person gives on the basis of their knowledge and practical experiences.

Today, a small child can access and navigate the Internet with ease, but at the same time miss out on the basic human traits of friendship, relationships and family values. Owing to excessive technology, children often miss out on the most enriching childhood experiences, such as playing outdoor games with friends, which apart from being immensely enjoyable and physically exhilarating, also develop traits such as teamwork and discipline at an early age.

As friendship is gradually being limited to virtual friends, it has harsh implications on the personality development of a child because the time spent with computers or mobile phones can never substitute for the holistic benefits of outdoor play. As for me, I think this age of rapid development has created at least as many problems as it has solved, if not more.

The reason is that the basic goal of life, which should be the pursuit of happiness has now been replaced by the pursuit of money. Money and happiness are considered analogous in our present society, but they actually are not so. In this mindless pursuit of money, nobody has time now to appreciate the beauty of life, which consists not of multi-billion dollar skyscrapers, but a simple act of kindness to someone at the time of need.

So, there is an urgent need to stop for a moment and think about where we are actually heading- is it development or is it destruction? Do we have to wait until people have grown so much apart from each other that we cannot see the suffering of our own species due to our mindless greed, or can we still mend our ways?

Based on your understanding of the passage, answer ANY EIGHT questions from the nine given below. (1 x 8 = 8)

(i) Give one point of evidence from the text that proves that it is like one side of the coin to say that life has changed owing to science.
Answer:
Different aspects of life have changed due to science. For example, values such as empathy and concern for our fellow human beings are gradually being eroded. So, it is only one side of the coin to say that life has changed owing to science.

(ii) State any one advantage of the internet mentioned in lines 8-10 in the passage.
Answer:
The Internet has enabled access to information and knowledge very easily.

(iii) What will be the consequence of our actions if we do not change our ways?
Answer:
If we do not change our ways, human beings will not feel each other’s suffering.

(iv) How does the internet hamper the holistic growth of a child?
Answer:
Because of excess use of technology, children miss childhood experiences like outdoor games. These activities develop traits such as teamwork and discipline at an early age. This hampers the holistic growth of the child.

(v) Rewrite the sentence replacing the underlined words with a word or phrase from lines 30 to 35 in the passage.

The relationship between a teacher and a student is similar in essence to that of a afatlter and a child.
Answer:
The relationship between a teacher and a student is analogous to that of a father and a child.

(vi) Why has the writer used the word ‘holistic’ to describe the benefits of outdoor play?
Answer:
Holistic means complete or whole. Outdoor games are enjoyable and also teach lessons in teamwork and discipline. They also provide physical exercise. As such, they are holistic.

(vii) Use a suitable phrase from lines 40-45 of the passage to complete this sentence.

The thief decided to after he listened to the sage’s advice on the fickleness of life.
Answer:
The thief decided to mend his ways after he listened to the sage’s advice on the fickleness of life.

(viii) True happiness lies in simple acts of kindness. What acts of kindness are expected from a school-going child?
Answer:
School-going children should be caring and kind. They should avoid being rude to others and help weak students in their studies. They should help old people and little children as much as they can.

(ix) What should be the basic goal of one’s life?
Answer:
The basic goal of one’s life is the pursuit of happiness.

Question 2.
Read the following passage. (5)

(1) Humour is infectious. The sound of roaring laughter is far more contagious than any cough, sniffle, or sneeze. When laughter is shared, it binds people together and increases happiness and intimacy. Best of all, this priceless medicine is fun, free, and easy to use.

(2) The ability to laugh, play, and have fun with others not only makes life more enjoyable but also helps you to solve problems. People who incorporate humour and play into their daily lives find that it renews them and all of their relationships.

(3) Life brings challenges that can either get the best of you or become playthings for your imagination. When you “become the problem” and take yourself too seriously, it can be hard to think out of the box and find new solutions. But when you play with the problem, you can often transform it into an opportunity for creative learning.

(4) Playing with problems seems to come naturally to children. When they are confused or afraid, they convert their problems into a game, giving them a sense of control and an opportunity to experiment with new solutions. Interacting with others in playful ways helps you retain this creative ability.

(5) As laughter, humour, and play become an integral part of your life, your creativity will flourish and new discoveries will occur to you daily. Humour takes you to a higher place where you can view the world from a more relaxed, positive, creative, joyful, and balanced perspective.

(A) On the basis of your reading the passage, make notes using recognisable abbreviations wherever necessary. Use a format you consider suitable, supply a suitable title. (3)
Answer:
(A) Laughter- A Priceless Medicine
1. Physical changes in body:
1.1. strengthens immne system
1.2. boosts energy , .
1.3. dimnshs pain
1.4. protects from stress.

2. Benefits of laughter.
2.1. makes life more enjyble
2.2. helps solve problems
2.3. connects with others

3. Laughter – An Integral Part of life
3.1. makes one creative
3.2. helps to balance prspctive
3.3. makes one more relaxed & positive

Key to Abbreviations
S.No.	Abbreviation	Word
1.	immne	immune
2.	dimnshs	diminishes
3.	enjyble	enjoyable
4.	prspetive	perspective
5.	&	and

(B) Make a short summary of the passage in about 50 words. (2)
Answer:
SUMMARY
Laughter is free, yet a priceless medicine. It makes the body healthy. Humour helps build better relationships. Problems are best solved by developing a child-like attitude of playing with them rather than brooding over them. Laughter should be an integral part of life to make it relaxed, positive and balanced.

Section – B
Writing and Grammar (12 Marks)

Question 3.
As President of the Residents Welfare Association of Mayur Colony, Delhi, design a poster in not more than 50 words for promoting cleanliness in the surroundings of your colony. (3)
Answer:
Commonly Made Error
Students tend to add unnecessary details to meet the word limit.

Answering Tip
Relevant details should be added. It should be such that the poster conveys the idea clearly.

Question 4.
Attempt ANY ONE from (A) and (B) given below.
(A) You are Meena Saxena of Kanpur. You have come across an advertisement of a coaching centre that prepares students for Pre Medical tests. Write a letter to the Director of the institute asking for information that you require before you decide to join it. (5)

Sky-High Institute Excellent Coaching Centre 25, High Street, Kanpur Are You Aspiring To Be A Doctor? We Are Here To Help You! Batches Starting From February Register Now! Contact: Tel: 23417689

Answer:

Note: No marks are to be awarded if only the format is given. Credit should be given for the candidate’s creativity in the presentation of ideas. Format – 01 Mark Sender’s address Date Receiver’s address Subject/heading Salutations Content – 03 Marks Expression – 02 Marks Letter to The Director, Sky-High Institute – Intent of writing the letter – Enquiry about the courses

7 Greater Colony,
Kanpur
21st January, 20XX
The Director
Sky-High Institute
25, High Street
Kanpur

Subject: Enquiry Regarding Coaching Classes.

Sir,
This is with reference to your advertisement in “The Daily Times’ dated 18th January, 20XX. I am a student of Class XII and am preparing for pre-medical test. I am interested in joining your esteemed institute, but before I do so, I would like to have the following information.

I would like to know the duration and the timings of the course. What is the fee structure and the mode of payment? I would also like to know the strength of each batch and the success rate of your institute. Do you offer any discount to students who have excelled in academics?

I would be grateful if you could give me a prompt response so that I can take timely decision to join your institute.

Thanking you

Yours sincerely,
Meena Saxena

(B) “Private cars should be banned in the congested commercial areas of the cities.” Write a debate in 120-150 words either for or against the motion. (5)
Answer:

Private Cars should be banned

Respected Judges and teachers,
Today, I stand before you to present my views for the motion on the topic, ‘Private cars should be banned in the congested commercial areas of the cities’. There is no end to the number of vehicles being driven around the city each day. Traffic jams have become a rather common problem and have only increased in the last ten years.

In my opinion, allowing private cars in congested commercial areas only adds to the problem. Long traffic jams do not only mean consumption of extra fuel but also more number of people suffering from respiratory diseases due to pollution. I strongly feel if people take public transport to work or hire a cab or car-pool and drive vehicles which run on CNG, the problem of constantly rising pollution and heavy traffic jams can be reduced drastically.

The Earth belongs to all of us and it is our duty to safeguard it and keep it clean for our future generations. Thus, first step towards protecting our environment would be to stop abusing the available amount of petroleum and avoid creating pollution.

Thank you for listening to my views patiently.

Question 5.
A. Complete the following sentences with an appropriate word. (1 x 4 = 4)
(i) Take as ____________ stamps as you need.
(ii) I ____________ (be) to the doctor. (Fill in the blank using the present perfect tense.)
Answer:
(i) many
Explanation: Since ‘stamps’ are countable, many will be used.

(ii) I have been to the doctor.
Explanation: In Present Perfect Tense has been have been is used. Here the subject is T. So, ‘have been1 will be used.

B. Rearrange the given words to make meaningful sentences,
(i) had/winning a medal/he/my/I told him/until/about
(ii) on/my bicycle/to/go/I used to/my school
Answer:

• He had not known about my winning a medal until I told him.
• I used to go to my school on my bicycle.

Section C
Literature (15 Marks)

Question 6.
Answer the following questions within 30-40 words each. (2 x 3 = 6)
(i) How can overfishing and decimation of forests prove harmful?
(ii) How does the rain justify its claim: ‘I am the Poem of Earth’?
(iii) What according to the poem, ’Childhood’, is involved in the process of growing up?
Answer:
(i) Overfishing may lead to stripping of fisheries. Then man will lose a rich source of protein. The decimation of forests will harm ecology. Moreover, several species of life that live in forest will face extinction.

(ii) The poem ‘The Voice of the Rain/describes the cycle of rain. It shows how the rain rises out of the sea and is transformed into rain drops in the heavens. Thus, it is the poem of the earth.

(iii) According to the poem, ‘Childhood’, the loss of childhood is involved in the process of growing up. This loss is compensated by some gains which come with adolescence.

Question 7.
Answer the following questions within 70-80 words each. (3 x 3 = 9)
(i) Who was Joe Morgan? Why had he been waiting for Dr. Andrew Manson?
OR
How did Einstein’s medical certificate prove useless?
(ii) Mothers should be respected not ignored. Examine the problem confronted by Mrs. Pearson under the light of this statement. Who helps her solve this problem and how?
(iii) Give a brief account of the author’s visit to the medical college at Darchen and the effect of the Tibetan medicines on him.
Answer:
(i) Joe Morgan was a driller in Blaenelly, a mining town. He was a big, strong and heavy middle-aged person. Joe and his wife Susan, who had been married for nearly twenty years and were expecting their first child. Joe was waiting for the doctor to help Susan in the delivery of the child.
OR
Einstein managed to get a medical certificate with the help of Yuri but before he could submit it to the head teacher, he was called by the teacher and was expelled from the school. The school authorities were of the opinion that he was turning rebellious. So the medical certificate proved useless.

(ii) Mrs. Pearson’s problem is that she is neglected by her husband and children. She is very fond of her family. She runs after them all the time, takes their orders as if she is their servant. She stays at home every night while they go out enjoying themselves. She takes no holiday.

They have come to believe that she is there simply to look after them and wait for them. So they take, no notice of her. Her neighbour, Mrs. Fitzgerald, helps her solve this problem. She advises Mrs. Pearson to assert her rights as the mistress of the house if she wants them to treat her properly.

(iii) At first, the author was impressed neither by the building of the medical college nor the Tibetan doctor. The building looked like a monastery and the doctor himself appeared like any other Tibetan in a thick-pullover and wooly hat. He had no white coat.

The author, then, explained his sleepless symptoms and sudden aversion to lying down. The Tibetan doctor diagnosed his illness as a cold and effect of the altitude. So, he gave him a five-day course of Tibetan medicines in fifteen screws of paper.

The medicines, which appeared to be quite ineffective, worked wonders and only after his first full day’s course, the author was able to sleep soundly at night. He woke up quite fresh the next day and recovered completely after the five-day course.

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 2 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B, and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks.
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the chokes in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

 SECTION – A
(Section A contains 3 questions of 2 marks each.)

Question 1.
How is a p-type semiconductor formed? Draw energy band of p-type semiconductors and name the majority charge carriers in it. (2)
Answer:
1. If trivalent impurity atoms of B, Al, or In are doped in a pure semiconductor of Silicon or Germanium, we get a p-type semiconductor. Holes are the majority charge carriers in a p-type semiconductor. The diagram of energy band of p-type semiconductors is given below.

Related Theory
There are two types of dopants used for doping the tetravaLent Silicon or Germanium.
1. Pentavalent dopants which have five valence electrons like Arsenic, Antimony, and Phosphorus.
2. Trivalent dopants have three valence electrons like Indium, Boron, and Aluminium.

Question 2.
Photons, with a continuous range of frequencies, are made to pass through a sample of rarefied hydrogen.
Transitions of electrons are shown in figure:

(A) Identify the spectral series, of the hydrogen emission spectrum corresponding to these lines.
(B) Which of these lines corresponding to absorption of radiation of maximum wavelength?
OR
Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions W₂ and W₂ (W₁ > W₂).
On what factors does the (i) slope and (ii) intercept of the lines depend? (2)
Answer:
(A) Transition I → Lyman series
Transition II → Balmer series
Transition III → Paschen series of spectral absorption lines in the continuous spectrum.

(B) Energy of electron in hydrogen atom
E_n = \(-\frac{13.6}{n^{2}}\) eV
E₁ = -13.6 eV
E₂ = – 3. 4 eV and similarly we can find E₃, E₄, ……………..
For transition III energy absorbed is:
ΔE = E₄ – E₃ = 0.66 eV

which is least out of three transitions. Transition III corresponds to the absorption of radiation of maximum wavelength.
OR
The graph of stopping potential Vs and frequency (v) for two metals 1 and 2 is shown in fig.

(A) Slope of graph tan θ= \(\frac{h}{e}\) and depends on h and e.
(B) Intersect of lines depend on the work function.

Question 3.
Distinguish between the metals and insulators on the basis of energy band theory. Why does a diamond behave as an insulator? (2)
Answer:
For metals, the valence band is completely filled and the conduction band have the possibilities,
either it is partially filled within extremely small energy gap between valence band and conduction band or it is empty, with two bands overlapping each other.

For insulators, the energy gap between the conduction and valence band are very large, also the conduction band is partially empty. When an electric field is applied across such a solid, the electrons find it difficult to acquire sufficient energy to jump to conduction band. There is a Large forbidden band of 6eV in diamond. It is difficult to excite the electrons from valence band to the conduction band. Due to the absence of free charges, diamond behaves as an insulator.

SECTION – B
(Section B consists of 8 questions of 3 marks each.)

Question 4.
(A) State Bohr’s postuLate to define stable orbits in hydrogen atoms. How does Louis de BrogUe’s hypothesis explain the stability of these orbits?
(B) A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 LeveL Estimate the frequency of the photon. (3)
Answer:
(A) Bohr’s postulate for stable orbits in hydrogen atom: An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of \(\frac{h}{2 \pi}\). where h is Planck’s constant.

If n is the principal quantum number of orbit, then an electron can revolve only in a certain orbits of definite radii. These are called stable orbits. Louis de Broglie’s explanation of stabiLity of orbits:

According to de Broglie. orbiting electron around the nucleus is associated with a stationary wave. Electron wave is a circular standing wave. Since destructive interference will occur if the standing wave does not close upon itself only those Louis de Brog lie waves exist for which the circumference of circular orbit Contains a whole number of wavelengths i.e., for orbit circumference of nth orbit as 2nπr_n
2πr_n = nλ
2πr_n = n \(\left(\frac{h}{m v}\right)\)
mvr_n = n \(\left(\frac{h}{2 \pi}\right)\)
which is the quantum condition proposed by Bohr.

(B) In ground state, n = 1
In excited state, n = 4
\(\frac{1}{\lambda}=R\left[\frac{1}{(1)^{2}}-\frac{1}{(4)^{2}}\right]\) where R is Rydberg constant
\(\frac{1}{\lambda}\) = 1.097 x 10⁷ x \(\frac{15}{16}\)
= 1.028 x 10⁷ m^-1
Frequency, v = \(\frac{c}{\lambda}\) = 3 x 10⁸ x 1.028 x 10⁷
= 3.09 x 10¹⁵ Hz.

Question 5.
(A) When p-n junction Is said to be reverse-biased, what will be the direction of flow of majority charge carriers?
(B) Which semiconductors are preferred to make LED and why?
(C) Give two advantages of using LED. (3)
Answer:
(A) A p-n junctión ¡s set to be reverse biased when its p-region is maintained at lower potential with respect to its n-region. i.e, p-end is connected with negative terminal, and n-end is connected with positive terminal In a reverse-biased p-n junction.
majority charge carriers move away from the junction.
(B) MateriaLs Like gaLLium phosphide (Gap) Gallium arsenide (GaAs), etc., are used.
They emit the maximum amount of energy in form of tight radiation.
(C) The following are the advantages of LEDs:
(i) They are operational at two voltages.
(ii) They are quick in action and their power consumption is Low.

Question 6.
What is meant by the size of nucleus? A nucleus of mass number 225 splits into two fresh nucLei having radius R = 1.1 x 10^-15 A^1/3 m, find the radii of the new nuclei formed. (3)
Answer:
The extremely small central core of the atom in which whole of the positive charge and practically the entire mass is confined is known as nucleus. Different nuclei differ in size. The radius of nuclei is directly proportional to A^1/3.
where A is its mass number.
Here A = 225, R = 1.1 x 10^-15 A^1/3 m
Let A₁ and A₂ be the respective mass number of two new nuclei formed.
A₁= \(\frac{3}{(3+2)}\) A
= \(\frac{3}{5}\) x 225
= 135
And,
A₂ = \(\frac{3}{(3+2)}\) A
=\(\frac{2}{5}\) x 225 = 90

R₁= 1.1 x 10^-15 A^1/3
= 1.1 x 10^-15 x 135^1/3
= 5.643 x 10^-15 m

R₂ = 1.1 x 10^-15 A^1/3
= 1.1 x 10^-15 x 90^1/3
= 4.93 x 10^-15 m.

Question 7.
State Huygens’ principLe. Using this principle draw a diagram to show how a plane wavefront incident at the interface of the two media gets refracted when It propagates from a rarer to a denser medium. Hence, verify Snell’s Law of refraction. (3)
Answer:
(A) Huygens’ Principle:
(i) Each point on the primary wavefront acts as a source of secondary wavelets, transferring out a disturbance in all directions in the same way as the original source of Light does.
(ii) The new position of the wavefront at any instant is the envelope of the secondary wavelets at that instant.

(B) Refraction on the basis of wave theory:
(i) Consider any point Q on the incident wavefront
(ii) Suppose when disturbance from point P on incident wavefront reaches point P’ on the refracted wavefront, the disturbance from point Q reaches Q’ on the refracting surface XY.

(iii) Since P’A’ represents the refracted wavefront, the time taken by light to travel from o point on incident wavefront to the corresponding point on refracted wavefront should always be the same. Now, time taken by Light to go from Q to Q will be

t = \(\frac{Q K}{c}+\frac{Q K}{v}\) ………………………… (i)

In right-angled ΔAQK,
∠QAK = i
∴ QK=AK sin i …………………………………. (ii)
In right-angled ΔP’Q’K.
∠Q’P’K = r
and Q’K = KP’ sin r …………………………. (iii)
Substituting (ii) and (iii) in equation (i).

The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e., given equation (iv) is independent of AK. It will happen so. if
\(\frac{\sin i}{c}-\frac{\sin r}{v}\) = 0
⇒ \(\frac{\sin i}{\sin r}=\frac{c}{v}\)
⇒ n = \(\frac{\sin i}{\sin r}\)
This is the Snells low for refraction of light.

Retated Theory
Wave theory explained Lows of reflection and refraction both.

Question 8.
Draw a labelled ray diagram of an astronomical telescope in the near-point adjustment position. A giant refracting telescope at an observatory has an objective Lens of Focal Length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens.
The diameter of the Moon is 3.48 x 10⁶ m and the radius of lunar orbit is 3.8 x 10⁶ m.
OR
ExpLain the refraction of Light through a glass-slab with a neat ray diagram. (3)
Answer:
Astronomical teLescope in near points adjustment

Here, f₀ = 15 m,
f_e=1 cm = 0.1 m
Diameter of moon,
d = 3.48 x 10⁶ m

Radius of lunar orbit
μ₀ = 3.8 x 10⁸ m
size of the image of moon I = ?
The angle subtended by the moon at the objective lens,
α = \(\frac{d}{\left|\mu_{0}\right|}\) ………………………….. (i)
And the angle subtended by the image of the moon at the objective lens,
β = \(\frac{I}{f_{0}}\) ……………………… (ii)

Both these angles in equations (i) and (ii) wift be equal.
∴ \(\frac{d}{\left|\mu_{0}\right|}=\frac{1}{f_{0}}\)
⇒ I = \(\frac{d \times f_{0}}{\left|\mu_{0}\right|}\)
= \(\frac{3.48 \times 10^{6} \times 15}{3.8 \times 10^{8}}\) m
= 13.73 x 10^-2 m
= 13.73 cm.
OR

On entering into the glass medium-light ray bends towards the normal that is tight ray gets refracted on entering the gLass medium. After getting refracted this ray now travels through the glass slab and comes Out of the glass slab by refraction from the other interface boundary. Since ray goes from glass medium to air it again gets refracted and bends away from normal.

The incident ray and the emergent ray are parallel to each other. i₁ is the angLe of incidence, r₁ is the angLe of
refraction and r₂ is the angle of emergence. The angle of incidence and angle of emergence are equal as emergent ray and incident ray are parallel to each other. When a Lightray is incident normally to the interface of two media then there is no bending of tight ray and it goes straight through the medium.

Question 9.
In the study of a photoeLectric effect the graph between the stopping potential V and frequency y of the incident radiation on two different metals P and Q is shown below:
(A) Which one of the two metals has higher threshold frequency?
(B) Determine the work function of the metal which has greater value.
(C) Find the maximum kinetic energy of electron emitted by light of frequency 8 x 10¹⁴ Hz for this metal. (3)
Answer:
(A) Q has higher threshoLd frequency.
(B) Work function of Q
Φ_o = hv₀
hv0 = (6.6 x 10^-34) x \(\frac{6 \times 10^{14}}{1.6 \times 10^{-19}}\) eV
= 2.5 eV

(C) K_max = h(v – v₀)
= \(\frac{6.6 \times 10^{-34}\left(8 \times 10^{14}-6 \times 10^{14}\right)}{1.6 \times 10^{-19}} \mathrm{eV}\)
= \(\frac{6.6 \times 10^{-34} \times 2 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV}\)
K_max=0.83eV.

Question 10.
(A) Monochromatic Light of wavelength 589 nm is incident from air on a water surface. If μ for water is 1.33, find the wavelength, frequency, and speed of the refracted light.
(B) A doubLe convex lens is made of a glass of refractive index 1.55, with both faces of the same radius of curvature.
Find the radius of curvature required, if the focal length is 20 cm. (3)
Answer:
(A) Given: m= 1.33
λ_a= 589 nm
We know that,
μ = \(\frac{C}{V}\)
v = \(\frac{c}{\mu}\)
= \(\frac{3 \times 10^{8}}{1.33}\)
Speed, v = 2.26 x 10⁸ m/s

Frequency remains same.
∴ v = \(\frac{c}{\lambda_{a}}=\frac{3 \times 10^{8}}{589 \times 10^{-9}} \)
Frequency, v = 5.09 x 10¹⁴ Hz
Wavelength.
λ_w = \(\frac{\lambda_{a}}{\mu} \)
= \(\frac{589 n m}{1.33}\)
= 442.8 nm

(B) Given: μ= 1.55.
20 cm
We know that,

R = 20 x 0.55 x 2
R= 22 cm.

Question 11.
Which constituent radiation of the electromagnetic spectrum is used
(A) in radar
(B) to photograph internal part of a human body, and
(C) for taking photographs of the sky during night and foggy conditions?
Give one reason for your answer in each case.
OR
(A) When a convex Lens of Focal Length 30 cm is in contact with concave tens of focal length 20 cm, find out if the system is converging or diverging.
(B) Obtain the expression for the angle of incidence of a ray of light which is incident on the face of a prism of
refracting angle A so that it suffers total internal reflection at the other face. (Given the refractive index of the glass of the prism is μ.)
Answer:
(A) Microwaves are used .in the operation of radar because microwaves have small wavelengths and are not bent by objects coming in its path. Due to it, these waves can be sent in a particular direction as a beam signal.

(B) X-rays are used to photograph internal parts of a human body because X-rays are a quite short wavelengths. They can easily pass-through flesh but not bones.
(i) Infrared rays are used for taking photographs of the sky during night and foggy conditions because these
can easily pass-through fog and dark.
OR
(A) f₁= f_convex= 30 cm
f₂ = f_concave = – 20 cm
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
= \(\frac{1}{30}-\frac{1}{20} \)
= \(\frac{2-3}{60}\)
\(\frac{1}{f}=-\frac{1}{60}\)
[∴ concave (diverging) lens)]

(B) μ = \(\frac{\sin \left(\frac{\delta_{\min }+A}{2}\right)}{\sin \frac{A}{2}}\)
Let C is the critica angle,
sin C = \(\frac{1}{\mu}\)

r₁+r₂ = A
r₁+C = A
r₁ = A – C
μ = \(\frac{\sin i}{\sin r_{1}}\)
⇒ sin i = μ sinr₁
sin i = μ sin(A-C)
i = sin^-1μ sin(A-C)

SECTION – C
(Section C consists one case study-based quote of 5 marks.)

Question 12.
If the setup of Young’s doubLe slit experiment is immersed in a liquid then the Wavelength of the Light gets changed. Since an immersion frequency remains unchanged, velocity decreases hence, waveLength decreases. Now, alt those parameters which are dependent on wave Length will be affected. If the fringe width is air in \(\frac{\lambda D}{d}\) then in Light it will be \(\frac{\lambda D}{\mu d}\) in the same way, the angular fringe width also changes and the position of cringes on-screen aIs changes. (3)

(A) In Young’s experiment when sodium Light of waveLength 5893A° is used, then 62 fringes are seen in the field of view. Instead, if violet Light of waveLength 4358A° is used, then the number of fringes that will be seen in the field of view will be:
(a) 54
(b) 64
(c) 74
(d) 84
at x = 0. The third maximum (taking the central maximum as zero maximum) will be at x equal to:
(a) 1.67 x 10^-2 m
(b) 1.5 x 10^-2 m
(c)O.5 10^-2m
(d) 5.0 10^-2 m

(C) In Young’s experiment, the ratio of maximum to minimum intensities of the fringe system is 4:1. The amplitudes of the coherent sources are in the ratio:
(a) 4:1
(b) 3:1
(c) 2:1
(d) 1:1

(D) The displacement of two Interfering Light waves are y₁ = 4sinωt and Y₂ = 3sinω \(\left(\frac{\omega t+\pi}{2}\right)\) .The ampLitude of the resultant wave is:
(a) 5
(b) 7
(c) 1
(d) 0
(E) In Young’s experiment, tight of waveLength 4000Å is used to produce bright fringes of width 0.6 mm, at a
dipped in a Liquid of refractive index 1.5, then the fringe width will be:
(a) 0.2 mm
(b) 0.3 mm
(c) 0.4 mm
(d) 1.2 mm (5)
Answer:
(A) (d) 84
Explanation: From Young’s double slit experiment,
λ₁ = 5893 Å
λ₂= 4359 Å
\(\frac{n_{1} \lambda_{1} D}{d}\) = \(\frac{n_{2} \lambda_{2} D}{d}\)
The above condition is total extent of fringes is constant for both wavelengths.
\(\frac{62 \times 5893 \times 10^{-10} \times \mathrm{D}}{d}=\frac{n_{2} \times 4359 \times 10^{-10} \times \mathrm{D}}{d}\)
n₂ = \(\frac{62 \times 5893}{4359}\)
= \(\frac{365366}{4359}\) = 83.8
n₂ = 84

Related Theory:
In Young’s double-slit, experiment the number of fringes ate inversely proportional to the fringe width.
(B) (b) 1.5 x 10^-2 m
Explanation: x = (n) \(\lambda \frac{\mathrm{D}}{d} \)
= 3 x 5000 x 10^-10 x \(\frac{2}{0.2 \times 10^{-3}}\)
= 1.5 x 10^-2 m = 1.5 cm

Caution:
Students are often confused about the number of slits. In Young’s double-slit experiment two slits are there.
(C) (b) 3:1

Explanation: \(\frac{I_{(m a x)}}{I_{(m i n)}}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)
⇒ \(\frac{4}{1}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}\)

Related Theory:
In Young’s experiment, intensity of tight is directly proportional to the square of the amplitude.
(D) (a) S
Explanation: Here,
Φ = \(\frac{\pi}{2}\)
Amplitude of first wave = A₁ = 4
Amplitude of second wave = A₂ = 3
Resultant = \(\sqrt{\left[(4)^{2}+(3)^{2}\right]}\)
= 16 + 9 = 5cm.

(E) (c) 0.4 mm
Explanation: Fringe width.

Caution:
Students are often confused in the relation of fringe width and wavelength of Light. Here, fringe width is
directly proportional of a wavelength of Light.