 CBSE X class science updated 70 most important questions for the board exam 2020 - Future Study Point # CBSE X class science updated 70 most important questions for the board exam 2020 # CBSE X class science 70 most important questions for the board exam 2020

Here in this post, we have updated a set of 70 most important questions as the most important science question bank for the class 10 board exam 2020 which are explained scientifically by the subject expert teacher. These updated most important science questions for 10 class board exams are selected as per the CBSE syllabus so you need not worry, we have prepared this set of 70 most important questions of science class 10 CBSE  in such a way that it will help you in your preparation of the CBSE board exams. In this question bank, we have answered 27 questions and the rest of the answers to the remaining questions, you can study through the link available here. In our website future study point, you can download free study material of science and maths in the form of pdf or you can study online. At the end of the post don’t forget to write your comment and please subscribe to our website so that you could get our every incoming post on your mobile. Best of luck for your CBSE board exam 2020. ## 70 most important science questions for 2020

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1.“We need to balance a skeletal chemical equation.” Give a reason to justify the statement.

Ans. We need to balance a chemical reaction because during chemical reaction neither the mass(atom) is destroyed nor it is created, it is known as conservation of mass. We also get the idea of through balancing chemical reaction how much reactants are needed to produce a particular amount of products.

Q2. Giving an example list two pieces of information that make a chemical equation more useful (informative).

Ans. ( 1) We can show the physical states of the reactants and products by highlighting the symbol (s),(l),(g) and (aq) beside the products and reactants inside the bracket which represents solid, liquid, gas and aqueous state respectively, for making the chemical equation more informative.

(ii) We can also show the solids(precipitation) by the downward arrow(↓) and gas by an upward arrow (↑) beside the reactants and products inside the brackets. The heat needed or evolved during the reaction is represented by the symbol Δ.

Ca(OH)2(aq) + CO2(↑) → CaCO3(↓) + H2O(l)

Consider the following chemical reaction
X + Barium chloride → Y + Sodium chloride
(White ppt)
(a) Identify ‘X’ and ‘Y’
(b) The type of reaction

Ans. The reactant X reacts with BaCl2 to form Sodium chloride and Y as a result of the exchange of Barium and chloride ion, so  X  should be a compound of Na, it may be Na2CO3 or Na2SO4 but white precipitation must be of BaSO4, therefore, X should be Na2SO4 and Y should be BaSO4.

The reaction is a double displacement reaction because the products so formed are a result of mutual exchange of ions and it is also a precipitation reaction since  BaSO4 is insoluble in NaCl so it is precipitated in the reaction.

Q3.Name the reducing agent in the following reaction:

State which is more reactive, Mn or Al and why?

Ans. Magnesium oxide is reduced to Mn with the help of aluminum, so aluminum is the reducing agent in the reaction, aluminum is more reactive because it forms aluminum oxide displacing magnesium.

Q5.(i) Write a balanced chemical equation for the process of photosynthesis.
(ii)When do desert plants take up carbon dioxide and perform photosynthesis?

Ans (i)

(ii) As we know there is a shortage of water in deserts, the plants in desert-adapted themselves accordingly, their leaves become Thorney and stems turn into thick leaf to minimize the loss of water due to hot and dry wind. The stomata of the desert plant open at night time and closed during the day so as to prevent the loss of water , at night these plants absorb carbon dioxide and use it for the process of photosynthesis during the day time.

Q6.A Name the type of chemical reaction represented by the following equation:

Ans. (i) In the reaction (i) calcium oxide combines with water molecule forms one compound calcium hydroxide , therefore it is a combination reaction. Since during the reaction heat is also evolved so it is exothermic reaction also.

(ii) In reaction (ii) the compound barium sulfate and aluminum chloride are formed due to the mutual exchange of ions of barium chloride and aluminum sulfate, therefore it is a double displacement reaction

(iii) In third reaction ferrous sulfate decomposed into ferrous oxide, sulfur dioxide, and sulfur trioxide, so it is decomposition reaction and because heat is absorbed during the reaction so it is an endothermic reaction also.

Q7.Write the chemical equation of the reaction in which the following changes have taken place with an example of each:
(i) Change in color
(ii) Change in temperature
(iii) Formation of a precipitate

Ans.(i) When iron filings are added to the solution of blue colored solution of copper sulfate, its color changes to light green due to the formation of ferrous sulfate.

CuSO4 +Fe → FeSO4 + Cu

(ii) When water is added to calcium oxide , calcium hydroxide is formed with the evolution of the heat

Cao + H2O → Ca(OH)2 + Heat

(iii) When carbon dioxide is passed through the lime water, white precipitation of calcium carbonate is formed.

Ca(OH)2 + CO2↑ → CaCO3↓ + H2O

Q8.   What is meant by the power of accommodation of the eye?

Ans. When we see the objects which are far away, the ciliary muscles of crystalline lens expands, due to which radius of curvature of the eye lens increases in order to focus the image of object at the retina and hence we become capable to see the objects and on the same way when we see the objects nearer to us, the ciliary muscles of the crystalline lens contracts and its radius of curvature decreases, focussing the image of the object at retina making us the object visible to us. The efficiency of our eye to see the far and near object is known as the power of accommodation of the eye.

Q9.   A person with a myopic eye cannot see objects beyond 1.2 m distance. What should be the type of corrective lens used to restore proper vision and find its focal length also?

Ans. The person can not see objects beyond 1.2 m distance means the far point of the given myopic eye is 1.2 m, the image of the object situated at a distance of beyond 1.2 m is formed near of the eye lens instead of the retina, so it should be treated by a diverging lens (concave lens) so that image could be formed at 1.2 m in front of the lens used. Therefore if the location of the object is at infinity its image could be seen at 1.2 m.

U = , V = 1.2 m

Applying the lens formula

f = – 1.2

Hence the focal length of the concave lens used to correct the vision of the given myopic eye is –1.2m.

Q10.   What is the far point and near point of the human eye with normal vision?

Ans. Far point of the human eye with normal vision is infinity and the near the point of the human eye with normal vision is 25 cm.

Q11.   A student has difficulty reading the blackboard while sitting in the Last row. What could be the defect the child is suffering from? How can it be corrected?

Ans. The child is unable to see the objects after a certain distance, so he is suffering from myopia, it can be corrected by applying a proper concave lense of such a power so that the images of all the objects lying beyond the far point of the eyes could be formed at the far point in front of the lens.

Q12.   The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia                                  (b) accommodation

(c) near-sightedness                      (d) far-sightedness

Ans. (b) accommodation

Q13.   The human eye forms the image of an object at its

(a) cornea.           (b) iris.

(c) pupil.              (d) retina.

Ans.(d) retina

Q14.   The least distance of distinct vision for a young adult with normal vision is about

(a) 25 m.           (b) 2.5 cm.

(c) 25 cm.           (d) 2.5 m.

Ans. (a) 25 cm`

Q15.   The change in focal length of an eye lens is caused by the action of the

(a) pupil.                                 (b) retina.

(c) ciliary muscles.           (d) iris.

Ans. (c) ciliary muscles

Q16.   A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Ans. A person needs a lens of power for correcting his distant vision = 5.5 D

The relation between focal length and power is

f ≈ –0.182 m = –18.2 cm

When power is 1.5 m

f = 0.67 m ≈ 67 cm

Hence for correcting the distant vision he needed a concave lens of the focal length of –18.2 cm and for correcting his near vision, he needed a convex lens of focal length 67 cm.

Q17.   The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Ans. Far point of a myopic person is 80 cm means the person is unable to see the objects clearly when the objects are placed beyond 80 cm, it occurs due to the image formation is held between the eye lens and retina so he is needed a diverging lens (concave lens) so that if the object is placed beyond 80 cm then the image could be formed by the lens at a distance of 80 cm in front of the applied lens.

V = –80 cm, U = ∞

Applying the lens formula

F = –80

The focal length of the concave lens should be of –80 cm

The relationship between power and focal length is

80 cm = 0.80 m

Therefore the person is needed to apply a concave lens of the power –1.25 D.

Q18.   Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is I m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Ans. The near point of a hypermetropic eye is = 1m = 100 cm, its meaning is that when object is placed closer than 100 cm, the image is formed beyond the retina, so it is needed a convex lens such that the image could be formed at a distance of 100 cm if object is placed at a distance of 25 cm in front of the lens since it is near point of the normal eye. V = –100 cm, U = –25 cm

The focal length of the lens is =100/3 cm =1/3 m

P = 1/F(in meter)

Therefore the power of the lens required to correct the defect is 3D of a convex lens

Q19.   Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Ans. A normal eye not able to see clearly the objects placed closer than 25 cm because it is the near point of a normal eye or it is a limit of the normal eye when the object is seen closer than 25 cm the image is formed beyond the retina and as a result, the image looks blurred.

Q20.   What happens to the image distance in the eye when we increase the distance of an object from the eye?

Q21.   Why do stars twinkle?

The light from the star refracted through the different layers of the atmosphere, so we see the virtual image of the star, the temperature of the atmospheric layers varies which results in changes in the position of the virtual image. This change of position appears to our eyes as a star are twinkling.

Q22.   Explain why the planets do not twinkle.

Answer. The planets are part of our solar system so they are very near to us, the light rays from them reach to us almost in the form of a beam so light rays through the atmosphere couldn’t be refracted that much as in the case of light ray of star, so the virtual and real image of the planet remains the same and that is why they don’t twinkle.

Q23.   Why does the sun appear reddish early in the morning?

Ans. The sun early in the morning is at the horizon, which is the maximum distance from us, light has to travel larger distance compared to other positions of the sun during the whole of the day, traversing this larger distance all the lower wavelength colors scattered away and highest wavelength red color reaches to us resulting the sun looks reddish.

Q24.   Why does the sky appear dark instead of blue to an astronaut?

Ans. The sky appears dark instead of blue to an astronaut because there is no atmosphere in the space, there are no gas molecules so the scattering of light does not take place and thus the sky looks black.

Q25.   What is the diameter of the human eye?

Ans. The diameter of the eye is 2.3 cm

Q26.   What is the function of the crystalline lens of the human eye?

Ans. The Christine lens is basically is an eye lens that is transparent and located behind the iris.  One-third of the total net refraction of light through the eye is taken place by the Christine lens. it is made of small flexible smooth muscles called ciliary muscles, these muscles change the shape of the lens. When we see nearby objects, these muscles constrict and radius of curvature of the lens decreases so that images could be focused on the retina and when we see the far objects these muscles dilate and radius of curvature of lens increases so that images could be focused on the retina, this action of eye is known as accommodation of eye

Q27.   In which type of eye defect far point of the eye gets reduced?

Ans. Myopia

If you want all the answers of the remaining questions on your mobile please subscribe to our website, we assure that you will get the answers with the complete explanation.

Q28.   Why do birds fly back to their nest in the evening?

Q29.   Why do you take time to find an object when you enter in the dimly lighted room from outside in the sun?

Q30.   Why does a ray of light splits when passed from prism?

Q31.   Why doesn’t the planet appear to be twinkling?

Q32.   Why we can’t see things very close to our eyes?

Q33.   When we see any object through the hot air over the fire, it appears to be wavy, moving slightly. Explain.

Q34.   Why does the sky appear blue on a clear day?

Q35.   What eye defect is hypermetropia? Describe with a ray diagram of how this defect of vision can be corrected by using an appropriate lens.

Q36.   A star sometimes appears brighter and some other times fainter. What is this effect called?  State the reason for this effect.

Q37.   A student cannot see a chart hanging on a wall placed at a distance of 3 m from him. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagrams for the (i) defect of vision and also (ii) for its correction

Q38.   Why is the red color selected for danger signal lights?

Q39.   (a) A person cannot read a newspaper placed nearer than 50 cm from his eyes. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate this defect. List its two possible causes. Draw a ray diagram to show how this defect may be corrected using a lens of appropriate focal length.

Q40.   Explain giving a reason why the sky appears blue to an observer from the surface of the earth? What will the color of the sky be for an astronaut staying in the international space station orbiting the earth? Justify your answer by giving a reason.

Q41.   (a) List three common refractive defects of vision. Suggest the way of correcting these defects.

(b) About 45 lac people in developing countries are suffering from corneal blindness. About 30 lac children below the age of 12 years suffering from this defect can be cured by replacing the defective cornea with the cornea of a donated eye. How and why can students of your age involve themselves to create awareness about this fact among people?

Q42.   With the help of a labeled diagram, explain why the sun appears reddish at the sun-rise and the sunset.

Q43.   (a) What is the dispersion of white light? What is the cause of this dispersion? Draw a diagram to show the dispersion of white light by a glass prism.

(b) a glass prism is able to produce a spectrum when white light passes through it but a glass slab does not produce any spectrum. Explain why?

Q44.Give reason for the following:

• Arteries are thick-walled blood vessels.
• Viens are thin-walled blood vessels
• Veins have valves in them.

Q45. Why is the energy needs in plants is very less as compared to animals? Explain?

Q46.Why cesium and gallium melt in our palm?

Q47.Why does magnesium ribbon start floating in hot water?

Q48.Complete the following chemical reactions:

Q49.  Explain why a ray of light passing through the center of curvature of concave minor gets reflected along the same path.

Q50. What is the important function of the presence of ozone in the earth’s atmosphere?

Q51.  The dark reaction of photosynthesis does not need light. Do plants undergo dark reactions at night? Explain.

Q52. How are we able to see distant and nearby objects clearly? Which part of the eye helps in changing the curvature of the lens? Why no image is formed at the blind spot?

Q53. (a) Which property of carbon leads to the formation of a large number of compounds? Define it.
(b) What is the functional group in the following molecules?

(i) CH3CH2CH2OH
(ii) CH3COOH

Q54. (a) Why magnification is taken negative for real images and positive for virtual images?

(b) Why convex mirror is used in rearview mirrors and not a concave mirror?

Q55. (a) What are ‘magnetic field lines’? How is the direction of a magnetic field at a point determined?

(b) Draw two field lines around a bar magnet along its length on its two sides and mark the field directions on them by arrow marks.

Q56. (a) Mention the pH range within which our body works. Explain how antacids
give relief from acidity. Write the name of one such antacid.
(b) Fresh milk has a pH of 6. How does the pH will change as it turns to curd? Explain your answer.
(c) A milkman adds a very small amount of baking soda to fresh milk. Why does this milk take a longer time to set as curd?
(d) Mention the nature of toothpaste. How do they prevent tooth decay?

Q57. Atoms of eight elements A. B. C. D, E, F. G and H have the same number of electronic shells but the different number of electrons in their outermost shell. It was found that elements A and G combine to form an ionic compound. This compound is added in a small amount to almost all vegetable dishes during cooking. Oxides of elements A and B are basic in nature while those of E and F are acidic. The oxide of D is almost neural. Based on the above information answer the following questions:

(i) To which group or period of the periodic table do the listed elements belong?

(ii) What would be the nature of the compound formed by a combination of elements B and F?

(iii) Which two of these elements could definitely be metals?

(iv) Which one of the eight elements is most likely to be found in the gaseous state at room temperature?

(v) If the number of electrons in the outermost shell of elements C and G are 3 and 7 respectively, write the formula of the compound formed by the combination of C and G.

Q58. Mention the essential material (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of the saponification reaction.

Q59. Write two precautions to be taken while identifying different parts of an embryo of a dicot seed.

Q60. A student is to conduct an experiment to show CO2 is released during respiration. List two precautions that he/she must take for obtaining correct observation.

Q61.What is the benefit of the residual volume of air in the respiratory process?

Q62.List two different functions of the pancreas in our body.

Q63.Name the plant Mendal used for his experiment. What type of progeny was obtained by Mendel in F1 and F2 generation when he crossed tall and short plants? Write the ratio he obtained in the F2 generation.

Q64.List two differences between acquired traits and inherited traits by giving an example of each.

Q65.Explain the following.

• Sodium chloride is an ionic compound that does not conduct electricity in the solid-state whereas it conducts electricity in the molten state as well as in an aqueous solution.
• Reactivity of aluminum decrease if it is dipped in nitric acid.
• Metals like calcium and magnesium are never found in their free state in nature.

Q66.What happens to the image distance in the eye when we increase the distance of an object from the eye.

Q67.Draw the structure of neurons and explain its function.

Q68.What are the major parts of the brain? Mention the function of different parts. What constitutes the central and peripheral nervous systems. How are the components of the central nervous system protected?

Q68.Mention one function of each hormone.

• Thyroxine
• Insulin
• Growth hormone
• Testosterone

Q69.Name various plant hormones. Also, they give their physiological effects on plant growth and development.

Q70.What is the reflex actions? Give two examples, explain a reflex arc.

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

### NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

### NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

Scroll to Top