Students can find that CBSE Previous Year Question Papers Class 10 Science with Solutions and CBSE Class 10 Science Question Paper 2018 (Delhi & Outside Delhi) effectively boost their confidence.
CBSE Class 10 Science Board Question Paper 2018 (Delhi & Outside Delhi) with Solutions
Time allowed: 3 hours
Maximum marks: 80
iii)
- The Question Paper comprises of two sections, A and B. You are to attempt both the sections.
- All questions are compulsory.
- All questions of Section A and all questions of Section B are to be attempted separately.
- Question numbers 1 to 2 in Section A are one mark questions. These are to be answered in one word or in one sentence.
- Question numbers 3 to 5 in Section A are two marks questions. These are to be answered in about 30 words each.
- Question numbers 6 to 15 in Section A are three marks questions. These are to be answered in about 50 words each.
- Question numbers 16 to 21 in Section A are five marks questions. These are to be answered in about 70 words each.
- Question numbers 22 to 27 in Section B are based on practical skills. Each questions is a two marks question. These are to be answered in brief.
Section – A
Question 1.
A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing white flowers. What will be the results in F
1
progeny?
Answer:
All plants will bear violet flowers in the F
1
progeny.
Question 2.
What colour change is observed on adding an alkaline solution in the solution of phenolphthalein indicator?
Answer:
Colourless solution of phenolphthalein changes to pink-coloured solution.
Question 3.
A compound ‘X’ on heating with excess cone, sulphuric acid at 443 K gives an unsaturated compound ‘Y’. ‘X’ also reacts with sodium metal to evolve a colourless gas ‘Z’. Identify ‘X’, ‘Y’ and ‘Z’. Write the equation of the chemical reaction of formation of ‘Y’ and also write the role of sulphuric acid in the reaction.
Answer:
X is CH
3
CH
2
OH (Ethanol).
Y is CH
2
= CH
2
(Ethene).
Z is H
2
(Hydrogen gas).
Question 4.
(a) Name one gustatory receptor and one olfactory receptor present in human beings.
(b) Write a and b in the given flow chart of neuron through which information travels as an electrical impulse.
Answer:
(a) Gustatory receptor present on tongue detects taste.
Olfactory receptor present in the nose detects smell.
Question 5.
If the image formed by a spherical mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a labelled ray diagram to support your answer.
Answer:
It is a convex mirror. Convex mirror always forms erect, virtual and diminished image for all positions of the object placed in front of it.
• Diagram:
Question 6.
Decomposition reactions require energy either in the form of heat or light or electricity for breaking down the reactants. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light and electricity.
Answer:
Thermal decomposition.
• When calcium carbonate is heated, it decomposes to give calcium oxide and carbon dioxide.
• When silver chloride is exposed to light it decomposes to form silver metal and chlorine gas.
• When electric current is passed through acidified water, it decomposes to give hydrogen gas and oxygen gas.
Question 7.
2 mL of sodium hydroxide solution is added to a few pieces of granulated zinc metal taken in a test tube. When the contents are warmed, a gas evolves which is bubbled through a soap solution before testing. Write the equation of the chemical reaction involved and the test to detect the gas. Name the gas which will be evolved when the same metal reacts with dilute solution of a strong acid.
OR
The pH of a salt used to make tasty and crispy pakoras is 9. Identify the salt and write a chemical equation for its formation. List its two uses.
Answer:
When sodium hydroxide solution is heated with granulated zinc metal, then Sodium zincate and hydrogen gas are formed.
Test for hydrogen. A splint is lit and held near the opening of the tube, then the stopper is removed to expose the splint to the gas. If the gas is flammable the mixture ignites. This test is most commonly used to identify hydrogen which burns with a pop sound. When zinc reacts with dil. sulphuric acid then zinc sulphate and hydrogen gas are formed.
OR
The salt used to make tasty and crispy pakoras is baking soda or sodium hydrogen carbonate (NaHCO
3
).
Sodium hydrogen carbonate is produced by reacting a cold and concentrated solution of sodium chloride (called brine) with ammonia and carbon dioxide.
Uses of baking soda:
(i) It is used for making baking powder which is used for baking bread, cakes, etc.
(ii) It is used as an ingredient in antacids to reduce the effect of excess acid in the stomach.
Question 8.
(a) Why are most carbon compounds poor conductors of electricity?
(b) Write the name and structure of a saturated compound in which the carbon atoms are arranged in a ring. Give the number of single bonds present in this compound.
Answer:
(a) Most of the carbon compounds are covalent compounds which do not conduct electricity because they do not contain ions.
(b) A saturated cyclic compound in which the carbon atoms are arranged in a ring is a Cyclo Hexane.
Question 9.
Name the hormones secreted by the following endocrine glands and specify one function of each:
(a) Thyroid
(b) Pituitary
(c) Pancreas
Answer:
| Gland | Hormone | Function |
| Thyroid | Thyroxine | It controls the rate of metabolism of fats, carbohydrates and proteins in the body. |
| Pituitary | Growth hormone | It controls the growth of human body like development of bones and muscles. |
| Pancreas | Insulin | It lowers the blood sugar level of the body. |
Question 10.
Write one main difference between asexual and sexual mode of reproduction. Which species is likely to have comparatively better chances of survival the one reproducing asexually or the one reproducing sexually? Give reason to justify your answer.
Answer:
| Asexual reproduction | Sexual reproduction |
| Only one organism is required. | Two separate individuals, male and female are required. |
| New organism produced is genetically similar to the parent organism. | New organism produced is genetically different from both. |
| All divisions involved are mitotic. | During gamete formation meiosis occurs. After fertilization, all divisions are mitotic. |
| It dose not help in evolution. | It helps in evolution. |
Encourage
Question 11.
State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum.
OR
What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens.
Answer:
Laws of refraction of light:
(i) First law. The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
(ii) Second law. The ratio of sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media. The second law of refraction is called Snell’s law.
Mathematically, it can be written as:
Absolute refractive index of a medium. When light goes from vacuum to another medium, then the value of refractive index is called absolute refractive index. It is defined as the ratio of the speed of light in vacuum or air to the speed of light in the medium.
Absolute refractive index (n) = \(\frac { Speed of light in vaccum }{ Speed of light in medium }\)
OR
Potver of a lens. The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it.
SI unit of power of lens is dioptre (D).
Given f = 40 cm = \(\frac { 40 }{ 100 }\)m = \(\frac { 2 }{ 5 }\)m
P = \(\frac { 1 }{ f(m) }\) = \(\frac { 1 }{ 2/5 }\) = \(\frac { 5 }{ 2 }\) = + 2.5 D;
Nature of lens is convex lens.
Given: f = -20cm = \(\frac { -20 }{ 100 }\)m = \(\frac { -1 }{ 5 }\)m
P = \(\frac { 1 }{ f(m) }\) = \(\frac { 1 }{ -1/5 }\) = \(\frac { -5 }{ 1 }\) = -5 D. Nature of lens is concave lens.
Question 12.
Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is
(i) 13.5 Ω,
(ii) 6 Ω ?
OR
(a) Write Joule’s law of heating.
(b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.
Answer:
Resultant resistance of R
1
and R
2
is R’
\(\frac{1}{\mathrm{R}^{\prime}}\) = \(\frac{1}{\mathrm{R}_1}\) + \(\frac{1}{\mathrm{R}_2}\)
….. [R
1
and R
2
are in parallel combination
⇒ \(\frac{1}{\mathrm{R}^{\prime}}\) = \(\frac { 1 }{ 9 }\) + \(\frac { 1 }{ 9 }\) = \(\frac{1+1}{9}\) = \(\frac { 2 }{ 9 }\)
∴ R’ = \(\frac { 9 }{ 2 }\) = 4.5 Ω
Resultant resistance of R’ and R
3
is R
R = R’ + R
3
…..[Series combination
(ii)
Resultant resistance of R
1
and R
2
is given by R’
R’ = R
1
+ R
2
= 9 + 9 = 18 Ω
Resultant resistance of R’ and R
3
is given by R
\(\frac{1}{R}\) = \(\frac{1}{\mathrm{R}^{\prime}}\) + \(\frac{1}{\mathrm{R}_3}\) = \(\frac{1}{18}\) + \(\frac{1}{9}\) = \(\frac{1+2}{18}\) = \(\frac{3}{18}\) = \(\frac{1}{6}\)
R = 6 Ω
OR
(a) Joule’s law of heating. When a current (I) flows through a resistor of resistance (R) for time (t), then the amount of heat produced (H) is given by
H = I
2
× R × t
This is known as Joule’s law of heating.
The heat produced in the wire is directly proportional to:
- square of current (I 2 )
- resistance of wire (R)
- time (t) for which current is passed.
(b) Given: Lamp (1) = 100 W; 220 V
Lamp (2) = 60 W; 220 V
I
1
= ? ; I
2
= ?
V
1
= V
1
= V = 220 V
P
1
= V
1
× I
1
100 = 200 × I
1
I
1
= \(\frac{100}{220}\) = \(\frac{5}{11}\) = 0.45A
P
2
= V
2
× I
2
60 = 220 × I
2
I
2
= \(\frac{60}{220}\) = 0.27A
Question 13.
(a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices? Give reason.
Answer:
(a) Factors on which the resistance of a conductor in the shape of a wire depends:
(i) The resistance of a straight conductor is directly proportional to its length.
R ∝ 1 …(i)
(ii) The resistance of a straight conductor is inversely proportional to its area of cross-section.
R ∝ \(\frac{l}{A}\)
From (i) and (ii)
R ∝ \(\frac{l}{A}\)
∴ R = ρ\(\frac{l}{A}\) ….[ρ = constant called resistivity
reuse,
(b) • Metals have free electrons which helps them in the conduction of electricity whereas glass has no free electrons therefore it is an insulator.
• The resistivity of glass is very high (= 10
10
to 10
14
Ω m) therefore it is a poor conductor of electricity whereas the resistivity of metals is low, therefore metals are good conductors of electricity.
(c) Alloys are commonly used in electrical heating devices because —
• the resistivities of alloys are much higher than those of the pure metals (from which they are made);
• an alloy does not undergo oxidation (or burn) easily even at high temperatures. Thus heating elements of electrical heating appliances such as electric iron, toaster, etc. are made of an alloy rather than a pure metal.
Question 14.
Students in a school listened to the news read in the morning assembly that the mountain of garbage in Delhi, suddenly exploded and various vehicles got buried under it. Several people were also injured and there was traffic jam all around. In the brain storming session the teacher also discussed this issue and asked the students to find out a solution to the problem of garbage. Finally they arrived at two main points – one is self management of the garbage we produce and the second is to generate less garbage at individual level.
(a) Suggest two measures to manage the garbage we produce.
(b) As an individual, what can we do to generate the least garbage? Give two points.
Answer:
(a) Measures to manage the garbage we produce:
• (i) Fruits and vegetable peels can be converted into paste and can be used as manure in potted plants and fields to increase soil fertility.
(ii) Use your own reusable cloth bags instead of accepting plastic bags from the stores to carry items.
(b) As an individual, we can adopt the following methods:
(i) Encourage recycling, it means that we should collect the used and discarded items of paper, plastic, glass and metals, and send them to the respective industries for making fresh paper, plastic, glass or metal objects.
which
Question 15.
What is a dam? Why do we seek to build large dams? While building large dams, which three main problems should particularly be addressed to maintain peace among local people? Mention them.
Answer:
Dams are man-made structures constructed across rivers with the aim of controlling, collecting and diverting the flow of water in rivers.
means,
(i) Social problems. Due to construction of high rise dams, a large number of human settlements are submerged in the water of large reservoirs formed by the dam and many people are rendered homeless.
(ii) Environmental problems. A vast variety of flora and fauna get submerged in the water of large reservoirs formed by the dam and this disturbs the ecological balance.
(iii) Economic problems. Construction of high rise dams involves the spending of huge amount of public money without the generation of proportionate benefits to the local community.
Question 16.
(a) Write the steps involved in the extraction of pure metals in the middle of the activity series from their carbonate ores.
(b) How is copper extracted from its sulphide ore? Explain the various steps supported by chemical equations. Draw labelled diagram for the electrolytic refining of copper.
Answer:
(a) • Some of the moderately reactive metals occur in nature as their carbonates. It is easier to obtain metals from their oxides (by reduction) than from carbonates.
• So before reduction the carbonate ore must be converted into metal oxide by the process of calcination.
• Calcination is the process in which a carbonate ore is heated strongly in the absence of air to convert it into metal oxide.
For example,
• The metal oxide obtained by calcination is converted to free metal by using reducing agents like carbon, aluminium, sodium or calcium. The reducing agent used depends on the chemical reactivity of the metal to be extracted.
For example,
Iron metal is extracted from its oxides by reduction with carbon. Tin and lead metals are also extracted by the reduction of their oxides with carbon.
The metals prepared by the reduction process usually contain some impurities and obtaining pure metals from impure metals is called refining of metals and most of the metals are refined by electrolytic refining.
(b) Copper can be extracted by heating its sulphide ore in the presence of excess of air. This process is called roasting.
Steps in the roasting of copper:
(i) The concentrated copper (I) sulphide ore is heated in air till a part of copper (I) sulphide is oxidised to copper (I) oxide.
(ii) When a good amount of copper (I) sulphide has been converted into copper (I) oxide, then the supply of air for roasting is stopped. In the absence of air, copper (I) oxide reacts with the remaining copper (I) sulphide to form copper metal and sulphur dioxide.
(iii) The obtained metal is then purified by electrolytic refining.
Question 17.
Write the chemical names of Na
2
CO
3
. 10H
2
O and Na
2
CO
3
. Write the significance of 10H
2
O molecules attached with the formula of a salt. With the help of chemical equations explain the method of preparation of Na
2
C0
3
. 10H
2
O. Also write two uses of Na
2
C0
3
. 10H
2
O.
Answer:
• Chemical name of Na
2
CO
3
. 10H
2
O: Sodium Carbonate Decahydrate (Washing soda)
• Na
2
Co
3
: Sodium Carbonate (Soda ash)
• 10H
2
O is the water of crystallization of sodium carbonate. Water of crystallization is the fixed number of water molecules present in one formula unit of a salt. Thus ’10’ water molecules are present in one formula unit of sodium carbonate.
Washing soda is produced by Solvay process, by the reaction of ammonical brine with CO
2
gas.
NaHCO
3
on heating produces Na
2
CO
3
(Sodium carbonate, called soda ash) which on recrystallization produces washing soda.
Washing soda is a basic salt.
Some important uses of washing soda:
(i) Washing soda is used in the glass, soap and paper industries.
(ii) It is used in the manufacture of borax compound.
(iii) It is used as a cleansing agent for domestic purposes.
(iv) It is used for removing permanent hardness of water.
Question 18.
(a) Mention any two components of blood.
(b) Trace the movement of oxygenated blood in the body.
(c) Write the function of valves present in between atria and ventricles.
(d) Write one structural difference between the composition of artery and veins.
OR
(a) Define excretion.
(b) Name the basic filtration unit present in the kidney.
(c) Draw excretory system in human beings and label the following organs of excretory system which perform following functions:
(i) form urine.
(ii) is a long tube which collects urine from kidney.
(iii) store urine until it is passed out.
Answer:
(a) Two components of blood:
if
possible,
we
| Arteries | Veins |
| 1. Arteries are thick walled. | Veins are thin walled. |
| 2. Arteries carry blood from the heart to different organs of the body. | Veins carry the blood from body organs to the heart. |
| Arteries do not have valves. | Veins have valves in them. |
| Arteries are deeply placed. | Veins are superficially placed. |
should
18. (a) The process of removal of toxic and nitrogenous wastes from the body of an organism is called excretion.
(b) The basic filtration unit present in the kidney is known as nephron.
(c) Human Excretory system:
| Function | Organ |
| (i) Form urine | Kidney |
| (ii) Long tube which collects urine from kidney | Ureter |
| (iii) Store urine until it is passed out | Urinary bladder |
Question 19.
(a) Write the function of following parts in human female reproductive system:
(i) Ovary
(ii) Oviduct
(iii) Uterus
(b) Describe in brief the structure and function of placenta.
Answer:
(a) (i) Ovary. Ovaries produce mature female sex cells called ‘ova’ or ‘eggs’ and also secrete certain female sex hormones.
(ii) Oviduct. The ovum released by an ovary goes into the oviduct and the fertilisation of ovum by a sperm takes place in the oviduct.
(iii) Uterus. The growth and development of a fertilised ovum into a baby takes place in the uterus.
reuse
Functions:
(i) It provides a large surface area required for the exchange of nutrients, oxygen and waste products between the embryo and the mother.
(ii) It also removes the metabolic waste from the embryo.
Question 20.
(a) A student is unable to see clearly the words written on the black board placed at a distance of approximately 3 m from him. Name the defect of vision the boy is suffering from. State the possible causes of this defect and explain the method of correcting it.
(b) Why do stars twinkle? Explain.
OR
(a) Write the function of each of the following parts of human eye:
(i) Cornea
(ii) Iris
(iii) Crystalline lens
(iv) Ciliary muscles
(b) What do you mean by power of accommodation of human eye lens? Write the values of near point and far point of a normal adult’s human eye.
Answer:
(a) The boy is suffering from nearsightedness or Myopia in which a person cannot see the distant objects clearly though he can see the nearby objects clearly.
Myopia is caused –
• due to high converging power of eye-lens (because of its short focal length); and
• due to elongation of the eyeball.
This defect can be corrected by placing a concave lens in front of the eye. Using a concave lens of suitable focal length in the spectacles of such a person can correct this defect.
(b) The stars appear very-very small, so stars can be considered to be point sources of light as they are far away from the earth. The continuously changing atmosphere is able to cause variations in the light coming from a point-sized star due to atmospheric refraction because of which the star appears to be twinkling.
OR
(a) (i) Cornea. The front part of the eye is called cornea. It is made of a transparent substance which is convex in shape. The light coming from objects enters the eye through the cornea.
the
same
things
(b) The ability of an eye to focus the distant objects as well as the nearby objects on the retina by changing the focai length or converging power of its lens is called power of accommodation.
A normal eye has a power of accommodation which enables objects as far as infinity and as close as 25 cm to be focussed on the retina.
Near Point = 30 cm; Far point = Infinity
Question 21.
(a) State Fleming’s left hand rule.
(b) Draw the magnetic field lines around a current carrying straight conductor.
(c) What type of magnetic field is observed inside of such type of solenoid?
Answer:
(a) Fleming’s left hand rule. According to Fleming’s left-hand rule, hold the forefinger, the centre finger and the thumb of your left hand at right angles to one another. Adjust your hand in such a way that the forefinger points in the direction of magnetic field and the centre finger points in the direction of current. Then the direction in which the thumb points, gives the direction of force acting on the conductor.
(b) (i) Strength of the magnetic field produced by a straight current carrying wire at a point is inversely proportional to the distance of that point from the wire.
(ii) Strength of the magnetic field is directly proportional to the current passing in the wire.
(c) Uniform magnetic field with parallel field lines.
Section – B
Question 22.
A student added few pieces of aluminium metal to two test tubes A and B containing aqueous solutions of iron sulphate and copper sulphate. In the second part of her experiment, she added iron metal to another test tubes C and D containing aqueous solutions of aluminium sulphate and copper sulphate.
In which test tube or test tubes will she observe colour change? On the basis of this experiment, state which one is the most reactive metal and why?
Answer:
In test tubes A, B and D, the change in colour takes place.
It means Al displaces Fe from (aq) iron sulphate and A1 also displaces copper from CuSO
4
solution. This shows that Alis more reactive that Fe and Cu.
In the second experiment – Fe being less reactive than aluminium, does not displace Al from aluminium sulphate solution but displaces Cu from CuSO
4
solution. Thus ‘Al’ is the most reactive metal.
Question 23.
What is observed when a solution of sodium sulphate is added to a solution of barium chloride taken in a test tube? Write equation for the chemical reaction involved and name the type of reaction in this case.
Answer:
When a solution of sodium sulphate is added to a solution of barium chloride, then a white precipitate of barium sulphate is formed along with sodium chloride solution. Hence, this reaction is precipitation reaction.
Question 24.
List the steps of preparation of temporary mount of a leaf peel to observe stomata.
Answer:
Step 1. The epidermal peel is taken from a freshly plucked leaf.
Step 2. These pieces are transferred with the help of paint brush into a petridish containing water.
Step 3. The leaf piece is transferred into another watch glass containing a dilute solution of safranin (a red stain).
Step 4. Specimen is transferred back into the petridish containing water to remove excess stain.
Step 5. Specimen is transferred to the slide with the help of a brush.
Step 6. Put a drop of glycerine on the peel and cover it gently with a cover slip by using a needle to avoid any kind of air bubble.
Step 7. Observe the slide under the low-power and high-power magnifications of the compound microscope.
Question 25.
Name the process by which an amoeba reproduces. Draw the various stages of its reproduction in a proper sequence.
OR
A student is viewing under a microscope a permanent slide showing various stages of asexual reproduction by budding in yeast. Draw diagrams of what he observes. (In proper sequence)
Answer:
Amoeba reproduces by binary fission.
Various stages of amoeba’s reproduction:
OR
The process of reproduction in yeast is budding.
Question 26.
An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the diagram. Also find the approximate ratio of size of the image to the size of the object.
Answer:
Given h
1
= +4 cm;
u = -30 cm;
f +20 cm
Using lens formula,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)
⇒ \(\frac { 1 }{ 20 }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ -30 }\)
\(\frac { 1 }{ v }\) – \(\frac { 1 }{ 20 }\) – \(\frac { 1 }{ 30 }\) = \(\frac { 3-2 }{ 60 }\) = \(\frac { 1 }{ 60 }\)
v = +60 cm
Now, \(\frac{h_2}{h_1}\) = \(\frac{v}{u}\) = \(\frac{60}{-30}\) = -2
⇒ \(\frac{h_2}{h_1}\) = \(\frac{2}{1}\) or h
2
: h
1
= 2 : 1
∴ Ratio of size of image to size of object is 2 : 1.
Question 27.
The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below:
| V (volts) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 4.0 | 5.0 |
| I (amperes) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.8 | 1.0 |
Plot a graph between current (I) and potential difference (V) and determine the resistance (R) of the resistor.
Answer:
Resistance = Slope of the graph
⇒ \(\frac { V }{ I }\) = R
R = \(\frac { 1 }{ 0.2 }\) = 5 Ω
