science cbse-x-2018 examination career … a mendelian experiment consisted of breeding pea plants...

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SCIENCE CBSE-X-2018 EXAMINATION CAREER POINT

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-5151200 Website : www.careerpoint.ac.in, Email: [email protected]

SUBJECT: - SCIENCE CLASS-X

PAPER-CODE:-31/1 (SET-1) Q.1 A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing

white flowers. What will be the result in F1 progeny? Sol. As per Mendel experiment, Monohybrid cross between:- Plants bearing violet flowers Plants bearing white flowers PP X pp (P1 generation) P P p p

All plants bearing Violet flowers, in F1 generation, because as per Mendel’s rule of “Law of dominance” all

the offsprings in F1 generation are having dominant traits Q.2 Write the energy conversion that takes place in a hydropower plant. Sol. Potential energy of water Kinetic/Mechanical energy in turbine Electric Energy Q.3 A compound ‘X’ on heating with excess conc. 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.

Sol. When ethanol is heated with conc. H2SO4 at 443K, it undergoes dehydration [removal of a molecule of water] to form ethene.

CH3CH2OH Conc.H2SO4 CH2= CH2+ H2O Heat [443K] Ethanol Ethene [X] [Y]

Here Sulphuric acid acts as dehydrating agent in this reaction.

2CH3CH2OH+2Na 2CH3CH2ONa + H2 Sodium ethoxide [z]

Here [x] = CH3CH2OH [y] = CH2=CH2

[z] = H2 gas

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Q.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. Sol. [a] gustatoryreceptors are helpful in Taste and found in tongue [Ex.:- Taste buds] Olfactory receptors are helpful in smell and found in nose. [b] a= cyton b= Axon So, electric impulse flow chart will be as follows:- Q.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. Sol. Convex mirror makes erect, virtual and diminished image of the object.

1. Object at infinity: When the object is at the infinity, a point sized image is formed at principal focus behind the convex mirror.

Properties of image: Image is highly diminished, virtual and erect.

2. Object between infinity and pole: When the object is between infinity and pole of a convex mirror, a diminished, virtual and erect image is formed between pole and focus behind the mirror. Properties of image: Image is diminished, virtual and erect.

a b End point of Neuron Dendrite

Dendrite Cyton Axon End point of Neuron

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Q.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.

Sol. [a] Decomposition by heat or thermal energy. CaCO3(s) Heat CaO (s) + CO2(g)

[b] Decomposition by light or photolysis 2AgCl (s) Sunlight 2Ag (s) + Cl2(g) [in presence of light] [c] Decomposition by electricity or electrolysis 2H2O (l) electricity 2H2(g) + O2(g)

Q.7 2 mL of sodium 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 14. Identify the salt and write a chemical equation

for its formation. List its two uses. Sol. If Sodium Hydroxide is added to granulated zinc metal in test tube and heated, sodium Zincate is formed

along with hydrogen gas. Zn + 2NaOH Na2ZnO2 + H2 Sodium Zincate Hydrogen gas. Detection:- Hydrogen gas can be detected by putting a burning splinter near the gas. The splinter will gives out a ‘Pop’

Sound (mini explosion), if it is hydrogen gas. Hydrogen gas will be evolved when Zn metal reacts with dilute solution of strong acid As, Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

OR Sol. The Soda commonly used in the Kitchen for making tasty crispy pakora is baking soda. It is a mild non-

Corrosive base. Chemically, it is sodium hydrogen Carbonate. [NaHCO3] The following reaction takes when it is heated during cooking. Manufacture of baking soda is shown in reaction given below:- NaCl + H2O + CO2 + NH3 NH4Cl + NaHCO3

The following reaction takes place when it is heated during cooking. 2NaHCO3 heat Na2CO3 + H2O + CO2 [Sodium hydrogen [Sodium Carbonate] Carbonate] Two Uses:- (1) It is used in Fire extinguisher. (2) It is also used as ingredient of antacid.

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Q.8 [a] Why are most carbon compound 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. Sol. [a] Because they do not dissociate to form ions and remain as molecules. The reason behind this is their

Covalent bonding. [b] Saturated ring compound Molecular formula C6H12 Structural formula

Line formula IUPAC Name- Cyclohexane [No. of Single bonds-18] Q.9 Name the hormones secreted by the following endocrine glands and specify one function of each : [a] Thyroid [b] Pituitary [c] Pancreas Sol. [a]Thyroid: Hormones secreted: thyroxin, triiodothyronin, and calcitonin. Functions: Thyroxine (T4) and triiodothyronine (T3): Regulates body metabolism and heat Calcitonin: Reduces blood calcium level. by enhanced deposition in bone.

[b] Pituitary: The pituitary gland has two components i.e., adenohypophysis (Anterior Part) and neurohypophysis(Posterior Part)

Anterior Part:- Hormones secreted- ACTH (Adrenocorticotropic hormone), which controls Adrenal gland secretions. TSH (Thyroid-stimulating hormone) which targets the Thyroid and is called Thyroxin LH/FSH (LH being in males, and FSH being in females) which affects the reproductive functioning. With

LH it helps regulate Testosterone. With FSH it produces Oestrogen. PRL (Prolactin) stimulates the breasts to produce milk. GH (Growth Hormone) this deals with every cell in the body. It helps growth and repair. MSH (Melanocyte-Stimulating hormone) action is unknown.

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Posterior Part:- Hormones secreted- ADH (Anti-diuretic hormone) Controls the blood and mineral levels in the body by affecting water retention

by the kidneys. Oxytocin - Affects uterine contractions in pregnancy and birth and also helps facilitate the release of breast

milk. [c] Pancreas: Hormones secreted: insulin and glucagon Functions: Insulin (β cells): Helps absorb glucose from blood to tissue and also release from liver Glucagon (α cells): Helps absorb glucose into liver to form glycogen

Q.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.

Sol. Difference between Sexual reproduction and Asexual reproduction:- Sexual reproduction just means combining genetic material from two parents. Asexual reproduction produces offspring genetically identical to the one parent. Sexually reproducing species is likely to have better chances of survival, because : Reason:- 1. During sexual reproduction, There might be errors in the DNA copying and lead to variations which might

promote survival advantage 2. In sexually reproducing organisms, natural selection might take place which may give survival advantage. In asexually reproducing organisms, DNA copying is errorless as they tend to preserve their similarities and

offspring are exact copies of parents. Q.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.

Sol. The Laws of Refraction:- 1. Incident ray, reflected ray, refracted ray and the normal of the system lie in the same plane. 2. Incident ray, coming from one medium to the boundary of another medium, is refracted with a rule derived

from a physicist Willebrord Snellius. He found that there is a constant relation between the angle of incident ray and angle of refracted ray. This constant is the refractive index of second medium relative to the first medium. He gives the final form of this equation like;

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The refracting index of a medium for a light of given wavelength may be defined as the ratio of the speed of

light of vacuum to its speed in that medium. Refractive index of a medium with the respect to vacuum is also called absolute refractive index.

Refractive index =

µ = OR

Power of lens is the ability to converge or divege light rays. It is defined as the reciprocal of its focal length in metres.

Its SI unit is Dioptre (D) where f should always be in metres

Power of lens = diopter (D)

Power of convex lens is positive as it has a real focus and its length measured is positive. So the student who is using lens of f =40cm has convex lens. Its power will be

P = 1/f P = 1/0.4 P = 2.5Diopte Power of concave lens is negative as it has a virtual focus length and the focal length measured will be

negative. So the student who is using lens of f= -20 cm has a concave lens. Its power will be P = 1/f P = –1/0.2 P = –5 Dioptre Q.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.

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Sol. Resistance of each wire = 9Ω (i) 13.5 Ω

+ –

R1

9Ω

9Ω R3

9Ω R2

R2 and R3 → It should connect in parallel and R1 should connect in series to them

Req = R1 +

⎭⎬⎫

⎩⎨⎧

+32 R

1R1

1

Req = 9 Ω +

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

+91

91

1

Req = 9 Ω + 29

Req = (9Ω + 4.5) Ω = 13.5 Ω (ii) 6 Ω

+ –

9Ω

R3

9Ω

R1

9Ω

R2

R2 and R3 → should connect in series and R1 connect in parallel to them.

Req =

132 R1

RR1

1

++

Req =

91

991

1

++

Req =

91

181

1

+

Req = 2/3

9

Req = 6 Ω

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OR (a) The joule's law shows the relationship between heat produced by a flowing electric current through a

conductor. Q = I2RT Where Q indicates amount of heat I shows electric current R is the amount of electric resistance in conductor T denotes time (b) P1 = 100 W, 220 V P2 = 60 W, 220 V

P1,R1

I1 P2,R2

I2 I

~

For 1st bulb P = VI 100 = 220 × I1

I1 = 220100 Amp.

For 2nd bulb P2 = VI2 60 = 220 I2

I2 = 22060 Amp.

I = I1 + I2

I = 22060

220100

+

I = 118 Amp.

Q.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. Sol. [a] The resistance of the conductor depends upon the following parameters 1. Physical parameter: it’s including length, area, bonding between the electron pair and type of material

used.

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• Resistance of a conductor is directly proportional with its resistivity. • Resistance of a conductor is directly proportional with the length of conductor. • Resistance of a conductor is inversely proportional with the cross section area of that conductor. • Bonding between the electron pair :

If electron bonding is strong → number of free electrons will be less → drift velocity of the electron will be less → conductivity will decrease → resistivity will increase → thus resistance will increase.

2. Temperature Resistance versus temperature relationship R = R0 (1+ α(T-T0) Here α → is known as temperature coefficient. Now If α is positive, resistance will increase. And if α is negative, resistance will decrease. [b] Electricity flows when there is a flow of electrons. Metals are good conductors (both of heat and

electricity) because at least one electron per atom is free: i.e., it is not tied to any particular atom, but is, instead, able to move freely throughout the metal. In good insulators, such as glass, all of the electrons are tightly bound to atoms (which are fixed), and so there are no free electrons.

[c] Alloys are used for making electrical heating devices due to the following reasons; [i] They have a higher resistivity as compared to pure metals. [ii] They have a wide range of high melting points. [iii] They do not burn in oxygen i.e. they don't oxidize readily at high temperatures. Kanthal, Nichrom are the two forms of the alloys widely used in the industry for heating elements Q.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 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. [c] List two values the teacher instilled in his students in this episode. Sol. Garbage produced can be managed by the following methods: [any two] 1. We can adopt the 3 R's for the minimal production of garbage. 2. Preparation of compost 3. Incineration 4. Landfill 5. Sewage treatment By following ways, we can generate the least garbage: 1. Recycle everything we can

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2. Ditch the plastic bags. 3. Start relying on reusable containers. 4. Start composting. 5. Learn to repair rather than discard. 6. Stop using disposable plates. 7. And at the very least, stop buying plastic water bottles. 8. Donate Clothes 9. Reduce Food Waste 10. Buy Things with Less Packaging 11. Purchase Items Made From Recycled Products 12. Stop buying individually packaged products. Values exhibited by teacher: The teacher has qualities of high degree of general awareness about garbage management and duty towards

the society and the nation. He suggested students to generate least garbage and describe various garbage management procedures.

Q.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. Sol. Dam: Dams are massive barriers built across rivers and streams to confine and utilize the flow of water for

human purposes such as irrigation and generation of hydroelectricity. We seek to build large dams in order to: 1. Store large quantity of water for irrigation and electricity generation. 2. Transfer water to different locations in adequate quantities. 3. Allow equitable distribution of water because people near the source grow water intensive crops. Construction of large dams causes: The building of dams poses social problems because of: 1. Displacement of large number of people without adequate compensation or rehabilitation. 2. Economic problems because huge inputs without proportionate benefits are done. 3. It poses environmental problems because of deforestation and loss of biodiversity. Q.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 labeled diagram for the electrolytic refining of copper. Sol. [a] The moderately reactive metals are extracted by the reduction of their oxides with carbon, aluminium,

Sodium or calcium. The concentrated ores Can be converted to metal oxide by the process of Calcination or roasting.

Metals can by extract from their carbonate ores by using calcinations process. Calcination – In this process the ore is heated to a high temperature in the absence of air, or where air does

not take part in the reaction. Usually carbonate ores are calcined. Eg. ZnCO3 (s) ZnO (s) + CO2(g)

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After that, metal oxide is reduced to obtain pure metal. The reduction is done in the presence of reducing agent as C, Al, Co

Reduction Eg. ZnO + C Zn + CO [ carbon monoxide] [b] Copper extracted from its Sulphide ore by roasting. Roasting – Sulphide ores are converted into oxides by heating strongly in the presence of excess air, so that oxygen gets

added to form the Corresponding oxides. Sulphur impurities escape as gas. Following are the steps for extraction of copper from its sulphide ore.

[i] Cu2S + 3O2 heat 2Cu2O (s) + 2SO2 (g) [ii] 2Cu2O (s) +Cu2S (s) heat 6Cu (s) +SO2 (g)

The Diagram shown below is of electrolytic refining of copper.

At Cathode Cu 2+ (aq) + 2e- Cu (s)

At anode Cu (s) Cu2+ (aq) + 2e -

Q.17 [a] The modern periodic table has been evolved through the early attempts of Dobereiner, Newland and

Mendeleev. List one advantage and one limitation of all the three attempts. [b] Name the scientist who first of showed that atomic number of an elements is a more fundamental

property than its atomic mass. [c] State Modern periodic law. Sol. [a] Advantages 1. Dobereiner triads – Johan woligang dobereiner tried to arrange the elements with same prosperities into

groups. He found out that some groups had only 3 elements and called them triads. He showed that when the 3 elements were written in their order of increasing atomic masses, the atomic mass of the middle element was the average of the Masses of the other 2 elements so the attempts of Dobereiner encouraged others chemist to correlate the properties of elements with their atomic masses.

Ex. Elements Atomic Mass L, - 6.9 Na - 23.0 K - 39.0

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2. Newland Octave Law – this law provided a basis for the classification of elements into groups of elements having similar properties. This law also provided a wider scope to arrange all known elements in to tabular form.

3. Mendeleev Periodic law- He left blank place which help scientist to discover new element. Disadvantages 1. Dobereiner Triads: - This law was valid only for a few group of elements known during that time. For example: - N, P, As doesn’t follow ‘triads’ law. 2. Newland law:- This law was applicable only up to calcium, as after calcium, every 8th elements did not

possess properties similar to that of first. 3. Mendeleev Law:- Position of isotopes could not be justified in modern periodic table. [b] ‘Henry Moseley’ [c] Modern Periodic Law:- Properties of elements are a periodic function of their atomic number. Q.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 the composition of 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 form kidney. [iii] store urine it is passed out. Sol. (a) Blood components:[any two] -Plasma consists mostly of water, with proteins, ions, nutrients, and wastes mixed in. -Red blood cells are responsible for carrying oxygen and carbon dioxide. -Platelets are responsible for blood clotting. -White blood cells are part of the immune system and function in immune response. (b) Path of movement of oxygenated blood in the body 1. Pulmonary Vein (oxygenated blood) 2. Left Atrium (oxygenated blood) 3. Left Ventricle (oxygenated blood) 4. Aorta (oxygenated blood) 5. Arteries-> Arterioles -> Capillaries (oxygenated blood) 6. Tissues and Cells (oxygen is released and carbon dioxide is picked up)

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(c) A heart valve normally allows blood to flow in only one direction through the heart. -The two atrioventricular (AV) valves, the mitral valve (bicuspid valve), and the tricuspid valve, which

are between the upper chambers (atria) and the lower chambers (ventricles). -These valves prevent backflow from the ventricles into the atria during systole. (d) Arteries have thick, elastic, muscular walls whereas veins have thin walls with few elastic fibers. Veins

have one-way valves whereas arteries do not have any valves. OR

(a) Excretion is the process by which metabolic wastes and other non-useful materials, such as faeces, are eliminated from an organism.

(b) Nephrons are the microscopic basic filtration units of the kidney (c)

Q.19 [a] writes 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. Sol. (a) Ovary – To produce the egg (ovum). To produce the hormones oestrogen and progesterone. Oviduct (fallopian tube): It is the site of fertilization. It carry egg (ovum) or fertilized ovum (zygote) to the uterus. Uterus – It is the site of development of embryo. It is made up of muscle tissue. It protects and nourishes the embryo. (b) Structure of placenta: Disc like structure embedded in the uterine wall. It has villi on the embryo’s side

and blood species on the mother’s side which surround the villi. Function of placenta: -Transports oxygen and nutrition from the mother’s blood to the embryo. -Removes the excretory substances from embryo into the mother’s blood.

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Q.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] Why does the sun appear reddish early in the morning? Will this phenomenon be observed by an astronaut on the Moon? Give reason to justify your answer.

Sol. [a] Myopia The defect of an eye in which it cannot see the distant objects clearly is called myopia. A person with

myopia can see nearby objects clearly. Student cannot see words written on the board which is far from him (3 m).

Myopia is caused due to: o High converging power of lens. Lens becomes thicker. o Eye-ball being too long. Elongation of the eye ball.

Due to high converging of the eye-lens the image is formed in front of the retina and a person cannot see clearly the distant objects. In another case, if the eye- ball is too long than the retina is at larger distance from the eye-lens. In this case also the image is formed in front of the retina even though the eye-lens has correct converging power.

Correction for myopic eye: Myopia or short-sightedness can be corrected by wearing spectacles containing concave lens. This is because

when a concave lens of suitable power is used for the myopic eye then the concave lens first diverge the parallel rays of light coming from distant object. Therefore, first a virtual image is formed at the far point of the myopic eye. Now since the rays of light appear to be coming from eye’s far point, they are easily focused by the eye-lens and image is formed on retina. Concave lens is used for myopic eye so as to decrease the converging power of the eye-lens.

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Formula for calculating power of concave lens to correct myopia is: 1/image distance (v)-1/object distance (u) = 1/focal length (f)

[b] Stars do not really twinkle; they just appear to twinkle when seen from the surface of Earth. The stars twinkle in the night sky because of the effects of our atmosphere. they’re so far away from Earth and appears as pinpoints. And it’s easy for Earth’s atmosphere to disturb the pinpoint light of a star. As a star’s light pierces our atmosphere, each single stream of starlight is refracted – caused to change direction, slightly – by the various temperature and density layers in Earth’s atmosphere. The light traveling a zigzag path to our eyes, instead of the straight path the light would travel if Earth didn’t have an atmosphere. Refraction by different layers of atmosphere causes the light from the star to twinkle when seen from the ground.

OR [a] 1. Cornea : it is a transparent spherical membrane covering the front of the eye. Because the cornea is as

smooth and clear as glass but is strong and durable, it helps the eye in two ways: It helps to shield the rest of the eye from germs, dust, and other harmful matter. The cornea shares this protective task with the eyelids, the eye socket, tears, and the sclera, or white part of the eye.

The cornea acts as the eye's outermost lens. It functions like a window that controls and focuses the entry of light into the eye. The cornea contributes between 65-75 percent of the eye's total focusing power.

The cornea also serves as a filter, screening out some of the most damaging ultraviolet (UV) wavelengths in sunlight. Without this protection, the lens and the retina would be highly susceptible to injury from UV radiation.

2. Iris: It is a colored diaphragm between the cornea and lens. Iris is made of muscles. They control the size of opening of pupil.

3. Crystalline lens: it is a transparent lens made of jelly like material of refractive index 1.396Lens lies just behind the pupil. Lens becomes thin to increase its focal length. This enables us to see distant objects clearly. To focus on nearer objects, lens becomes thick to decrease its focal length. But there is a limit. The minimum distance of clear vision is 25 cm. below this distance, we cannot see things clearly.

4. Ciliary muscle: these muscles hold the lens in the position. Ciliary Muscle is a circular muscle that relaxes or tightens the zonules to enable the lens to change shape for focusing. The zonules are fibers that hold the lens suspended in position and enable it to change shape during accommodation.

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[b] During the sunrise also in the sunset, the sun rays has to travel a greater distance through the atmosphere

because of being slanted. In the solar spectrum the colors having higher wavelengths like red do not scatter much but the shorter wavelength like violet, blue etc, get scattered easily. So in case of sunrise, when the rays that has to travel a longer distance, the smaller wavelengths like blue, violet, green etc are scattered off in the upper atmosphere and the ray reach earth consists of longer wavelengths of orange and red. Thus, sky appears reddish.

The same phenomenon will not be observed in the moon because of the absence of atmosphere on the surface of the moon. Hence there will be no scattered light and will appear dark.

Q.21 [a] state Fleming’s left hand rule. [b] Write the principle of working of an electric motor. [c] Explain the function of the following parts of an electric motor. (i) Armature (ii) brushes (iii) split ring

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Sol. [a] Whenever a current carrying conductor is placed in a magnetic field, the conductor experiences a force which is perpendicular to both the magnetic field and the direction of current. According to Fleming's left hand rule, if the thumb, fore-finger and middle finger of the left hand are stretched to be perpendicular to each other as shown in the illustration at left, and if the fore finger represents the direction of magnetic field, the middle finger represents the direction of current, then the thumb represents the direction of force. Fleming's left hand rule is applicable for motors

[b] When a rectangular coil carrying current is placed in a magnetic field, a torque acts on the coil which

rotates it continuously. When the coil rotates, the shaft attached to it also rotates and thus it is able to do mechanical work.

[c] (i) Armature: A D.C. motor consists of a rectangular coil made of insulated copper wire wound on a soft

iron core which experiences the force when it carries current in the magnetic field. This coil wound on the soft iron core forms the armature. The coil is mounted on an axle and is placed between the cylindrical concave poles of a magnet.

(ii) Brushes: Two small strips of carbon, known as brushes press slightly against the two split rings, and the

split rings rotate between the brushes. The carbon brushes are connected to a D.C. source. They conduct the electricity to the coil.

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(iii) Split Rings (commutator): A commutator is used to reverse the direction of flow of current. Commutator is a copper ring split into two parts C1 and C2. The split rings are insulated form each other and mounted on the axle of the motor. The two ends of the coil are soldered to these rings. They rotate along with the coil. Commutator rings are connected to a battery. The wires from the battery are not connected to the rings but to the brushes which are in contact with the rings. When the coil turns through 900, the brushes lose contact with the commutator and the current stops flowing through the coil. However the coil keeps turning because of its own momentum. Now when the coil turns through 1800, the sides get interchanged. As a result the commutator ring C1 is now in contact with brush B2 and commutator ring C2 is in contact with brush B1. Therefore, the current continues to flow in the same direction.

SECTION-B Q.22 A student added few pieces of aluminium metal to two test tubes a and b containing aqueous solution 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 aquesous 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.

Sol.

Metal Salt Solution in which added Colour change in Solution Reactivity

Aluminium (Al)

FeSO4 (A test tube)

Green colour of FeSO4 Disappears. Iron Metal is setting down at the bottom of the beaker

CuSO4

(B test tube)

Blue colour of CuSO4 Disappears. the brown colored Copper particles settle down at the bottom of test tube

Reactivity of Al is higher than Copper and Iron

Al2 (SO4)3 (C Test Tube)

Solution remain colourless (no change)

Fe cannot displace Al from Al2 (SO4)3 Solution Reactivity of Fe < Al

Iron

CuSO4 (D Test Tube)

Blue Colour of CuSO4 changes to light green coloured Reactivity of Fe > Cu

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Colour change test tube – A, B, D Most reactive Metal- Aluminium The reactivity series is a list of metals arranged in the order of their decreasing activities. According to reactivity series, Aluminium is more reactive than Cu, Fe metals. ‘Reactive metals displace less reactive metals from Solutions of their compounds’. Q.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. Sol. Na2SO4 (aq) + BaCl2 (aq) BaSO4 (S) + 2NaCl (aq) (Sodium Sulphate) (Barium Chloride) Barium Sulphate Sodium Chloride White Precipitate BaSO4 White Precipitate is formed NaCl Remain in the Solution In this reaction, exchange of ions between the reactant takes place so this reaction is ‘double displacement

reaction’ Q.24 List the steps of preparation of temporary mount of a leaf peel to observe stomata. Sol. Steps of preparation of temporary mount of leaf peel to observe stomata. 1. Remove a healthy leaf from the potted plant. 2. Remove a part of the peel from the lower surface of the leaf. You can do this by folding the leaf over and

gently pulling the peel apart using forceps. Keeps the peel in a watch glass containing water. 3. Add saffranin to the watch glass to stain the peel lightly. 4. After 2-3 minutes take out the peel and place it on a clean glass slide. 5. Put a drop of glycerin over the peel and place a clean coverslip gently over it with the help of a needle. 6. Remove the excess stain and glycerin with the help of blotting paper. 7. Observe the slide under the low-power and high-power magnifications of the compound microscope. Q.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)

Sol. Amoeba reproduces by the common asexual reproduction method called binary fission.

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OR

Q.26 An object of height 4.0 cm is placed at a distance of 30cm from the optical center ‘o’ of a convex lens of focal length 20cm. draw a ray diagram to find the position and size of the image formed. Mark optical center ‘o’ and principal focus ‘f’ on the diagram. Also find the approximate ratio of size of the image to the size of the object.

Sol. Height of object (hI) = 4 cm Distance of object (u) = –30 cm Focal length (f) = +20 cm By Len's formula

f1 =

u1

v1

−

+ 201 =

v1 – ⎟

⎠⎞

⎜⎝⎛−

301

– 201 =

v1 +

301

v1 =

301

201

−+

v1 =

30202030

×−

v = 10

6020×

v = 60 cm

60cm

O f2 2f22f1

f1

30cm

O

I

hh =

uv =

cm30cm60 = 2

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Q.27 The values of current (I) flowing thorough 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. Sol.

science cbse-x-2018 examination career … a mendelian experiment consisted of breeding pea plants...

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