CBSE Class 10 Science Previous Year Question Paper 2018 With Solution

CBSE Class 10 Science Previous year question paper 2018 With Solution

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31-1 (Science) SET-1

 

 

SET -1

Code No. 31/1

Series: TYM

Section- A

 

 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?

Ans:

  • A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing white flowers are crossed then the resulting F1 progeny will be consisting of all violet flowers.
  • Since, violet flowers is the dominant character while white flowers is the recessive character.

2.) Write the energy conversion that takes place in a hydropower plant?

Ans:

  • In hydropower plant, initially the potential energy of flowing water is converted into kinetic energy which is again stored as mechanical energy.
  • And finally, the mechanical energy is converted into electrical energy by using hydroelectric generator.

3.) A compound X on heating with excess conc. sulphuric acid at 443K 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.

Ans:

  • Here, X is the Ethyl alcohol CH3-CH2OH.

Y is the unsaturated compound ethene CH2=CH2.

And Z is the hydrogen gas H2.

  • When Ethyl alcohol on heating with excess sulphuric acid at 443K there will be dehydration of ethyl alcohol takes place and ethene with water is formed as given in following chemical reaction.

CH3-CH2OH -> CH2=CH2 + H2O

  • In this chemical reaction, sulphuric acid acts as a dehydrating agent which removes the water from ethanol.
  • Also, when ethanol reacts with sodium metal forms sodium ethoxide with the evolution of colourless gas hydrogen as shown in following chemical reaction.

2Na + 2CH3-CH2OH -> 2CH3CH2ONa + H2

 

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.

Ans:

a)

  • The gustatory receptor are those which brings the food and fluid outside the body to a gastrointestinal tract.
  • And olfactory receptor are those which has sense of smell i.e. olfaction.
  • In human beings, tongue is the gustatory receptor and nose is the olfactory receptor.

b)

  • In the given flow chart, the impulses are passed from dendrites to cell body and then axon to the end point of neuron.
  • Hence, here a is the cell body and b is the axon.

 

5.) If the image formed by a spherical mirror for all positions of the object placed in front of it always erect and diminished, what type of mirror is it? Draw a labelled diagram to support your answer.

Ans:

  • Convex mirror forms the image for all positions of the object placed in front of it always erect and diminished.
  • In following figure, the object is placed at infinity in front of the mirror then image will be formed at focus F behind the mirror which is highly diminished, point size, virtual and erect.

  • In the following figure, when the object is placed between infinity and pole P of the convex mirror then the image will be formed between pole P and focus F behind the mirror which is diminished, virtual and erect.

6.) Decomposition reaction requires 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.

Ans:

  • The Chemical reaction in which reactant molecule get broken into two or more product forms is called as decomposition reaction.
  • It occurs when energy is supplied in the form of heat, light and electricity.
  • When calcium carbonate i.e. limestone is heated it get decomposed into calcium oxide and carbon dioxide as shown below.

CaCO3 ->CaO + CO2

 

  • When silver bromide is decomposed to form silver and bromide in the presence of sunlight, is also the decomposition reaction as shown below.

  2AgBr -> 2Ag + Br2

 

  • NaCl get decomposed into Na and chlorine when electricity is passed through it is also the decomposition reaction as shown below.

2NaCl -> 2Na + Cl2

 

 

 7.) 2ml of sodium hydroxide solution is added to few pieces of granulated zinc metal taken in test tube. When the contents are warmed, a gas evolves which is bubbled through a soap solution before testing. Write the equation of chemical reaction involved and the test to do detect the gas. Name the gas which will be evolved when the same metal reacts with dilute solution of a string acid.

Ans:

  • When 2ml of sodium hydroxide solution is added to few pieces of granulated zinc metal taken in test tube then sodium zincate and hydrogen gas is released as shown in reaction below.

2NaOH + Zn -> Na2ZnO2 + H2

 

  • When contents are warmed by bringing the burner near the open end of test tube then it will burn with pop sound and hence we detect that the hydrogen gas is evolved.
  • Also, when zinc metal reacts with the strong acids like HCl or H2SO4 then also hydrogen gas or evolved as shown in following chemical reaction.

Zn + Dil. 2HCl -> ZnCl2 + H2

And

Zn + Dil. H2SO4 -> ZnSO4 + H2

 

 

 

Or 7.) The pH of salt used to make tasty and crispy pakoras is 14. Identify the salt and write a chemical equation for its formation. List it’s two uses.

Ans:

  • To make the tasty and crispy pakoras the salt used having pH 14 is the sodium bicarbonate NaHCO3.
  • When carbon dioxide gas is passed through mixture of sodium chloride, water and ammonia then sodium bicarbonate with NH4Cl is formed as given below.

NaCl + H2O + NH3 + CO2 -> NaHCO3 + NH4Cl

 

  • Uses of sodium bicarbonate or baking soda:
  • It is used to reduce the acidity in stomach.
  • It acts as antacid and hence used in the treatment of stomach upset and indigestion.
  • It is also used as baking soda in cooking.

 

8.) a) Why are most carbon compounds poor conductor of electricity?

b) Write the name and structure of saturated compound in which the carbon atoms are arranged in a ring. Give the number of single bonds present in this compound.

Ans:

a)

  • As we know that carbon compounds are formed by forming the covalent bonding in them. As there are no free electrons are available for conduction in carbon compounds hence they are the poor conductor of electricity.
  • Also carbon compounds doesn’t get ionised to form ions, due to which they can’t conduct electricity.

b)

  • There are many saturated compounds in which the carbon atoms are arranged in a ring they are cyclopropane, cyclobutane, cyclopentane and so on.
  • The following diagram shows their structure from which we can say that in cyclopropane there are 3 C-C single bonds and 6 C-H single bonds.
  • In cyclobutane there are 4 C-C single bonds and 8 C-H single bonds are present.
  • And in cyclopentane 5 C-C single bonds and 10 C-H single bonds are present.

 

9.) Name the hormones secreted by the following endocrine glands and specify one function of each.

a) Thyroid

b) Pituitary

c) Pancreas

Ans:

a) Thyroid gland:

The hormone secreted by Thyroid gland is the thyroxin which regulates carbohydrates, proteins and fat metabolism in the body in order to provide best balance for growth.

 

b) Pituitary gland:

The hormone secreted by Pituitary gland is growth hormone which regulates growth and development of the body.

 

c) Pancreas:

The hormone secreted by pancreas is insulin which helps in regulating blood sugar level.

 

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.

Ans:

  • The main difference between sexual and asexual reproduction is in asexual reproduction new generations are created from the single individual which are identical while in sexual reproduction the two individuals involves to create a new individual which has genetic variation.

 

  • Sexually reproducing species has better chances of survival as compared to asexually reproducing species because they are having genetic variation which survive them by adapting variations of the environment.

 

11.) State the laws of refraction of light. Explain the term absolute refractive index of the medium and write an expression to relate it with speed of light in vacuum.

Ans:

When a ray of light travels from one transparent medium to another transparent medium then it changes its direction of path travel is called as refraction of light.

The following are the laws of refraction of light:

1) The incident ray, refracted ray and the normal to the surface separating two media all lie in the same plane.

2) The ratio of sine of angle of incidence to the sine of angle of refraction is remains constant for the given media and colour.

This law is called as Snell’s law of refraction.

Hence, Sini/sinr = Constant

And the constant is called as refractive index of the second medium with respect to the first medium.

Absolute refractive index:

  • If the first medium is vacuum or air then the refractive index of medium 2 with respect to vacuum is called as the absolute refractive index of the medium.
  • And it is denoted by n2 only.
  • If speed of light in air is c and speed of light in medium is v then absolute refractive index of the medium is given by,

nm = speed of light in air/ speed of light in vacuum = c/v

 

 Or 11.) What is mean by power of lens? Write it’s SI unit. A student uses a lens of focal length 40cm and another of -20cm. Write the nature and power of each lens.

Ans:

  • The power of lens is the reciprocal of its focal length.
  • If f is the focal length then power of lens P is given by,

P= 1/f

  • The SI unit of power of lens is dioptre.
  • For a student using lens of focal length 40cm the power of lens will be

P= 1/f = 1/0.4m = 2.5D

  • As the power of lens is positive hence the lens would be convex lens.
  • For a student using lens of focal length -20cm the power of lens is given by,

P= 1/f = 1/-0.2m = -5D

  • As the power of lens is negative hence the lens used would be concave lens.

 

12.) Show how would you join three resistors, each of resistance 9 ohm so that the equivalent resistance of the combination is a) 13.5 ohm b) 6 ohm?

Ans: 

a) To get equivalent resistance as 13.5 ohm we have to connect two resistors in parallel and one is in series with them as shown in figure 1.

Hence,

1/Rp = 1/9 + 1/9= 0.22

Hence, Rp = 4.54 ohm

Re = 4.54 + 9 = 13.54 ohm

 

b) To get equivalent resistance as 6 ohm we have to connect two resistors in series first and then one resistor in parallel with them as shown in figure 2..

Hence,

Rs = 9 + 9 = 18 ohm

And

1/Re = 1/18 + 1/9 = 0.165

Hence, Re = 6.06 ohm

 

 Or 12.) a) Write Joules law of heating?

b) Two lamps, one rated 100W; 220V, and the other 60W; 220V are connected in parallel to electric main supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220V.

Ans:

Joule’s law of heating gives the heating effect of an electric current. According to Joule’s law of heating the heat produced in a resistor R is directly proportional to

  • Square of the current through the resistor R.
  • Resistance for a given current.
  • Time for which the current flows through the resistor.

Thus mathematically, Joule’s law is given by

H= I2Rt

 

b)

For bulb first:

P= 100W and V = 220V

We know that, P= V2*R

Hence, here R1 = V2/P = 48400/100 = 484 ohm

 

For bulb second:

P= 60W and V= 220 V

As we know that, P= V2*R

Hence, here R2 = V2/R = 48400/60= 806.66 ohm

 

Given that, the two bulbs are connected in parallel.

Hence we can write as,

1/Rp = 1/484 + 1/806.66

= 0.002066 + 0.001239

= 0.003305

Thus, Rp = 1/0.003305 = 302.57 ohm

The two bulbs are connected in parallel and voltage supply given is 220V.

Hence, current through two bulbs is given by,

I= V/R = 220/302.57 = 0.727 A

Thus, when these two bulbs are connected in parallel then the current through them will be 0.727 A.

 

13.) a) List the factors on which the resistance of conductor in the shape of 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.

Ans:

a)

  • The resistance of a conductor of shape of wire depends on the following factors:
  • The resistance of wire of conductor is directly proportional to the length of conductor.
  • The resistance of wire of conductor is inversely proportional to the area of cross section of the conductor.
  • And the resistance of wire of conductor also depends on the material of the conductor used.

b)

  • As we know that, in metals there are large number of free electrons which are easily available for conduction when some potential is applied to it.
  • While in glass materials there are no free electrons are available for conduction even if we have given some potential to it.
  • Hence, metals are good conductors than glass materials.

c)

  • Alloys are commonly used in electrical heating devices as they are having low conductivity and hence low melting point because of this they are used as heating devices.
  • While metals are having high electrical conductivity and also high melting point because of this reason alloys are mainly used in electrical heating devices.

14.) Students in a school listened to the news read in the morning assembly that the mountain of the garbage in Delhi, is suddenly exploded and various vehicles got buried under it. Several people were also injured and there was also traffic jam all around. In the baron storming session the teacher also discussed this issue and the asked the students to find out the the 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 the 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 the two values the teacher instilled in his students in this episode.

Ans:

a)

  • The measures suggested to manage the garbage we produce are as follows:
  • Before throwing the garbage outside first we have to separate them as biodegradable and non biodegradable wastes.
  • The waste from kitchen like food left, vegetables, fruits and other will be used as biodegradable wastes and can be composted to form compost.
  • While the other non degradable waste materials like plastic, bottles, metal sheets can be reused further.
  • So by recycling most of the wastes we can reuse them.
  • So we have to follow reuse, recycling and also we have to reduce the use of non biodegradable wastes like plastic.

b)

  • As an individual to generate least garbage we have to use paper bags or bags made from cloths instead of plastic material.
  • And also we have to reuse bio degradable waste fro producing compost.
  • We have to separate the wastes as biodegradable and non biodegradable and according to that we have to reuse and recycle them.
  • Then only there will be reduction in producing garbage at an individual level.

c)

  • The two values that teacher instilled in his students through this episode are follows:
  • Through this episode teacher instilled his students how we can least the garbage at individual level so that pollution will be reduced.
  • Also, the students get learned through this episode that everyone has responsibility to make environment pollution free.

15.) What is 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.

Ans:

  • Dams are constructed on the river to obstruct the flow of water and thereby to collect water in large reservoirs.
  • Following are the advantages of making large dams:
  • The large dams creates large amount of electricity as they possess huge potential energy.
  • Hydropower plants are based on the large dams construction.
  • Because of the above reasons large dams are produced.

The major problems occurred when large dams are constructed are as follows:

  • The dams are constructed only in those areas mostly in hilly terrain.
  • Due to construction of large dams, large areas of agricultural land and human habitation get sacrificed due to their submerging.
  • Also large eco-system are destroyed when they are submerged under the water in dams.
  • Also it creates the problem satisfactory rehabilitation of displaced people.

 

17.) a) The modern periodic table has been evolved through the early attempts of Dobereiner, Newlands and Mendeleev. List one advantage and one limitation of all the three attempts.

b) Name the scientist who first of all showed that atomic number of an element is a more fundamental property than its atomic mass.

c) State Modern periodic law.

Ans:

Dobereiner’s periodic table:

Advantage:

Dobereiner showed that when the three elements in a triad are arranged in their increasing atomic masses then the atomic mass of the middle element is the average of atomic masses of the other two elements.

Limitations:

The Dobereiner’s assumption is valid only for triads.

Newland period table:

Advantage:

According to Newlands octave rule, every eighth element had property similar to that of the first.

Limitations:

Newlands octave rule is applicable only upto the element Calcium after which doesn’t holds.

Mendeleev’s periodic table:

Advantage:

According to Mendeleev, the properties of elements are the periodic functions of their atomic masses.

Limitations:

But, he doesn’t gave the fixed position of hydrogen and also it was found that there were no periodic order is found in atomic masses.

b) Henry Moseley in 1913 showed that atomic number of an element is a more fundamental property than its atomic mass.

c) Modern periodic law can be stated as properties of elements are a periodic functions of their atomic number.

 

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.

Ans:

a)

The steps involved in the extraction of pure metals in the middle of the activity series from their carbonate ores are as follows:

1) Roasting:

In this step the sulphide ores are converted into oxides by strongly heating in the presence of excess air.

2ZnS + 3O2 -> 2ZnO + 2SO2

 

2) Calcination:

In this step the metal carbonate are converted into their oxides.

ZnCO3 ->ZnO + CO2

 

3) Reduction:

In this step metal oxides are reduced to the corresponding metals using suitable reducing agent such as carbon.

ZnO + C -> Zn + CO

 

b) Copper is extracted from its sulphide ore Cu2S which found in nature by heating it in air.

2Cu2S + 3O2 -> 2Cu2O + 2SO2

 

2Cu2O + Cu2S -> 6Cu + SO2

  • The following figure shows the electrolytic refining of copper.
  • In this process, impure metal is made the anode and a thin strip of pure metal acts as cathode.
  • And electrolyte used is the solution of metal salt. When current is passed through the electrolyte, the pure metal from anode get dissolved and hence an equivalent amount of pure metal is deposited on the cathode.
  • Soluble impurities are added to solution while the insoluble impurities get settled down at the bottom of the anode termed as anode mud.
  • The whole process is knows as electrolytic refining.

18.) a) Mention any two components of the blood.

b) Trace the movement of oxygenated blood in the human 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.

Ans:

a)

The components of blood are blood corpuscles which carries the oxygen and plasma which transports food, carbon dioxide and nitrogenous waste in dissolved form.

b)

  • The movement of oxygenated blood in the human body can be stated as follows.
  • The oxygenated blood from lungs reach to the left auricle through pulmonary vein then the blood is poured into the left ventricle and then through aorta it is supplied to the all body parts.

c)

  • The valves present in between atria and ventricles prevent the back flow of blood from ventricle to auricle when ventricles get contracted. These valves ensure the flow of blood only in one direction.

d)

  • Arteries have thick, elastic walls because blood emerges from the heart under high pressure which carry blood from heart to various organs of body.
  • Veins collect the blood from different organs and bring back to the heart and hence they not have thick walls because there the blood is not under pressure for long time and which prevent the backflow of blood.

 

Or 18.) Define excretion.

b) Name the basic filtration present in the kidney.

c) Draw excretory system in human beings and label the following organs of excretory system which performs following functions.

1) Form urine

2) is a long tube which collects urine from kidney.

3) store urine untill it is passed out.

Ans:

a)

The biological  process in which harmful metabolic wastes such as nitrogenous wastes in human beings and gaseous wastes in plants, are removed from the body is called as excretion.

b)

The basic filtration unit present in the kidney is the nephron or uriniferous tubule.

 

c) The following figure shows the labelled diagram of excretory system in human beings.

1) The urine is formed in the kidney as in figure.

2) Ureter is the long tube which collects urine from kidney as in figure.

3) Urinary bladder stores urine untill it is passed out as in figure.

 

 

19.) a) Write the function of the following parts in human female reproductive system.

1) Ovary

2) Oviduct

3) Uterus

b) Describe in brief the structure and function of placenta.

Ans:

a)

1) Ovary:

The female germ cells or eggs are formed in ovary.  Also it is responsible for producing female hormone oestrogen which is responsible for the development of secondary sexual characters in female.

 

2) Oviduct:

The egg formed in ovary is carried from ovary to the womb through the thin oviduct or fallopian tube which provides place for fertilization also.

 

3) Uterus:

The two oviducts unite into an elastic bag like structure called as uterus  which provide nutrition to the fertilized ovum I’m the foetus and it holds the foetus by supporting it till it become mature to born.

b)

  • The embryo gets nutrition from the mother’s blood with the help of some special tissue called as placenta which is in the form of disc and embedded in the uterine wall.
  • It has villi on the embryo’s side of the tissue while on the mother’s side are blood spaces which surrounds the villi.
  • It provides a large surface area for glucose and oxygen to pass from mother to embryo.
  • The waste generated by developing embryo are removed by transferring them into the mother’s blood through placenta.

 

20.) a) A student is unable to see clearly the words written on the blackboard placed at a distance of approximately 3m 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?

Ans:

a)

  • As the student is unable to see clearly the words written on the blackboard placed at a distance of approximately 3m from him means he is suffering from myopia.
  • Myopia is the disease of near-sightedness i.e. that person can see the nearby objects clearly but can’t see the distant objects clearly.
  • That means his far point lies nearer than infinity. Due to which the image is formed in front of retina instead of on the retina.
  • So he can see the objects which are upto few meters.
  • To correct him he can use concave lens of suitable power to form the image back on the retina.
  • The following figure shows the myopic eye in which image is formed in front of retina.
  • The following figure shows the correction for myopia with a concave lens.

Fig. myopic eye

 

Fig. correction in myopic eye using concave lens

b)

  • In atmosphere there are various layers and also their temperature is varying.
  • As we go upwards from the land the temperature will be decreased and the layer having high temperature become less dense than the layer having less temperature.
  • And hence hit air has low refractive index than cold air. Because of these variation in refractive index of atmospheric layers there is a refraction of light through atmosphere takes place.
  • The position of star as we see is its apparent position which is changing continuously due to which we observe like stars are twinkling.
  • And this is only due to he refraction of starlight through the earth’s atmosphere.

 

21.) a) state Fleming’s left hand rule.

b) Write the principle of working of an electric motor.

c) Explain the function of following parts of an electric motor.

1) Armature

2) Brushes

3) Split ring

Ans:

a)

  • According to Fleming’s left hand thumb rule, when we put a current carrying conductor inside the external magnetic filed it produces a force which is proportional to magnetic field and direction of current also.
  • Thus, if we stretched out left hand in such way that if the middle finger, index finger and the thumb are mutually perpendicular to each other then thumb gives the direction of motion, index finger gives the direction of magnetic field and the middle finger gives the direction of current.

b)

  • Electric motor is a rotating device which converts electrical energy into mechanical energy.
  • The principle in which electric motor works is stated as follows.

Principle:

When a current carrying coil is placed in the magnetic field then there will be torque produced in the coil which rotates the coil whose direction of rotation is given by Fleming’s left hand rule.

c)

1) Armature:

  • Armature is the rectangular coil insulated by copper wire and which is placed between two poles of magnet.
  • When current is passed through the coil then it rotates in the magnetic field whose direction is given by Fleming’s left hand rule.

2) Brushes:

There are two conducting stationary brushes through which current enters the armature coil.

3) Split ring:

The spilt rings are used to reverse the direction of current flowing through the armature coil.

 

Or 20.) a) Write the function of each part of the human eye.

1) Cornea

2) Iris

3) crystalline lens

4) 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.

Ans:

a)

1) Cornea:

  • The light enters through the cornea which is the thin membrane forming bulge at the front part of the eye which is also called as eyeball.
  • And the most of the refraction is takes place at the outer surface cornea due to large change in refractive index.

2) Iris:

  • The small part pupil controls the amount of light entering the eye. And the size of the pupil is controlled by the muscular part called as iris.

3) Crystalline lens:

  • Crystalline lens helps in adjusting the focal length in order to focus the object at different positions on retina.

4) ciliary muscles:

  • When the ciliary muscles get relaxed the lens becomes thinner and hence focal length increases due to which we can see the distant object clearly.
  • While when the ciliary muscles get contracted the lens becomes thick and hence focal length decreases due to which we can see the objects near to us clearly.

b)

  • The sun appears reddish at the time of sunrise because at that time the sun is very close to horizon due to which it has to travel longer distance through atmosphere in order to reach the observer on the earth.
  • During this, the blue and violet get scattered more and away from the path of light and hence cannot reach the observer directly.
  • The light reaching to observer is only red light and hence sun appears reddish at the time of sunrise or morning.
  • The phenomenon is scattering is not possible on the moon as there is no atmosphere to scatter the light.

 

Section B

 

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 tube 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.

Ans:

  • Given that, test tube A contains aqueous solution of FeSO4.
  • Test tube B contains aqueous solution of CuSO4.
  • Test tube C contains Al2(SO4)3 and test tube D contains aqueous solution of CuSO4.
  • When she add pieces of aluminium metal to test tube A and B then following chemical reaction takes place.

In test tube A:

Al + 3FeSO4 -> Al2(SO4)3 + 3Fe

In test tube B:

2Al + 3CuSO4 -> Al2(SO4)3 + 3Cu

  • In test tube A and B, the aluminium can displaces the iron and copper and hence there will be colour change in them.
  • When she adds iron metal to test tubes C and D then following chemical reaction takes place.
  • In test tube C:

2Fe + Al2(SO4)3 -> No reaction

 

In test tube D:

Fe + CuSO4 -> FeSO4 + Cu

  • In test tube C iron metal cannot displaces aluminium hence there will be no reaction.
  • While in test tube D iron metal displaces copper sulphate solution and hence there will be change in colour takes place in it.
  • Thus we can say that, aluminium displaces both the elements iron and copper hence it is more reactive element.

 

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 chemical reaction in this case.

Ans:

  • When a solution of sodium sulphate is added to a solution of barium chloride taken in test tube then there will be formation of insoluble precipitate of barium sulphate and sodium chloride is formed.
  • And hence this reaction is called as precipitation reaction.
  • The chemical equation for this reaction is as given below:

BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

 

24.) List the steps of preparation of temporary mount of a leaf peel to observe stomata.

Ans:

Following are the steps of preparation of a leaf peel to observe stomata:

Initially take a healthy leaf from the potted plant.

  • Now, fold the leaf over and slowly pull the peel part apart with the help of forceps. This removed part of the peel from lower surface is placed in a watch glass containing water.
  • Then we have to add some drops of safranin stain in the watch glass.
  • Leave the peel for 2-3 minutes in glass and then place it on a clean glass slide.
  • Now, we add a drop of glycerine over the peel by placing a clean coverslip slowly over it by using needle so that no air bubbles will be formed.
  • Now, by using blotting paper we remove the excess stain and glycerine.
  • Finally, we observe the slide under the low and high power magnification of compound microscope.
  • In this way, we have to follow the above mentioned steps for the preparation of temporary mount of leaf peel to observe stomata.

 

Or 25.) 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).

Ans:- The following figure shows the various stages of asexual reproduction by budding in yeast in proper sequence

 

26.) An object of height 4cm is placed at a distance of 30cm from the optical centre 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 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.

Ans:

Given that,

For convex lens:

Object height = 4cm

Object distance u= -30cm

Focal length f= 20cm

We know that, the lens formula is given by,

1/f = 1/v – 1/u

1/20= 1/v +1/30

Thus, 1/v = 1/20 -1/30

10/v = 1/2 – 1/3

10/v = 1/6

v/10= 6

Thus, v= 60cm

Hence we can say that, the image will be formed at a distance of 60 cm as shown in figure below.

We have,

Magnification of lens is given by,

m= v/u = image height/ object height

60/-30= image height/4

Image height = -8 cm

Here, the negative sign indicates that the image formed will be real and inverted as shown in figure

Hence, we can find the ratio of size of image to the size of object as

8/4= 2

 

27.) The values of current flowing through a given resistor of resistance R for the corresponding values of potential difference V across the resistor are as given below.

Plot a graph between current I and potential difference V and determine the resistance R of the resistor.

Ans:

We have taken two points on the graph A=(0.3,1.5) and B=(0.8, 4)

The slope of the graph gives value of resistance.

Hence,  slope = R= ∆V/∆I = (4-1.5)/ (0.8-0.3) = 2.5/ 0.5 = 25/5= 5ohm

Thus, the resistance R of the resistor is found to be 5 ohm.

 

 

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