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Punjab State Board PSEB 10th Class Science Book Solutions Chapter 9 Heredity and Evolution Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 9 Heredity and Evolution

PSEB 10th Class Science Guide Heredity and Evolution Textbook Questions and Answers

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as :
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw.
Answer:
(c) TtWW.

Question 2.
An example of homologous organs is :
(а) our arm and a dog’s fore-leg
(б) our teeth and an elephant’s tusks
(c) potato and runners of grass
(d) All of the above.
Answer:
(d) All of the above.

Question 3.
In evolutionary terms, we have more in common with :
(а) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
On this basis we cannot say that light eye colour is dominant or recessive until a cross is made between parent having light eye colour and another with dark eye colour. Only then it will be possible to predict the dominant or recessive nature of the gene.

Question 5.
How are the areas of study of evolution and classification interlinked?
Answer:
Evolution and classification are interlinked as evident from following points :

• Characteristics are shared by most of the organisms. The characteristic in the next level of classification will be shared by most and not by all.
• Cell designs also indicate this relationship.
• Groups formed during classification are related to their similarities.

Question 6.
Explain the terms homologous and analogous organs with example.
Answer:
Homologous organs: The organs of different classes have different forms because they have to perform different functions but their structures basically remain similar. Such organs are called homologous organs.

Example:

• Fore limbs of amphibians, birds and mammals have same fundamental structural plans but perform different functions.
• In plants, the homologous organs may be a thorn of Bougainvillea or a tendril of cucurbita both arising in axillary position.

Analogous organs: The organs are quite different in their structure and origin but similar in function. Such organs are known as analogous organs. The presence of analogous organs proves that different structures can be modified to perform a similar function. Analogy indicates convergent evolution.
Examples. The wings of insects and vertebrates perform the same function.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Make a chart or thermocol sheet showing the following monohybrid cross

Dominance of black coat colour in dogs

Question 8.
Explain the importance of fossils in deciding evolutionary relationship.
Answer:

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
Urey and Miller provided experimental evidence regarding origin of life from inanimate matter. They assembled an atmosphere similar to that, thought to exist on early earth.

In a spark flask they collected ammonia, methane and hydrogen sulphide, but no free oxygen over water at a temperature just below 100°C and sparks were passed through the mixture of gases to stimulate lightning. At the end they obtained organic molecules such as amino acid, urea, sugars. Amino acids which make up protein molecules. Thus they showed life originated from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable Variation than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
Genetic variations arise in nature as a result of following mechanism during sexual reproduction are more viable and raw materials of evolution.

• Crossing over during gamete formation.
• Random segregation of chromosome during meiosis at the time of gamete formation has decreased.
• Random rejoining of gametes having different genetic set up in the chromosomes during fertilisation.

Question 11.
How is equal genetic contribution of male and female parents ensured in the progeny?
Answer:
During sexual reproduction fusion of gametes having haploid set of chromosomes belonging to male and female parents ensure equal contribution.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes. The organism with useful variations will adapt and survive. Moreover they leave behind more offsprings and populations with such genetic variations will survive.

Science Guide for Class 10 PSEB Heredity and Evolution InText Questions and Answers

Question 1.
If a trait A exists in 10% of population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
In asexually reproducing organism trait B originated earlier. The variations in a population are only due to inaccuracies of DNA copying.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
The useful variation in individuals of a species will enable them to adapt according to the changes and new needs. Thus they will enable the survival of species.

Question 3.
How do Mendel’s experiments show that gene may be dominant or recessive?
Answer:
Mendel conducted experiments on garden pea plant selecting seven visible contrasting characters. He selected and crossed homozygous tall pea plant having the genotype TT with a homozygous dwarf pea plant having the genotype tt. Fx generation consists only of tall plants, having genotype Tt. Since they have an allele for dwarfness also, they are all hybrids. The expressed allele T for tallness is dominant over the unexpressed allele t for dwarfness. The fact that the allele for dwarfness is present in the F1 plants can be verified by interbreeding them when F2 progeny will consist of both tall and dwarf plants in the ratio of 3 : 1. On this basis he proposed “Law of Dominance.”

Question 4.
How do Mendel’s experiments proved that traits are inherited indepen dently?
Answer:
Mendel proposed a law on the basis of a dihybrid cross between two homozygous parents. He selected a dominant plant with round and yellow seeds and a recessive plant with wrinkled and green seeds, yields Fx offspring showing the dominant form of both traits, viz. round and yellow. Fx plants, on selling, produce F2 progeny with two parental and two new recombinant phenotypes, that is round yellow: round green: wrinkled yellow: wrinkled green in the ratio of 9 : 3 : 3: 1. This ratio is called Mendel’s dihybrid phenotypic ratio. The factors (genes) of different traits are independent of each other in their distribution into the gametes and in the progeny. This is Mendel’s law of independent assortment.

Question 5.
A man with blood group A married a person with blood group O. Their daughter has blood group O. Is this information enough to tell you which of the blood group trait A or O is dominant. Why or why not?
Answer:
As blood groups is hereditary character, the knowledge of blood groups of parents can give information about the possible blood groups of children and vice-versa.

In this case illustration is as follow :

In the above cross, 50 per cent of progeny will have A blood group and 50 per cent O blood group.
At the same time this data is insufficient. It is not mentioned father has homozygous or heterozygous A blood group. If it is homozygous A then 100 per cent of progeny will have A blood group as Gene IA is dominant over Gene I°.

Question 6.
How is the sex of child determined in human beings?
Answer:
Determination of the sex of child. Sex chromosomes determine sex in human beings. In males, there are 44 +
XY chromosomes, whereas, in female there are 44 + XX chromosomes. Here,
X and Y chromosomes determine sex in the human beings.

Sex determination in man (Note that all the eggs carry X-chromosome but one-half of the sperm carry an X-chromosome and one half carry a Y-chromosome)

Two types of gametes are formed in male, one type is having 50%
X-chromosome, whereas the other type is having Y-chromosome. In female, gametes are of one type and contain X-chromosome.

The females are homogametic. If male gamete having Y-chromosome (androsperm) undergoes fusion with female gamete having X-chromosome the zygote will have XY chromosomes and this gives rise to male child.

If the male gamete having Fig. 9.1. Sex determination in man (Note X-chromosome undergoes fusion with that all the eggs carry X-chromosome but female gamete having X-chromosome, one-half of the sperm carry an the zygote will be having XX-chromosome X-chromosome and one half carry a and this gives rise to female child. Y-chromosome)

Question 7.
What are different ways in which individuals with a particular trait may increase in a population?
Answer:
A particular population with specific traits will increase due to following reasons :

• Sexual reproduction which results into variations.
• Inheritance of variations.
• Natural Selection. The individuals with special traits survive the attack of their predators and multiply while the others will perish.
• Genetic drift provides diversity without any adaptation.

Question 8.
Why are traits acquired during life-time of an individual not inherited?
Answer:
Change in non-reproductive tissue (somatic cells) cannot be passed on to the DNA of germ cells. Thus the acquired trait will die with the death of individual. It is non- heritable and cannot be passed on to its progeny. Changes that occur in DNA of germ cells are inherited.

Question 9.
Why are the small number of surviving tigers is a cause of worry from the point of view of genetics?
Answer:
As the population of tigers is decreasing, there is loss of genes from the gene pool. There cannot be recombinations and variations. Hence no evolution. If number falls suddenly they may become extinct.

Question 10.
What factors could lead to the rise of new species?
Answer:
Factors leading to rise of new species

• Genetic variations
• Mutations
• Natural selection
• Reproductive isolation
• Origin of new species.

Question 11.
Will geographical isolation be a major factor in the speciation by a self- pollinating plant species? Why or why not?
Answer:
No, m self-pollinating species, geographical isolation will not play any role for speciation because the self-pollination is occurring on the same plant.

Question 12.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, as there is neither genetic drift nor gene flow play any role during speciation. Moreover asexual reproduction involves single parent and natural geographical barrier can occur between different organisms.

Question 13.
Give an example of characteristic being used to determine how close two species are in evolutionary terms.
Answer:
Homologous organs helps to identify the relationship between organisms. These characteristics in different organisms would be similar because they are inherited from a common ancestor. Example. Fore limbs of mammals having same basic structural plans in birds, reptiles and mammals however the functions get modified in different species.

Question 14.
Can the wing of butterfly and wing of a bat be considered homologous organs? Why or why not?
Answer:
Wings of insects and wings of birds have different basic structural plan and origin. They perform the same function. Thus they are analogous organs and not homologous organs.

Question 15.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are preserved remains, tracks or traces of organisms that lived in the past. Fossils have been found linking all major groups of vertebrates.

Significance of fossils

• Fossils are direct evidence in support of evolution.
• Living forms with simple organization appeared earlier than the complex forms. We can conclude this because fossils of lower layers of the earth are simple as compared to fossils of the upper layers.
• Several forms bearing intermediate characters indicate the transition from an earlier simple to a later complex.
• Fossils of Archaeopteryx serve as a missing link between reptiles and birds. This bird has wings and unlike birds, it had teeth and a long tail.
• On the basis of the fossil records, the complete evolutionary history of present-day horse has been studied.

Question 16.
Why are human beings which look so different from each other in terms of size, colour, and looks are said to be belonging to the same species?
Answer:

• DNA studies have shown that human beings belong to the same species.
• The number of chromosomes is the same.
• All have originated from a common ancestor.
• They interbreed among themselves to produce fertile young ones of their own kind.

Question 17.
In evolutionary terms can we say that which among bacteria, spiders, fish, and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Chimpanzees have a better body design as compared to the other three mentioned. They are better adapted for locomotion, communication, and thinking.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 13 Magnetic Effects of Electric Current Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

PSEB 10th Class Science Guide Magnetic Effects of Electric Current Textbook Questions and Answers

Question 1.
Which of the following correctly describes the magnetic field near a long wire? The field consists of :
(a) straight lines perpendicular to the wire.
(b) straight lines parallel to the wire.
(c) radial lines originating from the wire.
(d) concentric circle centred on the ware.

Answer:
(d) concentric circles centred on the wire (Figure)

Question 2.
The phenomenon of electromagnetic induction is :
(а) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating the coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called :
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator.

Question 4.
The essential difference between an AC generator and a DC generator is that:
(а) AC generator has an electromagnet while a DC generator has a permanent magnet.
(б) DC generator will generate higher voltage.
(c) AC generator will generate higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) AC generator has slip rings while DC generator has a commutator.

Question 5.
At the time of short circuit, the current in the circuit.
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electric energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire.
Answer:
(a) is false. It converts electric energy to mechanical energy.
(b) is true.
(c) is true.
(d) is false. Green is usually earth wire.

Question 7.
List three sources of magnetic field.
Answer:
Sources of Magnetic field are :

• Magnet.
• Current carrying conductor
• Current carrying solenoid.

Question 8.
How does a solenoid behave like a magnet? Can you determine north and south poles of current carrying solenoid with the help of bar magnet? Explain.
Answer:
Solenoid: It consists of a coil of a number of turns of insulated copper wire closely wound in the shape of a cylinder. Magnetic field around a current carrying solenoid is shown in Figure.

Field lines of the magnetic field inside and around a current carrying solenoid.

These magnetic lines due to current carrying solenoid appear to be similar to that of a bar magnet shown in Figure.

One end [right end] of solenoid behaves like north pole and the other end [left end] behaves like south pole. Magnetic field lines inside the solenoid are in the form of parallel straight lines. This means that the field is the same at all points inside the solenoid.

Field lines around a bar magnet.

A soft iron rod when placed inside the solenoid behaves, like an electromagnet.

Question 9.
When is, the force experienced in a magnetic field, the largest?
Answer:
When the field is perpendicular to current carrying conductor, the force experienced by a current carrying conductor placed in a magnetic field is largest.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam moving horizontally with back towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of the magnetic field?
Answer:
The magnetic field will be acting in vertically downward direction in accordance with Fleming’s left hand rule. [Direction of the current should be considered in a direction opposite to the direction in which the electrons move].

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in electric motor?
Answer:
Electric motor: It is a device which is used to convert electric energy into mechanical energy.
Principle, “When a current carrying coil is placed in a uniform magnetic field, it experiences a torque which rotates the coil.”

I indicates direction of current; F the direction field and M the direction of motion

Working: A direct current from a battery is passed through armature. The current flows in the coil along ABCD as shown in Figure (a). The limb AB of the coil experience downward and CD of the coil experience upward force in accordance with Fleming’s left hand rule. These two equal and opposite forces constitute a couple tending to rotate the coil in clockwise direction. After half the rotation, brush B₁ has contact with S₂ and brush B₂ with S₁. The direction of the current gets reversed. The current now flows along DCBA instead of along ABCD. Limb DC experiences downward and BA experiences an upward force in accordance with Fleming’s left-hand rule.

The process repeats itself and motion of armature becomes continuous after some time.
Functions of Split rings. They help in reversing the direction of current in the coil after every half rotation.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motor are used in battery operated toys, in tape recorder, in car fans, mixers, grinders, computers and variety of other electric appliances.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil; (ii) withdrawn from inside the coil; (iii) held stationary in the coil?
Answer:

• When the bar is pushed into the coil, there will be momentary deflection of galvanometer in one direction. This is so because when magnet is brought near the coil, magnetic lines linked with the coil increases, so that induced emf is produced which induces current in the coil.
• Faster, we push the magnet, more will be the deflection.
• When the bar magnet is withdrawn, there will again be momentary galvanometer deflection but in a direction opposite to that when magnet was pushed in. This time also current is induced in the coil.
• When the magnet is held stationary inside the coil, there will be no deflection in galvanometer. It is because no emf is induced and hence no current is induced in the coil.

Question 14.
Two circular coils A and B placed closed to each other. If the current in the coil A is changed, will some current be induced in coil B? Give reason.
Answer:
If the current in the coil is changed (switched on or switched off), then an electric current is induced in coil B. ‘

It is because when, plug in the key is introduced, current flows through the coil ‘A’ so that magnetic field is produced all round it. These magnetic lines produced in the coil ‘A’ will pass through the coil ‘B’ with the result induced emf and hence induced current is produced in the coil ‘B’ which is indicated by deflection of the galvanometer. Now when plug is removed from the key (switched off) the magnetic lines of force Jinked with the coil ‘B’ again change (decreases). This time again current is induced in the coil ‘B’.

Question 15.
When does an electric short circuit occur?
Answer:
Electric short-circuit occurs when :

• live wire incidently touches neutral or earth-wire.
• insulation around the current carrying wires is weak.
• insulation gets hardened by the excessive use.
• current passed through wire is more than its rating.

Question 16.
What is the function of earth wire? Why is it necessary to earth metallic appliances?
Answer:
Earth wire. It is used as a safety measure especially for electric appliance having metallic body. The metallic body of appliances like electric press, fans, toasters, refrigerators etc. are connected to earth wire which provides an easy path for current to go to the earth in case live wire touches the body of appliance incidently. The user will not suffer a severe electric shock in the event of touching a defective appliance.

Question 17.
State two properties of magnetic field lines.
Answer:

1. Magnetic field start from north and end at south.
2. They never intersect each other.

Science Guide for Class 10 PSEB Magnetic Effects of Electric Current InText Questions and Answers

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle is a small bar magnet with one end as north and the other end as south pole. It is a well known fact that similar poles repel each other and dissimilar poles attract each other. When N-pole of a bar magnet is brought near the compass, the north pole of compass gets repelled while its S-pole is attracted so that compass needle gets deflected.

Question 2.
Draw magnetic lines around a bar magnet.
Answer:
Place a bar magnet in the middle of a sheet of paper fixed on drawing board by using adhesive tape. Mark the boundary of the magnet with pencil. Place compass near north pole of the magnet when south pole of the compass will point towards north pole of the magnet. Mark the two points a, b at the two ends of the needle. Move the needle to a new position such that S-pole of compass needle is at b [position previously held by N-pole].

Repeat this till you reach south pole of the magnet (Figure).

Drawing magnetic field lines by compass needle.

Now Join the marked points on a paper by a smooth line which gives one magnetic line of force as shown in Figure. Repeat this procedure taking Figure Magnetic field around a bar different starting points.

Magnetic field around a bar magnet.

Question 3.
List the properties of magnetic lines of force.
Or
Write characteristics of magnetic field lines.
Answer:
Properties (characteristics) of magnetic lines of force are :

• These are the closed curves passing through the magnet. Outside the magnet lines of force start from north pole of the magnet and end at the south pole and inside the magnet, the direction of magnetic lines are from south pole to north pole.
• The two magnetic lines of force never intersect each other.
• They have a tendency to contract lengthwise which explains the attraction between opposite poles.
• The tangent at any point of the magnetic lines of force gives the direction of the field at that point.
• They exert lateral pressure upon each other, which explains repulsion between like poles.

Question 4.
Why two magnetic lines of forces never intersect each other?
Answer:
No two magnetic lines of force cross each other because, if they did so, then at the point of intersection; the compass needle would point towards two directions at the same time, which is not possible. Hence two magnetic lines never intersect each other.

Question 5.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply right hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
The direction of the magnetic field inside and outside loop is as shown in Figure. By applying right hand rule we find that the direction of magnetic field inside the loop is downward normally and outside the loop it is normal to the plane of paper.

Question 6.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
Uniform magnetic field is shown by equidistant and parallel lines as shown in Figure If the parallel lines are close to each other, the field is strong. The stronger the field, the closer are the lines.

Question 7.
Choose the correct option.
The magnetic field inside a long straight solenoid carrying current (a) is zero ;
(b) decreases as we move towards end ;
(c) increases as we move towards end;
(d) is same at all the points.
Answer:
(d) is correct. The magnetic field inside a long straight solenoid is same at all the points.

Question 8.
Which of the following property of proton can change while it moves freely in a magnetic field? [There may be more than one correct answer].

(a) mass
(b) speed
(c) velocity
(d) momentum.
Answer:
(c) Velocity and (d) Momentum.

Question 9.
In activity shown, how do you think the displacement of rod AB will be affected :
(i) if the current in rod AB is increased;
Answer:
Since force acting on the rod is directly proportional to the current passing through it. Therefore, the displacement will be increased when current is increased.

(ii) a stronger horse shoe magnet is used;
Answer:
Stronger the magnet, more will be the force and hence the displacement.

(iii) length of the rod AB is increased.
Answer:
Force is also directly proportional to the length of the rod. Hence rod will be displaced more if the length is increased.

Question 10.
A positively charged particle emitted from a nucleus alpha particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is :
(a) towards south
(b) towards east
(c) downward
(d) upward.
Answer:
(d) upward [In accordance with Fleming’s left hand rule]

Question 11.
State Fleming’s left hand rule.
Answer:
Fleming’s left hand rule. It states, “Stretch the thumb, fore finger and middle finger of your left hand such that they are mutually perpendicular to each other. If the first finger points in the direction of magnetic field and central (second) finger points towards the direction of current then thumb points towards the direction of motion as shown in Figure.

Flemmgs left hand rule for direction of force on current carrying conductor

Question 12.
What is the principle of an electric motor?
Answer:
Principle of Electric motor. Electric motor is based upon the principle that when a current carrying coil is placed in a uniform magnetic field, it experiences torque which rotates the coil.

Question 13.
What is the role of the split ring in an electric motor?
Answer:
The split rings act as a commutator in D.C. motor i.e., it reverses the direction of current through the circuit after every half cycle.

Question 14.
Explain different ways to induce current in a coil.
Answer:
There are mainly two methods to induce current in a coil.

Induced current due to relative motion between coil and magnet.

When the magnet is stationary [at rest as in Figure (a)], no induced current is produced.
1. When magnet is moving with its N-pole towards the coil, deflection in galvanometer is as shown in Figure (b). Direction of the current gets reversed when N-pole is withdrawn from inside the coil (going away from coil) as shown in Figure (c). Faster the magnet moves, more is the deflection and hence current in galvanometer.

Deflection produced in the galvanometer needle by the current is same when N-pole was moving down [Figure (6)] or when south pole is out of the coil [Figure (e)].

Current and hence deflection in the galvanometer needle will be again produced if magnet is kept at rest and coil is moved.

2. By changing current in the neighbouring circuit. Take a non¬conducting cylindrical tube (say of card board). Wind two set of coils I and. II on it as shown. Connect a battery and a key to the ends of coil I and a galvanometer to the ends of coil II. When plug is inserted in key K, there will be an instantaneous deflection m galvanometer even though there is no cell in this circuit. Now take out the plug from key K. An instantaneous large deflection in opposite direction to previous deflection will be produced in galvanometer. Thus current has been induced in coil II due to increase or decrease of current in coil I.

Induced Current by changing current m the neighbouring circuit.

Question 15.
State the principle of electric generator.
Answer:
Principle of Electric Gen¬erator. Electric generator is based upon Fleming’s right hand rule.

Right hand rule

Fleming’s right hand rule. Stretch the thumb, fore-finger, control (middle) finger of right hand so that they are perpendicular to each other. If fore-finger indicates the direction of magnetic field, the thumb shows the direction of motion of the conductor, the central (middle) finger will show the direction of induced current.

Question 16.
Name some sources of direct current.
Answer:
Sources of direct current are

• Cell
• Battery
• D.C. generator.

Question 17.
Which sources produce alternating current?
Answer:
Alternating current is produced by A.C. generators. There are hydro-generator and thermal generators.

Question 18.
Choose the correct option :
A rectangular coil of copper wires is rotated in magnetic field. The direction of induced current changes once in each:
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution.
Answer:
(c) half rotation.

Question 19.
Name two safety measures commonly used in electric circuits and appliances.
Or
Write the two safety measures commonly used in electric circuit appliances.
Answer:
Common Safety Measures used in Electric Circuits. Two most common safety measures used, are :

• A safety fuse of proper rating connected in a circuit prevents damage to the electric appliances and also the circuit due to overloading or short circuiting.
• Earth wire prevents possible electric shock when live wire incidently touches the body of appliance.

Question 20.
An electric oven of 2kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
Given : Power of oven (P) = 2 kW
= 2,000 W
Voltage (V) =220 volts
Current (I) =?

We know that, P = V x I
or I = \(\frac{\mathrm{P}}{\mathrm{V}}\)
Current (I) = \(\frac{2,000}{200}\)
= \(\frac{100}{11}\)
= 9.09 A

A current of 9.09 A will flow in the circuit. Since the current rating of circuit is 5 A, the fuse (of 5 A) rating if inserted in the circuit will burn up. If no fuse is placed in the circuit, there may be a fire.

Question 21.
What precautions should be taken to avoid the overloading of domestic electric circuits?
Answer:
Precautions to avoid overloading.

• Wires used for carrying current should be of proper current rating. Whereas wire of low current rating may be used for lighting electric bulbs, tubes, T.V. etc., and wires of higher current rating should be used for A.C., heating appliances etc.
• There should be a separate circuit for heating appliance.
• PVC of good quality should be used for insulating wires.
• Each circuit should have a fuse of proper rating.
• After every 3-4 years wires should be replaced by new wires of proper rating.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 3 Metals and Non-metals Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 3 Metals and Non-metals

PSEB 10th Class Science Guide Acids, Bases and Salts Textbook Questions and Answers

Question 1.
Which of the following pairs will give displacement reactions :
(а) NaCl solution and copper metal
(b) MgCl₂ solution and aluminium metal
(c) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal?
Answer:
(d) AgNO₃ solution and copper metal?

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting :
(a) applying grease
(b) applying paint
(c) applying a coating of zinc
(d) all of the above :
Answer:
(c) applying a coating of zinc

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be :
(a) calcium
(b) carbon
(c) silicon
(d) iron.
Answer:
(a) calcium

Question 4.
Food cans are coated with tin and not with zinc because :
(а) zinc is costlier than tin
(b) zinc has a higher melting point than tin
(c) zinc is more reactive than tin
(d) zinc is less reactive than tin.
Answer:
(c) zinc is more reactive than tin

Question 5.
You are given a hammer, a battery, a bulb, wires and a switch.
(i) How could you use them to distinguish between samples of metals and non-metals?
(ii) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Answer:
(i) Set up the electric circuit as shown below :

To distinguish between metals and non-metals on the basis of their electrical conductivity.

Insert the sample to be tested between clips A and B
If the bulbs glows, the sample is metal.
If the bulb does not glow, the sample is non-metal.
Thus metals are good conductors of electricity whereas non-metals are poor conductors of electricity.

(ii) If a substance produces a sound when struck beating with a hammer, it is a metal and if no sound is produced, it is a non-metal.
Metals are sonorous whereas non-metals are non-sonorous.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Amphoteric oxides. The metal oxide which react both with acids as well as bases to produce salts and water are known as amphoteric oxides.

Examples. Zinc oxide, ZnO Aluminium oxide, Al2O3

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:
Zinc and magnesium can displace hydrogen from dilute acids. Copper and silver cannot displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining a metal M, what would you take as the anode, the cathode and the electrolyte?
Answer:
During the electrorefining of metal, the impure metal is made as anode, a thin strip of pure metal M is made as cathode. The electrolyte used a soluble salt of metal M. to be refined.

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it as shown in figure

Collection of Gas
(a) What will be the action of gas on
(i) dry litmus paper?
Answer:
No action.

(ii) moist litmus paper?
Answer:
It turns moist litmus paper red and then bleaches it.

(b) Write a balanced chemical equation for the reaction taking place.
Answer:
S + O₂ → SO₂

Question 10.
State two ways to prevent the rusting of iron.
Answer:
The rusting can be prevented by

1. By painting, oiling, greasing, galvanising, chrome plate or anodising.
2. By forming alloys.

Question 11.
What type of oxides are formed when non-metals combine with oxygen?
Answer:
As they form two types of oxides :
(a) Neutral oxides such as CO, NO etc.
(b) Acidic oxides such as SO₂, CO₂ etc.

Question 12.
Give reasons :
(a) Platinum, gold and silver are used to make jewellery.
Answer:
Platinum, gold and silver are used to make jewellery because these are not attacked by air and moisture. They don’t undergo corrosion and retain their lustre for a long time. Also these metals are malleable and ductile.

(b) Sodium, potassium and lithium are stored under oil.
Answer:
Sodium, potassium and lithium are stored under oil because in contact with moist air containing carbon dioxide, these are covered with a carbonate layer.
e.g. 4Na + O₂ → 2Na₂O
Na₂O + H₂O → 2NaOH
2NaOH + CO₂ → Na₂CO₃ + H₂O

Also they react with water.
e.g. 2Na + 2H₂O → 2NaOH + H₂.
Hence, these metals are stored under oil.

(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.
Answer:
This is due to the reason that a thin sticking oxide layer of aluminium is formed on its surface which prevents further reaction. Also aluminium is a good conductor of heat and the oxide layer (Al₂O₃) is stable even at high temperatures.

(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.
Answer:
This is because it is easier to obtain a metal from its oxide as compared to its sulphide and carbonate ore.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
In tarnished copper vessel there is a layer of basic copper carbonate. This basic layer can be cleaned by using sour substances such as lemon or tamarind juice which are acidic in nature and dissolve the basic coat of copper carbonate

Question 14.
Differentiate betw een metals and non-metals.
Or
Differentiate between metals and non-metals on the basis of their chemical properties.
Answer:
Differences between metals and non-metals :

Metals	Non-Metals
1. They form basic oxides.	1. They form acidic or neutral oxides.
2. They displace hydrogen from dilute acids.	2. They don’t displace hydrogen from dilute acids.
3. Usually they don’t combine with hydrogen. Only a few reactive metals combine with hydrogen to form metal hydroxide which are electrovalent compounds.	3. They react with hydrogen to form hydrides which are covalent compounds.
4. They form chlorides which are electrovalent compounds.	4. They react with chlorine to form chlorides which are covalent compounds.

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?
Answer:
The solution used by goldsmith to bring back the glitter of old and dull gold ornaments was aqua regia. It is a freshly prepared mixture of cone. HCl and cone. HNOg in the ratio 3 : 1. It can dissolve gold,

Question 16.
Give the reasons why copper is used to make hot water tanks but steel (an alloy of iron) is not.
Answer:
This is because iron present in steel reacts with steam to form ferrosoferric oxide whereas copper has no action with water. As a result of it, the body of the steel tank becomes weaker and weaker in case of iron and not in case of copper.

Science Guide for Class 10 PSEB Metals and Non-metals InText Questions and Answers

Question 1.
Give an example of a metal which :
(i) is a liquid at room temperature
Answer:
Mercury

(ii) can be easily cut with a knife
Answer:
Sodium

(iii) it is the best conductor of heat
Answer:
Silver

(iv) is a poor conductor of heat.
Answer:
Lead.

Question 2.
Explain the meaning of malleable and ductile.
Answer:
Malleable: A substance is said to be malleable, if it can be beaten into sheets. e.g. metals are malleable.
Ductile: A substance is said to be ductile, if it can be drawn into wires e.g. the metals are ductile.

Question 3.
Why is sodium kept immersed in kerosene oil?
Answer:
This is because in contact with moist air containing carbon dioxide, it is covered with a carbonate layer.
4Na + O₂ → 2Na₂O
Na₂O + H₂O → 2NaOH
2NaOH + CO₂ → Na₂CO₃ + H₂O

Also sodium reacts with water.
2Na + 2H₂O → 2NaOH + H₂
Hence, sodium is kept immersed in kerosene oil.

Question 4.
Write equation for the reactions of
(i) iron with steam
Answer:

(ii) calcium and potassium with water.
Answer:
Ca + 2H₂O → Ca(OH)₂ + H₂
2K + 2H₂O → 2KOH + H₂

Question 5.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows :

Use the table above to answer the following questions about metals, A, B, C and D.
(i) Which is the most reactive metal?
Answer:
B is the most reactive metal.

(ii) What would you observe if B is added to a solution of copper (II) sulphate?
Answer:
B will displace copper from copper sulphate solution.

(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.
Answer:
The decreasing order of reactivity is B > A > C > D.

Question 6.
Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄.
Answer:
Hydrogen gas is produced when dilute hydrochloric acid is added to a reactive metal.

Question 7.
What would you observe when zinc is added to a solution of iron (II) sulphate? Write the chemical reaction that takes place.
Answer:
When zinc is added to a solution of iron (II) sulphate it will displace iron from it and light green colour of solution gradually fades away.

Question 8.
(i) Write the electron-dot structures for sodium, oxygen and magnesium.
Answer:
Electron-dot structures for sodium, oxygen and magnesium are

(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
Answer:
Formation of Na₂O:

(iii) What are the ions present in the above compounds?
Answer:

Question 9.
Why do ionic compounds have high melting points?
Answer:
In ionic compounds there are strong electrostatic forces of attraction between oppositely charged ion and a considerable amount of energy is required to break the strong interionic attraction.

Question 10.
Define the terms :
(a) mineral
Answer:
Mineral. The compounds of elements occurring in earth’s crust which are associated with earthly impurities are called minerals.

(b) ore
Answer:
Ore. An ore is a mineral from which metal can be extracted conveniently and economically.

(c) gangue.
Answer
Gangue. The earthly impurities such as sand, lime stone, rocks etc. associated with minerals and ores are collectively known as gangue or matrix.

Question 11.
Name two metals which are found in nature in the free state.
Answer:
Gold and Platinum.

Question 12.
What chemical process is used for obtaining a metal from its oxide?
Answer:
Reduction process.

Question 13.
Metallic oxides of zinc, magnesium and copper were heated with the following metals :

Metal	Zinc	Magnesium	Copper
Zinc oxide
Magnesium oxide
Copper oxide

In which cases will you find displacement reactions taking place?
Answer:
Zinc can displace copper from copper oxide

Magnesium can displace zinc from zinc oxide

Also Mg can displace copper from CuO

Question 14.
Which metals do not corrode easily?
Answer:
The metals which are not attacked by air and moisture don’t corrode easily.

Question 15.
What are alloys?
Answer:
Alloys. These are the homogeneous mixtures of two or more metals or metals and non-metals.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3); (- 1, 0); (2, – 4)
(ii) (- 5, – 1); (3, – 5); (5, 2)
Solution:
(i) Let vertices of the ∆ABC are A (2, 3); B(- 1, 0) and C (2, – 4)
Here x₁ = 2, x₂ = – 1 x₃ = 2
y₁ = 3, y₂ = 0, y₃ = – 4 .
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\) [2 × (0 + 4) – 1 × (- 4 – 3) + 2 × (3 – 0)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{21}{2}\)
= 10.5 sq units.

(ii) Let vertices of the ∆ABC are A (- 5, – 1); B (3, – 5) and C (5, 2)
Here x₁ = – 5, x₂ = 3, x₃ = 5
y₁ = – 1, y₂ = – 5, y₃ = 2
∴ Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [- 5 (- 5 – 2) + 3 (2 + 1) + 5 (- 1 + 5)]
= \(\frac{1}{2}\) [35 + 9 + 20]
= \(\frac{1}{2}\) × 64 = 32 sq units.

Question 2.
In each of the following find the value of ‘k’ for which the points are coimear.
(i) (7, – 2); (5, 1); (3, k)
(ii) (8, 1); (k, – 4); (2, – 5)
Solution:
(i) Let given points be A (7, – 2); B (5, 1) and C (5, k)
Here x₁ = 7, x₂ = 5, x₃ = 3
y₁ = – 2, y₂ = 1 y₃ = k
Three points are collinear iff
\(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) [7 (1 – k) + 5(k + 2) + 3(- 2 – 1)] = 0
or 7 – 7k + 5k +10 – 9 = 0
or – 2k + 8 = 0
or – 2k = – 8
or – k = \(\frac{-8}{-2}\) = 4 .
Hence k = 4.

(ii) Let given points be A (8, 1); B (k, – 4) and C(2, – 5)
Here x₁ = 8 x₂ = k, x₃ = 2
y₁ = 1, y = – 4, y = – 5
Three points are collinear iff
\(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)] = 0
or \(\frac{1}{2}\) [8 (- 4 + 5) + k (- 5 – 1) + 2 (1 + 4) = 0]
or 8 – 6k + 10 = 0
or – 6k = – 18 .
or k = \(\frac{-18}{-6}\) = 3.
Hence k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). FInd the ratio of the area of the triangle formed to the area of the given triangle.
Solution:
Let vertices of given triangle ABC are A(0, – 1); B (2, 1) and C (0, 3).
Also, D, E, F be the mid points of AB, BC, CA respectively.
Using mid point formula,
Coordinates of D = \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)\) = (1, 0)

Coordinates of E = \(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)\) = (1, 2)

Coordinates of F = \(\left(\frac{0+0}{2}, \frac{3-1}{2}\right)\) = (0, 1)

∴ Co-ordinates of the vertices of DEF are D (1, 0); E (1, 2); F (0,1).
Here x₁ = 1, x₂ = 1, x₃ = 0
y₁ = 0, y₂ = 2, y₃ = 1.
Area of ∆DEF = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [1 (2 – 1) + 1 (1 – 0) + 0 (0 – 2)]
= \(\frac{1}{2}\) [1 + 1 + 0] = \(\frac{2}{2}\) = 1.

In ∆ABC,
x₁ = 0, x₂ = 2, x₃ = 0
y₁ = – 1, y₂ = 1, y₃ = 3.
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [0 (1 – 3) + 2 (3 + 1) + 0 (- 1 – 1)]
= \(\frac{1}{2}\) [0 + 8 + 0] = \(\frac{8}{2}\) = 4
Required ratio = \(\frac{\text { Area of } \triangle \mathrm{DEF}}{\text { Area of } \triangle \mathrm{ABC}}\)
= \(\frac{1}{4}\)

Question 4.
Find the area of the quadrilateral whose vertices taken in order, are (- 4, – 2); (- 3, – 5); (3, – 2); (2, 3).
Solution:
Let co-ordinates of the given quadrilateral ABCD are A(- 4, – 2); B(-3, – 5); C(3, – 2) and D (2, 3).
Join AC then Quad. ABCD divides in two triangles
i.e. ∆ABC and ∆CDA

In ∆ABC
Here x₁ = – 4, x₂ = – 3, x₃ = 3
y₁ = – 2, y₂ = – 5, y₃ = – 2
Area of ∆ABC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [- 4 (5 + 2) + (- 3) (- 2 + 2) + 3 (- 2 + 5)]
= \(\frac{1}{2}\) [12 + 0 + 9] = \(\frac{21}{2}\) sq. units.

In ∆CDA
x₁ = 3, x₂ = 2, x₃ = – 4
y₁ = – 2, y₂ = 3, y₃= – 2
Area of ∆CDA = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [3 (3 + 2) + 2 (- 2 + 2) + (-4) (- 2 – 3)]
= \(\frac{1}{2}\) [20 + 15 + 0] = \(\frac{35}{2}\) sq. units.

Now, Area of quadritateral ABCD = (Area of ∆ABC) + (Area of ∆ACD)
= \(\frac{21}{2}+\frac{35}{2}=\frac{21+35}{2}\)
= \(\frac{56}{2}\) = 28 sq. units.

Question 5.
You have studied in Class IX, (Chapter 9, Q. 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).
Solution:
Given that coordinates of the vertices of ∆ABC are A(4, – 6); B (3, – 2) and C (5, 2)
Let CD is the median i.e. D is the mid point of AB which divides AABC into two pails i.e.

∆ADC and ∆CDB
Coordinates of D = \(\left(\frac{4+3}{2}, \frac{-6-3}{2}\right)\)
= \(\left(\frac{7}{2}, \frac{-8}{2}\right)\) = (3.5,- 4).

In ∆ADC
x₁ = 4, x₂ = 3.5, x₃ = 5
y₁ = – 6, y₂ = -4, y₃ = 2
Area of ∆ADC = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [4(—4—2)+3.5(2+6)÷5(—6+4)]
= \(\frac{1}{2}\) [- 24 + 28 – 101]
= \(\frac{1}{2}\) × -6
= 3 sq. units (∵ area cannot be negative).

In ∆CDB
x = 5, x = 35, x = 3
y = 2, y = – 4, y = – 2
Area of ∆CDB = \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\) [5 (- 4 + 2) + 3.5 (- 2 – 2) + 3 (2 + 4)]
= \(\frac{1}{2}\) [- 10 – 14 + 18]
= \(\frac{1}{2}\) × – 6 = – 3
= 3 sq. units(∵ area cannot be negalive)
From above discussion it is clear that area of ∆ADC = area of ∆CDB = 3 sq. units
Hence, a median of a triangle divides it into two triangles of equal areas.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students ?
Answer:
Frequency distribution table

From the frequency distribution table, it is very clear that the most common blood group is O and the rarest blood group is AB.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Answer:
Grouped frequency distribution table

From the frequency distribution table, we can conclude that for the majority of engineers, s i.e., 31 engineers, the distance from their residence to their place to work is 5 km or more than 5 km but less than 20 km. For some engineers, i.e., 5 engineers, this distance is less than 5 km. Still, for some engineers, i.e., 4 engineers, this distance is 20 km or more than 20 km but less than 35 km.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84-86, 86 – 88, etc.
Answer:

(ii) Which month or season do you think this data is about ?
Answer:
During 24 days out of 30 days, the relative humidity is 92 % or more than 92 %. This suggests that the data must have been collected during Monsoon.

(iii) What is the range of this data ?
Answer:
Range of the data
= The greatest observation – The least observation
= 99.2 – 84.9
= 14.3

Question 4.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:

(i) Represent the data given above by grouped frequency distribution table, taking the class intervals as 160 – 165, 165-170, etc.
Answer:
Grouped frequency distribution table

(ii) What can you conclude about their heights from the table?
Answer:
From the above frequency distribution, we can conclude that the height of 70 % students (35 students) is less than 165 cm while the height of only 10 % students (5 students) is 170 cm or more than that.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows :

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00-0.04, 0.04-0.08, and so on.
Answer:

(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Answer:
The concentration of sulphur dioxide was more than 0.11 ppm for 8 days (2 + 4 + 2).

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

Prepare a frequency distribution table for the data given above.
Answer:

Question 7.
The value of π up to 50 decimal places is given below:
3.1415926535897932384626433832795028 8419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
Answer:
Frequency distribution table

(ii) Which are the most and the least frequently occurring digits?
Answer:
The most frequently occurring digits are 3 and 9 (8 times each) and the least occurring digit is 0 (2 times).

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
Answer:

(ii) How many children watched television for 15 or more hours a week?
Answer:
Two children watched television for 15 or more hours a week.

Question 9.
A company manufactures car batteries of a Grouped frequency distribution table particular type. The lives (in years) of 40 such batteries were recorded as follows

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Answer:

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.1

Question 1.
Give five examples of data that you can collect from your day-to-day life.
Answer:
Five examples of data that can be collected from day-to-day life can be given as below:

1. Election results obtained from newspapers or TV
2. The number of different kinds of trees grown in our school.
3. Amounts of invoices of electricity for last one year at our home.
4. The number of students studying in different standards of our school.
5. Percentage of marks scored at last examination by the students in our class.

Question 2.
Classify the data in Q. 1 above as primary or secondary data.
Answer:
Among the five data given as the answer to Q. 1, data no.

Primary data:

• The number of different kinds of trees grown in our school.
• Amounts of invoices of electricity for last one year at our home.
• Percentage of marks scored at last examination by the students in our class.
• primary data which we can collect ourselves.

Secondary data:

• Election results obtained from newspapers or TV
• The number of students studying in different standards of our school.
• Secondary data as they are received from the sources of the newspapers or TV or the office of our school.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 13 Surface Areas and Volumes MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The total surface area of a cuboid with length 20 cm, breadth 15 cm and height 10 cm is
A. 1300
B. 650
C. 3000
D. 1500
Answer:
A. 1300

Question 2.
The lateral surface area of a cuboid with length 15 cm, breadth 8 cm and height 5 cm is ………………. cm².
A. 115
B. 230
C. 600
D. 300
Answer:
B. 230

Question 3.
The diameter of a cylinder is 7 cm and its curved surface area is 220 cm². Then, its height is ……………….. cm.
A. 35
B. 10
C. 44
D. 20
Answer:
B. 10

Question 4.
The total surface area of a closed cylinder with radius 3.5 cm and height 6.5 cm is ………………… cm².
A. 110
B. 220
C. 330
D. 440
Answer:
B. 220

Question 5.
The curved surface area of a cone is 880 cm². If its slant height is 20 cm, then its diameter is …………………… cm.
A. 14
B. 7
C. 3.5
D. 28
Answer:
D. 28

Question 6.
The curved surface area of a cone with diameter 14 cm and slant height 10 cm is ………………. cm².
A. 220
B. 1540
C. 110
D. 440
Answer:
A. 220

Question 7.
The height of a cone is 24 cm and its slant height is 25 cm. Then, its diameter is ………………. cm.
A. 14
B. 7
C. 4
D. 49
Answer:
A. 14

Question 8.
The circumference of the base of a cone is 44 cm and its slant height is 15 cm. Then, its curved surface area is ……………….. cm².
A. 14
B. 154
C. 330
D. 115
Answer:
C. 330

Question 9.
The diameter of a cone is 7 cm and its slant ! height is 16.5 cm. Then, its total surface area is …………………. cm².
A. 110
B. 220
C. 105
D. 154
Answer:
B. 220

Question 10.
Total surface area of a hemisphere with radius 7 cm is ………………….. cm².
A. 231
B. 115.5
C. 462
D. 154
Answer:
C. 462

Question 11.
Total surface area of a hemisphere is 72 cm².
Then, its curved surface area is ………………….. cm².
A. 24
B. 36
C. 48
D. 72
Answer:
C. 48

Question 12.
The surface area of a sphere is 616 cm².
Then, its radius is ……………. cm.
A. 6
B. 8
C. 7
D. 14
Answer:
C. 7

Question 13.
In a cuboid, the area of the face with sides length and breadth is 120 cm². If the height of the cuboid is 5 cm, then its volume is …………………… cm³
A. 120
B. 240
C. 600
D. 300
Answer:
C. 600

Question 14.
The volume of a cylinder is 2200 cm³ and its height is 7 cm. Then, the radius of the cylinder is ……………….. cm.
A. 5
B. 15
C. 10
D. 20
Answer:
C. 10

Question 15.
The radius and height of a cone are 7 cm and 3 cm respectively. Then, the volume of the cone is ……………… cm³.
A. 154
B. 168
C. 148
D. 462
Answer:
A. 154

Question 16.
The volume of a sphere is 4.5 π cm³. Then, its diameter is …………………. cm.
A. 3
B. 2
C. 1.5
D. 4
Answer:
A. 3

Question 17.
The ratio of radii of two cones is 2 : 3 and the ratio of their heights is 9:4. Then, the ratio of their volumes is
A. 1 : 1
B. 3 : 2
C. 1 : 3
D. 2 : 3
Answer:
A. 1 : 1

Question 18.
The circumference of the base of a cone is 44 cm and its height is 6 cm. Then, its volume is ………………… cm³.
A. 49
B. 98
C. 308
D. 154
Answer:
C. 308

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 15 Visualising Solid MCQ Questions

Multiple Choice Questions

Question 1.
Identify the side view for the given solid.

(a)

(b)

(c)

(d) None of these.
Answer:
(a)

Question 2.
Identify the front view for the given solid.


(d) None of these.
Answer:

Fill in the blanks :

Question 1.
The name of the figure is ……………..
Answer:
Cone

Question 2.
The number of edges of a cube are ……………..
Answer:
12

Question 3.
The number of faces of a cuboid are ……………..
Answer:
6

Question 4.
Opposite faces of a die always have a total of …………….. dots on there.
Answer:
7

Question 5.
The number of corners of a cube are ……………..
Answer:
8

Write True or False :

Question 1.
Cube is solid figure.
Answer:
True

Question 2.
Plane figure has three dimensions.
Answer:
False

Question 3.
Rectangle is a plane figure.
Answer:
True

Question 4.
A cylinder has two flat surfaces.
Answer:
True

Question 5.
A cube has twelve edges.
Answer:
True

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Translation

1. अपना बस्ता खोलो। – Open your bag.
2. अपना नाम बताओ। – Tell our name.
3. बातें मत करो। – Dont’t talk.

4. बुरी संगत से बचो। – Avoid bad company.
5. बड़ों का कहना मानो। – Obey your elders.
6. कूड़ा मत बिखराओ। – Don’t spread the litter.
7. अपने हाथ ऊपर उठाओ। – Hands up.
8. कभी झूठ मत बोलो। – Never tell a lie.
9. अपना काम पूरा करो। – Complete your work.
10. अपनी त्रुटियों को सुधारो। – Mend/Correct your mistakes.
11. मैं थका हुआ हूँ। – I am tired.
12. वह मूर्ख नहीं है। – He is not foolish.
13. मेरे माता-पिता जी शिक्षक हैं। – My parents are teachers.
14. हम भारतीय हैं। – We are Indians.
15. उसकी बहन बहुत बुद्धिमान है। – His/Her sister is very wise.
16. मुझे आप पर गर्व है। – I am proud of you.
17. पानी साफ है। – Water is clean.
18. वह बहुत स्वार्थी है। – He is very selfish.
19. घास हरी-भरी है। – Grass is green.
20. मैं आज स्वस्थ महसूस नहीं कर रहा हूँ। – I am not feeling well today.
21. मैं रोज़ स्कूल जाता हूँ। – I go to school daily.
22. मेरे माता जी स्वादिष्ट भोजन बनाते हैं। – My mother cooks tasty food.
23. सभी ने पाठ याद कर लिया है। – All have learnt their lesson.
24. मैंने अपना जन्मदिन मनाया। – I celebrated my birthday.
25. मेरे पिता जी बाज़ार जाएंगे। – My father will go to the market.
26. भगवान मेरी मदद करेगा। – God will help me.
27. वह तेज़ दौड़ रहा है। – He is running fast.
28. चपरासी घंटी नहीं बजा रहा है। – The peon is not ringing the bell.
29. छात्र पढ़ रहे होंगे। – The students will be studying.
30. किसान आराम नहीं कर रहे होंगे। – The farmers will not be taking rest.
31. सुशील ने अपना काम कर लिया है। – Susheel has done his work.
32. मेरे शिक्षक ने कापी चैक कर ली है। – The teacher has checked my notebook.
33. लड़कों ने कुर्सी नहीं तोड़ी है। – The boys have not broken the chair.
34. मुझे एक नौकरी मिल गई है। – I have got a job.
35. किसी ने मुझे बुलाया है। – Someone has called me.
36. मोहन ने एक गीत गाया है। – Mohan has sung a song.
37. लड़कियां कक्षा में चली गई हैं। – The girls have gone into the class.

38. पुलिस ने चोर को नहीं पकड़ा है। – The police have not caught the thief.
39. वह इस वर्ष पास नहीं हुआ है। – He has not passed this year.
40. मेरे चाचा जी ने मुझे तोहफे में घड़ी दी। – My uncle gave me a watch as a gift.
41. उसने सच नहीं बोला। – He did not speak the truth.
42. मुझे उस दिन बुखार था। – I was ill that day.
43. हमने डॉक्टर को बुला लिया था। – We had called in the doctor.
44. बूढ़े व्यक्ति ने घर बनाया। – The old man built a house.
45. हम बहुत मज़ा कर चुके थे। – We had a lot of fun.
46. सुधा ने खेल खेला था। – Sudha had played a game.
47. मेरी सहेली ने मेरी मदद की थी। – My friend had helped me.
48. बिल्ली ने दूध नहीं पिया। – The cat had not drunk milk.
49. सोनू ने आपको धक्का नहीं दिया था। – Sonu had not pushed you.
50. दादी ने अच्छी कहानी सुनाई। – The grandmother had told a good story.
51. आप कहाँ रहते हो? – Where do you live?
52. आपने झूठ क्यों बोला? – Why did you tell a lie?
53. आपने अपनी छुट्टी कहां बिताई? – Where did you pass your holiday.
54. कक्षा में कौन था? – Who was in the class?
55. आपका घनिष्ठ मित्र कौन है? – Who is your fast friend?
56. अध्यापिका ने क्या पढ़ाया? – What did the teacher teach?
57. उसने उत्तर क्यों नहीं दिया? – Why did he/she not reply?
58. आप कहाँ जा रहे हो? – Where are you going?
59. आप शोर क्यों मचा रहे हैं? – Why are you making a noise?
60. आप कब आ रहे हो? – When are you coming?
61. आप पुस्तक कब पढ़ रहे थे? – When were you reading the book?
62. आप कल क्या कर रहे थे? – What were you doing yesterday?
63. आप काम कब करोगे? – When will you do the work?
64. वह सवाल कैसे हल करेगी? – How will she solve the sum?
65. अब आप कहां जाओगे? – Where will you go now?
66. वह पार्टी पर क्यों नहीं आएगी? – Why will she not attend the party?
67. क्या आपने उत्तर लिख लिया है? – Have you written the answer?
68. क्या उसने पाठ याद कर लिया है? – Has he learnt the lesson?
69. उसने अपना घर क्यों बेच दिया? – Why did he sell his house?

70. क्या उन्होंने मैच जीत लिया था? – Had they won the match?
71. वह परीक्षा में किस तरह पास हो गया? – How did he pass the test?
72. क्या मैं अन्दर आ सकता हूँ? – May I come in?
73. क्या मैं बाहर जा सकता है। – May I go out?
74. क्या वह तैर सकती है? – Can she swim?
75. क्या मैं अब जा सकता हूँ? – May I go now?
76. अपने हाथ हमें भोजन से पहले धोने चाहिए। – We should wash our hands before meal.
77. मुझे अब सोना चाहिए। – I should go to bed now.
78. परिश्रम करो नहीं तो आप फेल हो जाओगे। – Work hard otherswise you will fail.
79. कृपया करके पृष्ठ पलटो। – Please turn the page.
80. शाबाश! आपने बहुत बढ़िया किया। – Well done!

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Formation of Adverb Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Formation of Adverb

Word – Adverb
Able – ably
Abrupt – Abruptly
Accident – Accidently
Active – Actively
Actual – Actually
Anger – Angrily
Annual – Annually
Awful – Awfully
Bad – Badly
Basic – Basically
Busy – Busily
Brief – Briefly
Bold – Boldly
Clear – Clearly
Close – Closly
Calm – Calmly
Certain – Certainly
Comfort – Comfortably

Dear – Dearly
Dead – Deadly
Easy – Easily
Different – Differently
Equal – Equally
Fair – Fairly
Entire – Entirely
Quiet – Quietly
Real – Reality
Rude – Rudely
Safe – Safely
Warm – Warmly
Virtual – Virtually
right – Tighdy
Tender – Tenderly
Wild – Wildly