Class 10 Science all exercises solutions

By:Chhatramani yadav

Force

a. Acceleration due to gravity:

Ans: The acceleration produced in a a freely falling body due to the influence of gravity is called Acceleration due to gravity.

Its symbol is “g” and its SI unit is meter per second (m/s2).

b. Weightlessness:

Ans: Weightlessness is the condition at which the apparent weight of a body is zero. Weightlessness is possible in the absence of gravity.

Ans:

Gravity

Gravitation

It is the force which pulls an object towards the centre of the earth.

It is the force with which two objects attract each other.

It depends upon mass and radius of planet.

It depends on the mass of two objects and distance between their centers.

Ans:

Mass

Weight

It is the quantity of matter contained in the body.

It is the force with which the object is attracted towards the centre of the earth.

It is scalar quantity

It is vector quantity.

Ans: Newton’s Law of Gravitation states that “The force of attraction between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.”

Ans:

When a coin and a feather are dropped from the roof of the building, the coin falls on the ground earlier because as the weight of coin is larger than that of the feather as well as the air resistance.

Ans:

Newton’s law of the gravitation holds true or applicable for all the objects present in this universe whether the objects be terrestrial or celestial. The gravitational force exists everywhere in this universe. Therefore Newton’s law of gravitation is called universe law.

Ans:

We have relation:

Weight=mass * gravity

As gravity decreases when the object is far from the surface of the earth so weight of the object also decrease when distance is increased.

Ans:

Weight depends upon gravity and gravity depends on the distance.

Weight=mass * gravity

As the stone is away from the earth surface the weight of the stone is less than the stone present in the bottom.

Ans:

We have relation:

Weight=mass * gravity

The value of gravity on the moon is six times less than that of the earth so the weight of object is less in moon than in earth.

Ans:

As there is less gravitational force on the moon it does not attract the atmosphere on its surface as a result there is no atmosphere on the surface of the moon.

Ans:

We have relation:

Weight = mass * gravity

When the body reaches the under the earth gravity decreases as a result the body will have less mass.

Ans:

The value of g is greater at the poles than that of the equator because gravity depends on radius of the earth and as the equator is nearer to the centre of earth the value is more at the equator than the poles.

Ans:

The person does not get injured when he falls from the parachute on the earth from the plane because there is presence of air resistance and the body does not fall freely.

Ans:

The relation of gravitational force between two heavenly bodies with the product of their masses is directional proportional and inversely proportional to the distance between their centers.

Ans:

Newton’s gravitational constant (G) is the gravitational existence between two bodies of unit masses placed at a unit distance from their centers.

The numerical value of gravitational constant (G) is 6.67*10-11 and its unit is Nm2/Kg2.

Ans: A coin and feather when dropped in vacuum falls simultaneously due to free fall as there is no external resistance in vacuum.

Ans:

The body will have more weight on the equator because gravity depends on distance as the equator is closer to the centre of earth than the poles so the body will have more weight on the equator.

Ans:

The attraction of 5Kg of iron to the earth is greater because the earth has more mass than the iron and the earth tends to pull the iron to its surface.

10.

Ans:

The statement is true because there occurs gravitational pull between apple and the earth. As the mass of apple is negligible as compared to the earth, the earth movement is not noticeable.

11.

Ans:

The value of G is 6.67*10-11 Nm2/kg2.

Gravity

Gravitation

It is the force which pulls an object towards the centre of the earth.

It is the force with which two objects attract each other.

It depends upon mass and radius of planet.

It depends on the mass of two objects and distance between their centers.

12.

Ans:

On the basis of his experiments , Galileo concluded that the acceleration of a an object falling freely towards the earth does not depend on the mass of the object .To verify this fact Robert Boyle performed an experiment in a vacuum. He put a feather and a coin in a tall jar and air from the jar was removed by using vacuum pump. When the jar was inverted both the coin and feather fell to the bottom simultaneously in the absence of medium.

13.

Ans:

According to the Newton’s law of Gravitation,

F =GMmd2GMmd2

=GMm(d2)2GMm(d2)2

=4∗GMmd24∗GMmd2

Or, F = 4 * (GMmd2GMmd2)

Hence, when the distance between the centers of the two body is halved the gravitational force increases by 4 times.

14.

Ans:

According to the Newton’s law of Gravitation,

F =G∗2M∗2md2G∗2M∗2md2

=4∗GMm(d2)24∗GMm(d2)2

=16∗GMmd216∗GMmd2

Or, F = 16 * (GMmd2GMmd2)

Hence, when the distance between the centers of the two body is doubled and their centers is halved the gravitational force increases by 16 times.

15.

Ans:

When an astronaut is in satellite they both are in freefall towards the earth. The astronaut does not exert any force on the side of the satellite. As a result the astronaut feels to be floating weightlessly though the force of gravity at the distance may not be zero. As a result the earth attracts the astronaut but negligibly.

16.

Ans:

The importance of Gravitational force is:

a. Existence of atmosphere on the surface of the earth.

b. Flowing of river.

17.

Ans:

When an object is falling towards the surface of the earth only under the influence of gravity without external resistance, the fall of the object is free fall.

The weight of the body increases when it falls under the influence of gravity as it comes closer to the core of earth.

Numerical Problems:

Solution:

Given

Mass of Sphere 1 (M) = 40 Kg

Mass of Sphere 2 (m) = 15 Kg

Force(F) = 9.8∗10−7N9.8∗10−7N

Distance between Sphere (d)= 20 cm=20/100 m = 0.2m

UniversalGravitational Constant(G) =?

We Know

Or,F=GMmr2F=GMmr2

Or, 9.8* 10-7 =G∗40∗150.22G∗40∗150.22

Hence, G is6.5 * 10-11N/m2

Solution:

Given

Mass of Earth (M) = M

Mass of heavenlybody (m) = M/2

Radius of Earth (R) = R

Radius of Heavenly Body (r) = R/2

Weight of stone on Earth (W) = 100N

Acceleration due to Gravity (g) =10 m/s2

We Know

First Part

On Earth

Weight = Mass *Gravity

Or, 100 = M*10

Or, M=10 Kg

Also,

g=GMR2GMR2

Or, 10 =(6.67∗10−7∗10)R2(6.67∗10−7∗10)R2

Or R=8.167 * 10-11m

Second Part

On heavenly body

Weight = Mass* Gravity

= mg

= 102∗(6.67∗10−11∗10)(8.167∗10−6)2102∗(6.67∗10−11∗10)(8.167∗10−6)2

= 200N

Hence, the stone weighs 200N in heavenly body.

Solution

Given

Mass of both bodyare same so,

Mass (M1) = 100 Kg

Mass (M2) = 100 Kg

Distance Between them (R) =1m

Universal GravitationalConstant (G) =?

Force of attraction (F) =?

First Part

We Know,

F=GMmr2GMmr2

Or, F =6.67∗10−7∗100∗100126.67∗10−7∗100∗10012

Hence, F= 6.67*10-3N

Second Part

No the force of attraction will not be different because G, M and d are all same in both case.

Solution

Given

Acceleration due to Gravity on moon (g) =1.6m/s2

Radius of moon (R) =1.7*106m

Mass of Body (M) =?

We know

Gravity = GMR2GMR2

Hence, Mass of Moon is6.93* 1022Kg

Solution

Given

Diameter of first lead sphere (D) = 20 cm

Radius (R1) = D2D2 cm = 10 cm = 0.1m

Diameter of second lead sphere (d) = 2 cm

Radius (R2) = d2d2 cm =1 cm = 0.01m

Distance between them (r) = 100 cm = 1 m

Force of attraction (F) =?

We Know

F=GMmr2GMmr2

=6.67∗10−11∗11500∗43∗227∗0.13∗11500∗43∗227∗0.013126.67∗10−11∗11500∗43∗227∗0.13∗11500∗43∗227∗0.01312(Density=MassVolumeMassVolume)and (Volume =43∗pi∗r3)43∗pi∗r3)

= 1.55 * 10-10 N

Hence, force of attraction is 1.55 * 10-10 N.

Solution

Given

Mass of both boys are same so,

Mass (M1) = 60 Kg

Mass (M2) = 60 Kg

Distance between them (r) =1m

Universal GravitationalConstant (G) =6.67* 10-11N/m2

Force (F) =?

We Know,

F=GMmr2GMmr2

Or F =(6.67∗10−11∗60∗60)12(6.67∗10−11∗60∗60)12

Hence, F = 2.4*10-7N

Solution

Given

Acceleration due to Gravity (g) =9.8m/s2

Radius of Earth(R) =6400Km = 6400*1000m = 6400000m

Mass of Earth (M) =?

We know

g=GMR2GMR2

Or, 9.8=(6.67∗10−11∗M)64000002(6.67∗10−11∗M)64000002

Hence, M= 6.018*1024 Kg

Solution

Given

Mass of Sun (M)=2* 1030 Kg

Mass of Earth (m) = 6* 1024 kg

Distance between them (r) = 1.5* 1011Kg

Universal Gravitational Constant (G) = 6.67* 10-11N/m2

Force (F) =?

We Know

F=GMmr2GMmr2

Or F= 6.67∗10−11∗2∗1030∗6∗1024(1.5∗1011)26.67∗10−11∗2∗1030∗6∗1024(1.5∗1011)2

Hence, F = 3.557*1022N

Solution

Given

Acceleration due to Gravity (g) = 9.8m/s2

Radius of Earth(R) = 6.4*106m

Height (h) = 8848m

Acceleration due to Gravity on Mt Everest Peak (g2) =?

We Know,

g2 =(RR+h)2∗g(RR+h)2∗g

Or, g2 = (6.4∗1066.4∗106+8848)2(6.4∗1066.4∗106+8848)2* 9.8

Hence, g2 is 9.77m/s2

10.

Solution

Given

Weight of Body (W) = 63 Kg

Radius of Earth(R) =6400Km = 6400*1000m = 6400000m

Height (h) = 3200Km = 3200000m

We know

Weight = Mass * Gravity

Or 63 = M * 10

Or M = 6.3 Kg

Also

g = GM(R+h)2GM(R+h)2

= 6.67∗10−11∗6.3(6400000+3200000)26.67∗10−11∗6.3(6400000+3200000)2

= 4.44 m/s2

Finally

F = M * g

= 6.3 * 4.44

= 28 N

Hence, Gravitational force in that height is 28 N

11.

Solution

Given

Mass of Object in earth (M) =200 Kg

Mass of object in moon (m) = ?

Now

In Earth

Weight (W) = M * g

= 200 * 10

= 2000N

Again

In moon

Weight (W) = M * g

Or, 2000= 200 * 10

Or, 2000 = m *1.67

Or, m = 1197.60 Kg

Hence, 1197.60 Kg can be lifted in the surface of moon.

12.

Solution

Given

Mass of Jupiter (M) =2*1027 Kg

Radius of Jupiter (R) =6.5*107m

Universal Gravitational Constant (G) =6.67*10-11N/m2

Acceleration due to gravity (g) =?

We Know,

g = GMr2GMr2

Or, g = 6.67∗10−11∗2∗1027(6.5∗107)26.67∗10−11∗2∗1027(6.5∗107)2

Hence, g is 31.57 m/s2

Second Part

Weight (W) =?

Mass (M) = 70 Kg

We Know,

W=mg

Or W=70*31.57

Or Weight = 2209.9

Hence, Weight of the person is 2209.9N.

Pressure

Ans:

Pressure is defined as the thrust on unit area of a surface. Its SI unit is Pascal. Upthrust is defined as the resultant thrust that a liquid uses to push up a body immersed in the liquid. Its SI unit is Newton.

Its SI unit is Newton.

Ans:

Thrust

Pressure

The force acting perpendicularly on a surface is called the thrust.

Pressure is defined as the thrust on unit area of a surface.

Its SI unit is Newton.

Its SI unit is Pascal.

Ans:

Foundation of buildings is made wider than the walls because pressure is greater in the depth of the building. To stand that much pressure foundation of buildings is made wider.

Ans:

Camel has flat soled feet which exerts less pressure due to the body weight of the camel where as horse does not have flat soled feet. As there is less pressure exerted by flat soled feet of camel, it can easily run faster than horses in deserts.

Ans:

Football player have to run and turn very fast. For this purpose the friction between the ground and shoes must be greater to balance the body. Therefore studs are

made on the sole of football players boot to increase the pressure on the ground which prevents the player from falling or sliding.

Ans:

We have,

Pressure = Force/ Area of the bottom

= F/A

If m be the mass then,

F= mg

Therefore P= mg/A

Or, P = Vdg/A

Or, P = Ahdg/A

Or, P = hdg.

Hence proved.

Ans:

Mass of the liquid and its depth affect the pressure contained in a vessel.

Ans:

A will have more pressure in its base and B will have pressure half of A if they are kept in same place.

Ans:

A will have more pressure because it is kept in Lumbini. This is due to gravity; Since Lumbini is nearer to the centre of earth than Pokhara.

Ans:

B will have more liquid pressure because the density of the salty water is greater than the water.

Ans:

We know that the liquid pressure is directly proportional to the height of the liquid column. The height of the liquid column on the down floor is more than that on the upper floor. So the speed of flow of water out of a tap of upper floor is less due to less pressure and that in the lower floor is more due to more pressure of water.

Ans:

We know that the liquid pressure is directly proportional to the height of the liquid column. So the blood pressure in human body is greater at the feet than the brain.

Ans:

In a deep sea the pressure becomes much more than the normal atmospheric pressure as pressure is directly proportional to the density of a liquid. If deep sea divers are not provided with special suit their body gets crushed due to high pressure. So deep sea divers need a special protective suit.

Ans:

When a train passes by rails, the rails exert greater pressure and it expands due to heat generated by the train. Since wood is bad conductor of heat wooden sleepers are used below the rails.

Ans:

When a body is fully immersed in a liquid the forces acting on the body are gravity and upthrust.

Ans:

Pascal’s law of liquid pressure states that, “The pressure is equally exerted perpendicularly on all sides as pressure is applied on a liquid kept in a closed container.”

Any two instruments based on Pascal’s Law are: Hydraulic Brake and Pistons.

10.

Ans:

It is easier to pull the bucket of water from a well until it is inside the water but difficult when it is out of the water because upthrust acts when the bucket is under water and gravity acts when the bucket is out of the water.

11.

Ans:

Pressure will be greater on the surface CD because of more depth.

Ans:

Upthrust will be greater in layer CD because of depth.

12.

Ans:

Upthrust is greater in the bottom than in the middle because of more depth in the bottom.

Ans:

Wall is made thicker in the bottom because there in more pressure in the bottom. In order to hold the wall the wall is made wider in the bottom.

13.

Ans:

Z will have more volume because density is inversely proportional to volume.

Ans:

Z will a have least mass as density is proportional to mass.

Ans:

Z will float on water because it has less density than others.

14.

Ans:

The tank containing 1000 liters will exerts more thrust at the bottom because it thrust is directly proportional to volume of liquid.

Ans:

Both will exert equal pressure in the bottom.

Ans:

The pressure will me more than before in 1000 liters than 500 liters.

15.

Ans:

Archimedes Principle

Law of Flotation

It states that,” When a body is partially or wholly immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced by it.”

It states that,” When a body floats in a liquid the weight of the floating body is equal to the weight of the liquid displaced by it.”

This principle is used to calculate upthrust or apparent weight of an object.

This law is used to construct hydrometer, sub marines, ships etc

16.

Ans:

Salt on mixing with water increases the density of water and hence the upthrust. As a result the egg can displace water equal to its weight and floats.

Ans:

Since the density of iron is greater than that of water the iron ball sinks as the weight of the wall is greater than the weight of the water displaced by it. But the ship made up of iron is designed in such a way that the relative density of ship becomes always less than that of water. Therefore the weight of water displaced becomes equal to the total weight of the ship and the ship floats.

Ans:

The density of sea water is more than that of river water. Thus for the same volume the sea water provides more upthrust as compared to the river water. Therefore it is easier for a man to swim in sea water than in river water.

17.

Ans:

This experiment is Archimedes principle.

18.

Ans:

Upthrust = 7N – 4N

= 3N

Also,

Force = mg

Or, 3 = mass * 10

Or Mass = 0.3 Kg

Ans:

This experiment is based on Archimedes principle.

Ans:

It states that,” When a body is partially or wholly immersed in a liquid, it experiences an upthrust equal to the weight of the liquid displaced by it.”

19.

Solution,

Weight of the ship (W) = 5 * 105 N

We have,

Force = mg

Or, Force = 5 * 105 * 10

Or, Force = 5 * 106 N

Hence, Force exerted by the ship is 5 * 106 N.

Numerical Problems:

Solution

Given

Height of Mercury Column (h)= 76cm = 76/100m = 0.76m

Density of Mercury (d) = 13600 Kg/m3

Acc due to gravity (g) = 9.8 m/s2

Pressure Exerted in mercury column (P) =?

We Know

P = dgh

= 13600 * 9.8 * 0.76

= 101292.8

= 1.01 * 105 Pascal

Hence Pressure is 1.01 * 105 Pascal

Solution

Given

Depth of rectangular tank (h) = 6m

Acc due to gravity (g) = 10 m/s2

Density of Water (d) = 1000 Kg/m3

Pressure (P) =?

We know

Pressure = dgh

= 1000 * 10 * 6

= 60000 Pascal

Hence pressure exerted by water is 60000 Pascal

Solution

Given

Depth of circular well = 5m

Depth of circular well having Water in it (h) = 5m – 2m = 3m

Acc due to gravity (g) = 9.8 m/s2

Density of Water (d) = 1000 Kg/m3

We Know

Pressure = dgh

= 1000 * 9.8 * 3

= 29400 Pa

= 2.94 * 104 Pa

Hence Pressure is 2.94 * 104 Pa

Solution

Given

Weight of Stone in Water (WW) = 18 N

Weight of Water displaced = 4 N

Weight of Stone in Air (WA) =?

We Know

Upthrust (U) = Weight of Water displaced

= 4 N

Also,

Upthrust = WA - WW

Or, 4 = WA – 18

Or, WA = 22N

Hence Weight of Stone in Air is 22N

Solution

Given

Height of Mercury Column (h) = 75cm = 75/100m = 0.75m

Density of Mercury (d) = 13600 Kg/m3

Acc due to gravity (g) = 10 m/s2

Pressure Exerted in mercury column (P) =?

We Know

P = dgh

= 13600 * 10* 0.75

= 102000

= 1.02 * 105 Pascal

Hence Pressure is 1.01 * 105 Pascal

Solution

Given

Volume of Wood (V1) = 0.2 m3

Density of Wood (d1) = 600 Kg/m3

Volume of Water (V2) =?

Density of liquid (d2) = 800 Kg/m3

We Know

According to Law of Flotation,

W1 = W2

Or, M1 * g = M2 * g

Or, d1 * v1 = d2 * v2(Density = Mass / Volume)

Or, 600 * v1 = 800 * v2

Or, v2v1v2v1= 3434( v2v1v2v1is the fraction under water)

Hence 3434part is under water.

Solution

Given

Mass of Wood (M1) = 24 Kg

Volume of Wood (V1) = 0.032 m3

Density of Wood (d1) =?

Density of water (d2) = 1000 Kg/m3

We Know

According to Law of Flotation,

W1 = W2

Or, M1 * g = M2 * g

Or, M1 = d2 * v2

Or, 24 = 1000 * *v2

Or, V2 = 0.024 m3

Also

Density of Wood = MassVolumeMassVolume

240.032240.032

= 750 Kg/m3

Hence the volume of block under water is 0.024 m3 and density of wood is 750 Kg/m3.

Solution

Given

In cubical oak

Length (l) = 15cm

Length of oak inside water (lin) = 10.5 cm

Density of water (d2) = 1 g/cm3

Density of oak (d2) =?

Now

Volume of oak inside water(V2) = (15 * 15 * 10.5) cm3

= 2362.5 cm3

Also,

Volume of water displaced (V1) = mw/dw = 2362.5 cm3

Or, mw = 2362.512362.51

Again

Density of oak (d2) = massofwatervolumeofoakmassofwatervolumeofoak

Or d2 = 2362.515∗15∗152362.515∗15∗15

Or d2 = 0.7 g/cm3

Hence density of oak is 0.7 g/cm3

Solution

Given

Area of Iceberg (A) = 1000cm2

Thickness (t) = 30cm2

Density of Ice (di) = 0.9 g/cm3

Density of Water (dw) = 1 g/cm3

Mass of displaced water (mw) =?

We Know

Volume of iceberg = A * t

= 1000 *30

= 30000 cm3

Also

Mass of iceberg (Mi) = Volume of iceberg * density of iceberg

= 30000 * 0.9

= 27000 g

We know

Weight of ice = weight of water displaced

Or, Mi * g = mw * g

Or, mw = 27000g = 27 kg

Hence, mass of water displaced is 27 kg.

10.

Solution

Given

Length of lower surface of rectangular body = 2m2

Height of rectangular body = (6-2) m2 = 4 m2

Volume of rectangular body = 4m * 2m2 = 8m3

Density of water (d) = 1000 Kg/m3

Upthrust (U) = d * g * v

= 1000 * 9.8 * 8

= 78400N

Hence, upthrust acted upon rectangular body is 78400N.

11.

Solution

Given

Volume of stone (V1) = 400 cm3 = 400/1000000 m3 = 4 * 10-4m3

Density of stone (d1) = 7.8 * 103 Kg/m3

Density of water (d2) = 1000 Kg/m3

Acceleration against gravity in air (g) = 10 m/s2

Weight of stone in air (WA) =?

Upthrust (U) =?

We Know

WA= m1g

= d1 * V1 * g

= 7.8 * 103 * 4 * 10-4 * 10

= 31.2N

Also

Up thrust (U) = d2gv1

= 1000 * 10 * 4 * 10-4

= 4 N

Hence Weight in air is 31.2N and upthrust is 4 N.

12.

Solution

Given

Weight of solid in air (WS) = 277.5 g

Weight of solid in liquid (Wl) =?

Density of liquid (dl) = 0.9g/cm3

Density of solid (dS) =?

Now,

Volume of solid (Vs) = WS / dS

= 277.5/ dScm3

Also,

Volume of water (vw) = mass of water displaced density

= 277.5−212.50.9277.5−212.50.9

= 6509cm36509cm3

Now

For sinking bodies

Vw = vs

Or, 227.5ds=6509227.5ds=6509

Or, ds = 3.84 g/cm3

Hence the density of solid is 3.84 g/cm3.

Heat

Ans:

Heat is defined as the sum of kinetic energy contained by the molecules of that object. The SI unit of heat is joule. Temperature is the degree of hotness or coldness of an object.

Ans:

Heat depends on mass of the object and average kinetic energy of the molecules.

Heat is directly proportional to the mass of object and average kinetic energy.

Ans:

Heat is the sum of the kinetic energy contained by the molecules of an object. Temperature is the average kinetic energy of the molecules of the body.

Ans:

Five effects of heat are:

a. It changes the state of matter

b. It changes the temperature of an object.

c. It changes the solubility of a substance.

d. It changes the size of an object.

e. It changes the color of the body.

Ans:

Due to the anomalous property of water fish and other aquatic animals survive in the water.

Ans:

Heat

Temperature

Heat is the sum of the kinetic energy contained by the molecules of an object.

Temperature is the average kinetic energy of the molecules of the body.

In SI system it is measured in Joule

In SI system it is measured in Kelvin

It is measured by calorimeter.

It is measured by thermometer.

Ans:

The specific heat capacity of a body is defined as the amount of heat required o raise the temperature of unit mass of that body by 10 C.

Ans:

One calorie of heat energy is the amount of heat energy required to raise the temperature of one gram of pure water through 10C.

Ans:

The unusual behavior of water that expands on cooling from 40C to 00C is called anomalous expansion of water.

Ans:

The amount of heat gained or lost by a body is equal to the product of the mass (m), the specific heat capacity (s) and the change in temperature (dt) of that body is called heat equation.

Q= msdt

Ans:

The specific heat capacity of water is 4200 JKg-1 0C-1 means that 4200 joule of heat energy is required to raise the temperature of 1 kilogram of water by 1 0C.

Ans:

The specific heat capacity of mercury is 138 JKg-1 0C-1 means that 138 joule of heat energy is required to raise the temperature of 1 kilogram of mercury by 1 0C.

Ans:

Water absorbs a large amount of heat from the engine as it has very high specific heat capacity. Due to this the engine does not get heated up when filled with water.

Ans:

When a person is suffering from fever a wet cloth is kept on his ore head to lower his body temperature. As the wet cloth contains water it can absorb a large amount of heat due to its high specific heat capacity. As a result the temperature of the patient decreases.

Ans:

When a beaker filled with water at 40C is cooled or heated, the water overflows from the beaker due to anomalous expansion of water. At 40C water shows abnormal behavior that is it expands so it overflows from the beaker.

Ans:

When hot water is poured in thick glass tumbler which cold there occur expansion and contraction of the glass due to which the glass breaks.

Ans:

Mercury is heated faster than the water because the specific heat capacity of mercury (140 JKg-1 0C-1) is less than water (4200 JKg-1 0C-1 ).

Ans:

When water is cooled to 00 c or below the surface water changes into ice since the density of ice is less than density of water, it floats in the surface. While in case of honey, there is decrease in the volume and increase in density when cooled so it solidifies from the bottom.

Ans:

Since heat loss or heat gained is directly proportional to the difference in temperature, when we go out, our body will loss more heat than when we are inside the room. So, we feel the air of the same room warmer.

Ans:

Since water has very high specific heat capacity, it can store a large quantity of heat energy for longer period of time in the water bag. Therefore, water is used in hot water bags.

10.

Ans:

C will gain maximum temperature because it has less specific heat capacity.

Ans:

A is suitable for cooling and heating purposes because it has high specific heat capacity.

Ans:

B will cool faster because object having less specific heat capacity cools or heats faster.

Ans:

C is suitable for thermometric liquid because it has less specific heat capacity than others.

11.

Ans:

Y has maximum heat because its specific heat capacity is high and liberates less heat.

Ans:

X has more mass because its specific heat capacity is less than others which can be seen from this relation:

m = Qs∗dtQs∗dt

Ans:

Specific heat capacity of Z is 470 JKg-1 0C-1means that 470 joule of heat is required to raise the temperature of body Z by 10C.

Ans:

Y will melt the wax for maximum depth because Y has large specific heat capacity and it liberates heat slowly than others.

Numerical Problems:

Solution

Given

Specific heat capacity of copper (s) = 380 JKg-1oc-1

Mass of copper (m)= 5 Kg

Thermal Capacity (C) =?

We Know

Thermal Capacity (C) = m * s

= 5 * 380

= 1900 J/0C

Hence, the thermal capacity of Copper is 1900 J/0C.

Solution

Given

SpecificHeat capacity of a substance (s) = 0.5238 cal g-10c-1

Now In SI system

We have

Q = msdt (where Q, m, and dt are in Kg, oc and Joule)

s = Qm∗dtQm∗dt

s = 0.5238 cal g-10c-1

= 0.5238 * 4.2 * 1000 (4.2J = 1 calorie and 1000 g = 1Kg)

= 2.1 * 103 JKg-1oc-1

Hence, 0.5238 calg-10c-1 in SI system is 2.1 * 103 JKg-1oc-1

Solution

Given

Mass of water (m) = 5 kg

Initial temperature of water (i) = 300c

Final temperature of water (f) = 1000c

Change in temperature (dt) = f-i

= (100 -30)oc

= 70oc

Specific heat capacity of water (s) = 4200 JKg-1oc-1

Heat energy required (Q) =?

We Know

Q = msdt

= 5 * 4200 * 70

= 1470000

= 1.47 * 106 J

Hence, 1.47 * 106 J heat energy isrequired.

Solution

Given

Mass of water (m) = 2 kg

Initial temperature of water (i) = 250c

Final temperature of water (f) = 500c

Change in temperature (dt) = f-i

= (50 -25)oc

= 25oc

Heat energy required (Q) = 2.1 * 105 J

Specific heat capacity of water (s) =?

We Know

Q = msdt

Or s = Qm∗dtQm∗dt

= 2.1∗1052∗252.1∗1052∗25

= 4200 JKg-1oc-1

Hence, Specific heat capacity of water is 4200 JKg-1oc-1

Solution

Given

First case

Mass of Paraffin (m1) = 1 kg

Heat (Q1) = 44000 J

Final temperature (dt) =200c

Specific heat capacity (s) = s

Q=msdt

Or s = Qm∗dtQm∗dt

Or s = 440001∗20440001∗20 …… (1)

Second Case

Mass of Paraffin (m2) = 5 kg

Final temperature (dt2) =100c

Specific heat capacity (s) = s

Heat (Q2) =?

Q2=m2sdt2

Or s = Q2m2∗dt2Q2m2∗dt2

Or s = Q25∗10Q25∗10…… (2)

Since Specific heat capacity of paraffin is same in both cases

So, equating equation (1) and (2)

440001∗20440001∗20 = Q25∗10Q25∗10

Or Q2 = 44000∗5∗102044000∗5∗1020

= 1.1 * 105 J

Hence, 1.1 * 105 J heat energy is required.

Solution

Given

Mass of pressure cooker (m) = 2 kg

Heat energy (Q) = 6400 J

Specific heat capacity of material (s) = 1000 JKg-1oc-1

Initial temperature (i) = 250c

Final temperature (f) =?

We Know

Q = msdt

Or Q = ms (f-i) (∴dt = f-i)

Or f = Qms+iQms+i

= 64002∗1000+2564002∗1000+25

= 28.20c

Hence, the final temperature is 28.20c

Solution

Given

First Case

Mass of water (m) = 300g = 0.3 Kg

Initial temperature of water (i) = 00c

Final temperature of water (f) = 400c

Change in temperature (dt) = f-i

= (40 -0o) c

= 40oc

Second Case

Mass of hot water added (m2) =?

Initial temperature of water (i) = 1000c

Final temperature of water (f) = 400c

Change in temperature (dt2) = i - f

= (100 – 40)oc

= 60oc

We Know

Heat lost = Heat Gained

Or msdt = m2sdt2

Or m2 = m * 40 /dt2

= 0.3 * 40 / 60

= 0.2 Kg

Hence, mass of hot water added is 0.2 Kg.

Solution

Given

First Case

Mass of metal (m) = 600 g = 0.6 kg

Change in temperature (dt) = (100 – 20)0c

= 800c

Specific heat capacity of metal (s) =?

Second Case

Mass of water (m2) = 300 g = 0.3 kg

Change in temperature (dt2) = (20 – 15)0c

= 50c

Specific heat capacity of water (s2) = 4200 JKg-1 oc-1

We Know

Heat lost = Heat Gained

Or msdt = m2s2dt2

0r s = m2s2dt2 / m* dt

Or s = 0.3∗4200∗50.6∗800.3∗4200∗50.6∗80

= 131.25JKg-1 oc-1

Hence, the specific heat capacity of metal is 131.25JKg-1 oc-1

Solution

Given

First Case

Mass of water (m) = 100 g = 0.1 Kg

Initial temperature of water (i) = 700c

Second Case

Mass of cold water (m2) = 200 g = 0.2 Kg

Initial temperature of cold water (i2) = 100c

Specific heat capacity of water (s) = 4200 JKg-1 oc-1

Final temperature of mixture (f) =?

We Know

Heat lost = Heat gained

Or, ms (f-i) = m2s(i2-f)

Or, 0.5 * (f-70) = (10- f)

Or, 0.5f – 35 = 10 – f

Or, f = 30

Hence, the final temperature of mixture is 300c.

10.

Solution

Given

Mass of block (m) = 1 kg

Power (P) = 48watt

Time taken (t) = 5 min = 5 * 60 sec = 14400 sec

Initial temperature (i) = 200c

Final temperature (f) = 500c

Change in temperature (dt) = i-f

= (50 -20)oc

= 30oc

We Know

Energy = Power * time

= 48 * 5 * 60

Or Energy = 14400 J

Also

Q = msdt

Or s = Qm∗dtQm∗dt

= 144001∗30144001∗30

= 480

Hence, Specific heat capacity of block is 480 JKg-1 oc-1

11.

Solution

Given

Mass of Water (m) = 20litres = 20 Kg

Initial temperature of water (i) = 50c

Final temperature of water (f) = 350c

Change in temperature (dt) = f-i

= (35-5)oc

= 30oc

Specific heat capacity of water (s) = 4200 JKg-1 oc-1

Heat required (Q) =?

We know

Q = msdt

= 20 * 4200 * 30

= 2.52 * 106 J

Hence, heat required is 2.52 * 106 J

12.

Solution

Given

First Case

Mass of cold water (m) =?

Initial temperature (i) = 150c

Final temperature (f) = 400c

Second Case

Mass of water (m2) = 60 kg

Initial temperature (i2) = 1000c

Final temperature (f) = 400c

We Know

Heat loss =Heat Gained

Or, ms (i-f) = m2s(i2-f)

Or, m * (40-15) = 60(100-40)

Or, m = 144Kg

Hence, mass of cold water is 144 Kg.

Light

Ans:

The power of lens is defined as the ability of a lens to converge or diverge the light rays falling on it.

Ans:

When rays parallel to the principal axis of a concave (or convex) lens appear to diverge (or converge) from a point then it is called principal focus.

Ans:

The image which can be obtained on the screen is called real image.

Ans:

Diverging lens is also called concave lens. It is thinner in the middle and thicker at the edges and has virtual focus.

Ans:

A convex lens is called converging lens because it converge rays of light passing through it at a point after refraction.

Ans:

A conave lens is called diverging lens because it diverge rays of light passing through it at a point after refraction.

Ans:

Concave Lens

Convex Lens

It is thicker in the middle and thinner at the edges.

It is thinner in the middle and thicker at the edges.

It converges rays of light passing through it at a point after refraction

It diverges rays of light passing through it at a point after refraction

It has a real focus

It has virtual focus

Ans:

Microscope

Telescope

It is used to see details of very small objects.

It is used to see very distant objects.

Objective lens of a microscope has shorter focal length than that of the eyepiece.

Eye piece of telescope has shorter focal length than that of objective lens.

Ans:

Near Point

Far Point

The nearest point up to which an object can be seen clearly by eyes is called near point.

He farthest point up to which objects can be seen clearly is called far point.

For normal eye it is 25 cm

For normal eye it is infinity.

Ans:

Camera

Eye

The focal length of the camera lens is constant

The focal length of the eye lens is variable

A permanent image is formed on the film

The image formed on the retina is not permanent

The camera has not vision defect

The eye has vision defect.

Ans:

It is used to capture picture permanently.

Ans:

It is used to see distant objects.

Ans:

It is used to see microscopic objects.

Ans:

It is used to magnify small objects.

Ans:

It diverge rays of light passing through it at a point after refraction.

Ans:

It converge rays of light passing through it at a point after refraction.

Ans:

When the object is kept between F and 2F, its image is formed beyond 2F on the other side which is real magnified and inverted.

Ans:

The formula for lens is:

$\frac{1}{{focal{\rm{\: }}length}}$ = $\frac{1}{{object{\rm{\: }}distance\left( u \right)}} + \frac{1}{{image{\rm{\: }}distance{\rm{\: }}\left( v \right)}}$

The formula for magnification of lens is:

Magnification (m) = $\frac{{\left( {height{\rm{\: }}of{\rm{\: }}image} \right)}}{{height{\rm{\: }}of{\rm{\: }}object}}$

Ans:

When the object is kept at the 2F the image of the same size is formed.

Ans:

The function of shutter in a camera is to protect the lens of the camera.

Ans:

Iris and pupil are responsible for light to enter in our eyes. The front part of the choroid forms color iris the iris has a hole at the centre called the pupil. The iris controls the amount of light entering the eye.

10.

Ans:

The defect in the picture is nearsightedness. To remove this defect concave lens of suitable lens should be used because it diverges beam of light.

11.

Ans:

It is defined as that type of defect of vision in which person cannot see distant objects clearly, but can see nearby objects without any difficulty.

To remove this defect concave lens of suitable lens should be used because it diverges beam of light.

12.

Ans:

It is defined as that type of defect of vision in which person cannot see nearby objects, but can see distant objects without any difficulty.

To remove this defect convex lens of suitable lens should be used because it converges beam of light.

13.

Ans:

Sanjay is suffering from farsightedness.

Ans:

He should use convex lens of suitable focal length.

14.

Ans:

Meriksha has problem of farsightedness. As ciliary muscle adjust the thickness of the eye lens.

15.

Ans:

The student is suffering from nearsightedness.

16.

Ans:

Objective lens has shorter focal length in compound microscope.

17.

Ans:

The defect is farsightedness.

Ans:

It can be removed by using convex lens of suitable lens focal length.

Ans:

Convex lens

Ans:

She should hold newspaper 25 cm to infinity away from her because person suffering from this defect cannot see objects nearer than 25 cm.

18.

Ans:

The defect in the picture is farsightedness. To remove this defect convex lens of suitable lens should be used because it converge beam of light.

19.

Ans:

Numerical Problems:

Solution

Given

In Convex Lens

Focal length (f) = 10 cm

Power (P) =?

We Know

Power (P) = $\frac{{100}}{{{\rm{focal\: length\: of\: lens}}}}{\rm{\: \: }}$

= $\frac{{100}}{{10}}$

= 10 dioptre

Hence, the power of lens is 10 dioptre.

Solution

Given

Image distance (v) = 80 cm

Object distance (u) = 40 cm

Size of object (O) = 2 cm

Size of real image (I)=?

We Know

$\frac{{{\rm{Size\: of\: real\: image}}}}{{{\rm{Size\: of\: object}}}} = \frac{{{\rm{Image\: distance}}}}{{{\rm{Object\: distance\: }}}}{\rm{\: }}$

Or, $\frac{{\rm{I}}}{{\rm{O}}} = \frac{{\rm{v}}}{{\rm{u}}}$

Or, $\frac{{\rm{I}}}{2} = \frac{{80}}{{40}}$

Or, I = 4 cm

Hence, the size of real image is 4 cm.

Solution

Given

Power of spectacles (P) = -2D

Since the power is -2D the student has problem of near sightedness, which means the student can see near object but cannot see far object. To remove this defect concave lens of suitable focal length has to be used.

Focal length of the lens (f) =?

We Know

Power (P) = $\frac{1}{{\rm{f}}}$

Or, -2 = $\frac{1}{{\rm{f}}}$

Or, f = -0.5 meter (-50 cm)

Hence, the focal length of lens is -0.5 m or -50 cm.

Solution

Given

Power of spectacles (P) = +1D

Focal length of the lens (f) =?

We Know

Power (P) = $\frac{1}{{\rm{f}}}$

Or, 1 = $\frac{1}{{\rm{f}}}$

Or, f = 1 meter (100 cm)

Nabin is suffering from far sightedness, which means he can see far object but cannot see near object. To remove this defect convex lens of suitable focal length has to be used.

Solution

Given

In Concave Lens

Focal length (f) = 25 cm

Power (P) =?

We Know

Power (P) = $\frac{{100}}{{{\rm{focal\: length\: of\: lens}}}}{\rm{\: \: }}$

= $\frac{{100}}{{25}}$

= 4 dioptre

Hence, the power of lens is 4 dioptre.

Solution

Given

In Concave lens

Object distance (u) = 20 cm

Focal length (f) = -20 cm (Because of Concave lens)

Image distance (v) =?

We Know

${\rm{\: }}\frac{1}{{\rm{f}}} = \frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}$

Or, $\frac{1}{{20}} = \frac{1}{{20}} + \frac{1}{{\rm{v}}}$

Or, $\frac{1}{{\rm{v}}} = - \frac{1}{{20}} - \frac{1}{{20}}$

Or, v = -10 cm

Hence, image distance is -10 cm.

Solution

Given

In Convex Lens

Object distance (u) = 20 cm

Focal length (f) = 15 cm

Now

For Image distance (v)

$\frac{1}{{\rm{f}}} = \frac{1}{{\rm{u}}} + \frac{1}{{\rm{v}}}$

Or, $\frac{1}{{15}} = \frac{1}{{20}} + \frac{1}{{\rm{v}}}$

Or, $\frac{1}{{\rm{v}}} = + \frac{1}{{15}} - \frac{1}{{20}}$

Or, v = 60 cm

Hence, image distance is 60 cm.

Since the value of focal length (f) is positive the image formed is real.

Real image are always inverted.

Electricity And Magnetism

Ans:

When pure copper plate is kept instead of iron nail, pure copper dissolves and deposits over the cathode. This concept is used for purification of metals.

Ans:

When the ionization, occurs copper ions moves towards the cathode.

Ans:

Oxygen gas is collected at anode.

Ans:

H2 so4 is mixed in water.

Ans:

Alternating current is one which changes its magnitude continuously and reverses its direction periodically and if the polarity of an electrical source does not change with tome it is called direct current.

Ans:

Fuse is a safety device to protect a circuit from excessive heating. A fuse wire is a short metal wire having low melting point made of an alloy of tin and lead. It is connected in series with a circuit.

Ans:

Series combination of Resistor

Parallel combination of Resistor

When a number of resistors in a circuit is connected from end to end in a circuit, it is called Series combination of Resistor

When a number of resistors in a circuit is connected between two common points, it is called Parallel combination of Resistor

Equivalent resistance is equal to the sum of individual resistance

The reciprocal of the equivalent resistance is equal to the sum of the reciprocal of individual resistance.

Ans:

Series combination of Cell

Parallel combination of Cell

In series combination of cells the total voltage is equal to the sum of the individual voltage of the cells

In parallel combination of cells the total voltage between any two points is always constant this is equal to the voltage of one cell.

The brightness of the bulb increases on increasing the number of cells.

The brightness of the bulb remains the same on increasing the number of cells.

Ans:

Electric appliances are connected in parallel in a domestic circuit because in parallel combination any devices can be used with our own will and if one of the device is damaged then it does nit affect the circuit connection.

Ans:

Switch and fuse should be connected in live wire because current flows in live wire and electric appliances get protected from high voltage current and electric shock.

Ans:

Wires of certain color are used for phase line, neural line and line for earthing. In household wiring red and brown is used as live wires, blue and black is used as neutral wire and green and yellow are used as earthing.

Ans:

Metallic bodies of electric appliances are connected to the earth wire because it protects the consumer from electric shock. If current leaks by technical mistakes or other causes on the metallic parts of heavy equipment, the current flows through the wire.

Ans:

Inert gases like argon, neon, nitrogen etc are filled in an electric bulb to prevent the filament from vaporizing when heated to a high temperature.

Ans:

Tungsten wire is used as filament in bulb because it has very high resistivity and has very high melting point.

Ans:

Filament is bright but not other wires of an electric bulb when electricity is passed through it because it has very high resistivity and melting point than other wires.

Ans:

An electric bulb is not filled with air because air contains oxygen, carbon dioxide etc and humidity which are not suitable gases and can react inside the bulb when current is passed through the circuit.

Ans:

Nichrome wire is used in a heater because it has very high resistivity and has high melting point and liberates heat.

Ans:

When voltage required is more than that of the e.m.f of a single cell a number of cell are connected in series connection.

Ans:

When current required is more than that delivered by a single cell a number of cells are connected in parallel connection.

Ans:

Live wire goes through a switch because in live wire current flows and the function of switch is to break down the current.

Ans:

The strength of electro magnet can be increased by inducing high voltage current and by increasing the number of turns in the magnet.

Ans:

KWh (Kilowatt hour) is defined as the amount of electric energy consumed by an electric appliance having a power rating of 1 kilowatt in 1 hour.

Joule represents a very small amount of energy. Thus it is inconvenient to use for trade purposes. In such situation we use a bigger unit of energy called kilowatt hour.

10.

Ans:

To avoid electric hazards a suitable fuse should be used. Earthing should be done in each electrical appliances and naked wire shouldn’t e used.

11.

Ans:

The fuse should be connected in A because when switch is closed current flows through the fuse which can protect from high voltage current.

12.

Ans:

Faradays laws of electromagnetic induction are:

1. Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.

2. The induced emf lasts as long as the change in the magnetic flux is taking place.

3. The magnitude of the induced emf is directly proportional to the rate of change of the magnetic flux.

13.

Ans:

When the wheel of the bicycle rotates, it causes the head of the dynamo to rotate, which in turn rotates the permanent magnet. Thus the magnetic flux passing through the coin changes and an emf is induced in it and current is produced.

When a current carrying a conductor is kept in a magnetic field, the conductor moves if it is allowed to do so. This is called motor effect.

14.

Ans:

Fleming’s right hand rule states that “If the first three fingers of the right hand are held mutually perpendicular to each other, with the index finger in the direction of the magnetic flux and the thumb in the direction of motion of coil, the middle finger points to the direction of the induced current.”

Fleming’s right hand rule is used to determine the direction of the induced current in a generator.

15.

Ans:

The current can be increased in the generator by increasing the number of turns in the coil, by increasing the strength of the magnetic field,by decreasing the distance between the coil and the magnet.

16.

Ans:

When an alternating emf is applied to the primary coil, a changing current flowing in it produces an alternating magnetic flux in it. This causes to change the magnetic flux linked with the secondary coil. An alternating emf is then induced in the secondary coil. It is called the principal of mutual inductance on which transformers are based.

17.

Ans:

Step up Transformer

Step down Transformer

It changes a low voltage AC to a high voltage AC of the same frequency.

It changes a high voltage AC to a low voltage AC of the same frequency.

The number of turns in the secondary coil is more than in the primary coil

The number of turns in the secondary coil is less than that in the primary coil.

More current flows in the primary

More current flows in the secondary.

Ans:

Electric Motor

Generator

It converts electrical energy into mechanical energy.

It converts mechanical energy into electrical energy.

It works on the principal of the motor effect

It works on the principal of electromagnetic induction

Fleming left hand rule gives the direction of motion of the conductor in an electric motor.

Fleming right hand rule gives the direction of induced current in a generator.

18.

Ans:

Transformers are used in voltage regulators for computer, television, air conditioner, trolley buses etc

19.

Ans:

The two forms in sequence are:

Electrical energy

Mechanical Energy

20.

Ans:

The two forms in sequence are:

Mechanical Energy

Electrical Energy

21.

Ans:

A simple experiment which illustrates the truth of Faradays law of electromagnetic induction is given below:

A Bicycle Dynamo:

22.

Ans:

The type of transformer in which turning of wire in the secondary coil is less than that of the primary coil is Step down transformer.

23.

a. bicycle dynamo

Ans:

b. step up transformer

Ans:

c. copper plating

Ans:

d. Electrolysis of water

Ans:

24.

A. Empty matchbox

B. Slip rings

C. Brushes

Ans:

Mechanical energy into electrical energy

Ans:

It works on the principal of Faradays Law of Electro magnetic induction.

Ans:

In electric motor electrical energy is converted into mechanical energy where as it is quite opposite in generator.

25.

Ans:

Deflection can be seen in the Galvanometer.

Ans:

The galvanometer does not deflect instead it remains stationary.

Ans:

The galvanometer deflects faster when the magnet is moved fast in and out of the coil.

26.

Ans:

Step down transformer is shown in the figure.

Ans:

A = Soft iron core

B = Primary Coil

C = AC source

D = Output

Ans:

It converts high voltage AC to a low voltage AC of the same frequency.

It is used as power sub stations to step down the voltage before its distribution to consumers.

Ans:

A is laminated because to separate the primary and secondary coil.

Numerical Problem:

Solution

Given

Power (P) = 1000W

Voltage (V) = 220V

Current (I) =?

We have

P = I * V

Or, 1000 = I * 220

Or, I = 4.54

Hence, the current is flowing in the circuit is 4.54 so 5A fuse is suitable for the circuit.

Solution

Given

For Bulbs

No of bulbs(N) =200

Power (P) = 60W = 0.06 kW

Time (T) = 6hrs

For Fans

No of fan(N) =50

Power (P) = 75W = 0.075 kW

Time (T) = 12hrs

Energy cost = Rs. 7 per unit

Energy bill for a month =?

Now

Electric Consumption by bulb (ECB) = P * t * N

= 0.06 * 6 * 200

= 72

Similarly,

Electric Consumption by fan (ECF) = P * t * N

= 0.075 * 12 * 50

= 45 unit

Total Unit Consumed = ECB + ECF

= 72 + 45

= 117 unit

Cost of Electricity = Total Unit Consumed * rate * 30 days

= 117 * 7 * 30

= Rs. 24570

Hence, the energy bill for a month is Rs. 24570.

Solution

Given

For Bulbs

No of bulbs (N) =50

Power (P) = 40W = 0.04 kW

Time (T) = 10hrs

For Fans

No of fan (N) =10

Power (P) = 50W = 0.050kW

Time (T) = 10hrs

Energy cost = Rs. 7 per unit

Energy bill for a month =?

Now

Electric Consumption by bulb (ECB) = P * t * N

= 0.04 * 10 * 50

= 20 unit

Similarly,

Electric Consumption by fan (ECF) = P * t * N

= 0.050 * 10 * 10

= 5 unit

Total Unit Consumed = ECB + ECF

= 20 + 5

= 25

Cost of Electricity = Total Unit Consumed * rate * 30 days

= 25 * 7 * 30

= Rs. 24570

Hence, the energy bill for a month is Rs. 5250.

Solution

Given

Power of heater (P) = 2 Kw

Time (T) = 5 hours

Cost = Rs 7.50

Now

Total Unit Consumed = P * t

= 2 * 5

= 10 unit

Also

Cost of lightening = Total Unit Consumed * Cost

= 10 * 7.50

= Rs 75

Hence, the total cost of lightening for 5 hours is Rs 75.

Solution

Given

Power of electric kettle (P) = 1000 w = 1 kW

Time (T) = 2 hrs

Total Units Consumed =?

Cost =?

Now

Total Units Consumed = P * T

= 1 * 2

= 2 unit

Also

Cost = Total Units Consumed * 10 days

= 2 * 10

= 20 units

Hence, 20 units is consumed by electric kettle.

Solution

Given

For Bulbs

Power (P) = 60W = 0.06 kW

Time (T) = 15hrs

For heater

Power (P) = 750W = 0.75 kW

Time (T) = 10hrs

Energy cost = Rs. 7 per unit

Energy bill for a month =?

Now

Electric Consumption by bulb (ECB) = P * t

= 0.06 * 15

= 0.9 units

Similarly,

Electric Consumption by heater (ECH) = P * t

= 0.75 * 10

= 7.5 units

Total Unit Consumed = ECB + ECFH

= 0.9 +7.5

= 8.4

Cost of Electricity = Total Unit Consumed * rate * 30 days

= 8.4 * 7 * 30

= Rs. 1764

Hence, the energy bill for a month is Rs.1764.

Solution

Given

Primary Voltage (V1) = 220 V

No of primary turns (N1) = 1000

Secondary Voltage (V2) = 110 V

No of secondary turns (N2) =?

We Know

V1 / V2 = N1 / N2

Or, 220110=1000N2220110=1000N2

Or N2 = 500 turns

Hence, 500 turns in secondary coil.

Solution

Given

Primary Voltage (V1) = 230 V

No of primary turns (N1) = 300

No of secondary turns (N2) = 2400

Secondary Voltage (V2) =?

We Know

V1 / V2 = N1 / N2

Or, 230V2=3002400230V2=3002400

Or V2 = 1840 volts

Hence, 1840 volts is the secondaryvoltage.

Solution

Given

No of primary turns (N1) = 200

No of secondary turns (N2) = 150

Secondary Voltage (V2) = 300 V

Primary Voltage (V1) =?

We Know

V1 / V2 = N1 / N2

Or, V1300=200150V1300=200150

Or V1 = 400 volts

Hence, 400 volts is the primary voltage.

10.

Solution

Given

No of primary turns (N1) = 1500

Primary Voltage (V1) = 220 V

No of secondary turns (N2) =?

Secondary Voltage (V2) = 12 V

We Know

V1 / V2 = N1 / N2

22012=1500N222012=1500N2

Or, N2 = 82 turns

Secondary Voltage (V2) = 24 V

We Know

V1 / V2 = N1 / N2

Or, 22024=1500N222024=1500N2

Or, N2 = 164 turns

Secondary Voltage (V2) = 6 V

We Know

V1 / V2 = N1 / N2

Or, 2206=1500N22206=1500N2

Or, N2 = 41 turns

Hence, when the secondary voltage is 12V,24V and 6V then the no of turns in secondary coils are 82,164 and 41 turns respectively.

11.

Solution

Given

Primary Voltage (VP) = 600 V

No of primary turns (N1) = 1000

No of secondary turns (N2) = 400

Secondary Voltage (VS) =?

We Know

VP / VS = N1 / N2

Or, 600Vs=1000400600Vs=1000400

Or V2 = 240 volts

Hence, 240 volts is the secondary voltage.

12.

Solution

Given

Primary Voltage (V1) = 220 V

No of primary turns (N1) = N1

No of secondary turns (N2) = N14N14

Secondary Voltage (V2) =?

We Know

V1 / V2 = N1 / N2

Or, 220V2=N1N14220V2=N1N14

Or V2 = 55 volts

Hence, 55 volts is the secondary voltage.

Classification of Elements

Ans:

A table in which elements is classified into various blocks, periods and group on the basis of their similarities and dissimilarities is called periodic table.

Ans:

Mendeleev’s periodic rule states that “The physical and the chemical properties of elements are a periodic function of their atomic weights.”

Ans:

The merits of Mendeleev’s periodic table are:

1. Incorrect atomic weights of some of the arranged elements were corrected.

2. Existence of some undiscovered elements was predicted and Mendeleev left gaps for them.

Ans:

The defects of Mendeleev’s periodic table are:

1. Hydrogen was placed in the group I with alkali metals like Li, Na etc but it could be also placed in the position of halogens.

2. The position of isotopes should be separated according to Mendeleev’s periodic rule but they were kept within the same group.

3. There were no suitable places for Lanthanides and Actinides series.

Ans:

Modern periodic rule states that “The physical and the chemical properties of elements are a periodic function of their atomic number.”

Ans:

The features of Modern Periodic Table are:

1. The wrong position of some elements like Argon and Potassium, Cobalt and Nickel were rearranged by their atomic number.

2. Isotopes of the same element can be placed within the same group due to the same atomic number.

3. The controversy of Hydrogen was explained.

4. Elements have been classified into 4 different blocks.

Ans:

Mendeleev’s Periodic Table

Modern Periodic Table

Here elements are arranged in increasing atomic weights.

Here elements are arranged in increasing atomic number.

There are 8 vertical columns which are groups of this Periodic Table.

There are 18 vertical columns which are groups of this Periodic Table

Elements having similar properties are placed separately.

Elements having similar properties are placed together within same group.

Ans:

Modern Periodic Table has many advantages like Elements are arranged in 4 different blocks, they are arranged in the increasing number of atomic number, position of alkali metals, alkali earth metals, halogens, inert gases etc are separated. So it is superior than Mendeleev’s periodic table.

Ans:

Group are the vertical columns of elements in periodic table.

Ans:

Periods are the horizontal rows of elements in the periodic table.

Ans:

A table in which elements is classified into various blocks, periods and group on the basis of their similarities and dissimilarities is called periodic table.

Ans:

Isotopes are the elements having same atomic number but different atomic weight.

Ans:

The different types of shell like s, p, d, f etc are called the sub shell.

Ans:

The elements of groups IB to VIIB and VIII are called transition elements. It is also known as d block elements.

10.

Ans:

Li, Na and K are placed in the same group of the periodic table because their outermost electron lies in p block.

11.

Ans:

Groups

Periods

The vertical columns of the periodic table are called groups

The horizontal rows of the periodic table are called periods.

The valence electrons of all elements in the same group are the same.

The valence electron of the elements in the same period increases from left to right.

The size of atoms increases from top to bottom.

The sixe of atoms decreases from left to right.

The reactivity of elements increases from top to bottom.

The reactivity of elements decreases from left to right.

12.

Ans:

An atom of argon consists of eight electrons in its outermost orbit. Due to this it does not take part in chemical reaction. So it is inactive or inert.

Ans:

The valency of Nitrogen is 3 means that has to gain 3 electrons from the atom to be in stable state.

Ans:

Chlorine is active non metal because it only has to gain one electron to be in stable state.

13.

Ans:

The two alkali earth metals are:

Beryllium and Calcium

14.

Ans:

The members of group IA alkali metals are :

Hydrogen

Lithium

Sodium

Potassium

Rubidium

Caesium

Francium

15.

Ans:

The valencies of the transition metal depend on the number of electron present in d –subshells.

16.

Ans:

The two most reactive non metals are: F and Cl

Ans:

The most reactive metal is Na because it has to lose only one electron to other atom to be in stable state which is one of the properties to be in metal group.

17.

Ans:

The names are Sodium, Neon, Chlorine and Hydrogen.

Ans:

B is inert because it does not have to gain or lose electron as it is already stable.

Ans:

The element which is placed in group VII A is Chlorine and the element which is placed in group I A is Hydrogen.

18.

Ans:

The valency of oxygen is 2, carbon is 4 and A is 1.

Ans:

7 electrons will be there in valence shell of W.

Ans:

C is the most reactive metal because its atomic size is bigger than others and it can lose electron easily.

Ans:

W is the most reactive non metal because its atomic size is small and it can attract electron easily.

Ans:

W= Fluorine

X= Chlorine

Y= Bromine

Z= Iodine

Ans:

The formula when MO compound reacts with K will be: MK

Where K = Sulphur

Ans:

C has larger size because it is he most reactive metal in all of them.

19.

Ans:

Valency

Period No

Group No

Metallic Nature

Block

Good

Better

VIIA

Non metal

VIIA

Non metal

Ans:

Na + Cl2 Nacl (Sodiumchloride)

K + F2 KF (Potassiumfloride)

Ans:

B and C are more reactive.

Acid, Base And Salt

Ans:

An acid is a substance which furnishes hydrogen ion when dissolved in water.

For Eg: HCl, HNO3, H2CO3 etc.

Ans:

Acids which are obtained from living organisms are called organic acid.

For Eg: Citric acid, Formic acid etc

Ans:

Strong acids are the acids that produce more concentration of hydrogen ions in aqueous solution.

For Eg: HCl, H2SO4, HNO3 etc

Ans:

pH value is the range of value from 1 to 14 which determine whether the substance is acid, base or salt.

Ans:

The salt formed by the complete displacement of hydrogen atoms by a metal or electropositive radical is called normal salt. Normal salt are neutral in nature.

For eg: NaCl, KNO3 etc

Ans:

A dyestuff extracted from lichens, that changes color when exposed to pH levels greater or less than certain critical levels.

Ans:

A pH meter is a standard scale which shows the quantity of H+ ions in an acid or a base. It is used to measure the degree of acidity, alkalinity or neutrality. The scale consists numbers 1 to 14 with their corresponding colors in the scale.

Ans:

Hydrochloric Acid → Chlorine + Hydrogen

2HCl → Cl2 + H2

Methyl Orange changes into Red because he compound (HCl) is acidic in nature.

The name of salt when treated with the compound is Calcium chloride.

Ca(OH)2 + 2HCl → CaCl2 +H2O

Ans:

Metallic oxides and hydroxides which are bitter in taste and give hydroxyl ions (OH-) if they dissolve in water are called base.

For Eg: Na2O, K2O, MgO etc

Ans:

Salt is defined as a compound which is formed of the partial or complete replacement of one or more hydrogen atoms of an acid by one or more metal atom or positive radical.

The characteristics of salt are:

1. Common salt (NaCl) is an essential constituent of our diet.

2. Most salts are insoluble in water but few are soluble in water.

3. The solution of salt can conduct electricity.

Ans:

Alkalis are water soluble bases. They give hydroxyl ions when dissolved in water.

For Eg: NaOH, KOH and Ca(OH)2.

Ans:

A Simple indicator is the indicator which indicates whether a solution is acidic or alkaline in nature. It cannot be utilized for determining the pH range or strength of acidic or alkaline solutions. For Eg: Litmus paper and Methyl Orange.

A universal indicator is the indicator which indicates the strength of the acidic or alkaline solution.

It can be utilized for determining the pH range or strength of acidic or alkaline solutions. For Eg: pH paper and pH solution.

Ans:

Acid

Base

Acid gives off hydrogen ions when dissolved in water

Base gives hydroxyl ion when dissolved in water

It turns blue litmus paper into red

It turns red litmus paper into blue

It has sour taste

It has bitter taste and soap feeling when touched.

Ans:

Strong Acid

Weak Acid

It under goes almost complete ionization to give more concentration of hydrogen.

It undergoes very less ionization to give less concentration of hydrogen ions.

Most of the minerals acids are strong acid.

All the organic acids and a few mineral acid are weak acid

It is more corrosive in nature. For eg Sulphuric acid

It is less corrosive in nature. For eg Acetic acid

Ans:

When an acid reacts with base it forms neutral substance ie salt and water. Such type of chemical reaction is called neutral reaction. For eg:

H2SO4 + Mg(OH)2 → MgSO4 + H2O

Acidic Basic Neutral Neutral

10.

Ans:

Acids are said to be proton donors because they lose proton to other atoms in chemical reaction. This makes them more electronegative substance as they will have more number of electrons than protons.

11.

Ans:

Hydrogen is an alkali which has no metal

12.

Ans:

A litmus paper is an indicator which identifies whether the given substance is acid, base or neutral by changing its color.

It is used in Chemical Laboratory

If the substance is acid litmus paper changes its color into red and if the substance is base litmus paper changes its color into blue.

13.

Ans:

Mg + H2SO4 → MgSO4 + H2

Ans:

HCl + Na2CO3 → NaCl + H2O + CO2

Ans:

HCl + NaOH → NaCl + H2O

Ans:

CuO + H2SO4 → CuSO4 +H2O

Ans:

Mg + O2 → MgO

H2SO4 → H2 + SO4

14.

Ans:

The three ways of making salt is:

1. By reaction of acid on metallic oxides.

MgO + H2SO4 → MgSO4 + H2O

2. By action of acid with metals.

Mg + 2HCl → MgCl2 + H2

3. By neutralizing of acid and hydroxide.

NaOH + HCl → NaCl + H2O

15.

Ans:

A universal indicator is a special kind of indicator which is used to measure the strength of acidity or alkalinity .It is prepared by several organic indicators of different of different colors. It is found in green blue solution or in the form of yellow litmus paper.

16.

Ans:

H2SO4

Ans:

H2CO3

Ans:

Ca(OH)2

Ans:

Na2CO3

17.

Ans:

The pH value of water, hydrochloric acid and ammonium sulphate is 7,1 and 10 respectively.

18.

Ans:

Substance

Methyl Orange

Phenolphthalein

Blue Litmus Paper

Red Litmus Paper

Sulphuric acid

Red

No change

Red

No change

Ammonia solution

Yellow

pink

No change

Blue

Salt Solution

Faint Orange

No change

19.

Ans:

During the formation of salt the radical derives from an acid is called acid radical and the radical obtained from a base is called basic radical.

For Eg:

NaOH + H2SO4 → Na2SO4 + H2O

In the salt Na2SO4, sodium radical (Na+) is obtained from the base (NaOH) and SO4--radical is obtained from the acid H2SO4. So Na+ and SO4—are called basic radical and acid radical respectively.

20.

Ans:

Mg + 2HCl → MgCl2 + H2

Ans:

MgO + H2SO4 → MgSO4 + H2O

Ans:

CaCO3 + 2HCl → CaCl2 +H2O + CO2

Chemical Reaction

Ans:

The reaction in which heat is absorbed from the surrounding is called endothermic reaction.

For Eg:

CaCO3 + Heat → CaO + CO2 A

Ans:

The sum of atomic weights of all the atoms of the same or different elements present in the molecule is called molecular weight.

Ans:

The molecular weight of a substance (elements or compounds) is expressed in term of gram which is called its gram molecular weight. It is also called gram mole or only mole.

Ans:

One mole of a substance is defined as a collection of 6.023*1023 particles of that substance.

Ans:

The exchange, synthesis or dissociation that occurs in the molecules during a chemical change is called chemical reaction.

Ans:

The reaction in which heat is released in the surrounding is called exothermic reaction.

For Eg:

C + O2 → CO2 + Heat

Ans:

The conditions required for chemical reaction are:

a. Simple Contact: When two elements are brought in contact they react. This is the simple condition under which chemical reaction takes place.

For E.g.

2Na + Cl2 → 2NaCl

b. Light: Light simulates the reactants for chemical reaction. When hydrogen react violently with chlorine in presence of diffused sunlight to produce hydrogen chloride.

For E.g.

H2 + Cl2 →Sunlight→Sunlight 2HCl

Ans:

The reaction in which two substances exchange ions is called Displacement reaction.

Ans:

When a single compound is formed by the direct combination of two or more reactants during a chemical reaction the process is called addition reaction or combination reaction.

For Eg:

2Na + Cl2 → 2NaCl

Ans:

The chemical reaction in which an atom in a molecule is replaced by another atom forming the end products is called single displacement reaction.

For Eg:

Cl2 + 2KI → 2KCl + I2

Ans:

When a single compound or substance is decomposed into two or more simple molecules in the form of products during chemical reaction, it is called decomposition reaction.

For Eg:

2Na + Cl2 → 2NaCl

Ans:

The reaction in which two compounds react together to form two other compounds by mutual exchange of their ions is called double displacement reaction.

For Eg:

AgNO3 + NaCl → AgCl + NaNO3

Ans:

The reaction in which acid and base react to give salt and water is called acid and base reaction.

For Eg:

NaOH + HCl → NaCl + H2O

Ans:

When a single compound is formed by the direct combination of two or more reactants during a chemical reaction the process is called combination reaction where as when a single compound or substance is decomposed into two or more simple molecules in the form of products during chemical reaction, it is called decomposition reaction. In combination reaction the reactant is more than the product while in decomposition reaction it is quite opposite.

Ans:

The electrolytic decomposition reaction is given below:

2H2O →Electrolysis→Electrolysis 2H2 + O2

10.

Ans:

The influencing factors of rate of change of chemical reaction are: Concentration, Temperature, Catalyst and nature of reactants.

11.

Ans:

Increase in temperature increases the rate of almost all chemical reactions while decrease in temperature decreases the reaction rate. The increase in temperature increases the kinetic energy of molecules. It causes the breaking of bonds n the reactant molecules and reaction takes place at a high speed.

12.

Ans:

The rate of reaction increases with the increase in the concentration of reactants. This means that the rate of reaction is directly proportional to the concentration of the reactants. For e.g. when dilute HCl is poured into a piece of calcium carbonate there is effervescence with the evolution of CO2.The effervescence is brisk when concentrated HCl is added. So it increases the rate of chemical reaction between CaCO3 and HCl.

13.

Ans:

The molecular weight of CO2 is 44 means that it is the sum of atomic weights of atoms in CO2.

14.

Ans:

A catalyst is defined as a chemical substance that changes the rate of chemical reaction itself without undergoing any permanent chemical change during the course of chemical reaction. They are of two types; they are positive catalyst and negative catalyst.

For Eg:

Manganese dioxide acts as a positive catalyst in the decomposition of hydrogen peroxide without heating. This is because it increases the decomposition rate of hydrogen peroxide. Also glycerin acts as a negative catalyst as it decreases the rate of reaction during the decomposition of hydrogen peroxide into water and oxygen.

2H2O2 →MnO2→MnO2 2H2O + O2

15.

2H2 + O2 → 2H2O

H2 + Cl2 → 2HCl

PbO + C → CO + Pb

3H2 + N2 → 2NH3

2KClO2 → 2KCl + 3O2

2Al + Cl2 → 2AlCl

Zn + 2HCl → ZnCl2 + H2

Mg + H2SO4 → MgSO4 + H2

Cu + H2SO4 → CuSO2 + H2O +S2O

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

NH4Cl + NaNO2 → Nacl + 2H2O + N2

2Al + H2SO4 → 2Al(SO4)3 + 3H2

Ca(OH)2 + CO2 → CaCO3 + H2O

NH3 + HCl → NH4Cl

Na2CO3 + SiO2 → 2NaSio3 + CO2

Na2CO3 + 3H2O → 2NaOH + 2H2CO3

4NH3 + 2H2SO4 → 2(NH4)2SO4

Fe + O2 → 2FeO

4Al + 3O2 → 2Al2O3

4P + 5O2 → 2P2O5

NH4Cl + Ca(OH)2 → CaCl2 + H2O +NH3

CaCO3 + HCl → CaCl2 + H2O + CO2

Mg(CO3)2 + Ca(OH)2 → CaCO3 +Mg(OH)2 + H2O

2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O

16.

CuCO3 → CuO + CO2

CH4 +2O2 → CO2 + H2O

6Hcl + 2Al → 2AlCl3 + H2

Mg(HCO3)2 → MgCo3 + H2O + CO2

2Al + 3H2So4 → Al2(SO4)3 + 3H2

HNO3 + NH3 → NH4NO3

Ca(OH)2 + CO2 → CaCO3 + H2O

2KClO3 → 2KCL + 3O2

2H2O2 → 2H2O + O2

17.

Mg(OH)2 + 2HNO3 MgNO3 +H2O

Zn + AgNO3 ZnNO3 + H

3Na + 3H2O 2NaOH + 3H2

Fe2O3 + 6HCl 2FeCl3 +3H20

Numerical Problems:

Solution

Here

The molecular weight of water (H20) = 2*1+16 = 18

∴ →Sunlight→SunlightGram molecular weight of water = 18 g

Thus,

1 mole of water = 18 g of water

0.5 mole of water = 0.5 * 18 g of water = 9g

Hence, the weight of 0.5 moles of water in grams is 9 g.

Solution

Given

The reaction is:

2KOH + H2SO4 → K2SO4+ H2O

2(39+16+1) (39*2+32+16*4)

122g 174g

Here,

122 g of KOH = 174g of K2SO4

1 g of KOH = 174122174122g of K2SO4

∴ 10 g of KOH = 174∗10122174∗10122 g of K2SO4

= 14.26 g of K2SO4

Hence, 10 g of KOH gives 31.07 g of K2SO4

Solution

Given

The reaction is:

MgCo3 → MgO + Co2

(24 +12 +16*3) (24 + 16) (12 +16*2)

84g 40g 44g

Here,

84g of MgCo3 gives 40g of MgO and 44g of Co2

1gof MgCo3 gives 40844084g of MgO and 44844484g of Co2

25gof MgCo3 gives 4084∗254084∗25g of MgO and 44844484*25 g of Co2

Hence, 25g of MgCo3 gives 11.9711.97g of MgO and 13.0913.09g of Co2.

Solution

Given

The reaction is:

2Fe2O3 → 2Fe + 3O2

2(2*52+3*16) 4(52)

304g 208g

1g 208304208304g

=6.06g =3.93g

Therefore, 10g of potassium chloride gives 6.06 g of potassium chloride and 3.96 g of Oxygen.

Solution

Given

The reaction is:

2Mg + O2 → 2MgO

2*24 16*2 2(24+16)

48g 32g 80 g

Now

48g of Mg gives 80g of MgO

1g of Mg gives 80408040g of MgO

10g of Mg gives 80∗104080∗1040g of MgO

= 16.6 g of MgO

Hence, 16.6 g of MgO is produced when 10g of Mg burns in air.

Solution

Given

The reaction is:

2H2O → 2H2 + O2

2*18 2*2 2*16

36g 4g 32g

1g 436436g 32363236g

7.2g 4∗7.2364∗7.236g 32∗7.23632∗7.236g

= 0.8g =6.4g

7.2g of water gives 0.8g of Hydrogen and 6.4 g of Oxygen.

Solution

Given

The reaction is:

2KClO3 → 2KCl + 3O2

2(39 + 35 +3*16) 2(39+35) 3(2*16)

244g 148g 96g

1g 148244148244g 9624496244g

10g 148∗10244148∗10244g 96∗1024496∗10244g

=6.06g =3.93g

10g of potassium chloride gives 6.06 g of potassium chloride and 3.96 g of Oxygen.

Some Gases

Ans:

Chemical Reaction:

CaCO3 + 2HCl → CaCl2 + H2O + CO2

Ans:

1. Carbon dioxide is used by all green plants during photosynthesis.

2. It is used as fire extinguishers.

3. It is used for making dry ice to preserve meat, fruits etc

4. It is also used in the process of carbonation in industries.

5. It is used for making soft drinks like beer, coca cola, soda etc

Ans:

The physical properties of Carbon dioxide are:

1. It is colorless, odourless and tasteless gas.

2. It is heavier than air.

3. It dissolves in water.

4. It turns wet blue litmus paper into red.

Ans:

Mol. Formula Chemical name

Slaked lime Ca(OH)2 Calcium Hydroxide

Lime Stone CaCO3 Calcite

Urea CO(NH2)2 Carbamide

Ans:

If we fill carbon dioxide in a balloon then it remains in he surface of the earth which proves that carbon dioxide is heavier than air.

Ans:

The ice obtained from cooling CO2 gas at about 780C is called dry ice.

It is used to preserve fruits and vegetables.

Ans:

Carbon dioxide covers the burning flame of fire making itself just like a blanket. As a result CO2 does not allow oxygen to pass and finally puts out the fire.

Ans:

When a burning match stick is put near the jar containing Carbon dioxide a pop sound is heard which indicates the presence of Carbon dioxide.

Ans:

When carbon dioxide is passed in lime water for a short time then calcium carbonate is formed.

Ca(OH)2 +CO2 → CaCO3 +H2O

10.

Ans:

When a burning magnesium ribbon is introduced into a gas jar filled with carbon dioxide, the reaction between magnesium and carbon dioxide takes place. As a result white colored magnesium oxide is produced along with the formation of black spots of carbon.

Reaction:

2Mg +CO2 → 2MgO + C

Ans:

When carbon dioxide is passed continuously into lime water for a long time, the milkyness disappears slowly due to the formation of water soluble calcium bicarbonate [Ca(HCO3)2].

Reaction:

CaCO3+ Co2 +H20 → Ca(HCO3)2

11.

Ans:

The process by which plants make the food with the help of sunlight.

Reaction:

6CO2 + 6H20 →sunlight→sunlight C6H12O6 + 6O2

12.

Ans:

Carbon dioxide is collected in laboratory by the upward displacement of air.

13.

Ans:

When moist blue litmus paper is inserted into the jar containing carbon dioxide gas it turns into red due to formation of carbonic acid.

14.

Ans:

Carbon dioxide is produced when limestone is reacted with hydrochloric acid.

Chemical Reaction:

CaCO3 + 2HCl → CaCl2 + H2O + CO2

When carbon dioxide is passed in lime water for a short time then calcium carbonate is formed.

Ca(OH)2 +CO2 → CaCO3 +H2O

15.

Ans:

Chemical Reaction in the preparation of Ammonia gas:

2NH2Cl + Ca(OH)2 → CaCl2 + 2H2O +2NH3

16.

Ans:

Ammonia is collected by the downward displacement of air because it is lighter than air.

17.

Ans:

The necessary conditions for the manufacture of Ammonia by Haber’s process are:

1. Iron as a catalyst

2. Molybdenum

3. Temperature at 5000C

4. High pressure 200-900

Equation:

N2 +3H2 →5000Fe\Mo200−900atm→200−900atm5000Fe\Mo 2NH3

18.

Ans:

Ammonia gas is not collected in water because it is highly soluble in water and forms ammonium hydroxide.

19.

Ans:

The physical properties of ammonia are:

1. It is colorless gas with pungent smell. It brings tears to our eyes.

2. It is highly soluble in water.

3. It is lighter than air.

4. It is neither combustible nor a supporter of combustion.

20.

Ans:

It shows that Ammonia is basic in nature.

21.

CO2 + H2O → H2CO3

Ca(HCO3)2 + 2HCl → CaCl2 + 2H20+2CO2

CaCO3 → CaO + CO2

2NH3 + CO2 → NH2-CO-NH2 + H2O

NH3 + HNO3 → NH4NO3

NH3 + H2SO4 → (NH4)2SO4

22.

Ans:

Liquor ammonia is strong ammonia solution.

The four uses of ammonia gas are:

1. It is used in the manufacture of chemical fertilizers.

2. It is used as cleansing agent for removing oils and grease.

3. It is used for developing the blue prints for maps.

4. Its solution is used as a laboratory reagent.

23.

Ans:

Water changes into pink color because of phenolphthalein and ammonia.

Ans:

Water enter the flask through the pipe as a fountain because ammonia is highly soluble in water.

Ans:

This experiment proves that ammonia is highly soluble in water.

Ans:

When kerosene is kept instead of water, no hydroxide compound is formed so it doesn’t change the color of phenolphthalein.

24.

Ans:

Ammoniun Sulphate can be prepared by reacting ammonia with sulphuric acid:

Reaction:

2NH3 + H2SO4 → (NH4)2SO4

25.

Ans:

A moist red litmus paper when inserted in the jar containing ammonia gas turns into blue.

Metals

Ans:

Ores are defined as the minerals from which metals can be extracted for commercial purposes. For Eg: Bauxite

Minerals are the natural, pure, inorganic substances which are crystalline. For Eg: Cuprites

Metallurgy is the complex and sequential process of extracting metals from their respective ones.

Ans:

Iron is very rarely found in free state but it occurs in a combined form. It is found in the body of living organisms.

The steps of metallurgy are:

a. Crushing of ore

b. Pulverization

c. Concentration of ore

d. Calcination and roasting

e. Extraction of metal from calcinated or roasted ore

f. Refining

Ans:

Iron is transition element and known as d block element. It belongs to group VII of the periodic table. The electronic configuration of iron is: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2,3d6

Ans:

The process of obtaining free metals from the calcinated or roasted ore is called extraction of metals.

First of all haematite ore is powdered and concentrated. The ore is separated from non magnetic substance by using magnetic method. Then the ore is washed with water, mixed with coke and lime stone, and heated strongly in a blast furnace by passing hot air from the bottom. As a result iron is formed in a grey solid which melts down in the furnace.

Ans:

The physical properties of metal are:

1. They are good conductor of heat and electricity.

2. They have high specific gravity.

3. They have high melting and boiling point.

4. They are malleable and ductile in nature.

5. They have metallic lusture.

Iron is called metal because it shows metallic property.

a. Haematite and Magnetite

b. Bauxite and Feldspar

c. Argentite and Silver copper glance

d. Cuprite and Copper pyrite

e. Alluvial soil and quartz veins.

Ans:

The external cover of aeroplane is made from aluminium instead of copper because it is light and it does not rust.

Ans:

First of all the crude bauxite is finely powered and heated with concentrated sodium hydroxide solution to produce sodium alumunate (NaAlo2). The insoluble impurities present in the sodium alumunate are removed by filtration. Sodium alumunate is then diluted with water and some hydrochloric acid is added to produce aluminium hydroxide [Al(OH)3]. Thus obtained aluminium hydroxide is filtered, washed dried and heated strongly to get pure aluminium.

Ans:

The three chemical properties of Aluminium is:

a. Action with nitrogen:

2Al + N2 → 2AlN (Aluminium Nitride)

b. Action with Halogen:

2Al + 3Cl2 → 2AlCl3(Aluminium tetra chloride)

c. Action with air:

4Al + 3O2 → 2Al2O3 + 3220 KJ

(Aluminium Hydroxide)

10.

Ans:

IRON:

1. It is used as making rods, pipes

2. It is used in the manufacture of steel

3. It is used in making household utensils

4. t is used as a catalyst in chemical reaction.

ALUMINIUM:

1. It is used in making household utensils

2. It is used in making silvery paints

3. It is used for making electric wires

4. It is used in making aluminium foils, aircraft body.

COPPER:

1. It is used for making electrical goods and cables

2. It is used for making coins, pieces of jewellery.

3. It is used for electroplating

4. It is used in electrolytic process.

GOLD

1. It is used for electroplating

2. It is used to prepare alloys.

3. It is used for making pieces of jewellery.

4. It is used as ornaments.

SILVER

1. It is used for making coins, pieces of jewellery.

2. Silver bromide is used in photography.

3. It is used in the preparation of silver salts and medicine.

4. It is used for silvering mirror.

11.

Ans:

Copper and the other elements like silver and gold of the group IB of the periodic table are collectively called coinage metals since they are used for making jewellery. They are different from other metal because they are used for making jewellery.

12.

Ans:

Coinage metals are grouped together in the periodic table because presence of d orbital in the valence shells.

13.

Ans:

First of all copper pyrite ore(CuFeS2) is powdered and then concentrated by froth floatation process to remove the impurities. The concentrated copper pyrites ore is roasted in air in a blast furnace in which copper sulphide (CuS) is oxidized into copper oxide(CuO) along with the evolution of sulphur dioxide gas. Here unoxidised copper sulphide reacts with copper oxide. As a result about 99.4% pure copper is obtained in molten state, which is further purified by electrolytic process.

14.

Ans:

Blister copper is about 99.4% pure copper obtained by heating the powder of chalcopyrite with air in a big furnace. It is refined by the process of electrolysis.

15.

Ans:

Gold is called noble metal. Gold is found in native state because it does not react with other elements in ordinary conditions.

16.

Ans:

Argentite is the principal ore of Silver. The ore is crushed, powdered and concentrated by froth flotation process. The concentrated ore is mixed with sodium cyanide solution and heated with hot air in a furnace which forms sodium argento-cyanide. Then zinc powder is mixed in the solution and stirred. In this way silver is obtained by filtration. Silver thus obtained is further purified by electrolytic refining.

The three uses of Silver are:

1. It is used for making coins, pieces of jewellery.

2. Silver bromide is used in photography.

3. It is used in the preparation of silver salts and medicine.

17.

Ans:

The three chemical properties of Silver are:

1. Action with Sulphur:

2Ag + S → Ag2S (Silver Sulphide)

2. Action with halogen:

2Ag + Cl → 2AgCl (Silver Chloride)

3. Action with nitric acid:

Ag + conc. 2HNO3 → 3AgNO3 + 2H2O +NO

(Silver Nitrate)

18.

Ans:

The alluvial sand containing gold is carried in a sluiceway system where a strong current of water is passed over it. A sluice way system is an inclined channel provided with ridges and groves at the base called riffles. The heavier gold particles are retained in the sluices while water carries the lighter particles of sand away. In this way gold is extracted from the alluvial sand.

19.

Ans:

The three chemical properties of Gold are:

1. Action with Acid:

2Au + 3HNO3 + 9HCl → 3NOCl + H2O +2AUCl3

(Aqua Regia) (Nitrosyl Chloride)

2. Action with Halogens:

2Au + 3Cl2 → 2AuCl3

3. Action with alkali Cyanide:

4Au + 8KCN + 2H2O → 4K[Au(CN)2] + 4KOH

(Potassium aurocyanide)

20.

Ans:

Aluminium is extracted from Bauxite ore and Iron is extracted from Haematite ore.

21.

Ans:

Aqua regia is the mixture of 3 parts of conc. HCl with 1 part of Conc. HNO3. We use metals for jewellery and decorations because it has shining property and is malleable and ductile.

22.

Ans:

The uses of aluminium foils are:

1. It is used to cover food.

2. It is used as body parts in aircraft.

3. It is used in pharmaceutical products.

4. It is used in cigarette.

23.

Ans:

The mixture is called Aqua Regia.

Chemical reaction:

2Ag + conc. H2SO4 → Ag2SO4 +H2O + SO2

24.

Fe + dil. H2SO4 → FeSO4 + H2

(Ferrous Sulphate)

Fe + S → FeS (Ferrous Sulphide)

Fe + CuSO4 → FeSO4 + Cu

(Ferrous Sulphate)

2Fe + conc.6H2SO4 → 3SO2+ Fe2 (SO4)3 + 6H2O

2Al + dil. 6HCl → 2AlCl3 + 3H2

2Al + conc. 6H2SO4 → Al2(SO4)3 + 3SO2+6H2O

2Al + 2NaOH + 2H2O → 2NaAlO2 +3H2

2Al + 3Cl2 → 2AlCl3

2Al + 3Br2 → 2AlBr3

2Al + 3I2 → 2AlI3

4Ag + conc. 4HCl + O2 → 4AgCl +2H2O

(Silver Chloride)

3Ag + dil. 4HNO3 → 3AgNO3 + 2H20 +NO

Ag+ conc. 2HNO3 → AgNO3 +H2O +NO2

Hydrocarbon And Their Derivatives

Ans:

The chemical compounds which are made by carbon and hydrogen are called hydrocarbons. For eg: Methane, Ethane etc

Ans:

Homologous series is defined as a group of organic compounds having similar structures and chemical properties in which the successive compounds differ by CH2 group.

Ans:

Organic chemistry is the branch of chemistry which deals with the study of hydrocarbons and their derivatives.

Ans:

Hydrocarbons are the compounds formed by the combination of hydrogen and carbon.

There are two types of hydro carbons. They are:

1. Saturated Hydrocarbons

2. Unsaturated Hydrocarbons

Ans:

IUPAC stand for International Union of Pure and Applied Chemistry.

Ans:

The hydrocarbons in which a carbon atoms are combined with single bonds is called saturated hydro carbons. For eg: Propane, Pentane etc.

Saturated Hydrocarbon

Unsaturated Hydrocarbon

In them the carbon atoms are bounded together by a single covalent bond.

In them the carbon atoms are bounded together by a triple or double covalent bond.

They have only one group which is alkane

They have two groups which is alkene and alkyne

They are less active

They are more active

Eg: Methane, propane

Eg: Ethene, Propene

Ans:

The hydrocarbons in which the carbon atoms are bounded together by a triple or double covalent bond are called unsaturated hydro carbons. For Eg: Ethylene, Acetylene etc

Ans:

An alkene is an unsaturated hydrocarbon in which the two carbon atoms are connected by a double bond. The double bond is formed by the sharing of two pairs of electrons. For eg: Methane, Propane etc

Ans:

Alkynes are an unsaturated hydrocarbon in which the two carbon atoms are connected by a triple bond. The triple bond is formed by the sharing of three pairs of electrons. For Eg: Acetylene, Methyl acetylene etc.

Ans:

Glycerol is the simplest trihydric alcohol. It occurs in almost all the natural animals fats and vegetable oils. It is proposed by hydrolysis of fat or oil in presence of alkali.

Ans:

Butane is the saturated hydrocarbon. It occurs in the natural gas and is found in petroleum mines. Butane is a fuel. I is also used as a raw material to manufacture synthetic rubber.

The hydrocarbon unit derived by the removal of one hydrogen atom from alkane is called alkyl group. For Eg: CH3(Methyl)

Ans:

The five uses of Methane are:

1. It is used to manufacture water gas or hydrogen gas.

2. It is used in the form of LPG for domestic use.

3. It is used for making printing ink.

4. It is used for making Carbon Black needed for paints and in rubber industries.

5. It is used as a gaseous fuel in industries and household works.

Ans:

Ethyl Alcohol

Ans:

The uses of alcohol are:

1. It is used for dry cleaning

2. It is used to make perfume, paint, medicine etc

3. It is used as alcoholic beverage.

4. It is used as fluids in thermometers.

a. C2H6

b. C3H8

c. C2H2

d. C4H6

e. C2H4

f. R-O-R Where R = CH3

g. C3H5(OH)3

h. C2H5OH

Ans:

The uses of ether are:

1. It is used as a solvent.

2. It is used as cooling agent.

3. It is used as purifying compounds by extracting.

10.

Ans:

The physical properties of ether are:

1. It is colorless volatile liquid with pleasant smell.

2. It is slightly soluble in water but readily soluble in alcohol.

3. Its freezing point is -1160C and boiling point is 350C

The physical properties of alcohol are:

1. It is colorless liquid but has taste.

2. It is soluble in water.

3. Its boiling point is 780C and freezing point is -1140C.

11.

Ans:

The characteristics of homologous series are:

1. All the members of the same series can be represented by the same formula.

2. All members of the series show similar chemical properties.

3. Each excessive member of a homologous series differs by CH2.

4. All the members of the same series have the same functional group like alcohol (-OH), ether (-O-) etc.

12.

Ans:

Functional group is defined as an atom of group of atoms which determines the chemical behavior of organic compounds.

The hydrocarbon unit derived by the removal of one hydrogen atom from alkane is called alkyl group.

13.

Ans:

The molecular formula of Butane is C4H10.

A common use of butane is:

It is used in LPG gas.

14.

Ans:

Acetylene is unsaturated hydrocarbon because the carbon atoms are bonded together by a double covalent bond. The two uses of ether are:

1. It is used as a solvent.

2. It is used as cooling agent.

15.

Ans:

Covalent bond is formed between the carbon and hydrogen. Because the bond between the carbons are covalent bond and covalent bonds are weak bond.

Materials Used In Daily Life

Ans:

Cement is the mixture of fine gray powder of calcium silicate and calcium aluminates.

Gypsum salt is added in the cement because it slows down the setting time of the cement.

a. Cement clinkers is a mixture of pea sized balls which consists of calcium silicate and calcium auminate .It is produces by heating a paste of calcium carbonate, clay and water in rotary kiln.

b. RCC stands for Reinforced Concrete Cement. It is the concrete having an iron frame inside it as a support. It is very strong material for construction.

c. Mortar is the mixture of cement, sand and water which is used as plastering material and to join bricks, stone etc.

d. Slurry is a material which makes the partical flow able.

e. Humus is a large group of natural organic compound found in the soil formed from the chemical and biological decomposition of plants and animals residues and from the synthetic activities of microorganisms.

f. Compost manure is a kind of natural fertilizer which is prepared by mixing of cattle dung, poultry droppings and crop residue like remnants of the straw, rotten vegetables etc

g. Monomer is a small organic molecule which can combine together in large number to form a big molecule called polymer. For Eg: Ethene, styrene etc

h. Pollution is the contamination of environment by harmful substances. For eg: Air pollution, Water pollution etc

i. When monomers are heated they attach with one another forming a long polymer. Tis is called polymerization.

j. Artificial fiber like nylon, polyester, acrylene and olefin that are prepared by chemical process are called synthetic fibers.

k. The fertilizers which contain more than one primary plant nutrients are called mixed fertilizers. For Eg: Ammonium hydrogen phosphate, potassium nitrate etc.

Ans:

The uses of Cement are:

1. Cement is used for making houses, buildings, roads etc

2. It is used for flooring and roofing.

3. RCC is used for framework of iron rods.

4. It is used for making pillars, bridges etc.

Ans:

Glass is an amorphous transparent homogenous mixture of silicates of alkali metals and silicates of alkaline earth metals.

Ordinary glass is the mixture of 50% silica, 15% sodium carbonate, 10% calcium carbonate and 25% glass piece.

Ordinary glass is used for making laboratory apparatus, windowpanes, light bulbs, glass sheets etc

Ans:

A mixture of silica and sodium carbonate or potassium carbonate while heating at 8000C melts and forms sodium silicate or potassium silicate. It is soluble in water, so it is called water glass.

Ans:

Soft Glass

Hard Glass

It is produced by heating sodium carbonate and calcium carbonate with silica.

It is obtained by heating potassium carbonate and calcium carbonate with silica.

It is used for making window panes, mirror etc

It is used for making heat resistant equipments.

Ans:

Colored glass is made by adding small amount of various metallic oxides to the hot molten raw material used to manufacture glass. Cobalt oxide is used in preparing blue glass.

A mixture of silica and sodium carbonate or potassium carbonate while heating at 8000C melts and forms sodium silicate or potassium silicate. It is soluble in water, so it is called water glass.

The homogenous mixture of sodium silicate, calcium silicate and boron silicate is called boro silicate glass or pyrex glass.

It is a polymer of styrene. It is a hard, transparent and light plastic. It is used for making toys, thermos flask, ceiling tiles etc.

Bakelite is made by condensation and polymerization of formaldehyde and carbonic acid. It is dark brown in color which is brittle and hard. It is used in electric fitting, to make handles of cook wares.

Ans:

Lead Crystal Glass is used for making eye lense.

Ans:

Glass is an amorphous transparent homogenous mixture of silicates of alkali metals and silicates of alkaline earth metals.

11.

Ans:

Ceramics is the science of pottery which my be defined as objects made from special type of clay, hydrated aluminium silicate, feldspar and silicates.

The hard, brittle and porous article are heated and coated with tin oxide or lead oxide which forms a thin layer over the surface. In this way glazing is done.

12.

Ans:

The uses of ceramics are:

1. Ceramics are used for making household pottery, jugs, tiles, bowls etc

2. It is used to make furnace, as insulators in television sets etc

3. They are used for making artificial teeth, bone joints and porcelain.

13.

Ans:

Glazed pottery is superior to earthen pots because they are hard, brittle and attractive. They are not affected by acids, alkali and other chemicals and they can withstand high temperature.

14.

Ans:

Artificial fiber like nylon, polyester, acrylene and olefin that are prepared by chemical process are called synthetic fibers.

15.

Ans:

Nylon is the first man made synthetic polymamide fibre made up of adipic acid and hexa methylene diamine. Nylon fibers are also called polyamide fibers. They are the first synthetic fibers. Nylon fibers are light in weight and strong.

The advantages of synthetic fibers are:

1. They are light in weight and strong.

2. They melt at low temperature.

3. They are used as fishing net, carpets etc

16.

Ans:

Plastics are the artificial materials made by the polymerization of carbon containing monomers.

The uses of plastics are:

1. They are used as shopping bags.

2. They can be used as insulators in wires.

3. They are used as waterproof.

4. It is used in surgical cases.

17.

Ans:

The disadvantages of plastics are:

1. It non biodegradable

2. It causes soil water and air pollution.

3. They are harmful to aquatic creatures.

18.

Ans:

Soap is the sodium salt of long chain fatty acids that has cleansing property in water.

The raw materials used for manufacturing soap are:

Animals fat or vegetable oil, sodium hydroxide, stearic acid, palmitic acid, oleic acid etc

19.

Ans:

Sodium carbonate or starch is addeded to make soap hard.

20.

Ans:

Detergents are the sodium salts of a long chain benzene sulphonic acid or long chain alkyl benzene sulphate which have more cleansing property in water.

Detergents are called the soap less soap because they have cleansing property in water like soap but their chemical nature is different from soap.

21.

Ans:

Pesticides are the chemical compounds which are used to control or kill the pest.

Excessive use of pesticides can destroy crops and the fertility of soil.

For Eg: Insecticides and Herbicides.

22.

Ans:

The disadvantages of pesticides are:

1. It kills crops and the fertility of soil if they are used excessively.

2. It also kills non targeted animals and insects.

23.

a. Dichloro diphenyl trichloroethane

b. Benzene Hexachloride

c. Polyvinyl chloride

24.

Ans:

The poisonous chemicals which are manmade and used to control or destroy harmful insects are called insecticides.

The advantages of insecticides are:

1. They kill or destroy harmful insects quickly.

2. They help to control several diseases by killing germs.

3. They can destroy all the stages of the lifecycle of harmful insets. As result there is increase in crop yield.

25.

Ans:

Soap

Detergent

It is a sodium or potassium salt of fatty acid.

It is a synthetic petrochemical obtained from hydrocarbon.

It is not suitable for washing purpose when the water is hard.

Synthetic detergent can be used for washing ever when the water is hard.

Soaps are biodegradable.

It is not biodegradable.

26.

Ans:

Organic fertilizers are made by decaying plants and waste products of animals. They do not cause chemical pollution.

Chemical fertilizers are made by using different chemicals. They cause chemical pollution.

27.

a. Potassium

Ans: Potassium Nitrate and Potassium Sulphate.

b. Nitrogen

Ans: Urea and Ammonium nitrate.

c. Phosphorous

Ans: Ammonium phosphate and Calcium super phosphate

28.

Ans:

The chemical fertilizer containing all plants nutrients ie. Nitrogen, phosphorous and potassium is called NPK fertilizer. NPK fertilizer is also called the complete fertilizer.

29.

Ans:

The advantages of fertilizer are:

1. It promotes fast growth and development of plants.

2. It helps to increase yield of crops.

3. It helps to resist disease.

The disadvantages of fertilizers are:

1. Low growth of plants

2. Reduce in disease resistant power of the plants.

3. The wilting and withering of leaves.

30.

Ans:

Nitrogen is important to plants because it promotes fast growth and development of plants. It helps to form a large amount of protein and chlorophyll.

31.

Ans:

Non biodegradable wastes are the poisonous substance which causes environment degradation. For Eg: Plastics and Detergent.

These substances are not degradable in the environment.

32.

Ans:

The environment pollution caused by various chemicals lie insecticides, chemical fertilizers, synthetic cleanser and other industrial chemicals is called chemical pollution.

The causes of environmental pollution are:

1. The use of insecticides and fertilizers.

2. Household waste and plastics.

3. Smoke from vehicles and factories.

4. Synthetic cleanser.

33.

Ans:

Humans are more responsible to degraded environment than other living beings on earth. Vehicles, big factories and large amount of chemicals we use cause great environmental pollution. To control this we have to take steps. We should dispose non biodegradable waste in proper place rather than throw them randomly. Old vehicles should be replaced with new one so that air will not get polluted by carbon monoxide released by them. Awareness campaign should be held so that people will get knowledge to protect the environment.

Classification Of Plants And Animals

Ans:

Classification is the process of grouping the living beings into various groups and sub groups on the basis of similar and dissimilar characteristics.

The two importance of classification are:

1. It makes the study of plants and animals easier and systematic.

2. It gives us the idea about the evolution of plants and animals.

Ans:

The great Swedish Biologist Carolus Linnaeus is considered as the father of taxonomy.

Ans: A genus is the group of closely related species.

Ans: A species is the group of closely related organisms having almost all similar characteristics which can interbreed freely and produce fertile offspring.

Ans: Nomenclature is the system of naming living organisms. The scientific way of assigning two names to an organism is called Binomial System of Nomenclature.

Ans: Taxonomy is defined as the branch of biological science that deals with the identification, nomenclature and classification of living organisms.

Ans:

Classification of organism is necessary because:

1. It makes the study of plants and animals easier and systematic.

2. It gives us the idea about the evolution of plants and animals.

The first letter of the generic name is written in a capital form and the specific name begins with a small letter. Both the generic names should be underlined when written by hand or typed and given in italics when printed.

For eg:

Homo sapiens

Zea mays

Ans:

The scientific way of assigning two names to an organism is called Binomial System of Nomenclature.

For eg:

Homo sapiens

Zea mays

Ans:

Phanerogams are the developed seed bearing plants.

The four characteristics are:

1. The body of these plants is divided into true root, stem and leaves.

2. Vascular tissue ie xylem and phloem are present.

3. The reproductive organs are multicellular and embryo develops from a fertilized egg.

4. Flowers/Cones and seed are present.

Ans:

Monocotyledons

Dicotyledons

A seed bears only one cotyledon

A seed bears two cotyledons

Fibrous root is present

Tap root is present

Distinct nodes and inter nodes are present

Distinct nodes and internodes are absent

Leaves are elongated having parallel venation.

Leaves are broad having reticulate venation.

a. PEA

Kingdom: Plant

Sub Kingdom: Phanerogams

Division: Phanerogams

Sub division: Angiosperm

Class: Dicotyledons

Type: Pea

b. PINE

Kingdom: Plant

Sub Kingdom: Phanerogams

Division: Spermatophyta

Sub division: Gymnosperm

Type: Pinus

c. WHEAT

Kingdom: Plant

Sub Kingdom: Phanerogams

Division: Spermatophyta

Sub division: Angiosperm

Class: Monocotyledons

Type: Wheat

d. MAIZE

Kingdom: Plant

Sub Kingdom: Phanerogams

Division: Spermatophyta

Sub division: Angiosperm

Class: Monocotyledons

Type: Maize

Ans:

A pine plant cannot be placed in sub division angiosperm they grow on dry land and they have naked seeds.

10.

1. Gymnosperm (Pine)

2. Angiosperm (Sunflower)

1. Angiosperm (Sugarcane)

2. Angiosperm (Pea)

11.

a. Spermatophyta (Mustard)

b. Spermatophyta (Maize)

12.

Ans:

13.

Ans:

The characteristics of Angiosperm are:

1. They bear developed flowers.

2. Their seed are enclosed in true fruits.

3. They may be herbs, shrubs and trees.

14.

Ans:

Vertebrates are the animals having backbone. For e.g. : Man and Bat

15.

Ans:

The classes of vertebrates are:

1. Mammalia (Man, Whale)

2. Aves (Crow, Duck)

3. Reptilai (Lizard, Snake)

4. Amphibia (Frog, Toad)

5. Pieces (Fish, Sea horse)

16.

Ans: They have moist skin and body is divided into head and trunk.

Ans: They have beak without teeth and bones are hollow.

Ans: Body is covered with hairs and it is divided into head and trunk.

Ans: They have backbone and body is covered with scales.

17.

Ans: Hemichordata are the notochord in the anterior region of the body.

Ans: The act of rendering fertile.

Ans: Cold blooded animals are the animals which can change their body temperature according to their habitat.

18.

Ans:

Birds evolved from reptiles and still show many reptilian characters so they are called glorified reptiles.

The characters of aves:

1. Their body is streamlined and divided into head, neck, trunk and tail.

2. Their body is covered with feathers, scales are present in hind legs only.

19.

Ans:

Name: Frog

Phylum: Chordata

Their body is divided into head and trunk.

Their skin is usually thin, soft and moist.

Ans:

Name: Sea Horse

Phylum: Chordata

Their body is long and streamlined to reduce water resistance.

Body is covered with waterproof scales.

20.

Ans:

Bat belongs to class mammalian because they have characteristics of mammal. Their body is divided into head, neck and trunk. They give birth to young ones. They are viviparous and warm blooded animals and their heart is divided into four chambers.

21.

Ans:

Bats

Birds

They are viviparous

They are oviparous

A pair of external ears(pinnae) are present.

Pinnae are absent.

Body is covered with hair.

Body is covered with feathers.

Ans:

Pigeon

Frog

Body is covered with moist feathers.

Body is covered with moist skin

Locomotion takes place with the help of wings.

Locomotion takes place with the help of limbs.

Heart is four chambered.

Heart is three chambered.

Body is divided into head and trunk and tail.

Body is divided into head and trunk.

Ans:

Seahorse

Horse

They are oviparous

They are viviparous

They are cold blooded animals

They are warm blooded animals

Locomotion takes place with the help of fins.

Locomotion takes place with the help of limbs.

Heart is two chambered.

Heart is four chambered.

Ans:

Frog

Lizard

Body is covered with moist skin

Body is covered with hard, dry and horny scales.

Locomotion takes place with the help of limbs.

Heart is three chambered.

Heart is four chambered.

Body is divided into head and trunk.

Body is divided into head, tail, neck and trunk.

22.

Ans:

23.

Ans:

a. Whale

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Mammalia

b. Frog

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Amphibia

c. Snake

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Reptilia

d. Bat

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Mammalia

e. Dolphin

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Mammalia

f. Rohu

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Pieces

g. Lizard

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Reptilia

h. Rat

Kingdom: Animalia

Sub Kingdom: Metazoa

Phylum: Chordata

Sub Phylum: Vertebrata

Class: Mammalia

24.

Ans: Fish

Locomotion takes place with the help of fins.

Heart is two chambered.

Ans: Seahorse

Locomotion takes place with the help of fins.

Heart is two chambered.

Ans: Seahorse

Locomotion takes place with the help of fins.

Heart is two chambered.

25.

Ans:

The characteristics of vertebrates:

1. Notochord is present either in the early stage or throughout the life.

2. They have a dorsal tubular hollow nerve tube.

3. They have pharyngeal gill slits.

4. They have post natal tail.

5. Their eyes are originated from brain not from skin.

26.

Ans:

The five characteristics of amphibian are:

1. Their body is divided into head and trunk.

2. Their skin is usually thin, soft and moist.

3. They have pairs of pentadactyle limbs.

4. They have three chambered hearts with two auricles and one ventricle.

5. Their tadpole stage is provided with a tail.

Virus

Ans:

Viruses can exist outside the living cell for a long period of time but they cannot reproduce outside the living cell. Since the viruses can reproduce only inside the specific cell of the living host, they are called obligatory parasites.

Ans:

Any three diseases caused by virus are:

1. Diarrhea

2. Typhoid

3. Common cold

Ans:

The two disadvantages of virus to animals are:

1. It can damage the cells of animals.

2. It causes diseases.

Ans:

The causative agent of mumps is infected saliva.

Symptoms:

a. Intermittent fever.

b. Swelling of the salivary glands located on the sides of face below the ears and pain.

Preventive Measures:

a. Administration of MMR vaccine.

b. Covering mouth and nose while coughing, sneezing and talking.

The causative agent of measles is paramyxo virus.

Symptoms:

a. Runny nose, watery eyes

b. Appearance of small red spots.

Preventive Measures:

a. Administration of MMR vaccine.

b. Covering mouth and nose while coughing, sneezing and talking.

High Fever and Respiratory tract become affected.

Fever is intermittent and testes.

Ans:

The person infected with HIV slowly loses his immunity against other diseases. So if the patient suffers from any diseases it cannot be treated due to lack of immunity. The virus does not kill he person but it reduces fighting capability against other diseases. As a result the patient suffers from various diseases and finally dies due to various diseases. So we can say that HIV itself does not kill human beings but it helps other diseases.

Ans:

The two symptoms of AIDS are:

1. Unexpected high fever at interval rate for more than a week.

2. Rapid loss of body weight.

Ans:

The three preventive measures of AIDS are:

1. Avoid bisexual or homosexual relations and prostitutes.

2. Avoid the use of intravenous drug.

3. Use of contra septic devices.

Ans: Mode of transmission of virus is the medium through which virus transfer from one body to another like air, water, blood, physical contact etc.

Ans: The agents who transfer virus from one body to another are the causative agents.

10.

Ans:

X: DNA

Y: Tail fibers

Z: Head

Ans:

It helps in the transfer of genes.

Ans:

Head protects DNA.

11.

Ans:

The two living characteristics of virus are:

1. It can reproduce inside the cell.

2. It contains genetic characteristics which are transmitted to offspring.

The two non living characteristics of virus are:

1. It does not have cellular structure.

2. It does not perform metabolic activities.

12.

Ans:

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/15_12.jpg

13.

Ans:

Capsid is the protein coat in the head of bacteriophage virus which protects DNA inside it.

The body of bacteriophage virus can be divided into two parts that is head and tail. The ear is hexagonal in shape which consists of a protein called capsid. The capsid surrounds the genetic material that is DNA. The cylindrical tail consists of end plate and tail fibers. The tail fibers are adapted to stick it the surface of host body.

14.

Ans:

The importance of Virus is:

1. It consists of genetic materials which are transmitted to their offspring’s.

2. Some viruses contain enzymes or vitamins eg. Small pox virus contains enzymes like riboflavin and biotin.

Stimulation And Reaction

Ans:

The environmental change that brings about a response in the organism is called stimulus. For eg: Removal of hand from a hot thing.

The response of the organism to the stimulus is called reaction. For Eg: closing the leaflets to touch me not plant.

Ans:

Animals use both nervous system and endocrine system. So they can respond quickly to a stimulus. But nervous system is absent in plants and they use only endocrine system for coordination. Therefore plants cannot respond quickly to a stimulus.

Ans:

The movement of an entire cell or an organism in response to a stimulus is called taxis.

The movement of parts of plants (roots, steam or leaves) in response to an external stimulus is called tropism. It is a directional movement of the plant caused by their growth.

Ans: Positive Photo tropism

Ans: Negative Photo tropism

Ans: Negative chemo tropism

Ans: The plant will grow towards the glass window.

Ans: To let the plant grow straight the roof should be removed.

Ans:

The root grows towards the soil because of Negative Photo tropism.

Ans:

The system that receives the stimulus transmits it to the other parts of the body which show the corresponding effects are known as nervous system.

The spinal chord is a long, soft and white jelly like substance present in the neural canal of the vertebral column.

Ans:

The brain is located in sub arachnid cavity. The three meanings of brain are: Cerebrum, Cerebellum and Brain stem.

Ans:

Medulla oblongata is the lowest part of the brain and arises from there.

Ans: The functions of cerebrum are:

1. It is the centre of intelligence, memory, imagination and emotions.

2. It controls the functions of other part of the brain.

Ans: The functions of cerebellum are:

1. It maintains the equilibrium and controls the posture of the body.

2. It makes body movement smooth, steady and coordinate.

Ans: The functions of Medulla Oblongata are:

1. It receives and integrates signal from the spinal cord and sends resulting impulses to the cerebrum and cerebellum.

2. It contain different centers that regulate heart beat rate, blood pressure, breathing, vomiting and some involuntary action.

10.

Ans:

The cells that transmit the message from one part of the body to the other are known as nerve cells or neurons.

The structures of neuron are:

1. Afferent or sensory nerves: They carry impulses from various parts of the body to the brain or to the spinal cord.

2. Efferent or motor nerves: They carry impulses from the brain or the spinal cord to various parts of the body.

3. Mixed nerves: Sometimes axons of both motor and sensor neurons from a nerve which is called mixed nerves. All the spinal nerves are of mixed type.

11.

Ans:

Spinal cord is protected cylindrical structure that arises from the medulla oblongata and passes through the neural canal of the vertebral column.

A spontaneous, mechanical and automatic response to a stimulus controlled by the spinal chord without the involvement of the brain is called reflex action. For eg: Immediate withdrawal of hand after touching a hot object or pin prick.

12.

Ans:

A spontaneous, mechanical and automatic response to a stimulus controlled by the spinal chord without the involvement of the brain is called reflex action.

Reflex arc is the path way followed by the sensory and motor nerves in a reflex action.

13.

Ans:

The function of sympathetic nervous system is to increase the heart beat, the blood pressure and the blood flow to the brain. And the function of parasympathetic nervous system is to reverse the action of sympathetic nervous system and restore the normalcy of the organs.

14.

Ans:

The spinal nerves arises from the spinal cord and spread to different parts of the body. There are thirty one pairs of spinal nerves in the human beings. Out of them 8 pairs are cervical, 12 pairs are thoracic, 5 pairs are lumbar, 5 pairs are sacral and one pair is coccygeal. They are mainly responsible for reflex actions of the body.

The cranial nerves arise from the brain and terminate inside I, except the vagus. The vagus is connected to the alimentary canal. There are 12 pairs of cranial nerves in human being. They control the activities of eyes, ears, tounge etc. Three pairs are cranial nerves are sensory, 5 pairs are motor and four pairs are of mixed types.

15.

Ans:

Sensory Nerves

Motor Nerves

It conducts impulses from the receptor to the spinal cord

It conducts impulses from interneuron to effectors organs

17.

Ans:

Endocrine glands are: Pituitary gland, thyroid gland, parathyroid gland, adrenal gland etc

Exocrine glands are: Salivary gland, tears glands, liver etc

Heterocrine glands are: Pancreas and Gonads.

18.

Ans:

Pituitary gland controls and stimulates many activities of body and the secretion of other endocrine glands. So it is also called master gland.

19.

Ans:

Adrenal gland is situated on the top of each kidney. Adrenalin hormone secreted by the adrenal gland at the time of emergency prepares the body t face and emergency situation for flight, fright or fight. So adrenalin is called an emergency hormone.

20.

Ans:

Exocrine Glands

Endocrine Glands

They are ducted glands.

They are ductless glands.

The secret juices like mucus, saliva, tear etc

They secrete hormones.

Their secretions are related life processes like respiration, digestion etc

Their secretions are responsible for control and coordination of growth.

21.

Ans:

Hormones are called the chemical messenger of the body because it carries chemicals to all parts of the body through blood circulation to bring about the harmonious working of the body.

22.

Ans:

Pituitary gland is located in the skull. Its secretions are: Growth hormone and stimulation hormone. The effects of hypo secretion of this hormone are Dwarfism, high B.P. and more urination.

23.

Ans:

Thyroid gland is located in throat. The diseases caused by the deficiency of iron in diet are retardation and drying of skin.

24.

Ans:

Pancreas has an endocrine part to produce hormones and an exocrine part to produce enzymes. So pancreas is called heterocrine gland. The function of insulin is for the maintenance of sugar level in blood and the function of glucagon is to supply sugar in blood.

25.

Ans:

A = Pituitary gland. Secretion = Growth hormone and stimulation hormone. Function = Controls the growth of the body.

B = Thyroid gland. Secretion = Thyroxine. Function = Growth of the body and stimulates cellular metabolism.

C = Adrenal Gland. Secretion = Adrenalin, Aldosterone and Hydrosterone. Function = Maintains blood pressure.

D = Pancreas. Secretion = Insulin, Glucagon. Function = Maintenance of sugar level in the blood.

E = Ovary. Secretion = Oestrogen and Progesterone. Function = Growth of mammary gland.

F = Testes. Secretion = Testrsterone. Function = Production of sperms

Cell Division

Ans:

The biological process in which a parent cell divides into two or four daughter cells is called cell division.

Mitosis division occurs in somatic or vegetative cells, so it is also called somatic cell division.

Ans:

A = Prophase

B = Telophase II

C = Anaphase

Ans:

Animal Cell

Plant Cell

Centriole is present and divides into two parts

Centriole is absent

Aster formation takes place from centrioles

Aster formation does not take place

The formed spindle fibres are of astral type.

The formed spindle fibres are anastral type.

Ans:

A = Daughter Chromosome

B = Aster rays

1. The centromere of each chromosome divides into two parts so that two sister chromosomes or daughter chromosomes are formed.

2. Sister chromosome migrates towards the opposite pole by the contraction of spindle fibres.

Ans: It is defined as the process of dividing nucleus during cell division. It completes in four stages: they are Prophase, Metaphase, Anaphase and Telophase.

Ans: It is defined as the process of dividing cytoplasm during cell division. In animal cells it occurs by construction method and in plant cells it occurs by cell plate method.

Ans: The pairing of homologous chromosomes is called synapsis. It takes place in the zygotene sub stage of prophase I of meiosis cell division.

Ans: The X shaped point after crossing over is called chiasmata.

Ans:

A = Pole

B = Centromere

C = Spindle fiber

D = Chromatid

1. The nuclear membrane and the nucleous disappear and the formation of spindle fibres is completed.

2. Chromosomes become short, thick and very clear.

Ans:

Prophase of mitotic cell division

Prophase of meiotic cell division

Pairing of homologous chromosome does not take place.

Pairing of homologous chromosomes takes place.

Crossing over does not take place.

Crossing over takes place.

Ans:

Telophase of mitosis

Telophase of meiosis

The number of chromosomes in daughter cells is equal to the number of chromosomes in the mother cell.

The number of chromosomes in daughter cell is reduced to half to that of motor cell.

Each homologous pair of chromosome is transferred in a daughter cell.

Each chromosome of a homologous pair is transferred in a daughter cell.

Ans:

Chromosome

Chromatid

It is thread like microscopic structures present in the nucleus of a cell which contains genes.

It is the part of chromosomes which are obtained when the cell developes.

Ans:

Meiosis and mitosis are similar in Telophase and are different in Prophase, Metaphase and anaphase.

The pairing of homologous series, crossing over, the placement of chromosomes and division of centromeres are different in above three cells.

Ans:

A = Diploten

B = Pachytene

C = Zygotene

10.

Ans:

The process of exchanging genetic materials or chromosomal segments between the two non sister chromatids of a homologous pair is called crossing over.

Its importance is that it helps in the exchange of genes or small segment of chromosome.

11.

Ans:

12.

Ans:

Chromosomes are made up of proteins and DNA. Each chromosome consists of two strands or threads called chromatids.

In Pairs:

Sugarcane: 40

Frog: 13

Rice: 12

Human: 23

Dog: 39

Onion: 8

Rat: 20

13.

Ans:

The significance of Mitosis cell division are:

1. It helps in growth and development of the body.

2. There is no crossing over.

3. The whole process completes in one sequence or phase.

The significance of Meiosis cell division are:

1. It helps to form gametes and pores.

2. There is crossing over.

3. The whole process completes in successive divisions, which occur one after another.

Reproduction In Plants Through Spores

Ans:

The plant body of mushroom consists of two parts. They are mycelium and fruiting body. Mushroom is non vascular, multi cellular and saprophytic fungus.

Ans:

Mushroom grows during the rainy season on the soil rich in dead and decaying organic matter, damp places and truck of trees.

Ans:

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/19_3.jpg

Ans:

The advantages of mushroom are:

a. Mushroom is taken as a nutritious food because it contains proteins, vitamins and minerals in a remarkable amount.

b. Mushroom helps to prevent high blood pressure, blood cholesterol, heart disease etc

c. Mushrooms are cultivated from business point of view.

Ans:

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/19_5.jpg

Ans:

Mushroom is a non vascular, multi cellular, saprophytic fungus. Generally it grows in dark, moist and shady places on rotten logs of woods, tree trunks, decaying organic matter and in rich soil. It gets food from decomposing dead and decaying organic matters. So it is called saprophytic fungus.

Ans: The vegetative body of mushroom which is thin thread like structure is called mycelium. The mycelium absorbs soluble food materials from the substratum.

Ans:

Hyphae are the fine thread like structure in the substratum.

Ans:

Gills are the under surface of the pileus bears plate like structures or thin vertical stripes in the radiating rows.

Ans:

The spores produced by gills of mushroom are called basidiospore.

A hypha that has two nuclei is called dikaryotic cell.

Ans:

The three edible mushrooms are:

a. Agaricius

b. Pleutrotus

c. Volvariella

Ans:

The four poisonous mushrooms are:

a. Amanita verna

b. Amanita phalloides

c. Agaricus silvicola

d. Agaricus xanthodermis

10.

Ans:

Fern plants grow in moist, shady and cold places. The steam of fern is called rhizome.

11.

Ans:

Fern plants are kept under tracheophyta because it is differentiated into roots, steam and leaves. These plants show the alternation of generations with the sporophyte which is more prominent.

12.

Ans:

The systemic position of fern plant are: Adult sporophyte, Fertile pinna, V.S. of fertile pinnule, Young sporangium, Spore germination, Prothallus, Aechegonium, Fertilization, Young Embryo, Young Sporophyte.

13.

Ans:

A cluster of sporangia on the edge or underside of fern frond is called sorus.

The type of root in fern plant is called Rhizome.

14.

Ans:

Sporophyte is the most developed stage in the life cycle of fern plant. The plant body consists of rhizoids, rhizome and leaves. The sporophyte consist well developed vascular tissue and reproduces through spores.

15.

Ans:

Xylem and phloem are the conducting tissue in vascular plants.

16.

Ans:

Rhzoids are the fine hair like structures that originate irregularly from the base of rhizome. They are highly branched, small and usually brown.

Ans:

It is the flask shaped female reproductive organ of prothallus.It produces ovum and is located at the centre of prothallus below the notch.

Ans:

The young gametophyte of a liverwort or peat moss is called prothallus.

Ans:

Oosphere is the fusion of one of the antherozoids and an egg.

17.

Ans:

The life cycle of fern plant is completed by the alternation of two distinct generations which are as follows:

a. Sporophytic generation: The fern plant is called sporophyte because it bears spores and reproduces asexually by means of spores. Under the favorable condition the spores in the soil undergo germination.

b. Gametophytic generation: In the early stage of germination the spores grow in the form of elongated tube like structure which continues to grow and finally forms a heart shaped structure called prothallus. The prothallus decays and the young sporophyte grow into a fern plant.

18.

Ans:

The life cycle of fern plant is completed by the alternation of two distinct generations which are as follows:

19.

Ans:

A fern plant produces spores so it is called sorophyte.

20.

Ans:

The prothallus constitutes the gametophyte of the fern, Which is a haploid stage. It is independent and its mode of nutrition is autotropic. Prothallus has several thick cells in the middle region behind the apical notch. It is a heart shaped, thin, green, flat and multicellular structure.

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/19_20.jpg

21.

Ans:

Gametophyte is the stage of fern plant which produces both male and female gametes. It reproduces by sexual method. It is haploid stage.

Antheridium is a dome shaped male reproductive organ of prothallus. It produces antherozoids. It is located on the lower side of prothallus along with rhizoids.

22.

Ans:

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/19_22.png

23.

Ans:

Malic acid is found in the neck part of the fern plant.

24.

Ans:

Gametophyte

Sporophyte

Gametophyte is the stage of fern plant which produces both male and female gametes.

Sporophyte is the prime stage of fern plant which produces spores.

It reproduces by sexual method.

It reproduces by asexual method.

It is haploid stage.

It is diploid stage.

25.

Ans:

The nutrition found in mushrooms are protein, vitamins etc.

a. Some mushrooms like toadstools which are highly poisonous if mistakenly taken can cause peoples death.

b. Boeltus causes pain, vomiting and diarrhea in humans.

Asexual And Sexual Reproduction

Ans:

A biological process in which living organisms produce their own kind by asexual and sexual method is called reproduction. Living organism produce their own kinds to maintain the life of their species on the earth.

Ans:

The types of reproduction are:

a. Asexual reproduction

b. Sexual reproduction

The types of asexual reproduction are:

a. Fission

b. Budding

c. Fragmentation

d. Sporulation

e. Vegetative propagation

Ans: The process of asexual reproduction in which a parent organism divides into two or more daughter organism in called fission. For eg: amoeba and paramecium.

Ans: The method of asexual reproduction which takes place by the formation of a bud is called budding. For eg: Hydra and yeast.

Ans: Fragmentation is the process in which the elongated body of an organism breaks into two or more fragments due to various agencies like heat, wind etc. For eg: Tapeworm and starfish.

Ans: The method of asexual reproduction which takes place by the formation of spores is called sporulation. For eg: Mucor and Moss.

Ans:

Vegetative propagation is the method of asexual reproduction in which new plants are produces by cutting, sowing or grafting of the vegetative parts of plant like root, stem or leaves. Any four points to show the significance of vegetative propagation are:

a. The plants which do not produce viable seeds can be easily propagated by this method. For eg: Banana, Rose etc

b. The plants produces by vegetative propagation beats flowers and fruit earlier than those produced from seeds. For eg: Orange, lemon and mango etc

c. It is a cheap and rapid method relative to a long period of seed dormancy.

d. It gives rise to genetically uniform population, the clone.

Ans:

The three methods of vegetative propagation are:

a. Roots: Some plants like sweet [otato, Dahlia, guava, mint etc reproduce asexually by means of roots. Roots of these plants contain adventitious buds. When such roots are planted in the soil, new plants grow.

b. Stem: The stems of some plant like potatoes, cannas, turmeric, ginger, onion, garlic etc grow in different forms below the soil. The stems of such plants are modified for food. They are called underground stems. They are like roots in appearance but are different from the roots by nodes, inter nodes and buds. When the stems with nodes are planted in the soil, they develop into new plants.

c. Leaves: The leaves of Bryophyllum are thick and fleshy. They bear buds in the marginal notches. When leaves fall and come in contact with moist soil, new plantlets develop on it. They grow into new small plants. Begonia also reproduces by this method.

Ans: The threes methods of artificial vegetative propagation are:

a. By cutting: In this method plants are propagated by cutting small pieces of the steam. When the pieces are placed in soil, roots emerge from the nodes.

b. Layering: In this method one of the lower branches of the plant is bent and covered partially with soil. The part inside the soil develops roots after three to four months.

c. Tissue Culture: It is a modern technique of vegetative propagation. It issue culture a small piece of tissue from a plant is kept in a container with essential nutrients under proper conditions.

Ans:

1. (a) shows vegetative propagation.

2. The name of the animal is Amoeba. It shows fission.

3. Budding is shown in (c). Budding is a common asexual reproduction in which an organism forms a small outgrowth called bud at its one side.

4. (a) is the simples method of reproduction. The process of asexual reproduction in which a parent organism divides into two or more daughter organism in called fission. For eg: amoeba and paramecium.

5. The name of the plant is Mucor.

A = Sporangium

B = Spores being shed

C = Stalk

D = Rhizoids

Ans: Reproduction is the outstanding characteristic of living organisms by which every kind of living organism multiplies to form new individuals of its own kind.

Ans: Vegetative propagation is advantageous to farmers because the plants bear flowers and fruit earlier than those produced from seeds. It is cheap and rapidly grows to a long period of seed dormancy.

Ans:

Gametes are haploid and zygotes are diploid because gametes are formed by meiosis cell division and zygote is formed by the fusion of nuclei of male and female gametes.

Ans:

The fertilization in flowering plants involves the fusion of two male gametes separately ie one male gamete with the egg cell (ovum) and the male gamete with the secondary nucleus. Therefore the fertilization in flowering plants is called double fertilization.

Ans:

Earthworm is called hermaphrodite animal because it contains the combination of two opposite sex in single body and can reproduce itself.

Ans:

Conjugation is the simplest method of sexual reproduction. In this method of reproduction two isogametes are formed in two different cells of organism.

In chlamydomonas two individuals form similar structured gametes inside them.

10.

Ans:

The complex process in which reproduction takes place by the fusion of male gamete and female gamete is called sexual reproduction.

Asexual Reproduction

Sexual Reproduction

The reproduction which takes place without fusion of a male gamete and a female gamete is called Asexual Reproduction

The reproduction which takes place with fusion of a male gamete and a female gamete is called sexual Reproduction

This process is common in primitive plants and animals

This process is common in higher plants and animals

Only single parent is involved in asexual reproduction

Both parents is involved in asexual reproduction

11.

Ans:

The flora part which helps in sexual reproduction in plants is calyx, corolla, androecium and gynoecium.

Calyx protects the inner part of the plant. Corolla attracts insects for pollination. Androecium forms pollen grains from male gametes and gynoecium forms ovaries.

12.

Ans:

Ovum

Zygote

It is a female gamete having haploid chromosomes.

It is a fertilized ovum having diploid chromosomes.

It is formed by oogenesis.

It is formed after fertilization.

Ans:

Spores

Gametes

They are the microscopic asexual reproduction bodies covered by a hard protective coat which grow into new organism under favorable conditions

Gametes are the haploid and are formed by the meiosis cell division.

Eg: Mucor, moss ans fern.

Eg: Male gamete and female gamete

Ans:

Pollination

Fertilization

The process of transference of pollen grains from anther to the stigma of a flower is called pollination.

The process of formation of a zygote by the fusion of male gamete and female gamete is called fertilization.

This process is found only in the flowering plants.

This process is commonly in both flowering plants and animals that reproduce by sexual method.

Ans:

Self Pollination

Cross Pollination

External agents of pollination are not required.

External agents of pollination like insects, wind, water is required.

It does not help in variation.

It helps in variation.

This process is commonly in bisexual flowers.

This process is commonly in both bisexual flowers and unisexual flowers.

Ans:

External Fertilization

Internal Fertilization

The fusion of a male gamete and a female gamete takes place outside the body of female.

The fusion of a male gamete and a female gamete takes place inside the body of female.

Mating of male and female is not necessary.

Mating of male and female is necessary.

This process is common in fish and frogs.

This process is common in reptiles, birds and mammals.

Spores

Zygote

They are the microscopic asexual reproduction bodies covered by a hard protective coat which grow into new organism under favorable conditions

It is a fertilized ovum having diploid chromosomes.

Eg: Mucor, moss ans fern.

It is formed after fertilization.

13.

Ans:

In angiosperms, gametes are formed in specialized organs called flowers. Usually a flower is borne on a stalk called pedicel. It has an upper swollen region called thalamus. A typical flower has four whorls of the flower. They are calyx, corolla, androecium and gynoecium. Androecium consists of varing number of stamens and gynoecium is composed of ovary, style and stigma. They fuse to each other and reproduction takes place.

14.

Ans:

In multicellular animals testes produce sperms and ovaries produce ova. Asperm is male gamete and an ovum is female gamete. The male and female gametes unite to form a diploid cell called zygote.

15.

Ans:

The significance of sexual reproduction are:

a. The rare of a species is continued but the daughter organism genetically differs from the parents.

b. Since there are variations, it contributes to the evolution of the species.

c. Offspring formed by this method are of more resistant type.

16.

Ans:

a = Pollen tube

b = Ovule

c = Polar nuclei

d = Embryo

e = Sperm cell

f = Stigma

17.

Ans:

A = Male gamete

B = Female gamete

C = Mitosis

D = Fertilization

Blood Circulation System In Human Body

Share6

Ans:

The circulation of blood through the entire body is called blood circulation. The various parts of circulatory system are: blood, blood vessels and heart.

Ans:

The importance of blood circulation is:

a. It carries nutrients throughout the entire body.

b. It supplies oxygen.

Ans:

Blood is a fluid connective tissue, which contains of 55% plasma and 45% blood cells. The components of blood are:

a. Plasma:

1. It transports digested food to different organs and tissues of the body.

2. It removes the major portion of carbon dioxide from the tissue.

b. Blood cells:

1. It consist red blood cell and white blood cell.

2. It covers about 45% of the total volume of blood.

Ans:

RBCs

WBCs

They are oval, biconcave and non- nucleated.

They are round or amoeboid and nucleated.

They are small in size.

They are large in size.

They help in respiration.

They help in the defense of the body.

Ans:

Vein

Artery

It carries blood from body organs to the heart.

It carries blood away from the heart to different body organs.

Blood pressure in it is mostly deoxygenated.

The arteries carry oxygenated blood

The wall of the vein is thin and less elastic.

The wall of the artery is thick and more elastic.

Ans:

Pulmonary Artery

Pulmonary Vein

It carries deoxygenated blood to the lungs for purification.

It carries oxygenated blood to the left auricle.

It arises from the right ventricle.

It arises from the lungs.

It is guarded by the semi lunar valves at its base in the heart.

The openings of pulmonary veins into the left auricle are not guarded by valve.

Ans:

Bicuspid valve

Tricuspid valve

The bicuspid valve guards the opening of the left auricle into the left ventricle.

The tricuspid valve guards the right auriculo-ventricular aperture.

This valve consists of two membranous flaps or cusps.

This valve consists of three membranous flaps or cusps.

Ans:

Auricles

Ventricles

They are thin walled upper chambers of heart.

They are thick walled lower chambers of heart.

They receive blood from the different parts of the body.

Ventricles pump blood to the various parts of body.

Ans:

The different functions of blood are:

a. It is red colored, viscous and complex tissue fluid.

b. It is salty in taste.

c. The specific gravity of blood is 1.05-1.06.

Ans:

Plasma is a transparent straw-colored fluid part of blood. Plasma forms about 55 percent of the blood volume.

Its functions are:

a. It transports digested food to the various parts of body.

b. It regulates the amount of water and other chemicals in the body.

c. It regulates the body temperature.

Ans:

b = WBC

c = Platelets

Ans:

Platelets(b) play an important role in the coagulation of blood or clotting of blood.

RBC(d) help in respiration.

Ans:

B is related to disease resistance power.

Ans:

A contains 55% of blood.

Ans:

F(Left ventricle) pumps blood to different part of the body.

Ans:

K(Pulmonary vein) brings blood from the lungs to the heart.

Ans:

H(Tricuspid valve) blocks the back flow of blood from the right ventricle to the right auricle.

Ans:

F(Left ventricle) has the thickest part of the wall.

10.

Ans:

Molecules of glucose, water, oxygen, hormones etc exchange takes place between blood capillaries and tissues.

11.

Ans:

The circulation of blood between the heart and different parts of the body except he lungs is called systemic circulation.

The circulation of blood between the heart and the lungs is called pulmonary circulation.

12.

Ans:

There are three types of blood vessels:

a. Arteries

b. Veins

c. Capillaries

a. Arteries:

The blood vessels which carry the blood away from the heart are called arteries.

The blood vessels which bring the blood into the heart from different parts of the body are called veins.

They connect the arteries to the veins.

13.

Ans:

The number of heart beat in one minute is called heart beat rate. It is faster while walking, during exercise, excitement, fever etc.

14.

Ans: Blood is red in color because there is presence of hemoglobin.

Ans: The left ventricle posses a thicker wall than that of the right ventricle because left ventricle pumps blood to different parts of he body.

Ans: The arteries are thicker than the veins because blood flows faster in arteries than in veins.

Ans:

There are no valves in arteries because the flow of blood is faster and blood does not flow back.

Ans:

Plasma is very important component of blood because it contains 55% of the blood.

Ans:

WBCs increase in blood causes blood cancer.

Ans:

Anemia occurs if there is decrease of RBCs in blood.

15.

Ans:

B = Pulmonary Veins

E = Aorta

Ans:

A receives and pumps blood to heart.

Ans:

Oxygenated blood and Deoxygenated blood flows in ‘E’ and ‘F’.

Ans:

Pulmonary veins occurs in the heart and A.

Heredity

Share6

Ans:

Heredity is the phenomenon by which living organisms transmit parental characteristics to their offspring.

Genetics is the branch of biological science which deals with the nature and behavior of genes. Genetics is also defined as the biological science of heredity and variation.

Ans:

Gregor Johann Mendal is an Austrian monk who is considered as the father of genetics because he introduced the concept of genes as the basic unit of heredity for the first time.

Ans:

Mendel selected Pea plants for his study because they contain a large number of contrasting character in pairs such as tall and dwarf, round and wrinkled, smooth and constricted etc. They have short life cycle and be cultivated easily.

Ans:

No Mendel would not succeed if he had used frog instead of pea plant because frog do no have short life cycle, they do not produce large amount of offspring and they are difficult to handle.

Ans:

The characteristics which are prominent and appear in successive generations are called dominant characteristics. For eg: In the cross pollination between tall tree plant and dwarf pea plant tall is dominant characteristics.

The suppressed characteristics which remain unexpressed in successive generation are called recessive characteristics. For eg: In the cross pollination between tall pea plant and dwarf pea plant dwarf is recessive characteristics.

Ans:

The external appearance of an organism for a contrasting character is called phenotype. It is expressed in words for eg: Tall, dwarf etc

Ans:

The genetic make up or genetic constitution of an organism is called genotype. It is expressed in letters eg: TT, Tt etc.

Ans:

Hybrids are the organisms produced after cross fertilization between two genetically different organisms. Hybrids are of various types for eg: Mono hybrids, dihybrids etc

Ans:

Alleles are one form of a gene. There may be two or more alleles of a gene. In genetics different alleles are donated by letters eg. TT, Tt etc.

Ans:

The cross involving only one pair of contrasting characteristics is called monohybrid cross.

Ans:

The cross involving two pairs of contrasting characteristics is called di-hybrid cross.

Ans:

The cross involving more than two pairs of contrasting characteristics is called poly-hybrid cross.

Ans:

Definite external character of an individual of a cross is called its unit character. Each organism consists of its own type of characters. For eg: a rose plant has its own type of leaf, flower, stem etc which act as unit character for it.

Ans:

Mutation is defined as sudden chromosomal change by very new and different characters from parents is seen in offspring’s. For eg six fingered limb, cut at lip by birth etc.

Ans:

Genes are tiny units of heredity located in chromosomes. They are arranged chromosomes in a linear fashion. Each gene occupies a fixed position in fixed chromosomes. Genes determine the physical, anatomical and physiological characters of organisms. These characters are transmitted from generation to generation.

Ans:

100% and 75% of dominant character of a pea plant was found in F1 and F2 generations respectively.

Ans:

The character of Blue Eye is dominant because the black eyed parents are not pure i.e. they are hybrid.

10.

Ans:

According to Mendels experiments, ‘Law of dominance’ states that in crossing between pure organisms for contrasting characteristics of a pair, only one characteristics of the pair appears in the first filial generation.

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/21_10.png

11.

Ans:

It states that” The two members of a pair of factors separate during the formation of gametes. They do not blend with each other but segregate out into different gametes. Thus, any gamete is not impure.

12.

Ans: The phenol type of the first generation is RY.

Ans: The ratio of phenotype in the F2 generation is 9:3:3:1.

13.

Ans:

14.

Ans:

According to Mendel’s law of dominance in crossing between pure organisms for contrasting characteristics of the pair only one characteristic of the pair appears in the first filial generation.

http://www.khullakitab.com/uploads/kcfinder-2.51/upload/images/Class%2010/Science/Solutions/Graded%20Modern%20Science/21_14.png

15.

Ans:

The phenotypic ratio of Red: White flowered pea plant is 3:1

The genotypic ratio of Red: Hybrid: White flowered pea plant is 1:2:1

16.

Ans:

Nucleic acids are the acid involving in replicating, preserving and expression of every heredity characters in a living cell.

DNA

RNA

It is mainly found in chromosomes.

It is found in nucleus and cytoplasm.

It is made up of deoxyribose sugar.

It is made up of ribose sugar.

Thymine is present and uracil is absent.

Uracil is present and thymine is absent.

It is double stranded helically coiled macro molecule.

It is a single stranded macro molecule.

Ecosystem

Ans:

Ecosystem is defined as the structural, functional and self sustaining unit of biosphere, which consists of biotic and abiotic components.

If the number of tertiary consumers increased more than necessary then they will consumers primary and secondary consumers more than usual so it causes imbalance in the ecosystem.

Ans:

The two types of components of ecosystem are:

a. Aquatic Ecosystem

b. Terrestrial Ecosystem

Ans:

Pyramid of biomass is the graphic representation of biomass of organisms present in unit area of various tropic levels of a food chain. In this pyramid, the biomass of producers is placed at the base and that of producers is kept at the apex. The pyramid of biomass of an aquatic ecosystem is inverted because of the progressive increase in the biomass of organisms from the second tropic level to the final tropic level.

Ans:

Green plants can synthesize their own food by photosynthesis. So they are called producers.

Ans:

Farmers are suggested to plant leguminous plants at least in the duration of 3-4 years because root nodules of leguminous plants contain symbiotic bacteria like Rhizobium. Those bacteria convert atmospheric nitrogen into nitrates and hence increase the fertility of soil.

Ans:

Rain with thunder in falgun-chaitra is good for cultivation because the nitrogen present in the atmosphere reacts with oxygen during lightening in the sky and produces oxides of nitrogen. Those oxides of nitrogen get dissolved in rain water forming dilute nitric acid. This nitric acid reacts with alkalis of the soil and forms nitrates. This process helps to increase the fertility of the soil by nitrogen fixation.

Ans:

Producers are the green plants which can prepare food themselves. Green plants can prepare their own food due to the presence of chlorophyll. For eg: Sunflower, mango plant.

Ans:

Consumers are the heterotropes that depend on green plants directly or indirectly for their food. For eg: Dog, Lion etc.

Ans:

Decomposers are the saprophytic organisms that feed on dead and decaying organic matters and decompose them into simpler forms. For eg: Bacteria, Fungi etc

Ans:

Herbivores are the animals that feed on green plants. They cannot survive without green plants and provide food for carnivores. For eg: Deer, Rabbit etc

Ans:

A floating sea organism which obtain energy by photosynthesis.

Ans:

Deforestation is the process of clearance of forest. Deforestation affects the bond between the soil and its plants and if there is raining then it sweeps all the small plants. So its affects land ecosystem.

Ans:

The consumers of pond ecosystem are: Sharks, Dolphin, and Whales etc.

10.

Ans:

Nitrogen cycle is defined as the cyclic process in which nitrogen is circulated continuously through the living and non living components of a biosphere. The nitrogen present in the atmosphere is turned nitrates with the help of leguminous plant so it is helpful in nitrogen cycle.

11.

Ans:

The bacteria which convert atmospheric nitrogen into nitrates in soil are nitrogen fixing bacteria. The bacteria which converts ammonia into nitrites and nitrites into nitrates are called nitrifying bacteria. For eg: Nitrosomonas, Nitrobacter etc

12.

Ans:

Ammonification is the conversion of complex organic compounds into ammonia gas. Various bacteria and fungi are involved in this process.

13.

Ans:

Carbon cycle is defined as the cyclic process in which carbon element is circulated continuously through the living and non living components of biosphere.

14.

Ans:

15.

Ans:

Decomposers are the saprophytic organisms that feed on dead and decaying organic matters and decompose them into simpler forms. For eg: Bacteria, Fungi etc. Micrococcus is the example of denitrifying bacteria.

16.

Ans:

Grassland ecosystem is an example of terrestrial ecosystem. The grassland is an open land with grasses. Abiotic and biotic components of grassland are the two type of ecosystem.

17.

Ans:

If all the snakes in grassland ecosystem are killed then the number of primary consumers will decline slowly as snakes feed on primary consumers. It brings chaos in the ecosystem.

18.

Ans:

Trees and plants need water to survive and they store water. They return the used water by the respiration and transpiration. Remaining water is released after the death of plants through the process of decay.

19.

Ans: The producer is trees.

Ans: Rabbit and Deer are secondary consumer and Lion is tertiary consumer.

Ans: Trees should be in the large number to maintain the ecosystem.

20.

Ans:

At the base of ecological pyramid Producers are represented and Top consumers are present in the top.

21.

Ans:

The exchange of biogeochemical between living and non living components of the biosphere is called biogeochemical cycle. The four important biogeochemical cycles in the biosphere are:

a. Water Cycle

b. Oxygen Cycle

c. Carbon Cycle

d. Nitrogen Cycle

22.

Ans:

The bacteria like Azotobacter and Rhizobium convert atmospheric nitrogen into nitrates and these nitrates are consumed by plants. This process is called nitrogen fixation.

23.

Ans:

The microorganisms that play key role in nitrogen fixation are: Rhizobium and Azotobacter.

24.

Ans:

Nitrogen Fixation

Nitrification

It is the conversion of atmospheric nitrogen into nitrates in soil.

It is the conversion of ammonia into nitrites and nitrites into nitrates.

Bacteria like Azotobater, Rhizobium etc

Bacteria like Nitrosomanas, Nitrobacter etc.

History Of Earth

Ans:

The name of hypothesis given about the origin of Earth is:

a. Old planetesimal hypothesis

b. Nebular hypothesis

c. New planetesimal hypothesis

d. Tidal hypothesis

e. Dust cloud hypothesis

Ans:

A German philosopher named Immanuel Kant proposed nebular hypothesis about the origin of the solar system in 1796 AD, and it was improved by a French astronomer named Laplace later in 1796 AD. According to this hypothesis the sun and the planets were formed from a large whirling cloud of gases and dust. When the cloud cooled and grew smaller, it began to spin faster. As the surface of the cloud cooled by radiation, a ring of matter was formed at the equatorial region of it. The ring was escaped out from the surface of the main mass and the first planet was formed. This process was repeated again and again till the entire solar system was formed and the remaining mass was the sun.

Ans:

English astronomers Sir James Jeans and Sir Harold Jeffrey proposed the tidal hypothesis in 1917 AD. They also supposed that a passing comet had exerted a tidal pull upon the gaseous mass but the effect was to cause a long filament of gas to be drawn from the mass. The outer part of the filament escaped in to space the inner part came back into the gaseous mass and the middle part formed a series of round structures of different sizes which were the planets and the remaining mass was the sun.

Ans:

This hypothesis proposed that the sun and the planets were formed from a large cloud of gases and dust. The light of the star pushed the atoms of the gases and dust to form large particles. These large particles were attracted to each other by the pull of gravity and they began to crowd together. Eventually a huge ball of material formed the sun and the solar energy produced in it by the nuclear fusion.

Ans:

A time scale that covers the earth history from its origin to the present is called geological time scale. The age of the earth has been estimated about 4.6 billions years.

Ans:

Archean Eon extended from 200 to 3800 millions years ago. In this eon life was originated on the earth. Proterozoic Eon was extended from 570 million years ago to about 2500 million years ago. In this eon invertebrates and cryptogams were evolved. Phanerozoic Eon began about 570 million years ago and it is still continuing a present. In this eon vertebrates and phanerogams are developed.

Ans:

Proterozoic Eon was extended from 570 million years ago to about 2500 million years ago. In this eon invertebrates and cryptogams were evolved.

Ans:

Mammals and Birds were originated and gymnosperms were dominant plants. Dinosaurs were at their peak.

In Mesozoic era primates were originated.

10.

Ans: Palaeozoic Era began about 570 million years ago and ended about 250 million years ago. This era is further divided into seven periods.

Ans: Cenozoic Era began about 65 millions years ago and is still continuing. It is divided into two periods. The wide variety of plants and animals that we know today came into existence during Cenozoic Era.

Ans: Mesozoic era began about 250 million years ago and ended about 65 million years ago. This era is further divided into three periods. Dinosaurs roamed the earth in this era. This era was the time of reptiles. Bats and birds were originated in this era.

Ans:

A fossil is the mark or hardening remains of a plant or animal that lived thousand or millions of years ago. Some fossils are leaves, shells or skeletons that were preserved after a plant or animal died.

11.

Ans:

Palaeozoic Era

Mesozoic Era

The duration of this era was 54 crore years ago to 25 crore years ago.

The duration of this era was 25 crore years ago to 6 crore 55 lakh years ago.

Dinosaurs were not present in this era.

There was dominance of dinosaurs in this era.

This era is divided into six periods.

This era is divided into three periods.

12.

Ans:

Fossils are mainly formed due to following reasons:

a. If the animal has hard body pats, the chance of fossilization increases.

b. The organism must be covered by the protective materials shortly after the death.

13.

Ans:

Fossils can be identified by following ways:

a. The impression of a whole or a part of any organ of dead body remains in hard ground or stone.

b. The whole skeleton of animals and plants are pressed into rocks.

c. Fossils are identified by morphology.

Organisms are fossilized in the form of cast in sediments. The water inside the sediment dissolves the hard parts, leaving a hollow space in the sediments. The hollow space is called mold, the space is then filled with minerals which harden to form a cast.

14.

Ans:

An epoch is defined as the length of time which is a division of a period in which important changes or events happened. Examples: Eocene epoch, Miocene epoch.

A geological period is defined as a length of time which is a division of era. Example: Jurassic period, Cretaceous period.

15.

Ans:

In Cenozoic era rhinoceros, camels, elephant and man were evolved.

16.

Ans:

The name of petroleum products are:

a. Liquefied petroleum gas

b. Petrol ether

c. Diesel

d. Kerosene

e. Grease

f. Vaseline

g. Paraffin wax

h. Petrol

17.

Fossils are found in sedimentary rocks.

18.

Ans:

The fuels which are formed in nature by the decomposition and preservation of organic matters under high temperature and pressure for a long time are called fossils.

Fossils are very limited in the earth and are non renewable source of energy. The cause air pollution when they are burnt.

19.

Ans:

The list of types of coal is:

a. Liquefied

b. Sub bituminous

c. Bituminous

d. Anthracite/Asphalt

20.

Ans:

In the ancient time plants and animals were buried under the earth crust due to different geological phenomenon like earthquake, volcano etc. Then the remnants were pressed under high pressure and temperature for a long time. When a whole part of a plant remains in swamp land in the form of a fossil for a long time it is converted into coal.

21.

Ans:

The importance of coal are:

a. It is used as fuel in railway transportation, industries, brick factories, iron etc.

b. It is used to generate electricity, for manufacturing petrol like natural gas etc.

c. It is also used for making various organic compounds like benzene, phenol etc.

22.

Ans:

Mineral oil is the mixture of hydrocarbon and other substances like oxygen, nitrogen, sulphur etc.

The mineral oil is formed after the compression of fossils under sedimentary rocks for millions of year. For eg. Petroleum products.

23.

Ans:

Earth was originated 4.5 billion years ago where as human were evolved 2 million years ago. This was verified by the help of fossils of dinosaurs which were found on the earth crust. The dinosaurs evolved before human beings.

24.

Ans:

The periods of Mesozoic era are:

a. Cretaceous

b. Jurassic

c. Triassic

The chief animals were: Dinosaurs, Birds

The periods of Cenozoic era are:

a. Quaternary

b. Tertiary

The chief animals were: Man, Elephants, Camels and Rhinoceros.

25.

Ans:

According to the palaeontological evidence there was no suitable condition for living organisms to live in the land. Some singled cell organisms were originated in the oceans.

Ans:

Coal and minerals are not formed on the earth surface because they need high temperature and pressure to form which is not present in the crust of earth.

Ans:

The study of fossils of dinosaur suggests that it was dated back in era of Mesozoic era. So we can conclude that dinosaurs were extinct in that era.

Ans:

Life was not possible in Azoic era because it was the first era on earth and the temperature and climate was not favorable of living creatures.

Atmosphere

Share2

Ans:

The names of different layers of atmosphere are:

a. Troposphere

b. Stratosphere

c. Mesosphere

d. Thermosphere

e. Exosphere

Troposphere is the lowermost layer of atmosphere. It is also the thinnest layer and has the thickness of 16 km. At the equator warm air is found which makes this layer expanded.

Ans:

Atmospheric layer are divided on the basis of their thickness. Ionosphere is very important in communications because radio waves can travel to different parts of the world by bouncing off it.

Ans:

Ozone layer is found in stratosphere. It is extended from 16 km to 50 km above the earth surface. Its thickness is 34 Km. This layer is clear and cloudless and jet planes fly in this layer.

Ans:

The upper layer of the stratosphere is called stratopause. The upper boundary of the mesosphere is called menopause.

Ans:

The upper most layer of stratosphere contains ozone layer. This thick layer of air absorbs most of the solar radiation. Due to absorption of solar radiations, the temperature of the stratosphere increases as the increase in altitude.

Ans: The layer 3 is Mesosphere and layer 5 is exosphere.

Ans: Layer 2 has maximum ozone layer.

Ans: Thermosphere has ionized particles of ions.

Ans: Mesosphere is the coldest layer of the atmosphere.

Ans: Troposphere has maximum pressure.

Ans: Thermosphere as temperature of -1100C and 12000C.

Ans: Jet planes fly in Stratosphere.

Ans:

The layer of the atmosphere above the stratosphere is called mesosphere. It is extended from 50 to 80 Km above the earth. It consists of N, O and O2. Its thickness is 30 Km. The layer of the atmosphere above the thermosphere is called exosphere. It is extended beyond 720 Km above the earth’s surface. In this layer of the atmosphere, gravity is so weak and there are so few atoms that some escape into the space. The layer of the atmosphere above the mesosphere is called thermosphere. It is also called ionosphere. It is extended from 80 to 720 Km above the earth. Its thickness is about 640 Km. This layer contains charged ions.

Ans:

The layer of pale-bluish gas which is found in the uppermost layer of stratosphere is called ozonosphere. This thick layer of air absorbs most of the solar radiations and shields the earth from the harmful ultraviolet radiation.

Ans:

The full form of CFCs is Chlorofluorocarbon. Chlorofluorocarbon depletes ozone layer because they dissociate nascent chlorine when they are influenced by ultraviolet rays in the stratosphere.

10.

Ans:

The thinning of the layer of the ozone is called ozone layer depletion. It occurs when CFC dissociate nascent chlorine when they are influenced by ultraviolet rays in the stratosphere.

11.

Ans:

The list of compounds which depletes ozone layer are CFC, methyl bromide, carbon tetrachloride and methyl chloroform.

13.

Ans: Ozone layer protects earth from harmful radiation and if there is ozone layer depletion then rays of sun enter earth without any obstacle which causes the rise in temperature of earth.

14.

Ans: Te cause of the above factor is ozone layer depletion.

Ans: The use of CFCs gas should be reduced.

Ans: Eye problem is another cause.

15.

Ans:

The three points to conserve ozone layer are:

a. The production and use of CFC should be banned.

b. The alternatives for the CFCs should be developed.

c. Release of oxides of nitrogen should be avoided.

Acid rain is the rain containing small amounts of acids in it formed from gases like sulphur dioxide and nitrogen oxides present in atmosphere.

16.

Ans:

The artificial green house in gardens using glass is called artificial green house. It is warmer inside a green house at night it traps the rays of sun in day time.

17.

Ans:

Earth act as a green house because there are many layers above the earth surface which traps the radiation coming out from the sun and makes the earth warm.

18.

Ans:

Green house effect helps to grow off seasonal fruits and crops where as it has disadvantage too. There is an increase in the rate of evaporation from the sea as a result of the increase in the temperature of the atmosphere. The polar icecaps and glaciers start to melt. As a result the sea level increases.

19.

Ans:

The disadvantages of green house effect are:

a. The increase in gas that affects the green house in the atmosphere brings about a change in weather slowly.

b. As a result of the green house effect the temperature increases and it affects the water cycle.

c. . There is an increase in the rate of evaporation from the sea as a result of the increase in the temperature of the atmosphere. The polar icecaps and glaciers start to melt. As a result the sea level increases.

20.

Ans:

Industrial gases are the gases produced by the industries. For eg: Sulphur dioxide, carbon monoxide etc.

The adverse effect o these gases are:

a. Dust, smoke, lead etc released from factories cause air pollution.

b. It affects mental functional as well.

c. It causes global warming which affects plants and animals.

21.

Ans:

22.

Ans: It is because the troposphere is the lowermost layer of the atmosphere in which we live. We exchange biogeochemical like nitrogen, oxygen, carbon and water from the nature.

Ans: The thermosphere is very important in communications because radio waves can travel to different parts of the world by bouncing off it.

Ans:

Ozone is useful because it does not let the harmful radiation of sun to come to earth as it acts a protective shield around the earth. It is also harmful because it causes throat burning and when it exceeds that amount it can destroy the lungs very badly.

Ans:

The use of CFCs must be reduced because CFC dissociate nascent chlorine when they are influenced by ultraviolet rays in the stratosphere which causes ozone laye depletion.

Ans:

Green house is the process of tapping the radiation of the sun and making the atmosphere warm. It causes global warming as it increases the temperature of the earth.

Ans:

It is warmer in the green house than the surroundings because green house traps the rays of sun and increases the temperature.

Ans:

There are many layer of atmosphere above the earth surface which traps the rays of sun and make the surrounding warm. So it acts as a green house effect.

Ans:

Summer vegetables can be grown inside the green house because inside green house the temperature is warm as it traps the radiation coming out from the sun.

Ans:

We should conserve ozone layer because it protects us from the harmful radiation coming out from the sun. It acts as a green house effect which makes earth warm.

Ans:

HFCs should be used instead of CFCs and HCFCs because HFCs are not harmful as other above. They do not deplete the ozone layer.

Universe

Share3

Ans:

The huge space which contains everything that exists is called the universe. Astronomical unit is used to measure the distance between the heavenly masses.

Ans:

The sun is a medium sized star. It is light yellow in color. It is the largest body in the solar system. Its diameter is 1392000 Km and it is situated at about 2.5*104 light years away from the galactic centre of the Milky way galaxy. The sun provides heat and light by nuclear fusion.

Ans:

Milky Way galaxy is our home galaxy. It contains 100 billion stars in it. It is spiral in shape. It is situated at about 2.5*104 light years away from the galactic centre of the Milky Way galaxy.

Ans:

The distance that the light travels in one year through the vacuum is called light year. In one year light travels about 9.46*1015 m.

Ans:

All the stars and constellations of a galaxy revolve around the imaginary central point of a galaxy called galactic centre.

Ans:

The sun takes 2.5*108 years to revolve around the galactic centre. This time is called cosmic year.

Ans:

Mercury is considered as the hottest planet. The surface temperature of the mercury is about 3330F. It is the nearest planet from the sun.

Ans:

Planets

Stars

They revolve around the sun.

They do not revolve around the sun.

They are non luminous objects.

They are luminous objects.

They do not twinkle at night.

They twinkle at night.

Ans:

Light Year

Cosmic Year

The distance that the light travels in one year through the vacuum is called light year.

The sun takes 2.5*108 years to revolve around the galactic centre. This time is called cosmic year.

It is measured in meter.

It is measured in years.

Ans:

Galaxy

Constellation

A galaxy is a collection of billions of stars.

A constellation is a collection of only very few stars.

It does not form a definite pattern which resemble the shape of an animal, Human etc

It appears to resemble the shape of an animal, human etc

There are about 100 billion galaxies in the universe.

There are about 88 constellations named so far.

Ans:

The example of Spiral galaxy is Milky Way. The example of oval galaxy is Fornix. The example of irregular galaxy is NCG 6822.

Ans:

Velocity of light = about 3*108 m/s

1 Year = 365 days, 1 day = 24 hrs = 60 min and 1 min = 60 seconds

Now,

The distance covered by the light in one year = 365*24*60*60*3*108 m

= 9.45*1015 m

Therefore, 1 light year = 9.45*1012 Km.

Ans:

Asteroids are also known as minor planets or planetoids or baby planet. Most of the asteroids revolve around the sun between the orbits of Mars and the Jupiter. Most of the asteroids are irregular in shape. The size of the bodies varies from hungered of km to objects less than 1 Km in diameter. They are composed of various rocks and metallic substances.

10.

Ans:

Some comets disappear forever because when comets approach the sun they lose some ice, dust particles and gases each time. These comets finally change into asteroids. Some comets take hundreds of years to revolve around the sun. Sa scientist cannot see the comet in his life time.

A meteor is a piece of rock in outer space that travels very fast and burns with a bright light as it enters the earth atmosphere. It is also called shooting star or falling star.

Meteorite is the meteor in the space which hits the earth. A meteor which does not burn completely on entering the earth atmosphere and lands on the earth is called meteorite.

11.

Ans:

When the hydrogen present in a star completely converts into helium, the fusion reaction stops as the core of the star and the protostar. This way reaction provides enormous energy to the newly born star. In this way stars are formed.

12.

Ans:

A star which consists of a matter mainly in the form of neutron is called a neutron star. A neutron star is formed from the collapsed core of supernova.

13.

Ans:

Black hole has an extremely large gravitational field. Due to this when matter enters in the field of black hole it gets pulled towards itself so no matter can escape from black hole.

Ans:

The tail of the comet does not always remain in existence because the tail is formed due to the evaporation of ice present in the comet and this happens only when it is close to the sun.

14.

Ans:

There are numerous clouds of dust and gases in the space. When the clouds of dust and gases in the space attain a certain size, they start to contract due to effect of gravitation acting among them. Finally, these clouds are compressed to a highly condensed mass. As a result a protostar is formed. The protostar is a beginning star. The matters in the protostar condense again due to gravity, the clouds heat up and the temperature as wall as pressure inside the protostar also increases. Finally the core of the protostar becomes very hot and gains vey high pressure. In such condition nuclear fusion reaction takes place as the core of the protostar. This reaction provides enormous energy to the newly born star. In this way stars are formed.

15.

Ans:

If the remnant of supernova explosion is more than three times the mass of the sun, the star collapses and forms a black hole.

16.

Ans:

The core of a young star is made up of hot gases like hydrogen and helium. Pressure and temperature is high in the core of a star.

Ans:

A delicate equilibrium is the equilibrium in which gravity pulling inward and the pressure of radiating heat outward are balanced. It makes the size of a star constant.

Ans:

A white dwarf star is a small dense star which has exhausted all its nuclear energy and shines only due to some retained heat energy.

Ans:

A protostar is the pre state of a star which has a dense hot core and it is now close to becoming a star. About 105 years is needed to form a star.

17.

Ans:

Star

Constellation

A star is made up of hot gases like hydrogen and helium.

A constellation is a collection of only very few stars.

A star is a self luminous body. It has its own source of light.

It appears to resemble the shape of an animal, human etc.

Ans:

Light Year

Cosmic Year

The distance that the light travels in one year through the vacuum is called light year.

The sun takes 2.5*108 years to revolve around the galactic centre. This time is called cosmic year.

It is measured in meter.

It is measured in years.

Ans:

Comets

Asteroid

A comet is made up of gas, ice and dust particles.

The irregular pieces of rocks revolving round the sun between the orbits of Mars and Jupiter are called Asteroids.

A comet can be seen as a ball with a long glowing tail when it approaches the sun.

They are bigger than comets.

Ans:

Meteors

Meteorites

A meteor is a piece of rock in outer space that travels very fast and burns with a bright light as it enters the earth atmosphere.

Meteorite is the meteor in the space which hits the earth. A meteor which does not burn completely on entering the earth atmosphere and lands on the earth is called meteorite.

A meteor does not affect life and property on the earth.

A meteorites affect.

18.

Ans:

Galaxy

Milky Way, Andromeda

Planet

Jupiter, Mars, Saturn, Venus, Neptune

Comet

Enke, Shoemaker Levy, Temple Tuttle

Satellite

Moon, Phobos, Europa, Titan

Asteroid

Ceres, Juno, Pallus, Vesta, Eros

Constellation

Ursa Major, Ursa Minor, Sagittarius, Aquarius, Draco, Cassiopeia, Cepheus.

Help CY

Adsense

Class 10 Science all exercises solutions

You may like these posts

Post a Comment

Top Post Ad

Below Post Ad

Adsense

Ads Area

Popular Post

Questions of 1st Term - Class 10 - Gov. of Nepal

Answers of Maths in SEE 2075 - PDF

Arduino As A Thermometer

Question of 2nd Term - Class 10 - Gov. of Nepal

Total Pageviews

Labels

Categories

Random Posts

Popular Posts

Questions of 1st Term - Class 10 - Gov. of Nepal

Answers of Maths in SEE 2075 - PDF

Arduino As A Thermometer

Question of 2nd Term - Class 10 - Gov. of Nepal

Footer Copyright

Contact form