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Grade IXScience

Exam Preparation Booklet

Chapter wise Important Questions & Solutions

Motion_Questions

Q. No.1

Study the two given tables related to the types of motion and answer the following questions. Table A (Types of motion)

Types of Motion Definition

Uniform motion A body has uniform motion if an object covers equaldistance in equal interval of time

Non-uniform motion A body has non-uniform motion if an object travels unequaldistance in equal interval of time

Table B (Time vs position of a car)

Time in minute Position from origin0 0 m2 2 m4 4 m6 6 m8 7 m10 10 m12 12 m14 14 m

(a) Refer to table B, the motion of car is uniform or non-uniform?(b) What would be the nature of the distance-time graph of car for the first 6 minutes?(c) The average speed of a car is(i)

(ii)

(iii)

(iv) (d) Which of the following is an example of uniform motion?(i) Ball falling towards earth(ii) Slowing car(iii) Boy running at a constant speed(iv) Kicking of a football

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. Acceleration is defined as the rate of change of velocity. It is a vector quantity and its SI unit is ms-2. Velocity can bechanged by changing the magnitude of speed or direction of motion or both. When an object moves in a circular pathat a constant speed, it is called uniform circular motion. Speed of an object (v) in a circular path of radius r is: where t is the time.Since the object is moving in a circular path, its direction of motion is changing continuously, thus, the velocity of anobject changes continuously. As velocity is changing continuously, motion along the circular path is acceleratedmotion. Some of the examples of uniform circular motion are:(i) the motion of moon and earth.(ii) satellites in a circular orbit around the earth.(iii) motion of the tip of seconds' hand.

0 m

min

1 m

min

3. 92 m

min

4. 64 m

min

v = 2πr

t

(a) Name the physical quantity that remains constant during uniform circular motion.(b) If an object is moving in a uniform circular motion, then the magnitude of the acceleration is _______. (c) A runner completes a circular path in 1 minute. If the radius of the circular path is 105 m, find his speed. (d) What is the acceleration for the given velocity-time graph?

Q. No.3

Assertion: Motion of earth around the sun is an example of circular motion.Reason: Circular motion is the one in which distance of an object is same from a fixed point.

A. Both Assertion and Reason are true and Reason is correct explanation of the Assertion.B. Both Assertion and Reason are true and Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: An object after moving in a certain path returns at its initial position, then the displacement is zero but thedistance travelled is not zero.Reason: Distance is the length of the actual path travelled by an object while displacement is the shortest pathbetween the initial and final position.


Q. No.5

A bullet of mass 15 g is fired horizontally with a velocity of 100 m/s. It strikes a stationary wooden block and comes torest in 0.02 s. The distance of penetration of bullet into the block will be :

A. 1.0 mB. 1.5 mC. 2.0 mD. 4.0 m

Q. No.6

A boy starts from rest moves with constant acceleration is a straight line. The ratio of distance covered by him during4th second to that covered in 4 seconds is:

A.

B.

C.

D. the given information is insufficient

Q. No.7

Distance-time graph of a body in motion is given by the below-shown graph. Speed of the body is:

A.

B. 1C.

D. 3

Q. No.8

The area between the speed-time graph and time gives:

A. DistanceB. speedC. velocityD. acceleration

Q. No.9

Two bodies in a circular path of radii 1:2 take the same time to complete their circular paths. The ratio of their linearspeeds is:

A. 1:2B. 1:1C. 2:1D. 3:1

7

16

116

7

1

√3

√3

Q. No.10

What is the significance of the area under the velocity-time graph?

Motion_SolutionsSolution 1

(a) Since the car is not covering the equal distance in equal interval of time (for 6 to 8 minute and 8 to 10 min), the motionof a car is non-uniform.

(b) For the first six minutes, the distance-time graph of a car would be straight-line inclined to the time axis.

(c)

Hence, the correct answer is option (ii).

(d) Ball falling towards earth (increasing speed), slowing car (decreasing speed), kicking a football (increasing speed offootball) are all non-uniform motion because speed in all these cases in changing. Boy running at constant speed is anexample of uniform motion.

Hence, the correct answer is option (iii).

Solution 2

(a) Speed of an object remains constant during uniform circular motion.

(b) If an object is moving in a uniform circular motion, then the magnitude of the acceleration is constant.

(c) Given, time, t = 1 minute = 60 s radius of circular path, r = 105 m

(d) From the given velocity-time graph, velocity of an object is constant.

Acceleration is zero, as velocity is constant.

Solution 3

Circular motion is the one in which distance of an object is same from a fixed point. Motion of earth around the sun is anexample of circular motion because distance of earth from sun is constant.Therefore, both assertion and reason are true and reason is correct explanation of the assertion.

Average speed = =

= 1

total distance

total time1414

mmin

v =

=

= 11 ms−1

Speed, 2πr

t

2×22×105

7×1×60

Acceleration =change in velocity

time

Hence, the correct answer is option A.

Solution 4

Distance is the length of the actual path traveled by an object while displacement is the shortest path between the initialand final position. When an object after moving in a certain path returns at its initial position, then the displacement iszero because the initial and final position is same but the distance traveled is not zero because an object has coveredsome path and length of that path is greater than zero.Therefore, both assertion and reason are true and reason is correct explanation of the assertion.


Solution 5

Given, the mass of the bullet, m = 15 g initial velocity, u = 100 m/s final velocity, v = 0 m/s time, t = 0.02 sWe can find the distance of penetration (s) using equations of motion

and

So, the distance of penetration of bullet will be 1 m.


Solution 6

Given, initial velocity, u = 0 acceleration = a (constant)Since, boy is moving is straight line, distance = displacement = s


Solution 7

using, v = u + at

0 = 100 + 0. 02 × a

a = − = −5000 100

0.02m

s2

s = ut + at2

= 100 × 0. 02 + × (−5000) × 0. 02 × 0. 02

= 2 − 1 = 1 m

12

12

From second equation of motion, s = ut + at2 (where t is time)

For t = 4 s, s4 = 0 × 4 + × a × 42

= 8a

12

12

For t = 3 s, s3 = 0 × 5 + × a × 32

= 4. 5a

Distance travelled in 4th second = s4 − s3

= 8a − 4. 5a = 3. 5a

12

Ratio of distance covered during 4th second to that of covered in 4 seconds =

= =

s4−s3

s4

3.5a

8a7

16

For the given distance-time graph,

Hence, the correct answer is option C.

Solution 8

The area between the speed-time graph and time gives distance traveled by an object.


Solution 9

Speed of an object (v) in a circular path of radius r is:

where t is the time


Solution 10

Area under the velocity-time graph gives displacement of an object.

= tan 60°

= √3

speed = = slopedistancetime

Distance = speed × time

v = 2πr

t

= × (given, time is same)

=

=

v1 : v2 = 1 : 2

v1

v2

2πr1

t

t

2πr2

v1

v2

r1

r2

12

Force and Laws of Motion_Questions

Q. No.1

Study the two given tables related to force and momentum and answer the following questions. Table A (Force and Momentum)

Definition Formula SI Unit

Force

Force is a push orpull that changes ortends to changes the

state of rest or ofuniform motion or

direction of motion ofan object.

N (Newton)

Momentum

Momentum is definedas the quantity of

motion contained in abody.

Ns (Newtonsecond)

Table B (data of some objects)

Mass Initial velocity Final velocity TimeBicycle X 0 ms-1 4 ms-1 6 s

Man 60 kg 0 ms-1 2 ms-1 6 s

Car 240 kg 0 ms-1 15 ms-1 6 s

Bus 500 kg 0 ms-1 12 ms-1 6 s

(a) Find the value of X if the momentum of a bicycle is 4 Ns?(b) Refer to table A, find force in terms of mass, velocity, and time.(c) Refer to table B, amount of force experienced by the car is:(i) 0 N(ii) 240 N(iii) 600 N(iv) 3600 N(d) If the value of X is 4 kg, maximum momentum is possessed by:(i) Bicycle(ii) Man(iii) Car(iv) Bus

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. Force is a push or pull that changes or tends to changes the state of rest or of uniform motion or direction of motion ofan object. Detailed qualitative and quantitative ideas about force were given by Newton in the form of three laws.Newton's first law of motion: Every body remains in a state of rest or of uniform motion in a straight line, unless it iscompelled by some external force. This law gives only a qualitative idea about force. Newton's second law of motion: This law gives a quantitative idea about force. This law states that the rate ofchange of momentum of an object is proportional to the applied unbalanced force in the direction of the force.Newton's third law of motion: Whenever one body exerts a force on another body, the second body exerts an equaland opposite force on the first body.(a) What is the relation between SI and CGS unit of force?(b) Define the momentum of a body? (c) Newton's first law of motion is also called Gallileo's law of _______. (d) If the mass of a body is 5 kg. Find the amount of force required to accelerate it to 10 ms-2?

Force = mass × acceleration =change in momentum

time

Momemtum = mass × velocity

Q. No.3

Assertion: A stone drops from a certain height and hits the floor. The force experienced by the floor is equal to therate of change of momentum of stone.Reason: The floor does not move because the action and reaction forces, being equal and opposite cancels eachother.


Q. No.4

Assertion: A ball of mass m strikes a wall with speed v and rebounds at the same speed. Change in momentum, inthis case, is zero. Reason: Change in momentum = force time


Q. No.5

A wooden cube of mass 4 kg is kept on a frictionless surface. After some time, jet releasing water strike at a rate of 1kg/s and at a speed of 8 m/s. The initial acceleration of the block will be:

A. 2 m/s2

B. 4 m/s2

C. 8 m/s2

D. 16 m/s2

Q. No.6

Two cars of different mass initially at rest, are acted upon by the same force for the same time. Which of the followingis same for both the cars:

A. velocityB. momentumC. accelerationD. kinetic energy

Q. No.7

×

Which of the following is the correct relation between SI and CGS unit of force?

A.

B.

C.

D.

Q. No.8

Given the velocity-time graph, we can find the force exerted on an object, if

A. mass of an object is givenB. the velocity of an object is givenC. acceleration of an object is givenD. distance traveled by an object is given

Q. No.9

Action and reaction forces

A. act on same bodyB. act on the different bodiesC. same in magnitude and directionD. nothing could be said

Q. No.10

Draw the graph between momentum and mass keeping velocity constant.

1 N = 105 Dynes

1 N = 10−5 Dynes

1 N = 103 Dynes

1 N = 10−3 Dynes

Force and Laws of Motion_SolutionsSolution 1

(a) Given, mass of bicycle = X the momentum of a bicycle = 4 Ns Velocity = 4 ms-1

(b)

(c) Mass of car = 240 kg Initial velocity = 0 ms-1

Final velocity = 15 ms-1

time = 6 s


(d) Momentum of bicycle = 4 х 4 = 16 Ns Momentum of man = 60 х 2 = 120 Ns Momentum of car = 240 х 15 = 3600 Ns Momentum of bus = 500 х 12 = 6000 Ns

maximum momentum is possessed by bus.

Hence, the correct answer is option (iv).

Solution 2

(a) SI unit of force is Newton and CGS unit of force is dyne.

Therefore, 1 N = 105 dynes

(b) Momentum is defined as the quantity of motion contained in a body. (c) Newton's first law of motion is also called Gallileo's law of inertia.

(d) Given, mass of the body, m = 5 kg acceleration, a = 10 ms-2

Momemtum = mass × velocity

4 = X × 4

X = 1 kg

Force = mass × acceleration

=

=

change in momentum

time

mass×(final velocity−initial velocity)

time

Force = mass × acceleration =

=

= 600 N

change in momentum

time

240×(15−0)

6

1 N = 1 kgms−2

= 1000 g × 100 cms−2

= 100000 gcms−2

= 105 dynes (1 gcms−2 = 1 dyne)

Therefore, the amount of force required is 50 N.

Solution 3

Force experienced by floor can be found by using Newton's second law of motion.

From Newton's third law of motion, for every action, there is an equal and opposite reaction. Action (force) and reaction(force) acts on two different bodies. Since action and reaction acts on two different bodies, these forces can not canceleach other.Therefore, assertion is true but reason is false.


Solution 4

Given that the mass of the ball is m and the velocity of the ball is v.The initial momentum of ball = mvAfter striking the wall, the direction of the ball changes. Velocity after striking the wall = -vFinal momentum of the ball = -mvChange in momentum = final momentum - initial momentum = - mv - mv = -2mv

Also, from Newton's second law of motion,

Therefore, the assertion is false but the reason is true.

Hence, the correct answer is option D.

Solution 5

Mass of given wooden block, m = 4 kgIf, F is the force applied on this block, then from Newton's second law,F = ma (where a is the acceleration of block by application of force) = 4aForce on the block is applied by jet releasing water. So, F = rate of mass striking block speed of jet F = 1 8 = 8 N

Since force experienced by a wooden block = force applied by jet releasing water 4a = 8 a = 2 m/s2


Solution 6

= ma

= 5 × 10

= 50 N

Force, F

Force =change in momentum

time

force =

change in momentum = force × time

change in momentum

time

××

= mA

initial velocity, uA = 0 (given)

final velocity = vA

Let for car A, mass

= mB

initial velocity, uB = 0 (given)

final velocity = vB

Let for car B, mass

Hence, the correct answer is option B.

Solution 7

SI unit of force is N (Newtons) and CGS unit of force is dyne.


Solution 8

From the slope of the velocity-time graph, we can find the acceleration of an object. Once the acceleration is known andmass of an object is given, we can easily find the force exerted on an object from Newton's second law of motion.


Solution 9

Action and reaction forces act on different bodies. These are equal in magnitude and opposite in direction.


Solution 10

We know that momentum is given by, p = mv, where m is the mass and v is the velocity.If the velocity of an object is constant, then p ∝ m, means, for different bodies, if the mass is more, momentum will be more and if the mass is less, momentum is less.

The correct graph will be represented as:

As force × time = change in momentum

Since, force × time is same for both the cars

so, change in momentum for car A = change in momentum for car B

mAvA − mAuA = mAvA − mBuB

mAvA = mAvA (as uA = uB = 0)

so, final momentum of car A = final momentum of car B

1 N = 1 kg × 1 ms−2

= 1000 g × 100 cms−2

= 100000 gcm−2

= 105 Dynes

Gravitation_Questions

Q. No.1

Study the two given tables related to laws of floatation and answer the following questions. Table A (Laws of Floatation)

Condition Density of body > Density of liquid The body will sink in the liquid

Density of body = Density of liquid The body will remain in equilibrium inside the liquid. Nopart of the body remains above the surface of the water.

Density of body < Density of liquid The body will float in such a way that a portion of an objectis inside the liquid.

Table B (Density of some substances)

Substance DensityWood 800 kgm-3

Water 1000 kgm-3

Glass 2500 kgm-3

Iron 7800 kgm-3

Silver 10800 kgm-3

Mercury 13600 kgm-3

(a) Define density of a substance?(b) Refer to table B, which of the objects will float in water?(c) Relative density of iron is:(i) 0.78(ii) 1 (iii) 7.8(iv) 7800(d) Which of the substances will sink in liquid mercury?(i) Wood(ii) Iron(iii) Glass(iv) none of these

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. Seeing the sky in night time always amazes us and motivates us to know how the interaction of celestial bodies. Inancient times, people were trying to find the reasons of moon orbiting the earth, the motion of stars in the sky. GalileoGalilei was the first one who realized that all celestial objects fall freely at the same rate. Another legend Newtonformulated these observations in mathematical form. Newton stated that every object attracts every other object witha force which is proportional to the product of their masses and inversely proportional to the square of the distancebetween them.

G = universal gravitational constantM = mass of first objectm = mass of second objectr = distance between objects Some of the characteristics of this force are:It is always attractive in nature.It is applicable to all objects, whether the bodies are big or small.

Force, F = GMm

r2

(a) What is the value of G in SI units?(b) What is the acceleration due to gravity on earth in terms of mass and radius of earth?(c) Two objects of equal mass exert a gravitational force of 10 N on each other. What will be the gravitational forcebetween them, if both of them are dipped in water (density of water = 1000 kgm-3 ) ? (d) If we double the distance between two bodies, how will the gravitational force change?

Q. No.3

Assertion: When an object is immersed in a fluid, fluid exerts an upward force on an object.Reason: Difference in pressure is the cause of buoyancy.


Q. No.4

Assertion: If the distance between two objects of mass m each becomes half, the gravitational force between themwill become one fourth.Reason: Gravitational force follows inverse square law.


Q. No.5

A red planet has mass 0.2 times the mass of earth and radius is of the earth. The ratio of acceleration due togravity on planet surface to earth surface is:

A. 2 : 5B. 4 : 5C. 5 : 4D. 5 : 2

Q. No.6

SI unit of weight is:

A. KgB. NC. quintalD. tonnes

1

2

Q. No.7

If the density of aluminum is 2700 kgm-3, then its relative density is:

A. 2.7B. 27C. 270D. 2700

Q. No.8

An object of mass 60 g has a volume of 20 cm3. If this object is placed in water of density 1 gcm-3, it will

A. sinkB. completely immersedC. float with a part outside waterD. given information is insufficient

Q. No.9

Thrust is a

A. scalar quantityB. vector quantityC. tangential forceD. none of these

Q. No.10

What is the acceleration of a stone of mass 50 g falling freely from 50 m high building?

Gravitation_SolutionsSolution 1

(a) Density of a substance: It is defined as mass per unit volume.

(b) Since density of wood is less than density of water, it will float when placed in water.

(c)


(d) As density of mercury is more than the density of wood, iron, and glass, all these substances will float in liquidmercury.


Solution 2

(a)

(b)

G = universal gravitational constantM = mass of earthR = radius of earth

(c) The gravitational force between two objects does not depend on medium between them. If two objects of equal massare dipped in water, they still exert a gravitational force of 10 N on each other.

(d)

Where r is the distance between objects.If the distance between two bodies is doubled, then the force will become one fourth as gravitational force obeys inversesquare law.

Solution 3

When an object is placed in the fluid, a pressure is exerted on the object. We know that pressure increases as depth offluid increases. Therefore, the bottom part of the object will face more pressure and the upper part will face less pressure,this pressure difference will give rise to buoyancy.Therefore, both Assertion and Reason are true and Reason is correct explanation of the Assertion.


Solution 4

The gravitational force between two objects of mass m each is given by

Where G is universal gravitational constant and r is the distance between objects.

Density =mass of the substance

volume of the substance

Relative density of iron =

=

= 7. 8

Relative density =Density of a substance

Density of water

Density of iron

Density of water

7800

1000

Universal gravitational constant, G = 6. 67 × 10−11 Nm2 kg−2

Acceleration due to gravity, g = GM

R2

F ∝Force, 1

r2

F = Gm2

r2

Since gravitational force follows inverse square law, if the distance between objects becomes half, then force will becomefour times.Therefore, assertion is false but reason is true.


Solution 5

Let the mass of earth = meand radius of earth = re

for the red planet, mr = 0.2me

so, gr : ge = 4 : 5


Solution 6

Weight of a body is the force with which it is attracted towards the centre of the earth and is given by:W = mgwhere m is mass of an object and g is acceleration due to gravity at that point.SI unit of weight is same as that of force and is N (Newton).Hence, the correct answer is option B.

Solution 7

Given the density of aluminum = 2700 kgm-3


Solution 8

Give, the mass of object = 60 gThe volume of object = 20 cm3

Since, Density of object > Density of water, object will sink when placed in water.


acceleration due to gravity on earth surface, ge = G (G is universal constant)me

re2

rr =

gr = G

gr = G

re

2mr

rr2

0.2×me

( )2re

2

= G ×

= 0. 2 × 4= 0. 8

=

gr

ge

0.2×me

( )2re

2

1

G

re2

me

4

5

Relative density=

= (density of water = 1000 )

=2. 7

density of object

denstiy of water at 4°C

2700

1000

kg

m3

Density of object =

=

= 3 gcm−3

mass of object

volume of object

60

20

Solution 9

Thrust is the force acting on an object perpendicular to the surface. Since force is a vector quantity, thrust is also a vectorquantity.


Solution 10

Since stone is falling freely, its acceleration will be equal to, a = 9.8 ms-2.

Work and Energy_Questions

Q. No.1

Study the two given tables related to energy and power and answer the following questions.

Table A (Energy and Power) Energy Power

Definition Energy is the ability to do work Power is the rate of doing work.

Formula Energy = work = force × displacement

SI unit (J) Joule (W) Watt

Table B ( Raw Data of some objects/appliances)

Object/appliance Data Physical quantity

Bicycle mass = 20 kg, initial speed = 2 ms-1, final speed =4 ms-1, time = 2 s

Energy

Radio set Energy consumed = 3 kWh in time = 50 hours Power(a) Name the physical quantity whose unit is kWh?(b) Energy transferred by a 50 W bulb in 5 minutes is ________.(c) Refers to table B, moving bicycle has:(i) kinetic energy(ii) potential energy(iii) the sum of kinetic energy and potential energy(iv) the difference between kinetic energy and potential energy(d) Refers to table B, the value of power for radio set is:(i) 0.06 W(ii) 0.6 W(iii) 60 W(iv) 60 kW

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.Energy exists in many different forms. The beauty of energy is that it can be converted from one form to another formof energy.When energy is converted, the total energy remains unchanged. This transformation of energy is known as the law ofconservation of energy.This law of conservation of energy is valid in all situations and for all kinds of transformations.From here, it can be concluded that energy can neither be created nor be destroyed.Some examples are:Freely falling body: When a body falls freely from a height, its potential energy gets converted into kinetic energy.

Hydroelectric powerhouse: In it, water is allowed to fall from a great height so that its potential energy is convertedinto kinetic energy. This kinetic energy is converted into electrical energy with the help of generators.

(a) Sum of kinetic energy and potential energy of an object is called _______.(b) Name the energy changes involved when a battery lights a bulb.(c) What happens to kinetic energy if its velocity is doubled?(d) Joule is the unit of work or energy?

Power = =work done

time taken

energy consumed

time taken

potential energy → kinetic energy

→ electrical energypotential energy → kinetic energy

Q. No.3

Assertion: Kinetic energy of a body remains constant in case of uniform circular motion.Reason: In a uniform circular motion, linear momentum does not change with time.


Q. No.4

Assertion: Kinetic energy of a body becomes nine times when its speed is tripled.Reason: Kinetic energy is proportional to the square of the speed of a body.


Q. No.5

A ball starts sliding from point P on a frictionless surface with an initial velocity of 4 ms-1. Velocity of the ball at pointR is:(g = 10 ms-2)

A. 4 ms-1

B. 6 ms-1

C. 6√2 ms-1

D. 12 ms-1

Q. No.6

Two bodies of mass m and 4m are moving with the same velocities, the ratio of their kinetic energy is:

A. 4 : 1B. 2 : 1C. 1 : 2

D. 1 : 4

Q. No.7

A body at rest can have

A. momentumB. kinetic energyC. potential energyD. all of these

Q. No.8

100 units of energy is equivalent to

A. 100 JB.

C.

D.

Q. No.9

1 kW is equal to

A. 1000 WB.

C.

D. all of these

Q. No.10

Give an example of negative work done.

3. 6 × 104 J

3. 6 × 106 J

3. 6 × 108 J

106 milli watt

10−3 mega watt

Work and Energy_SolutionsSolution 1

(a) kWh is the commercial unit of energy. (b) Power = 50 W Time = 5 minutes = 300 s

Energy transferred by a 50 W bulb in 5 minutes is 15000 J.

(c) The speed of bicycle is increasing or changing. Due to this moving bicycle possesses kinetic energy.

Hence, the correct answer is option (i).

(d) Given, energy consumed = 3 kWh Time = 50 hours


Solution 2

(a) Sum of kinetic energy and potential energy of an object is called total mechanical energy.

(b) When a battery lights a bulb, then the chemical energy of the battery is converted into light energy.

(c)

where m is the mass v is the velocity if velocity is doubled, kinetic energy will become four times.

(d) An object having capabilities to do work is said to possess energy.Whenever work is done, an equal amount of energy is consumed. Thus, work and energy have the same units and Joule(J) is the SI unit for both.

Solution 3

In a uniform circular motion, the speed of the body does not change with time. Due to this kinetic energy of the bodyremains constant as

The velocity of a body changes continuously due to the change in direction of motion.Linear momentum = mass velocityDue to change in velocity, the linear momentum of a body changes with time.Therefore, assertion is true but reason is false.


Energy = power × time

= 50 × 300 = 15000 J

Power =

=

= 0. 06 kW = 60 W

Energy

time

3 kWh

50 h

= mv2

or, KE ∝ v2

Kinetic energy 12

kinetic energy = × mass × (speed)212

×

Solution 4

The kinetic energy of a body is given by,

if the speed of a body is tripled, its kinetic energy will become nine times.Therefore, both assertion and reason are true and reason is correct explanation of the assertion.


Solution 5

The velocity of the ball at point R can be calculated by applying the conservation of mechanical energy (since the surfaceis frictionless).If m is the mass of the ball,At point PvP is the velocity and hP is the heightTotal mechanical energy at P = Kinetic Energy + Potential Energy

At point RvR is the velocity and hR is the heightTotal mechanical energy at R = Kinetic Energy + Potential Energy

Mechanical energy at P = mechanical energy at R

Velocity at point R, vR = 6√2 ms-1


Solution 6

KE = mv2

where, m = mass of body

v = speed of body

12

= mvP2 + mghP

= × m × 42 + m × 10 × 6

= 8m + 60m = 68m

1212

= mvR2 + mghR

= × m × vR2 + m × 10 × 3. 2

= × m × vR2 + 32m

1212

12

68m = × m × vR2 + 32m

× m × vR2 = 36m

vR2 = 72

vR = 6√2 ms−1

12

12

If m1 is the mass, v1 is the velocity and KE1 is the kinetic energy of the first body and m2 is the mass, v2 is the velocityand KE2 is the kinetic energy of the second body, then


Solution 7

When a body is kept on the top of a roof, it possesses potential energy (due to height).As potential energy, PE = mghm = mass of an objectg = acceleration due to gravityh = height at which object is kept


Solution 8

1 unit = 1 kWh100 units = 100 kWh


Solution 9

1 kW = 1000 W

Kinetic energy = mv2

where, m = mass of object

v = velocity of object

12

= ×

=

Given, v1 = v2 (as velocity of both objects is same)

KE1

KE2

1×m1×v12

22

1×m2×v22

m1×v12

m2×v22

=

= =

KE1

KE2

m1

m2

m

4m

1

4

1 kWh = 3. 6 × 106 J

100 kWh = 100 × 3. 6 × 106 J

100 kWh = 3. 6 × 108 J

so, 100 units = 3. 6 × 108 J

1 Watt = 1000 milli watt1 kW = 1000 W = 1000 × 1000 milli watt

= 106 milli watt

1 Watt = 10−6 mega watt

1 kW = 1000 W

= 1000 × 10−6 mega watt

= 103−6 = 10−3 mega watt

Thus, 1 KW = 1000 W = 106 milli watt = 10−3 mega watt


Solution 10

When a force is in direction opposite to the direction of motion, work done in such a case is negative.

Sound_Questions

Q. No.1

Study the two given tables related to frequency range and answer the following questions. Table A (Frequency range)

Frequency rangeInfrasonic 0-20 HzAudible 20 Hz-20 kHz

Ultrasonic > 20 kHz Table B ( Frequency range of some objects/phenomena)

Object/phenomena frequency rangeEarthquake less than 20 HZ

Vibrating simple pendulum less than 20 HzDogs X

Dolphins more than 20 kHzProduced by SONAR ultrasonic frequency

(a) Why we cannot hear vibrations of a simple pendulum?(b) What is the frequency range dogs can hear?(c) An earthquake produces(i) low-frequency audible sound(ii) high-frequency audible sound(iii) low-frequency infrasonic sound(iv) low-frequency ultrasonic sound(d) The time period corresponding to 10 kHZ frequency is(i) 10 s(ii) 1 s(iii) 0.001 s(iv) 100 μs

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studied concepts. A wave is a vibratory disturbance in a medium that carries energy from one point to another. In wave motion, the disturbanceis carried forward.There are two types of waves (on the basis of the direction of vibration of medium particles): longitudinal and transversewaves.Longitudinal waves: A wave in which the particles of the medium vibrate back and forth in the same direction in which thewave is moving. In it, particles do not move from one place to another but they simply oscillate back and forth about theirposition of rest. Example: Sound wavesTransverse waves: A wave in which the particles of the medium vibrates at the right angle to the direction in which the waveis moving. Example: when we drop a pebble in a pond, the waves we see on the water surface is an example of transversewaves.Waves are also classified into two types based on the medium requirement for propagation:Mechanical waves: A waves which need a material medium (solid, liquid or gas) for its propagation. A mechanical wavecannot travel through a vacuum. Example sound waves in airNon-mechanical waves: These waves do not require a material medium for its propagation. These waves can travelthrough vacuum also. Example light waves (we are able to see sunlight traveling from the sun to earth)(a) Name the wave which travels in the form of crests and troughs?(b) ___________ are the regions of low pressure in longitudinal waves. (c) Define frequency.(d) Is air a material medium?

Q. No.3

Assertion: The frequency of a source of sound is 50 Hz. It means it vibrates 3000 times in a minute.Reason: Frequency is the number of vibrations completed per unit time.


Q. No.4

Assertion: In SONAR (Sound Navigation and Ranging), ultrasonic waves are used to measure distance, direction and speedof underwater objects.Reason: Ultrasounds are the low frequency (frequency less than 20 Hz) waves.


Q. No.5

A sound wave travels from north to south, in which direction do the particles of air move:

A. east-westB. north-southC. right-leftD. none of the above

Q. No.6

A.

B.C.D.

Q. No.7

Identify incorrect statement related to the range of hearing for some organisms:

A. The frequency of vibrations of simple pendulum lies in the audible range of humans.B. Bats search out prey in the night by emitting and detecting reflections of ultrasonic waves.C. Rats play games by producing ultrasonic waves.D. Rhinoceroses communicate using infrasonic frequency.

Q. No.8

The velocity of sound in air is V when its frequency is ν. Velocity when its frequency changes to 4ν is :

V

4

V

4V

8V

A ship sends out an ultrasound that returns from the seabed and is detected after 2 s. If the speed of ultrasound throughwater is 1531 ms-1, the distance of seabed from the ship is:

A. 0.15 kmB. 1.53 kmC. 15.3 kmD. 1531 km

Q. No.9

The time period of a wave having a wavelength 1.5 m and frequency 100 Hz is

A. 0.01 sB. 0.1 sC. 100 sD. 150 s

Q. No.10

Can we hear ultrasound waves?

Sound_SolutionsSolution 1

(a) Vibrations of a simple pendulum are in the frequency range of 0-20 Hz. Since the frequency range of vibrating simplependulum is less than the hearing range of humans, we are not able to hear them. (b) Dogs can hear frequency upto 25 kHz.

(c) An earthquake produces low-frequency infrasonic sound.


(d) Given, Frequency = 10 kHz =10000 Hz

Time period = 0.0001 s = 100 μs


Solution 2

(a) Transverse waves travel in the form of crests and troughs.

(b) Rarefactions are the regions of low pressure in longitudinal waves. (c) The number of vibrations made by a particle in unit time is called frequency.

(d) Yes, air is a material medium as it contains a mixture of gases like nitrogen, oxygen, carbon dioxide and others.

Solution 3

Frequency is the number of vibrations completed per unit time. If frequency of a source of sound is 50 Hz. It meansnumber of vibrations in one second is 50. So in one minute (60 seconds), the number of vibrations is 50 60 = 3000.Therefore, both assertion and reason are true and reason is correct explanation of the assertion.


Solution 4

Ultrasounds are the high frequency (frequency more than 20 kHz) waves. In SONAR (Sound Navigation and Ranging),ultrasonic waves are used to measure distance, direction and speed of underwater objects. Method is based on echo.Therefore, assertion is true but reason is false.


Solution 5

Sound waves are longitudinal waves in which back and forth vibrations of the air layers are in the same direction as themovement of a sound wave. So, air particles would also vibrate in north-south direction.


Time period =

=

= 0. 0001 s = 10−4 s

1

Frequency

1

10000

1 s = 106 μs

10−4 s = 10−4 × 106 μs

= 102 μs

= 100 μs

×

Solution 6

The velocity of a sound wave is independent of frequency in a medium. So, velocity will remains same on changingfrequency.


Solution 7

The frequency of vibrations of simple pendulum lies in the infrasonic range. Humans cannot hear infrasonic frequencyvibrations. Thus, these vibrations are inaudible for us.


Solution 8

Time between transmission and detection = 2 sSpeed of ultrasound in water = 1531 ms-1

The distance of seabed from the ship = 1531 m = 1.53 km

Hence the correct answer is option B.

Solution 9

Give, wavelength = 1.5 mFrequency = 100 Hz

The time period of a given wave is 0.01 s.


Solution 10

No, because of the frequency of ultrasound waves is more than the hearing range of humans.

Distance traveled by ultrasound = 2 × distance of seabed

= 2 × d

where d is the distance of seabed from the ship

2d = 1531 × 2

d = 1531 m

using, distance = speed × time

Time period =

=

= 0. 01 s

1

Frequency

1

100

Matter in Our Surroundings_Questions

Q. No.1

Study the two given tables related to temperature conversions and answer the following questions. Table A (Conversion formulae)

S.No. Temperature scales involved Formula of conversion1 Kelvin and Celcius K = 273.15 + C

2 Fahrenheit and Celcius

Table B (Unknown conversions)

S.No. Value on Temperature Scale- I Value on Temperature Scale -II1 -273.15 C "X" K2 -40 F "Y" C

(a) Find "X"(b) Find "Y"(c) What of the following temperatures is called "absolute zero"?(i) 0 K(ii) -273.15 C(iii) -459.67 F(iv) All of the above(d) Under which of the below mentioned ambient pressure values, the boiling point of water is going to be maximum?(i) 100 kPa(ii) 150 kPa(iii) 200 kPa(iv) 250 kPa

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. An underlying viral illness is what leads to viral fever. Human beings can be affected by a huge variety of viralinfections, from the common cold to flu. Recent studies have shown that many people in the country have beenaffected by dengue fever, which in some cases has also led to catastrophic outcomes. A similar case happened withX, who was admitted to a nearby hospital in ABC city due to viral fever. Using a clinical thermometer, the head doctor,Dr.Y, found that X's body temperature was 2°C more than the normal human body temperature. (a) What is the normal human body temperature in the Celcius scale?(b) Which scale of temperature measurement is also called the absolute temperature scale?(c) Calculate the equivalent body temperature of X on the absolute temperature scale. (d) Find the temperature at which the Celcius and the Fahrenheit scale of temperature measurement show the samereading.

Q. No.3

Assertion: Water at 0 C is capable of causing more effective cooling as compared to ice at the same temperature.Reason: Water at 0 C is having extra energy as compared to ice at the same temperature, in the form of latent heatof fusion.

°

°F = ( )°C + 329

5

°

° °

°°

°°

A. Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: Dry ice disappears after some time on its own.Reason: Solid CO2 is a sublimable substance.

A. Both Assertion and Reason are true and Reason is the correct explanation of the Assertion. B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion. C. Assertion is true but Reason is false. D. Assertion is false but Reason is true.

Q. No.5

Which of the following temperature conversions is incorrect?

A.B.C.D.

Q. No.6

Which of the following statements regarding latent heat is incorrect?

A.Latent heat of vapourization can be defined as the amount of energy required to convert unit mass of liquid into gas, atatmospheric pressure, at its boiling point, without causing a change in temperature. B. Water at is capable of causing more effective cooling as compared to ice at . C. Steam at is capable of causing more severe burns as compared to water at . D.Latent heat of fusion can be defined as the amount of energy required to convert unit mass of solid into liquid, atatmospheric pressure, at its melting point, without causing a change in temperature.

Q. No.7

X, a local resident of Kochi in Kerala, visited Delhi recently. He observed that clothes in Delhi dried up must faster ascompared to Kochi. Which of the following alternatives is the correct explanation behind the faster drying of clothes?

A. Due to slower wind speed in Delhi as compared to Kerala. B. Because the temperature of Delhi is lower than the temperature of Kochi. C. Due to lower water vapour content in the surrounding air in Delhi as compared to that in Kochi. D. Because humidity in Delhi is more than that in Kochi.

Q. No.8

32. 8°C is equivalent to 305. 80 K

489 K is equivalent to 216°C

17°C is equivalent to 62. 6°F

98. 6°F is equivalent to 38. 96°C

0°C 0°C

100°C 100°C

Which of the following is not an example of matter?(i) Love(ii) Wave(iii) Cold(iv) Light

A. Both (i) and (iii)B. Both (ii) and (iv)C. All of the above. D. None of the above.

Q. No.9

In order to demonstrate an experiment in the class, so as to prove the properties of matter, the teacher took a beakerfilled with 500 mL water and added 2 tablespoons of sugar in it. The resulting solution was left to stand still for aduration of 2 hours. Meanwhile, the students were asked to predict the change that shall be incurred in the level of thesolution, after 2 hours. Which of the following predictions made by the students in context to the change in the level of water is correct?

A. The water level doubles.B. The water level reduces to half.C. The water level remains the same.D. The water level triples.

Q. No.10

Give one difference between evaporation and boiling.

Matter in Our Surroundings_SolutionsSolution 1

(a) "X" is 0 K. (b) "Y" is -40 C. -40 degrees is the temperature at which Celcius and Fahrenheit are equal. (c) Absolute zero is the lowest possible temperature. It is equal to 0 K or -273.15 C or -459.67 F . Hence, the correct answer is option (iv). (d) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to or greaterthan the atmospheric pressure or the ambient pressure. The lower the ambient pressure, the lower will be the boilingpoint and vice-versa. Therefore, at 250 kPa the boiling point of water is going to be the maximum. Hence, the correct answer is option (iv).

Solution 2

(a) The normal temperature of a human body in the Celcius scale is equal to 37 C. (b) Kelvin scale of temperature measurement is also called as the absolute temperature scale. (c) The normal temperature of a human body = 37 C Ramadhir's body temperature = 37 C + 2 C = 39 C

C = ( C + 273) K

39 C = 273 + 39 = 312 K (d) is the temperature at which the Celcius and the Fahrenheit scale of temperature measurement show thesame reading.

Solution 3

Ice at 0 C is capable of causing more effective cooling as compared to water at the same temperature. This is becausewater at 0 C is having extra energy as compared to ice at the same temperature, in the form of latent heat of fusion. Dueto higher energy content water is not capable of providing effective cooling. Therefore, Assertion is false but Reason is true. Hence, the correct answer is option D.

Solution 4

Dry ice, also called as solid CO2 is a sublimable substance, which can directly convert into the gaseous state from thesolid-state, without actually passing through the liquid state. Therefore, both Assertion and Reason are true and Reason is the correct explanation of the Assertion. Hence, the correct answer is option A.

Solution 5

°

° °

°

°

° ° °

° °

∴ °

−40∘C

°°

Using the formula,

°F = ( )°C + 32

On solving, we get

98. 6°F is equivalent to 37°C

9

5

Typesetting math: 100%


Solution 6

Ice at is capable of causing more effective cooling as compared to water at . This is because water at ishas more energy compared to ice, in the form of latent heat of fusion. Hence, the correct answer is option B.

Solution 7

The rate of evaporation is (i) directly proportional to surface area exposed for evaporation. (ii) directly proportional to the wind speed. (iii) directly proportional to the temperature. (iv) inversely proportional to humidity (water vapour content in the surrounding air) As Kochi in Kerala is located in a coastal area, the humidity at Kochi is higher as compared to that in Delhi. Due to higherwater vapour content in the surrounding air in Kochi, as compared to Delhi, clothes in Delhi dried up faster as comparedto that in Kochi. Hence, the correct answer is option C.

Solution 8

Feelings, emotions and adjectives are not examples of matter. Further, wave and light are also not examples of matter. Hence, the correct answer is option C.

Solution 9

There shall be no increase in the level of the solution. This happens because the sugar particles will occupy the vacantspaces present between the water molecules. It is an indication of the particulate nature of matter. Sugar molecules,divide themselves into smaller and smaller particles, thereby occupying the vacant spaces present between the watermolecules. Hence, the correct answer is option C.

Solution 10

Evaporation is a surface phenomenon, whereas boiling is a bulk phenomenon.

0°C 0°C 0°C


Is Matter Around Us Pure?_Questions

Q. No.1

Study the given table related to solubility and answer the following questions. Table of Solubilty Solubility in g (per 100 g of H2O)Name of Salt 10 C 20 C 30 C 40 CAluminium chloride 44.90 g 45.80 g 46.60 g 47.30 g Barium nitrate 6.77 g 9.02 g 11.50 g 14.10 g Calcium chloride 64.70 g 74.50 g 100 g 128 g (a) Define solubility. How does the solubility of solids in liquids change due to change in pressure?(b) Write the chemical formula of (i) Aluminium chloride(ii) Barium nitrate(c) Calculate the mass of crystals of calcium chloride that would separate out of the solution, if the saturated solutionof calcium chloride at 40 C is allowed to cool to room temperature.(i) 64.70 g(ii) 74.50 g (iii) 53.50 g (iv) 100 g (d) How does the solubility of a solid in liquid change with the change in temperature?(i) The solubility of a solid in liquid decreases with the increase in temperature.(ii) The solubility of a solid in liquid increases with the increase in temperature.(iii) The solubility of a solid in liquid doesn't depend on temperature.(iv) The solubility of a solid in liquid first increases and then decreases with the increase in temperature.

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. Air is a true solution and thus, a homogeneous mixture, having a uniform composition throughout. The percentagecomposition by volume of the three major constituents of air, i.e., N2, O2 and Ar are 78.1%, 20.9% and 0.9%,respectively. Dinitrogen gas has the maximum boiling point of -196 C, followed by argon gas having a boiling point of-186 C, and oxygen gas has the least boiling point of -183 C. The constituents of air are separated by the process offractional distillation. In this process, firstly the air is compressed and cooled by increasing pressure and decreasingtemperature. As a result, liquid air is obtained that is allowed to warm up slowly in a tall fractional distillation column.Finally based on their respective boiling points, the gases present in the air get separated at different heights.(a) Why is it necessary to pass the gas initially through a filtration unit?(b) Why does the temperature of the air increase after it is passed through the compression chamber?(c) Why is it important to separate out water vapour and carbon-dioxide in the separating chamber?(d) Why is the cooled air allowed to expand quickly into a chamber through a jet?

Q. No.3

Assertion: Evaporation is a better technique of separation than crystallisation.Reason: Some solids may decompose on heating.

° ° ° °

°

°

° °


Q. No.4

Assertion: Colloidal solutions cannot be separated by filtration.Reason: Size of colloidal particles is greater than the pore size of filter paper.


Q. No.5

A solution is made by mixing 30 g of KCl in enough amount of water, in order to make a solution weighing 140 g.Calculate the concentration of the solution in terms of mass by mass percentage.

A. 21.42%B. 22.72%C. 24.12% D. 27.27%

Q. No.6

A chemistry student placed an empty beaker on a weighing machine. As a result, the weighing machine showed areading of 100 g. On adding water to this empty beaker, the reading changed to 600 g. Then the student added 100 gof NaOH pellets into the beaker. Find the concentration of the resulting solution in terms of mass by masspercentage?

A. 16.67%B. 0.16%C. 20.00%D. 0.20%

Q. No.7

Which of the following example(s) have the dispersed phase in the liquid state?(i) Sponge(ii) Mist(iii) Cheese(iv) Milk of magnesia

A. Only (i)B. Only (ii)C. Both (ii) and (iii)D. Only (iv)

Q. No.8

Find the incorrect property of suspensions, amongst the below-mentioned properties.

A. Suspension is a heterogeneous mixture.B. The particles of a suspension can be seen by the naked eye.C. The solute particles settle down when a suspension is left undisturbed, that is, a suspension is stable.D. The particles of a suspension scatter a beam of light passing through it and make its path visible.

Q. No.9

Which of the following is an emulsion?

A. Shaving creamB. Face creamC. PumiceD. Butter

Q. No.10

Define mixture.

Is Matter Around Us Pure?_SolutionsSolution 1

(a) The maximum amount of solute that can be dissolved in 100 g of a solvent at a fixed temperature, in order to form asaturated solution is called the solubility of that solute in that solvent at that temperature. Pressure has no effect on the solubility of solids in liquids. (b) (i) AlCl3 (ii) Ba(NO3)2 (c) The mass of crystals of calcium chloride that would separate out of the solution, if the saturated solution of calciumchloride at 40 C is allowed to cool to room temperature, i.e., (20 C) is : 128 g - 74.50 g = 53.50 g Hence, the correct answer is option (iii). (d) The solubility of a solid in liquid increases with the increase in temperature. Hence, the correct answer is option (ii).

Solution 2

(a) Passing the gas through filter helps to remove the dust particles present in the air, that would otherwise cause ahindrance in the distillation process. (b) When we increase the pressure of a fixed mass of gas (air in this case), at a fixed volume, the temperature of the gasincreases. This is because the pressure of a fixed mass of gas at constant volume is directly proportional to temperature. (c) Water vapour would convert into ice and carbon-dioxide would convert into dry-ice, thus, blocking the pipeline. Hence,it is important to remove them, prior to feeding the liquid air into the fractional distillation column. (d) When liquid air is allowed to expand quickly into the fractional distillation column, through the jet, it cools down the gaseven more.

Solution 3

Crystallisation is a better technique of separation than evaporation. This is because some solids may decompose onheating. Therefore, Assertion is false but Reason is true. Hence, the correct answer is option D.

Solution 4

Colloidal solutions cannot be separated by filtration. This is because the size of colloidal particles is smaller than the poresize of filter paper. Therefore, Assertion is true but Reason is false. Hence, the correct answer is option C.

Solution 5

Given: Mass of solute (KCl) = 30 g Mass of solution = 140 g We know,

° °



Solution 6


Solution 7

The dispersed phase of sponge is gas. The dispersed phase of mist is liquid. The dispersed phase of cheese is liquid. The dispersed phase of milk of magnesia is solid. Hence, the correct answer is option C.

Solution 8

The solute particles settle down when a suspension is left undisturbed, that is, a suspension is unstable. Hence, the correct answer is option C.

Solution 9

Face cream is an emulsion, having both dispersed phase and dispersing medium in the liquid phase. Hence, the correct answer is option B.

Solution 10

A Mixture is formed due to the combination of more than one kind of pure substance (element/compound).

Mass percentage of solution= × 100

= × 100

=21. 42%

Mass of solute

Mass of solution

30

140

Mass of beaker + mass of water = 600 g

∴ Mass of water (solvent)= 500 g

Mass of NaOH pellets (solute) = 100 g

Mass % of solution = × 100

= × 100

= 16. 67%

mass of solute

mass of solution

100

500 + 100


Atoms and Molecules_Questions

Q. No.1

Study the two given tables related to the cleansing agent (soaps) and answer the following questions. Table A

S.No. Type of Soap Molecular formula Molecular Mass1. Sodium palmitate 278 amu2. Sodium stearate 306 amu3. Sodium oleate 304 amu Table B

S.No. Type of SoapRatio by mass of

combining elements (C:H:O:Na)

Mass % ofCarbon

Mass % ofHydrogen

Mass % ofOxygen

Mass % ofSodium

1 Sodium palmitate "X" ? 11.15% 11.51% 8.27%2 Sodium stearate "W" 70.58% 11.43% 10.45% 7.51%3 Sodium oleate "Y" 71.05% 10.85% 10.52% ?

(a) The mass % of carbon in sodium palmitate is:(i) 69.60%(ii) 67.66%(iii) 4.31%(iv) 69.06%(b) The mass % of sodium in sodium oleate is :(i) 7.65%(ii) 7.56%(iii) 56.7%(iv) 65.7%(c) The values of "X", "W" and "Y", respectively are :(i) 192:31:32:23, 216:35:32:23, 216:33:32:23(ii) 192:31:32:23, 216:33:32:23, 216:35:32:23(iii) 216:33:32:23, 216:35:32:23, 192:31:32:23(iv) 216:33:32:23, 192:31:32:23, 216:35:32:23(d) The ratio by mass of the combining atoms in sucrose in the order C:H:O is:(i) 72:11:88(ii) 6:12:6(iii) 12:22:11(iv) 6:1:8

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. X, Y and Z are three very close friends, studying in three different schools in three different parts of the country. Theteacher at their respective schools just finished teaching the chapter "atoms and molecules". Being extremely inclinedtowards the chemistry subject all of them decided to calculate the ratio by mass of the combining elements in tapwater and convey the result to each other via video call.(a) Do you think all of them would have obtained the same result? Why?(b) If yes, then name and state the underlying principle behind this phenomenon.(c) Name the scientist who gave this principle. (d) Do you think the result would have been the same in case the three friends would have selected any othercompound, instead of water?

C15H31 COONa

C17H35 COONa

C17H33 COONa


Q. No.3

Assertion: One mole of ozone gas contains 6.022 1023 number of oxygen atoms.Reason: One ozone molecule contains three oxygen atoms.

A. Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: The chemical formula of aluminium oxide is Al2O3 and not AlO.Reason: Aluminium ion is a divalent ion and not a trivalent ion.


Q. No.5

Which of the following has the maximum number of oxygen atoms?(i) 1 g of O atoms(ii) 1 g of O2 gas(iii) 1 g of O3 gas

A. (i)B. (ii)C. All of the above. D. (iii)

Q. No.6

The molar mass of which of the following compounds have been calculated incorrectly?(i) The molar mass of nitric acid (HNO3) is 69 g/mol.(ii) The molar mass of glucose (C6H12O6) is 188 g/mol. (iii) The molar mass of sugar (C12H22O11) is 362 g/mol.(iv) The molar mass of sodium hydrogen carbonate (NaHCO3) is 82 g/mol.

A. Both (i) and (ii)B. Both (ii) and (iii)C. Both (iii) and (iv)D. All of the above.

Q. No.7

×


X went to the market to purchase 3 moles of table sugar (C12H22O11). If the cost of sugar is Rs 40/kg, then find thecost of 3 moles of sugar.

A. Rs 41B. Rs. 68 C. Rs. 80D. Rs. 120

Q. No.8

Calculate the total number of oxygen atoms present in 1.1 g of CO2 gas.

A. 3.011 1023

B. 3.011 1022

C. 6.022 1023

D. 12.044 1023

Q. No.9

The mass of an atom of a monoatomic element 'P' is . Find the molar mass of the element.

A. 63.47 g/molB. 68.92 g/molC. 64.75 g/molD. 66.94 g/mol

Q. No.10

Find the atomicity of a phosphorus molecule.

×

×

×

×

10. 54 × 10−23 g


Atoms and Molecules_SolutionsSolution 1

The correct table with all the solutions has been provided below:

S.No. Type of SoapRatio by mass of combining

elements (C:H:O:Na)

Mass % ofCarbon

Mass % of Hydrogen

Mass % ofOxygen

Mass % ofSodium

1 Sodium palmitate "X" = 192:31:32:23 69.06% 11.15% 11.51% 8.27%2 Sodium stearate "W" = 216:35:32:23 70.58% 11.43% 10.45% 7.51%3 Sodium oleate "Y" = 216:33:32:23 71.05% 10.85% 10.52% 7.56%

(a) The mass % of carbon in sodium palmitate is 69.06% Hence, the correct answer is option (iv) (b) The mass % of sodium in sodium oleate is 7.56% Hence, the correct answer is option (ii) (c) The values of "X", "W" and "Y", respectively are 192:31:32:23, 216:35:32:23, 216:33:32:23 (d) The ratio by mass of the combining atoms in sucrose in the order C: H: O is 72:11:88

Solution 2

(a) The formula of water is H2O. In H2O, the combining atoms are hydrogen and oxygen. The ratio by mass of the combining elements is equal to 1:8. Hence, all three friends are going to obtain the same result. (b) The underlying principle behind this phenomenon is the Law of definite proportion. According to this law, irrespective of the source of a compound and irrespective of the fact whether it is naturally obtained orchemically synthesized, the ratio or proportion in which the individual elements combine to form the compound remains the same. (c) The Law of definite proportion was given by a scientist named Proust. (d) Yes, the result would have been the same even if the three friends had selected any other compound. This is because the ratioby mass in which the constituent elements of a compound combine is always fixed, irrespective of its source.

Solution 3

1 mole of ozone gas contains 3 6.022 1023 number of oxygen atoms, i.e, 3 NA number of oxygen atoms. This is because 1ozone molecule contains 3 oxygen atoms. Therefore, Assertion is false but Reason is true. Hence, the correct answer is option D.

Solution 4

Aluminium ion is a trivalent ion as it is having 3+ charge. Oxygen ion has 2- charge, thus, aluminium ion and oxygen ion combinetogether to form Al2O3. Therefore, Assertion is true but Reason is false. Hence, the correct answer is option C.

Solution 5

× × ×


Solution 6

The correct molar mass of the mentioned compounds have been listed below: (i) The molar mass of nitric acid (HNO3) is 63 g/mol. (ii) The molar mass of glucose (C6H12O6) is 180 g/mol. (iii) The molar mass of sugar (C12H22O11) is 342 g/mol. (iv) The molar mass of sodium hydrogen carbonate (NaHCO3) is 84 g/mol. Hence, the correct answer is option D.

Solution 7

Mass of 1 mole of table sugar = 342 g ( the molar mass of C12H22O11 is 342 g/mol) The mass of 3 moles of table sugar = 3 342 = 1026 g

Cost of 1000 g (1 kg) of sugar = Rs 40

Cost of 1026 g of sugar =

= Rs 41.04 Thus, the cost of 3 moles of sugar = Rs 41 (approximately) Hence, the correct answer is option A.

Solution 8

1 molecule of CO2 = 2 atoms of O

6.022 1023 molecules of CO2 (44 g of CO2) = 12.044 1023 atoms of O

1.1 g of CO2 = atoms of O

= 3.011 1022 atoms of O Thus, the total number of atoms present in 1.1 g of CO2 = 3.011 1022 Hence, the correct answer is option B.

Solution 9

The mass of 1 atom of element P = 10.54 10-23 g The mass of 6.022 1023 atoms of the element P = = 63.47 g The molar mass of element P = 63.47 g/mol


Solution 10

A phosphorus molecule is tetratomic (polyatomic). Thus, the atomicity of a phosphorus molecule is 4.

∵

∴ ×

∴

× 102640

1000

× ×

∴ × 1. 112.044×10

23

44

×

×

×

∴ × 10. 54 × 10−23

× 6. 022 × 1023

∴

Structure of the Atom_Questions

Q. No.1

Study the given table related to the isotopes and answer the following questions.Assume the following: 1. Mass of electron is negligible 2. Mass of proton = Mass of neutron Table A (Isotopes of Calcium)

Isotope Number Total mass of nucleons(g)

Abundance(%)

1st 96.941

2nd 0.647

3rd 0.135

4th 2.086

5th 0.004

6th 0.187

Table B (Isotopes of Silicon) IsotopesNumber

Total mass of nucleons (g)

Abundance (%)

1st 92.230

2nd 4.683

3rd 5. 01 × 10−23

3.087

(a) Using Table A find the average atomic mass of calcium.(b) Using Table B find the average atomic mass of silicon.(c) Chlorine occurs in nature mainly in the form of two isotopes, namely, and . Out of every 500 chlorine atoms, 375atoms will be of the type . The average atomic mass of chlorine is(i) 35 u(ii) 37 u (iii) 35.75 u (iv) 35.5 u(d) Silicon occurs in nature in mainly three isotopic forms, namely, . The relative abundance of is 92.23% and thatof is 4.67%. The average atomic mass of silicon is (i) 28 u(ii) 29 u (iii) 30 u (iv) 28.10 u

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. Metals like sodium and potassium can easily catch fire if they come in contact with air or water. This is the reason whythey are stored under kerosene. Like these metals, there are also a few non-metals that are highly reactive. X is anexample of such a type of highly reactive non-metal. As X can easily ignite at room temperature, it is storedunderwater. A form of X is widely used on the striking surface of a matchbox. (a) Find X.(b) Write the valency of X.(c) Write it's electronic configuration.Typesetting math: 100%

(d) Find the total number of nucleons present in an atom of X.

Q. No.3

Assertion: Mass number is calculated by ignoring the mass of electrons.Reason: The mass of an electron is approximately 2000 times the mass a proton.


Q. No.4

Assertion: Most of the alpha particles in Rutherford's experiment passed straight through the gold foil.Reason: Most of the space inside an atom is empty.


Q. No.5

Which of the following pair of species is not having an equal number of electrons?(i) N3- and O2-

(ii) F- and Na+

(iii) N3- and Na+

(iv) Mg2+ and Ne

A. (i)B. (ii)C. (iii)D. None of the above

Q. No.6

Which of the following element is having multiple valencies?

A. Phosphorus B. NeonC. BerylliumD. Silicon

Q. No.7Typesetting math: 100%

Calculate the total number of protons present in 1.4 mg of N2 gas.

A. 2.101020 protonsB. 4.211020 protons C. 8.431020 protonsD. 6.0221023 protons

Q. No.8

Which of the following statement is incorrect in context with the conclusions/observations made by Rutherford, afterhis alpha particle scattering experiment?

A. Most of the alpha particles passed undeviated, as most of the space inside an atom is empty. B.Some of the particles were deviated by large angles as the positive charge is present in a small region, and not uniformlyspread all over (in accordance with J.J. Thompson's atomic model). C.Very few (about 1 in 20,000) particles were deflected back by large angles in the range of , indicating that the size of thepositively charged body is negligible as compared to the total size of the atom. D.Most of the space inside an atom is empty. Electrons revolve around the nucleus in this space, with very high speed, incircular paths, just like planets revolve around the sun.

Q. No.9

Four statements regarding the application of isotopes have been stated as follows:(i) An isotope of uranium is used as a fuel in nuclear reactors.(ii) An isotope of iodine is used in the treatment of cancer.(iii) An isotope of cobalt is used in the treatment of goitre.(iv) An isotope of sodium is used to detect the presence of blood clots.Which of the following statements mentioned above is incorrect?

A. Both (i) and (ii)B. (i), (ii) and (iv)C. Only (iii)D. Both (ii) and (iii)

Q. No.10

Define isotopes.


Structure of the Atom_SolutionsSolution 1

(a)

Number of nucleons = Total mass of nucelons in g

Mass of 1 proton in g or Mass of 1 neutron in g =Total mass of nucleons in g

1 . 67 × 10 - 24 g

=

Table A (Isotopes of Calcium)

IsotopeNumber

Total mass of nucleons (g) Number of Nucleons Abundance

(%)1st 6. 68 × 10 - 23 40 96.941

2nd 7. 014 × 10 - 23 42 0.647

3rd 7. 181 × 10 - 23 43 0.135

4th 7. 348 × 10 - 23 44 2.086

5th 7. 682 × 10 - 23 46 0.004

6th 8. 016 × 10 - 23 48 0.187

Average atomic mass of Calcium =

1

100 × ( (96. 941 × 40) + (0. 647 × 42) + (0. 135 × 43) + (2. 086 × 44) + (0. 004 × 46) + (0. 187 × 48) ) = 40.1156 u

(b) Number of nucleons = Total mass of nucelons in g

Mass of 1 proton in g or Mass of 1 neutron in g =Total mass of nucleons in g

1 . 67 × 10 - 24 g

=

Table B (Isotopes of Silicon)

IsotopesNumber

Total mass of nucleons (g)

Numberof

Nucleons

Abundance (%)

1st 4. 676 × 10 - 23 28 92.230

2nd 4. 843 × 10 - 23 29 4.683

3rd5. 01 × 10 - 23

5. 01 × 10−23 30 3.087

Average atomic mass of silicon = 1

100 × 92. 230 × 28 + 4. 683 × 29 + 3. 087 × 30

= 28.1085 u

(c) The abundance of Cl3517 isotope =

375

500× 100 = 75%

The abundance of Cl37

17 isotope = (100 - 75)% = 25%

∴ The average atomic mass of chlorine atom = 75

100 × 35 + 25

100 × 37 = 35. 5 u

Total mass of nucelons in g

Mass of 1 proton in g or Mass of 1 neutron in g

Total mass of nucleons in g

1.67×10−24 g

Total mass of nucelons in g

Mass of 1 proton in g or Mass of 1 neutron in g

Total mass of nucleons in g

1.67×10−24 g

( ( ) ( ) ( ) )

Hence, the correct answer is option (iv) (d) The relative abundance of Si30

14 = 100-(92.23 + 4.67) = 3.1%

∴ The average atomic mass of Silicon = 92 . 23

100× 28 +

4 . 67

100× 29 +

3 . 1

100× 30 = 28. 10 u

Therefore, the average atomic mass of Silicon = 28.10 u Hence, the correct answer is option (iv)

Solution 2

(a) X is phosphorus. It is a highly reactive non-metal, stored underwater. Red phosphorus is used on the striking surfaceof a matchbox. (b) X can have multiple valencies, i.e., both 3 and 5. (c) The atomic number of phosphorus is 15. ∴ The electronic configuration of phosphorus is: 2, 8, 5 (d) The mass number of phosphorus is 31. ∴ The total number of nucleons present in an atom of X = 31.

Solution 3

The mass number is calculated by ignoring the mass of electrons because the mass of an electron is negligible ascompared to the mass of a proton. Further, the mass of a proton is approximately 2000 times the mass an electron. Therefore, Assertion is true but Reason is false. Hence, the correct answer is option C.

Solution 4

The reason why most of the alpha particles in Rutherford's experiment passed straight through the gold foil, without anybending or deviation was because most of the space inside an atom is empty. Therefore, both Assertion and Reason are true and Reason is the correct explanation of the Assertion. Hence, the correct answer is option A.

Solution 5

All the mentioned species, N3-, O2-, F-, Na+, Mg2+ and Ne are having an equal number of electrons. All the species are having 10 electrons. Hence the correct answer is option D.

Solution 6

The element phosphorus is having multiple valencies, i.e., 3 and 5. Hence, the correct answer is option A.

Solution 7

1 N atom contains 7 protons ( ∵ atomic number of Nitrogen is 7 ) ∴ 1 molecule of N2 gas will contain 14 protons ( ∵ 1 molecule of N2 gas contains 2 N atoms )

28 g of N2 gas = NA number of molecules of N2 gas ( ∵ molar mass of N2 gas is 28 g/mol)

∴ 1.4 × 10-3 g (1.4 mg) of N2 gas = NA

28× 1. 4 × 10 - 3 number of molecules of N2 gas

⇒

NA

28 × 1. 4 × 10 - 3 molecules of N2 gas = 14 ×NA × 1 . 4 × 10 - 3

28 number of protons ( ∵ 1 molecule of N2 gas = 14

protons) = 4. 21 × 1020 protons Hence, the correct answer is option B.

Solution 8

Some of the particles were deviated by small angles as the positive charge is present in a small region, and not uniformlyspread all over (in accordance with J.J. Thompson's atomic model). Hence, the correct answer is option B.

Solution 9

Iodine isotope is widely used by doctors to find out the rate at which iodine is taken up by the thyroid gland in the body.This helps in the treatment of goitre. An isotope of cobalt is used in the treatment of cancer. Hence, the correct answer is option D.

Solution 10

Isotopes are the atoms of the same element having the same atomic number, but different mass number.

( )

The Fundamental Unit of Life_Questions

Q. No.1

Study the given table related to the cell organelles and answer the following questions. Table-A(cell organelles and presence/absence of membrane )

Organelle Membrane-bound(Yes/no)Ribosomes no

Golgi apparatus yesChloroplast yes

Nucleus yesMitochondria yes

Table-B(cell organelles and their functions)

Cell organelle FunctionRibosomes Site of protein synthesis

Golgi apparatus XChloroplast Pigment required for photosynthesis

Nucleus Site for ribosome synthesisMitochondria Y

(a) What is the name given to the organelle involved in the packaging function?(b) Which organelle contains the genetic material of the cell?(c) Which cell organelle provides energy in the form of ATP for vital activities of the living cell? (i) Endoplasmic reticulum (ii) Mitochondrion (iii) Golgi apparatus (iv) Nucleus(d) Which labeled structure contains chromosomes and plays a central role in cellular reproduction?

(i) I (ii) II (iii) III

(iv) IV

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.In plant cells, there is rigid outer covering called 'X' lying outside another membrane 'Y'. In most of the plants, this'X' is composed of 'Z' providing structural strength to plants.(a) What is 'X'?(b) Write any one function of 'X'.(c) Identify 'Y' and 'Z'.

(d) What is the function of 'Y' in plants?

Q. No.3

Assertion: The presence of a cell wall is a characteristic feature of plant cells.Reason: The plant cell wall is composed of peptidoglycans.

A. Both Assertion and Reason are true and Reason is correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: Eukaryotic cells have more DNA than prokaryotic cells. Reason: Eukaryotes are genetically more complex than prokaryotes.


Q. No.5

Which of the following will result in the swelling of the cell?

A.the concentration of water molecules in the surrounding medium is higher than water molecules concentration inside thecellB. the concentration of water molecules in the cell is higher than the concentration of water molecules outsideC. the concentration of water molecules is the same in the cell and in the surrounding mediumD. the concentration of water molecules does not matter

Q. No.6

Which of the following is the component of chromosomes?

A. RNAB. ProteinC. DNAD. DNA and protein

Q. No.7

Select a pair of appropriate option that is not a function of ribosomes?I. It helps in the manufacture of starch molecules.II. It helps in the manufacture of hormones.III. It helps in the manufacture of enzymes.IV. It helps in the manufacture of protein molecules.

A. II and IIIB. III and IVC. IV and ID. I and II

Q. No.8

Which of the following cell organelle is involved in forming complex sugars from simple sugars?

A. nucleusB. ribosomesC. golgi apparatusD. lysosomes

Q. No.9

"Endoplasmic reticulum is an organelle that consists of membranous tubes, sacs, and flattened channels that form adistinct compartment in the cytoplasm" Read the above statement carefully and select the correct functionthat holds true for endoplasmic reticulum?

A. Protection of cellB. Synthesis of proteinC. Light−energy conversionD. Storage of starch

Q. No.10

Name two structures found only in plant cells and not in animal cells.

The Fundamental Unit of Life_SolutionsSolution 1

(a) The organelle involved in the packaging function is Golgi complex. These are called dictyosomes in plant cells. (b) The nucleus contains chromatin network. Chromatin networks are thread-like structures present inside the nucleus.The chromatin divides during cell division and is seen as rod-like chromosomes. Chromosomes are composed ofDNA(Deoxyribonucleic acid) and proteins. The DNA stores information required for all life processes. The entire segmentof the DNA present is not always functional. The functional regions are called genes. (c) Mitochondria are the sites of cellular respiration. They provide energy in the form of ATP for vital activities of the livingcells. Hence, the correct answer is option (ii). (d) In the given diagram, the label I represent the nucleus. It contains chromosomes, which are visible only at the time ofcell division. The chromosomes contain DNA and proteins. Since DNA contains genes, it plays an essential role in theinheritance of characteristics from parents to offsprings. The nucleus controls all cellular functions and plays a vital role incell reproduction. Hence, the correct answer is option (i) .

Solution 2

(a) In plants, 'X' is cell wall. (b) 'X' (cell wall) is a protective envelop of plant cells that provides them a definite shape and also helps to withstand withthe external media without bursting.(c) 'Y' is plasma membrane and 'Z' is cellulose. (d) Plasma membrane is semi-permeable and it regulates the movement of molecules across the cell.

Solution 3

All plant cells contain a cell wall while animal cells lack cell wall. The plant cell wall is composed of cellulose and itprovides structural strength to the plants. Hence, the correct answer is option C.

Solution 4

In eukaryotic cells, there are membrane-bound organelles and chromosomes present in the nucleus are composed ofDNA and proteins. In prokaryotic cells, there is no membrane-bound organelle and proteins are absent in the nucleus.Therefore, eukaryotic cells have more DNA than prokaryotic cells. Hence, the correct answer is option A.

Solution 5

Osmosis is a spontaneous process where the movement of solvent molecules occurs into a region of higher soluteconcentration from lower solute concentration through a semi-permeable membrane, so as to equalize the soluteconcentration on the two sides. When the movement of the solvent takes place from outside to inside, the process iscalled endosmosis. It occurs in hypotonic solution and causes the swelling of the cell. Hence, the correct answer is option A.

Solution 6

Chromosomes are thread-like structures usually present in the nucleus which becomes visible only during cell division.Each chromosome is made up of two components, DNA and proteins. Hence, the correct option is answer D.

Solution 7

Ribosomes are dense, spherical and granular particles which remain freely in the matrix or remain attached to theendoplasmic reticulum. They are not bounded by a membrane and are present in both prokaryotic and eukaryotic cells.They play an important role in the synthesis of proteins. All enzymes are proteins but all hormones are not proteins.


Solution 8

The Golgi apparatus arises from the membrane of the smooth endoplasmic reticulum, which in turn originates from therough endoplasmic reticulum. They store, modify, package and condense the proteins synthesized in the ribosomes.They also synthesize some polysaccharides for the cell membrane. Hence, the correct answer is option C.

Solution 9

The storage of starch and light-energy conversion is done by plastids in plant cells. Hence, alternatives A and C areincorrect. A cell is protected by its cell wall. Hence, alternative D is incorrect. An endoplasmic reticulum (ER) that hasribosomes attached to it is known as a rough endoplasmic reticulum. It is responsible for the transport of proteins. Hence, the correct answer is option B.

Solution 10

The cell wall and plastids are found only in plant cells and not in animal cells.

Tissues_Questions

Q. No.1

Study the two given tables related to connective tissue and answer the following questions. Table-A(connective tissue and their number or percentage)

Formed element Number or percentageA. Erythrocytes(RBCs) 4-6million/mm3

B. Leucocytes(WBCs) 6000-9000/mm3

(i)Agranulocytes Lymphocytes 30-35%Monocytes 3-7%(ii) Granulocytes Neutrophils 55-60%Eosinophils 2-5%Basophils 0-1%C. Platelets 200,000-400,000/mm 3

(a) What will happen if all the blood platelets are removed from the blood?(b) Which type of WBC is most abundant in lymph?(c) Which of the following proteins is/are present in blood plasma? (i) Albumins (ii) Globulins (iii) Fibrinogen (iv) All of the above (d) Blood is a type of connective tissue. Which of the following statements about the functions of blood is correct? (i) It helps in transporting gases and nutrients. (ii) It helps in connecting bones to muscles. (iii) It helps in the digestion of food. (iv) It helps in the storage of fats.

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts. The growth of plants occurs only in certain specific regions. This is because of the presence of dividing tissue knownas 'X'. These tissues are present at different locations of the plant. New cells produced by meristem are initially likethose of meristem itself, but as they grow and mature, their characteristics slowly change and they becomedifferentiated as parts of other tissues. (a) Identify 'X'.(b) What are the three kinds of meristematic tissues that plants require for complete growth?(c) Write the function of any one of the meristems.(d) Which meristem helps in increment of the girth of stem or root?

Q. No.3

Assertion: Phloem transports the food materials usually from the leaves to other parts of the plant.Reason: Sieve tube elements, companion cells, phloem parenchyma, and phloem fibres are the components ofphloem.

A. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: Stomach and intestine of our body has columnar epithelium.Reason: Secretion and absorption is conferred by the Columnar epithelium.


Q. No.5

Select the correct option of tissue that contains dead cells.

A. SclerenchymaB. ParenchymaC. CollenchymaD. Epithelial tissue

Q. No.6

Which of the following statement is wrong?

A. Collenchymatous tissues are irregularly thickened at corners.B. Parenchymatous tissues have intercellular spaces.C. Meristematic tissues lack vacuoles in their early stages.D. Apical and intercalary meristems are permanent tissues.

Q. No.7

Which of the following meristem is responsible for the increment in girth of the stem?

A. intercalary meristemB. apical meristemC. vertical meristemD. lateral meristem

Q. No.8

Which of the elements does not have perforated cell wall?

A. Companion cellsB. TracheidsC. VesselsD. Sieve tubes

Q. No.9

What type of epithelial cells are responsible for absorbing the digested food materials in the intestine?

A. Cuboidal epitheliumB. Spindle fibresC. Stratified squamous epitheliumD. Columnar epithelium

Q. No.10

Which type of epithelium is present in the organs where exchange of substances takes place?

Tissues_SolutionsSolution 1

(a) Blood platelets are responsible for the release of thromboplastin necessary for blood clotting at the site for injury. Inthe absence of blood platelets, blood clotting will not occur after an injury, and bleeding will continue from the wound. Thismay lead to the death of the injured person. (b) Lymphocytes are WBCs that are most abundant in the lymph. (c) Plasma constitutes almost 55% of the total volume of blood. Soluble proteins such as albumins, globulins, andfibrinogen are present in plasma. Hence, the correct answer is option (iv). (d) Blood is a fluid connective tissue. It helps in transporting gases, hormones, nutrients, and wastes materials to variousbody parts. Hence, the correct answer is option (i).

Solution 2

(a) 'X' is known as meristematic tissue. (b) The three kinds of tissues required by plants for their complete growth are − apical meristem, intercalary meristem,and lateral meristem. (c) Apical meristem − It helps the plant to increase lengthwise by increasing the growth of tips i.e., root tips and stem tips. (d) Lateral meristem helps in increment of the girth of stem or root.

Solution 3

The food prepared as a result of photosynthesis is transported from leaves to different parts of the plant with the help ofphloem. In angiosperm, phloem is composed of sieve tube elements, companion cells, phloem parenchyma, and phloemfibres. Hence, the correct answer is option B.

Solution 4

The columnar epithelium is composed of a single layer of tall and slender cells. They are found in the lining of thestomach and intestine and helps in secretion and absorption. Hence, the correct answer is A.

Solution 5

Sclerenchyma cells are the permanent tissues present in the plants. They provide hardness and stiffness to the plant andare composed of dead cells. This tissue is present in stems around vascular bundles in the veins of leaves and in thehard covering of seeds and nuts. e.g., the husk of the coconut is made up of sclerenchymatous tissues. The cells ofsclerenchyma are long narrow and thickened due to lignin. Hence, the correct answer is option A.

Solution 6

Apical and intercalary meristem are responsible for primary growth (increase in height) and secondary growth (increase indiameter) respectively and are classified under meristematic tissues. Hence, the correct answer is option D.

Solution 7

The girth of the stem increases due to lateral meristematic tissues. They are found beneath the bark (called corkcambium) and in vascular bundles of dicot roots and stems (called vascular cambium) as thin layers. This increase in thediameter and girth of the plant is called secondary growth. Hence, the correct answer is option D.

Solution 8

Tracheids and vessels are xylem elements and are concerned with the transport of water. They are long tube-likestructures with partially or completely dissolved walls to form water pipes (in vessels) and pits in the cell wall (intracheids) for conducting water. Sieve tubes are slender tube-like structures with their end walls perforated by numerouspores and are called sieve plates. Hence, the correct answer is option A.

Solution 9

The columnar epithelium consists of pillar-like cells with their nuclei towards the base. They form the lining of thestomach, small intestine, and colon, forming the mucous membrane. Their main function is absorption (e.g., stomach,intestine) and secretion (e.g., mucous by goblet cells). Hence, the correct answer is option D.

Solution 10

Squamous epithelium is present in the organs where exchange of substances takes place.

Diversity in Living Organisms_Questions

Q. No.1

Study the two given tables related to the classification of plants and answer the following questions. Table-A(Basis of classification and its description)

Basis of classification Description

Differentiation of body Some plants are well-differentiated into roots, stem, andleaves while some plants don't show clear differentiation.

Presence and absence of vascularbundles

Vascular tissues(xylem and phloem) are present in some plant andabsent in others.

Seed producing habit Some plants produce seeds and some don not produce seeds.

Type of seed produced Some plants produce naked seeds and some produce seeds enclosedin fruits.

Table-B(Divisions and examples)

Divisions ExamplesThallophyta Ulothryx, Cladophora, SpirogyraPteridophyta Ferns, Horsetails, and Lycophytes

Bryophyta Liverworts, Hornworts, and MossesAngiosperms Apple, Brinjal, and Guava

Gymnosperms Conifers, Cycads, Ginkgo(a) State any two general features of division Gymnosperms.(b) Name the group of plants that do not contain well-differentiated body components.(c) Bryophytes, as well as pteridophytes, belong to Kingdom Plantae. Pteridophytes differ from bryophytes becausethe former (i) produce seeds (ii) produce flowers (iii) have vascular tissues (iv) have rhizoids(d) Gymnosperms and angiosperms are higher groups of plants. Which structure is present in angiosperms butabsent in gymnosperms? (i) Seeds (ii) Fruits (iii) Leaves (iv) Flowers

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.Eukaryotic organisms are those organisms that consist of membrane-bound organelles. Kingdom 'X' consists of alleukaryotic organisms. Some of these organisms use appendages that are hair-like structures such as cilia andflagella for their movement. They have both kinds of nutrition i.e autotrophic as well as heterotrophic.(a) Identify the kingdom 'X'.(b) Give some examples of organisms included in this kingdom.(c) What is the cellularity of the organisms included in this kingdom?(d) Name any one disease that is caused by organisms of this kingdom.

Q. No.3

Assertion: Two kingdom classification was not sufficient.Reason: The majority of organisms did not fell into either of the categories in two kingdom classification.


Q. No.4

Assertion: Euglena cannot be classified on the basis of the two-kingdom system.Reason: Euglena is a plant as it contains chlorophyll.


Q. No.5

Select the appropriate option amongst the following that has specialized tissue for the conduction of water?(i) Thallophyta(ii) Bryophyta(iii) Pteridophyta (iv) Gymnosperms

A. (i) and (ii)B. (ii) and (iii) C. (i) and (iv)D. (iii) and (iv)

Q. No.6

Which of the following is not a criterion for classification of living organisms?

A. Membrane-bound nucleus and cell organellesB. Ability to produce one’s own foodC. Height of the plantD. The body design of the organism

Q. No.7

Which of the following is not a characteristic of protochordata?

A. Bilateral symmetry and coelomB. Presence of notochord

C. Presence of the circulatory systemD. Jointed legs

Q. No.8

Which of the following scientist introduced the system of scientific nomenclature of organisms?

A. Ernst HaeckelB. Robert WhittakerC. Carolus LinnaeusD. Robert Whittaker

Q. No.9

Ashi and Anand observed an animal in their garden, Ashi called it as an insect while Anand said it was an earthworm.Choose the appropriate character from the following which confirms that it as an insect.

A. Bilateral symmetrical bodyB. Body with little segmentationC. Cylindrical bodyD. Body with jointed legs

Q. No.10

Euglena is also called as plant-animals. Give reason.

Diversity in Living Organisms_SolutionsSolution 1

(a) Two features of division Gymnosperms: (i) Vascular bundles are present, but the xylem lacks vessels and the phloem lacks companion cells. (ii) Naked seeds are produced; they are not enclosed inside the fruits. (b) Thallophyta (algae) are the group of plants that do not contain well-differentiated body components. (c) Pteridophytes have vascular tissues. Bryophytes are lower plants that lack vascular tissues. Hence, the correct answer is option (iii). (d) Angiosperms are plants that produce fruits, whereas gymnosperms do not produce fruits. Hence, the correct answer is option (ii).

Solution 2

(a) 'X' is the Kingdom Protista. (b) Kingdom Protista includes organisms like Amoeba, Parameocium and Euglena. (c) All organisms included in this kingdom are unicellular. (d) Malaria is caused by a protist called Plasmodium.

Solution 3

Two kingdom classification (founded by Linnaeus) failed to distinguish between the eukaryotes and prokaryotes,unicellular and multicellular organisms, photosynthetic and non-photosynthetic organisms. Hence, the two kingdomclassification used for a long time was found inadequate. Hence, the correct answer is option A.

Solution 4

Euglena is studied as a plant as well as an animal. Plant characters of Euglena are the presence of chloroplasts withchlorophyll and holophytic (photosynthetic) nutrition in presence of light while animal characters of Euglena are thepresence of a protein-rich layer(pellicle) and flagella for movement. Hence, the correct answer is option B.

Solution 5

Pteridophyta are the oldest vascular plants and their body is differentiated into on aerial shoot system and anunderground root system. All ferns fall under this group. Both thallophytes and bryophytes are non-vascular plants i.e.,they do not have xylem and phloem. Hence, the correct answer is option D.

Solution 6

Classification deals with the arrangement of organisms or groups of organisms into categories. The living form varies inshape, size, structure, and function, etc. The variation in their appearance, body design, and behaviours form the basis ofclassification. Hence, the correct answer is option C.

Solution 7

Jointed legs are characteristics of arthropods so it is not a character of protochordata. Hence, the correct answer is option D.

Solution 8

Carolus Linnaeus (a Swedish botanist) is considered the father of Taxonomy. He formulated the method of namingorganisms called the binomial system of nomenclature. Hence, the correct answer is option C.

Solution 9

The most striking feature of arthropods i.e., insects is the presence of jointed appendages or the presence of jointed legs. Hence, the correct answer is option D.

Solution 10

Euglena is also called plant-animal because it has characteristics of both plants and animals. Euglena has chloroplasts,so in the presence of sunlight, they perform photosynthesis and are autotrophic. In dark, they are completelyheterotrophic. They lack cell wall but have flagellum, and eyespot, etc.

Why Do We Fall Ill?_Questions

Q. No.1

Study the given table related to the diseases and their causative agents and answer the following questions. Table A( Diseases and their causative agents)

Column I Column IITuberculosis Mycobacterium tuberculosis

Malaria PlasmodiumRingworm Microsporum

Amoebiasis Entamoeba histolyticaAIDS P

Cholera Vibrio cholerae Table B( Diseases and their causative agents)

Disease PathogenSARS Virus

Kala-azar QAcne R

Rabies VirusH1N1 S

(a) What is the difference between symptoms and signs of disease?(b) Tuberculosis is a chronic disease, while SARS is an acute disease. Justify.(c) Which of the following statement is correct for P and Q? (i) P is a bacterial disease while Q is a protozoal disease (ii) P is a viral(retroviral) disease while Q is a protozoal disease (iii) P is a fungal disease while Q is a protozoal disease (iv) P is a viral disease while Q is a fungal disease(d) Select the correct statement for R and S. (i) R is a bacterial disease while S is a viral disease (ii) R is a viral disease while S is a protozoal disease (iii) Ris a fungal disease while S is a protozoal disease (iv) Both R and S are bacterial diseases

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.Any disease is characterized by various signs and symptoms. Diseases can be infectious or non-infectious. Infectiousdiseases are those that spread through infectious agents that include bacteria, viruses, protozoans, and fungi. Non-infectious diseases are not transmissible directly from one person to another. There are medicines that help to inhibitinfections caused by bacteria. These medicines are known as 'X'. They do this by killing the bacteria or by keepingthem from copying themselves or reproducing.(a) Name the protozoan causing kala-azar.(b) Which microorganism is responsible for sleeping sickness?(c) Identify 'X'.(d) Why 'X' does not work against viral infections?

Q. No.3

Assertion: Sleeping sickness is a disease caused by infection with protozoan.Reason: Trypanosoma is a protozoan that causes sleeping sickness.


Q. No.4

Assertion: Leishmania is the protozoan organism that causes acne.Reason: The protozoans are oval-shaped, and each has one long whip-like structure.


Q. No.5

The following are examples of viral diseases except-

A. TyphoidB. InfluenzaC. AIDSD. Dengue

Q. No.6

Which of the following can be transmitted by contact with an infected person?

A. Blood cancerB. High blood pressureC. Genetic abnormalitiesD. Common cold

Q. No.7

Making anti-bacterial drugs is easier than making anti-viral medicines as

A. viruses are on the borderline of living and non-livingB. viruses have very few biochemical mechanisms of their ownC. viruses have a protein coatD. viruses make use of host machinery

Q. No.8

Which of the following is the causative agent of kala-azar?

A. BacteriaB. TrypanosomaC. AscarisD. Leishmania

Q. No.9

Which of the following options is correct regarding vectors?

A. infected personB. diseased plantsC. animals carry the infecting agents from sick person to another healthy personD. microorganisms which cause many diseases

Q. No.10

Mention any two diseases caused by protozoans. What are their causative organisms?

Why Do We Fall Ill?_SolutionsSolution 1

(a) Symptoms are deviations from normal function or feeling which is apparent to a patient, reflecting the presence of anunusual state such as pain, cramps, etc. Signs are the definite indications of a disease that are understood by the doctorand help in diagnosis. (b) Tuberculosis is a chronic disease. A chronic disease lasts for a long time and shows less intense symptoms at thebeginning. The symptoms intensify slowly. Acute disease is a type of disease which though lasts for a short time butshows intense symptoms right from the beginning. SARS and pneumonia are examples of an acute disease. (c) (ii) P is a viral(retroviral) disease while Q is a protozoal disease (d) (i) R is a bacterial disease while S is a viral disease

Solution 2

(a) Leishmania is a protozoan that causes kala-azar. (b) Trypanosoma is a protozoan responsible for sleeping sickness. (c) 'X' is antibiotics. (d) Antibiotics act by blocking certain biochemical pathways that are essential for the survival of bacteria like cell wallformation, protein synthesis, etc. Viruses do not possess any cellular machinery of their own to function. Hence, it isdifficult to find virus-specific targets. Therefore, antibiotics are not effective against viral infections.

Solution 3

Sleeping sickness is a disease caused by the infection of protozoans called Trypanosoma. Hence, the correct answer is option A.

Solution 4

The disease called kala-azar is caused by the protozoans(Leishmania). The protozoans are oval-shaped, and each hasone long whip-like structure. The acne is caused by the infection of bacteria (Staphylococci). Hence, the correct answer is option D.

Solution 5

Typhoid is a disease caused by a bacterial(Salmonella typhi). It is transmitted by the ingestion of contaminated food andwater with faeces of infected persons. While, HIV, dengue and influenza are viral diseases. Hence, the correct answer is option A.

Solution 6

The transmission of microbes can occur through a little droplet, sneezed out by an infected person. A person close to aninfected person can inhale these droplets containing disease-causing microbes and may become infected. Hence, the correct answer is option D.

Solution 7

The virus uses the host cell to replicate. So, it makes difficult to find the exact target where the drug can interfere with thevirus without harming the host cell. Hence, the correct answer is option D.

Solution 8

Leishmania is a protozoan which causes the disease kala-azar. Hence, the correct answer is option D.

Solution 9

Vectors or intermediaries are the ones that carry infectious agents from a sick person to a healthy person. For example,mosquitoes, flies, and rats, etc.


Solution 10

Sleeping sickness and malaria are protozoal diseases caused by Trypanosoma gambiense andPlasmodium, respectively.

Natural Resources_Questions

Q. No.1

Study the two given tables related to biogeochemical cycles and answer the following questions. Table-A(Biogeochemical cycles and biotic process)

Biogeochemical cycles Biotic processesCarbon cycle Photosynthesis, respiration, and decompositionOxygen cycle Photosynthesis, respiration

Nitrogen cycle Biological nitrogen fixation, nitrification, denitrification

Table-B(Biogeochemical cycles and abiotic processes)Biogeochemical cycles Abiotic processes

Carbon cycle The burning of fossil fuelsOxygen cycle CombustionNitrogen cycle Nitrogen fixation by lightning

(a) Define nitrification and denitrification.(b) Name the processes by which water present on the earth’s surface enters the atmosphere during the water cycle.(c) The process that transports water from the surface of Earth to the atmosphere is known as _______________. (i) condensation (ii) precipitation (iii) evaporation (iv) runoff(d) Which of the following agents is responsible for soil erosion? (i) Solar radiations (ii) Soil organisms (iii) River water (iv) Plant roots

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.'X' is a very abundant element on Earth. It is found in the elemental form in the atmosphere to the extent of 'Y'. 'X'occurs in the combined form in the Earth’s crust and also in the air in the elemental form. 'X' from the atmosphere isused up in three processes, namely combustion, respiration and in the formation of oxides of nitrogen. 'X' is returnedto the atmosphere by one major process, that is, photosynthesis.(a) Identify 'X'.(b) What is the significance of 'X' in our body?(c) What amount is 'Y'?(d) Mention any one process that takes place in the absence of 'X'.

Q. No.3

Assertion: Ozone layer depletion causes the rocks to disintegrate directly. Reason: Ozone layer depletion occurs because of greenhouse gases.

A. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.

C. Assertion is true but Reason is false.D. Assertion is false but Reason is true.

Q. No.4

Assertion: Fertilizers and pesticides are harmful to the soil.Reason: They make soil porous and allows water and air to penetrate into the deep underground.


Q. No.5

The presence of which of the following abort the process of nitrogen-fixation by bacteria?

A. dihydrogenB. elemental form of nitrogenC. elemental form of oxygenD. water

Q. No.6

The formation of wind is not influenced by which of the following factors?

A. Variations in temperatureB. Coastal regionsC. Variations in pressureD. Mountain ranges

Q. No.7

What would be affected if there is marked temperature changes in the aquatic environment?

A. availability of nutrientsB. process of digestion in animalsC. more growth of aquatic plantsD. breeding of animals

Q. No.8

The air in the atmosphere is responsible for controlling

A. the amount of rainfall on the earthB. revolution of the Earth around the SunC. the earth’s atmospheric temperature

D. the occurrence of high tides

Q. No.9

The release of pollutants like oxides of nitrogen and sulfur in the atmosphere can result in ______________.

A. floodB. droughtC. acid rainD. tropical cyclone

Q. No.10

What are detritivores? Give anyone example.

Natural Resources_SolutionsSolution 1

(a) The process by which, the nitrogen of the atmosphere transforms into nitrates, is called nitrification. The conversion ofnitrates back into nitrogen is called denitrification. (b) The processes by which water present on the earth’s surface enters the atmosphere are evaporation andtranspiration. (c) Sun’s energy heats up water present in rivers, seas, and oceans. Solar energy turns water into water vapor. Thisprocess is known as evaporation. Hence, evaporation of water transports water from the surface of Earth to theatmosphere. Hence, the correct answer is option (iii). (d) Running water carries soil sediments along with it as it flows. During floods, river water carries away the valuabletopsoil. This process is called soil erosion. Hence, the correct answer is option (iii).

Solution 2

(a) 'X' is oxygen. (b) 'X' is an essential component of most biological molecules like carbohydrates, proteins, nucleic acids and fats (orlipids) in our body. (c) 'Y' is 21% (d) The process of nitrogen-fixation by bacteria takes place in the absence of oxygen.

Solution 3

The main cause of ozone depletion and the ozone hole is manufactured chemicals, especially manufactured halocarbonrefrigerants, solvents, propellants and foam blowing agents (chlorofluorocarbons (CFCs), HCFCs, halons), referred toas ozone-depleting substances (ODS). The ozone layer and the gases that deplete the ozone layer do not cause rocks todisintegrate. Hence, the correct answer is option D.

Solution 4

Fertilizers and pesticides are harmful to the soil as they kill the microorganism involved in the recycling of nutrientsrequired by plants to grow. Hence, the correct answer is option C.

Solution 5

Certain bacteria such as Azotobacter and Rhizobium convert atmospheric nitrogen to water-soluble nitrates. This processis known as nitrogen fixation. There are some enzymes required for the biofixation of nitrogen, called nitrogenases.These nitrogenases are susceptible to be destroyed in aerobic conditions. Some bacteria also stop producingnitrogenase in the presence of oxygen. Hence, the correct answer is option C.

Solution 6

The wind is the movement of air caused by the uneven heating of the Earth by the sun. Coastal regions do not influencethe formation of winds. Hence, the correct answer is option B.

Solution 7

Most of the aquatic animals are cold-blooded, i.e, their body temperature may vary according to the environment.Moreover, the eggs and larvae of aquatic animals are highly susceptible to temperature changes. Hence, the correct answer is option D.

Solution 8

Air is a bad conductor of heat. Therefore, the presence of air in the atmosphere prevents the sudden decrease orincrease in temperatures on the Earth. Hence, the correct answer is option C.

Solution 9

Fossil fuels such as petroleum and coal contain small amounts of sulfur and nitrogen. Nitrogen and sulfur also burn alongwith these fossil fuels. As a result, oxides of nitrogen and sulfur are formed. These oxides react with water vapour in theatmosphere to form acids. These acids fall on the ground as rain. Hence, the correct answer is option C.

Solution 10

The organisms which consume detritus are called as detritivores. For example,earthworm.

Improvement in Food Resources_Questions

Q. No.1

Study the given table related to nutrient management by plants and answer the following questions. Table-A(nutrients in air and water)

nutrients in air nutrients in watercarbon hydrogen oxygen oxygen

Table-B(nutrients in soil)X Y

nitrogen ironphosphorus manganese potassium boron

calcium zincmagnesium copper

sulphur chlorine(a) What will be the effect of deficiency of these nutrients in plants? (b) What measures should be taken to increase the yield of crops?(c) Select the correct statement regarding X and Y. (i) X are macronutrients while Y are micronutrients (ii) X are micronutrients while Y are macronutrients (iii) Both X and Y are micronutrients (iv) Both X and Y are macronutrients(d) Which of the following is the correct definition for X? (i) The nutrients are used by plants in small quantities and are called macronutrients. (ii) The nutrients are used by plants in large quantities and are called micronutrients. (iii) The nutrients are used by plants in large quantities and are called macronutrients. (iv) None of the above

Q. No.2

Answer the given questions on the basis of your understanding of the following paragraph and the related studiedconcepts.Different crops require different climatic conditions, temperatures, and photoperiods for their growth and completion oflife cycle. Photoperiods are related to the duration of sunlight. The growth of plants and flowering is dependent onsunlight. There are some crops, which are grown in the rainy season, called 'X' while some of the crops are grown inthe winter season, called 'Y'.(a) Identify 'X' and give some examples.(b) Identify 'Y' and give some examples.(c) Mention the practices involved in farming.(d) What are the major groups of activities required for improving the yield of crop?

Q. No.3

Assertion: Kharif crops are grown in monsoon season.Reason: Rice, maize, and groundnut are examples of Kharif crops.

A. Both Assertion and Reason are true, and Reason is the correct explanation of the Assertion.B. Both Assertion and Reason are true but Reason is not the correct explanation of the Assertion.C. Assertion is true but Reason is false.

D. Assertion is false but Reason is true.

Q. No.4

Assertion: Macronuntrients are required in large quantities.Reason: Iron, Magnesium, and Boron are examples of macronutrients.


Q. No.5

Which of the following statement is wrong?

A. Increasing food production without compromising with environmental quality is called sustainable agriculture.B. Blue revolution is meant for an increase in fish productionC. White revolution is meant for increase in milk productionD. None of the above

Q. No.6

Read the given statement carefully and answer the question.

Green manures help in increasing the fertility of the soil. Which of the following statements about green manure is correct?

A. Green manure enriches the soil with hydrogen and oxygen.B. Plants such as rose and lantana are used as green manure.C. Green manure enriches the soil with zinc and copper.D. Plants such as guar and sun hemp are used as green manure.

Q. No.7

Which of the following pairs of statements are correct? (i) Crossing between genetically dissimilar plants is called hybridisation. (ii) Interspecific hybridisation is the cross between two varieties. (iii) The genetically modified crop is obtained by introducing genes of the desired character into a plant. (iv) Intervarietal hybridisation is the cross between plants of two species.

A. (ii) and (iv)B. (i) and (iii)C. (ii) and (iii)D. (iii) and (iv)

Q. No.8

Which of the following pair of statements are correct regarding manures?

(i) Large quantities of organic matter and small quantities of nutrients are there in manure.(ii) The water-holding capacity of sandy soil is also increased by manures.(iii) It helps in draining out of the excess of water from clayey soil.(iv) Its excessive use pollutes the environment because it is made of animal excretory waste.

A. (iii) and (iv)B. (ii) and (iii)C. (i) and (ii)D. (i) and (iii)

Q. No.9

Which one of the following species of honey bee is an Indian species?

A. Apis indicaB. Apis melliferaC. Apis floraeD. Apis dorsata

Q. No.10

Name two freshwater fishes of India.

Improvement in Food Resources_SolutionsSolution 1

(a) The deficiency of these nutrients affects physiological processes in plants including reproduction, growth, andsusceptibility to diseases. (b) To increase the yield, the soil can be enriched by supplying these nutrients in the form of manure and fertilizers. (c) (i) X are macronutrients while Y are micronutrients (d) (iii) The nutrients are used by plants in large quantities and are called macronutrients.

Solution 2

(a) 'X' is kharif crops and paddy, soyabean, maize, and cotton are some examples. (b) 'Y' is rabi crops and wheat, gram, peas, and linseed are some examples. (c) The practices involved in farming can be divided into three stages. The first is the choice of seeds for planting. Thesecond is the nurturing of the crop plants. The third is the protection of the growing and harvested crops from loss. (d) The major groups of activities for improving crop yields can be classified as crop variety improvement, productionimprovement, and protection management.

Solution 3

Different crops need different climatic conditions, photoperiod and temperature for the proper growth and development.Kharif crops are grown in the rainy season i.e from June to October. For example, rice, maize, groundnut, cotton, andblack gram, etc. Hence, the correct answer is option B.

Solution 4

Macronutrients are the essential elements that are required by the plant in large quantities(more than 1mg/gm of drymatter). They help in the formation of different cellular structures in plants. Examples of macronutrients are Nitrogen,Phosphorus, Potassium, etc. Iron, magnesium and boron are micronutrients. Hence, the correct answer is option C.

Solution 5

Sustainable agriculture is food production without compromising environmental quality. White Revolution in India, resultedin making India the largest producer of milk and milk products. Blue Revolution rapidly increased fish production in smallponds and water bodies, a boon to small farmers. Hence, the correct answer is option D.

Solution 6

During the preparation of green manure, leguminous plants such as guar and sun hemp are first sown and then mulchedby ploughing. Rhizobium, which is a nitrogen-fixing bacterium, is present in the root nodules of leguminous plants. It fixesnitrogen in the soil. After the formation of green manure, the soil is enriched with nitrogen and phosphorus. Hence, the correct answer is option D.

Solution 7

Hybridization refers to crossing between genetically dissimilar plants. It may be inter varietal, interspecific andintergeneric. In this process, two crops of good characters (desired character) are selected and crossed to obtain a newcrop having desired characters of parental crops. This method of hybridization improves crops with respect to yield,disease resistance, pest resistance, etc. Genetically modified crops (GM crops) are plants used in agriculture, the DNA ofwhich has been modified using genetic engineering methods. Hence, the correct answer is option B.

Solution 8

Manures are advantageous than fertilizers because it improves soil texture and water-retaining capacity of the soil. Sincemanure is not nutrient specific so it replenishes all the nutrients of the soil. It increases the number of useful micro-organisms in the soil. It does not change the chemical nature of soil so it does not harm the fertility of the soil. Hence, the correct answer is option C.

Solution 9

The Indian species of honey bee is Apis indica. Hence, the correct answer is option A.

Solution 10

Rohu and Catla are two freshwater fishes of India.

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