Q1.Which is not a source of fresh water?

(a) Glaciers and ice sheets
(b) Groundwater
(c) Surface run off
(d) Oceans
Answer: (d) Oceans
EXPLANATION: -The water in the oceans is saline and not fresh. The water in theocean collects all thesalt and minerals from all the rivers
that flow into it.

Q2.According to Falkan Mark, water stress occurs when:

(a) water availability is less than 1000
cubic metre per person per day.
(b) there is no water scarcity.
(c) there is flood.
(d) water availability is more than 1000
cubic metre per person per day.
Answer:(a) water availability is less than 1000 cubic metre per person
per day.

EXPLANATION: -According to Falken Mark, a Swedish expert, water stress occurs whenwater
availability is less than1,000 cubic metre per person per day.

Q3.Nangal River Valley Project is made on the river:

(a) Sutlej-Beas
(b) Ravi-Chenab
(c) Ganga
(d) Son
Answer: (a) Sutlej-Beas
EXPLANATION: -Bhakra Dam is a oncrete gravity dam on the Sutlej River in Bilaspur, Himachal Pradesh in northern

Q4.Hirakud Dam is constructed on
the river:
(a) Ganga
(b) Manjira
(c) Manas
(d) Mahanadi
Answer:(d) Mahanadi
EXPLANATION: -Only 15 kms. northo f Sambalpur, the longest earthen
dam of the world stands in its lone majesty across the great river
Mahanadi, which drains an area of 1,33,090 Sq.

Q5.Water of Bhakra Nangal Project is being used mainly for:

(a) hydel power and irrigation
(b) fish breeding and navigation
(c) industrial use
(d) flood control
Answer:(a) hydel power and
irrigation
EXPLANATION: -Its main function is to turn the turbines of power Dam but it also supplies water to the
Bhakra irrigation canals. Power houses have been built to generate hydroelectricity from water of the
Satluj River

Q6. Why is water scarcity mainly caused?

A. Water pollution
B. excessive use and unequal
access to water
C. Water management
D. Using to utility

Q7.The diversion channels seen in the Western Himalayas are called:

(a) Guls or Kuls
(b) Khadins
(c) Johads
(d) Recharge pits
Answer: (a) Guls or Kuls
EXPLANATION: -In hills and mountainous regions,specially in western Himalayas, people build
diversion channels called guls and kuls for irrigational purposes .

Q8.Agricultural fields which are used as rainfed storage structures are called:

(a) Kuls
(b) Khadins/Johads
(c) Recharge pits
(d) None of the above
Answer:(b) Khadins/Johads
EXPLANATION: -In the arid and semi-arid regions, agricultural fields were converted into rain fed storage
structures that allowed the water to sand and moisten the soil, these water storage structures are known
as Khadins in Jaisalmer (Rajasthan) and Johads in other parts of Rajasthan.

Q9.Underground tanks seen in Rajasthan to store rainwater for drinking is called:

(a) Tankas
(b) Khadin
(c) Ponds
(d) Kuls
Answer:(a) Tankas
EXPLANATION: -Taanka, also known as kund or kundi, is the simplest and most common water

Q10.Bamboo drip irrigation system is prevalent in:

(a) Manipur
(b) Meghalaya
(c) Mizoram
(d) Madhya Pradesh
Answer:(b) Meghalaya
EXPLANATION: -Bamboo Drip Irrigation of the Noartheastern Hills

Q11.The only State which has made rooftop rainwater harvesting structure compulsory to all the houses is:

(a) Andhra Pradesh
(b) Karnataka
(c) Tamil Nadu
(d) West Bengal

Answer: (c) Tamil Nadu
EXPLANATION: -Therefore, Tamil Nadu is the correct answer, since Tamil Nadu is the first and the only
state in India, which has made rooftop rainwater harvesting structures and is compulsory to all
the houses across the state.

Q12.The remote village that has earned the rare distinction of being
rich in rainwater?

(a) Gari
(b) Kaza
(c) Gendathur
(d) none of the above
Answer:(c) Gendathur

EXPLANATION: -In Gendathur, a remote backward village in Mysore,
Karnataka,villagers have installed rainwater harvesting system to meettheir water needs

Q13.The major source of fresh water in India is

(a) rainfall
(b) ground water
(c) atmospheric water
(d) ocean water
Answer: (b) ground water

EXPLANATION: -Ground water and surface waters are the major sources of Fresh water in India.
Ground water is the water present beneath the earth’s surface which
can be extracted through various ways.

Q14.Rana Pratap Sagar Dam is located in

(a) Odisha
(b) Uttarakhand
(c) Rajasthan
(d) Andhra Pradesh
Answer: (c) Rajasthan
EXPLANATION: -The Ranapratap
Sagar Dam is a gravity masonry dam
of 53.8 metres (177 ft) height built on
the Chambal River at Rawatbhata in
Rajasthan in India.

Q15.Nagarjuna Sagar Dam is built on which river?

(a) Clenab
(b) Mahanadi
(c) Krishna
(d) Satluj
Answer: (c)Krishna
EXPLANATION: -Nagarjuna Sagar Dam is a masonry dam across the Krishna
River at Nagarjuna Sagar which straddles the border between Nalgonda
district in Telangana and Guntur district in Andhra Pradesh.

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IMPORTANT QUESTION :Water Resources

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1. Question 1. An object is placed at a distance of 0.25 m
   in front of a plane mirror. The distance between the object and image will be
   (a) 0.25 m
   (b) 1.0 m
   (c) 0.5 m
   (d) 0.125 m

Answer: c
Explanation:
(c) Distance between object and image = 0.25 + 0.25 = 0.5 m

 Question 2. The angle of incidence for a ray of light having zero reflection angle is
(a) 0
(b) 30°
(c) 45°
(d) 90°

Answer: a
Explanation:
(a) For reflecting surface ∠i = ∠r

Question 3. For a real object, which of the following can produce a real image?
(a) Plane mirror
(b) Concave mirror
(c) Concave lens
(d) Convex mirror

Answer: b
Explanation:
(b) Only concave mirror can produces a real image for the any position of object between its focus and infinity.

Question 4. Which of the following mirror is used by a dentist to examine a small cavity?
(a) Convex mirror
(b) Plane mirror
(c) Concave mirror
(d) Combination of convex and concave mirror

Answer: c
Explanation:
(c) Concave mirror forms erect and enlarged image when held close to the cavity.

Question 5. An object at a distance of 30 cm from a
concave mirror gets its image at the same point. The focal length of the mirror is
(a) – 30 cm
(b) 30 cm
(c) – 15 cm
(d) +15 cm

Answer: c
Explanation:
(c) When object is placed at 2F, the image formed by concave mirror is also at 2F.
So 2F = -30 or F = -15 cm.

 Question 6. An object at a distance of + 15 cm is slowly moved towards the pole of a convex mirror. The image will get
(a) shortened and real
(b) enlarged and real
(c) enlarge and virtual
(d) diminished and virtual

Answer: d
Explanation:
(d) Convex mirror always formed virtual and diminished image.

Question 7. A concave mirror of radius 30 cm is placed in water. It’s focal length in air and water differ by
(a) 15
(b) 20
(c) 30
(d) 0

Answer: d
Explanation:
(d) The focal length of spherical mirror does not depends on the surrounding medium.

Question 8. A concave mirror of focal length 20 cm forms an image having twice the size of object. For the virtual position of object, the position of object will be at
(a) 25 cm
(b) 40 cm
(c) 10 cm
(d) At infinity

Answer: c
Explanation:
(c) For virtual image,

Question 9. The image formed by concave mirror is real, inverted and of the same size as that of the object. The position of object should be
(a) at the focus
(b) at the centre of curvature
(c) between focus and centre of curvature
(d) beyond centre of curvature

Answer: c
Explanation:
(c) When object lies at C of a concave mirror, image is also formed at ‘C’ and having same size real and inverted.

Question 10. The nature of the image formed by concave mirror when the object is placed between the focus (F) and centre of curvature (C) of the mirror observed by us is
(a) real, inverted and diminished
(b) virtual, erect and smaller in size
(c) real, inverted and enlarged
(d) virtual, upright and enlarged

Answer:
Explanation:
(c) When object lies between C and F, the real, inverted and enlarged image is formed beyondC.

Question 11. The nature of image formed by a convex mirror when the object distance from the mirror is less than the distance between pole and focal point (F) of the mirror would be
(a) real, inverted and diminished in size
(b) real, inverted and enlarged in size
(c) virtual, upright and diminished in size
(d) virtual, upright and enlarged in size

Answer:
Explanation:
(c) Convex mirror always forms a virtual, erect diminished image irrespective of the position of object in front of it.

Question 12. If a man’s face is 25 cm in front of concave shaving mirror producing erect image 1.5 times the size of face, focal length of the mirror would be
(a) 75 cm
(b) 25 cm
(c) 15 cm
(d) 60 cm

Answer: a
Explanation:
(a) In concave shaving mirror, virtual erect and large size image, behind the mirror is obtained, using

Hence, focal length of concave mirror is 75 cm.

 Question 13. As light travels from a rarer to a denser medium it will have
(a) increased velocity
(b) decreased velocity
(c) decreased wavelength
(d) both (b) and (c)

Answer:
Explanation:
(d) When light ray travel from rarer to denser medium, its velocity and wavelength both decrease as v = vλ.

Question 14. The angle of incidence i and refraction r are equal in a transparent slab when the value of i is
(a) 0°
(b) 45°
(c) 90°
(d) depend on the material of the slab

Answer:
Explanation:
(a) When the incident ray falls normally on the glass slab, it will refracted without deviation, i.e. along the normal in the glass slab. So, ∠i = ∠r = 0

 Question 15. The refractive index of transparent medium is greater than one because
(a) Speed of light in vacuum < speed of light in tansparent medium (b) Speed of light in vacuum > speed of light in tansparent medium
(c) Speed flight in vacuum = speed of light in tansparent medium
(d) Frequency of light wave changes when it moves from rarer to denser medium

Answer: b
Explanation:

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1. The sequence of gradual changes which takes place in the primitive organisms over millions of years in which new species are produced is known as

(a) Evolution

(b) Heredity

(c) Generation

(d) Inheritance

Answer. (a) Evolution

Explanation: Evolution is a series of gradual changes in an organism, wherein in the end, the organism is evolved and advanced.

2. If a normal cell of human body contains 46 pairs of chromosomes then the numbers of chromosomes in a sex cell of a human being is most likely to be:

(a) 60

(b) 23

(c) 22

(d) 40

Answer. (b) 23

Explanation: The sex gametes are always haploid. They have 23 chromosomes.

3. In the human blood grouping, the four basic blood types are type A, type B, type AB, and type O. They are:

(a) Simple dominant and recessive traits

(b) Co-dominant traits

(c) Recessive traits

(d) Inherited traits

Ans: (b) Co-dominant traits

Explanation: A co-dominant trait is the one in which neither allele is dominant or recessive and both are expressed equally.

4. A pregnant woman has an equal chance of her baby being blood group A or blood group AB. Which one of the following shows the possible genotypes of the woman and the father of her child?

(a) I^A I^A and I^B I^O

(b) I^A I^B and I^B I^O

(c) I^A I^O and I^B I^O

(d) I^O I^B and I^A I^O

Answer. (a) I^A I^A and I^B I^O

Explanation:

	IB	IO
IA	IA IB  means AB	IA IO means A
IA	IA IB means AB	IA IO means A

Ratio of AB: A= 1:1. Hence, first answer is correct.

5. What will be the number of chromosomes present in each gamete produced by the plants if the palisade cells of a species of plant contain 28 chromosomes in all?

(a) 56

(b) 28

(c) 14

(d) 4

Answer. (c) 14

Explanation: Gamete cells are always haploid.

6. The following results were obtained by a scientist who crossed the F₁ generation of pure-breeding parents for round and wrinkled seeds.

Dominants trait	Recessive trait	No. of F2 offspring
Round seeds	Wrinkled seeds	7524

From these results, it can be concluded that the actual number of round seeds he obtained was:

(a) 1881

(b) 22572

(c) 2508

(d) 5643

Answer. (d) 5643

Explanation: The actual number of seeds obtained were 5643.

7. A cross between a tall plant (TT) and short plant (tt) resulted in progeny that were all tall plants as:

(a) Tallness is the dominant trait

(b) Shortness is the dominant trait

(c) Tallness is the recessive trait

(d) Height of plant is not governed by gene t or t

Answer. (a) Tallness is the dominant trait

Explanation: The best traits are always superior. For instance, tallness and round seeds are dominant traits.

8. In peas, a pure tall plant (TT) is crossed with a pure short plant (tt). The ratio of pure tall plants to pure short plants in F₂ generation will be:

(a) 1 : 3

(b) 3 : 1

(c) 1 : 1

(d) 2 : 1

Answer. (c) 1 : 1

Explanation: The ratio of TT and tt plants of F₂ generation will be the same.

9. In human males, all the chromosomes are paired perfectly except one. These unpaired chromosomes are:

(i) Large chromosome

(ii) Small chromosome

(iii) Y chromosome

(iv) X chromosome

(a) (i) and (ii)

(b) (iii) and (ii)

(c) (iii) and (iv)

(d) (ii) and (iv)

Answer. (c) (iii) and (iv)

Explanation: X and Y are sex chromosomes.

10. Which of the following determines the sex of a child?

(a) The length of the mother’s pregnancy

(b) The length of time between ovulation and copulation

(c) The presence of an X chromosome in an ovum

(d) The presence of a Y chromosome in a sperm

Answer. (d) The presence of a Y chromosome in a sperm

Explanation: The male sex chromosome is the deciding factor for the gender of the child. If the X part fertilizes with the ovum, girl is born. If Y fertilizes, then a boy is born.

11. Which is the one characteristic of the parents that can be inherited by their children?

(a) Deep scar on chin

(b) Snub nose

(c) Technique of swimming

(d) Cut nose

Answer. (b) Snub nose

Explanation: A dominant inherent character can only be inherited by the children.

12. What could be the reason for the fossil of an organism to be found in the deeper layers of the earth?

(a) The extinction of organism has occurred few years back

(b) The extinction of organism has occurred thousands of years ago

(c) The position of fossil in the layers of earth is not related to its time of extinction

(d) Time of extinction cannot be determined.

Answer. (b) The extinction of organism has occurred thousands of years ago

Explanation: The deeper the fossil is embedded in the earth, the more likely it is to be very old.

13. What is the ancient name for all human beings?

(a) Monkey

(b) Chimpanzee

(c) Homo sapiens

(d) Invertebrates

Answer. (c) Homo sapiens

Explanation: The scientific name of human is Homo sapiens.

14. The organs present in two organisms indicate that they are derived from the same ancestor are:

(a) Analogous Organs

(b) Respiratory Organs

(c) Sense organs

(d) Homologous Organs

Answer. (d) Homologous Organs

Explanation: The structures which are similar in their morphology, anatomy, and embryology but dissimilar in their functions are homologous organs.

15. Which of the following pair of organ is not homologous?

(a) Forelimbs in humans and lizard

(b) Forelimbs in lizard and frog

(c) Wings in butterfly and bat

(d) None of these

Answer. (c) Wings in butterfly and bat

Explanation: Wings of a bat and a butterfly are considered as analogous organs.

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1. Which among the following is a unicellular organism that reproduces by budding:

(a) Hydra

(b) Planaria

(c) Yeast

(d) Spirogyra

Answer. (c) Spirogyra

Answer : Yeast is a unicellular organism which reproduces by budding. Spirogyra reproduces by fragmentation; planaria reproduces by regeneration and; hydra (a multicellular organism) reproduces by budding.

2. Which among the following does not reproduce by spore formation:

(a) Penicillium fungus

(b) Yeast fungus

(c) Mucor fungus

(d) Rhizopus fungus

Answer. (b) Yeast fungus

In spore formation method structures called sporangia produces tiny cells called spores. when the spores come in contact with a moist surface , it develops into new individual. example – rhizopus , mucor , penicilium etc. AMOEBA , PARAMAECIUM , PLASMODIUM , YEAST , HYDRA SPYROGYRA ARE NOT PRODUCED BY SPORE FORMATION.

3. The rapid spreading of bread mould on slices of bread are due to:

(i) Presence of large number of spores in air

(ii) Presence of large number of thread-like branched hyphae

(iii) Presence of moisture and nutrients

(iv) Formation of round shaped sporangia

(a) (i) and (iii)

(b) (ii) and (iv)

(c) (i) and (ii)

(d) (iii) and (iv)

Answer. (a) (i) and (iii)

4. The asexual reproduction in the Spirogyra involves:

(a) Breaking up of filaments into smaller bits

(b) Division of a cell into many cells

(c) Division of a cell into two cells

(d) Formation of a large number of buds

Answer. (b) Division of a cell into many cells

Explanation: In spirogyra, asexual reproduction takes place through the process of fragmentation. Spirogyra, a filamentous alga, breaks down its filament into smaller parts, which will grow completely to form a new organism.

5. Reason for the greater similarities among the offsprings produced by asexual reproduction, is:

(i) Asexual reproduction involves only one parent

(ii) Asexual reproduction involves two parents

(iii) Asexual reproduction involves gametes

(iv) Asexual reproduction does not involve gametes

(a) (i) and (ii)

(b) (i) and (iii)

(c) (ii) and (iv)

(d) (i) and (iv)

Answer. (d) (i) and (iv)

6. The process of the division of cell into several cells during reproduction in Plasmodium is termed as:

(a) Fragmentation

(b) Budding

(c) Multiple fission

(d) Binary fission

Answer. (c) Multiple fission

Binary fission is a form of asexual reproduction and cell division where the cell divides into two separate cells.
Budding is a form of asexual reproduction in which a new organism develops from an outgrowth or bud due to cell division at one particular site.
Fragmentation is a form of asexual reproduction in which an organism is split into fragments.
Multiple fission is the process where the nucleus divides into a number of daughter nuclei followed by the division of the cell body into an equal number of parts. This kind of division takes place in Plasmodium.
Thus, the correct answer is option (C), ‘Multiple fission’

7. The number of chromosomes in parents and offsprings of a particular species remains constant due to:

(a) Doubling of chromosomes after zygote formation

(b) Halving of chromosomes during gamete formation

(c) Doubling of chromosomes after gamete formation

(d) Halving of chromosomes after gamete formation

Answer. (b) Halving of chromosomes during gamete formation

Answer: (b) The number of chromosomes in parents and offsprings of a particular species remains constant due to halving of chromosome during gamete formation. The gametes are special type of cells which contain only half the amount of DNA as compared to normal cells of an organism.

8. A Planaria worm is cut horizontally in the middle into two halves P and Q such that the part P contains the whole head of the worm. Another Planaria worm is cut vertically into two halves R and S in such a way that both the cut pieces R and S contain half head each. Which of the cut pieces of the two Planaria worms could regenerate to form the complete respective worms?

(a) Only P

(b) Only R and S

(c) P, Rand S

(d) P, Q, R and S

Answer. (d) P, Q, R and S

In Planaria, each body piece can regenerate into a full, new organism.

9. The number of chromosomes in both parents and offsprings of a particular species remains constant because:
(a) Chromosomes get doubled after zygote formation

(b) Chromosomes get doubled after gamete formation

(c) Chromosomes get halved during gamete formation

(d) Chromosomes get halved after gamete formation

Answer. (c) Chromosomes get halved during gamete formation

Explanation: The number of chromosomes in parents and offsprings of a particular species remains constant due to the halving of chromosomes under the meiosis process during gamete formation.

10. The figure given alongside shows the human male reproductive organs. Which structures make sperms and seminal fluid?

(a) V makes sperms and X makes seminal fluid

(b) W makes sperms and Y makes seminal fluid

(c) X makes sperms and W makes seminal fluid

(d) Y makes sperms and V makes seminal fluid

Answer. (d) Y makes sperms and V makes seminal fluid

Y represents the testes that produce sperms and V represents the prostate gland that produces seminal fluid.

11. An organism capable of reproducing by two asexual reproduction methods one similar to the reproduction in yeast and the other similar to the reproduction in Planaria is:

(a) Spirogyra

(b) Hydra

(c) Bryophyllum

(d) Paramecium

Answer. (b) Hydra

Hydra can reproduce by the method of budding similar to yeast, and also by regeneration as in the case of Planaria.

12. Among the following select the statements that are true regarding the sexual reproduction in flowering plants?

(i) Fertilisation is a compulsory event

(ii) It always results in the formation of zygote

(iii) Offsprings formed are clones

(iv) It requires two types of gametes

(a) (i) nad (iv)

(b) (i), (ii) and (iii)

(c) (i), (ii) and (iv)

(d) (ii), (iii) and (iv)

Answer. (c) (i), (ii) and (iv)

13. Which among the following are not the functions of testes at puberty?

(i) Formation of germ cells

(ii) Secretion of testosterone

(iii) Development of placenta

(iv) Secretion of estrogen

(a) (i) and (ii)

(b) (i) and (iii)

(c) (ii) and (iv)

(d) (iii) and (iv)

Answer. (d) (iii) and (iv)

14. Which out of the following processes does not lead to the formation of clones:

(a) Fertilisation

(b) Fission

(c) Tissue culture

(d) Fragmentation

Answer. (a) Fertilisation

Fertilisation is the fusion of two gametes in a sexual reproduction. It produces genetically different offsprings and does not lead to the formation of clones.

15. The ratio of number of chromosomes in a human zygote and a human sperm is:

(a) 2 : 1

(b) 3 : 1

(c) 1 : 2

(d) 1 : 3

Answer. (a) 2 : 1

Therefore, 23 chromosomes (sperm cell, n) + 23 chromosomes (egg cell, n) = 46 chromosomes (zygote); or 23 pair of chromosomes (zygote, 2n). The answer is 2:1

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1. The movement of a plant part in response to the force of attraction exerted by the earth is called:

(a) Hydrotropism

(b) Geotropism

(c) Chemotropism

(d) Phototropism

Answer. (b) Geotropism

2. A big tree falls in a forest, but its roots are still in contact with the soil. The branches of this fallen tree grow straight up (vertically). This happens in response to:

(a) Water and light

(b) Water and minerals

(c) Gravity and water

(d) Light and gravity

Answer. (d) Light and gravity

3. The main function of the plant hormone called abscisic acid is to:

(a) Increase the length of cells

(b) Promote cell division

(c) Inhibit growth

(d) Promote growth of stem and roots

Answer. (c) Inhibit growth

4. The growth of tendrils in pea plants is due to the:

(a) Effect of sunlight on the tendril cells facing the sun

(b) Effect of gravity on the part of tendril hanging down towards the earth

(c) Rapid cell division and elongation in tendril cells that are away from the support

(d) Rapid cell division and elongation in tendril cells in contact with the support

Answer. (c) Rapid cell division and elongation in tendril cells that are away from the support

5. The plant hormone which triggers the fall of mature leaves and fruits from the plant body is:

(a) Auxin

(b) Gibberellin

(c) Abscisic acid

(d) Cytokinin

Answer. (c) Abscisic acid

6. The stimulus in the process of thigmotropism is:

(a) Touch

(b) Gravity

(c) Light

(d) Chemical

Answer. (a) Touch

7. A growing seedling is kept in a dark room. A burning lamp is placed near to it for a few days. The top part of seedling bends towards the burning candle. This is an example of:

(a) Chemotropism

(b) Hydrotropism

(c) Phototropism

(d) Geotropism

Answer. (c) Phototropism

8. Dandelion flowers open the petals in bright light during the daytime but close the petals in dark at night. This response of dandelion flowers to light is called:

(a) Phototropism

(b) Thigmonasty

(c) Chemotropism

(d) Photonasty

Answer. (d) Photonasty

9. The number of pairs of nerves which arises from the spinal cord is:

(a) 21

(b) 31

(c) 41

(d) 51

Answer. (b) 31

10. Iodine is necessary for the synthesis of which of the following hormone?

(a) Adrenaline

(b) Auxin

(c) Thyroxine

(d) Insulin

Answer. (c) Thyroxine

11. Which of the following controls the involuntary actions in the body?

(a) Medulla in forebrain

(b) Medulla in hindbrain

(c) Medulla in spinal cord

(d) Medulla in midbrain

Answer. (b) Medulla in hindbrain

12. Which of the following control and regulate the life process?

(a) Reproductive and endocrine systems

(b) Respiratory and nervous systems

(c) Endocrine and digestive systems

(d) Nervous and endocrine systems

Answer. (d) Nervous and endocrine systems

13. A doctor advised a person to take injection of insulin because his:

(a) Blood pressure was high

(b) Heart beat was high

(c) Blood sugar was high

(d) Thyroxine level in blood was high

Answer. (c) Blood sugar was high

14. The dramatic changes in body features associated with puberty are mainly because of the secretions of:

(a) Estrogen from testes and testosterone from ovary

(b) Estrogen from adrenal gland and testosterone from pituitary gland

(c) Testosterone from testes and estrogen from ovary

(d) Testosterone from thyroid gland and estrogen from pituitary gland

Answer. (c) Testosterone from testes and estrogen from ovary

15. Electrical impulse travels in a neuron from:

(a) Dendrite → axon → axon end → cell body

(b) Cell body → dendrite → axon → axon end

(c) Dendrite → cell body → axon → axon end

(d) Axon end →axon → cell body → dendrite

Answer. (c) Dendrite → cell body → axon → axon end

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Q1. A dice is thrown. Find the probability of getting an even number.

(A) 2/3

(B) 1

(C) 5/6

(D) 1/2

Answer:  (D)

Explanation: Total number of cases = 6 (1,2,3,4,5,6)

There are three even numbers 2,4,6

Therefore probability of getting an even number is:

P (even) = 3/6

⇒ P (even) = 1/2

Q2. Two coins are thrown at the same time. Find the probability of getting both heads.

(A) 3/4

(B) 1/4

(C) 1/2

(D) 0

Answer:  (B)

Explanation: Since two coins are tossed, therefore total number of cases = 2² = 4

Therefore, probability of getting heads in both coins is:

∴ P (head) = 1/4

Q3. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:

(A) 1/10

(B) 3/10

(C) 1/9

(D) 4/9

Answer:  (C)

Explanation: Total cases = 36

Total cases in which sum of 9 can be obtained are:

(5, 4), (4, 5), (6, 3), (3, 6)

∴ P (9) = 4/36 = 1/9

Q4. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.

(A) 3/4

(B) 27/50

(C) 1/4

(D) 29/100

Answer:  (C)

Explanation: Total prime numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

That means 25 out of 100

So probability is:

P (prime) = 25/100

⇒ P (prime) = 1/4

Q5. A bag contains 5 red balls and some blue balls .If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:

(A) 5

(B) 10

(C) 15

(D) 20

Answer:  (B)

Explanation: Let the number of blue balls be x

Then total number of balls will be 5 + x.

According to question,

x/(5 + x) = 2 X (5/5+x)

⇒ x = 10

Q6.  A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

(A) 143/150

(B) 147/150

(C) 1/25

(D) 1/50

Answer:  (B)

Explanation:

P (non-defective bulb) = 1 – P (Defective bulb)

= 1 – (12/600)

= (600 – 12)/600

= 588/600

= 147/150

Q7. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.

(A) 9/100

(B) 1/10

(C) 3/10

(D) 19/100

Answer:  (B)

Explanation: The perfect square numbers between 2 to 101 are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Total numbers from 2 to 101 =100

So probability of getting a card with perfect square number is:

P (perfect square) = 10/100

⇒ P (perfect square) = 1/10

Q8. What is the probability of getting 53 Mondays in a leap year?

(A) 1/7

(B) 53/366

(C) 2/7

(D) 7/366

Answer:  (C)

Explanation: With 366 days, the number of weeks in a year is

366/7 = 52 (2/7)

i.e., 52 complete weeks which contains 52 Mondays,

Now 2 days of the year are remaining.

These two days can be

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are 7 pairs, in which Monday occurs in 2 pairs,

So probability is:

P (53 Monday) = 2/7

Q9. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red suit.

(A) 1/26

(B) 3/26

(C) 7/52

(D) 1/13

Answer:  (A)

Explanation: There are total 4 kings in 52 cards, 2 of red colour and 2 of black colour

Therefore, Probability of getting a king of red suit is:

P (King of red suit) = 2/52

⇒ P (King of red suit) = 1/26

Q10. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1,2,3……12 ,then the probability that it will point to an odd number is:

(A) 1/6

(B) 1/12

(C) 7/12

(D) 5/12

Answer:  (A)

Explanation: The odd numbers in 1,2,3……..12 are:

1,3,5,7,9,11

Therefore probability that an odd number will come is:

P (odd number) = 6/12

⇒ P (odd number) = 1/2

Q11. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Aryan wins if all the tosses give the same result i.e. three heads or three tails and loses otherwise. Then the probability that Aryan will lose the game.

(A) 3/4

(B) 1/2

(C) 1

(D) 1/4

Answer:  (A)

Explanation: Total outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Favourable outcomes for losing game are

HHT, HTH, THH, HTT, THT, TTH

Therefore probability of losing the game is:

P (Losing the game) = 6/8

⇒ P (Losing the game) = 3/4

Q12. Riya and Kajal are friends. Probability that both will have the same birthday isthe same birthday is:

(A) 364/365

(B) 31/365

(C) 1/365

(D) 1/133225

Answer:  (C)

Explanation:

Riya may have any one of 365 days of the year as her birthday. Similarly Kajal may have any one of 365days as her birthday.

Total number of ways in which Riya and Kajal may have their birthday are:

365 × 365

Then Riya and Kajal may have same birthday on any one of 365 days.

Therefore number of ways in which Riya and Kajal may have same birthday are:

= 365/365 X 365

= 1/365

Q13. A number x is chosen at random from the numbers -2, -1, 0 , 1, 2. Then the probability that x² < 2 is?

(A) 1/5

(B) 2/5

(C) 3/5

(D) 4/5

Answer:  (C)

Explanation: We have 5 numbers −2,−1,0,1,2

Whose squares are 4,1,0,1,4

So square of 3 numbers is less than 2

Therefore Probability is:

P (x² < 2) = 3/5

Q14. A jar contains 24 marbles. Some are red and others are white. If a marble is drawn at random from the jar, the probability that it is red is 2/3, then the number of white marbles in the jar is:

(A) 10

(B) 6

(C) 8

(D) 7

Answer:  (C)

Explanation: Let the number of white marbles be x.

Since only two colour marbles are present, and total probability we know of all the events is equal to 1.

P (white) = 1 – P (red)

x/24 = 1 – (2/3)

⇒ x/24 = 1/3

⇒ x = 8

So there are 8 white marbles.

Q15. A number is selected at random from first 50 natural numbers. Then the probability that it is a multiple of 3 and 4 is:

(A) 7/50

(B) 4/25

(C) 1/25

(D) 2/25

Answer:  (D)

Explanation: The numbers that are multiple of 3(from first 50 natural numbers) are:

3, 6, 9, 12, 15, 18………………..48

The numbers that are multiple of 4 (from first 50 natural numbers) are:

4, 8, 12, 16…………………….48

The numbers that are multiples of 3 and 4 both are the multiples of 3×4=12 as both 3 and 4 are co-prime.

So common multiples are:

12, 24, 36, 48

Therefore probability is:

P (multiple of 3 and 4) = 4/50

⇒ P (multiple of 3 and 4) = 2/25

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(A) lower limits of the classes

(B) upper limits of the classes

(C) midpoints of the classes

(D) frequencies of the class marks.

Answer:  (C)

Explanation: We know that d_i = x_i – a_i. i.ed_i’s are the deviations from the midpoints of the classes.

Q2. While computing mean of the grouped data, we assume that the frequencies are:

(A) evenly distributed over all the classes                                         

(B) centered at the class marks of the classes

(C) centered at the upper limits of the classes

(D) centered at the lower limits of the classes

Answer:  (B)

Explanation: In computing the mean of grouped data, the frequencies are centred at the class marks of the classes

(A) 0

(B) – 1

(C) 1

(D) 2

Answer:  (A)

Explanation:

Q4. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency of a grouped data gives its:

(A) Mean

(B) Median

(C) Mode

(D) All of these

Answer:  (B)

Explanation: Since the intersection point of less than type ogive and more than ogive gives the median on the abscissa.

Q5. For the following distribution,

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of lower limits of median class and modal class is:

(A) 15

(B) 25

(C) 30

(D) 35

Answer:  (B)

Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Now N/2 = 66/2 = 33 which lies in the interval 10 – 15.Therefore lower limit of the median class is 10.

The highest frequency is 20 which lies in the interval 15 – 20. Therefore, lower limit of modal class is 15.

Hence required sum is 10 + 15 = 25

Q6.  If the arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x = ?

(A) 1

(B) 2

(C) 6

(D) 4

Answer:  (D)

Explanation:

According to question

Q7. If the mean of first n natural numbers is 5n/9, then n =?

(A) 6

(B) 7

(C) 9

(D) 10

Answer:  (C)

Explanation:

But according to question,

Q8. If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by:

(A) 2

(B) 1.5

(C) 1

(D) 0.5

Answer:  (D)

Explanation: We have

30, 34, 35, 36, 37, 38, 39, 40

The data has 8 observations, so there are two middle terms, 4^th and 5^th term i.e. 36 and 37.

The median is the mean of both these terms.

Median = (36 + 37)/2

Median = 36.5

When 35 is removed from given data as 30, 35, 36, 37, 38, 39, 40 then the number of observations becomes 7.

Now the median is the middle most i.e 4^th term which is equal to 37.

Therefore median is increased by 37 – 36.5 = 0.5

Q9. The Median when it is given that mode and mean are 8 and 9 respectively, is:

 (A) 8.57

(B) 8.67

 (C) 8.97

(D) 9.24

Answer:  (B)

Explanation: By Empirical formula:

Mode = 3median – 2 mean

8 = 3medain – 2 X 9

8 = 3median – 18

3median = 8 + 18

Median = 26/3

Median = 8.67

(A) 3

(B) 4

(C) 5

(D) 6

Answer:  (D)

Explanation: According to question,

Q11. In a hospital, weights of new born babies were recorded, for one month. Data is as shown:

Weight of new born baby (in kg)	1.4 – 1.8	1.8 – 2.2	2.2 – 2.6	2.6 – 3.0
No of babies	3	15	6	1

Then the median weight is:

(A) 2kg

(B) 2.03kg

(C) 2.05 kg

(D) 2.08 kg

Answer:  (C)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	Cumulative Frequency (cf)
1.4-1.8	3	1.6	3
1.8-2.2	15	2	18
2.2-2.6	6	2.4	24
2.6-3.0	1	2.8	25

Since N/2 = 25/2 = 12.5

12.5 is near to cumulative frequency value 18

So median class interval is 1.8 – 2.2

∴Median = l + [(N/2 – cf)/f]/h

Here

Hence median weight is 2.05 kg.

Q12. In a small scale industry, salaries of employees are given in the following distribution table:

Salary (in Rs.)	4000 – 5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000
Number of employees	20	60	100	50	80	90

Then the mean salary of the employee is:

(A) Rs. 7350

(B) Rs.  7400

(C) Rs. 7450

(D) Rs. 7500

Answer: (C)

Explanation:

Therefore mean is:

Q13.  For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are:

Number of days	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40	TOTAL
Total Number of students	15	16	x	8	y	8	6	4	70

(A) x = 4 and y = 3

(B) x = 7 and y = 7

(C) x = 3 and y = 4

(D) x = 7 and y = 6

Answer:  (D)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi
0-5	15	2.5	37.5
5 – 10	16	7.5	120
10 – 15	x	12.5	12.5x
15 – 20	8	17.5	140
20 – 25	y	22.5	22.5y
25 -30	8	27.5	220
30 – 35	6	32.5	195
35 – 40	4	37.5	150
TOTAL	70		12.5x+22.5y+862.5

mean = (12.5x + 22.5y + 862.5)/70

⇒ 15.5 = (12.5x +22.5y + 862.5)/70

⇒ 15.5 X 70 = 12.5x +22.5y + 862.5

⇒ 12.5x + 22.5y = 222.5

⇒ 125x + 225y = 2225

⇒ 5x + 9y = 89                        …..(i)

Also,

x + y + 57 = 70

x + y = 13         ……(ii)

Multiplying equation (ii) by 5 and then subtracting from (i) as,

Substituting the value of y in equation (ii), we get

x + y = 13

⇒ x + 6 = 13

⇒ x = 7

Hence x = 7 and y = 6

Q14. Pocket expenses of a class in a college are shown in the following frequency distribution:

Pocket expenses	0-200	200-400	400-600	600-800	800-1000	1000-1200	1200-1400
Number of students	33	74	170	88	76	44	25

Then the median for the above data is:

(A) 485.07

(B) 486.01

(C) 487.06

(D) 489.03

Answer:  (C)

Explanation:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi	cf
0-200	33	100	3300	33
200-400	74	300	22200	107
400-600	170	500	85000	277
600-800	88	700	61600	365
800-1000	76	900	68400	441
1000-1200	44	1100	48400	485
1200-1400	25	1300	32500	510
	510		321400

Since N/2 = 510/2 = 255

255 is near to cumulative frequency value 277.

So median class interval is 400-600

Here,

l = 400

N/2 = 255

cf = 107

f = 170

h = 100

Therefore,

Answer:  (B)

Explanation: We have

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1. A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) two cylinders.

Answer: (A)

Explanation: The shape of a sharpened pencil is:

2. A cone is cut through a plane parallel to its base and then the cone that is for medon one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer:  (A)

Explanation: Observe figure

3. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

Answer:  (C)

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

4. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) 2h cm

Answer: (B)

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

5. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8th space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142244

(B) 142396

(C) 142496

(D) 142596

Answer:  (A)

Explanation:

6. A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively, is melted and recast into the form of a cone with base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

Answer:  (B)

Explanation:Since volume will remain same, therefore,

7. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

Answer:  (A)

Explanation: Since volume will remain the same, therefore,

8. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(A) 4πr²

(B) 6πr²

(C) 3πr²

(D) 8πr²

Answer:  (A)

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr² + 2πr² = 4πr²

9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(A) 4πrh + 4πr²

(B) 4πrh − 4πr²

(C) 4πrh + 2πr²           

(D) 4πrh − 2πr²

Answer:  (C)

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr².

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr²

= 4πrh + 2πr²

10. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(A) 4950 cm²

(B) 4951 cm²

(C) 4952 cm²

(D) 4953 cm²

Answer: (A)

Explanation:

Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm²

11. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm³

(B) 0.35 cm³

(C) 0.34 cm³

(D) 0.33 cm³

Answer:  (A)

Explanation:

Since diameter of the cylinder = diameter of the hemisphere = 0.5cm

Radius of cylinder r = radius of hemisphere r = 0.5/2 = 0.25 cm

Observe the figure,

Total length of capsule = 2cm

Capacity of capsule is:

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer:  (C)

Explanation:

Therefore diameter of each solid sphere = 2cm

13. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer:  (A)

Explanation: Since shape of bucket is like Frustum,

Therefore, volume of bucket

14. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

Answer: (D)

Explanation: According to question,

Therefore ratio of surface area is:

15. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is:

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer:  (B)

Explanation: According to question,

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1. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

(A) R₁ + R₂ = R

(B) R₁² + R₂² = R²

(C) R₁ + R₂ < R

(D) R₁² + R₂² < R²

Answer: (B)

Explanation: According to given condition,

Area of circle = Area of first circle + Area of second circle

2. If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

Answer:  (B)

Explanation: According to given condition

Circumference of a circle = Perimeter of square

Hence Area of the circle > Area of the square

3. Area of the largest triangle that can be inscribed in a semi-circle of radius r units, in square units is:

(A) r²

(B) 1/2r²

(C) 2 r²

(D) √2r²

Answer:  (A)

Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.

Consider the following figure:

4. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:

(A) 22:7

(B) 14:11

(C) 7:22

(D) 11:14

Answer: (B)

Explanation: Perimeter of circle = Perimeter of square

5. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(A) 10 m

(B) 15 m

(C) 20 m

(D) 24 m

Answer: (A)

Explanation: Area of first circular park, whose diameter is 16m

= πr² = π (16/2)² = 64π m²

Area of second circular park, whose diameter is 12m

= πr² = π (12/2)² = 36π m²

According to question,

Area of new circular park =

6. The area of the circle that can be inscribed in a square of side 6 cm is

(A) 36 π cm²

(B) 18 π cm²

(C) 12 π cm²

(D) 9 π cm²

Answer:  (D)

Explanation: Given,

Side of square = 6 cm

Diameter of a circle = side of square = 6cm

Therefore, Radius of circle = 3cm

Area of circle

= πr²

= π (3)²

= 9π cm²

7. The area of the square that can be inscribed in a circle of radius 8 cm is

(A) 256 cm²

(B) 128 cm²

(C) 642 cm²

(D) 64 cm²

Answer:  (B)

Explanation: Radius of circle = 8 cm

Diameter of circle = 16 cm = diagonal of the square

Therefore side of square = diagonal/√2

= 16/√2

Therefore, are of square is = (side)² = (16/√2)²

= 256/2

= 128 cm²

8. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is

(A) 56 cm

(B) 42 cm

(C) 28 cm

(D) 16 cm

Answer:  (C)

Explanation: According to question,

Circumference of circle = Circumference of first circle + Circumference of second circle

πD = πd₁ + πd₂

D = 36 + 20

D = 56cm

9. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm respectively, is

(A) 31 cm

(B) 25 cm

(C) 62 cm

(D) 50 cm

Answer:  (D)

Explanation: According to question

Therefore diameter = 2 × 25 = 50cm

10. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then

(A) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

(B) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.

(C) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.

(D) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.

Answer:  (A)

Explanation: According to Question,

11. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

(A) 300

(B) 400

(C) 450

(D) 500

Answer: (D)

Explanation:

12. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:

(A) 154 m²

(B) 156 m²

(C) 158 m²

(D) 160 m²

Answer:  (A)

Explanation:Figure according to question is:

Area of the field in which cow can graze= Area of a sector AFEG

= (θ/360) X πr²

= (90/360) X π (14)²

= (1/4) X (22/7) X 196

= 154 m²

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13. The area of the shaded region in Fig., where arcs drawn with centres P, Q, Rand S intersect in pairs at mid-points A, B, C and D of the sides PQ, QR, RS and SP, respectively of a square PQRS, is:

(A) 25.25 cm²

(B) 27.45 cm²

(C) 29.65 cm²

(D) 30.96 cm²

Answer:  (D)

Explanation:

14. Area of a sector of central angle 120° of a circle is 3π cm². Then the length of the corresponding arc of this sector is:

(A) 5.8cm

(B) 6.1cm

(C) 6.3cm

(D) 6.8cm

Answer:  (C)

Explanation:

Given that

Area of a sector of central angle 120° of a circle is 3π cm²

15. A round table cover has six equal designs as shown in the figure. If the radius of thecover is 28 cm, then the cost of making the design at the rate of Rs. 0.35 per cm² is:

(A) Rs.146.50

(B) Rs.148.75

(C) Rs.152.25

(D) Rs.154.75

Answer:  (B)

Explanation: The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius of circle.

Area of given hexagon = Area of 6 equilateral triangles.

= 6 X (√3/4) X (side)²

= 6 X (√3/4) X (28)²

= 1999.2 cm²                           (Taking √3 = 1.7)        

Area of circle = πr²

= π × 28²

= 2464 cm²

So, area of designed portion = 2464 – 1999.2 = 464.8 cm²

Cost of making design = 464.8 × 0.35

= Rs. 162.68

Important Link

Quick Revision Notes : Areas Related to Circles

NCERT Solution : Areas Related to Circles

MCQs: Areas Related to Circles

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