Author name: Prasanna

These NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Control and Coordination NCERT Solutions for Class 10 Science Chapter 7

Class 10 Science Chapter 7 Control and Coordination InText Questions and Answers

In-text Questions (Page 119)

Question 1.
What is the difference between a reflex action and walking ?
Answer:
Reflex action: “A spontneous; automatic and mechanical response to a stimulus acting on a specific receptor without the will of an animal”. For e.g. blinking of eyes, coughing, yanwing, sneezng etc.

Walking : Hind brain : It has three regions : cerebellum, pons varolii, medulla oblongata.
(a) Cerebellum : maintains the posture and equilibrium.
(b) Pons varolii : Controls some aspects of respiration.
(c) Medulla oblongata: Controls rate of heart-beat, breathing movements, swallowing, coughing, sneezing vomitting etc.

Question 2.
What happens at the synapse between two neurons ?
Answer:
Both axon and dendrites arise from cell body. The functional junction between neurons is called synapse. The synapse is a gap between a pair of adjacent neurons. The functional unit of nervous system is neurons. Cyton contains a nucleus within the cytoplasm and Nissl granules formed of ribo nucleic acid (RNA) and thread like fibres called neurofibrils of cytons within brain and spinal cord called a nucleus.

Question 3.
Which part of the brain maintains posture and equilibrium of the body ?
Answer:
Cerebellum, maintains the posture and equilibrium of the body.

Question 4.
How do we detect the smell of an agarbatti (increase sticks) ?
Answer:
Olfactory Lobes : These receives impulses from olfactor receptors which are organs of smell in the nose with the help of the organs we can detect the smell of an agarbatti.

Question 5.
What is the role of the brain in reflex action ?
Answer:
At the time of reflex action, the action is stimulated by the spinal card, but information is also sent to the brain. The main work is done by the spinal card.

In-text Questions (Page 122)

Question 1.
What are the plant hormones ?
Answer:
The function of control and coordination in plants in performed by chemical substances known as plant hormones. For example, Auxins, Gibberellins, Cytokinin, Abscisic acid and Ethylene, These hormones are plant growth hormones whose function are :

• Auxins: It promotes cell enlargement and cell differentiation in plants. It also promotes fruit growth.
• Cytokinins : It promotes cell-division in plants. It helps in breaking dormancy of seeds and buds. Cytokinins activate the opening of stomata and also promote fruit growth.
• Gibberellins : It promotes cell enlargement and cell differentiation in the absence of auxins It helps in breaking dormancy in seeds and buds and also promotes fruit growth.
• Abscisic Acid : It supports dormancy in seeds and buds. It supports closing of stomata. It also helps in falling of leaves, floral parts of fruits etc.
• Ethylene; It supports in ripening of fruits and also helps in breaking and bud dormancy

Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
The movement of leaves of the sensitive plant is seismonastic movement occur in response to touch (shock) is best seen in ‘Touch-me-not’ plant (Mimosa pudica). It is called ‘Lajwanti’ or ‘Chhuimui’ plant. If we touch the leaves of the Mimosa plant with our finger we find that all its leaves fold up and droop after sometime, the leaves regain their original status. It is called tropic movement. The movement of a shoot towards light is called phototropic movement.

Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Growth promoters: These stimulate the plant growth e.g. auxins, gibberellins, cytokinins and ethylene.

Question 4.
How do auxins promote the growth of a tendril around a support?
Answer:
Auxins generally in habit flowering. Auxins regulate some of the important plant growth movements, e.g. phototropism and geotropism. These promote cell enlargement and cell differentiation in plants. Auxins induce apical dominance, e.g. the influence of apical bud in suppressing the growth of lateral buds. Like the pea plant climb up other plants/fences by means of tendrils. These tendrils are sensitive to touch. When they come in contact with any support the part of the tendril in contact with the object does not grow as rapidly as the part of the tendril away from the object

Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
‘Hydro’ means water. Thus, hydrotropism is response of plants towards water.
To design the experiment–Self doing

In-text Questions (Page 125)

Question 1.
How does chemical coordination take place in animals?
Answer:
The integration of nervous system and hormonal system is brought about by hypothalamus. The hypothalamus is part of the brain. It secretes neurohormones which regulate the secretion of pituitary hormones. The pituitary hormones in term regulate the growth and secretions of their endocrine glands. In this way, both the system control and coordination various activities of the body.

Question 2.
Why is the use of iodized salt advisable ?
Answer:
Iodine is necessary for the thyroid gland to make thyroxin hormone. Thyroxin regulates carbohydrate, protein and fat metabolism in the body so as to provide the best balance for growth. Iodine is essential for the synthesis of thyroxin. Incase of iodine is deficient in our diet, there is a possibility that might suffer from goitre. The symptom of this disease is a swollen neck.

Question 3.
How does our body respond when adrenaline is secreted into the blood ?
Answer:
Adrenaline is also termed as emergency hormone. In normal situations, these hormones are secreted in small amount.

However, when a person faces stress or danger these are secreted in large amounts to prepare the body to face emergency situation These increase the rate of heartbeat and breathing blood pressure basal metabolic rate, and sugar level in blood. Because of the above roles of these hormones in fight/flight reaction. The adrenal above role are known as the gland of emergency.

Question 4.
Why are some patients of diabetes treated by given injections of insulin ?
Answer:
To control the digestion sugar (glucose) into glycogen and lowering the level of blood glucose in human beings. That’s why some patients of diabetes are treated by giving injection of insulin.

Class 10 Science Chapter 7 Control and Coordination Textbook Questions and Answers

Page No. 125

Question 1.
Which of the following is a plant hormone ?
Answer:
(d) Cytokinin

Question 2.
The gap between two neurons is called a:
Answer:
(b) Synapse

Question 3.
The brain is responsible for
Answer:
(d) All of the above.

Question 4.
What is the function of receptors in our body ? Think of situations where receptors do not work properly. What problems are likely to arise ?
Answer:
A receptor is a nerve cell/group of nerve cells which 15 sensitive to a specific stimulus/to a specific change in the environment. The animals have different sensory receptors/sense organs for receiving different stimuli such as thermo receptors heat and cold photoreceptors for light. phono recepto sound olfactory receptors for smell, gustato receptors for taste and tang receptors for touch. The receptors transmits nerve impulses to central nervous system. The latter transmits motor impulses to appropriate effectors (muscles/ glands) which produce suitable responses. If receptors do not work properly the animals could not receive the different stimuli such as thermo, heat, cold, light, sound, smell, taste and touch from the environment. The above said problems will rise up.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:
Structure of Neuron: It is the structural and functional unit of nervous system.

A neuron consists of three parts
(i) Cell body (ii) Dendrites (iii) Axon
(i) Cell body: The cell body of a neuron is also called cyton/ soma. It has cytoplasm called neuroplasm, it has mitochondria, Golgiapparatus, neurofibrils, neurotubules, special granules called nissl’s gramules. It concerns with metabolic maintenance and growth.

(ii) Dendrites (Singular dendron): These are several short, tapering much branched protoplasmic processes stretching out from the cell body of a neuron. The conduct nerve impulses towards the cell body.

(iii) Axon: It is a single, very long, cylindrical protoplasmic process (nerve fibre) of uniform diameter arising from the cell body The cell membrane of the axon is called axotemma and its cytoplasm is termed axoplasm. Neurons transmit messages in the form of nerve impulses.

Special properties of Neurons.

• They do not divide.
• From shortly after birth, new neurons do not develop
• They are not repaired when injured.
• They are only glucose as a respiratory substrate.
• They die of deprived of oxygen for over five minutes.

Question 6.
How does phototropism occur in plants ?
Answer:
Plants are autotrophs i.e., they manufacture their own food in the presence of sunlight. Therefore, they respond to light by growing towards it. Plants also turn their leaves towards the sun to ensure that the latter get maximum sunlight.
Requirement: Two potted plants.

Method: Take two potted plants. Keep one plant in the open so that it receives the sunlight coming from above: place the other plant in a room near the window in such a way that it receives sunlight from one side only i.e. through window. After some days observe both the plants.

Observation: The first plant (which was kept in the open) has grown up straight towards light. The second plant which was kept in the room and receiving light from one side has grown by bending towards the light.

Explanation and conclusion: This movement of the plant part stem) is caused by the action of auxin hormone. The auxin hormone is formed by the meristematic tissue at the tip of stem.

(i) Auxin spreads uniformly Down the stem in plant kept in the open and receiving sunlight from above. Due to presence of auxin equally on the both sides the stem grows up straight because both the sides of the stem show growth at the same space.

(ii) The second plant is receiving light only from one side through the window. In this case the auxin hormone moves from tip of stem to be concentrated more on the side not receiving light.

Due to presence of more auxin hormone the shady side of stem grows faster than the side of stem receiving light. As a result, the stem bends towards the direction of light. This experiment shows that the stem of the plant responds to light by showing growth movement towards light (positive phototropism).

Question 7.
Which signals will get disrupted in case of a spinal cord injury ?
Answer:
It loses sensation when the nerve supplying it remains stopped/sudden response to a stimulus is called reflex action.

Question 8.
How does chemical coordination occur in plants ?
Answer:
In plants, control and co-ordination take place with the help of plant hormones (phytohormones). These are naturally occurring chemical substances secreted by specific parts of plants such as shoot apex, root apex and leaves. They control various activities and responses of plants such as elongation of stem, flowering, phototropism, geotropism etc.

Question 9.
What is need for a system of control and co-ordination in an organism ?
Answer:
The integration of nervous system and hormonal system is brought about by neuro hormones which regulates the secretion of pituitary hormones. The pituitary hormones in turn, regulate growth and secretions of other endocrine glands. In this way both the systems control and co-ordination various activities of body.

Question 10.
How are involuntary actions and reflex actions different from each other ?
Answer:
An involuntary action, on the other hand is performed by the animal without its will. It is very quick and the animal has no choice in it. Therefore, the same stimulus always gets the same response, for e.g. the hand foot is withdrawn every time it is suddenly pinched or pricked with a needle or touched by a hot object.

Reflex action: “A reflex action is a spontaneous, automatic and mechanical response to a stimulus acting on a specific receptor without the will of an animal is called reflex action.”

For e.g. blinking of eyes, coughing, yawing, sneezing movement of diaphragm during respiration etc.

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and co-ordination in animals.
Answer:

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and movement in our legs?
Answer:
Seismonastic movements : Such movement occur in response to touch (shock). Seismonastic movement are best see in ‘touch me not’ plant (Mimosa-pundica). It is also called ‘lajwanti’ or Chhuimui plant. It we touch the leaves of mimosa plant with our finger, we find that all its leaves fold up and droop. After sometimes the leaves will remian their original status.

Mechanism of Reflex action: A reflex action works in this way.

• Receptor organ : It receives the stimulus and initiates a sensory nerve impulse.
• Sensory nerve fibre: It conducts impulses from receptor to the spinal cord.
• Spinal cord : It acts as modulator and changes sensor impulse into the motor impulse.
• Motor Nerve fibre : It conducts motor nerve impulse from spinal cord to effectors.
• Effector organ : It gives the response. This produces either the movement of some muscles/ secretion from a gland.

Relflexs travels in the following sequence:
Stimulus → Receptor organs → Sensory nerve → Spinal cord → Motor nerve → effector organ → Response to stimulus

Class 10 Science Chapter 7 Control and Coordination Textbook Activities

Activity 7.1 (Page 115)

Question 1.
Put some sugar in your mouth. How does it taste ?
Answer:
When we put some sugar in our mouth, our tongue starts watering and it taste sweet.

Question 2.
Block your nose by pressing it between your thumb and index finger. Now eat sugar again. Is there any difference is its taste ?
Answer:
The receptors are usually located in out sense organs such as the inner ear, the nose, the tongue and so on. So gustatory receptors will detect taste while olfactor receptors will detect smell and deficiency of oxygen.

Question 3.
While eating lunch, block your nose in the same way and notice if you carefully appreciate the taste of your food which you are eating.
Answer:
No, we do not appreciate the taste of food.

Activity 7.2 (Page 121)

• Fill a conical flask with water.
• Cover the neck of flask with a wire mesh.
• Keep two or three freshly germinated bean seeds on the wire mesh.
• Take a cardboard box which is open from one side.
• Keep the flask in the box in such a manner that the open side of the box faces light coming from a window. (Fig.)
• After two or three days, you will notice that the shoots bend towards light and roots away from light.

• Now turn the flask so that the sheets are away from light and the roots towards light. Leave it undisturbed in this condition for a few days.

Question 1.
Have the old parts or the shoot and root changed direction ?
Answer:
Yes, the old part of the shoot have moved towards the light. The roots parts have moved way from the light.

Question 2.
Are there differences in the direction of the new growth ?
Answer:
The new growth is opposite.

Question 3.
What can we conclude from this activity ?
Answer:
The shoot system show positive response to light and root system shows negative response to light.

Activity 7.3 (Page 123)

• Look at Fig.
• Identify the endocrine glands mentioned in the figure.
• Some of these glands have been discussed in the text. Consult books in the library and discuss with your teachers to find out about the functions of other glands.

Endocrine glands in human beings in male and female
Gland: A cell, a tissue, or an organ which secretes certain useful chemical compounds required for particular functions is called a gland. There are three types of gland.
(i) Exocrine glands: These glands have duets for discharging their secretions on the body surface or into the body.

(ii) Endocrine glands: These glands lack duets and pass their secretion into the surroundings blood for transport to the site of action. They are also called ductless gland/glands of internal secretion. Their secretions are known as hormones/internal section for e-g. Pituitary, thyroid, parathyroid glands, adrenal secret glands etc.

(iii) Heterocrine glands: These glands consists of both exocrine and exocrine tissue.The exocrine tissue sends its secretion/ products by way of a diet. The endocrine tissue discharges its secretion into the blood. For eg pancreas and gonads (Testes and ovaries) Human body possesses large number of endocrine glands called which elaborate store and release chemical messengers hormones. These endocrine glands are;

(i) Hypothalamus: Pituitary (hypophysis). Thyroid gland, parathyroid gland, pancreas, adrenal glands, pineal glands. Thymus gland, Testes (males) and ovaries in (females).
1. Hypothalamus: It is situated at the base of the brain and is composed of nervous tissue. The
neurosecretary cells of the hypothalamus secrete several neurohormones called Releasing hormones (RH) and inhibiting factors/hormones.
2. Pituitary gland (Hypophysis): Pituitary gland is present just below the brain. It is a small, red-grey, pea-shaped attached to the hypothalamus of the brain by a stalk-or in fundibulum.

Pituitary gland consists of three lobes :
(a) Anterior lobe (b) Intermediate lobe (c) Posterior lobe.
All three lobes of the pituitary secretes separate hormones.
A. Anterior lobe of pituitary: It produce six hormones.
(i) Follicle-Stimulating Hormone [FSHI]: It stimualtes sperm formation in male and growth of ovarian follicles in female.
(ii) Luteinising Hormone [LH]: In male, it induces the testes to produce male sex hormones named androgens (Testosterone).
(iii) Thyroid-Stimulating Hormone [TSH]: It stimulates growth of thyroid gland and production of thyroid hormones.
(iv) Adreno cortico Trophic Hormones [ACTH]: It stimulates the adrenal cortex to grow and secrete its hormones.
(v) Somatotrophic/Growth Hormones (STH/ GHI): It stimulates growth and development of all tissues by accelerating protein Synthesis and cell division, and by retaining calcium in the body.
(vi) Protactin Hormone [PH] or Luteotrophic Hormones: It Stimulates the growth of milk glands during pregnancy and the secretion of milk after delivery of the child.

B. Intermediate lobe of Pituitary: It secretes a single hormone named melanocyte stimulating hormone (MSH). The hormone stimulates the synthesis of black pigment melanin in the skin, and also causes dispersal of melamin granules in the pigment cells.

C. Posterior lobe of Pituitary :It stores and releases two hormones.
(i) Oxytocin [OT]; It induces contractions of smooth muscles of the uterus during the birth of the young one and myoepithelia cells of the mammary glands to cause release of milk during sucking by the infant.
(ii) Vasopressin: It is also called antidiuretic hormone [ADH] it decreases the loss of water in the urine by increasing the reabsorption of water in the distal convoluted tubules.

3. Thyroid gland: It is the largest endocrine gland. It is a bilobed structure situated in the neck region. Hie thyroid gland secretes three hormones.
(i) Thyroxine (T₄)¹ (ii) Triodothyromine (T₃)² (iii) Calcitonin

4. Parathyroid glands: There are four small flat; oval glands situated on the posterior surface of the thyroid gland, two in each lobe of the thyroid. They secrete parathormone [PTH]. It is also called collip’s hormones and regulate the calcium-phosphorus balance in the. blood.

5. Adrenal glands: These are a pair of glands situated on upper side of each kidney. Therefore they are also called syprarenals.
(a) Adrenal cortex : It is firm, outer pale yellowish pink region essential for life. It secretes three hormones mineral corticoids-The help in control the sodium and Potassium.
(i) Gluco corticoids : They regulate the metabolism of carbohydrate proteins and fats.
(ii) Sex corticoids : They stimulate the development of secondary sexual characters both in males and females.
(b) Adrenal medulla : It is an internal, soft, dark reddish tissue of adrenal gland. It enables the animal to face physical emotional stress.

6. Pancreas: Pancreas lies below the stomach. It is an elongated and emotional stress yellowish gland. It consists largely of lobules that secrete pancreatic ice. The lobules are cells, called Islets of Langerhans’ which reproduce hormones, Islets of Langerhans secretes two hormons.
(a) Insulin by B cells
(b) Glucagon b o cells

7. Testes : In males, a pair of testes are extra abdominal in position and located in scrotum. Testes secrete male sex hormones.

8. Ovaries : A pair of ovaries lies in the abdomen in females. They secrete three female sex hormone
(i) Estrogen (ii) Progesterone (in) Relaxin.

9. Pineal: It is a very small, reddish-grey; vascular, solid body lying between the two cerebral hemispheres of the brain. It secretes melatonin hormone.

10. Thymus gland: It is situated in the upper chest near the front side of the heart. It secretes thymosin hormone which stimulates the development and differentiation of lymphocytes and thereby increasing resistance to infection.

Class 10 Science Chapter 7 Control and Coordination Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name the structural and functional unit of nervous system.
Answer:
Neuron.

Question 2.
Write full form of (a) PNS (b) CNS.
Answer:
(a) PNS-Peripheral nervous system
(b) CNS-Central nervous system.

Question 3.
Mention the two parts of CNS.
Answer:
CNS is divided into two parts

Question 4.
What is the weight of fully grown brain ?
Answer:
1200-1400 gm.

Question 5.
Name the box in which brain is situated ?
Answer:
Cranium

Short Answer Type Questions

Question 1.
What is Cranium ? What is its function ?
Answer:
Cranium is a bony covering that encloses the brain. Its function is to protect the brain from mechanical injury.

Question 2.
What are Meanings ? What is their function ?
Answer:
Meanings are the three membrances that surround the brain. The space between these membrance is filled with cerebral spinal fluid. These membranes and the cerebro spinal fluid protect the brain from mechanical shocks.

Question 3.
Name the main parts of nervous system of grasshopper.
Answer:
The nervous system in grasshopper consists of a bilobe cerebral ganglion (brain) in the head region, long verbral nerve cord extending from brain to almost end of the body, several ganglia at specific regions and nerves extending from ganglia and spreading to various parts of the body.

Long Answer Type Question

Question 1.
How do we detect that we are touching a hot object ?
Answer:
All information from our environment is detected by the specialised tips of some nerve cells. These receptors are annually located in our sence organs, such as the inner ear, the nose, the tongue etc. The nerves from all over the body meet in a bundle in the spinal cord on their way to the brain. For example, when we touch a hot object, the receptor send a signal to the spinal card, the spinal card responses quickly and send signals to remove the hand from the hot object.

Multiple Choice Question

Question 1.
The nervous tissue is made up of an organised networks
(a) Cells
(b) Nervous
(c) Muscles
(d) None of the above
Answer:
(b) Nervous

Question 2.
Reflex action is mainly controlled by the-
(a) Spinal cord
(b) brain
(c) muscles
(d) skull
Answer:
(a) Spinal cord

Question 3.
Writing talking etc, are
(a) voiunatary activities
(b) involuntary actions
(c) (c) can be (a) or (b)
(d) none of the above
Answer:
(a) voiunatary activities

Question 4.
The main thinking part of the brain is-
(a) hind-brain
(b) mid-brain
(c) fore brain
(d) skull
Answer:
(c) fore brain

Question 5.
The brain is protected by the
(a) skull
(b) mouth
(c) head
(d) teeth
Answer:
(a) skull

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6

Question 1.
In the given figure, PS is the bisector of ∠QPR of ∆PQR. Prove that \(\frac { QS }{ SR } =\frac { PQ }{ PR } \)

Solution:

Question 2.
In the given figure, D is a point on hypotenuse AC of ∆ABC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i) DM² = DN . MC
(ii) DN² = DM . AN
Solution:
Given ∆ABC is right angled at B, DM ⊥ BC and DN ⊥ AB.
Construcation : Join BD.
To Prove: (i) DM² = DN. MC
(ii) DN² = DM. AN.
Proof:
(i) Consider ∆BDC
∠BDC = 90°
⇒ BDM + MDC = 90° … (i)
Also in AMCD
∠MCD + ∠MDC = 90° … (ii)
(∵ DMB = 90° by exterior theorem)
From (i) and (ii) ∠MCD = ∠BDM … (iii)
In ∆s BMD and CMD
∠CMD = ∠BMD (90° each)
∠MCD = ∠MDB (from (iii)
∴ BMD ~ DMC

Proof (ii): Now consider similar triangles BND and AND
We
\(\frac { DN }{ DM }\) = \(\frac { AN }{ DN }\)
DN² = DM.MN (Using BN = DM)

Question 3.
In the given figure, ABc is triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC²  + 2BC . BD

Solution:
In right angle triangle ADC. ∠D = 90°
By Pythagoras theorem.
AC² = AD² + DC²
or AC² = AD² + (BD + BC)² (DC – BD + BC)
or AC² = AD² + BD² + BC² + 2BD.BC
(a + b)² = a² + b² + 2ab)]
Now in ∆ADB ∠D produced right angle.
AB² = AD² + BD²
AD² = AB² – BD² … (ii)
From (i) and (ii)
AC² = AB² – BD² + BD² + BC² + ²BC. BD
⇒ AC² = AB² + BC² + 2BC. BD

Question 4.
In the given figure, ABC is atriangle in which ∠ABC 90° and AD ⊥ CB. Prove that AC² = AB² + BC² – 2BC . BD

Solution:

Question 5.
In the given figure, Ad is a median of a triangle ABC and AM ⊥ BC. Prove that.

Solution:

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
ABCD is a parallelogram and AC and BD are its diagonals.
By the property of parallelogram
∆ODC ~ ∆OAB
∆OAD ~ ∆OCB

Question 7.
In the given figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC ~ ∆DPB
(ii) AP . PB = CP . DP

Given : Let AB and CD be the chords intersecting at P.
To Prove: (i) ∆AFC ~ ∆DPB
(ii) AP. PB = CP. DP
Proof:
(i) In ∆PAC and ∆PDB. We have
∠PAC = ∠PDB (∠s in the same segement)
∠APC = ∠BPD (Vertically opp. ∠s)
∴ ∆APC ~∆DPB (Proved)

(ii) ∆APC ~∆DPB
\(\frac { AP }{ DP }\) = \(\frac { CP }{ PB }\)
⇒ AP. PB = CP. DP (Proved)

Question 8.
In the given figure, two chords Ab and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆PAC ~ ∆PDB
(ii) PA . PB = PC . PD

Solution:

Question 9.
In the given figure, D is a point on side BC of ∆ABC, such that \(\frac{B D}{C D}=\frac{A B}{A C}\) Prove that AD is the bisector of ∆BAC.

Solution:
Given: A ∆ABC, in which D is a point on BC such that
\(\frac{B D}{C D}=\frac{A B}{A C}\)
To Prove: AD is the bisector of ∠BAC.
Construction: Produce BA to E such that AE = AC. Join EC.
Proof: In ∆ACE, we have

Question 10.
Nazima is fly fishing in a stream. The trip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the trip of the rod. Assuming that her string (from the trip of the rod to the fly) is that, how much string does she have out (see the figure)? If she pills in the string at the rate of 5 cm per second, what will be the  horizontal distance of the fly from her after 12 seconds?

Solution:
ABC is a right triangle right angled at C by Pythagoras theorem
AB² = AC² + BC²
AB² = (1.8)² + (2.4)²
AB² = 9
AB = 3 m.
Hence 3 m the distance is
Now after the pull the vertical distance is 1.8 – 0.6 = 1.2 m
Now
∠ABC= ∠PRQ = 90°
∠B = ∠Q (The angle from water)
∆ABC ~ ∆PQR

These NCERT Solutions for Class 10 Science Chapter 8 How Do Organisms Reproduce Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

How Do Organisms Reproduce NCERT Solutions for Class 10 Science Chapter 8

Class 10 Science Chapter 8 How Do Organisms Reproduce InText Questions and Answers

In-text Questions (Page 128)

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
Since, DNA copying is accompanied by the creation of an additional cellular apparatus, a cell divides to give rise to two cells. These two cells are similar, but are they likely to be absolutely identical and it will depend on how accurately the copying reactions involved occur. No biochemical reaction is absolutely reliable. Therefore, it is only to be expected that the process of coying the DNA will have some variations each time.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:
If some variation were to be present in a few individuals in these populations, there would be some chance for them to survive. If there were a population of bacteria living in temperate waters and if the water temperature were to be increased by global warming most of these bacteria would die, but the few variants resistant to heat would survive and grow further. Variation is thus useful for the survival of species over time.

In-text Questions (Page 133)

Question 1.
How does binary fission differ from multiple fission?
Answer:
Binary fission : This method of reproduction occurs in favourable conditions like enough food, water, temperature, etc. In this process of asexual reproduction, the nuclear division takes place for example Amoeba, Paramecium, Euglena is bacteria, blue green algae in multi-cellular animals like planaria.

Multiple fission: Many daughter individuals are formed from single parent is called multiple fission. This method of reproduction occurs in unfavourable conditions. The unicellular organism develops a protective covering the cyst over the cell. On return of farourable conditions the cyst breaks and daughter cells are liberated for e.g. Plasmodium.

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
It is common method of reproduction in some bacteria and fungi Rhizopus, Mucor. During spore formation a structure called sporangium develops from the fungal hyphae. The nucleus divides several times within the sporangium and each nucleus with a small amount of cytoplasm develop into a spore. The spores are released; on falling on the suitable substratum they germinate to form new hyphae,

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Mode of reproduction that depend on the involvement of two individuals before a new generation can be created. Bulls alone cannot produce new calves nor can hens alone produce new Chicks. In such cases both sexes, males and females are needed to produce new generations.

Question 4.
Why is vegetative propagation practised for growing some types of plants?
Answer:
This type of propagation also makes possible that have lost the capacity to produce seeds. Another advantage of vegetative propagation is that all plants produced are genetically similar enough to the parent plant to have all its charateristics. This property of vegetative propagation is used in methods such as layering or grafting to grow many plants like sugar can, roses, grapes for agricultural purposes, for e.g. Banana, Orange, Rose, Jamine, etc.

Question 5.
Why is DNA copying an essential part of the process of reproduction?
Answer:
The creation of two cells from one involves copying of the DNA as well as of the cellular apparatus. The DNA copying mechanism, as we have noted cannot be absolutely accurate; and the resultant errors are a source of variations is populations of organisms. Every individual organism cannot be protected by variations, but in population.

In-text Questions (Page 140)

Question 1.
How is the process of pollination different from Fertilization?
Answer:
Pollination: It is the transfer of pollen grains from the anther of a flower to the stigma.
Fertilization: The process of nuclear fusion of the make and one egg nucleus is termed as fertilization.

Question 2.
What is the role of seminal vesicles and the prostate gland?
Answer:

• Seminal Vesicles: There is one pair of seminal vesicle. They lie at the base of urinary bladder above prostate gland. They secrate a viscous fluid for nourishment of sperms.
• Prostate Gland : There is one prostate gland. It surrounds first part of urethra. It secretes alkaline fluid which is discharged into urettra.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
Secondary sexual characters in human females include.

• Growth of breasts.
• Growth of pubic hair and extra hair in the armpit.
• Hair on beard, moustache and chest are lacking
• Broadening of pelvis.
• Initiation of menstruation and ovulation.
• Increase in the subantaneous, particularly in thighs shoulders, Buttocks and face.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
1. Ovaries: Each ovary is of the shape and size of almond. Ovaries are attached to the uterus by ovarian ligaments and also to the oviducts by ovarian fimbriate. Ovaries produce ovary and female hormones estrogen and progesterone,

2. Oviducts: A pair of muscular and internally ciliated tube the oviducts/fallopian tubes.

3. Uterus: It is a muscular; pear shaped hollow organ in pelvic cavity in between urinary bladder (infront). It is held by three ligaments. The upper part of uterus into which oviduct opens is broad and is called body. Its lower narrow constricted part is the neck which contains a cervical canal that leads into vagina; Uterus receives fertilized ovum which develops here into excuate an infant (child), taking about 36 weeks (9 months).

4. Vagina: It is also a muscular tube opening above in the reproduction uterus through internal orifice. Vagina is also known as birth canal since follicles is delivered through vagina.

5. Proliferative/follicular phase: This extends from 6th to 14 day about 10 days. This phase is stimulated by ‘FSH’ of. Pituitary gland due to which growth and development of Griffin follicles takes place. Within each follicle lies an ovum that develops and mature; but usually one ovum matures.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
Intrauterine Contraceptive Device [IUCD] – These are contraceptive devices made of copper, plastic/stainless steel. A copper-T is inserted into uterus by a practicing doctor or a skilled nurse and left in place It prevents implantation in the uterus. It stop the unwanted birth and pregnancies but will not help protecting her from sexually transmitted diseases.

Class 10 Science Chapter 8 How Do Organisms Reproduce Textbook Questions and Answers

Page No. 141

Question 1.
Asexual reproduction takes place through budding in.
(a) amoeba
(b) Yeast
(c) plasmodium
(d) Leishmania
Answer:
(b) Yeast

Question 2.
Which of the following is not a part of the female reproduction system in human beings?
(a) Ovary
(b) Uterus
(c) Vas deferens
(d) Fallopian tube
Answer:
(c) Vas deferens.

Question 3.
The anther contains
(a) sepals
(b) ovules
(c) Carpel
(d) pollen grains.
Answer:
(d) pollen grains.

Question 4.
What are the advantages of sexual reproduction over a sexual reproduction?
Answer:
Advantages of sexual reproducing over asexual reproduction : Sexual reproduction has several advantages over asexual reproduction because it involves (i) Fusion of male and female gametes coming from male and female organisms. Since the fusing gametes come from two different and sexually distinct individuals the off springs exhibit diversity of characters and (ii) Meiosis during gametogenesis provides opportunities for new combination of genes. It plays a prominent role in the origin of new species and leads to variation required for evolution.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
Testes : Testes (Singular-Testis) are the male gonads. They produce both male gametes i.e. Spermatozoa and male sex hormone testosterone (Primary sex organ), testes are suspended in thin pouch of skin called scrotal sac located outside the main body cavity (extra abdominal in position in between the thighs). The testis of man produce sperms from puberty onwards, throughout his life.

Question 6.
Why does menstruation Occur?
Answer:
The menstrual cycle is the cycle of events the place in genital tract, ovaries and uterus and marked by menstrual flow in every 28 days. The days are numbered from the first day of the blood flow in the menstrual Period, starting from first fourteen day is the proliferating period when ovarian follicle grow into mature follicles, At ovulation the mature ovum is release from the ovary on 4th day. During the next day which is also known as secretary phase, the inner wall of uterna thicknes and the ovum moves into fallopian tube. If fertilization does not occur the oocyte under goes antolysis and blood vessels. Rupture causing bleeding. This process is called menstruation/menstral flow. This phase lasts for about four days.

Question 7.
Draw a labelled diagram of the longitudinal section on a flower.
Answer:

Question 8.
What are the different methods of Contraception ?
Answer:
The contraceptive method fall in a number of categories one category is the creation of mechanical barrier so that the sperm does not reach the egg Condom on the penis or similar coverings worn in the vagina can serve this purpose. Another category of contraceptives acts by changing the hormonal balance of the body so that eggs are not released and fertilization cannot occur. As orally pills taken to be commonly drugs can change hormonal balances, they can case side effects too. Other contraceptive devices such as the loop or the copper T are placed in the uterus to prevent pregnancy.

Question 9.
How are the modes for reproduction different unicellular and multicellular organisms?
Answer:
(a) Binary fission : This is the division of parent cell into two small equal sized identical daughter individuals. The two daughter individuals then grow into adult organisms so, fission of single parent cell results in the formation of two new unicellaular organisms. Hence it is called binary fission. For eg. Amoeba, Panamedum.

(b) Budding : The production of new individuals from an out growth of the parent individual. For e.g. in yeast, some protozoans and certain lower animals, e.g. Hydra, in multicellular organisms such as Hydra a bulging on the body appears as a result of repeated mitotic divisions in the cells. It results in the formation of a lateral outgrowth called bud

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Every individual organism cannot be protected by variations but in a population variators are useful for ensuring the survival of the species. It would therefore make sense it organization came up with reproductive modes that allowed more and more variation to be generated.

Question 11.
What could be the reasons for adopting contraceptive methods?
Answer:
Reason for adopting contraceptive method:

• To avoid pregnancy.
• To save from sexual diseases.

Class 10 Science Chapter 8 How Do Organisms Reproduce Textbook Activities

Activity 8.1 (Page 129)

• Dissolve about 10 gm of sugar in 100 mL of water.
• Take 20 mL of this solution in a test tube and add a pinch of yeast granules to it.
• Put a cotton plug on the mouth of the test tube and keep it in a warm place.
• After 1 or 2 hours, put a small drop of yeast culture from the test tube on a slide and cover it with a coverslip.

Question 1.
Observe the slide under a microscope.
Answer:
The yeast starts to grow.

Activity 8.2 (Page 129)

• Wet a slice of bread, and keep it in a cool, moist and dark place.
• Observe the surface of the slice with a magnifying glass.

Question 1.
Record your observations for a week.
Answer:
The bread turns green in colour due to the presence of mound in it.

Activity 8.3 (Page 129)

• Observe a permanent slide of Amoeba under a microscope.
• Similarly observe another permanent slide of Amoeba showing binary fission.
• Now, compare the observations of both the slides

Solution : Binary fission : This method of reproduction occurs is favourable conditions like enough food, water, temperature, etc. In this process of asexual reproduction, the nuclear division takes place first followed by the appearance of a constriction in the cell membrane which gradually increases inwards and divides the cytoplasm into two parts, finally two daughter cells are formed. There are genetically and morphologically similar e.g. Like Amoeba, Paramecium, etc.

Activity 8.4 (Page 130)

• Collect water from a lake or pond that appears dark green and contains filamentous structures.
• Put one or two filaments on a slide.
• Put a drop of glycerine on these filaments and cover it with a coverslip.
• Observe the slide under a microscope.

Question 1.
Can you identify different tissues in the Spirogyra filaments?
Answer:
The spirogyra, breaks up into smaller pieces upon maturation. These pieces or fragments grow into new individuals. This is not true for all multicellular organisms. They cannot simply divide cell by cell. The reason is that many multicellular organisms as we have seen are not simply a random collection of cells.

Activity 8.5 (Page 132)

• Cut the potato into small pieces such that some pieces contain a notch or bud and some do not.
• Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
• Observe changes taking place in these potato pieces over the next few days. Make sure that the cotton is kept moistened.

Question 1.
Take a potato and observe its surface. Can notches be seen?
Answer:
Yes, the notches can be seen

Question 2.
Which are the potato pieces that give rise to fresh green shoots and roots?
Answer:
Those pieces of potato given rise to freshgreen shoots and roots in which cotton was kept moistened.

Activity 8.6 (Page 132)

• Select a money-plant.
• Cut some pieces such that they contain at least one leaf.
• Cut out some other portions between two leaves.
• Dip one end of all the pieces in water and observe over the next few days.

Question 1.
Which ones grow and give rise to fresh leaves?
What can you conclude from your observations?
Answer:
The portion that contain one leaf grows. The plant needs the essential parts to grow.

Activity 8.7 (Page 135)

• Soak a few seeds of Bengal gram (chana) and keep them overnight.

• Drain the excess water and cover the seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry
• Cut open the seeds carefully and observe the different parts.
• Compare your observations with the Fig, and see if you can identify all the parts.

Question 1.
Have you ever observe any flower part still persisting in the fruit ?
Answer:
Yes, we can identify the cotyledon, plumule and the radicle.

Class 10 Science Chapter 8 How Do Organisms Reproduce Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Define reproduction.
Answer:
Reproduction is the production of new individuals of the same species produced by existing organisms.

Question 2.
Name the two types of reproduction.
Answer:
A sexual reproduction and sexual reproduction.

Question 3.
Give one example of animal which reproduces asexually by budding
Answer:
Hydra.

Question 4.
Define parthenogenesis.
Answer:
Parthenogenesis is the development of an organisms from egg without fertilization.

Question 5.
Define unisexual organisms.
Answer:
The organisms which bear both male and female sex organism.

Short Answer Type Questions

Question 1.
What methods will you use for growing Jasmine and rose plants?
Answer:
For growing Jasmine plant we will use layering whereas for growing rose plant we will use cutting. In layering the development of adventitious roots is induced before the plant part is detached by covering a branch of plant with moist soil, whereas in cutting a plant part is used, maybe stem and part of it is immersed in the soil it then forms adventitious roots.

Question 2.
Define parturition. What is the approximate weight of the new bom child?
Answer:
Parturition is the birth of fully developed foetus on completion of gestation. The newborn child weighs about 3.5 kg.

Question 3.
What is the role of hypothalamus and pituitary in human reproduction?
Answer:
Hypothalamus and pituitary release certain hormones which control the functioning of ovary and other reproductive organs.

Long Answer Type Question

Question 1.
With the help of a neat tabelled diagram describe the sexual reproduction in plants.
Answer:
Diagram please see on page 176.
A typical angiospermous flower consists of four whorls of floral appendages attached on the receptacle. The receptacle is the of the flower stalk (Pedical). The four whorls of floral appendages are
1. Calyx 2. Corolla 3. Androecium 4. Gynoecium (Pistil).

1. Calyx (sepals): It is the outermost whorl of floral leaves aled sepals. Sepals are generally green in colour and protective in function.
2. Corolla (Petals): It is the collection of petals. Petals are generally large, showy and brightly coloured to attract the insect pollinators, Calyx and corolla are non-essential parts of the flower because they are not directly involved in reproduction.
3. Androecium: It is the collection of stamens. The male reproductive organs of the flower each stamen consists of anther and filament. Anthers are bilobed and contain four pollen sac. The pollen grains are made inside the pollen sacs.
4. Gynoecium (Pistil): It is the collection of carpels. The female reproductive organs of the flower.

Multiple Choice Questions

Question 1.
In which of the following organism multiple fission takes place?
(a) Bacteria
(b) Plasmodium
(c) Planaria
(d) Paramecium
Answer:
(b) Plasmodium

Question 2.
Grafting is successful in
(a) Dicots
(b) Monocots
(c) Both
(d) None of these
Answer:
(a) Dicots

Question 3.
In drone honey bees develop by :
(a) Fertilization
(b) Parthenocarpy
(c) Parthenogenesis
(d) Fission
Answer:
(c) Parthenogenesis

Question 4.
Which of the following organism is hermaphrodite?
(a) Rabbit
(b) Hydra
(c) Earthworm
(d) Both (a) & (b)
Answer:
(d) Both (a) & (b)

Question 5.
Which of the following statement is incorrect?
(a) Ovule change into seed
(b) Ovary changes into fruit
(c) Self pollination takes place in maize
(d) Each member of calyx whorl is called sepal.
Answer:
(c) Self pollination takes place in maize

These NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Heredity and Evolution NCERT Solutions for Class 10 Science Chapter 9

Class 10 Science Chapter 9 Heredity and Evolution InText Questions and Answers

In-text Questions (Page 143)

Question 1.
If a trait A exits in 10% of a population of an asexually reproducing specie’s and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier ?
Answer:
Both the traits arise in the same position, trait A will exist 10% and trait B will be exist in 60% ratio at the same time. For the segeregation of traits sexual reproduction (crossing of) is essential without sexual reproduction traits will not be segregated.

Question 2.
How does the creation of variations in a species ensure survival ?
Answer:
Creation of variations in a species survive by the synthetic theory of Evolution in which organ origin of species is based on the interaction of genetic variation and natural selection

In-text Questions (Page 147)

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive ?
Answer:
When plants which are dissimilar in given character (such as flower colour) are crossed; the two characters thus brought together differ markedly in ability to express themselves in the resulting hybrid plant.

When the pure red flowered plant is crossed with a pure white flowered plant; all the offspring of F1 generation resemble the red parent. Such a character which appears (as a red flower colour) was termed by Mendel as ‘dominent’ whereas the other character which fails to appear in offspring (as the white flower colour) was termed ‘recessive’.

Question 2.
How do Mendel’s experiments show that traits are inherited independently ?
Answer:
According the Mendel’s Law of segregation traits are inherited independently, When two of the red flowered plants of F1 generation are crossed; both red flowered and white flowered plants appear in the off spring of the F2 generation.

Question 3.
A man with blood group ‘A’ marries a woman with Blood group ‘O’ and their daughter has blood group ‘O’. Is this information enough to tell you which of the traits-blood group A or O is dominant? Why or why not?
Answer:
Blood group ‘O’ is dominant because in F1 generation only dominant traits appear.

Question 4.
How is the sex of the child determined in human beings?
Answer:
Sex determination takes place in human being as follows:
The human male has one x chromosome and oney chromosome. That half the sperms will have x chromosomes and the other half sperms will have y chromosomes. A human female has two x chromosomes (but no y chromosomes) That, is; all the ova (or eggs) will have only x chromosomes. The sex of a child depends on what happens at fertilization.

(a) If sperm carrying x chromosome fertilizes an ovum (or egg) which carries x chromosome; then the child born will be a girl (or female). This is because the child will have xx combination of sex chromosome.

(b) If a sperm carrying y chromosome fertilizes an ovum (or egg) which carries x chromosomes then the child born will be a boy (or male) this is because the child will have xy combination of chromosomes.

In-text Questions (Page 150)

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
Different ways in which individuals with a particular trait may increase in a population:

• They can increase by hybridisation.
• They can increase by natural selection.
• They can increase by struggle for existence.
• They can increase by adaptation or survival of fittest.

Question 2.
Why are traits acquired during the lifetime of an individual not inherited ?
Answer:
Traits acquired during lifetime are not inherited because these changes are somatic changes means there do not have any influence of changes in the genes, genetic (D.N.A.) changes are only inherited changes. Somatic (changes in body) are not at all e.g.

A rat whose tail has been cut does not give rise to a rat without tail.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics ?
Answer:
Small numbers of tigers a cause of worry from the point of view of genetics because it the variety of tigers are less than the recombination of genes will be in less quantity, then there are no chances for the new varieties. Number of individual’s should be more for the new varieties.

In-text Questions (Page 151)

Question 1.
What factors could lead to the rise of a new species ?
Answer:
Following factors could lead to rise a new species.

• It should be some changes in genes (D.N.A) (mutation)
• Change in chromosome structure and number.
• Genetic recombination.
• Natural selection.
• Reproductive isolation.

Question 2.
Will geographical isolation be a major factor in the separation of a self pollinating plant species ? Why or why not?
Answer:
Yes, because the plant needs some organism for Pollination.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually ? Why or Why not ?
Answer:
Geographical isolation is not a major actor in the speciation of an organism that reproduces asexually because only those changes are inherited which are in the ‘genes’ (D.N.A.), isomatic changes are not inherited to the next generation. Asexual reproduction is totally somatic, there is no any crossing over at the time of asexual reproduction. Thus it will not be any speciation during a sexual reproduction.

In-text Questions (Page 156)

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
Way of tracing evolutionary relationships depends on the original idea that changes in D.N.A. during reproduction are the basic events in reproduction. If that is the case then comparing the D.N.A. of different species should give us a direct estimate of how much the D.N.A. has changed during the formation of these species.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not ?
Answer:
Wings of a butterfly and the wings of a bat cannot be considered homologous organs because the origin of these wing are different and the function of these wings are same.

Homologous organs : Those organs which have same origin and different function are called as homologous organs e.g., arm of man of mar and wing of a bird.

Question 3.
What are fossils ? What do they tell us about the Process of evolution ?
Answer:
Fossils : The remains (or impressions) of dead animals or plants lived in the remote past are known as fossils. Fossils provide evidence for evolution; a fossil bird called Archaeopteryx look like a bird but it has many other features which are found in reptiles. This is because Archaeopteryx has feathered wings like those of birds but teeth and tail like those of reptiles.

Archaeopteryx 1s, therefore, a connecting link between the reptiles and birds; and hence suggest that the birds have evolved from the reptiles. Thus fossils provide the evidence that the present animals (plants) have originated from the pre existing ones through the process of continous evolution.

In-text Questions (Page 158)

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species ?
Answer:
All human beings who look so different from each other in terms of size; colour and looks said to by products of the same species because all human beings contain same number of chromosomes and their genetic cells (sperm and ova) and somatic cells, there are some somatic differences in the human beings but the genetic structure is same in all the human beings.

Question 2.
In evolutionary terms, can we say which among bacteria, Spiders, fish and chimpanzees have a better body design ? Why or why not ?
Answer:
In evolutionary terms we can say that chimpanzees have a better body design. Among all of above, chimpanzees have better by design because it has more complex body and it has more sensitive organs in order of evolution.

Class 10 Science Chapter 9 Heredity and Evolution Textbook Questions and Answers

Page no. 159

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

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

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

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive ? Why or why not ?
Answer:
The light coloured eyes of the children is due to dominant genetic character. Because in first generation according to the Menden’s Law of dominance only dominant characters appeared.

Question 5.
How are the areas of study-evolution and classification-interlinkerd ?
Answer:
Evolution of the organisms and classification of the organisms are interlinked because both the processes are in same direction. Both the processes move from simpler one to complex ones. Evolution and classification of organisms never move backwardly.

Question 6.
Explain the terms analogous and homologous organs with examples
Answer:
Analogous organs : Analogous organs are those organs whose structure (origin) is different and function same eg wings of insects, birds etc.

Homologous organs : Those organs whose origin are same and function different called as homologous organs, eg Forelimbs of a frog bird and a man show the same basic design (or basic structure) of bones but they work differently. Such organs are called as homologous organs.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
To find the dominant coat colour in dogs the DNA Sequence has to be checked.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Fossils provide the evidence that present animals (and plants) have originated from the previously existing ones. Turough the process. Of continously evolution fossils are the connecting links between many species and phylums.

Question 9.
What evidence do we have for the origin of life from inanimate matter ?
Answer:
For the origin of life from inanimate matter are amino acids. From simple elements, amino acids are formed. These amino acids part in the formation of genes. These genes like chemicals are suppose to be ancient living beings. So the amino acids are important for the origin of life.

Question 10.
Explain how sexual reproduction gives rise to more variable variation than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually ?
Answer:
In sexual reproduction two individuals are involved, one Male and the other female. Both may have different blood groups and the off springs produced may also vary in blood group. This way sexual reproduction gives rise to more variable variation. Change in DNA sequence and blood group can affect the evolution of their organism that produce sexually. For example, if any partner Suffering from an infection diseases, the off spring might also be affected by this disease.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Equal contribution of male and female parents is ensured in the progeny because both parents are involved in the formation of the zygote which leads to the formation of the of spring.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement ? Why or why not.
Answer:
No, evolution cannot be said to progress from lower form to higher forms. Rather, evolution seems to have given rise to more complete body designs even while the simple body designs continue to flourish.

Class 10 Science Chapter 9 Heredity and Evolution Textbook Activities

Activity 9.1 (Page 143)

Question 1.
Observe the ears of all students in the class. Prepare a list of students having free or attached ear lobes and calculate the percentage of students having each (Fig.) Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest possible rule for the inheritance of earlobes types.
Answer:
There are forty students in the class. Out of forty 12 students have attached earlobes and 28 have free earlobes. It means 30% of the students have attached and 70% students have free earlobes.

Fig. : (A) free and (B) attached earlobes. The lowest parts of the ear. called the earlobe. Is closely attached to the side of the head in some of us, and not in other. Free and attached earlobes are two variants found in human populations.

Activity 9.2 (Page 144)

Question 1.
In Fig, what experiment would we do to confirm that the F2 generation did in fact have a 1 : 2 : 1 ratio of TT, Tt and tt trait combinations ?
Answer:
In this explanation, both TT and Tt are tali plants, while only tt is a short plant. In other words, a single copy of ‘T’ is enough to make the plant tail, while both copies have to be ‘t’ for the plant to be short. Traits like ‘T’ are called dominant traits, while those that behave like ‘t’ are called recessive traits.

Class 10 Science Chapter 9 Heredity and Evolution Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is Heredity ?
Answer:
Heredity means the continuity of features from one generation to another. In other words, heredity can be defined as the resemblances among individuals related by descent.

Question 2.
Name the plant on which Mendel performed his experiment.
Answer:
Mendel performed his experiments on garden pea.

Question 3.
Define variation.
Answer:
Variation can be defined as the occurrence of differences among the individuals.

Question 4.
Define a gene.
Answer:
A gene is a segment of D.N.A. on a chromosome occupying specific position and performs specific functions.

Question 5.
Write the expanded form of D.N.A.
Answer:
The expanded form of D.N.A. is De-oxyribo nucleic Acid.

Short Answer Type Questions

Question 1.
What is a sex chromosome ?
Answer:
The pair of chromosomes which determines the sex of an individual is called sex chromosomes. In males, the sex chromosome are xy and the gametes they produced are the two types: x and y bearing. On the other hand, in females the sex chromos are xx and the gametes they produce are the only one type : X bearing

Question 2.
How sex is determined in human beings ?
Answer:
In male individuals, one x-sex chromosome and one y sex chromosomes are present. In female individuals, the part of X sex chromosome is present.

Question 3.
Define homologous organs?
Answer:
Homologous organs are those organs which are formed on the same fundamental and structural plan but they differ in their shape because they have to function differently.

Long Answer Type Question

Question 1.
Describe the principle of inheritance as explained by Mendel.
Answer:
The principle of inheritance was first of all explained by scientist named Mendel studied the inheritance of contrasting characters (or traits) like tallness, dwarfness etc. of the garden pea plant (Pisumsatum) in various generations of garden pea and came to the conclusion that the various contrasting characters or traits of the pea plant (like tallness; dwarfness etc.) are controlled by certain factors. Mendel considered these factors as the carriers of hereditary information from one generation of the pea plants to the next. We now know that the Mendel’s factors which were considered to be the carriers of hereditary information are actually genes. The term ‘gene’ was coined by a scientist named Johnson in 1909.

Multiple Choice Questions

Question 1.
Laws of inheritance were formulated by
(a) Strass Burger
(b) Mendel
(c) Watson and Crick
(d) Darwin
Answer:
(b) Mendel

Question 2.
Which was the material for Mendel’s work ?
(a) Drosophila
(b) Neuropteran
(c) Pea plants
(d) None of these
Answer:
(c) Pea plants

Question 3.
A tall pea plant when crossed with a dwarf pea plant it produces tall offspring. This proves that tallness is:
(a) Dominant factor
(b) Recessive factor
(c) Hybrid factor
(d) Non-genetic factor
Answer:
(a) Dominant factor

Question 4.
What is the name of discipline which deals with studies about inheritance of characters?
(a) Cylology
(b) Evolution
(c) Genetics
(d) Embryology
Answer:
(c) Genetics

Question 5.
According to Mendels law of independent assortment in a dihybrid cross:
(a) It is possible to calculate the rate of reproduction in animals.
(b) The F generation contains 16 phenotypes.
(c) Only one among 16 offsprings of F, generation exhibit both recessive characters.
(d) F2 generation contains 4 genotypes in the ratio of 9 : 3 : 3 : 1.
Answer:
(c) Only one among 16 offsprings of F, generation exhibit both recessive characters.

These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

Question 1.
Determine the ratio, in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Solution:
Given line is 2x + y – 4 = 0 … (i)
Given point A (2, – 2)
Putting x = 2, and y = 2 in equation (i)
4 – 2 – 4 = 0 = 0
Given point B (3, 7)
Putting x = 3, y = 7 in equation (i)
6 + 7 – 4 = 9 = 9
Ratio = \(\frac { 2 }{ 9 }\) = 2 : 9

Question 2.
Find a relation between x and y, if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
We know that

which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
We know that equation of circle is
x² + y² + 2gx + 2fy + c = 0
where (- g, – f) are the centre of circle.
(6, – 6) passes through the circle
36 + 36 + 12g + 2f(- 6) + r = 0
72 + 12g – 12f + c = 0 … (1)
(3, – 7) passes through the circle
(3)² + (- 7)² + 2g (3) + 2f (- 7) + c = 0
9 + 49 + 6g – 14f+ c = 0
58 + 6g – 14f + c = 0 …. (2)
(3, 3) passes through the circle
(3)² + (3)² + 2g (3) + 2f(3) + c = 0
18 + 6g + 6f + c = 0 … (3)
From (2) and (3),

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
ABCD is a square
AB = BC
(AB)² = (BC)²

Hence the coordinates are (1. 0) and (1, 4)

Question 5.
The class X students school in krishnanagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR, if C is the origin?
Also, calculate the areas of the triangles in these cases. What do you observe?
Solution:
(i) Coordinates are
P (4, 6), Q (3, 2), R (6, 5) taking AD and AB as coordinate axis.
Area of ∆PQR

(ii) Coordinates are (12, 2) (13, 6) (10,3) taking CB and CD as coordinate axis.
Area of ∆PQR

Area of ∆PQR is same in both regions.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively. such that \(\frac { AD }{ AB } =\frac { AE }{ AC } =\frac { 1 }{ 4 } \). calculate the area of the ∆ADE and compare it with the area of ∆ABC.
Solution:

Area of  ∆ADE

Question 7.
Let A(4, 2), B(6,5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meters BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD, such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively, such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note: The points which is common to all the three medians is called centroid and this point divides each median in the ratio 2 : 1]
(v) If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC, find the coordinates of the centroid of the triangles.
Solution:
(i) D is median from A

(ii) Let the coordinate of the point P be (x₁, y₁)

(iii) Coordinate of E

(iv) The coordinate of the points are
P(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\)), Q(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\)), R(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\))
P, Q and R the same point

(v) Given A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the coordinates of the vertices of ∆ ABC

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1), P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
P is the mind point of AB.

In quadrilateral PQRS each side are \(\frac{\sqrt{61}}{2}\) and the diagonal is distinct. Hence is a rhombus.

These NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Light Reflection and Refraction NCERT Solutions for Class 10 Science Chapter 10

Class 10 Science Chapter 10 Light Reflection and Refraction InText Questions and Answers

In-text Questions (Page 168)

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis at which all parallel rays to the principal axis meets is called the principal focus of the concave mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Focal length (l)

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror.

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Convex mirrors are preferred because they always give an erect, through diminished, image. Also, they have a wider field of view as they are curved outwards.

In-text Questions (Page 171)

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
R = 2f
∴ f = \(\frac{R}{2}=\frac{32 \mathrm{~cm}}{2}\) = 16 cm

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Magnification (m) = \(\frac{h^{1}}{h}=-\frac{v}{u}\)
3 = – = \(-\frac{v}{(+4)}=-\frac{v}{4}\)
u = (-3) × 4 = – 12 cm
The image is located in front of the mirror at 12 cm from the pole of the concave mirror.

In-text Questions (Page 176)

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
The light ray bend towards the normal. When light travels from rarer (air) to denser (water) medium it bends towards the normal.

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10s m/sec.
Answer:
The refractive index of the medium ‘nm’ is given by

Question 3.
Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:
Higher will be the refractive index, higher will be the optical density of the material and lower refractive index means lower will be the optical density.

Diamond has the highest optical density (Refractive index 2.42) and air has the lowest optical density, (refractive index 1.0003)

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.
Answer:
The refractive index of kerosene, turpentine and water as follows :
Kerosene – 1.44
Water – 1.33
Turpentine – 1.47
The light travel fastest in water because it has lowest refractive index.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
Refractive index of the medium (or material) is the ratio of the speed of light in air and speed of light in medium S.e.

The refractive index of diamond of means that the ratio of the speed of light in air and the speed of light in diamond is 2.42.

In-text Questions (Page 184)

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
We know that

Distance of the image from the lens = + 50 cm.
Now by lens formula

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:

Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Questions and Answers

Page no. 185

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) At twice the focal length

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave
(b) convex
(c) The mirror is concave and the lens is convex.
(d) The mirror is convex, but the lens is concave.
Answer:
(a) Both are concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.
Answer:
(c) Convex only

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
(d) A concave lens of focal length 5 cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
To obtain an erect image of an object it should be placed between ‘P’ and ‘F’ i.e., between pole and pricipal focus. So the range of the distance of the object from the mirror will be greater than zero and less than fifteen.

The image will be virtual and formed behind the mirror. The image is larger than the object.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Headlight of a car: Concave mirror is used in head lights of a car to get a powerful parallel beams of light.
(b) Side/rear-view mirror of a vehicle: Convex mirros are commonly used because they always given an erect, through diminished image and they also have a wider field of view.
(c) Solar furnace : Large concave mirrors are used in the solar furnace to concentrate sunlight to produce heat in the furnace.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
When one half of a convex lens is covered by a black paper it may or may not be form a complete image. The image formation is based on the position of the object placed in front of the lens.
Experimently we observed that.

• When object is placed at infinity highly diminished and point-sized real and inverted image is formed.
• When object is placed beyond 2F₁ at 2F₁ between F₁ and 2F₁, and at focus F₁ not a complete image is formed.
• When the object is placed between focus F₁ and optical centre ‘O’ and enlarged, virtual and erect image is formed.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
Given h = 5 cm
u = -25 cm
f = 10 cm
To find v = ?
h₁ = ?
From the lens formula

The positive sign of ‘v’ means, image is formed at a distance of 50/3 cm to the right of the optical centre of the lens.
now m= \(\frac{h^{1}}{h}=\frac{v}{u}\)

The negative sign of h’ suggests that the image is real, inverted and diminished.

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Given v = -10 cm
f = -15 cm
u = ?


\(\frac{1}{u}=\frac{-3+2}{30 \mathrm{~cm}}=-\frac{1}{30 \mathrm{~cm}}\)
or, u = -30 cm

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Given u = -10 cm
f = -15 cm
v = ?
By mirror formula

Because v is positive, so image will form behind the mirror. The image will be erect.

So the size of image will be 0.6 times the object size.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification (m)

So, Height of the image (h’) = Height of object (h)
‘+1′ magnification means the height of the object is equal to the height of the image and the image is virtual and erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given h = 5.0 cm
R = 30 cm
u = – 20 cm
v = ?
h’ = ?
We know that R = 2f

So the image will form behind the mirror and it will be virtual and erect.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Given h = 7.0 cm
u = -27 cm
f = -18 cm
v = ?
h’ = ?
From the mirror fourmla :

= \(\frac{(-54 \mathrm{~cm}) \times 7 \mathrm{~cm}}{(-27 \mathrm{~cm})}\)
= -14 cm
So the image is inverted, real and of 14 cm height and lies in the direction opposite to the incident light ray.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given p = -2.0 D
f = ?
We know that p = \(\frac{1}{f}\)
f = \(\frac{1}{P}=\frac{1}{(-2.0 D)}\) = -0.5 m
This is a concave lens.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:

Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Activities

Activity 10.1 (Page 161)

• Take a large shining spoon. Try to view your face in its curved surface,
• Do you get the image? Is it smaller or larger?
• Move the spoon slowly away from your face. Observe the image. How does it change?
• Reverse the spoon and repeat the Activity. How does the image look like now?
•  Compare the characteristics of the image on the two surfaces.

Observation:

• Yes we get the image. It is a virtual image. It is smaller when spoon is moved slowly away from the face/the size of the image decreases.
• In reverse spoon. We observe an inverted image i.e., real image. When spoon is moved slowly from the face the size of the image decreases.

Activity 10.2 (Page 162)

CAUTION : Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.

• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light

Question 1.
Hold the mirror and the paper m the same position for a few minutes. What do you observe? Why?
Answer:
Observation : The paper at first begins to burn producing smoke. The light rays from the sun is converged at a point, as a sharp, bright spot by the mirror. The heat produced due to the concentration of sunlight ignites the paper.

Activity 10.3 (Page 163)

You have already learnt a way of determining the focal length of a concave mirror. In Activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun., it was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror.

• Take a concave mirror. Find out its approximate focal length in the way described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a Table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will now correspond to the positions of the points P» F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus Flies mid-way between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
• Note down and tabulate your observations.

Observation:

Activity 10.4 (Page 166)

• Draw neat ray diagrams for each position of the object shown in Table 10.1.
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.7.
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.

Observation : The result of activity 10.3 in table 10.1.
The rays diagram obtained are similar to the rays diagram given in Fig,

Activity 10.5 (Page 167)

• Take a convex mirror. Hold it in one hand
• Hold a pencil in the upright position in the other hand.

Question 1.
Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
Answer:
Image is erect. Image is diminished.

Question 2.
Move the pencil away from the mirror slowly. Does the image become smaller or larger?
Answer:
Image become smaller; when the pencil is moved away from the mirror slowly.

Question 3.
Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
Answer:
If object is moved away from the mirror, image moves closer to the focus.

Activity 10.6 (Page 167)

• Observe the image of a distant object, say a distant tree, in a plane mirror.

Question 1.
Could you see a full-length image?
Answer:
We see full-length image of a tall building or tree in a small convex mirror.

Question 2.
Try with plane mirrors of different sizes. Did you see the entire object in the image?
Answer:
A full-length image is observed in plane mirror (in different type of mirror).

Question 3.
Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
Answer:
A diminished image is observed which is virtual and erect by using a convex mirror.

Question 4.
Now try using a convex mirror. Did you succeed? Exp1air your observations with reason.
Answer:
In concave mirror, high!y diminished point sized. real and in.erted image s formed when object is placed at infinity, but in convex mirror highly diminished point-sized virtual and erect image is formed when object is placed at infinity, this image is formed at the focus F, behind the mirror.

Activity 10.7 (Page 172)

• Place a coin at the bottom of a bucket filled with water.
• With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Repeat the Activity. Why did you not succeed in doing it in one go?
• Ask your friends to do this. Compare your experience with theirs.

Observation: No we cannot succeed in picking up the coin. This is because of the fact the rays of light traveling upward from the coin of the buket changes their direction due to refraction as they reach the surface of the water

Activity 10.8 (Page 172)

• Place a large shallow bowl on a table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.

Question 1.
Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?
Answer:
Observation: The coin becomes visible again on pouring water into the bowl. The coin appears slightly raised above its actual position due to refraction of light.

Activity 10.9 (Page 172)

• Draw a thick straight line in ink, over a sheet of white paper placed on a table.
• Place a glass slab over the line in such a way that one of its edges makes an angle with the line.

Question 1.
Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
Answer:
We observe that the line inside the slab appears to be bent at the edges.

Question 2.
Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
Answer:
We find that there is no bending of the line.

Question 3.
Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
Answer:
We find the portion of tire line between the edges of the slab appears to be raised.

Activity 10.10 (Page 173)

• Fix a sheet of white paper on a drawing board using drawing pins,
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’, in Fig.

In this Activity, you will note, the light ray has changed its direction at points O and O’. Note that both the points O and O’ lie on surfaces separating two transparent media. Draw a perpendicular NN’ to AB at O and another perpendicular MM’ to CD at O’. The light ray at point O has entered from a rarer medium to a denser medium, that is, from air to glass. Note that the light ray has bent towards the normal. At O’, the light ray has entered from glass to air, that is, from a denser medium to a rarer medium The light here has bent away from the normal. Compare the angle of incidence with the angle of refraction at both refracting surfaces AB and CD.

In Fig., EO is the incident ray. OO’ is the refracted ray and O’ H is the emergent ray. We observe that the emergent ray is parallel to the direction of the incident ray. The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray. However, the light ray is shifted sideward slightly.

Activity 10.11 (Page 177)

CAUTION: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.

Question 1.
Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 10.2.
Answer:
Observation : The paper begins to burn producing smoke. It may even catch fire after a while. Why does this happen? The light from the Sun constitutes parallel rays of light. These rays were converged by the lens at the sharp bright spot formed on the paper. The bright spot you got on the paper is a real image of the Sun.

Activity 10.12 (Page 178)

• Take a convex lens. Find its approximate focal length in a way described in Activity 10.11.
• Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
• Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.
• The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F2 and 2F₂, respectively.
• Place a burning candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
• Note down the nature, position and relative size of the image.
• Repeat this Activity by placing object just behind 2F₁ between F₁ and 2F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observation: Nature, position and relative size of the image formed by a convex lens for various positions of the object.

Activity 10.13 (Page 179)

• Take a concave lens. Place it on a lens stand.
• Place a burning candle on one side of the lens.
• Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not. observe the image directly through the lens.
• Note down the nature, relative size and approximate position of the image.
• Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens.

Observation; Nature, position and relative size of the image formed by a concave lens for various positions of the object.

Class 10 Science Chapter 10 Light Reflection and Refraction Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is the unit of power of a lens ?
Answer:
Dioptre.

Question 2.
What is the relationship between radius of curvature (R) and focal length) (f).
Answer:
R = 2f.

Question 3.
An incident ray makes an angle of 60° with the surface of a plane mirror. What is the angle of reflection ?
Answer:
Angle of reflection = 90° – 60° = 30°

Question 4.
Given an example of a converging lens.
Answer:
Convex lens.

Question 5.
Give an example of a diverging lens.
Answer:
Concave lens.

Short Answer Type Questions

Question 1.
Why does diamond sparkle ?
Answer:
Diamond sparkle due to total internal reflation of light. Diamond has very high value of refractive index i.e., 2.47 hence the critical angle for diamond air interface is only 24.4°. When a light ray falls at any face of diamond crystal at an angle more than critical angle, it comes back in the same medium and strike an other face. So this way light trapped inside and sparkling takes place.

Question 2.
Why the sun appears red while rising or setting ?
Answer:
While rising or setting, the sun is on the horizon and the sun light has to travel much larger distance through the air than at mid day, so, larger amount of red light is scattered at the rising or setting of sun and the light received by us is purely in red colour, due to which it appears red.

Question 3.
What are optically denser and optically rarer medium ?
Answer:
A medium which has larger refractive index is optically denser and which has lower value of refractive index is optically rarer medium.

For example diamond is denser because is has refractive index 2.42 while air is rarer because it has refractive index 1.0003.

Long Answer Type Question

Question 1.
What are the new cartesian sign conventions for the reflection by spherical mirrors ?
Answer:

• The object is on the left side of the miror. i. e., light is incident on the mirror from the left hand side of the mirror.
• All the distances parallel to the principal axis are measured from the pole of the spherical mirror.
• The distances measured in the direction of incident lights are taken as positive.
• The distances measured in the direction opposite to the direction of incident light are taken as negative.
• The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive.
• The heights measured downwards and perpendicular to the principal axis of the mirror are taken as negative.

Multiple Choice Questions

Question 1.
One Dioptre is the power of lens of focal length.
(a) 5 cm
(b) 1000 cm
(c) 1 m
(d) 12 m
Answer:
(c) 1 m

Question 2.
When ray of light enters a glass slab from air:
(a) its frequency decreases
(b) its frequency increases
(c) its wavelength decreaces
(d) its wavelength increaces
Answer:
(c) its wavelength decreaces

Question 3.
The refractive index of glas is 1.5, the velocity of light in glass is:
(a) 2.5 × 10¹⁰ cm/sec
(b) 2 × 10¹⁰ cm/sec
(c) 3.5 × 10¹⁰ cm/sec
(d) 1.5 × 10¹⁰ cm/sec
Answer:
(b) 2 × 10¹⁰ cm/sec

Question 4.
The focal length of lens whose power is -1.5 D is:
(a) +2.5 m
(b) -66.6 cm
(c) -2.5 m
(d) +66.6 cm
Answer:
(b) -66.6 cm

Question 5.
The magnification of a concave mirror.
(a) Does not depend on its focal length.
(b) Is equal to the ratio of the object distance to the image distance.
(c) Is equal to the ratio of image distance of the object distance.
(d) Does depend on its focal length
Answer:
(c) Is equal to the ratio of image distance of the object distance.