Month: October 2024

Students must go through these JAC Class 10 Science Notes Chapter 7 Control and Coordination to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 7 Control and Coordination

→ The movement is shown as a response to a change in the environment by the organism.

→ Each kind of a change in the environment evokes an appropriate movement as a response.

→ Some movements are growth related while others are not.

→ Nervous system and hormones bring about control and coordination of the bodies.

→ Neuron is a structural and functional unit of nervous tissue.

→ Cell body (cyton), dendrites and axon are the structural parts of neuron.

→ The responses of the nervous system can be classified as reflex action, voluntary action and involuntary action.

→ Reflex action is an involuntary response to external stimuli without the knowledge of voluntary centres of brain.

→ Reflex arc is a connection between the input (sensory) nerve and the output (motor) nerve along with spinal cord.

→ Voluntary actions occur under the control of will of an animal.

→ The nervous system uses electrical impulses to transmit messages.

→ Human nervous system :

• Central nervous system is made-up of brain and spinal cord.
• Peripheral nervous system is made-up of cranial nerves and spinal nerves.

Brain has three major parts:

• Fore-brain
• mid-brain and
• hind-brain.

→ Brain is protected in bony box and spinal cord is protected in a vertebral column.

→ The simplest form of movement at the cellular level is the movement of muscle cell.

→ Plants have neither a nervous system nor muscles.

→ Plants show two different types of movements – one dependent on growth and other independent of growth.

→ Touch-me-not is a plant of the Mimosa family whose leaves move very quickly in response to touch.

→ Tendrils of a pea plant are sensitive to touch.

→ Environmental factors such as light, gravity, water, etc. change the directions of growing part of a plant. These are called directional or tropic movements.

→ Chemical coordination is seen in both plants and animals.

→ Auxin, gibberellin and cytokinin are growth promoting hormones of plants. Abscisic acid is a plant hormone which inhibits growth.

→ In animals the chemical coordination is due to neurotransmitters and hormones.

→ An endocrine gland is the ductless gland that secretes hormones.

→ Hormones are directly poured into blood and are transported to their functional site.

→ Pituitary gland, pineal gland, thyroid gland, parathyroid gland, thymus gland, adrenal gland, pancreas and testis or ovary are endocrine glands in human body.

→ The timing and amount of hormone secretion are regulated by feedback mechanisms.

Jharkhand Board JAC Class 10 Science Important Questions Chapter 7 Control and Coordination Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 7 Control and Coordination

Additional Questions and Answers

Question 1.
Distinguish between :
(1) Response in Plants and Response in Animals
Answer:

Response in Plants	Response in Animals
1. Plants do not possess nervous system but only possess hormones for expressing their response.	1. Animals possess both, the nervous system and endocrine system for expressing their response.
2. The response in plants is not rapid and needs more time to be observed.	2. The response in animals is rapid and seen immediately.
3. The response in plants is limited.	3. The response in animals is not limited.
4. There are no muscular tissues in plants to show the response.	4. There are muscular tissues in animals through which the response is shown.
5. In plants, there is no specific tissue for the transmission of information to different parts of the body.	5. In animals there is specific nervous tissue for the transmission of information to different parts of the body.

(2) Nervous System and Endocrinal System
Answer:

Nervous System	Endocrinal System
1. Its structural and functional unit is a neuron (nerve cell).	1. Its functional unit is a hormone.
2. In human body it comprises central nervous system, peripheral nervous system and autonomous nervous system.	2. In human body it comprises of various endocrine glands.
3. Its function is to collect the information, its analysis and interpretation and convey the responsive message to the motor organs.	3. Its function is to produce stimulatory or inhibitory effect on the tissues or organs through the medium of hormones.
4. The coordination that occurs through the nervous system is very rapid as the impulses pass through the nerve fibres.	4. The coordination that occurs through the endocrine system is a relatively slow process as the hormone flows through the blood stream to reach the target organ.

(3) Cerebrum and Cerebellum
Answer:

Cerebrum	Cerebellum
1. It is a major part of the fore-brain.	1. It is a part of the hind-brain.
2. It coordinates thoughts and various other senses.	2. It coordinates the functions of voluntary muscles and thereby maintains the body equilibrium.
3. It is the largest and most complex part of the brain.	3. It is a part lying behind the cerebrum, on the dorsal side beneath the pons.

(4) Plant hormones and Animal hormones
Answer:

Plant hormones	Animal hormones
1. Plant hormones are secreted by plant cells but no specific glands are there.	1. Animal hormones are secreted from endocrine glands.
2. Plant hormones are either growth promoting or growth inhibiting.	2. There is no inhibitor hormone for growth in animals.
3. Plant hormones reach to their target site by simple diffusion.	3. Animal hormones reach to their target site through blood circulation.
4. Secretion of it is not regulated by feedback mechanism.	4. Secretion of some hormones regulated by feedback mechanism.

Question 2.
Give scientific reasons for the following statements:
(1) Response to stimuli is the characteristic of every living organism.
Answer:
All living organisms experience the effects of changes in the atmosphere that surrounds them and tend to respond differently against them.

The living organisms show a slow or rapid response to stimuli such as heat, cold, sound, touch, pressure, etc. Responses are expressed against these stimuli through hormones in plants and nervous system as well as hormones in animals, e.g., The plants bend in the direction of light. Man shows shivering effect in severe cold and perspiration in hot season.

Thus, the response to stimuli is the characteristic of every living organisms.

(2) Unlike animals, the plants do not show immediate response.
Answer:
All organisms show response to stimuli. However, the animals possess nervous system and s sense organs, which are absent in plants. Animals use the nervous system as well as endocrine system for control and coordination of all their activities, while the plants possess only hormones to coordinate their activities. The hormones diffuse from cell to cell quite slowly in plants while most animals have blood for rapid transport of hormones. Because of all these reasons, the plants do not show immediate response.

(3) The roots in plants grow against the direction of light.
Answer:
The roots in plants show positive tropic movement in the direction of water and gravitational force, i.e., The roots show positive hydrotropism and positive geotropism.

Thus, the roots bend in the direction Thus, the roots bend in the direction of water and gravitational force. So it grows against the direction of light.

(4) The central nervous system is very well protected.
Answer:
The central nervous system comprises of a brain and a spinal cord.

Brain is contained in a fluid-filled balloon which absorbs mechanical shocks and thereby protect it from any severe injuries. Moreover, the brain is enclosed in a strong bony box – the cranium of the skull and the spinal cord is enclosed in a long and strong bony vertebral column. Thus, the central nervous system (of the human) is very well protected.

(5) The foot is very suddenly lifted off as soon as it comes in contact with a burning coal.
Answer:
It is a reflex action – In this phenomenon, the impulse commencing from the sensory organ (the skin of the foot) enters into the spinal cord through the sensory nerve fibres. The impulse is analysed in the spinal cord and the responsive motor impulse is transmitted through the motor nerve fibres to the effector organ (the foot muscles) to contract and thereby show the response. Sp the foot is very suddenly lifted off as soon as it comes in contact with a burning coal.

(6) The hormones secreted from the endocrine glands are present everywhere in the body.
Answer:
The hormones are secreted from the endocrine glands which are ductless glands. These glands are richly supplied with blood. The secretions are directly poured in the blood stream. As the blood circulates in all the parts of the body, these hormones are carried and hence present everywhere in the body.

(7) The diabetic patient is given injections of insulin.
Answer:
It is necessary to maintain a definite blood sugar level in the human body. In the patients of diabetes the blood sugar level remains high due to the deficiency of insulin hormone from the pancreas. This higher blood sugar level causes several harmful effects.

To prevent the diabetic patient from such harmful effects the blood sugar level is required to be maintained at certain definite level. Insulin is a hormone that reduces blood sugar level. Hence, the diabetic patient is given injections of insulin.

(8) It is advisable to take iodized salt in daily food.
Answer:
The hormone thyroxin is secreted from the thyroid gland. It is an iodized hormone (rich in iodine). Thyroxin is not formed when there is deficiency of iodine in the blood. As a result, the condition called hypothyrodism occurs in which the size of the thyroid gland gets gradually enlarged, this disease is called goitre. The iodized table salt provides proper amount of iodine for the formation of thyroxin in the thyroid gland.

Hence, it is advisable to take iodized salt in daily food.

Question 3.
Carefully observe the given diagram id answer the questions related with it:

Questions :

1. Identify ‘a’ and state the function of it.
2. At which region in diagram chemicals are released?
3. Identify ‘c’ and state from which it originates.

Answer:

1. a – Dendrite.
   Function: Information received at the end of its tip.
2. At d – region (nerve ending) in diagram chemicals are released.
3. c – Axon. It originates from cell body.

Questions :

1. Identify ‘a and state its function.
2. Where is information stored in brain? Mention its alphabet.
3. State any two functions of part ‘d’
4. Identify ‘c’ State the name of special structure formed by it which show quick response.

Answer:

1. a-Cranium (bony box).
   Function : It protects brain.
2. b – Cerebrum
3. d – Cerebellum. It is responsible for precision of voluntary actions and maintaining the posture and balance of the body.
4. c – Spinal cord. For quick response, the reflex arc is formed by it.

Questions:
(1) State the name of ‘a’ and ‘b’ with location.
(2) State any two functions of ‘b’
(3) In which condition, ‘a’ is stimulated and give the name of its secretion.
Answer:
(1)

Name	Location
a. Adrenal gland	At the anterior of kidneys
b. Pancreas	Under the stomach in abdomen

(2) Functions of ‘b’ (Pancreas) : Secretes insulin and regulates blood sugar level. Secretes pancreatic juice and help in digestion.

(3) In a fear or tense condition ‘a’ (adrenal gland) is stimulated and secretes adrenaline hormone.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Name the two systems of control and coordination in higher animals.
Answer:
The nervous system and the endocrine system are the two systems of control and coordination in higher animals.

(2) Name the three components of a nerve cell.
Answer:
The three components of a nerve cell s are a cyton, an axon and one or more dendrons.

(3) Name the most important part of the s human brain.
Answer:
The most important part of the human brain is cerebrum which is the largest part of the brain.

(4) State one function of cerebellum.
Answer:

Part	Function
Cerebellum	Coordination of movement of the and maintenance of body equilibrium.

(5) Give the function of medulla.
Answer:
Function of medulla: It maintains the rhythm of various involuntary processess such as breathing, heartbeats, peristalsis of the alimentary s canal, etc.

(6) Name the structural and functional unit of nervous system.
Answer:
The neuron (nerve cell) is the structural and functional unit of nervous system.

(7) Name one hormone secreted by the pituitary gland.
Answer:
One of the hormones secreted by the pituitary gland is GH (Growth Hormone).

(8) Give example of the movement of a plant part which is caused by the loss of water.
Answer:
Sensitive plant: Touch-me-not (Mimosa ( pudica).

(9) What is the response of roots to gravity? What is this phenomenon known as?
Answer:
A root give positive response to gravity, this phenomenon is called positive geotropism.

(10) What is the response of stem to light? What is this phenomenon known as?
Answer:
A stem moves towards light. This phenomenon is known as positive phototropism.

(11) What is an impulse?
Answer:
Information that is transmitted through the nerves in form of electro-chemical signals are called impulse.

(12) What is gustatory receptors? Where they located?
Answer:
The receptors which detect taste are called gustatory receptors. They are located on the surface of tongue.

(13) By what is nervous tissue made-up of? What is its special ability?
Answer:
The nervous tissue is made-up of an organised network of neurons. It is specialised for conducting information via electrical impulses.

(14) State any two examples of movement in order to protect ourselves.
Answer:
When bright light is focussed on eyes, we constrict pupils, we pull our hand when we touch a hot object. These are the examples of movement in order to protect ourselves.

(15) What are components of peripheral nervous system?
Answer:
The peripheral nervous system consists of cranial nerves arising from brain and spinal nerves arising from the spinal cord.

(16) Which separate areas are there in fore brain?
Answer:
There are separate areas specialised for hearing, smell, sight, etc. in fore-brain.

(17) Why thinking is called a complex activity?
Answer:
Thinking is called a complex activity, because it is bound to involve a complicated interaction of many nerve impulses from many, neurons.

(18) How do we know that we have eaten enough?
Answer:
The sensation of feeling full is because of a centre that is associated with hunger, located in separate part of the fore-brain.

(19) Which involuntary actions are controlled by medulla in the hind-brain?
Answer:
Involuntary actions such as blood pressure, salivation, vomiting are controlled by medulla in the hind-brain.

(20) How do animal muscles move?
Answer:
When a nerve impulse reaches the muscle, the muscle fibre can move due to stimulation.

(21) How does a muscle cell move?
Answer:
Muscle cells move by changing their shape either shortening or elongating.

(22) How do muscle cells change their shape?
Answer:
Muscle cells have special proteins that change both their shape and arrangement in cells, so muscle cells change their shape.

(23) How does leaves of chhui-mui respond to touch stimulus?
Answer:
The leaves of the sensitive plant chhui- mui move very quickly in response to touch stimulus.

(24) Why are hormones called chemical messengers?
Answer:
Hormones are called chemical messengers because they carry information in the form of chemicals that regulate the biological processes of body.

(25) Which plant hormone inhibits growth? State its effect.
Answer:
Abscisic acid – a plant hormone that inhibits growth. It affects wilting of leaves.

(26) Why is chemical signal required along with electrical impulses in higher animals?
Answer:
In higher animals, electrical impulses bring immediate response to only those cells that are connected by nervous tissue while chemical signals reaching to each and every cells of body.

(27) Who constitutes a second way of control and coordination in our body?
Answer:
Hormones, secreted by endocrine system constitute a second way of control and coordination in our body.

(28) Where is auxin synthesised and where does it diffuses?
Answer:
Auxin is synthesised at the shoot tip and it diffuses towards the side of the shoot which is in shade.

(29) Secretion of which hormone in males and in females is responsible for pubertal changes?
Answer:
Secretion of testosterone in males and estrogen in females causes changes associated with puberty.

(30) State the secretory site and function of growth hormone releasing factor.
Answer:
Growth hormone releasing factor : Secretory site : Hypothalamus.
Function: Stimulates the pituitary gland to release growth hormone.

Question 2.
Define : OR Explain the terms :
(1) Stimulus
Answer:
The changes, that take place in the external environment of living organisms (plants, animals, microorganisms, etc.) and induce definite types of reactions or responses, are called stimuli (singular – stimulus).

(2) Response
Answer:
The reaction of living organisms against the stimulus induced by the change in the external environment is called response.

(3) Coordination
Answer:
Different organs of the body jointly s function systematically against the stimuli and react to express proper response against the £ stimuli is called coordination.

(4) Tropism
Answer:
The growth related movement in the plant organs, which are induced as a response ” to directional stimulus is called tropism.

(5) Hormone
Answer:
A chemical messenger, synthesised or secreted in extremely small quantities by endocrine

(6) Endocrine gland
Answer:
The ductless glands that secret hormones c are called endocrine gland.

(7) Receptors
Answer:
The specialised structures that receive stimuli from external environment are called t receptors.

(8) Central Nervous System
Answer:
A controlling and coordinating system of body which consists of brain and spinal cord is called central nervous system.

(9) Reflex arc
Answer:
A connection between the input (sensory) nerve and output (motor) nerve along with spinal cord is called reflex arc.
OR
It is a path of reflex action which consists of receptor, sensory neuron, relay neuron and motor neuron connected to effector organ.

Thus a neural pathway for the sensory and motor messages that pass through the spinal cord forms the reflex arc. A very rapid response is shown through it.

The following example can clarify the meaning of reflex arc:
Suppose by mistake and unknowingly one touches a hot object by one’s hand. One would withdraw the hand all of a sudden without giving a slightest thought. Here the hot object is the source of stimulus.

This stimulus activates the sensory nerve fibre in the hand and carry that impulse to the spinal cord. The sensory centres in the spinal cord receive the stimulus and transmits the response to the motor centre of the spinal cord. This motor message is transmitted through the motor nerve fibre to definite muscles of the hand, which upon contraction, withdraws the hand. The hand or its muscles act as effector organ. This entire neural path from the sensory or receptor organ, to the effector organ is a reflex arc.

(10) Synapse
Answer:
In the arrangement of two consecutive neurons the axon fibre endings of one neuron and s the dendrite endings of the next neuron having a microscopic gap is called synapse.

Question 3.
Fill in the blanks :

1. The plants coordinate their behaviour against the environmental changes by using ………………..
2. The responses of plants are not rapid for want of ………………..
3. ……………….. is a growth-inhibitor hormone of plants.
4. The stem exhibits ……………….. geotropism and ……………….. phototropism.
5. The tendrils of pea plants are the example of ………………..
6. The olfactory receptors will detect ………………..
7. ……………….. have evolved in animals as efficient ways of functioning in the absence of true thought processes.
8. ……………….. part of brain for learning process and part of brain responsible for memory.
9. The cells of sensitive plant change shape by changing the ……………….. in them resulting in swelling or shrinking.
10. Cytokinins promotes ……………….. in plants.
11. ……………….. hormone has the target organ heart.
12. The timing and amount of animal hormone released are regulated by …………………

Answer:

1. hormones
2. nervous system
3. Abscisic acid
4. negative, positive
5. thigmotropism
6. smell
7. Reflex arc
8. Cerebellum, cerebrum
9. amount of water
10. cell division
11. Adrenaline
12. feedback mechanism

Question 4.
State whether the following statements are true or false:

1. Many involuntary actions are controlled by the mid-brain and hind-brain.
2. Brain is never involved in reflex action.
3. The communication between the peripheral nervous system and the other parts of the body is facilitated by the central nervous system.
4. Neuromuscular junction is synapse like gap between nerve ending and muscle fibre.
5. When we have cold, efficiency of olfactory receptors reduces.
6. Nerve cells have special proteins that change their shape and arrangement for conduction of impulse.
7. Touch-me-not, a sensitive plant is of the Mimosa family.
8. The movement of sunflowers in response to day or night is quite fast.
9. The sensitive plants detect the touch though there is no nervous tissue or any muscle tissue.
10. Gibberellin gives signal to plant to stop the growth.
11. Hypothalamus plays an important role in the release of many hormones from pituitary gland.
12. Chemical coordination is seen in both plants and animals.

Answer:

1. True
2. False
3. False
4. True
5. True
6. False
7. True
8. False
9. True
10. False
11. True
12. True

Question 5.
Match the following:
(1)

Column I	Column II
1. Fore-brain	a. Balance of body
2. Medulla	b. Reflex arc
3. Cerebellum	c. Main thinking part
4. Spinal cord	d. Salivation

Answer:
(1 → c), (2 → d), (3 → a), (4 → b).

(2)

Column I	Column II
1. Insulin	a. Testes
2. Testosterone	b. Pancreas
3. Growth hormone	c. Ovaries
4. Estrogen	d. Pituitary

Answer:
(1 → b), (2 → a), (3 → d), (4 → c).

(3)

Column I	Column II
1. Auxin	a. Wilting of leaves
2. Gibber ellin	b. Promotes cell-division
3. Cytokinin	c. Helps in stem-growth
4. Abscisic acid	d. Phototropism

Answer:
(1 → d), (2 → c), (3 → b), (4 → a).

(4)

Column I	Column II
1. Adrenaline	a. Regulates blood sugar level
2. Thyroxin	b. Increases breathing, heart beats.
3. Insulin	c. Regulates menstrual cycle.
4. Estrogen	d. Regulates metabolism for body growth.

Answer:
(1 → b), (2 → d), (3 → a), (4 → c).

Question 6.
Chart-diagram based questions:
1. Which of the following diagram is correct? Why?
Answer:

Diagram (a) is correct, because roots show positive geotropism and stem shows negative geotropism.

2. Label (a), (b), (c) and (d) in the given figure showing the pathway of thermal impulse.

Answer:
(a) Sensory neuron
(b) Relay neuron
(c) Motor neuron
(d) Effector = muscle in skin

3. Identify (a), (b), (c) and (d) in the given figure. Give one name of the hormone secreted I; from each of any two of them.

Answer:
(a) Pineal gland – Melatonin hormone
(b) Pituitary gland – Growth hormone
(c) Thyroid gland – Thyroxine
(d) Thymus gland – Thymocine

4. A graph shows change after a lunch of a healthy individual whose diet is rich in sweets.

What you explain from it?
Answer:
When the sugar level rises in blood, pancreas releases more insulin to regulate blood sugar level. As blood sugar level falls, insulin secretion is reduced by feedback mechanism.

Question 7.
Select the correct alternative from those given below each question:
1. The roots of a plant are ………………..
A. positive phototropic, but negative geotropic
B. negative geotropic, but negative phototropic
C. negative phototropic, but positive hydrotropic
D. negative hydrotropic, but positive phototropic
Answer:
C. negative phototropic, but positive hydrotropic

2. The diagram shows a plant which has received light from one side only. Which characteristics are shown by the plant?

A. Excretion and growth
B. Response and reproduction
C. Growth and response
D. Reproduction and nutrition
Answer:
C. Growth and response

3. The growth of a pollen tube towards the ovule is caused by
A. phototropism
B. hydrotropism
C. geotropism
D. chemotropism
Answer:
D. chemotropism

4. For the synthesis of which of the following hormone is iodine necessary?
A. Adrenaline
B. Auxin
C. Thyroxin
D. Insulin
Answer:
C. Thyroxin

5. Which of the following hormone prepares our body for action in emergency situations?
A. Testosterone
B. Growth hormone
C. Adrenaline
D. Insulin
Answer:
C. Adrenaline

6. Which is male sex hormone?
A. Estrogen
B. Adrenaline
C. Testosterone
D. Progesterone
Answer:
C. Testosterone

7. Which of the following endocrine gland does not occur as a pair in the human body?
A. Adrenal
B. Pituitary
C. Testis
D. Ovary
Answer:
B. Pituitary

8. Which of the following helps in maintaining posture and balance of the human body?
A. Cerebrum
B. Cerebellum
C. Medulla
D. Pons
Answer:
B. Cerebellum

9. Which of the following plant shows immediate response to touch by its leaves?
A. Sunflower
B. Pea
C. Mimosa
D. None of the given
Answer:
C. Mimosa

10. By whom are the continuous heartbeats controlled?
OR
Where are the regulatory centres for the blood pressure located?
A. Cerebrum
B. Cerebellum
C. Mid-brain
D. Medulla
Answer:
D. Medulla

11. Which of the following plant hormone helps in the growth of the stem?
A. Auxin
B. Gibberellin
C. Cytokinin
D. Abscisic acid
Answer:
A. Auxin, Gibberellin

12. Through whom does the impulse enter into the cyton?
A. Dendrite
B. Axons
C. Both A and B
D. None of the given
Answer:
A. Dendrite

13. Which is the largest and the most complex part of the human brain?
A. Medulla
B. Cerebellum
C. Hypothalamus
D. Cerebrum
Answer:
B. Cerebellum

14. Who regulates the involuntary reflexes such as coughing, sneezing, hickup, vomiting, etc.?
A. Medulla
B. Cerebellum
C. Hypothalamus
D. Cerebrum
Answer:
A. Medulla

15. The deficiency of which hormone causes diabetes?
A. Estrogen
B. Thyroxin
C. Adrenaline
D. Insulin
Answer:
D. Insulin

16. Where is the arrangement in the body for i reflex action?
A. In medulla oblongata
B. In spinal cord
C. In pons
D. In heart
Answer:
B. In spinal cord

17. What is the main function of endocrine system in animals?
A. Coordination
B. Combination
C. Regulatory
D. None of the given
Answer:
A. Coordination

18. Which ovarian hormone regulates menstrual cycle in women?
A. Testosteron
B. Estrogen
C. Thyroxin
D. None of these
Answer:
B. Estrogen

19. Which gland is stimulated to release its <; secretion in a scary situation in squirrels?
A. Adrenal
B. Pituitary
C. Thyroid
D. Hypothalamus
Answer:
A. Adrenal

20. “Withdrawal of hand when unknowingly rose prickle pricks the hand”, which is this process?
A. Autonomous reaction
B. Reflex action
C. Thigmotropism
D. None of the given
Answer:
B. Reflex action

21. Hypothalamus is a part of ……………..
A. Fore-brain
B. Spinal cord
C. Muscle tissue
D. Cerebellum
Answer:
A. Fore-brain

22. Who secretes releasing hormones?
A. Pituitary gland
B. Hypothalamus
C. Autonomous Nervous System
D. Thalamus
Answer:
B. Hypothalamus

23. Whose excessive secretion causes the body to look like a gorilla?
A. Thyroxin
B. Growth hormone
C. Adrenaline
D. All of the given
Answer:
B. Growth hormone

24. Which disease takes place when there is a increase of sugar in the blood and in the mine?
A. Dwarfism
B. Goitre
C. Diabetes
D. Both A and B
Answer:
C. Diabetes

25. Statement A: Adrenaline diverts the blood to skeletal muscles of squirrel.
Reason R: Squirrel relies not only on electrical impulses but also on chemical signals.
Which option is correct for Statement A and Reason R?
A. Both A and R are correct and R is explanation of A.
B. Both A and R are correct, but R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
B. Both A and R are correct, but R is not explanation of A.

26. Statement A: The fore-brain is the main thinking part of the brain.
Reason R: Thinking is a complex activity that S involves a complicated interaction? of many nerve impulses.
Which option is correct for Statement A and i Reason R?
A. Both A and R are correct and R is explanation of A.?
B. Both A and R are correct, but R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
A. Both A and R are correct and R is explanation of A.?

27. Statement A: Auxin helps the cells to grow longer and plant appears to bend s towards light.
Reason R: Auxin diffuses towards shady side s of shoot from its tip.
Which option is correct for Statement A and Reason R?
A. Both A and R are correct and R is explanation of A.
B. Both A and R are correct, but R is not? explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
Answer:
A. Both A and R are correct and R is explanation of A.

28. How many endocrine glands from following are not in pairs?
Pancreas, adrenal, thyroid, testes, pituitary?
A. 2
B. 3
C. 4
D. 5
Answer:
B. 3

29. Find incorrect statement for cerebrum.?
A. It receives sensory impulses from various receptors.
B. It has areas where information stored.
C. It passes information to motor areas which control the movement of voluntary muscles.?
D. It maintains posture and balance of the body.
Answer:
D. It maintains posture and balance of the body.

30. Which is sensitive to touch?
A. Human skin
B. Tendrils of pea
C. Leaves of mimosa
D. Flowers of Sunflower?
Answer:
A. Human skin, Tendrils of pea, Leaves of mimosa

Question 8.
Answer as directed : (Miscellaneous)
(1) What is full form of CNS?
Answer:
CNS – Central Nervous System

(2) Give the correct sequence of conduction of impulse.
Answer:
Correct sequence of conduction of impulse : Dendrite → cell body → axon → nerve ending → synapse → dendrite

(3) Trace the sequence of events which occur when a bright light is focused on your eyes?
Answer:
Stimulus (bright light) → photo receptors of eyes
↓
pupil ← motor ← mid – ← sensory nerves constrict nerve brain /neurons (neuron)

(4) Find mismatched pair :
(i) Iodine – Functioning of thyroid gland
(ii) Insulin – Regulates blood sugar level
(iii) Hypothalamus – Regulates the secretion of pituitary
(iv) Estrogen-Obstructs menstrual cycle
Answer:
(iv) Estrogen-Obstructs menstrual cycle

(5) Identify me : I am a hormone generally for an emergency, increases breathing rate but reduce blood flow to digestive system and skin.
Answer:
Adrenaline

(6) State the correct sequence of impulse for spinal reflex.
Answer:
Receptors in skin → sensory neuron → spinal cord motor neuron effector (muscles)

(7) Who am I? I am present in greater concentration areas of rapid cell division, such as in fruits and seeds.
Answer:
Cytokinin

(8 ) Movement dependent on growth : Tendrils of pea :: Movement independent of growth : ………………
Answer:
Drooping of leaves of chhui-mui

(9) Which event is indicated in the diagram? Which hormone is responsible for it?

Answer:
Positive phototropism in shoot is indicated in the diagram. Auxin is responsible for it.

(10) Deficiency of growth hormone: ……………… ::
Deficiency of ……………… : diabetes
Answer:
Dwarfism, Insulin

(11) Identify me : I am controlling your daily activities such as walking, riding a bicycle, picking up object, etc.
Answer:
Cerebellum

(12) Find mismatched pair:
(i) Growth related movement – slower
(ii) Movement in sunflower in response to day or night – quite slow
(iii) Movement of our leg-very slow
Answer:
(iii) Movement of our leg-very slow

(13) Give the scientific terms used to represent the following:
(i) Bending of a shoot towards light
Answer:
Phototropism

(ii) Growing of roots towards the earth
Answer:
Geotropism

(iii) Growing of a pollen tube towards ovule
Answer:
Chemotropism

(iv) Bending of roots towards water
Answer:
Hydrotropism

(v) Winding of tendril around a support
Answer:
Thigmotropism

(14) Bony box : Brain
……………… : Spinal cord
Answer:
Backbone (vertebral column)

Fill a and b to make the correct sequence.
Answer:
a – growth hormone releasing factor
b – pituitary gland

Value Based Questions With Answers

Question 1.
You saw a tragic road accident resulting in death of two persons. The reason for that is bike rider tried to overtake the truck from wrong-side with overspeed. He was not wearing a helmet that caused head injury.

You were scared because you also have same habit to drive activa without wearing helmet. Your father often warned you that to drive vehicle without a driving licence is a crime.
Questions :

1. Which part of brain is involved into learning how to drive vehicle?
2. Injury to which part of brain leads to instant death of an individual? Why?
3. Which gland became more active when you saw the accident? Which hormone was poured in blood? What changes did you feel in body?

Answer:

1. Cerebrum and Cerebellum
2. Medulla oblongata. Because all involuntary activities are controlled by it. Injury to that part disturbs all involuntary functions and leads to instant death.
3. Adrenal gland became more active. Adrenaline s hormone was poured in blood. Perspiration, increased heartbeats and breathing rate, etc. were the changes felt in the body.

Question 2.
Your father-mother are in a age-group at forty plus. On their routine blood test, some disturbance in their blood tests were reported, s Your family doctor advised your father to walk? regularly and consume a diet containing low s sugar. Doctor insisted your mother to take 25 mg elthroxine tablet daily.

Questions:

1. From your study, what do you think about s the report of the blood test of your father?
2. Which gland is not functioning properly and which hormone is not secreted in s appropriate amount in your father’s body?
3. Which gland is affected in your mother’s? body?
4. Why elthroxine tablet is suggested by the S doctor?

Answer:

1. Blood sugar level may be more than the normal level in the blood of father.
2. Pancreas, insulin
3. Thyroid gland
4. Elthroxine is a synthetic thyroxine. It fulfills? the deficiency of thyroxine.

Question 3.
You were visiting your friend’s house. You observed some indoor plants in drawing room. You noticed that shoots of all plants slightly moved towards open window. Aunty told you that she had often changed the position of s the arrangement of plants. But in any position shoots of plants showed same behaviour.

Questions:

1. Which movements are shown by plants?
2. Which hormone is responsible for such movement in plant?
3. How such hormone is functioning in plant body?
4. Which arrangement do you suggest for plants to show straight growth of shoot?

Answer:

1. Phototropism
2. Auxin
3. Auxin is synthesised at shoot tip and it diffuses towards shady side of the shoot. It stimulates the cells to grow longer on the side of the shoot which is away from light.
4. Generally plants that are arranged in open area where they get direct sunlight, may grow straight shoots.

Question 4.
Note five different situations you have experienced in which your adrenal gland had become overactive and had given you a ‘fight or rim away’ response.

The situations in which your adrenal gland had become overactive and had given a ‘fight or run away’ response.
(1) While driving an autovehicle without having a license, the police attempted to stop your vehicle and you increased your speed in an attempt to run away.

(2) Your classteacher caught you telling lie, even though you had not done your homework assigned.

(3) At the annual examination of Std. IX, you felt that the question paper of Science and Technology set at the examination was tough and you were afraid in the examination hall and started perspiring profusely.

(4) You went to see movie in the cinema hall by bunking the school. School authority informed about your absence in the classroom, to your parents. You tried to tell a lie at home and you were caught.

(5) In the World Cup Cricket final match Sehwag and Sachin were out with very low scores and you became tense, thinking that this prestigeous match will be lost.

[Note : In the conditions of stress or fear, there is somewhat over secretion of adrenaline from the adrenal glands.]

Practical Skill Based Questions With Answers

Question 1.
Four different students of your class observed network of neurons in a slide under microscope. They drew a diagram of synaptic junction.
From your knowledge.
Questions:

• Which of the following figures shows the correct pathway for the conduction of impulse?
• What is synaptic junction?
• Which other junction do you know that is some what similar to synapse?
  

Answer:

• Diagram (c) shows correct pathway for the conduction of impulse.
• A microscopic gap between nerve ending of axon of a neuron and dendrites of next neuron is called synaptic junction.
• Neuromuscular junction.

Question 2.
Your subject teacher arranged a potted plant as shown in figure (a). Next day turn its position horizontally as shown in figure (b). After few days, you observed the plant as shown in figure (c).

Questions :

• State the direction of growth of root and stem in the plant.
• What happens when the pot along with the plant is placed as in fig. (b)?
• Why the growth of the entire plant as shown in fig. (b) does not occur parallel to the soil surface?
• Which type of movement does this experiment explain?

Answer:

• The growth of root is negative phototropic and positive geotropic. The growth of stem is positive phototropic.
• When the plant was placed as in fig. (b). both stem and root were initially horizontal.
• The entire plant growth does not occur parallel to the soil surface because stem grows towards light and away from gravity.
• This experiment explains tropic movements due to growth in plant.

Question 3.
Observe figs, (a), (b). (c) and (d).

• Determine on the basis of your observations whether the movement occurring in the plant is growth based or not?
• State the type of movement occurred herein.
  

Answer:
In figures (a), (b) and (c) the movements are’ growth related and in fig. (d) the movement is not related to growth.

Memory Map:

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 4 द्विघात समीकरण Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 4 द्विघात समीकरण

लयूत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
एक मोटर बोट जिसकी स्थिर जल में चाल 18 किमी / घण्टा है। वह वोट 12 किमी धारा के प्रतिकूल जाने में समय धारा के अनुकूल जाने की अपेक्षा अधिक लेती है। धारा की चाल ज्ञात कीजिए।
हल :
माना कि धारा की चाल x किमी / घण्टा है।
धारा की अनुकूल में मोटर बोट की चाल = (18 + x) किमी / घण्टा
धारा की प्रतिकूल जाने में मोटर बोट की चाल = (18 – x) किमी / घण्टा
धारा की प्रतिकूल में जाने में लगा समय = \(\frac{12}{18-x}\) घण्टे
[∵ समय = दूरी / चाल]
इसी प्रकार अनुकूल जाने में लगा समय = \(\frac{12}{18+x}\) घण्टे
प्रश्नानुसार,
\(\frac{12}{18-x}-\frac{12}{18+x}\) = \(\frac{1}{2}\)
2 × 12 [(18 + x) – (18 – x)] = (18 + x) (18 – x)
24 [18 + x – 18 + x] = 324 – x²
24 × 2x = 324 – x²
48x + x² – 324 = 0
⇒ x² + 48x – 324 = 0
श्रीधराचार्य सूत्र से :
x = \(\frac{-48 \pm \sqrt{48^2+1296}}{2}\)
= \(\frac{-48 \pm \sqrt{3600}}{2}\)
= \(\frac{-48 \pm 60}{2}\)
= 6 या – 54
∵ x धारा की चाल है और यह ऋणात्मक नहीं हो सकती है। इसलिए हम – 54 को छोड़ देते है।
∴ धारा की चाल x = 6 किमी / घण्टा है।

प्रश्न 2.
एक रेलगाड़ी 63 किमी दूरी सामान्य चाल से तथा 72 किमी दूरी सामान्य चाल से 6 किमी / घण्टा ज्यादा चाल से तय करती है। यदि रेलगाड़ी यात्रा को पूरा करने में 3 घण्टे का समय लेती है, तो उसकी सामान्य चाल क्या है ?
हल :
माना कि रेलगाड़ी की चाल x किमी / घण्टा है। तब रेलगाड़ी की बढ़ी हुई चाल = (x + 6) किमी / घण्टा ।

⇒ 45x + 126 = x² + 6x
⇒ x² + 6x – 45x – 126 = 0
⇒ x² – 39x – 126 = 0
⇒ x² – (42 – 3)x – 126 = 0
⇒ x² – 42x + 3x – 126 = 0
⇒ x (x – 42) + 3 (x – 42) = 0
⇒ (x – 42) (x + 3) = 0
यदि x – 42 = 0 हो, तो x = 42
या x + 3 = 0 हो, तो x = – 3
∵ x रेलगाड़ी की सामान्य चाल है तथा यह कभी भी ऋणात्मक नहीं हो सकती है।
∴ x = – 3 को छोड़ देते हैं।
अत: x = 42
इसलिए रेलगाड़ी की सामान्य चाल = 42 किमी/ घण्टा ।

प्रश्न 3.
9000 रुपये को समान रूप से कुछ लोगों में बाँटा गया। यदि 20 व्यक्ति और होते तो प्रत्येक व्यक्ति को 160 रुपये कम मिलते। व्यक्तियों की वास्तविक संख्या कितनी है?
हल :
माना व्यक्तियों की वास्तविक संख्या है।
∴ प्रत्येक व्यक्ति का हिस्सा = \(\frac{9000}{x}\)
यदि व्यक्तियों की संख्या 20 और बढ़ायी जाए तो प्रति व्यक्ति का हिस्सा = \(\frac{9000}{x+20}\)
चूँकि 20 व्यक्ति बढ़ने पर प्रति व्यक्ति का हिस्सा 160 रुपये कम हो जाता हैं ।

⇒ 9000 × 20 = 160x (x + 20)
⇒ x² + 20x – 1125 = 0
⇒ x² + 45x – 25x – 1125 = 0
⇒ x(x + 45) – 25(x + 45) = 0
⇒ (x + 45)(x – 25) = 0
अर्थात् x + 45 = 0 या फिर x – 25 = 0
x = – 45 या x = 25
∵ व्यक्तियों की संख्या ऋणात्मक नहीं हो सकती है।
अतः व्यक्तियों की वास्तविक संख्या = 25 है।

प्रश्न 4.
17 मीटर व्यास वाले एक वृत्ताकार पार्क की परिसीमा के एक बिन्दु पर एक खम्भा इस प्रकार गाड़ना है कि इस पार्क के एक व्यास के दोनों अन्त बिन्दुओं पर बने फाटकों A और B से खम्भे की दूरियों का अन्तर 7 मीटर हो क्या ऐसा करना सम्भव है ? यदि है, तो दोनों फाटकों से कितनी दूरियों पर खम्भा गाड़ना है ?
हल :
माना कि बिन्दु C पर खम्भा गाड़ा गया है तथा फाटक B से बिन्दु C की दूरी = x मीटर

प्रश्नानुसार, फाटक B और A से खम्भे की दूरियों का अन्तर 7 मी. है।
∴ AC = (x + 7) मी.
चूँकि AB व्यास है।
अतः ∠ACB = 90°
(∵ अर्द्धवृत्त में बना कोण समकोण होता है)
अब समकोण त्रिभुज ACB में,
AB² = AC² + BC²
(पाइथागोरस प्रमेय से)
⇒ (17)² = (x + 7)² + x²
⇒ 289 = x² + 49 + 14x + x²
⇒ 289 = 2x² + 14x + 49
⇒ 2x² + 14x + 49 – 289 = 0
⇒ 2x² + 14x – 240 = 0
⇒ x² + 7x – 120 = 0
अब b² – 4ac = (7)² – 4 × 1 × (120)
= 49 + 280
= 529 > 0
अतः दिए गए द्विघात समीकरण के दो वास्तविक मूल है और इसलिए खम्भे को पार्क की परिसीमा पर गाड़ा जा सकता है।
अब x² + 7x – 120 = 0
⇒ x² + (15 – 8)x – 120 = 0
⇒ x² + 15x – 8x – 120 = 0
⇒ (x² + 15x) (8x + 120) = 0
⇒ x(x + 15) – 8 (x + 15) = 0
⇒ (x + 15 ) (x – 8 ) = 0
⇒ x + 15= 0 और x 8 = 0
⇒ x = – 15 और x = 8
∵ x खम्भे और फाटक B के बीच की दूरी है। अतः यह कभी भी ऋणात्मक नहीं हो सकती है।
अतः x = – 12 को छोड़ देते हैं।
अतः x = 8 मी और x + 7 = 8 + 7 = 15 मी
अतः खम्भे से फाटक की दूरी 15 मीटर तथा फाटक B से 8 मीटर है।

प्रश्न 5.
500 किमी की एक हवाई उड़ान में एक वायुयान खराब मौसम के कारण धीनी गति से चला। पूरी उड़ान की औसत चाल 200 किमी / घंटा घट गई तथा उड़ान का समय 30 मिनट बढ़ गया। उड़ान का मूल समय ज्ञात कीजिए ।
हल :
माना वायुयान की सामान्य चाल x किमी / घंटा है।
वायुयान की घटी चाल = (x – 200) किमी / घंटा
दूरी = 600 किमी
600 किमी दूरी तय करने में वायुयान को लगा समय = दूरी / चाल = \(\frac{600}{x}\)घंटे
नई चाल से 600 किमी दूरी तय करने में लगा समय = \(\frac{600}{x-200}\)घंटे
प्रश्नानुसार,

⇒ x² – 200x = 2400
⇒ x² – 200x – 2400 = 0
⇒ x² – 600x + 400x – 2400 = 0
⇒ x(x – 600) + 400(x – 600) = 0
⇒ (x – 600) (x + 400) = 0
यदि x – 600 = 0 तो x = 600
और यदि x + 400 = 0, तो x = – 400
∵ वायुयान की चाल ऋणात्मक नहीं तो सकती, इसलिए x ≠ – 400
∴ वायुयान की सामान्य चाल = 600 किमी / घंटा
अतः उड़ान का मूल समय = दूरी / चाल = \(\frac{600}{600}\) = 1 घंटा

प्रश्न 6.
एक रेलगाड़ी 480 किमी की दूरी एकसमान चाल से चलती है। यदि उसकी चाल 8 किमी / घंटा कम हो, यह उसी दूरी को तय करने में 3 घंटे अधिक लेती है। रेलगाड़ी की मूल चाल ज्ञात कीजिए।
हल :
माना रेलगाड़ी की मूल चाल किमी/घंटा है।
तब रेलगाड़ी की नई घटी हुई चाल (x – 8) किमी / घंटा है।
दूरी = 480 किमी
प्रश्नानुसार,
\(\frac{480}{x-8}-\frac{480}{x}\) = 3
[∵ समय = दूरी / चाल]
⇒ \(\frac{480[x-(x-8)]}{x(x-8)}\) = 3
⇒ 480 × 8 = 3x (x – 8)
⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
⇒ x² – 40x + 32x – 1280 = 0
⇒ x(x – 40) + 32(x – 40) = 0
⇒ (x – 40) (x + 32) = 0
यदि x – 40 = 0 तो x = 40
और यदि x + 32 = 0 तो x = – 32, जो कि असम्भव हैं, क्योंकि चाल ऋणात्मक नहीं हो सकती।
अतः रेलगाड़ी की मूल चाल 40 किमी/घंटा है।

प्रश्न 7.
एक तेज चलने वाली रेलगाड़ी 600 किमी की यात्रा में एक धीमी चलने वाली रेलगाड़ी से 3 घंटे कम समय लेती है। यदि धीमी चलने वाली रेलगाड़ी की चाल, तेज चलने वाली रेलगाड़ी की चाल से 10 किमी / घंटा कम है, तो दोनों गाड़ी की चाल ज्ञात कीजिए।
हल :
माना तेज चलने वाली रेलगाड़ी की चाल x किमी / घंटा है,
तब
धीमी चलने वाली रेलगाड़ी की चाल (x – 10) किमी / घंटा होगी।
तेज चलने वाली गाड़ी द्वारा 600 किमी की यात्रा में लिया गया समय = \(\frac{600}{x}\) घंटे
धीमी चलने वाली गाड़ी द्वारा 600 किमी. की यात्रा में लिया गया समय = \(\frac{600}{x-10}\) घंटे
प्रश्नानुसार,
\(\frac{600}{x-10}-\frac{600}{x}\) = 3
⇒ \(\frac{600[(x-(x-10)]}{x(x-10)}\) = 3
⇒ 600 × [(x – (x + 10)] = 3x(x – 10)
⇒ 6000 = 3x(x – 10)
⇒ x(x – 10) = 2000
⇒ x² – 10x – 2000 = 0
⇒ x² – 50x + 40x – 2000 = 0
⇒ x (x – 50) + 40 (x – 50) = 0
⇒ (x – 50) (x + 40) = 0
यदि x – 50 = 0, तो x = 50
और यदि x + 40 = 0 तो x = – 40; जो कि असम्भव है।
अत: तेज चलने वाली रेलगाड़ी की चाल = 50 किमी / घंटा
तथा धीमी चलने वाली रेलगाड़ी की चाल = (50 – 10) किमी / घंटा = 40 किमी / घंटा

प्रश्न 8.
दो प्राकृत संख्याओं का अंतर 5 है, तथा उनके प्रतिलोमों का अंतर \(\frac{1}{10}\) है। संख्याएँ ज्ञात कीजिए।
हल :
माना दो प्राकृत संख्याएँ और x + 5 हैं।
प्रश्नानुसार,
⇒ \(\frac{1}{x}-\frac{1}{x+5}\) = \(\frac{1}{10}\)
⇒ \(\frac{(x+5)-x}{x(x+5)}\) = \(\frac{1}{10}\)
⇒ \(\frac{5}{x^2+5 x}\) = \(\frac{1}{10}\)
⇒ x² + 5x = 50
⇒ x² + 5x – 50 =0
⇒ x² + 10x – 5x – 50 = 0
⇒ x(x + 10) – 5(x + 10) = 0
⇒ (x + 10) (x – 5) = 0
यदि x + 100, तो x = – 10; जो कि असम्भव है, क्योंकि कि प्राकृत संख्याएँ ऋणात्मक नहीं होती।
यदि x – 5 = 0, तो x = 5
तथा x + 5 = 5 + 5 = 10
अतः दो प्राकृत संख्याएँ 5 और 10 है।

प्रश्न 9.
एक व्यक्ति के पास एक टूर (यात्रा) के दौरान खर्च के लिए ₹ 4200 हैं। यदि वह अपना दूर 3 दिन बढ़ा दे उसे अपना प्रतिदिन का व्यय ₹70 कम करना पड़ता है। उसके दूर की मूल अवधि ज्ञात कीजिए।
हल :
माना व्यक्ति के टूर की अवधि दिन है।
प्रतिदिन का व्यय = \(\frac{4200}{x}\)
यदि दूर 3 दिन बढ़ा दिया जाए, तब प्रतिदिन का व्यय = \(\frac{4200}{x+3}\)
प्रश्नानुसार,
\(\frac{4200}{x}-\frac{4200}{x+3}\) = 70
⇒ \(\frac{4200[x+3-x]}{x(x+3)}\) = 70
⇒ 4200 × 3 = 70x (x + 3)
⇒ \(\frac{4200 \times 3}{70}\) = x² + 3x
⇒ x² + 3x – 180 = 0
⇒ x + 15x – 12x – 180 = 0
⇒ x(x + 15) – 12 (x + 15) = 0
⇒ (x + 15) (x – 12) = 0
यदि x + 15 = 0, तो x = – 15; जो कि असम्भव है, क्योंकि दिनों की संख्या ऋणात्मक नहीं हो सकती।
और यदि x – 12 = 0, तो x = 12
अतः यात्रा की अवधि 12 दिन है।

प्रश्न 10.
तीन क्रमागत धनपूर्णांक ऐसे हैं कि पहले के वर्ग तथा अन्य दो कि गुणनफल को जोड़ने पर 46 प्राप्त होता है। पूर्णांक ज्ञात कीजिए।
हल :
माना तीन क्रमागत धनपूर्णांक x (x + 1) तथा (x + 2) हैं।
प्रश्नानुसार,
x² + (x + 1)(x + 2) = 46
x² + x + 2x + x + 2 = 46
2x² + 3x + 2 – 46 = 0
2x² + 3x – 44 = 0
2x² + 11x – 8x – 44 = 0
x(2x + 11) – 4(2x + 11) = 0
(2x + 11) (x – 4) = 0
यदि 2x + 11 = 0, तो x = – \(\frac{11}{2}\), असम्भव क्योंकि ऋणात्मक है।
यदि x – 4 = 0, तो x = 4
∴ x + 1 = 4 + 1 = 5 और x + 2 = 4 + 2 = 6
अतः अभीष्ट धनपूर्णांक संख्याएँ 4, 5 और 6 हैं।

प्रश्न 11.
कुछ विद्यार्थियों ने पिकनिक पर जाने की योजना बनाई खाने का कुल बजट ₹ 2000 रखा गया परन्तु 5 विद्यार्थियों के न आने पर प्रति विद्यार्थी खाने पर खर्च ₹ 20 बढ़ गया। कितने विद्यार्थी पिकनिक पर गए तथा प्रत्येक विद्यार्थी ने खाने के लिए कितनी राशि दी ?
हल :
माना पिकनिक पर x विद्यार्थी गए।
कुल धन राशि = ₹ 2000
∴ प्रत्येक विद्यार्थी द्वारा दी गई राशि = \(\frac{2000}{x}\)
यदि 5 विद्यार्थी न आते तब प्रत्येक विद्यार्थी द्वारा दी गई राशि = \(\frac{2000}{x-5}\)
प्रश्नानुसार,
\(\frac{2000}{x-5}-\frac{2000}{x}\) = 20
\(\frac{2000[x-(x-5)]}{x(x-5)}\) = 20
2000 × 5 = 20x (x – 5)
500 = x² – 5x
x² – 5x – 500 = 0
x² – 25x + 20x – 500 = 0
x – (x – 25) + 20(x – 25) = 0
(x – 25) (x + 20) = 0
यदि x – 25 = 0, तो x = 25
और यदि x + 20 = 0 तो x = – 20 (असम्भव)
अतः पिकनिक पर गए विद्यार्थी = 25
प्रत्येक विद्यार्थी द्वारा दी गई राशि = ₹ \(\frac{2000}{25}\) = ₹ 80

प्रश्न 12.
एक ऐसे आयताकार पार्क को बनाना है जिसकी चौड़ाई उसकी लम्बाई से 3 मी कम है। इसका क्षेत्रफल पहले से निर्मित समद्विबाहु त्रिभुजाकार पार्क जिसका आधार आयाताकार पार्क की चौड़ाई के बराबर तथा ऊँचाई 12 मी. है, से 4 वर्ग मी. अधिक हो। इस पार्क की लम्बाई और चौड़ाई ज्ञात कीजिए।
हल :
माना पार्क की लम्बाई x मी है।
पार्क की चौड़ाई (x – 3) मी होगी।
आयताकार पार्क का क्षेत्रफल = x (x – 3) वर्ग मी समद्विबाहु त्रिभुजाकार पार्क का आधार = आयताकार पार्क की चौड़ाई = (x – 3) मी ऊँचाई = 12 मी
समद्विबाहु त्रिभुजाकार पार्क का आधार
= \(\frac{1}{2}\) × आधार × ऊँचाई
= \(\frac{1}{2}\) × (x – 3) × 12
= 6 (x – 3) वर्ग मी
प्रश्नानुसार,
आयताकार पार्क का क्षेत्रफल = त्रिभुजाकार पार्क का क्षेत्रफल + 4
x(x – 3) = 6(x – 3) + 4
x² – 3x = 6x – 18 + 4
x² – 3x – 6x + 14 = 0
x² – 9x + 14 = 0
x² – 7x – 2x + 14 = 0
x(x – 7) – 2(x – 7) = 0
(x – 7)(x – 2) = 0
यदि x – 7 = 0, तो x = 7
और यदि x – 2 = 0 तो x = 2 जो कि असम्भव है, क्योंकि अगर लम्बाई 2 मी. है तो चौड़ाई (x – 3) मी = – 1 मी होगी जो कि ऋणात्मक है।
अतः पार्क की लम्बाई = 7 मी
तथा चौड़ाई = (7 – 3) मी = 4 मी

प्रश्न 13.
कुछ लम्बाई वाले एक कपड़े की कुल लागत ₹ 200 है। यदि यह कपड़ा 5 मीटर अधिक लम्बा हो तथा प्रत्येक मीटर कपड़े की लागत ₹ 2 कम हो, तो कपड़े की कुल लागत में कोई परिवर्तन नहीं होगा। कपड़े का वास्तविक प्रति मीटर मूल्य ज्ञात कीजिए तथा कपड़े की लम्बाई भी ज्ञात कीजिए।
हल :
माना, कपड़े की लम्बाई x मी हैं तथा कपड़े के प्रत्येक मीटर का मूल्य y है।
प्रश्नानुसार,
xy = 200 ……..(i)
यदि कपड़े की लम्बाई 5 मी अधिक होतो तथा प्रत्येक मीटर का मूल्य ₹2 कम होता तो,
(x + 5)(y – 2) = 200
xy – 2x + 5y – 10 = 200 ……..(ii)
समीकरण (i) व (ii) को बराबर करने पर,
xy = xy – 2x + 5y – 10
2x – 5y = 10 ……..(iii)
2x – 5 × \(\frac{200}{x}\) = – 10 (समीकरण (i) से y = \(\frac{200}{x}\))
2x – \(\frac{1000}{x}\) = – 10
2x² – 1000 = – 10x
2x² + 10x – 1000 = 0
x² + 5x – 500 = 0
x² + 25x – 20x – 500 = 0
x(x + 25) – 20(x + 25) = 0
(x – 20)(x + 25) = 0
यदि x – 20 = 0, तो x = 20
और x + 25 = 0 तो x = – 25; जो कि असम्भव है, क्योंकि लम्बाई ऋणात्मक नहीं होती।
अब x × y = 200
20 × y = 200
y = 10
अतः कपड़े की लम्बाई 20 मीटर है तथा प्रत्येक मीटर का मूल्य ₹ 10 है।

प्रश्न 14.
दो पानी के नल एक साथ एक टैंक को 1\(\frac{7}{8}\) घंटों में भर सकते हैं। बड़े व्यास वाला नल टैंक को भरने में, कम व्यास वाले नल से 2 घंटे कम समय लेता है। प्रत्येक नल द्वारा अलग से टैंक को भरने का समय ज्ञात कीजिए।
हल :
माना छोटा नल टैंक को भरने में x घंटे लेता है।
∴ बड़ा नल टैंक को भरने में (x – 2) घंटे लेगा।
अब, 1 घंटे में बड़े नल द्वारा भरा गया पानी = \(\frac{1}{x-2}\)
1 घंटे में छोटे नल द्वारा भरा गया पानी = \(\frac{1}{x}\)
तथा 1 घंटे में दोनों नल द्वारा भरा गया पानी = \(\frac{8}{15}\)
\(\frac{1}{x}+\frac{1}{x-2}\) = \(\frac{8}{15}\)
\(\frac{x-2+x}{x(x-2)}\) = \(\frac{8}{15}\)
\(\frac{2 x-2}{x(x-2)}\) = \(\frac{8}{15}\)
\(\frac{x-1}{x(x-2)}\) = \(\frac{4}{15}\)
15x – 15 = 4x² – 8x
4x² – 8x – 15x + 15 = 0
4x² – 23x + 15 = 0
4x² – 20x – 3x + 15 = 0
4x(x – 5) – 3(x – 5) = 0
(x – 5) (4x – 3) = 0
यदि x – 5 = 0, तो x = 5
और यदि 4x – 3 = 0, तो x = \(\frac{3}{4}\) जो कि अमान्य है।
अतः छोटा नल को टैंक भरने में 5 घण्टे लगेंगे एवं बड़े नल को 3 घण्टे लगेंगे।

प्रश्न 15.
निम्न समीकरणों के मूल ज्ञात कीजिए :
(i) 8x² – 2x – 3 = 0
(ii) 14x² + 17x – 6 = 0
(iii) 3x² – 4\(\sqrt{3}\)x + 4 = 0
(iv) x² + 5x – (a² + a – 6) = 0
हल :
(i) दिया है,
8x² – 2x – 3 = 0
⇒ 8x² – 6x + 4x – 3 = 0
⇒ 2x(4x – 3) + 1(4x – 3) = 0
⇒ (4x – 3)(2x + 1) = 0
यदि 4x – 3 = 0, तो x = \(\frac{3}{4}\)
और यदि 2x + 1 = 0, तो x = –\(\frac{1}{2}\)
अत: x = \(\frac{3}{4}\), –\(\frac{1}{2}\)

(ii) दिया है, 14x² + 17x – 6 = 0
⇒ 14x² + 21x – 4x – 6 = 0
⇒ 7x(2x + 3) – 2(2x + 3) = 0
⇒ (2x + 3)(7x – 2) = 0
यदि 2x + 3 = 0, x = – \(\frac{3}{2}\)
और यदि 7x – 2 = 0, तो x = \(\frac{2}{7}\)
अत: x = –\(\frac{3}{2}\), \(\frac{2}{7}\)

(iii) दिया है,
3x² – 4\(\sqrt{3}\)x + 4 = 0
3x² – 2\(\sqrt{3}\)x – 2\(\sqrt{3}\)x + 4 = 0
\(\sqrt{3}\)x(\(\sqrt{3}\) – 2) – 2(\(\sqrt{3}\) – 2) = 0
(\(\sqrt{3}\) – 2) (\(\sqrt{3}\) – 2) = 0
यदि \(\sqrt{3}\)x – 2 = 0 तो x = \(\frac{2}{\sqrt{3}}\)
अतः x = \(\frac{2}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)

(iv) दिया है,
x² + 5x – (a² + a – 6) = 0
यहाँ, a² + a – 6 = a² + 3a – 2a – 6
= a(a + 3) – 2(a + 3)
= (a + 3)(a – 2)
∴ x² + 5x – (a – 3)(a – 2) = 0
x² + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0
x[x + a + 3] – (a – 2)[x + a + 3] = 0
(x – a + 2)(x + a + 3) = 0
यदि x – a + 2 = 0, x = a – 2
और यदि x + a + 3 = 0, तो x = – (a + 3)
अत: x = (a – 2) और – (a + 3)

प्रश्न 16.
x के लिए हल कीजिए:

हल :

⇒ (4x + 1)(5x + 1) = 3(x² + 5x + 4)
⇒ 20x² + 4x + 5x + 1 = 3x² + 15x + 12
⇒ 17x² – 6x – 11 = 0
⇒ 17x² – 17x + 11x – 11 = 0
⇒ 17x(x – 1) + 11(x – 1) = 0
⇒ (x – 1)(17x + 11) = 0
⇒ x = 1 या x = \(\frac{-11}{17}\)
दिया है x ≠ 1, अतः x = \(\frac{-11}{17}\)

(ii) दिया है, \(\frac{1}{2 x-3}+\frac{1}{x-5}\) = 1\(\frac{1}{9}\)
\(\frac{x-5+2 x-3}{(2 x-3)(x-5)}\) = \(\frac{10}{9}\)
\(\frac{3 x-8}{2 x^2-13+15}\) = \(\frac{10}{9}\)
9(3x – 8) = 10 (2x² – 13x + 15)
27x – 72 = 20x² – 130x + 150
20x² – 157x + 222 = 0
20x² – 120x – 37x + 222 = 0
20x(x – 6) – 37 (x – 6) = 0
(x – 6) (20x – 37) = 0
(x – 6 ) = 0 या 20x – 37 = 0
अत: x = 6 या x = \(\frac{37}{20}\)

(iii)

5x² + 2x + 2 = 2(2x² – x – 1)
5x² + 2x + 2 = 4x² – 2x – 2
x² + 4x + 4 = 0
(x + 2)² = 0
या तो x + 2 = 0 या x + 2 = 0
अतः x = – 2, -2

प्रश्न 17.
दो नल एक साथ टैंक को 3\(\frac{1}{13}\) घण्टे में भर सकते हैं। यदि एक नल टैंक को भरने में दूसरे नल से 3 घण्टे अधिक लेता है, तो प्रत्येक नल टैंक को भरने में कितना समय लेगा ?
हल :
माना कि टैंक एक नल से x घण्टे में भरता है। अत: टैंक दूसरे नल से (x + 3) घण्टे में भरेगा।
टैंक को दोनों नल एक साथ भरते हैं = 3\(\frac{1}{13}\)
= \(\frac{40}{13}\) घण्टे में
प्रश्नानुसार,

13x² + 39x = 80x + 120
13x² – 41x – 120 = 0
13x² – 65x + 24x – 120 = 0
13x(x – 5) + 24 (x – 5) = 0
(x – 5)(13x + 24) = 0
यदि x – 5 = 0 तो x = 5
और यदि 13x + 24 = 0 तो x = – \(\frac{24}{13}\), जो कि संभव नहीं है।
अतः एक नल टैंक भरेगा = 5 घण्टे में।
और दूसरा नल टैंक भरेगा = (x – 3) = 8 घण्टे में ।

प्रश्न 18.
एक रेलगाड़ी 300 किमी की दूरी एकसमान चाल से तय करती है। यदि रेलगाड़ी की चाल 5 किमी / घंटा बढ़ा दी जाए, तो यात्रा में 2 घंटे कम समय लगता है। रेलगाड़ी की मूल चाल ज्ञात कीजिए।
हल :
माना, रेलगाड़ी की मूल चाल = x किमी / घंटा
रेलगाड़ी की नई चाल = (x + 5) किमी / घंटा
दूरी = 300 किमी
प्रश्नानुसार,
\(\frac{300}{x}-\frac{300}{x+5}\) = 2
⇒ \(\frac{300(x+5-x)}{(x)(x+5)}\) = 2
⇒ 1500 = 2(x² + 5x)
⇒ 1500 = 2x² + 10x
⇒ 2x² + 10x – 1500 = 0
⇒ x² + 5x – 750 = 0
⇒ x² + 30x – 25x – 750 = 0
⇒ x(x + 30) – 25 (x + 30) = 0
⇒ (x + 30)(x – 25) = 0
यदि x + 30 = 0 तो x = – 30, जो अमान्य है।
और यदि x – 25 = 0 तो x = 25
अतः रेलगाड़ी की मूल चाल = 25 किमी / घण्टा ।

प्रश्न 19.
A एक कार्य को करने में B से 6 दिन कम लेता है। यदि A और B दोनों एक साथ काम करते हुए इसे 4 दिन में कर सकते हैं, तो B इस कार्य को समाप्त करने में कितने दिन लेगा?
हल :
माना B कार्य को x दिन में समाप्त कर सकता है। तो A कार्य को (x – 6) दिन में समाप्त करेगा।
दोनों मिलकर कार्य 4 दिन में समाप्त करते हैं।
प्रश्नानुसार, \(\frac{1}{x}+\frac{1}{x-6}\) = \(\frac{1}{4}\)
⇒ \(\frac{x-6+x}{(x)(x-6)}\) = \(\frac{1}{4}\)
⇒ 4(2x – 6) = x² – 6x
⇒ 8x – 24 = x² – 6x
⇒ x² – 14x + 24 = 0
⇒ x² – 12x – 2x + 24 = 0
⇒ x (x – 12) – 2 (x – 12) = 0
⇒ (x – 12) (x – 2) = 0
यदि x – 12 = 0 तो x = 12
और यदि x – 2 = 0 तो x = 2 (अमान्य हैं।)
अतः B कार्य को करने में 12 दिन लेगा।
A कार्य को करने में 6 दिन लेगा।

प्रश्न 20.
एक नाव की शांत जब में चाल 15 किमी / घंटा है। यह नाव 30 किमी धारा के विपरीत दिशा में जाकर पुन: उसी जगह 4 घंटे 30 मिनट में वापस लौट आती है। धारा की चाल ज्ञात कीजिए।
हल :
माना धारा की चाल x किमी / घंटा है।
प्रश्नानुसार,

अतः धारा की चाल 5 किमी / घण्टा ।

प्रश्न 21.
एक आयताकार खेल का विकर्ण उसकी छोटी भुजा से मी अधिक लम्बा है। यदि बड़ी भुजा छोटी भुजा से 20 मी. अधिक हो, तो खेत की भुजाएँ ज्ञात कीजिए।
हल :
माना खेत की छोटी भुजा = x मी
∴ खेत की बड़ी भुजा = (x + 20) मी
और खेत का विकर्ण = (x + 40) मी
पाइथागोरस प्रमेय से,
(विकर्ण)² = शेष दोनों भुजाओं के वर्गों का योग
⇒ (x + 40)² = x² + (x + 20)²
⇒ x² + 1600 + 80x = x² + x² + 400 + 40x
⇒ x² – 2x² + 80x – 40 + 1600 – 400 = 0
⇒ – x² – 40x + 1200 = 0
⇒ x² + 40x – 1200 = 0
⇒ x² + 60x – 20x – 1200 = 0
⇒ x(x + 60) – 20(x + 60) = 0
⇒ (x + 60) (x – 20) = 0
यदि x + 60 = 0, तो x = – 60 जो कि असम्भव है, क्योंकि भुजा की लम्बाई ऋणात्मक नहीं हो सकती।
और यदि x – 20 = 0 तो x = 20
अतः खेत की छोटी भुजा = 20 मी
तथा खेत की बड़ी भुजा = (20 + 20) मी = 40 मी.

प्रश्न 22.
एक आयताकार खेत का विकर्ण उसकी छोटी भुजा से 25 मी अधिक लंबा है। यदि बड़ी भुजा छोटी भुजा से 23 मी अधिक है, तो खेत की भुजाएँ ज्ञात कीजिए ।
हल :
माना रेलगाड़ी की चाल x किमी / घंटा है।
300 किमी दूरी तय करने में लगा समय = \(\frac{300}{x}\) घंटे
रेलगाड़ी की नई चाल = (x + 10) किमी / घंटा
नई चाल से 300 किमी दूरी तय करने में लगा समय = \(\frac{300}{x+10}\) घंटे
प्रश्नानुसार,
\(\frac{300}{x}-\frac{300}{x+10}\) = 1
⇒ \(\frac{300[x+10-x]}{x(x+10)}\) = 1
⇒ 300 × 10 = x(x + 10)
⇒ x² + 10x – 3000 = 0
⇒ x² + 60x – 50x – 3000 = 0
⇒ x(x + 60) – 50 (x + 60) = 0
⇒ (x + 60) (x – 50) = 0
यदि x + 60 = 0, तो x = – 60; असम्भव क्योंकि चाल ऋणात्मक नहीं होती।
और यदि x – 50 = 0 तो x = 50
अतः रेलगाड़ी की चाल = 50 किमी / घंटा

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. ऐसा व्यंजक जिसमें चर की घात ……………….. होती है, द्विघात बहुपद कहलाता है।
2. समीकरण x² + bx + c = 0 के मूल बराबर हैं, यदि ………………. है।
3. द्विघात समीकरण 3x² – 2x + 4 = 0 के मूल ……………….. होंगे।
4. समीकरण 9x² = 10 का मानक रूप ………………….. है।
5. प्रत्येक द्विघात समीकरण के अधिक से अधिक ………………. वास्तविक मूल हो सकते हैं।

हल :

1. 2
2. b² – 4c = 0
3. काल्पनिक
4. 9x² + 0x – 10 = 0
5. दो

निम्न में से सत्य / असत्य कथन छाँटिए :

प्रश्न (ख)

1. द्विघात समीकरण ax² + bx + c = 0 का विविक्तर, D = b² + 4ac होता है।
2. यदि b² – 4ac > 0 तथा पूर्ण वर्ग हो, तो मूल वास्तविक, परिमेय तथा असमान होते हैं।
3. यदि b² – 4ac > 0 तथा पूर्ण वर्ग न हो, तो मूल वास्तविक, अपरिमेय तथा असमान होते हैं।
4. यदि b² – 4ac < 0 तो मूल काल्पनिक होते हैं।
5. यदि b² – 4ac = 0 तो मूल काल्पनिक, परिमेय तथा असमान होते हैं।

हल :

1. असत्य,
2. सत्य,
3. सत्य,
4. सत्य,
5. असत्य

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
बहुपद x² – 3x – m (m + 3) के शून्यक हैं:
(A) m, m + 3
(B) – m, m + 3
(C) m, – (m + 3)
(D) – m, – (m + 3)
हल :
शून्यक के लिए : x² – 3x – m(m + 3) = 0
x² – (3 + m – m)x – m(m + 3) = 0
x² – (m + 3)x + mx – m(m + 3) = 0
x[x – (m + 3)] + m[x – (m + 3)] = 0
[x – (m + 3)] (x + m) = 0
यदि [x – (m + 3)] = 0, तो x = m + 3
यदि x + m = 0, तो x = – m
अत: विकल्प (B) सही है।

प्रश्न 2.
द्विघात समीमरण x² – 0.04 = 0 के मूल हैं:
(A) ± 0.2
(B) ± 0.02
(C) 0.4
(D) 2
हल :
दिया है, x² – 0.04 = 0
x² = 0.04 = \(\frac{4}{100}\)
x = ± \(\sqrt{\frac{4}{100}}\)
= ± \(\frac{2}{10}\) = ± 0.2
अत: सही विकल्प (A) है।

प्रश्न 3.
λ का वह मान जिसके लिए (x² + 4x + λ) एक पूर्ण वर्ग है, है:
(A) 16
(B) 9
(C) 1
(D) 4
हल :
दिया है, x² + 4x + λ
= (x)² + 2(2) x + λ
= (x)² + 2(2) x + (2)²
= (x + 2)², जो कि एक पूर्ण वर्ग है।
∴ λ = (2)² = 4
अत: सही विकल्प (D) है।

प्रश्न 4.
द्विघात समीकरण x² – 4x + k = 0 के दो भिन्न वास्तविक मूल होंगे, यदि
(A) k = 4
(B) k > 4
(C) k = 16
(D) k < 4
हल :
दो भिन्न वास्तविक मूलों के लिए प्रतिबन्ध :
B² – 4AC > 0
⇒ (-4)² – 4 × 1 × k > 0
⇒ 16 – 4k > 0
⇒ – 4k > – 16
⇒ k < 4
अत: सही विकल्प (D) है।

प्रश्न 5.
यदि द्विघात समीकरण 2x² + kx + 2 = 0 के मूल समान हों, तो का मान है :
(A) 4
(B) ± 4
(C) – 4
(D) = 0
हल :
मूलों के समान होने के लिए प्रतिबन्ध :
B² – 4AC = 0
⇒ (k)² – 4 × 2 × 2 = 0
⇒ k² – 16 = 0
⇒ k² = 16
⇒ k = ± 4
अत: सही विकल्प (B) है।

प्रश्न 6.
द्विघात समीकरण 2x² – 4x + 3 = 0 के मूल हैं:
(A) वास्तविक तथा बराबर
(B) वास्तविक तथा भिन्न
(C) वास्तविक नहीं
(D) वास्तविक
हल :
यहाँ B² – 4AC = (-4)² – 4 × 2 × 3
= 16 – 24 = – 8
⇒ B² – 4AC < 0
अत: सही विकल्प (C) है।

प्रश्न 7.
समीकरण ax² + bx + c = 0, a ≠ 0 के मूल वास्तविक नहीं होंगे यदि :
(A) b² < 4ac
(B) b² > 4ac
(C) b² = 4ac
(D) b = 4ac
हल :
सही विकल्प (A) है।

प्रश्न 8.
द्विघात समीकरण px² + qx + r =0, p ≠ 0 के मूल समान होंगे यदि :
(A) p² < Apr
(B) p² > 4qr
(C) p² = 4pr
(D) p² = 4qr
हल :
सही विकल्प (C) है।

प्रश्न 9.
द्विघात समीकरण 2x² – x – 6 = 0 के मूल है :
(A) – 2, \(\frac{3}{2}\)
(B) 2, – \(\frac{3}{2}\)
(C) – 2, – \(\frac{3}{2}\)
(D) 2, \(\frac{3}{2}\)
हल :
दिया गया द्विघात समीकरण है :
2x² – x – 6 = 0
⇒ 2x² – (4 – 3) x – 6 = 0
⇒ 2x² – 4x + 3x – 6 = 0
⇒ (2x² – 4x) + (3x – 6) = 0
⇒ 2x(x – 2) + 3(x – 2) = 0
⇒ (x – 2) (2x + 3) = 0
⇒ x = – 2 = 0 या 2x + 3 = 0
⇒ x = 2 या x = \(\frac{-3}{2}\)
अत: विकल्प (B) सही है।

प्रश्न 10.
द्विघात समीकरण 2x² – \(\sqrt{5}\)x + 1 = 0 के :
(A) दो भिन्न वास्तविक मूल है
(B) दो बराबर वास्तविक मूल है
(C) कोई वास्तविक मूल नहीं है
(D) दो से अधिक वास्तविक मूल है।
हल :
सही विकल्प (C) है।

प्रश्न 11.
यदि f(x) = ax² + bx + c और c = \(\frac{b^2}{4 a}\) हो, तो f(x) = 0 के मूल होंगे :
(A) वास्तविक और बराबर
(B) वास्तविक और असमान
(C) वास्तविक नहीं
(D) इनमें से कोई नहीं
हल :
सही विकल्प (A) है।

प्रश्न 12.
यदि p(x) एक बहुपद में चर x इस प्रकार है कि p(h) = 0 हो तो h, p(x) का
(A) मूल है
(B) गुणनखण्ड है
(C) शून्यक है
(D) इनमें से कोई नहीं ।
हल :
सही विकल्प (C) है।

प्रश्न 13.
बहुपद p(x) में x इस तरह कि p (h) = 0 हो, तो p(x), p(x – h) :
(A) मूलं है.
(B) गुणनखण्ड है
(C) शून्यक है
(D) इनमें से कोई नहीं
हल:
सही विकल्प (B) है।

प्रश्न 14.
दो संख्याएँ ज्ञात कीजिए जिनका योग 27 और गुणनफल 182 हो :
(A) 13 और 14
(C) 17 और 10
(B) 15 और 12
(D) इनमें से कोई नहीं
हल :
सही विकल्प (A) है।

प्रश्न 15.
यदि द्विघात समीकरण x² (a² + b) + 2x( ac + bad) + (c2 + 2) = 0 के मूल बराबर हों तो :
(A) ad = – bc
(B) ad = ± bc
(C) ad = bc
(D) इनमें से कोई नहीं
हल :
सही विकल्प (A) है।

Jharkhand Board JAC Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 1 Chemical Reactions and Equations

Jharkhand Board Class 10 Science Chemical Reactions and Equations Textbook Questions and Answers

Question 1.
Which of the statements about the reaction below are incorrect?
2PbO(s) + C(s) → 2Pb(s) + CO₂(g)
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.
(i) (a) and (b)
(ii) (a) and (c)
(iii) (a), (b) and (c) (iv) all
Answer:
(i) (a) and (b) s

Question 2.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
The above reaction is an example of a…
(a) combination reaction.
(b) double displacement reaction.
(c) decomposition reaction.
(d) displacement reaction.
Answer:
(d) displacement reaction.

Question 3.
What happens when dilute hydrochloric acid is added to iron fillings?
(a) Hydrogen gas and iron chloride are s produced.
(b) Chlorine gas and iron hydroxide are produced.
(c) No reaction takes place.
(d) Iron salt and water are produced,
Answer:
Hydrogen gas and iron chloride are produced.

Question 4.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
A chemical reaction in which atoms of $ different elements of reactant and product side are equal then the chemical equation S is called balanced chemical equation.

Now, according to the law of conservation of mass, matter can neither be created, nor be destroyed. Hence in a chemical reaction, mass of reactants S should be equal to the mass of products.

Number of atoms of each element should be equal on both side of the reaction for keeping mass constant.
Hence, it is essential to balance the chemical equation.

Question 5.
Translate the following statements into chemical equations and then balance them:
(a) Hydrogen gas combine with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassiam metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer:

Question 6.
Balance the following chemical equations:
(a) HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
(b) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Answer:
(a) 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
(b) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question 7.
Write the balanced chemical equations for the following reactions :
(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride
Answer:

Question 8.
Write the balanced chemical equation for the following and identify the type of reaction in each case :
(a) Potassium bromide (aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide (s)
(b) Zinc carbonate(s) → Zinc oxide (s) + Carbon dioxide (g)
(c) Hydrogen (g) + Chlorine (g) → Hydrogen chloride (g)
(d) Magnesium (s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)
Answer:
(a) 2KBr(aq) + Bal₂(aq) → 2KI(aq) + BaBr₂(s)
The given reaction is a double displacement reaction.

Given reaction is a thermal decomposition reaction.

(c) H₂(g) + Cl₂(g) → 2HCl(g)
Given reaction is a combination reaction.

(d) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Given reaction is a displacement reaction.

Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:
Exothermic reaction : A chemical reaction in which heat energy is evolved during the formation of product is called an exothermic reaction.

For example, Combustion of natural gas (methane) is an exothermic reaction.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) + heat

Endothermic reaction : A chemical reaction in which heat is absorbed during the formation of products is called an endothermic reaction.
For example, Decomposition of silver chloride in presence of sunlight is called an endothermic reaction.

Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:
We need energy for staying alive. We get S this energy from the diet (food), we eat. Food is broken down into simple substances during digestion.

For example, Rice, potatoes and bread consists of carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and provides energy. This reaction (process) is called respiration. Thus, energy is released during respiration process, it is called an exothermic reaction.
C₆H₁₂O₆(aq) + 6O₂(aq) → 6CO₂(aq) + 6H₂O(1) + Energy

Question 11.
Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.
Answer:
In decomposition reaction, a single molecule of reactant by absorbing heat is broken down into two or more products (atoms), while in combination reaction, opposite phenomenon to decomposition reaction is observed. In combination reaction, two or more substances (elements or compounds) combine to form a single product with heat change.

Decomposition reactions :
AB + Energy → A + B

Combination reactions :
A + B > AB + Energy
C(s) + O₂(g) → CO₂(g) + Energy

Question 12.
Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Answer:

Question 13.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:
In displacement reaction, more reactive element displaces less reactive element from its solution. For example, In case of Zn and Ag, Zn being more reactive than Ag, it displaces Ag from its solution of AgNO₃.
(i) Zn(s) + 2AgNO₃(aq) → Zn(NO₃)₂(aq) + 2Ag(s)
Similarly, Fe being more reactive displaces Cu from CuSO₄ solution.

(ii) Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
In double displacement reaction, two / compounds exchange their ions to form s two new compounds.
For example,

• Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl(aq)
• 2KBr(aq) + BaI₂(aq) → 2KI(aq) + BaBr₂(s)

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
Answer:
Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
Chemical reaction, in which reactants react to form insoluble precipitate is called precipitation reaction.
For example,

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each:
(a) Oxidation (b) Reduction
Answer:
(a) Oxidation: It is a reaction in which atom or molecule gains oxygen or loses hydrogen.
For example,
C + O₂ → CO₂
2Cu + O₂ → 2CuO
In above reactions, ‘C’ and ‘Cu’ undergo Oxidation.

(b) Reduction : It is a reaction in which atom or molecule loses oxygen or gains hydrogen.
For example,
CuO + H₂ → Cu + H₂0
CO₂ + H₂ → CO + H₂O
In above reactions, ‘CuO’ and ‘CO₂’ undergo reduction.

Question 17.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:
Here, element ‘X’ is copper (Cu). When it is heated in air, it forms black coloured copper oxide (CuO).
For example,

In this reaction, Cu is oxidised to CuO.

Question 18.
Why do we apply paint on iron articles?
Answer:
Iron objects undergo rusting due to metal corrosion; hence, surface of iron is coated with paint to prevent it from rusting. As a result, iron does not come in contact with air. Thus, iron objects remain safe for longer period and rusting does not occur.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
Food items containing oil and fat reacts with oxygen and become rancid. Such food items have bad smell and are harmful to health. Hence, to prevent the spoilage of food items, they are flushed with unreactive gas like nitrogen, which acts as an antioxidant.

Question 20.
Explain the following terms with one example each:
(a) Corrosion
(b) Rancidity
Answer:
(a) Corrosion:
Metal gets corroded in the presence of acid and moisture. This process is called corrosion. Rusting of iron, tarnishing of silver, green coating of salt on the utensils of copper are the examples of corrosion.

• Due to corrosion, iron gets covered with a red brown substance (powder) which is called rusting of iron (or corrosion).
• Corrosion causes damage to the articles (objects) made from iron such as car-bodies, bridges, iron railings, ships, etc. There is a huge expenditure to control corrosion.
• Rusting can be prevented by painting the surface of iron or coating the surface of iron with more reactive metal (Zn).
• Prevention of corrosion by coating iron surface with zinc is called galvanisation.

(b) Rancidity:
When food containing oil and fat is kept in an open air, it undergo oxidation and become rancid; due to which their smell and taste changes. This process is called rancidity.

Usually, to avoid rancidity in food containing oil and fat, substances are added which prevent oxidation. Such substances are known as antioxidant substances. Food kept in air-tight containers helps to slow down its oxidation and rancidity is retarded.

Manufacturers of potato chips flush bags of chips with gas such as inactive nitrogen gas as antioxidant to prevent the oxidation of potato chips.

Jharkhand Board Class 10 Science Chemical Reactions and Equations InText Questions and Answers

Question 1.
Why should a magnesium ribbon be cleaned before burning in air?
Answer:
Magnesium ribbon is more reactive and when it is exposed to air, a layer of MgO is formed on its surface.
2Mg(s) + O₂(g) → 2MgO(s)
This layer of MgO is removed by rubbing with sand paper before burning in air. Therefore, magnesium ribbon easily reacts with oxygen. Thus, magnesium ribbon is cleaned before burning in air.

Question 2.
Write the balanced equation for the following chemical reactions :
(1) Hydrogen + Chlorine → Hydrogen chloride
(2) Barium chloride + Aluminium sulphate Barium sulphate + Aluminium chloride
(3) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
(1) H₂(g) + Cl₂(g) 2HCl(g)
(2) 3BaCl₂(aq) + Al₂(SO₄)₃(aq) → 3BaSO₄(s) + 2AlCl₃(aq)
(3) 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Question 3.
Write a balanced chemical equation with physical state symbols for the following reactions :
(1) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
(2) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.
Answer:

Question 4.
A solution of a substance ‘X’ is used for whitewashing.
(1) Name the substance ‘X’ and write its s formula.
(2) Write the reaction of the substance ‘X’ named in (1) above with water.
Answer:
(1) Substance X is calcium oxide. Its formula is CaO.

(2) Calcium oxide reacts vigorously with water and forms slaked lime (Ca(OH)₂) and liberates large amount of heat. It is an exothermic reaction.
CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat

Question 5.
Why is the amount of gas collected s in one of the test tubes in activity 1.7 double of the amount collected in the other? Name these gases.
Answer:
Hydrogen and oxygen are obtained separately during the electrolysis of water. Water is made-up of two parts of hydrogen and one part of oxygen. Hence, two parts of hydrogen gas in one test tube and one part of oxygen l gas in another test tube are obtained during the electrolysis of water. Hence, hydrogen and oxygen gases are obtained in the ratio of 2 : 1 by volume.

Question 6.
Why does the colour of copper sulphate solution change, when an iron nail ( is dipped in it?
Answer:
When an iron nail is dipped in copper s sulphate solution, iron being more reactive than l copper, it displaces copper from the solution, s As a result, iron sulphate is formed, which is green in colour. Thus, colour of copper sulphate s solution changes.

Question 7.
Give an example of double displacement reaction other than the one ? given in activity 1.10.
Answer:
When sodium carbonate reacts with calcium chloride, it forms a precipitate of calcium S carbonate and sodium chloride. In this reaction, ( exchange of carbonate and chloride ions form two 5 new compounds.

This is double displacement reaction.

Question 8.
Identify the substances that are oxidised and the substances that are reduced in the following reactions:
(1) 4Na(s) + O₂(g) → 2Na₂O(s)
(2) CuO(s) + H₂(g) → Cu(s) + H₂O(1)
Answer:
(1) 4Na(s) + O₂(g) → 2Na₂O(s)
In this reaction, sodium (Na) metal is oxidised to Na₂0 and O₂ is reduced.

(2) CuO(s) + H₂(g) → Cu(s) + H₂O(1)
In this reaction, CuO is reduced to Cu, while H₂ is oxidised to H₂O.

Activity 1.1 [T. B. Pg. 1]

Aim : To study the burning of magnesium ribbon in air.

Caution: It is necessary that this activity should be performed in the presence of a teacher. For safety purpose, teacher and student should wear goggles.

Activity:

• Take approximately 3-4 cm long magnesium ribbon and make it clean by rubbing it with sand paper.
• Hold it with a pair of tongs and heat on the flame of burner or spirit lamp and the ash being formed collects in the watch-glass as shown in the figure 1.1.
• Collected ash in the watch-glass is of magnesium oxide.
• Burn the magnesium ribbon. Keeping it away as far as possible from your eyes.
  

Questions:

Question 1.
Why is magnesium ribbon selected?
Answer:
Magnesium ribbon is highly reactive and burns easily in air.

Question 2.
What is the colour of magnesium ribbon initially?
Answer:
Magnesium ribbon, initially, rubbed with sand paper appears silvery white.

Question 3.
Which type of flame is formed during the burning of magnesium ribbon?
Answer:
Magnesium ribbon burns with dazzling white flame in the flame of burner.

Question 4.
What is the composition of ash collected in watch-glass?
Answer:
White ash collected in watch-glass is of magnesium oxide (MgO).
Note: Due to O₂ and N₂ present in air, MgO is formed as a major product, moreover traces of Mg₃N₂ is also obtained.

Question 5.
Write the chemical reaction of forming the magnesium oxide.
Answer:

Activity 1.2 [T.B.Pg.2]

Aim : To study the reaction between lead nitrate and potassium iodide.

Activity:

• Take lead nitrate solution in a test tube.
• Add the solution of potassium iodide in it.

Questions :

Question 1.
What is the colour of lead nitrate solution?
Answer:
Lead nitrate solution is colourless.

Question 2.
What is the colour of an aqueous solution of potassium iodide?
Answer:
The solution of potassium iodide is colourless.

Question 3.
Write the balanced chemical equation for the reaction that takes place between lead nitrate and potassium iodide.
Answer:
Pb(NO₃)₂(aq) + 2KI(aq) → PbI₃(s) + 2KNO₃

Question 4.
What is the colour of PbI₂?
Answer:
PbI₂ is yellow.

Question 5.
Identify the type of the above reaction.
Answer:
This reaction is a double displacement reaction.

Activity 1.3 [T. B. Pg. 2]

Aim : To study the reaction between zinc metal and dilute sulphuric acid.

Caution: Use the acid with care.

Activity:

• Take a conical flask.
• Add a piece of zinc granules in it.
• Then add dilute hydrochloric acid or dilute sulphuric acid.
  

Questions :

Question 1.
What appears around the zinc granules?
Answer:
The granules of zinc react with dilute hydrochloric acid evolving hydrogen gas. Hence, bubbles of hydrogen gas appear around granules of zinc.

Question 2.
What happens to the conical flask?
Answer:
Conical flask becomes hot which can be felt on touching to it.

Question 3.
State the reaction occurring between the pieces of zinc and dilute HC1.
Answer:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Question 4.
Which type of reaction takes place between the pieces of zinc and dilute HCl?
Answer:
Reaction taking place between zinc granules and dilute HCl is an exothermic reaction.

Activity 1.4 [T. B. Pg. 6]

Aim: To study the reaction between calcium oxide and water.

Activity:

• Take some quick lime (Calcium oxide – CaO) in a beaker. Add water to it slowly.
• Touch the beaker as shown in the figure 1.3.
  

Questions:

Question 1.
What is formed by reaction of quick lime with water? Write reaction.
Answer:
The reaction between quick lime and water forms Ca(OH)₂.
CaO(s) + H₂O(l) → Ca(OH)₂(aq)

Question 2.
What is called the reaction occurring between quick lime and water?
Answer:
Reaction occurring between slaked lime and water is called slaking of lime.

Question 3.
What do you feel by touching the beaker outside?
Answer:
By touching the beaker, it is observed that the beaker becomes warm since, heat is evolved during reaction.

Question 4.
What is slaked lime?
Answer:
Calcium hydroxide (Ca(OH)₂) is known as slaked lime.

Activity 1.5 [T. B. Pg. 8]

Aim: To study the decomposition of ferrous sulphate on heating

Activity:

• Take approximately 2 g of ferrous sulphate crystals in a dry boiling tube.
• Heat the boiling tube over the flame of a burner.
• Observe the colour of the ferrous sulphate crystals carefully during the heating.
• Observe the colour of the crystals after heating it.
  

Questions :

Question 1.
What is the colour of crystals of ferrous sulphate?
Answer:
The crystals of ferrous sulphate are greenish (or faint or light green).

Question 2.
Which colour is observed on heating the crystals of ferrous sulphate?
Answer:
On heating the ferrous sulphate, it is decomposed into Fe2O₃ and green colour changes to (reddish) brown.

Question 3.
On heating ferrous sulphate in boiling tube, a gas evolved which has a characteristic smell. What is its reason?
Answer:
On heating ferrous sulphate, a gas having specific smell is evolved due to combustion of sulphur. The combustion of sulphur forms SO₂(g) and SO₃(g) which possesses strong irritating smell.

Activity 1.6 [T. B. Pg. 8]

Aim : To study the decomposition of lead nitrate.

Activity:

• Take about 2 g of lead nitrate powder in a boiling tube.
• Hold the boiling tube with a pair of tongs and heat it over the flame of a burner.
  

Questions :

Question 1.
Which gas is evolved from boiling tube on heating lead nitrate?
Answer:
Nitrogen dioxide (NO₂) is evolved from the boiling tube on heating lead nitrate. It is brown coloured gas.

Question 2.
Write the equation of reaction of heating lead nitrate.
Answer:

Activity 1.7 [T. B. Pg. 9]

Aim : To study electrolysis of water.

Activity:

• Take a plastic mug. Drill two holes at the base of the mug. Fix two rubber corks in it; as shown in the figure 1.6.
• Arrange the: test tubes in inverted position such that carbon electrodes remains in it as shown in the figure.
• Add water in a mug such that electrodes are immersed.
• Add a few drops of dilute sulphuric acid to water.
• Connect electrodes to a 6 volt battery.
• Now, switch on the current and leave the apparatus undisturbed for some time.
  

Questions :

Question 1.
What is an electrode?
Answer:
Platinum or carbon rod which remains dipped in the solution of electrolyte and on the surface of which chemical reactions occur is known as an electrode.

Question 2.
Which gas is evolved at anode during electrolysis of water?
Answer:
Oxygen gas is evolved at anode during electrolysis of water.

Question 3.
Which gas is evolved at cathode during electrolysis of water?
Answer:
Hydrogen gas is evolved at cathode during elecrolysis of water.

Question 4.
Are the volumes of the gases collected during electrolysis are same?
Answer:
No

Question 5.
State the chemical equation of reaction of electrolysis of water.
Answer:

Question 6.
What happens when O₂(g) and H₂[g) evolved are brought close to the burning candle?
Answer:
O₂(g) does not burn, while H₂(g) burns with popping sound.

Activity 1.8 [T. B. Pg. 9]

Aim : To study photochemical decomposition of silver chloride.

Activity:

• Take about 2 g of silver chloride in a china dish.
• Put this china dish in sunlight for some-time.
  

Questions :

Question 1.
What was the colour of silver chloride before exposure to sunlight?
Answer:
Silver chloride was white in colour before its exposure to sunlight.

Question 2.
What is the change in colour of silver chloride during its exposure to sunlight for sometime?
Answer:
When silver chloride is exposed to sunlight for sometime, its white colour changes to grey.

Question 3.
State tire chemical equation of decomposition reaction of silver chloride and silver bromide in presence of sunlight.
Answer:

Question 4.
Does the decomposition of silver bromide occur in sunlight?
Answer:
Yes

Question 5.
State the uses of AgCl and AgBr.
Answer:
AgCl and AgBr are used in black and white photography in films.

Activity 1.9 [T. B. Pg. 10]

Aim : To study the displacement reaction taking place between iron nail and solution of copper sulphate.

Activity:

• Take three iron nails and clean their surface by rubbing them with a sand paper.
• Take two test tubes labelled as (A) and (B). Take about 10 mL solution of copper sulphate in each test tube.
• Tie two iron nails with a thread and immerse them in copper sulphate solution for about 20 minutes.
  Keep one iron nail aside for comparison.
• Take out the iron nails from the copper sulphate solution after 20 minutes.
• Compare the colour of both iron nails with the nail kept aside.
• Compare the intensity of the colour of copper sulphate solutions of both the test tubes, (A) and (B).
  

Questions:

Question 1.
What would be the colour of iron nail placed in the solution of CuSO₄ ?
Answer:
Iron nail become brownish in colour.

Question 2.
What would be the change in colour of solution of copper sulphate ?
Answer:
Blue colour of copper sulphate solution fades (or becomes light blue).

Question 3.
Which reaction takes place when iron nail is dipped in the solution of copper sulphate ?
Answer:

Question 4.
What type of chemical reaction occur, when iron nail is dipped in copper sulphate solution?
Answer:
Reaction of iron nail with the solution of copper sulphate is a displacement reaction.

Activity 1.10 [T. B. Pg. 11]

Aim: To study double displacement reaction between barium chloride and sodium sulphate.

Activity:

• Take about 3 mb of sodium sulphate solution in a test tube.
• In another test tube, take about 3 mL of barium chloride solution.
• Mix the two solutions as shown in the figure.
  

Questions:

Question 1.
What is the colour of sodium sulphate and barium chloride solutions?
Answer:
Sodium sulphate and barium chloride are colourless solutions.

Question 2.
Which precipitate is obtained by mixing the solution of sodium sulphate and barium chloride ? Mention its colour.
Answer:
When solutions of sodium sulphate and barium chloride are mixed, a precipitate of barium sulphate (BaSO₄) is formed, which is white in colour.

Question 3.
Write a balanced chemical equation for the reaction between barium chloride and sodium sulphate.
Answer:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

Question 4.
What type of chemical reaction takes place between barium chloride and sodium sulphate?
Answer:
A chemical reaction taking place between barium chloride and sodium sulphate is a double displacement reaction.

Activity 1.11 [T. B. Pg. 12]

Aim: To study oxidation of copper to copper oxide.

Activity:

• Take 1 g of copper powder in a china dish and heat it as shown in the figure 1.10.
  

Questions:

Question 1.
What happens on heating the copper powder?
Answer:
Copper powder, on heating forms copper (II) oxide on its surface.

Question 2.
State the colour of copper oxide.
Answer:
The colour of copper oxide is black.

Question 3.
What type of reaction is represented by the reaction of formation of copper oxide from copper?
Answer:
The reaction of formation of copper oxide from copper is an oxidation reaction.

Question 4.
Which product is obtained by passing hydrogen gas over hot copper oxide?
Answer:
Copper is obtained by passing hydrogen gas over hot copper oxide.

Question 5.
Name the reaction of forming copper from copper oxide.
Answer:
The reaction to form copper from copper oxide is a reduction reaction.

Jharkhand Board JAC Class 10 Science Important Questions Chapter 1 Chemical Reactions and Equations Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 1 Chemical Reactions and Equations

Additional Questions and Answers

Question 1.
Answer the following questions :
(1) State the type of reaction in the following :
(i) Vegetable matter changing into compost.
(ii) Burning of natural gas
(iii) Adding water to quick lime to form slaked lime.
Answer:
(i) Vegetable matter changing into compost is an exothermic and decomposition reaction.
(ii) Burning of natural gas is an exothermic reaction.
(iii) Adding water to quick lime to form slaked lime is combination reaction, which is exothermic.

(2) Identify the substance which acts as an oxidising agent and reducing agent in the following reaction :
(i) H₂(g) + Cl₂(g) → 2HCl(g)
(ii) 2Al(s) + Cr₂O₃(s) > Al₂O₃(s) + 2Cr(s)
Answer:
Cl₂ in reaction (i) and Cr₂O₃ in reaction (ii), acts as an oxidising agent.
While, H₂, in reaction (i) and Al, in reaction (ii) acts as a reducing agent.

(3) AgNO₃(aq) + NaCl(aq) → AgCl(s) ↓ + NaNO₃(aq)
FeS(s) + H₂SO₄(aq) → FeSO₄ + H₂S ↑
In above chemical equations, two different type of arrows (4- and t) are shown with products. What does these arrows represent?
Answer:
↓ Indicate the formation of insoluble ↑ substance (precipitate) while ↑ represent the evolution of gas.

(4) Distinguish between :
1. Endothermic reaction and Exothermic reaction
Answer:

Endothermic reaction	Exothermic reaction
1. A chemical reaction in which heat energy is absorbed during the formation of product is called an endothermic reaction.	1. A chemical reaction in which heat energy is evolved during the formation of product is called an exothermic reaction.
2. For example, reaction of barium hydroxide with ammonium chloride is an endothermic reaction.	2. For example, reaction of quick lime with water is an exothermic reaction.

2. Oxidation reaction and Reduction reaction
Answer:

Oxidation reaction	Reduction reaction
1. A chemical reaction in which substance gains oxygen or loses hydrogen is called an oxidation reaction.	1.  A chemical reaction in which substance gains hydrogen or loses oxygen is called a reduction reaction.
2. For example, C + O2 → CO2 2Cu + O2 → 2CuO	2. For example, CuO + H2 → Cu + H2O CO2 + H2 → CO + H2O

(5) What happens when carbon dioxide and water react in the same ratio?
Answer:
When six molecules of carbon dioxide and six molecules of water undergo reaction, it forms glucose with evolution of oxygen gas.

(6) How can the black coating of copper oxide be removed chemically?
Answer:
The black coating of copper oxide can be removed chemically by passing the hydrogen gas over heated copper oxide. As a result, the black coating turns brown in colour as oxygen is removed by hydrogen.
CuO + H₂ → Cu + H₂O

(7) When quick lime is added to water, a hissing sound is produced. Write the chemical reaction and state the type of reaction that takes place.
Answer:
CaO + H₂O → Ca(OH)₂ + heat
This type of reaction is exothermic and combination.

(8) Write the name of the products obtained and type of reaction given below:
Na₂SO₄ + BaCl₂ → …………. + …………..
Answer:

(9) Reactants A and B react together and forms zinc chloride and hydrogen gas. Identify A and B. Write the chemical equation.
Answer:
The reactant A is a zinc metal and the reactant B is hydrochloric acid.
Zn + 2HCl → ZnCl₂ + H₂

(10) State the ascending order of reactivity for Cu, Ag and Fe metals based on the reaction given below:
(i) Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
(ii) Cu(s) + FeSO₄(aq) → No reaction
(iii) Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
(iv) 2Ag(s) + CU(NO₃)₂ → No reaction
Answer:
Based on the given reaction, ascending order of reactivity of Cu, Ag and Fe is as follows :
Ag < Cu < Fe

(11) State the true and false option for the chemical reaction given below:
3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)
(i) Fe is being oxidised.
(ii) Water is being reduced.
(iii) Water acts as reducing agent.
(iv) Water acts as oxidising agent.
Answer:
Here, (i), (ii) and (iv) are true options, while (iii) is a false option.

(12) Choose the true and false statements for the reaction CuO + H₂ → Cu + H₂O.
(i) CuO is an oxidising agent.
(ii) H₂ is being oxidised.
(iii) The reaction is a displacement reaction.
(iv) The valency of Cu is not changing.
Answer:
Statements, (i), (ii) and (iii) are true, while statement (iv) is false.

(13) Three test tubes are taken and marked as ‘X’, ‘Y’ and ‘Z’. In test tube X, iron nail is dipped in water. In test tube Y, iron nail is dipped in mixture of water and oil. In test tube Z, iron nail is added with dry CaCl₂. In which test tube, the iron nail will rust? Why?
Answer:
The iron nail will rust in test tube X, because it provides the essential moist conditions for rusting since both moisture as well as air is present in the test tube.

(14) A metal ‘M’ kept in air turns green and when it is heated, it turns black. Name the metal and the compound formed in both cases.
Answer:
Metal M is copper. Green compound is due to formation of copper carbonate and black coloured compound is due to formation of copper oxide.

(15) State the equations of zinc and lead, where it displaces copper from its compounds.
Answer:

(16) Write balanced equations for thermal decomposition and photochemical decomposition reactions.
Answer:
Thermal decomposition reaction :

Photochemical decomposition reaction :

(17) Name the methods used to prevent rusting.
Answer:

• Oiling of metals
• Applying paints
• Preparation of an alloy and
• Electroplating or galvanising.

(18) Why does most of the metal articles become dull, when left in an open air?
Answer:
Metal articles left in an open air reacts with the gases of atmosphere (air) and forms a ( layer of oxide compound on their surface which make metal dull and the lustre is lost.

For example, aluminium metal reacts readily l with oxygen of air and forms a layer of aluminium oxide and the surface of Al becomes dull.

(19) Give two examples of a reaction which is both endothermic and decomposition in nature.
Answer:

(20) What is called rancidity? What is the general name of chemicals which are added to fat and oil containing food to prevent the rancidity?
Answer:
When oil and fat containing food is i exposed to air, it reacts with oxygen and gets? oxidised. As a result, food become rancid and S their taste and smell changes. This process is called rancidity.

The general name of the chemicals which are added to food to prevent its rancidity are known as antioxidants. For example, Nitrogen gas is an antioxidant substance.

(21) Why is photosynthesis considered as an endothermic reaction?
Answer:
In photosynthesis reaction, energy is required to form glucose from carbon dioxide and water. Energy, in the form of sunlight is used for s photosynthesis. Hence, this reaction is called an endothermic reaction.

(22) What is meant by electrolysis? Mention its two uses with examples.
Answer:
When electric current is passed through an aqueous solution or molten ionic compound, } the ions of the compound form products on electrodes by electrolytic oxidation – reduction. This process is called an electrolysis.

Uses:
1. Gas and metal can be obtained from an aqueous solution of salt.
For example, Electrolysis of an aqueous solution of NaCl produces Na metal and Cl₂gas.

2. The coating (protective layer) of one metal can be formed on the surface of another metal.
For example, Gold plating can be carried out on the surface of silver articles.
Zn layer can be formed on the surface of iron.

(23) Write the definition of displacement reaction. Give one example of it and explain, how is it different from double displacement reaction.
Answer:
The reaction in which more reactive metal displaces the less reactive metal from its compound is called as displacement reaction.
For example,
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
This reaction is different than double displacement reaction. In double displacement reaction ions get exchanged in two compounds.
For example,
Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl(aq)
In short, one ion gets exchanged in displacement reaction, while two ions are exchanged in double displacement reaction.

(24) How does a chemical equation makes a reaction more informative?
Answer:
For making a chemical equation more informative –

• The physical states of the reactants and products are represented by symbols (g) for gas, (l) for liquid, (s) for solid and (aq) for aqueous solution.
• Specific conditions like temperature, pressure, catalyst are shown above/below the arrow in the equation.

(25) Give the balanced chemical equations for the following :
(i) Reaction used in black and white photography
(ii) Reaction of oxidation of glucose
(iii) Formation of water from H₂ and O₂
Answer:

(ii) C₆H₁₂O₆(s) + 6CO₂(g) → 6CO₂(g) + 6H₂O(g) + Energy
(iii) 2H₂(g) + O₂(g) → 2H₂O(l)

(26) What is a redox reaction? When a magnesium ribbon is burnt in air, it burns with a dazzling white flame and white powder is formed. Is magnesium oxidised or reduced? Why?
Answer:
A reaction in which oxidation and reduction reactions occur simultaneously is called redox reaction.
2Mg(s) + O₂(g) → 2MgO(s)
In this reaction, magnesium combine with oxygen. Hence, it is oxidised to MgO.

(27) Name different types of chemical reactions; and explain it with suitable example.
Answer:
Name of different types of chemical reactions with suitable example are given below :
1. Combination reaction : Two or more than two reactants combine together to form a single product.
For example, CaO(s) + H₂O(l) → Ca(OH)₂(aq)

2. Decomposition reaction : A single reactant, on heating decomposes to form more than one product.
For example,

3. Displacement reaction : More reactive metal displaces the less reactive metal from its compound.
For example,
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

4. Double displacement reaction : In this an exchange of ions occurs between two compounds.
For example,
Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2NaCl(aq)

5. Oxidation reaction : A reaction in which oxygen is added or hydrogen is removed is called an oxidation reaction.
For example, C(s) + O₂(g) → CO₂(g)

6. Reduction reaction : A reaction in which hydrogen is added or oxygen is removed is called reduction reaction.
For example, CuO + H₂ → Cu + H₂O

7. Precipitation reaction : In this reaction, insoluble salt is formed.
For example,

(28) (i) What is called rancidity?
(ii) Suggest two methods to reduce the effect of rancidity.
(iii) How is corrosion different from rusting?
Answer:
(i) Rancidity : Oil and fat containing food when exposed to air gets oxidised and it becomes rancid due to which its taste and smell change. This reaction is called rancidity. Such food items are not suitable for consumption.

(ii) To reduce the problem of rancidity –

• Keep the food in a closed container.
• Use antioxidants.

(iii) Corrosion is observed in all the metals, when exposed to air; and a layer of compound s is formed due to reaction of metal with moisture, acid and gases present in atmosphere.

Rusting is a process in which iron reacts with t air and moisture to form brownish red coloured powder called rust. Due to corrosion, metal lose their lustre and become dull while due to rusting iron metal is > slowly destroyed.

(29) Water contains hydrogen and oxygen in the ratio of 2:1. Prove it by an activity.
Answer:
Take a plastic mug. Drill two holes at its base and fit rubber stoppers into these holes. S Insert carbon electrodes in these rubber stoppers as shown in the figure.

• Connect electrodes to a 6 volt battery.
• Fill the mug with water such that the electrodes are immersed and add a few drops of dilute sulphuric acid to the water to make it conducting.
• Take two test tubes filled with water and invert them over the two carbon electrodes.
• Switch on the current and leave the apparatus undisturbed for sometime.
  
• You will observe the formation of bubbles at both the electrodes. These bubbles displace water of the test tubes.
• The gas collected in test tube attached to cathode is twice in volume than the gas collected at anode.
• Once the test tubes are filled with the gases, remove them carefully.
• The gases can be tested by bringing a burning candle close to the mouth of each test tube.

A gas at anode make the candle to light s brightly indicating it is oxygen while a gas at cathode burns with popping sound indicating it is hydrogen.

(30) What is meant by chemical equation ? Explain its characteristics with example.
Answer:
Symbolic representation of chemical reaction using symbols and formulae of the d substances involved in it is called chemical equation.

Characteristics :
1. Substances which are involved in the chemical reaction are known as reactants and S they are written on the left side of the arrow (→) sign.

2. Substances which are formed at the end of the chemical reaction are known as products s and they are written on the right side of the arrow (→) sign.

3. In chemical equation, the physical states of the reactants and products are mentioned, such as (s) for solid, (l) for liquid, (g) for gas and (aq) > for an aqueous solution. These notations (s), (1), (g) and (aq) are shown at right (just after their formula) in bracket with small letters.
For example,
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

4. In chemical equation, the gaseous product is represented by putting an arrow pointing upwards (↑) and an insoluble product (Precipitate) is represented by putting an arrow pointing downwards (↓).
For example,

• C(s) + O₂(g) → CO₂(g) or CO₂(↑)
• AgNO₃(aq) + NaCl(aq) → NaNO₃(aq) + AgCl(s) or AgCl(↓)

5. Some chemical reactions take place under specific reaction conditions such as temperature, pressure, catalyst which are shown above/below the arrow in the equation.
For example,

6. According to the law of conservation of mass, chemical equation should be written as balanced equation.

Question 2.
Give scientific reasons for the following statements:
(1) Chemical equation must be balanced.
Answer:
In balanced chemical equation, the number of atoms of each element remains the same on both sides. Now, according to the law of conservation of mass, matter can neither be created nor be destroyed in a chemical reaction. Hence, the total mass of reactants and products must be equal.

As a result, it is essential to keep the number of atoms equal in both sides of chemical reaction. Thus, it is the basic need of balancing the chemical equation.

(2) Magnesium ribbon should be cleaned before burning in air.
Answer:
Magnesium ribbon is more reactive and when it is exposed to air, a layer of MgO is formed on its surface.
2Mg(s) + O₂(g) → 2MgO(s)
This layer of MgO is removed by rubbing with sand paper before burning in air. Therefore, magnesium ribbon easily reacts with oxygen. Thus, magnesium ribbon is cleaned before burning in air.

(3) Lime water (solution of slaked lime) is used for whitewashing the walls of the house.
Answer:
Solution of lime water – Ca(OH)₂ is used for whitewashing the walls of the house.
It is prepared by slaking of lime as shown s in the following reaction :

Solution of lime water is applied to the walls of the house, then it reacts slowly with the CO₂ gas of air and a thin layer of CaCO₃ is formed on walls after two to three days. As a result walls appear shining white.

(4) Respiration is an exothermic reaction.
Answer:
We need energy for staying alive. We get S this energy from the diet (food), we eat. Food is broken down into simple substances during digestion.

For example, Rice, potatoes and bread consists of carbohydrates. These carbohydrates are broken down to form glucose. This glucose combines with oxygen in the cells of our body and provides energy. This reaction (process) is called respiration. Thus, energy is released during respiration process, it is called an exothermic reaction.
C₆H₁₂O₆(aq) + 6O₂(aq) → 6CO₂(aq) + 6H₂O(l) + Energy

(5) Food items should be stored in air tight closed containers.
Answer:
When food items are left exposed to air, they react with oxygen and become rancid. As a result taste and smell of the food change, which is harmful to health. Thus, food items should be stored in air tight closed containers to avoid rancidity.

(6) A thin layer of zinc is applied on the plates of steamer to prevent its rusting.
Answer:
The plates of steamer consists of iron. After sometime, due to presence of water and dissolved oxygen, its surface get covered with a reddish brown flaky substance called rust.

Hence, to prevent the rusting of iron, a layer of more reactive metal Zn is applied on its surface.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Write the chemical reaction taking place between lead nitrate and potassium iodide. Identify the type of reaction.
Answer:
Equation for reaction :
Pb(NO₃)₂(s) + 2KI(s) → PbI₂(aq) + 2KNO₃(aq)
It is a double displacement reaction.

(2) What is meant by skeletal equation? Give an example.
Answer:
Equation in which chemical formulae of the substances are represented briefly is called skeletal equation.
In skeletal equation, the reaction is not balanced and the mass is not the same on both sides of equation.
For example, Mg + O₂ → MgO

(3 ) To which sides of a chemical equation the reactants and products are represented?
Answer:
In chemical equation, reactants are written on left side of an arrow (→), while products are written on right side of an arrow (→).

(4 ) Write the name and molecular formula of the substance used in whitewashing the walls of the house.
Answer:
Calcium hydroxide is used for white washing the walls of the house. Its molecular formula is Ca(OH)₂.

(5) Blue crystals of copper sulphate on heating in a dry test tube become colourless. Why?
Answer:
The blue colour of copper sulphate is due to five water molecules attached with it. On heating, it loses water molecules; and become colourless. [It is converted into anhydrous white CuSO₄.]

(6) Why does the solution of AgNOa stored in dark brown coloured bottle?
Answer:
The solution of AgNOs is decomposed in presence of sunlight hence it is stored in dark brown coloured bottle; which prevent the passage of light through the solution.

(7) What is meant by reversible reaction?
Answer:
The reaction in which the reactants have a tendency to form products while the products have a tendency to reform the reactants is called a reversible reaction.

(8) State the law of conservation of energy.
Answer:
Matter can neither be created nor be destroyed, (or Energy can neither be created nor be destroyed but can be converted from one form to another).

(9) Which carbohydrate substances, on decomposition form glucose?
Answer:
Carbohydrates such as rice, potatoes and bread, etc. form glucose on their decomposition.

(10) Which substance is reduced in the following reaction?
Reaction : 8Al + 3Fe₃O₄ → 4Al₂O₃ + 9Fe
Answer:
In this reaction, oxygen is removed from Fe₃O₄. Hence Fe₃O₄ is reduced.

(11) Which compound is used in black and white photography?
Answer:
Silver bromide or silver chloride is used in black and white photography.

(12) What characteristics of food containing oil and fat change when exposed to air for a long time?
Answer:
When food containing oil and fat are exposed to air for a long time then its taste and smell get changed due to rancidity.

Question 2.
Fill in the blanks :

1. The colour of lead nitrate powder is …………….
2. Respiration is an ……………. type of reaction.
3. Matter can neither be created, nor be destroyed. It is called the law of …………….
4. Quick lime is also known as …………….
5. Burning of coal and formation of water is ……………. type of reaction.
6. During electrolysis of water, electrodes are connected with ……………. V of battery.
7. When iron nail is dipped in a solution of copper sulphate, then, the colour of the solution becomes ……………. after sometime.
8. Reaction,
   
   is a ……………. reaction.
9. ……………. is used in black and white photography.
10. Reducing agent undergoes …………….
11. Pb(s) + ……………. → PbCl₂(aq) + Cu(s)
12. ……………. is a more reactive metal among Fe and Mg.

Answer:

1. white
2. exothermic
3. conservation of mass
4. Ca(OH)₂ or Calcium hydroxide
5. combination
6. 6
7. Pale green
8. decomposition
9. AgBr or AgCl
10. oxidation
11. CuCl₂
12. Mg

Question 3.
State whether the following statements are true or false:

1. Hydrogen gas was discovered by Henry Cavendish.
2. (a) PbS(s) + (b) O₂(g) (c) PbO(s) + (d) SO₂(g).
   In this reaction, the values of co-efficient (a), (b), (c) and (d), in balanced equation are taken 2, 3, 2, 2 respectively.
3. Oxidising agent undergoes oxidation in the reaction.
4. Reducing agent gives H₂ or receives O₂.
5. In reaction, MnO₄^2- → MnOa + MnO₄^1- MnO₄^2- acts both as oxidising agent as well as reducing agent.
6. Fe acts as an oxidising agent in the following reaction :
   Fe + CuSO₄ → FeSO₄ + Cu
7. Heating of calcium carbonate is an endothermic reaction.
8. In the reaction, 3Mg + N₂ → Mg₃N₂, Mg undergo oxidation.
9. Oxidation and reduction does not occur in redox reaction.
10. In decomposition reaction, single reactant forms two or more products.
11. Precipitate obtained in precipitation reaction is soluble in water.
12. In rancidity, taste and smell of food item changes.
13. The colour of KI is orange-grey.
14. Reaction of zinc metal with dilute HCl liberates H₂ gas.
15. Mg is highly inactive metal.
16. Ca(OH)₂ is also known as lime water.
17. Burning of natural gas is an exothermic reaction.
18. The proportion by volume of H₂ and O₂ gases obtained at electrodes during electrolysis of water is 1 : 1.
19. Reaction, Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl is an example of double displacement reaction.
20. MgO + H₂ → Mg + H₂O is an example of oxidation reaction.

Answer:

1. True
2. True
3. False
4. True
5. True
6. False
7. True
8. True
9. False
10. True
11. False
12. True
13. False
14. True
15. False
16. True
17. True
18. False
19. True
20. False

Question 4.
(A) Choose the correct option from those given below each question:
1. 6g of hydrogen is burnt in the presence of excess oxygen. The mass of water formed is :
A. 54 g
B. 108 g
C. 36g
D. 18 g
Answer:
54 g
Hint:

6 gm = 3 moles
From equation,
2 moles H₂ = 2 moles H₂O
∴ 3 moles H₂ = 3 moles H₂O
Now, Mass of H₂O = mole x Molecular wt.
= 3 x 18 = 54g

2. Which information is not obtained by the balanced chemical equation?
A. Physical states of reactants and products.
B. Symbols and formulae of all the substances involved in a particular reaction.
C. Number of atoms/molecules of the reactants and products formed.
D. Whether a particular reaction is actually feasible or not.
Answer:
D. Whether a particular reaction is actually feasible or not.

3. Which of the following are exothermic reactions?
(i) Evaporation of water
(ii) Dilution of H₂SO₄
(iii) Reaction of water with quick lime
(iv) Sublimation of crystalline camphor
A. (i) and (ii)
B. (iii) and (iv)
C. (i) and (iv)
D. (ii) and (iii)
Answer:
D. (ii) and (iii)

4. Reaction : 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) is an example of:
(i) displacement reaction
(ii) combustion reaction
(iii) redox reaction
(iv) neutralisation reaction
A. (i) and (iv)
B. (ii) and (iii)
C. (iii) and (iv)
D. (i) and (ii)
Answer:
B. (ii) and (iii)

5. Which of the following reaction occurs in whitewashing of walls?
A. 2Ca + O₂ → 2CaO
B. CaO + H₂O → Ca(OH)₂ + heat
C. Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Answer:
C. Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

6. What happens when crystals of lead nitrate are heated strongly in a dry test tube?
A. Crystals melt immediately.
B. Brown fumes are obtained.
C. White fumes get formed in the test tube.
D. Yellow precipitates are obtained.
Answer:
B. Brown fumes are obtained.

7. Dilute hydrochloric acid is added to test tube containing pieces of zinc. The following observations are recorded. Identify the correct observation.
A. The surface of metal becomes lustrous.
B. The reaction mixture becomes milky.
C. Odour of pungent smelling gas is experienced.
D. A colourless and odourless gas is formed.
Answer:
D. A colourless and odourless gas is formed.

8. Rancidity can be prevented by…
A. adding antioxidants.
B. storing the food in freeze.
C. keeping food away from the light.
D. All of the given.
Answer:
D. All of the given.

9. Which of the following is not a single displacement reaction?
A. CuO + H₂ → H₂O + Cu
B. Zn + CuSO₄ → ZnSO₄ + Cu
C. 4NH₃ + 5O₂ → 4NO + 6H₂O
D. Zn + 2HCl → ZnCl₂ + H₂
Answer:
C. 4NH₃ + 5O₂ → 4NO + 6H₂O

10. An element X on exposure to moist air turns reddish brown and new compound Y is formed. Identify X and Y.
A. X = Fe, Y = Fe₂O₃
B. X = Ag, Y = Ag₂S
C. X = Cu, Y = CuO
D. X = Al, Y = Al₂O₃
Answer:
A. X = Fe, Y = Fe₂O₃

11. Identify the reducing agent in the following reaction :
Reaction : 3O₂(g) + 2H₂S(g) → 2H₂O(l) + 2SO₂(g).
A. O₂
B. H₂S
C. H₂O
D. SO₂
Answer:
B. H₂S

12. Both H₂ and CO₂ gases are …
A. heavier than air.
B. colourless.
C. acidic in nature.
D. soluable in water.
Answer:
B. colourless.

13. Decomposition of lead (II) nitrate forms lead (II) oxide, nitrogen dioxide and oxygen gas. What is the value of co-efficient of nitrogen dioxide in the balanced equation?
A. 1
B. 2
C. 3
D. 4
Answer:
D. 4

14. When reddish brown copper metal is heated, it forms a black solid surface. Which of the following statement is incorrect?
A. Black solid substance is CuO.
B. It is redox reaction.
C. It is precipitation reaction.
D. Copper undergo oxidation.
Answer:
C. It is precipitation reaction.

15. Silver chloride is stored in dark coloured bottle because …
A. it is a white solid.
B. it gives redox reaction.
C. to avoid the effect of sunlight.
D. None of these
Answer:
C. to avoid the effect of sunlight.

16. On immersing the Zn rod in the solution of copper sulphate, you will observe …
A. deposition of Cu on Zn.
B. deposition of Zn on Cu.
C. Cu²⁺ oxidises.
D. blue coloured solution become more dark.
Answer:
B. deposition of Zn on Cu.

17. The reaction of H₂ gas with oxygen gas forms water. This reaction is an example of:
A. combination reaction
B. redox reaction
C. an exothermic reaction
D. all of these reactions
Answer:
D. all of these reactions

18. For the reaction, CuO + H₂ → Cu + H₂O, choose the correct statement.
A. CuO is an oxidising agent.
B. H₂ undergo oxidation.
C. It is a displacement reaction.
D. All of the given
Answer:
D. All of the given

19. Select the proper option for the following statements :
Statement 1 : Burning of magnesium ribbon in air is a redox reaction.
Statement 2 : Oxidation number of oxygen in its metal oxide is -1.
A. Statement 1 is correct.
B. Statement 2 is incorrect.
C. Statement 1 is correct, but statement 2 is incorrect.
D. Statement 1 and statement 2 both are incorrect.
Answer:
C. Statement 1 is correct, but statement 2 is incorrect.

(B) Choose more than one correct options from those given below each question:
1. Identify the type of reaction for
2Al + Cr₂O₃ → Al₂O₃ + 2Cr
(i) Oxidation
(ii) Reduction
(iii) Redox
A. Only (i)
B. Only (ii)
C. (i), (ii), (iii)
D. None of these

2. What is correct for redox reaction?
(i) Reducing agent undergoes oxidation.
(ii) Reducing agent undergoes reduction.
(iii) Reduction reaction always occurs at cathode.
(iv) Oxidising agent undergo oxidation.
A. Only (i), (ii)
B. Only (i), (iii)
C. Only (iii)
D. Given all

3. Quick lime when added in water produces hissing sound. State the type of reaction.
(i) Combination
(ii) Endothermic
(iii) Exothermic
(iv) Redox
A. Only (i)
B. Only (iii)
C. Only (i), (ii), (iii)
D. Only (iv)
Answer:
1. (i), (ii), (iii)
2. Only (i), (iii)
3. Only (i), (ii), (iii)

Question 5.
Answer the following questions in one word:

1. State the formula of lime.
2. Write the formula of rust.
3. Name the reaction observed during rancidity of food.
4. Write the formula of lead sulphate.
5. What is obtained by the combustion of methane?
6. Mention the respective proportion by volume of H₂(g) and O₂(g) obtained during electrolysis of water.
7. Which coloured ash is obtained by burning 5 magnesium ribbon in air?
8. What is called the insoluble substance formed during the chemical reaction?
9. Name two metals which do not corrode, s
10. Which gas burns with popping sound?
11. Name the product formed, when silver bromide is exposed to sunlight.
12. Which compound is used to detect the formation of carbon dioxide gas?
13. Name the gas evolved when lead nitrate is 5 heated.
14. Write the example of decomposition reaction which occurs in nature.
15. Write the name of chemical used in black and white photography. I
16. Which ions are present in barium sulphate?
17. Mention one use of quick lime.
18. Give one example of decomposition reaction in which solid and gas are formed as products.
19. Give an example of antioxidant.
20. Write the reaction in which hydrogen acts as a reducing agent.
21. Name the reaction in which two compounds exchange their ions to form two new compounds.
22. What would be the effect on lime water, when carbon dioxide gas is passed through it?
23. State the molecular formula and chemical S name of iron (III) oxide.
24. State different forms of energy required for breaking down the molecules of reactant in decomposition reaction.
25. What does (II) indicate in iron (II) oxide?
26. Give two examples of noble metals.
27. Mention two examples of exothermic reactions.
28. When does the breaking and forming of bonds of chemical substances take place?
29. Name the reaction which forms insoluble salts.
30. Write the formula for two oxides of sulphur.
31. Identify the type of reaction for the following reaction :
    
32. On immersing an iron nail in CuSO₄ solution for few minutes, what will you observe?

Answer:

1. CaO
2. Fe₂O₃.XH₂O
3. Oxidation
4. PbSO₄
5. CO₂ and H₂O
6. 2 : 1
7. White
8. Precipitate
9. Gold (Au) and Platinum (Pt)
10. Hydrogen gas (H₂(g))
11. Silver and Bromine
12. Calcium hydroxide (Ca(OH)₂)
13. Nitrogen dioxide and Oxygen gas
14. Rottening of fruits and vegetables
15. Silver bromide
16. Ba²⁺ and SO₄^2-
17. In manufacture of cement
18. 
19. Nitrogen gas
20. CuO + H₂ → Cu + H₂O
21. Double displacement reaction
22. Lime water turns milky
23. Molecular formula : Fe₂O₃ Chemical name : Ferric Oxide
24. Heat, light and electricity
25. Valency of Fe (Oxidation number)
26. Au and Pt
27. Respiration and Water added to lime
28. In chemical reaction
29. Precipitation reaction
30. SO₂ and SO₃
31. Decomposition and Endothermic
32. Solution turns light green

Question 6.
Match the following:
(1)

Column I (Type)	Column II (Reactions)
1. Neutralisation reaction	a. Zn + Cu2+ → Zn2+ + Cu
2. Redox reaction	b. CaCO3 → CaO + CO2
	c. HCl + NaOH → NaCl + H2O

Answer:
(1 – c), (2 – a).

(2)

Column I (Type)	Column II (Molecular Formula)
1. Quick lime	a. CaCO3
2. Slaked lime	b. CaO
	c. Ca(OH2)

Answer:
(1 – b), (2 – c).

(3)

Column I (Colour)	Column II (substance)
1. Green	a. Copper sulphate
2. Blue	b. Barium sulphate
	c. Ferrous sulphate

Answer:
(1 – c), (2 – a).

(4)

Column I (Reaction)	Column II (Type)
1. CaO + H2O → Ca(OH)2 + heat	a. Displacement
2. Zn + CuSO4 → ZnSO24 + Cu	b. Double displacement

Answer:
(1 – c), (2 – a).

Question 7.
Complete the following chemical reactions:

(1) Fe + CuSO₄ → ……………… + ………………
(2) K2SO₄ + BaBr₂ → ……………… + ………………
(3) Pb(NO₃)₂ + 2KI → ……………… + ………………
(4) Zn + 2HCl → ……………… + ………………

(13) Fe₂O₃ + 2Al > + → ……………… + ………………
(14) 2PbO + C → ……………… + ………………
(15) 2HNO₃ + Ca(OH)₂ → ……………… + ………………

Answer:

(1) FeSO₄ + Cu
(2) 2KBr + BaSO₄
(3) Pbl₂ + 2KNOs
(4) ZnCl₂ + H₂
(5) C₆H₁₂O₆ + 6O₂
(6) 2NaOH + H₂
(7) Ca(OH)₂
(8) 2PbO + 4NO₂ + O₂
(9) 2Ag + Cl₂
(10) AgCl + NaNO₃
(11) Cu + H₂O
(12) CaCO₃ + 2NaCl
(13) Al₂O₃ + 2Fe
(14) 2Pb + CO₂
(15) Ca(NO₃)₂ + H₂O

Question 8.
Carefully observe the given diagram and answer the questions realated with it:
(1) Which gas is collected in a test tube at anode as shown in figure?

Answer:
Oxygen gas

(2) Does the reaction occur in the test tube as shown in the figure? State the change in colour.

Answer:
Reaction does not occur.
Colour change does not take place.
Colour of the solution remains green.

(3) Take a small amount of ammonium chloride in a beaker as shown in figure. Add water to it. What change do you observe in temperature? Which type of reaction is this?

Answer:
Temperature decreases. Reaction is an endothermic.

(4) As shown in figure, the strong smell of which gas will the student experience?

Answer:
SO₂ Sulphur dioxide.

Value Based Questions With Answers

Question 1.
Hitarthi saw her grandmother storing pickles in ceramic pots. She also knew that pickles should not be stored in metal containers. Her friend often brought pickles wrapped in aluminium foil. Hitarthi advised her not to wrap pickles in foil.

• Why pickles should not be stored in metal containers?
• Which substances present in pickles reacts with the metal containers?
• What value of Hitarthi is seen in the above act?

Answer:

• Pickles reacts with the metals like copper and aluminium, hence, it should not be stored in metal containers.
• Pickles are sour in taste and contains acidic nature. Thus, they react with the metal containers.
• Hitarthi showed the value of awareness and responsible citizen.

Question 2.
Surekha was very upset as her silver jewellary had turned black; and lost its lustre. Her : father is a science teacher and he washed and cleaned the jewellary using toothpaste and s brought the shine back.

• Why does the silver jewellary tarnishes when left open?
• How had the toothpaste got the shine of silver back?
• What value of Surekha’s father is seen in this act?

Answer:

• Silver jewellary reacts with gases and moisture of air due to which it tarnishes and get corroded.
• Toothpaste reacts with black coating of silver sulphide and removes it.
• Surekha’s father showed the value of responsible behaviour and teaching art.

Question 3.
Rakesh visited the government hospital to meet his cousin. He observed the medicine in dark bottles were not stored properly. They were not kept away from light and heat. Rakesh immediately reported the same to the medical superintendent and made sure that all medicines are stored properly.

• Why are some medicines stored at cool places in dark bottles?
• Why do some medicines are kept in refrigerator?
• What value of Rakesh is seen in this act?

Answer:

• Some medicines when exposed to sunlight and high temperature, their composition changes completely. Thus, to prevent the change in composition, they are stored at cool places in dark bottle.
• Some medicines decompose and become dangerous and toxic at temperature little higher than room temperature. Hence, they are stored in refrigerator.
• Rakesh expressed the value of responsible and aware citizen.

Practical Skill Based Questions With Answers

Question 1.
What would happen if you add zinc coated iron nail into the solution of copper sulphate?
Give reason for your prediction.
Answer:
The zinc coated iron nail will react with the CuSO₄ and the blue colour of the S solution gradually fade and after sometime solution will turn green because, now, iron nail will react with the copper sulphate, This happens because iron and zinc both displace the copper of copper sulphate solution because iron and zinc both are s more reactive than copper.

Question 2.
How will you differentiate sodium metal and zinc metal given in the test tubes in the laboratory? You are advised not to touch any s of the metals. Identify the type of reaction? used.
Answer:
First of all, I add water in both the test t tubes and observe the reaction. The test tube in which vigorous reaction takes place s contains Na metal, which reacts and forms H₂(g) gas in the test tube.
2Na + 2H₂O → 2NaOH + H₂
This is combination and exothermic reaction.

Question 3.
State the example of the combination reaction > that is also exothermic by nature. How will you s show its exothermic nature in the laboratory?
Answer:
Reaction of quick lime (calcium oxide) with water is highly exothermic and liberates more heat which can be measured by a thermometer in the laboratory.

Question 4.
The teacher wishes to show in laboratory that all the crystals will decompose to give water on heating, which is collected on the upper surface inside the test tube. Name any four compounds the teacher should use in the laboratory to perform this activity.
Answer:
The teacher can use the following four compounds :

• Blue vitrol, i.e., Copper sulphate (Blue crystalline salt)
• Green vitrol, i.e., Iron sulphate (Green crystalline salt)
• Sodium carbonate (White coloured crystals)
• Calcium sulphate (White coloured crystals)

Question 5.
A student wants to study the decomposition reaction of iron sulphate in laboratory. What care the student should take and why?
Answer:
The student should wear lab coat, hand gloves and safety goggles because heating of test tube containing iron sulphate may cause burn; Moreover sulphur dioxide gas released during decomposition of iron sulphate has strong chocking smell. Hence, the student should use mask and keep the exhaust fan on in the laboratory.

Memory Map:

Jharkhand Board JAC Class 10 Science Solutions Chapter 2 Acids, Bases and Salts Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 2 Acids, Bases and Salts

Jharkhand Board Class 10 Science Acids, Bases and Salts Textbook Questions and Answers

Question 1.
A solution turns red litmus blue; its pH is likely to be…
(a) 1
(b) 4
(c) 5
(d) 10
Answer:
10

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime water milky. The solution contains …
(a) NaCl
(b) HCl
(c) LiCl
(d) KCl
Answer:
HCl

Question 3.
10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HC1. If we take 20 mL of the same solution of NaOH, the amount of HC1 solution (the same solution as before) required to neutralise it will be…
(a)4mL
(b) 8 mL
(c) 12mL
(d) 16 mL
Answer:
16 mL
[Hint: Here, 10 mL NaOH neutralises 8 mL HCl
∴ 20 mL NaOH is neutralised by = \(\frac{20 \mathrm{~mL} \times 8 \mathrm{~mL}}{10 \mathrm{~mL}}\) = 16 mL]

Question 4.
Which one of the following types of medicines is used for treating indigestion?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic
Answer:
Antacid

Question 5.
Write word equations and then balanced equations for the reaction taking place when…
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron filings.
Answer:
(a) Zinc granules + Dilute sulphuric acid → Zinc sulphate + Hydrogen gas
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

(b) Magnesium + Dilute hydrochloric acid → Magnesium chloride + Hydrogen gas
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

(c) Aluminium + Dilute sulphuric acid → Aluminium sulphate + Hydrogen gas
2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)3(aq) + 3H₂(g)

(d) Iron + Dilute hydrochloric acid → Iron chloride + Hydrogen gas
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

Question 6.
Compounds such as alcohols and glucose also contain hydrogen but are not categorised as acids. Describe an activity to prove it.
Answer:
Arrange the apparatus as shown in figure 2.3. Add an aqueous solution of alcohol (ethanol) in a beaker and record the observation. Thereafter, add solution of glucose instead of alcohol and record the observation.

Observation : Bulb does not glow for both the solutions, which indicate that electric current is not flowing through both the solutions.

This experiment shows that ethanol and glucose do not ionise and do not release H⁺ ions while in the solution of acids, H⁺ ions are released and allow an electric current to flow through the solution.

Thus, eventhough alcohol and glucose contain hydrogen but are not categorised as acids.

Question 7.
Why does distilled water not conduct : electricity whereas rain water does ?
Answer:
Distilled water is a pure water and it does not contain ions while rain water contains impurities like acid, which provides ions and they help in carrying electricity.

Question 8.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids cannot release H⁺(aq) ions in absence of water. H⁺ ions are responsible for acidic character. So acids cannot release H⁺(aq) ions in absence of water and it does not exhibit acidic character.

Question 9.
Five solutions A, B, C, D and E, when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively. Which solution is …
(a) neutral? (b) strongly alkaline? (c) strongly acidic? (d) weakly acidic? (e) weakly alkaline?
Arrange the pH in increasing order of hydrogen-ion concentration.
Answer:
(i)

• Solution ‘D’ is neutral, as its pH value is 7.
• Solution ‘C’ is strongly alkaline, as its pH value is 11 (Highest pH).
• Solution ‘B’ is strongly acidic as its pH value is 1 (lowest pH).
• Solution ‘A’ is weakly acidic as its pH value is 4. S
• Solution ‘E’ is weakly alkaline as its pH value is 9.

(ii) Increasing order of H⁺ ion concentration based on the pH values is as follows :
pH : 11 < 9 < 7 < 4 < 1
Concentration of H⁺ ion = [H⁺ (aq)] :
10^-11M < 10^-9M < 10^-7M < 10^-4M < 10^-1M
Solution :
C < E < D < A < B

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A; while acetic acid (CH₃COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?
Answer:
Hydrochloric acid (HCl) is a stronger l acid than acetic acid (CH₃COOH). Hence, < more vigorous fizzing will occur in test tube A. Here, HC1 being a strong acid s undergoes complete ionisation giving more H⁺ ions.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.
Answer:
When milk changes into curd, it produces lactic acid. Lactic acid is more acidic than milk, therefore, the value of pH decreases and curd tastes sour.

Question 12.
A milk man adds a very small amount of baking soda to fresh milk.
(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?
(b) Why does this milk take a long time to set as curd ?
Answer:
(a) When a very small amount of baking soda is added to fresh milk, then the pH value of the milk increases beyond 6, since baking soda possesses basic character and alkaline medium does not allow milk to turn sour quickly.

(b) Milk turns basic by adding a small quantity of baking soda in it and lactic acid of milk is neutralised. Therefore, milk takes a longer time to set as a curd.

Question 13.
Plaster of parts should be stored in a moisture-proof container. Explain why?
Answer:
Plaster of paris should be stored in a moisture-proof container, because it absorbs moisture and sets into hard solid mass called gypsum. As a result, it loses the characteristics of plaster of paris.

Question 14.
What is a neutralisation reaction ? Give two examples.
Answer:
The reaction between an acid and a base to form a salt and water is called a neutralisation I; reaction.
For example,

In short, neutralisation reaction can be S’ represented as follows:
Acid + Base → Salt + Water

Question 15.
Give two important uses of washing soda and baking soda.
Answer:
Uses of washing soda:

• Washing soda is used in glass, soap, textile and paper industry.
• It is used for removing permanent hardness of water.

Uses of baking soda :

• Baking soda is used as an antacid and as disinfectant.
• It is used in soda-acid fire-extinguishers.

Jharkhand Board Class 10 Science Acids, Bases and Salts InText Questions and Answers

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:
Take three test tubes and label them as A, B and C. Add one drop of solution from test tubes A, B and C on the red litmus paper separately. The solution of a test tube which turns red litmus paper to blue contains a base.

Now, tire remaining two test tubes contain acid and distilled water. Use the above blue litmus paper to test the solution of remaining two test tubes. Add one drop of solution from two test tubes on blue litmus paper separately. The solution of a test tube which turns blue litmus paper to red contains an acid.

The solution of a test tube which does not change either red or blue litmus paper contains distilled water. Thus, solutions of three test tubes can be tested.

Question 2.
Why should curd and sour substances not to be kept in brass and copper vessels?
Answer:
Curd and sour substances are acidic in nature, which react with brass and copper vessels and form toxic substances; which are harmful to the human body.
Therefore, curd and other sour substances should not be kept in brass and copper vessels.

Question 3.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:
Reaction of a metal with an acid liberates hydrogen gas.
For example,
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
The presence of hydrogen gas can be tested by bringing a burning candle or match-stick near the mouth of the test tube, in which hydrogen gas is collected. The hydrogen gas burns with a popping sound.

Question 4.
A metal compound ‘A’ reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction, if one of the compounds formed is calcium chloride.
Answer:
The compound of metal ‘A’ is CaCO₃.
Gas formed is CO₂.
Balanced chemical equation:
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(1)

Question 5.
Why do HCl, HNO₃, etc. show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
HCl, HNO₃, etc. release H⁺ions in their aqueous solutions. Hence, their solutions show acidic character. While alcohol and glucose do not release H⁺ ions in their aqueous solutions. So, their aqueous solutions are not acidic.

Question 6.
Why does an aqueous solution of an acid conduct electricity?
Answer:
when an acid dissolves in water to form a solution, it ionises forming ions, which carry electricity.

Question 7.
Why does dry HCl gas not change the colour of the dry litmus paper?
Answer:
Dry HCl gas does not release H⁺ ions, so it does not possess acidic character and it does not impart any effect on dry litmus paper. As a result, the colour of the dry litmus paper does not change.

Question 8.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid.
Answer:
When water is added to a concentrated ? acid for dilution, the heat generated may cause the mixture to splash out and may cause burns. The glass container may also break due to excessive local heating.

Therefore, for dilution of concentrated acid, instead of adding water to acid, acid is added s slowly to water with constant stirring. Hence, liberated heat energy during dilution will spread in water and does not cause harm.

Question 9.
How is the concentration of hydronium ion (H₃O⁺) affected when a solution of an acid s is diluted?
Answer:
when the acid solution is diluted, the concentration of hydronium ions (H₃O⁺) per unit volume decreases.

Question 9.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide ?
Answer:
when excess of water is added to the solution of sodium hydroxide (NaOH), i.e., solution is diluted, the concentration of hydroxide ions (OH^–) per unit volume decreases.

Question 10.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
The pH value of solution A is 6.
∴ It is acidic and has higher concentration of hydrogen ions. (about 10^-6 M)
The pH value of solution B is 8.
∴ It is basic and having lower concentration of hydrogen ions. (about 10^-8M)

Question 11.
What effect does the concentration of H⁺(aq) ions have on the nature of the solution?
Answer:
A solution has higher concentration of H⁺(aq) ions (more than 10^-7) is acidic in nature and the solution has lower concentration of H⁺ ions (less than 10^-7 M) is basic in nature.

Question 12.
Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?
Answer:
Yes, basic solutions also possesses H⁺ ions, but due to more number of OH^– ions than H⁺ ions they are basic in nature.

Question 13.
Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
The soil having pH value less than 6.5 is called acidic soil. Now to make this soil neutral, the farmers add basic substances such as quick lime, slaked lime or chalk to the soil.

Question 14.
What is the common name of the compound CaOCl₂?
Answer:
The common name of CaOCl₂ is bleaching powder.

Question 15.
Name the substance which on treatment with chlorine yields bleaching powder.
Answer:
Calcium hydroxide (Ca(OH)₂) reacts with chlorine to form bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 16.
Name the sodium compound which is used for softening hard water.
Answer:
Sodium carbonate – Na₂CO₃.

Question 17.
What will happen if a solution of sodium hydrogencarbonate is heated? Give the equation of the reaction involved.
Answer:
The solution of sodium hydrogen- carbonate, on heating forms sodium carbonate, water and carbon dioxide.

Question 18.
Write an equation to show the reaction between plaster of paris and water.
Answer:

Activity 2.1 [T. B. Pg. 18]

Aim : To test acids and bases in the laboratory.

Activity:
1. Collect the solutions of hydrochloric acid (HCl), sulphuric acid (H₂SO₄), nitric acid (HNO₃), acetic acid (CH₃COOH), sodium hydroxide (NaOH), potassium hydroxide (KOH), calcium hydroxide (Ca(OH)₂), magnesium hydroxide (Mg(OH)₂) and ammonium hydroxide (NH₄OH) from the laboratory.

2. Put a drop of each of the above solution on a watch-glass and test with a drop of the indicators as shown in Table 1.

3. Observe the colour change with red litmus paper, blue litmus paper, phenolphthalein and methyl orange solution for each of the sample solution.

4. Write your observations in Table 1.

Sample solution	Red litmus solution (or paper)	Blue litmus solution (or paper)	Phenolphthalein solution	Methyl orange solution
HCl	No colour change	Red	Colourless	Red
H2SO4	No colour change	Red	Colourless	Red
HNO3	No colour change	Red	Colourless	Red
CH3COOH	No colour change	Red	Colourless	Red
NaOH	Blue	No colour change	Pink	Yellow
KOH	Blue	No colour change	Pink	Yellow
Ca(OH)2	Blue	No colour change	Pink	Yellow
Mg(OH)2	Blue	No colour change	Pink	Yellow
NH4OH	Blue	No colour change	Pink	Yellow

Activity 2.2 [T. B. Pg. 18-19]

Aim: To test acids and bases using olfactory > indicators.

Activity:

• Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases.
• Take two of these cloth strips and check their odour.
• Keep them on a clean surface and put a few drops of dilute HCl solution on one strip and a few drops of dilute NaOH solution on the other.
• Rinse both cloth strips with water and again check their odour.
• Note your observations.
• Now take some dilute vanilla extract and clove oil and check their odour.
• Take some dilute HC1 solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla extract to both test tubes and shake well. Check the odour once again and record changes in odour, if any.
• Similarly, test the change in the odour of clove oil with dilute HC1 and dilute NaOH solutions and record your observations.

Questions :

Question 1.
Which among the vanilla extract, onion and clove oil can be used as olfactory indicator?
Answer:
Vanilla extract, onion and clove oil can be used as olfactory indicators.

Question 2.
What change in odour do you observe when vanilla extract, onion and clove oil are mixed with dilute HCl and dilute NaOH solutions separately?
Answer:
When dilute HCl solution is added in vanilla extract, onion and clove oil separately, their odour remains unchanged, but adding dilute NaOH solution separately in vanilla extract, onion and clove oil, their odour disappears.

Activity 2.3 [T .B. Pg. 19]

Aim: Testing of hydrogen gas formed by the reaction of zinc granules with dilute sulphuric acid (H₂SO₄).
Caution: This activity needs the teacher’s assistance.
Activity:

• Arrange the apparatus as shown in the figure 2.1.
• Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
• What do you observe around the surface of zinc granules?
• Pass the gas being evolved through the soap solution.
• Why are bubbles formed in the soap solution?
• Take a burning candle near a gas filled in bubbles.
• What do you observe?
• Repeat this activity with some more acids like HCl, HNO₃ and CH₃COOH.
• Are the observations in all the cases the same or different?
  
• In above reactions, metal displace hydrogen gas from acid.
• In short, when metal reacts with an acid forms salt and evolves H₂ gas.
  Metal + Acid → Salt + Hydrogen gas

Questions :

Question 1.
What do you observe around the surface l of zinc granules?
Answer:
Hydrogen gas is bubbled around the surface of zinc granules.

Question 2.
Why are bubbles formed in the soap? solution?
Answer:
Zinc granules reacts with H₂SO₄ forming hydrogen gas which, when passed through the solution of soap form bubbles.

Question 3.
What happens, when a burning candle is brought near the hydrogen gas?
Answer:
When a burning candle is brought near the s hydrogen gas, it burns with a popping sound.

Question 4.
Does zinc metal liberate hydrogen gas with dilute HCl, dilute HNO₃ and CH₃COOH solution? Mention it with equations of? chemical reactions.
Answer:

Question 5.
Write the balanced chemical equation of ( reaction of zinc metal with dilute H₂SO₄.
Answer:

Activity 2.4 [T. B. Pg. 20]

Aim : To study the reaction of a base with metal.

Activity:

• Take a few pieces of granulated zinc metal in a test tube.
• Add about mL of sodium hydroxide solution in it and warm the mixture of the test tube.
• Repeat the rest of the steps as in Activity 2.3 and record your observations.

Questions :

Question 1.
Which gas is liberated when zinc metal reacts with sodium hydroxide solution?
Answer:
Hydrogen gas

Question 2.
State the molecular formula of sodium zincate.
Answer:
Na₂ZnO₂

Question 3.
Give the balanced chemical equation for the reaction of zinc with sodium hydroxide solution.
Answer:
Zn(s) + NaOH(aq) → Na₂ZnO₂(s) + H₂(g)

Activity 2.5 [T. B. Pg. 20]

Aim : To study the reaction of metal carbonates and metal hydrogencarbonates with acids.

Activity :

• Take two test tubes, label them as A and B.
• Take about 0.5 g of sodium carbonate (Na₂CO₃ in test tube A 0.5 g of sodium hydrogencarbonate (NaHCO₃) in test tube B.
• Add about 2mL of dilute HCl to both the test tubes.
• What do you observe?
• Pass the carbon dioxide gas formed into calcium hydroxide solution as shown in figure 2.2 and record your observations.
  

Questions :

Question 1.
Which gas is evolved, when sodium carbonate and sodium hydrogencarbonate are treated with HCl?
Answer:
Carbon dioxide gas

Question 2.
Write the general word equation for the above activity.
Answer:

Question 3.
What is the common name of Ca(OH)?
Answer:
Lime water

Question 4.
What is the solubility of Ca(HCO₃)₂ in water?
Answer:
Ca(HCO₃)₂ is highly soluble in water.

Activity 2.6 [T. B. Pg. 21]

Aim : To study the reaction takes place between acids and bases.

Activity:

• Take about mL of dilute NaOH solution in a test tube. Add two drops of phenolphthalein and observe the colour change.
• Add dilute HCl solution to the above solution drop by drop.
• Is there any colour change for the reaction mixture?
• Why did the colour of phenolphthalein change after the addition of an acid?
• Now add a few drops of NaOH to the above mixture.
• Does the pink colour of phenolphthalein reappear?

Questions:

Question 1.
What happens when an indicator phenolphthalein is added to NaOH solution?
Answer:
NaOH solution becomes pink.

Question 2.
What happens when few drops of HCl are added in the solution of NaOH and phenolphthalein indicator?
Answer:
Pink colour disappears.

Question 3.
Write only the name of the reaction occurring between acid and base.
Answer:
Neutralisation reaction.

Question 4.
Write an equation for a neutralisation reaction.
Answer:
NaOH(aq) + HCl(aq) → NaCl(aq) + HO(l)

Activity 2.7 [T. B. Pg. 21]

Aim : To study the reaction of metal oxide with an acid.

Activity:

• Take a small amount of copper oxide in a beaker.
• Add dilute hydrochloric acid slowly while stirring.
• Note the colour of the solution. What has happened to the copper oxide?

Questions:

Question 1.
State the colour of powder of copper oxide.
Answer:
The powder of copper oxide is black.

Question 2.
What would be the colour of solution when dilute hydrochloric acid is added to copper oxide?
Answer:
Blue-green.

Question 3.
State the reason of blue-green colour of the solution.
Answer:
The blue-green colour of the solution is due to the formation of copper (II) chloride in the reaction between copper oxide and hydrochloric acid.

Question 4.
Write the molecular formula of copper (II) chloride.
Answer:
CuCl₂

Question 5.
State the general equation of reaction that occurs between a metal oxide and an acid.
Answer:
Metal oxide + Acid → Salt + Water

Activity 2.8 [T. B. Pg. 22]

Aim : To study the conduction of electricity by aqueous solutions of acids and bases.

Activity:

• Arrange the apparatus as shown in the figure 2.3.
  
• In the given arrangement, when solutions of different compounds are added in the beaker separately, following observations were recorded :

Solution	Bulb	Reason
1. HCl	Glows	H+ ions released
2. H2SO4	Glows	H+ ions released
3. Glucose (C6H12O6)	Does not	H+ or OH– ions are not released.
4. Alcohol (C2H5OH)	glow	H+ or OH– ions are not released.
5. NaOH	Does not glow	OH– ions released
6. Ca(OH)2	Glows	OH– ions released

Questions :

Question 1.
What does the glowing of bulb indicate?
Answer:
Glowing of bulb indicates that electric current is passing through the solution.

Question 2.
By whom is the electric current carried through the solution?
Answer:
The electric current is carried through the solution by ions.

Question 3.
Mention the ions responsible for acidic and basic character of the compound.
Answer:
H⁺ ions in an aqueous solution are responsible for acidic character, while OH” ions in an aqueous solution are responsible for basic character.

Activity 2.9 [T.B.Pg.23]

Aim : To test that dry HC1 gas is not acidic but its aqueous solution is acidic.

Activity:

• Take about 1 g solid NaCl in a clean and dry test tube and set up the apparatus as shown in figure .4.
• Add some concentrated sulphuric acid to the test tube.
• What do you observe? Is there a gas coming out of the delivery tube?
• Test the gas evolved successively with dry and wet blue litmus paper.
• In which case does the litmus paper change colour?
• On the basis of the above activity, what do you infer about the acidic character of: (i) dry HCl gas and (ii) HCl solution?
  

Questions:

Question 1.
Does the separation of H⁺ ions from HCl molecules occur in the absence of water?
Answer:
No

Question 2.
How are the hydrogen ions always represented?
Answer:
Hydrogen ions are always represented by H⁺(aq) or H₃O⁺ (hydronium ion).

Question 3.
Which ion is responsible to represent the basic character of an aqueous solution?
Answer:
OH⁺ ion.

Question 4.
Mention the solubility of alkali (base) in water.
Answer:
Alkalis (base) are soluble in water.

Question 5.
Give two examples of alkali (base).
Answer:
NaOH, KOH

Question 6.
State the properties of alkalis (base).
Answer:
Alkalis are soapy in touch, bitter in taste and corrosive.

Question 7.
State the general equation for neutralisation reaction.
Answer:
Acid + Base → Salt + Water
or
HX + MOH → MX + H₂O

Activity 2.10 [T. B. Pg. 24]

Aim: To demonstrate that dilution of a strong acid and a strong base is exothermic.

Activity:

• Take 10 mL water in a beaker.
• Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
• Touch the base of the beaker.
• Is there a change in temperature?
• Is this an exothermic or an endothermic process?
• Repeat the above activity with sodium hydroxide pellets and record your observations.

Questions:

Question 1.
Name the process of dissolving an acid or a base in water.
Answer:
Exothermic reaction

Question 2.
What change occurs in concentration of solution by adding water to an acid or a base?
Answer:
Concentration of solution decreases.

Activity 2.11 [T. B. Pg. 26]

Aim : To test the pH values of given solutions, s

Activity:
Test the pH values of the following solutions :

• Saliva (before meal)
• Saliva (after meal)
• Lemon juice
• Colourless aerated drink
• Carrot juice
• Coffee
• Tomato juice
• Tap water
• 1 M NaOH
• 1 M HCl

Solution	Colour of pH paper	Approxi-mate pH value	Nature of substance
1. Saliva (before meal)	Light green	7.4	Basic
2. Saliva (after meal)	Light yellow	5.8	Acidic
3. Lemon juice	Pink-red	2.5	Acidic
4. Colourless aerated drink	Light yellow	6.0	Acidic
5. Carrot juice	Light orange	4.0	Acidic
6. Coffee	Orange yellow	5.0	Acidic
7. Tomato juice	Dark orange	4.1	Acidic
8. Tap water	Green	7.0	Neutral
9. 1M NaOH	Dark blue	13-14	Basic
10. 1MHCl	Red	1.0	Acidic

Activity 2.12 [T. B. Pg. 27]

Aim: To find the pH value of the soil.

Activity:

• Take about g soil in a test tube and add 5 mL water to it.
• Shake the test tube, filter the contents and collect the filtrate in a test tube.
• Check the pH of this filtrate with the help of universal indicator paper.

Questions :

Question 1.
Filtrate in a test tube turns red litmus to blue. What does it indicate about sample of soil?
Answer:
The sample of soil possess basic property.

Question 2.
What would be the property of soil having pH value less than 5.6?
Answer:
The soil possess acidic property.

Question 3.
How will you neutralise the acidic and basic soil?
Answer:
Add base like lime to neutralise the acidic soil and add acid like gypsum to neutralise the basic soil.

Question 4.
What should be the pH value of soil for healthy growth and development of plants?
Answer:
The soil should have pH value between 6.5 to 7.3.

Activity 2.13 [T. B. Pg. 28 – 29]

Aim: Identify the acids and bases required to form the given salts.

Activity:
Some salts are given in the following table. Identify the required acids and bases to form the salts :
Table 3

Name of the Salt	Formula of the salt	Acid required	Base required
1. Potassium sulphate	K2SO4	H2SO4	KOH
2. Sodium sulphate	Na2SO4	H2SO4	NaOH
3. Calcium sulphate	CaSO4	H2SO4	Ca(OH)2
4. Magnesium sulphate	MgSO4	H2SO4	Mg(OH)2
5. Copper sulphate	CuSO4	H2SO4	CU(OH)2
6. Sodium chloride	NaCl	HCl	NaOH
7. Sodium nitrate	NaNO3	HNO3	NaOH
8. Sodium carbonate	Na2CO3	H2CO3	NaOH
9. Ammonium chloride	NH4Cl	HCl	NH4OH

Questions:

Question 1.
What is called family of salt?
Answer:
Salts having the same positive or negative radicals (ions) are said to be family of salt.

Question 2.
Give examples of family of sodium salts.
Answer:
NaCl, NaSO₄, Na₂CO₃, etc. are family of sodium salts.

Question 3.
State the examples of family of chloride salts.
Answer:
NaCl. KCl, NH₄Cl, etc. are family of chloride salts.

Activity 2.14 [T. B. Pg. 29]

Aim: To find the pH and test the solubility of salt in water.

Activity:

• Check the solubility of given salts (in Table 4) in distilled water. Write the approximate value of their pH and mention the acids and bases from which the salts are formed.
  Table 4
  

Activity 2.15 [T. B. Pg. 32]

Aim: To remove the water of crystallisation.

Activity:

• Heat a few crystals of copper sulphate in a dry boiling tube.
• What is the colour of copper sulphate after heating?
• Do you observe water droplets in the boiling tube? Where have these come from?
• Add 2-3 drops of water on the sample of copper sulphate obtained after heating and observe it.
• Is the blue colour of copper sulphate restored?
  

Questions:

Question 1.
State the colour of copper sulphate after heating.
Answer:
It turns white from blue.

Question 2.
Do you notice the water droplets in the boiling tube?
Answer:
Yes. Copper sulphate crystals which seem to be dry contain water of crystalisation. On heating the copper sulphate crystals, water molecules are removed from the crystal and appear in the boiling tube.

Question 3.
Does the blue colour of copper sulphate solution reappear on adding 2 to 3 drops of water after heating it?
Answer:
Yes.

Question 4.
What is the chemical formula of hydrated copper sulphate?
Answer:
CuSO₄.5H₂O

Students should go through these JAC Class 10 Maths Notes Chapter 1 वास्तविक संख्याएँ will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 1 वास्तविक संख्याएँ

भूमिका :
हमने पिछली कक्षाओं में प्राकृतिक संख्याओं, पूर्ण संख्याओं, परिमेय संख्याओं अपरिमेय संख्याओं व वास्तविक संख्याओं के विषय में पढ़ा है। हमने इनके गुणों के विषय में भी पढ़ा है स्मरण करें कि वास्तविक संख्याएँ परिमेय संख्याओं और अपरिमेय संख्याओं का समूह होती हैं।

इस अध्याय में हम \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) आदि की अपरिमेयता सिद्ध करने, परिमेय संख्याओं का दशमलव प्रसार, सांत आवर्ती दशमलवं प्रसार, यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग, अंकगणित की आधारभूत प्रमेव यूक्लिड विभाजन प्रमेविका आदि के विषय में अध्ययन करेंगे।
→ वास्तविक संख्याएँ (Real Numbers) : परिमेय संख्या और अपरिमेय संख्याओं के समूह को वास्तविक संख्याएँ कहते हैं।
जैसे 3, 5, \(\sqrt{7}\), \(\frac{-\sqrt{3}}{2}\) आदि।

→ परिमेय संख्याएँ (Rational numbers): वे संख्याएँ जिनका दशमलव प्रसार सांत अथवा असांत आवर्ती होता परिमेय संख्याएँ कहलाती हैं। इन्हें \(\frac{p}{q}\) (जहाँ q ≠ 0) के रूप में लिखा जा सकता है।
जैसे- \(\frac{2}{4}\), 3, \(\frac{1}{3}\) आदि।

→ अपरिमेय संख्याएँ (Irrational numbers): वे संख्याएँ जिनका दशमलव प्रसार असांत तथा अनावर्ती होता है. अपरिमेय संख्याएँ कहलाती हैं।
जैसे- \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt[3]{4}\), π आदि ।

→ पूर्णांक (Integers): शून्य प्राकृतिक संख्याओं एवम् ऋणात्मक प्राकृतिक संख्याओं के समूह की पूणांक कहते हैं।
जैसे- -9, -8, 0, 5, 4, आदि।

→ सम संख्याएँ (Even numbers): वे संख्याएँ जो 2 से विभाजित होती हैं, सम संख्याएँ कहलाती है।
जैसे- 2, 4, 6, 8, 10 ….. (सम संख्याओं में इकाई का अंक 0, 2, 4, 6, 8 होता है।)

→ विषम संख्याएँ (Odd numbers): वे संख्याएँ जो 2 से विभाजित नहीं होती हैं, विषम संख्याएँ कहलाती हैं।
जैसे- 1, 3, 5, 7, 9, …. (विषम संख्याओं में इकाई का अंक 1, 3, 5, 9 होता है।)

→ अभाज्य संख्याएँ (Prime numbers): वे संख्याएँ जिनके सिर्फ दो गुणनखंड (1 व स्वयं) होते हैं, अभाज्य संख्याएँ कहलाती हैं अथवा वे संख्याएँ जो या स्वयं से विभाजित होती हैं, अभाज्य संख्या कहलाती हैं।
जैसे- 2, 5, 7, 11, 13, ….. ( सबसे छोटी अभाज्य संख्या 2 होती है।)

→ भाज्य संख्याएँ (Composite numbers): वे संख्याएँ जिनके और स्वयं के अलावा कम से कम एक और “गुणनखण्ड हो, भाज्य संख्याएँ कहलाती हैं।
जैसे- 4, 6, 8, 9, 10, 12, …
महत्वपूर्ण बिन्दु: 1 न तो अभाज्य संख्या है और न ही भाज्य संख्या।

→ सांत दशमलव प्रसार (Terminating decimal expansion): जिन परिमेय संख्याओं के हर 2ⁿ × 5^m के रूप में होते हैं, उनके दशमलव प्रसार सांत होते हैं।
जैसे \(\frac{2}{4}\) = 0.5, \(\frac{1}{5}\), 0.2, \(\frac{10}{25}=\frac{2}{5}\) = 0·4

→ असांत आवर्ती दशमलव प्रसार (Non-terminating recurring decimal expansion): जिन परिमेय संख्याओं के हर 2ⁿ × 5^m के रूप में नहीं होते हैं, उन परिमेय संख्याओं के दशमलव प्रसार असांत आवर्ती होते हैं।
जैसे \(\frac{1}{3}\) = 0.333, …. \(\frac{17}{6}\) = 2.8333…

→ लघुत्तम समापवर्त्य (LCM): दी गयी संख्याओं के छोटे से छोटे सार्वगुणज को लघुत्तम समापवर्त्य कहते हैं।

→ महत्तम समापवर्त्य (HCF): दी गयी संख्याओं के बड़े से बड़े सार्व गुणनखंड को महत्तम समापवर्तक कहते हैं।

पूर्णांकों की भाजकता तथा धनात्मक पूर्णांकों के दो महत्वपूर्ण गुण-
(i) यूक्लिड विभाजन एल्गोरिथ्म ( कलन विधि) (Euclid’s Division Algorithm ): यूक्लिड विभाजन एल्गोरिथ्म का प्रयोग इस अध्याय में दो धनात्मक पूर्णांकों के महत्तम समापवर्तक (HCF) परिकलित करने में करेंगे।
(ii) अंकगणित की आधारभूत प्रमेय (Fundamental Theorem of Arithmetic ): अंकगणित की आधारभूत प्रमेय का प्रयोग दो अनुप्रयोगों में करेंगे:
(i) प्रथम अनुप्रयोग में कुछ संख्याओं जैसे \(\sqrt{2}\), \(\sqrt{3}\) और \(\sqrt{5}\) आदि की अपरिमेयता सिद्ध करने में करेंगे। (ii) किसी दी गई संख्या के अभाज्य गुणनखण्ड ज्ञात करने में करेंगे।

यूक्लिड विभाजन प्रमेविका :
यूक्लिड प्रथम यूनानी गणितज्ञ थे, जिन्होंने समतल ज्यामिति के अध्ययन हेतु एक नई विचारधारा को प्रारम्भ किया जिसमें से एक यूक्लिड विभाजन प्रमेयिका है। इसके अनुसार एक धनात्मक पूर्णांक a को किसी अन्य धनात्मक पूर्णांक b से विभाजित करने पर भागफल q और शेषफल r प्राप्त होता है तथा शेषफल r या तो शून्य होता है या भाजक b से छोटा होता है अर्थात् 0 ≤ r < b होता है।

यही यूक्लिड की विभाजन प्रमेयिका है।
इसे औपचारिक रूप से निम्न प्रकार व्यक्त कर सकते हैं:
यदि a और b दो धनात्मक पूर्णांक हैं तो दो ऐसे अद्वितीय पूर्णांक q और r विद्यमान होते हैं कि
a = bq + r, जहाँ 0 ≤ r < b
ध्यान रहे कि या शून्य भी हो सकते हैं।
साधारण शब्दों में, भाज्य (a) = भाजक (b) × भागफल (q) + शेषफल (r)

अंकगणित की आधारभूत प्रमेय :
प्रत्येक भाज्य संख्या को अभाज्य संख्याओं के एक गुणनफल के रूप में व्यक्त (गुणनखंडित) किया जा सकता है तथा यह गुणनखंडन अभाज्य गुणनखंडों के आने वाले क्रम के बिना अद्वितीय होता है अर्थात् इस पर कोई ध्यान दिए बिना कि अभाज्य गुणनखंड किस क्रम में आ रहे हैं।

उदाहरण के लिए :
44 एक भाज्य संख्या है। इसे 2 × 2 × 11 अथवा 11 × 2 × 2 के रूप में लिखा जा सकता है। यदि हम 2 × 2 × 11 अथवा 11 × 2 × 2 के क्रम पर ध्यान नहीं देते अर्थात् यदि 2 × 2 × 11 और 11 × 2 × 2 में कोई अन्तर नहीं है, तो यह गुणनखण्ड अद्वितीय भी है।
दो संख्याओं के म. स. तथा ल.स. में सम्बन्ध- दो संख्याओं के म. स. तथा ल.स. का गुणनफल उन संख्याओं के गुणनफल के बराबर होता है। अर्थात्
म.स. (H.C.F.) × ल.स. (L.C.M.) = एक संख्या (a) × दूसरी संख्या (b)
यदि दो संख्याएँ a और b हों, तो
H.C.F. × L.C.M. = a × b
इस मुख्य सम्बन्ध की सहायता से निम्नांकित सम्बन्ध भी लिखे जा सकते हैं-
(i) H.C.F. = \(\frac{a \times b}{\text { L.C.M }}\)
(ii) L.C.M. = \(\frac{a \times b}{\text { H.C.F. }}\)
(iii) a = \(\frac{\text { H.C.F. } \times \text { L.C.M. }}{b}\)
(iv) b = \(\frac{\text { H.C.F. } \times \text { L.C.M. }}{a}\)

अपरिमेय संख्याओं का पुनभ्रमण :
अपरिमेय संख्याएँ: वे संख्याएँ जिनका दशमलव प्रसार असान्त (Non-terminating) और अनावर्ती (Non-repeating) हो, अपरिमेय संख्याएँ कहलाती हैं। इन संख्याओं को \(\frac{p}{q}\) के रूप में व्यक्त नहीं किया जा सकता; जहाँ p और q पूर्णांक है और q ≠ 0.
उदाहरण : \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{15}\), π, \(\frac{-\sqrt{2}}{\sqrt{3}}\), 0.10110111011110…….. इत्यादि।
हम विरोधाभास विधि द्वारा संख्याओं की अपरिमेयता को सिद्ध करते हैं।

परिमेय संख्याओं और उनके दशमलव प्रसारों का पुनभ्रमण :
कोई भी परिमेय संख्या जिसका दशमलव प्रसार सांत है, उसे हम एक ऐसी परिमेय संख्या के रूप में लिख सकते हैं जिसका हर 10 की कोई घात होती है अर्थात् कोई भी 10 की धनात्मक घात को 2 और 5 की घातों के गुणनफल के रूप में व्यक्त कर सकते हैं।
जैसे 10ⁿ = (2 × 5)ⁿ ⇒ 2ⁿ × 5ⁿ
10² = 2² × 5²
10³ = 2³ × 5³ इत्यादि
अतः जिन परिमेय संख्याओं के दशमलव प्रसार सांत होते हैं, उनके हर 2ⁿ × 5^m के रूप में होते हैं।

Students must go through these JAC Class 10 Science Notes Chapter 9 Heredity and Evolution to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 9 Heredity and Evolution

→ Variation: The differences in the characters or traits shown by the individuals of same species are called variations.

→ Variations arise during the process of reproduction. These variations help the survival of the individuals.
Depending on the nature of variations, different individuals have different kinds of advantages.

→ Heredity: A process of transmission of characters generation after generation is called heredity. It is a process by which parental characters are transmitted to offsprings.

→ Dominant and recessive trait: A trait which expresses itself in the presence of its contrasting trait is termed as dominant trait whereas recessive trait remains unexpressed in presence of dominant one.

→ Trait: It is a distinguishing feature of a person’s character.

• In simple term, trait is the way in which the characters of an organism are expressed.
• Examples : Blond hair, blue eyes, attached earlobes, tallness, etc.

→ Factor: The unit, which is responsible for the heredity or regulating the expressions of a character is called a factor.

• The factors responsible for an expression of a trait are always in a pair.
• The factor as suggested by Mendel, is now known as gene in the modern genetics.
• The factor (gene) is represented by the English alphabet.

→ Gregor Johann Mendel: He conducted a series of experiments on garden pea plant (Pisum sativum) in order to study heredity.
The results of the Mendel’s experiments explain the mechanism of inheritance.

→ Gene : A section of DNA (nucleotide sequence) that provides information for the synthesis of a specific protein is called gene.

→ Sex determination: The phenomenon of determining the sex of an individual of a species is called sex determination.

• There are different strategies e.g., temperature, chromosomes, genes, hormones, etc. responsible for determining the sex in certain animals.
• In certain animals, the sex determination is not genetically determined, e.g., the Snails can change their sex.
• In human beings in each cell there are 22 pairs of autosomes and 1 pair of sex chromosomes. The 23rd pair is either XX as in females or XY as in males.

→ Genetic drift: A change in the frequency of some genes in a population which provides diversity without any survival advantage is called genetic drift.

→ Evolution: A process of slow, gradual and progressive changes by which complex form of organisms are produced from simple form of organisms over a long period of time is called evolution.

→ Charles Darwin conducted various experiments that led him to formulate the theory of natural selection which states that evolution proceeds due to natural selection.

→ Acquired trait: The characteristics of the organisms, which develop as an interaction with the environment and which are not hereditary are called acquired characteristics or traits, e.g., weight loss achieved by dieting.

→ Inherited trait: The characteristic of the organism, that comes into existence by the changes in DNA structure is called an inherited trait, e.g., skin colour, eye colour, colour of flower, height of plant, etc.

→ Speciation: The phenomenon of origin of a new species from the pre-existing one due to reproductive isolation of a part of its population is called speciation.

→ Speciation is caused due to genetic drift, natural selection, geographical and reproductive isolation.

→ Gene flow: The transfer of genes or alleles between inbreeding population of a particular species is called gene flow.

→ Evidences of evolution:

Homologous Organs	Analogous Organs	Fossils
Structurally similar but may be different functionally.	Basically different in structure and appear similar in their look and also in the function.	Organs preserved in the earth crust and obtained therefrom as samples or impressions
Example : Forelimbs of wall lizard and forelimbs of human being	Example : Wings of birds and wings of bats	Example : Ammonites, Trilobites

• Evolution can be studied with the help of living species and fossils.
• Evolutionary relationships are traced in the classification of organisms.
• Organs or features may be adapted to new functions during course of evolution, e.g., feathers have been initially evolved for. warmth and later adopted for flight.

→ Evolution through artificial selection: Farmers have obtained newer varieties of cabbage from the wild variety of cabbage by artificial selection. Cauliflower, Broccoli, Kohlrabi and Kale species are apparently different from their wild ancestral cabbage.

→ Human evolution: Study of the evolution of human beings indicates that all of us belong to a single species Homo sapiens. Our genetic footprints can be traced back to our African roots.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 त्रिभुज Ex 6.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 त्रिभुज Exercise 6.2

प्रश्न 1.
आकृति (i) और (ii) में DE || BC है। (i) में, EC और (ii) में AD ज्ञात कीजिए।

अथवा
यदि ΔABC में DE || BC है, AD = 1.5 सेमी, BD = 3 सेमी तथा AE = 1 सेमी हो, तो EC ज्ञात कीजिए।
हल:
(i) ΔABC में,
DE || BC (दिया है)
(आधारभूत समानुपातिक प्रमेय से)
\(\frac{A D}{D B}=\frac{A E}{E C}\)
⇒ \(\frac{1.5}{3}=\frac{1}{E C}\)
⇒ EC = \(\frac{3}{1.5}\)
∴ EC = 2 सेमी

(ii) ΔABC में,
DE || BC (आधारभूत समानुपातिक प्रमेय से)

प्रश्न 2.
किसी ΔPQR की भुजाओं PQ और PR पर क्रमशः बिन्दु E और F स्थित हैं। निम्नलिखित में से प्रत्येक स्थिति के लिए, बताइए कि क्या EF || QR है:
(i) PE = 3.9 सेमी, EQ = 3 सेमी, PF = 3.6 सेमी और FR = 2.4 सेमी।
(ii) PE = 4 सेमी, QE = 4.5 सेमी, PF = 8 सेमी और RF = 9 सेमी।
(iii) PQ = 1.28 सेमी, PR = 2.56 सेमी, PE = 0.18 सेमी और PF = 0.36 सेमी।
हल:
ΔPQR में दो बिन्दु E और F क्रमश: PQ और PR भुजाओं पर स्थित हैं।

(i) PE = 3.9 सेमी, EQ = 3 सेमी, PF = 3.6 सेमी और FR = 2.4 सेमी,

अत: EF, QR के समान्तर नहीं है।

(ii) PE = 4 सेमी, QE = 4.5 सेमी, PF = 8 सेमी और RF = 9 सेमी,
\(\frac{P E}{Q E}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}\) …(1)
तथा \(\frac{P F}{R F}=\frac{8}{9}\) …(2)
समीकरण (1) व (2) से,
\(\frac{P E}{Q E}=\frac{P F}{R F}\)
(आधारभूत आनुपातिकता प्रमेय के विलोम से)
अत: EF || QR

(iii) PQ = 1.28 सेमी, PR = 2.56 सेमी, PE = 0.18 सेमी, PF = 0.36 सेमी
EQ = PQ – PE
= 1.28 – 0.18 = 1.10 सेमी
FR = PR – PF
= 2.56 – 0.36 = 2.20 सेमी

समीकरण (1) व (2) से,
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
(आधारभूत आनुपातिकता प्रमेय के विलोम से)
अत: EF || QR

प्रश्न 3.
निम्न आकृति में, यदि LM || CB और LN || CD हो, तो सिद्ध कीजिए कि \(\frac{A M}{A B}=\frac{A N}{A D}\) है।

हल:
ΔABC में,
ML || BC (दिया है)
∴ \(\frac{A M}{M B}=\frac{A L}{L C}\) …(i)
(आधारभूत समानुपातिक प्रमेय से)
∴ पुन: ΔADC में,
LN || DC (दिया है)
\(\frac{A N}{N D}=\frac{A L}{L C}\) …(ii)
(आधारभूत आनुपातिकता प्रमेय से)

प्रश्न 4.
निम्न चित्र में, DE || AC और DF || AE है। सिद्ध कीजिए कि \(\frac{B F}{F E}=\frac{B E}{E C}\) है।

हल:
दिया है : ΔABC में भुजा AB पर एक बिन्दु D हैं और भुजा BC पर दो बिन्दु E व F हैं। रेखाखण्ड DF, DE व AE खींचे गये हैं।
सिद्ध करना है : \(\frac{B F}{F E}=\frac{B E}{E C}\)
उपपत्ति : ΔBCA में, DE || AC (दिया है)
∴ \(\frac{B E}{E C}=\frac{B D}{D A}\) …(i)
(आधारभूत समानुपातिक प्रमेय से)
पुन: ΔBEA में, DF || AE (दिया है)
∴ \(\frac{B F}{F E}=\frac{B D}{D A}\) …(ii)
(आधारभूत समानुपातिक प्रमेय से)
समीकरण (i) व (ii) से,
\(\frac{B F}{F E}=\frac{B E}{E C}\) इति सिद्धम्।

प्रश्न 5.
निम्न चित्र में, DE || OQ और DF || OR है। दर्शाइए कि EF || QR है।

हल:
दिया है दी गई आकृति में DE || OQ तथा DF || OR है।
सिद्ध करना है : EF || QR
उपपत्ति : ΔPOQ में,
DE || OQ
\(\frac{P E}{E Q}=\frac{P D}{D O}\) …(i)
(आधारभूत आनुपातिकता प्रमेव से)
पुन: ΔPOR में,
DF || OR
\(\frac{P F}{F R}=\frac{P D}{D O}\) …(ii)
(आधारभूत आनुपातिकता प्रमेव से)
समीकरण (i) व (ii) से,
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
अब ΔPQR में,
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
(आधारभूत अनुपातिकता प्रमेय के विलोम से)
∴ EF || QR इति सिद्धम्।

प्रश्न 6.
निम्न चित्र में क्रमश: OP, OQ और OR पर स्थित बिन्दु A, B और C इस प्रकार हैं कि AB || PQ और AC || PR है। दर्शाइए कि BC || QR है।

हल:
दिया है : ΔPQR में बिन्दु A, B और C क्रमश: OP, OQ और OR पर इस प्रकार स्थित हैं कि AB || PQ और AC || PR
सिद्ध करना है : BC || QR
उपपत्ति : ΔPQO में,
AB || PQ (दिया है)
\(\frac{O A}{A P}=\frac{O B}{B Q}\) …(i)
(आधारभूत समानुपातिक प्रमेय से)
पुन: ΔPRO में,
AC || PR
\(\frac{O A}{A P}=\frac{O C}{C R}\) …(ii)
(आधारभूत समानुपातिक प्रमेय से)
समीकरण (i) व (ii) से,
\(\frac{O B}{B Q}=\frac{O C}{C R}\)
(आधारभूत समानुपातिक प्रमेय के विलोम से)
ΔCQR में, BC || QR. इति सिद्धम्।

प्रश्न 7.
प्रमेय 6.1 का प्रयोग करते हुए सिद्ध कीजिए कि एक त्रिभुज की एक भुजा के मध्य बिन्दु से होकर दूसरी भुजा के समान्तर खींची गई रेखा तीसरी भुजा को समद्विभाजित करती है। (याद कीजिए कि आप इसे कक्षा IX में सिद्ध कर चुके हैं।)
हल:
दिया है : ΔABC में; D, AB का मध्य- बिन्दु है अर्थात् AD = DB है।
BC के समान्तर रेखा l, AB व AC को क्रमश: D तथा E बिन्दु पर प्रतिच्छेद करती है।
सिद्ध करना है : E, AC का मध्य- बिन्दु है।

उपपत्ति: ∵ D, AB का मध्य बिन्दु है (दिया है)
∴ AD = DB
\(\frac{A D}{B D}=1\) …(i)
ΔABC में DE || BC
\(\frac{A D}{D B}=\frac{A E}{E C}\)
(आधारभूत समानुपातिक प्रमेय से)
\(1=\frac{A E}{E C}\)
[समी. (i) के प्रयोग से]
AE = EC
∴ E, AC का मध्यबिन्दु है। इति सिद्धम्।

प्रश्न 8.
प्रमेय 6.2 का प्रयोग करते हुए सिद्ध कीजिए कि एक त्रिभुज की किन्हीं दो भुजाओं के मध्य-बिन्दुओं को मिलाने वाली रेखा तीसरी भुजा के समान्तर होती है। (याद कीजिए कि आप कक्षा IX में ऐसा कर चुके हैं।)
हल:
दिया है ΔABC में, AB तथा AC के मध्य-बिन्दु क्रमश: D और E हैं अर्थात् AD = BD और AE = EC हैं। D को E से मिलाया।

सिद्ध करना है: DE || BC
उपपत्ति D, AB का मध्य बिन्दु है
∴ AD = BD (दिया है)
⇒ \(\frac{A D}{B D}=1\) …(i)
E, AC का मध्य- बिन्दु है।
∴ AE = EC
⇒ \(\frac{A E}{E C}=1\) …(ii)
समीकरण (i) व (ii) से,
⇒ \(\frac{A D}{B D}=\frac{A E}{E C}\)
(आधारभूत समानुपातिकता प्रमेय के विलोम से)
∴ DE || BC इति सिद्धम्।

प्रश्न 9.
ABCD एक समलम्ब है जिसमें AB || DC है तथा इसके विकर्ण परस्पर बिन्दु O पर प्रतिच्छेद करते हैं। दर्शाइए कि \(\frac{A O}{B O}=\frac{C O}{D O}\) हैं।

हल:
दिया है : समलम्ब चतुर्भुज ABCD है जिसमें AC और BD दो विकर्ण हैं, जो परस्पर O बिन्दु पर काटते हैं।
सिद्ध करना है : \(\frac{A O}{B O}=\frac{C O}{D O}\)
रचना : O से जाती हुई OE || CD खींची।
उपपत्ति: ΔADC में,
OE || DC
\(\frac{A E}{E D}=\frac{A O}{C O}\) …(i)
(आधारभूत समानुपातिक प्रमेय से)
समलम्ब चतुर्भुज ABCD में,
AB || CD
∴ OE || CD (रचना से)
OE || AB
अब ΔADB में,
OE || AB
\(\frac{E D}{A E}=\frac{D O}{B O}\)
⇒ \(\frac{A E}{E D}=\frac{B O}{D O}\) …(ii)
समीकरण (i) व समीकरण (ii) से,
\(\frac{A O}{C O}=\frac{B O}{D O}\)
⇒ AO × DO = BO × CO
⇒ \(\frac{A O}{B O}=\frac{C O}{D O}\) इति सिद्धम्।

प्रश्न 10.
एक चतुर्भुज ABCD के विकर्ण परस्पर हिन्दु O पर इस प्रकार प्रतिच्छेद करते हैं कि \(\frac{A O}{B O}=\frac{C O}{D O}\) है। दर्शाइए कि ABCD एक समलम्ब है।

हल:
दिया है ABCD एक चतुर्भुज है जिसके विकर्णं AC तथा BD बिन्दु O पर एक दूसरे को इस प्रकार प्रतिच्छेद करते हैं कि
\(\frac{A O}{B O}=\frac{C O}{D O}\)
सिद्ध करना है : ABCD एक समलम्ब है।
रचना : O से OE || DC खींची।
उपपत्ति: ΔBDC में,
OE || DC
\(\frac{B O}{D O}=\frac{B E}{E C}\) …(i)
परन्तु दिया है, \(\frac{A O}{B O}=\frac{C O}{D O}\)
⇒ \(\frac{A O}{C O}=\frac{B O}{D O}\) … (ii)
समीकरण (i) व (ii) से,
\(\frac{A O}{C O}=\frac{B E}{E C}\)
⇒ \(\frac{C O}{A O}=\frac{E C}{B E}\)
∴ OE || AB
(आधारभूत आनुपातिक प्रमेय के विलोम से)
इसी प्रकार, OE || CD
⇒ AB || CD
अत: ABCD एक समलम्ब है। इति सिद्धम्।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 2 Acids, Bases and Salts Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 2 Acids, Bases and Salts

Additional Questions and Answers

Question 1.
Answer the following questions:
(1) How are acids and bases formed and from what?
Answer:
The reaction of oxides of non-metal with water forms an acid.

In short, Oxide of non-metal + Water → Acid
The reaction of oxides of metal with water forms a base.

In short, Oxide of metal + Water → Base

(2) Explain : Strong acid and weak acid
Answer:
Strong acid: The mineral acids like hydrochloric acid (HCl), sulphuric acid (H₂SO₄) and nitric acid
(HNO₃) when dissolved in water, they get completely ionised.
The acid which gets ionised completely when dissolved in water is known as strong acid.
For example,

Thus, in 1 M HCl aqueous solution, concentrations of H₃O⁺ and Cl- are 1M.

Thus, in 1 M H₂SO₄ aqueous solution, concentration of H₃O⁺ and SO₄^2- are 2M and 1M respectively, because two moles HsO⁺ are formed by ionisation of one mole H₂SO₄.

In the aqueous solution of strong acid, total solute substance is in the form of ions, i.e., the solute substance is ionised to H₃O⁺.

Weak acid: The organic acids like acetic acid (in vinegar), lactic acid (in curd, buttermilk), citric acid (in lemon, orange), tartaric acid (in tamarind), oxalic acid (in tomato) when dissolved S in water they get partially ionised.

The acid which gets ionised partially or incompletely when dissolved in water is known as weak acid.
For example, The concentration of H₃O⁺ is not 1M in 1M CH₃COOH aqueous solution but it is very less, (approximately 2 to 3 %).

(3) Explain: Strong acid and weak base
Answer:
Strong base:The base which is ionised ) completely when dissolved in water is called strong base. For example, sodium hydroxide (NaOH), potassium hydroxide (KOH).

In an aqueous basic solution, total solute substance is in the form OH^– ions because of its complete ionisation, i.e., the total solute substance is ionised in the form of OH^–.
For example,

Thus, in 1M NaOH aqueous solution, the concentrations of Na⁺ and OH^– are 1M.

Thus, in 1 M Mg(OH)₂ aqueous solution, the concentrations of Mg²⁺ and OH^– are 1M and 2M respectively. Thus, by ionisation of lmole Mg(OH)₂, 2 moles OH^– are formed.

Weak base:
The base when dissolved in water is ionised or dissociated partially or incompletely is known as weak base. For example, ammonium hydroxide (NH₄OH).

In the aqueous solution of weak base, very less amount (approximately 2 to 3%) of solute is ionised to OH^–. For example, In aqueous solution of 1M NH₄OH, the concentration of OH^– is not 1M but it is very less than 1 M.

(4) Explain the chemical properties of acid.
Answer:
For the chemical reaction of acid, H⁺ s or H₃O⁺ present in its aqueous solution is responsible.

(i) Reaction of acid with metal : By the reaction of acid with metal, a salt of corresponding metal and dihydrogen gas are produced.
For example,

In short,
Acid + Metal → Salt of metal + Dihydrogen gas

(ii) Reaction of acid with base : Salt and water are formed by the reaction of acid with base. This reaction is called neutralisation reaction.

In short, Acid + Base → Salt + Water

(iii) Reaction of acid with metal oxide: The reaction of acid with metal oxide forms salt and water. For example,

In short, Acid + Metal oxide → Salt + Water

(iv) Reaction of acid with metal carbonate or metal hydrogencarbonate: Most of the acids produce salt, water and carbon dioxide by reaction with metal carbonate or metal hydrogencarbonate.
For example,

In short, Acid + Metal carbonate / Metal hydrogencarbonate → Salt + Water + CO₂(g)

(5) Explain chemical properties of base.
Answer:
For a chemical reaction of base, OH^– present in its aqueous solution is responsible.

(i) Reaction of base with acid : Reaction of a base with an acid forms salt and water. This reaction is called neutralisation.
For example,

• NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
• 2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)

In short, Base + Acid → Salt + Water

(ii) Reaction of base with non-metal oxide : Reaction of base with oxides of non-metal forms salt and water.
For example,

In short,
Base + Oxide of non-metal → Salt + Water

(iii) Reaction of a base with some metals : By the reaction of a strong base like sodium hydroxide and potassium hydroxide with certain amphoteric metals (Zn, Al) gives salt and hydrogen gas.
For example,

(6) “The aqueous solution of the salt produced by neutralisation of weak acid and strong base possesses basic nature, while aqueous solution of salt produced by neutralisation of weak base and strong acid possesses acidic nature.” Explain.
Answer:

Hence, salt Na₂CO₃ is formed by the reaction of strong base and weak acid and its aqueous solution contains higher concentration of OH^– ions than the concentration of H⁺ ions. Therefore, solution possesses basic nature.

Here, salt NH₄Cl is formed by the reaction of strong acid and weak base, and its aqueous solution contains higher concentration of H⁺(aq) ions than the concentration of OH^–(aq) ions. Therefore solution possesses acidic nature.

(7) What is meant chemically by baking soda and baking powder? What happens if baking soda instead of baking powder is added, in making cake?
Answer:
Sodium hydrogencarbonate (NaHCO₃) is a baking soda, while baking powder is a mixture of NaHCO₃ and tartaric acid. For making cake, if only baking soda is used, then, on heating it produces sodium carbonate and the cake will taste bitter. To avoid this, appropriate amount of tartaric acid is added. Baking soda forms sodium salt of an acid which does not affect the taste of cake.

(8) Compare the acidity of two different acidic solutions A and B having different pH values.
Answer:
The comparison of two different acidic solutions A and B is shown in the table below:

(Thus, it can be said from the above table that if the difference in pH of two acidic aqueous solutions is x, then solution having less pH will possess H₃O⁺ concentration 10^x times or antilog x times more acidic.)

Question 2.
Give scientific reasons for the following statements:
(1) An aqueous solution of FeCl₃ is acidic.
Answer:
FeCl₃ is a salt; when it is dissolved in water, it forms strong acid (HCl) and weak base (Fe(OH)₃). Hence, concentration of H⁺ (aq) s ions is more than the concentration of OH^– ions? in the solution. Therefore solution shows acidic behaviour. As a result, an aqueous solution of FeCl₃ is acidic in nature.

(2) An aqueous solution of CH₃COONa is basic.
Answer:
CH₃COONa is a salt, which produces l strong base (NaOH) and weak acid (CH₃COOH) in water. Hence, concentration of OH^–(aq) is more in comparison with the concentration of H⁺(aq) s in the solution and hence solution shows basic behaviour. As a result, an aqueous solution of CH₃COONa is basic in nature.

(3) An aqueous solution of NaCl is neutral.
Answer:
NaCl is a salt and it produces strong acid (HCl) and strong base (NaOH) in water. Hence, concentration of H⁺(aq) ions and concentration of OH^–(aq) ions are equal in the solution and solution shows neutral behaviour. Therefore an aqueous solution of NaCl is neutral.

(4) Curd and sour substances should not be kept in brass and copper vessels.
Answer:
Curd and sour substances are acidic in nature, which react with brass and copper vessels and form toxic substances; which are harmful to the human body. Therefore, curd and other sour substances should not be kept in brass and copper vessels

(5) Distilled water does not conduct electricity, while rain water does it. Explain.
Answer:
Distilled water is a pure water and it does not contain ions while rain water contains impurities like acid, which provides ions and they help in carrying electricity.

Question 3.
Solve the following numericals :
(1) Calculate the concentration of OH^– ion in an aqueous solution having pH value 8.
Solution:
pH + pOH = 14
∴ pOH = 14 – pH = 14 – 8 = 6
Now, pOH = – log₁₀ [OH^–]
∴ log₁₀[OH^–] = – pOH
∴ [OH^–] = Antilog – pOH
= Antilog – 6.0
= 1 x 10^-6M

(2) Calculate the molarity of an aqueous solution prepared by dissolving 2 mol of HC1 in water to form 500 mL solution.
Solution:
Molarity (M) = \(\frac { mole }{ litre }\)
= \(\frac { 2 }{ 0.5 }\)
= 4 M

(3) Calculate the pH of 0.01 M HCl solution.
Solution:
HCl is a strong acid, hence it ionises completely.

∴ [H⁺(aq)] = 0.01 M = 1 x 10^-2M
Now, pH = – log₁₀ [H⁺(aq)]
pH = – log₁₀ [1 x 10^-2]
= – [2.0000]
∴ pH = 2

(4) 50 mL KOH solution neutralise completely 5 mL HNOa solution. What volume of HNOs is required for complete neutralisation of 15 mL of KOH solution?
Solution:
50 mL KOH solution neutralise 5 mL HNO₃ solution
∴ 15mLKOH solution will neutralise,
\(\frac { 15×5 }{ 0.5 }\) = 1.5 mL HNO₃ solution.

(5) How many times would be an aqueous solution of pH = 2 be more acidic than aqueous solution of pH = 4?
Solution:
For solution having pH = 4,
PH = – log₁₀[H₃O⁺ ]
∴ – log₁₀[H₃O⁺] = 4
∴ log₁₀[H₃O⁺] = – 4
∴ [H₃O⁺] = 10^-4M
Similarly, [H₃O⁺] for solution having pH = 2 = 10^-2M.
Now,

Thus, the concentration of [H₃O⁺] in concentrated solution of pH = 2 will be 100 times more than that of pH = 4, i.e., it will be 100 times more acidic.

(6) How many times would be concentrated aqueous solution having pH = 11.9 be more basic as compared to aqueous solution having pH = 8?
Solution:
pOH of basic aqueous solution having pH = 8 will be 14-8 = 6.
∴ [OH^–] = 1 x 10^-6M
Similarly, pOH of basic aqueous solution having pH 11.9 will be 14 – 11.9 = 2.1.
Now, pOH = – log₁₀[OH^–]
2.1 = – log₁₀ [OH^–]
– (2.1) = log₁₀[OH^–]
∴ log₁₀[OH^–] = \(\overline{3}\).9
∴ [OH^–] = antilog \(\overline{3}\).9000
= 7.943 x 10^-3 M
Now,

Thus, the solution having pH 11.9 will have 7943 times more basicity than solution having pH 8.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) What is the difference of water molecules between gypsum and plaster of paris?
Answer:
There are two molecules of water in gypsum. Plaster of paris contain \(\frac { 1 }{ 2 }\) (half) molecule of water. Thus, the difference of water molecules = 2 – \(\frac { 1 }{ 2 }\) = 1.5 molecules.

(2) What is soda lime? State the importance of lime in it.
Answer:
Mixture of caustic soda (NaOH) and lime (CaO) is called soda lime. In soda lime, lime is useful to absorb moisture during reaction.

(3) How does the soda-acid fire-extinguisher extinguish the fire?
Answer:
Soda-acid fire-extinguisher extinguishes the fire by stopping the contact of air.

(4) Write a balanced chemical equation for the reaction that occurs during chlor-alkali reaction.
Answer:
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

(5) Arrange the water, hydrochloric acid and acetic acid in descending order of their acidity.
Answer:
Descending order of acidity:
Hydrochloric acid > Acetic acid > Water

(6) What is meant by dilution?
Answer:
Mixing of an acid or base with water results in decrease in concentration of acid or base or ions per unit volume. This process is called dilution.

(7) Write the name and molecular formula of product obtained by the reaction of zinc with sodium hydroxide.
Answer:
Sodium zincate – Na₂ZnO₂.

(8) What are properties of the oxides of metal and non-metal usually?
Answer:
The oxides of metallic elements are basic, while oxides of non-metals are acidic.

(9) Which acid is formed when milk turns into curd?
Answer:
During the formation of curd from milk, lactic acid is formed.

(10) Which ions are responsible for acidic and basic properties of the solution?
Answer:
H⁺(aq) ions in solution are responsible for acidic properties and OH^–aq) ions in solution S are responsible for basic properties of the solution.

(11) Does the solution of urea conduct electricity? Why?
Answer:
The solution of urea does not conduct electricity, because urea does not possess ions in solution.

(12) Give two examples of strong alkali (base).
Answer:
NaOH and KOH

(13) State the nature of saliva before and after the meal.
Answer:
The nature of saliva before meal is basic s but after meal it becomes acidic.

(14) What is an acid rain?
Answer:
When the pH of rain water is less than 5.6, it is called an acid rain.

(15) What pH range is required to a soil for healthy growth and development of plants?
Answer:
The soil with pH in the range of 6.5 to 7.3 is appropriate for healthy growth and development of plants.

(16) Why does red ant bite (sting) cause pain and irritation?
Answer:
Red ant bite (sting) injects formic acid s in our body, which causes pain and irritation.

(17) Two solutions have pH value of 5 and 9 respectively. Which solution will be more basic in nature? Why?
Answer:
Solution with pH value 9 will be more basic, because the concentration of H⁺ ions in it is less or the concentration of OH^– ions is more.

Question 2.
Define :
(1) Olfactory indicator
Answer:
Substances, which changes their odour s (smell) in presence of an acid or a base are known as olfactory indicators.

(2) Neutralisation reaction
Answer:
The reaction between an acid and a base to give a salt and water is known as a neutralisation reaction.

(3) Dilution process
Answer:
The process in which mixing an acid or base with water results in decrease in the concentration of ions (H₃O⁺ or OH^–) per unit volume is called dilution process.

(4) pH scale
Answer:
A scale for measuring the concentration of hydrogen ions in a solution is called pH scale.

(5) Family of salts
Answer:
Salts having the same positive or negative radicals are called family of salts.

(6) Rock salt
Answer:
The solid deposits of salts dissolved in sea water in the form of large crystals which turn brown due to impurities are known as rock salt.

Question 3.
Fill in the blanks :

1. Neutralisation reaction occur between acid and base forms ………………. and ……………….
2. The pH value of acidic solution is lower than ……………….
3. An aqueous solution having pH value 4 is less ………………. than an aqueous solution having pH 2.
4. Milk of magnesia is used as ……………….
5. Molecular formula of gypsum is ……………….
6. Butter milk possesses ………………. character.
7. The farmers add ………………. to the acidic soil to neutralise it.
8. Litmus is a ………………. indicator.
9. MgO is a ………………. oxide.
10. Generally, ………………. paper is used with universal indicator to measure the pH value of the solution.
11. The atmosphere of ………………. planet is made-up of thick white and yellowish clouds of sulphuric acid.
12. The stinging hair of nettle leaves injects ………………. into the skin causing burning pain.

Answer:

1. salt, water
2. 7
3. acidic
4. an antacid
5. CaSO₄.2H₂O
6. acidic
7. lime (CaO)
8. natural
9. basic
10. impregnated
11. venus
12. methanoic acid

Question 4.
State whether the following statements true or false:

1. The molecular formula of sodium zincate is Na₂Zn(OH)₄.
2. An aqueous solution of washing soda possesses acidic nature.
3. Bacteria present in the mouth produce base by degradation of food particles left in the mouth after eating.
4. The pH value of the blood is greater than 7.
5. Orange contains citric acid.
6. OH^–(aq) ions in the solution are responsible for chemical reaction of base.
7. Phenolphthalein is a natural indicator.
8. Ca(HCO₃)₂ is soluble in water.
9. Cl₂O₇ is a basic oxide.
10. Solutions of glucose and alcohol do not conduct electricity.
11. Hydrogen ions are formed in HCl in presence of water.
12. Dissolution of acids and bases in water is an endothermic reaction.
13. The p in pH stands for ‘potenz’ in German.
14. Salts of strong acids and strong bases are neutral with pH value 7.
15. Baking soda is used for faster cooking.

Answer:

1. False
2. False
3. False
4. True
5. True
6. True
7. False
8. True
9. False
10. True
11. True
12. False
13. True
14. True
15. True

Question 5.
Choose the correct option from those given below each question:
1. When does tooth decay occur?
A. When the pH of inner part of mouth is lower than 5.5.
B. When the pH of inner part of mouth is higher than 5.5.
C. When the pH of inner part of mouth is 5.5.
D. When the pH of inner part of mouth is 7.0.
Answer:
A. When the pH of inner part of mouth is lower than 5.5.

2. Which of the following solution is most basic?
A. pH = 8.2
B. pH = 9.3
C. pH = 11.5
D. pH = 10.6
Answer:
C. pH = 11.5
Hint: For acidic solution, pH < 7 and pOH > 7.
For basic solution, pH > 7 and pOH < 7.

3. What is the pH value of an aqueous solution of NH₄CI?
A. pH = 7
B. pH > 7
C. pH < 7
D. pH = 0
Answer:
C. pH < 7
Hint: An aqueous solution of NH₄Cl is acidic. Hence, its pH < 7.

4. Which of the following is a strong acid?
A. Acetic acid
B. Citrus acid
C. Oxalic acid
D. Nitric acid
Answer:
D. Nitric acid

5. The pH values of aqueous solutions A, B, C and D are 1.9, 2.5, 2.1 and 3.0 respectively; then what will be the correct order of their acidic strength?
A. A < C < B < D
B. D < C < B < A
C. D < B < C < A
D. D > C > B > A
Answer:
D. D > C > B > A
Hint : Aqueous solution with lower pH value is more acidic.

6. Which of the following solutions is neutral?
A. Juice of citrus fruits
B. Lemon juice
C. Solution of washing soda
D. An aqueous solution of salt – NaCl
Answer:
D. An aqueous solution of salt – NaCl

7. What will be the value of pH, if blue litmus paper turns red in aqueous solution?
A. Between 0 to 7
B. Between 7 to 14
C. 14
D. 0
Answer:
A. Between 0 to 7

8. The outer layer of the teeth is made-up of…
A. Calcium phosphate
B. Calcium nitrate
C. Potassium phosphate
D. Sodium phosphate
Answer:
A. Calcium phosphate

9. The pH of aqueous solution(s) of which salt is 7?
A. Na₂CO₃
B. CH₃COONa
C. NH₄Cl
D. KNO₃
Answer:
D. KNO₃
Hint: Since KNOs is a salt of strong acid and strong base.

10. CaCl₂ + X → CaSO₄ + 2NaCl, what is X?
A. Na₂SO₄
B. CaSO₃
C. Na₂SO₃
D. CaSO₂
Answer:
A. Na₂SO₄

11. Which of the following pair is not appropriate?
A. Citrus fruits – citric acid
B. Curd – lactic acid
C. Sting of red ant – methanoic acid
D. Tomatoes – tartaric acid
Answer:
Tomatoes – tartaric acid
Hint: Tomatoes contains citric acid.

12. Which of the following aqueous solutions contains higher proportion of OH^– ions?
A. NaCl
B. Na₂SO₄
C. CH₃COONa
D. Equal in all
Answer:
CH₃COONa
Hint:Due to weak acid (CH₃COOH) and strong base (NaOH) forms in an aqueous solution of CH₃COONa.

13. Which of the following substance does not produce carbon dioxide, when it is treated with dilute acid?
A. Marble
B. Lime stone
C. Lime
D. Baking soda
Answer:
C. Lime

Question 6.
Match the following:
(1)

Column I	Column II
1. Strong acid	p. NaOH
2. Strong base	q. HNO3
3. Salt	r. NaCl
4. Amphoteric	s. H2O

Answer:
(1 – q), (2 – p), (3 – r), (4 – s).

(2)

Column I	Column II
1. Sodium carbonate	p. NaNO3
2. Sodium bicarbonate	q NaNO2
3. Sodium nitrate	r. Na2CO3
4. Sodium nitrite	s. NaHCO3

Answer:
(1 – r), (2 – s), (3 – p), (4 – q).

(3)

Column I	Column II
1. Melittin	p. Alkaline soil
2. Calcium phosphate	q. Baking soda
3. The value of pH is more than 7.3	r. A polypeptide containing 26 amino acids
4. Antacid	s. Outer layer of teeth

Answer:
(1 – r), (2 – s), (3 – p), (4 – q).

Question 7.
Answer the following questions in one word:

1. State the chemical formula of caustic potash.
2. Write the composition of aqua regia.
3. Write the molecular formula of soda ash.
4. Which substance reacts with chlorine to form bleaching powder?
5. Which acid is present in orange?
6. State the pH value of human blood.
7. At normal temperature, substance kept in atmosphere loses the water of crystallisation. By which name the substance is known?
8. Name the reaction in which base is added in acid.
9. What is called a base which is soluble in water?
10. State the chemical name of salt which is useful and important raw material of food.

Answer:

1. KOH
2. 1 part concentrated HNO₃ + 3 parts concentrated HCl
3. Na₂CO₃
4. Ca(OH)₂
5. Ascorbic acid
6. 7.36 to 7.42 (about 7.4)
7. Hydrated salt
8. Neutralisation reaction
9. Alkali
10. Sodium chloride

Question 8.
Complete the following reactions :

1. CaO(s) + H₂O(l) →
2. 2HNO₃(aq) + Ca(OH)₂(aq) →
3. 2HNO₃(aq) + CaO(s) →
4. HCl(aq) + NaHCO₃(aq) →
5. 2NaOH(aq) + H₂CO₃(aq) →
6. CH₃COOH(aq) + NaOH(aq) →
7. HNO₃(aq) + KOH(aq) →
8. BaCl₂(aq) + H₂SO₄(aq) →

Answer:

1. Ca(OH)₂(aq)
2. Ca(NOs)₂(aq) + 2H₂O(l)
3. Ca(NO₃)₂(aq) + H₂O(l)
4. NaCl(aq) + H₂O(l) + CO₂(g)
5. Na₂CO₃(aq) + 2H₂O(l)
6. CH₃COONa(aq) + H₂O(l)
7. KNO₃(aq) + H₂O(l)
8. BaSO₄(s) + 2HCl(aq)

Value Based Questions With Answers

Question 1.
Ramesh’s father is a farmer and he is disappointed as he cannot grow any crop on his farmland, because the land has become highly basic in nature. It happens due to paper industry nearby dumping its waste water into canals; that supply water to the farmland.

Ramesh and other youth in the village told the owner of the paper industry about the pollution caused by the paper industry.

• What type of land is suitable for the growth of crops?
• Use of fertilizers also causes change in the pH of soil. How do farmers overcome this problem?
• What values of Ramesh are reflected in the above case?

Answer:

• The farmland should neither be acidic nor basic, it should be neutral for the growth of crops.
• If the soil is acidic then compounds like lime should be added to the soil to neutralise it. If the soil is basic, then the acidic salt is added to neutralise it.
• Ramesh showed the values of aware citizen and group work.

Question 2.
Rita’s mother had severe pain when a honeybee stung her on hand. Rita’s grandmother tried to relieve the pain by rubbing a iron metal on the affected area. But, Rita took the baking powder from the kitchen and applied on her mother’s hand which quickly relieved the pain.

• Why do honeybee stinging causes pain?
• How does baking powder relieve the beesting pain?
• What values of Rita is seen in the above act?

Answer:

• Honeybee sting releases an acid that is injected in body and causes pain.
• Baking powder is basic in nature, it neutralised the acid released in the body by bee-sting and relieves the pain.
• Rita showed the values of prompt, proactive and responsible behaviour.

Question 3.
Bharat’s friend Jaydeep is very fond of coffee. He drinks two cups of coffee everyday in the school and complain the problem of stomach pain very often. Bharat advised him not to drink coffee in the morning.

• What was the cause of stomach pain?
• What would be the pH of stomach juices after consuming coffee?
• What values of Bharat are reflected in the above act?

Answer:

• Two cups of coffee (excess) causes acidity which results into stomach pain.
• The pH of stomach after consuming coffee is 5.
• Bharat showed the values of concern, caring and awareness.

Question 4.
Ashish noticed that his friend always brought sweets in his lunch-box, due to which his tooth decayed. Ashish suggested his friend to eat less sweets in school hours to avoid tooth? decay.

• Why do tooth decay on eating sweets?
• What is the pH of mouth after eating sweets?
• What value of Ashish is seen in the above act?

Answer:

• On eating sweets, the particles of sweets get stuck in the teeth. On the sweet particles bacteria grows and produces acid. This acid causes tooth decay.
• The pH of the mouth after eating sweets lowers and remains in the range of 2 to 6.
• Ashish showed sincerity, responsibility and? care for others.

Practical Skill Based Questions With Answers

Question 1.
In the laboratory, a test tube rack is placed with test tubes containing some acids in it. How will you classify these acids?
Answer:
To classify the acids, we should test their pH and group them as strong or weak acids.

Question 2.
What is the pH of water? Does it change on heating the water? Why?
Answer:
Generally, the pH of pure water is 7 at 25 °C temperature. On heating water, the pH value will change, it decreases, because on heating water molecules will ionise more and as a result concentration of H⁺ ions and OH^– ions increase, which results in decreasing pH. But water remains neutral.

Question 3.
A test was conducted in the laboratory to find the pH of different cold drinks. Draw the observation table to collect the data for this experiment and predict the result for the same.
Answer:
Observation table of collected data :

Sr. No.	pH of drink-1	pH of drink-2	pH of drink-3	pH of drink-4
1.
2.
3.

Prediction:
The pH of the dark cold drink will be higher, as it contains more caffeine which is more acidic and the cold drink which releases more bubbles on opening is highly acidic and will have low pH as the carbonated water is present in it.

Memory Map

Jharkhand Board JAC Class 10 Science Important Questions Chapter 10 प्रकाश-परावर्तन तथा अपवर्तन Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 10 प्रकाश-परावर्तन तथा अपवर्तन

अतिलयु उत्तरीय प्रश्न

प्रश्न 1.
गोलीय दर्पण की वक्रता-त्रिज्या तथा फोकस-दूरी में सम्बन्ध लिखिए।
उत्तर:
वक्रता त्रिज्या (R) = 2 x फोकस दूरी (f)।

प्रश्न 2.
अपने चेहरे का सीधा एवं बड़ा प्रतिबिम्ब देखने के लिए किस प्रकार के दर्पण का प्रयोग कीजिएगा?
उत्तर:
अवतल दर्पण का।

प्रश्न 3.
किस प्रकार के दर्पणों से केवल आभासी प्रतिबिम्ब बनते हैं?
उत्तर:
समतल दर्पणों तथा उत्तल दर्पणों से केवल आभासी प्रतिबिम्ब बनते हैं।

प्रश्न 4.
एक दर्पण, वस्तु के सापेक्ष सीधा व आकार में छोटा प्रतिबिम्ब बनाता है। यह किस प्रकार का दर्पण है? प्रतिबिम्ब वास्तविक है अथवा आभासी।
उत्तर:
उत्तल दर्पण (आभासी)।

प्रश्न 5.
किस प्रकार के दर्पण में रेखीय आवर्धन 1 से अधिक प्राप्त हो सकता है?
उत्तर:
अवतल दर्पण में।

प्रश्न 6.
प्रतिबिम्ब को पर्दे पर लेने के लिए किस प्रकार के दर्पण का प्रयोग कीजिएगा?
उत्तर:
अवतल दर्पण का।

प्रश्न 7.
समतल दर्पण की फोकस-दूरी कितनी होगी?
उत्तर:
अनन्त (∞) क्योंकि समान्तर आपतित किरणें परावर्तन के पश्चात् समान्तर ही रहती हैं-अर्थात् अनन्त पर ही मिलती हैं।

प्रश्न 8.
दर्पण का मुख्य अक्ष क्या होता है?
उत्तर:
दर्पण के ध्रुव एवं वक्रता-केन्द्र से गुजरने वाली रेखा को मुख्य अक्ष कहते हैं।

प्रश्न 9.
समान्तर किरणों को उदपसरित करने के लिए किस प्रकार का दर्पण उपयोगी होता है?
उत्तर:
उत्तल दर्पण।

प्रश्न 10. अवतल दर्पण के लिए u, v तथा f में सम्बन्ध लिखिए।
उत्तर:
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

प्रश्न 11.
समतल दर्पण में तथा उत्तल दर्पण में बने प्रतिबिंबों में क्या समानताएँ होती हैं?
उत्तर:

• दोनों में प्रतिबिम्ब दर्पण के पीछे बनते हैं।
• दोनों में बने प्रतिबिम्ब आभासी तथा सीधे होते हैं।

प्रश्न 12.
संलग्न चित्रों में आपतित एवं परावर्तित किरणों का मार्ग देखकर बताइए कि चित्र में प्रदर्शित बॉक्स में किस प्रकार का दर्पण स्थित हो सकता है?

उत्तर:

• उत्तल दर्पण (समान्तर आपतित किरणें परावर्तन के बाद अपसरित होती हैं)।
• समतल दर्पण (समान्तर आपतित किरणें समान्तर बनी रहती हैं)।
• अवतल दर्पण (समान्तर आपतित किरणें परावर्तन के बाद अपसरित होती हैं)।

प्रश्न 13.
परावर्तन किसे कहते हैं?
उत्तर:
जब प्रकाश किसी चिकने व चमकदार पृष्ठ पर पड़ता है तो इसका अधिकांश भाग विभिन्न दिशाओं में लौट आता है। इस घटना को प्रकाश का परावर्तन कहते हैं।

प्रश्न 14.
दर्पण कितने प्रकार के होते हैं?
उत्तर:
दर्पण मुख्यतः दो प्रकार के होते हैं-
(1) समतल दर्पण
(2) गोलीय दर्पण –

• अवतल दर्पण
• उत्तल दर्पण।

प्रश्न 15.
अवतल तथा उत्तल दर्पण में अन्तर समझाइए।
उत्तर:
अवतल दर्पण में बाह्य उभरे हुए गोलीय पृष्ठ पर पॉलिश कर दी जाती है जिससे प्रकाश का परावर्तन आन्तरिक पृष्ठ (अवतल) पर होता है। जबकि उत्तल दर्पण में भैंसे हुए गोलीय पृष्ठ पर पॉलिश कर दी जाती है जिससे प्रकाश का परावर्तन बाह्य पृष्ठ (उत्तल) पर होता है।

प्रश्न 16.
अवतल दर्पण के फोकस की परिभाषा लिखिए।
उत्तर:
अवतल दर्पण के मुख्य अक्ष के समान्तर आने वाली सभी प्रकाश किरणें दर्पण से परावर्तन के पश्चात् मुख्य अक्ष पर स्थित जिस बिन्दु पर एकत्रित होती हैं वह बिन्दु अवतल दर्पण का मुख्य फोकस कहलाता है।

प्रश्न 17.
लम्बन से क्या तात्पर्य है?
उत्तर:
लम्बन (विस्थापनाभास) – यदि दो वस्तुओं ( या प्रतिबिम्बों) को एक सीध में विभिन्न स्थितियों में रखकर देखें तो वे सम्पाती प्रतीत होती हैं। लेकिन आँख को दायें या बायें हटाने पर दोनों वस्तुएँ एक दूसरे से अलग होती हुई प्रतीत होती हैं। आँख से दूर वाली वस्तु (या प्रतिबिम्ब) पास वाली वस्तु की अपेक्षा उसी दिशा हटती दिखाई देती है जिस दिशा में आँख हटाते हैं। इस प्रकार आँख को इधर-उधर हटाने पर दो वस्तुओं के बीच सापेक्ष विस्थापन को लम्बन (या विस्थाना भास) कहते हैं।

प्रश्न 18.
गोलीय दर्पण के लिए 40 व में सम्बन्ध लिखिए।
अथवा
दर्पण सूत्र लिखिए एवं प्रत्येक संकेत का अर्थ लिखिए।
उत्तर:
गोलीय दर्पण के लिए u, v व f में निम्न सम्बन्ध को दर्पण सूत्र कहते हैं-
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
जहाँ
u = दर्पण से वस्तु की दूरी
v = दर्पण से प्रतिबिम्ब की दूरी
f = दर्पण की फोकस दूरी

प्रश्न 19.
गोलीय दर्पण कितने प्रकार के होते हैं? नाम लिखिए।
उत्तर:
गोलीय दर्पण दो प्रकार के होते हैं-

• अवतल दर्पण
• उत्तल दर्पण।

प्रश्न 20.
एक समतल दर्पण की फोकस दूरी तथा उसकी वक्रता त्रिज्या में सम्बन्ध लिखिए।
उत्तर:
एक समतल दर्पण की फोकस दूरी तथा उसकी वक्रता त्रिज्या दोनों ही अनन्त होती हैं।

प्रश्न 21.
एक अवतल दर्पण को यदि पानी में रख दिया जाये तो उसकी फोकस दूरी में क्या परिवर्तन होगा?
उत्तर:
कोई परिवर्तन नहीं होगा।

प्रश्न 22.
यदि एक वस्तु का प्रतिबिम्ब सदैव छोटा व आभासी बनता है तो दर्पण कैसा होगा?
उत्तर:
यदि वस्तु का प्रतिबिम्ब सदैव छोटा व आभासी बनता है तो दर्पण उत्तल होगा।

प्रश्न 23.
एक दर्पण में हम अपने चेहरे का सीधा व बड़ा प्रतिबिम्ब देख रहे हैं तो बताइए कि दर्पण कैसा होगा?
उत्तर:
अवतल दर्पण।

प्रश्न 24.
एक अवतल दर्पण के वक्रता केन्द्र पर एक वस्तु रखी है। इस वस्तु के प्रतिबिम्ब का रेखीय आवर्धन कितना होगा?
उत्तर:
जब वस्तु अवतल दर्पण के वक्रता केन्द्र पर होती है तो प्रतिबिम्ब का आकार वस्तु के आकार के बराबर, वास्तविक तथा उलटा होता है अतः प्रतिबिम्ब का रेखीय आवर्धन 1 होगा।

प्रश्न 25.
किसी अवतल दर्पण के ध्रुव तथा फोकस मध्य स्थित वस्तु के सापेक्ष उसके प्रतिबिम्ब की प्रकृति कैसी होगी?
उत्तर:
किसी अवतल दर्पण के ध्रुव तथा फोकस के मध्य स्थित वस्तु के सापेक्ष उसका प्रतिबिम्ब आभासी, सीधा व बड़ा होगा।

प्रश्न 26.
एक अवतल दर्पण में हम अपना वास्तविक तथा अपने आकार का प्रतिबिम्ब देख रहे हैं तो बताइए कि हमारी स्थिति क्या है?
उत्तर:
यदि हम अवतल दर्पण में अपना वास्तविक तथा अपने आकार का प्रतिबिम्ब देख रहे हैं तो हम अवतल दर्पण के वक्रता केन्द्र पर स्थित हैं।

प्रश्न 27.
अच्छी शेव के लिए कौन-सा दर्पण प्रयुक्त करना ठीक रहता है।
उत्तर:
अच्छी शेव करने के लिए अधिक फोकस दूरी का अवतल दर्पण ठीक रहता है।

प्रश्न 28.
दाढ़ी बनाने के लिए अवतल दर्पण का उपयोग क्यों किया जाता है?
उत्तर:
जब किसी वस्तु को अवतल दर्पण के सामने उसके ध्रुव एवं फोकस के बीच रखा जाता है तो उसका सीधा, आभासी एवं वस्तु से बड़ा प्रतिबिम्ब बनता है जो दाढ़ी बनाने के लिए उपयुक्त होता है।

प्रश्न 29.
उत्तल दर्पण के कोई दो उपयोग लिखिए।
उत्तर:
उत्तल दर्पण के दो उपयोग निम्नलिखित हैं-

• मोटर वाहनों में पीछे का ट्रैफिक देखने में।
• सड़क या चौराहे पर लगी बत्तियों में परावर्तक के रूप में।

प्रश्न 30.
अवतल दर्पण के दो उपयोग लिखिए।
उत्तर:
अवतल दर्पण के उपयोग निम्नलिखित हैं-

• टार्च, कार, मोटर, रेलवे इंजन के हैडलाइट में परावर्तक के रूप में।
• चिकित्सा में डॉक्टरों द्वारा कान, नाक, गला की जाँच करने में।

प्रश्न 31.
रैखिक आवर्धन से क्या तात्पर्य है? गोलीय दर्पण के लिए आवर्धन सूत्र लिखिए।
उत्तर:
रैखिक आवर्धन- यदि वस्तु एवं प्रतिबिम्ब की लम्बाई मुख्य अक्ष के लम्बवत् नापी जाये तो प्रतिबिम्ब की लम्बाई एवं वस्तु की लम्बाई के अनुपात को रेखीय आवर्धन कहते हैं।

प्रश्न 32.
प्रकाश क्या है? इसके प्राकृतिक स्रोत का नाम लिखिए।
उत्तर:
प्रकाश ऊर्जा का वह रूप है जिसकी सहायता से नेत्र द्वारा देखने की अनुभूति होती है। प्रकाश का प्राकृतिक स्रोत सूर्य है।

प्रश्न 33.
किसी अवतल दर्पण की फोकस दूरी = 20 सेमी है, उसकी वक्रता त्रिज्या ज्ञात कीजिए।
उत्तर:
दिया है अवतल दर्पण की-
फोकस दूरी (f) = – 20 सेमी
वक्रता त्रिज्या (R) = ?
∵ R = 2 x f
∴ R = 2 x (- 20) = – 40 सेमी
अतः अवतल दर्पण की वक्रता त्रिज्या 40 सेमी होगी।

प्रश्न 34.
यदि कोई अवतल दर्पण एक ऐसे खोखले गोले का भाग है जिसकी त्रिज्या 40 सेमी है, तो अवतल दर्पण की फोकस दूरी क्या होगी?
उत्तर:
दिया है अवतल दर्पण की-
वक्रता त्रिज्या (R) = 40 सेमी.
फोकस दूरी (f) =?
∵ R = 2 x f
⇒ f = \(\frac{R}{2}=\frac{-40}{2}\) = – 20 सेमी
अतः अवतल दर्पण की फोकस दूरी 20 सेमी है।

प्रश्न 35.
प्रकाश की एक किरण वायु से आती हुई स्वच्छ जल के स्थिर तल पर आपतित होती है। यह किरण के तल पर परावर्तित होगी या अपवर्तित या दोनों।
उत्तर:
जल के तल से प्रकाश का कुछ अंश परावर्तित तथा अधिकांश अपवर्तित होगा।

प्रश्न 36.
एक माध्यम (1) में किसी प्रकाश किरण का आपतन कोण 35° तथा दूसरे माध्यम (2) में उसी किरण का अपवर्तन कोण 45° है। दोनों में से किस माध्यम का निरपेक्ष अपवर्तनांक अधिक है?
उत्तर:
माध्यम (1) का।

प्रश्न 37.
किसी माध्यम का निरपेक्ष अपवर्तनांक, वायु के सापेक्ष उस माध्यम के अपवर्तनांक के लगभग बराबर माना जाता है। वायु का निरपेक्ष अपवर्तनांक लगभग कितना होगा?
उत्तर:
वायु का निरपेक्ष अपवर्तनांक लगभग 1 होगा।

प्रश्न 38.
दो माध्यमों A तथा B के अपवर्तनांक n_A तथा n_B हैं। इनमें से किस माध्यम से किस माध्यम में जाने पर प्रकाश का पूर्ण-आंतरिक परावर्तन संभव है यदि N_A > n_B?
उत्तर:
माध्यम A से माध्यम B की ओर।

प्रश्न 39.
प्रकाश के ‘स्पेक्ट्रम’ से क्या तात्पर्य है?
उत्तर:
किसी मिश्रित प्रकाश के संयोजी वर्णों (रंगों) को उनके तरंगदैघ्यों के आरोही अथवा अवरोही क्रम को प्रकाश का स्पेक्ट्रम कहते हैं।

प्रश्न 40.
सौर प्रकाश के स्पेक्ट्रम के रंगों को अपवर्तनांक बढ़ते हुए क्रम में लिखिए।
उत्तर:
लाल (Red), नारंगी (Orange), पीला (Yel-low), हरा (Green), आसमानी (Blue), नीला (Indigo), बैंगनी (Violet)।

प्रश्न 41.
किसी प्रिज्म के न्यूनतम विचलन कोण, प्रिज्म कोण एवं प्रिज्म के पदार्थ के अपवर्तनों में सम्बन्ध लिखिए।
उत्तर:
यदि प्रिज्म कोण A, न्यूनतम विचलन कोण δm तथा अपवर्तनांक n हो तो
n = \(\frac{\sin [(A+\delta m) / 2]}{\sin [A / 2]}\)

प्रश्न 42.
श्वेत प्रकाश के किन रंगों के लिए काँच का अपवर्तनांक अधिकतम और न्यूनतम होता है?
उत्तर:
बैंगनी रंग श्वेत प्रकाश के लिए काँच अपवर्तनांक अधिकतम और लाल के लिए न्यूनतम होता है।

प्रश्न 43.
पूर्ण परावर्तक प्रिज्म क्या है?
उत्तर:
एक ऐसा काँच का प्रिज्म जो समकोण समद्विबाहु त्रिभुजाकार होता है, वह पूर्ण परावर्तक प्रिज्म होता है जो पूर्ण आन्तरिक परावर्तन के सिद्धान्त पर कार्य करता है।

प्रश्न 44.
पूर्ण परावर्तन प्रिज्म के दो उपयोग लिखो।
उत्तर:
पूर्ण परावर्तक प्रिज्म का उपयोग प्रकाशिक यन्त्रों में होता है-

• 90° के विचलन के लिए पेरिस्कोप में।
• 180° के विचलन के लिए बायनोकुलर में।

प्रश्न 45.
प्रकाशिक तन्तु क्या है?
उत्तर:
प्रकाशिक तन्तु पूर्ण आंतरिक परावर्तन के सिद्धान्त पर आधारित ऐसी युक्ति है जिसके द्वारा प्रकाशिक संकेत को इसकी तीव्रता में बिना क्षय हुए एक स्थान से दूसरे स्थान तक वक्रीय पथ से स्थानांतरित किया जा सकता है।

प्रश्न 46.
लेंस किसे कहते हैं?
उत्तर:
दो गोलीय तलों में घिरे पारदर्शीय माध्यम को लेंस कहते हैं जिसका एक तल गोलीय व दूसरा तल उत्तल, अवतल अथवा समतल होता है।

प्रश्न 47.
लेंस के प्रकारों के नाम लिखो।
उत्तर:
लेंस मुख्यतः दो प्रकार के होते हैं-

• उत्तल लेंस (अभिसारी लेंस)
• अवतल लेंस (अपसारी लेंस)

प्रश्न 48.
अभिसारी और अपसारी लेंस की विशेषता क्या है?
उत्तर:

• अभिसारी लेंस की विशेषता जब प्रकाश की समान्तर किरण पुंज अभिसारी लेंसों में से होकर गुजरती है तो लेंस से अपवर्तन के पश्चात् उन्हें एक बिन्दु पर केन्द्रित कर देता है।
• अपसारी लेंस की विशेषता – जब प्रकाश की समान्तर किरण पुंज अपसारी लेंसों में से होकर गुजरती है तो लेंस से अपवर्तन के पश्चात् उन्हें एक बिन्दु से अपसरित करता हुआ प्रतीत होती है।

प्रश्न 49.
उत्तल लेंस के उपयोग लिखो।
उत्तर:
उत्तल लेंस के निम्नलिखित उपयोग हैं-

• प्रकाशिक यन्त्रों, जैसे दूरदर्शी, कैमरा आदि।
• दूर दृष्टि दोष दूर करने हेतु चश्में में।

लघु उत्तरीय प्रश्न

प्रश्न 1.
आवश्यक किरण आरेख बनाकर निम्नलिखित की परिभाषा दीजिए-
(i) वक्रता केन्द्र तथा वक्रता त्रिज्या,
(ii) अवतल दर्पण का मुख्य फोकस तथा फोकस दूरी
(iii) उत्तल दर्पण का मुख्य फोकस फोकस दूरी।
उत्तर:
(i) वक्रता केन्द्र तथा वक्रता त्रिज्या वक्रता केन्द्र (Centre of Curvature) – गोलीय दर्पण जिस खोखले गोले का भाग होता है, उसके केन्द्र को वक्रता केन्द्र कहा जाता है। चित्र में इसे C अक्षर से प्रदर्शित किया गया है।

वक्रता त्रिज्या (Radius of Curvature)- गोलीय दर्पण का भाग होता है, उसकी त्रिज्या को गोलीय दर्पण की वक्रता त्रिज्या कहते हैं अतः दर्पण के किसी भी बिन्दु को वक्रता केन्द्र से मिलाने वाली रेखा की लम्बाई को वक्रता त्रिज्या कहते हैं। सामान्यतः यह दूरी वक्रता केन्द्र (C) से ध्रुव (P) तक नापी जाती है अत: राशि (L/R) को दर्पण की वक्रता कहते हैं।

(ii) अवतल दर्पण का मुख्य फोकस तथा फोकस – दूरी (Focal Length and Principal Focus of Concave Mirror) – अवतल मुख्य अक्ष के समान्तर आपतित किरणें परावर्तन के पश्चात् दर्पण के मुख्य अक्ष पर स्थित जिस बिन्दु से होकर जाती हैं- उसे मुख्य फोकस कहते हैं। दर्पण के ध्रुव से मुख्य फोकस की दूरी को फोकस दूरी कहते हैं।

(iii) उत्तल दर्पण का मुख्य फोकल तथा फोकस दूरी (Focal Length and Principal Focus Convex Mirror) – उत्तल दर्पण के मुख्य अक्ष के समान्तर आपतित किरणें परावर्तन के पश्चात् मुख्य अक्ष के जिस बिन्दु अपसरित होती प्रतीत होती हैं। उसे मुख्य फोकस कहते हैं।

प्रश्न 2.
‘प्रतिबिम्ब’ से क्या तात्पर्य है? आभासी तथा वास्तविक प्रतिबिम्ब में क्या अन्तर होता है?
उत्तर:
प्रतिबिम्ब (Image) – दर्पण के तल से परावर्तन के पश्चात् प्रकाश किरणें, जिस बिन्दु पर अभिसरित (converge) होती हैं अथवा जिस बिन्दु से अपसरित (diverge) होती प्रतीत होती हैं, उसे वस्तु का प्रतिबिम्ब कहते हैं [चित्र (क) तथा (ख)]

(i) वास्तविक प्रतिबिम्ब (Real Image) – यदि परावर्तन के पश्चात् किरणें वास्तव में किसी बिन्दु पर अभिसरित होती हैं, , तो इसे वास्तविक प्रतिबिम्ब कहते हैं। चित्र (क) में 1, वस्तु O का वास्तविक प्रतिबिम्ब है। यदि स्थान I पर कोई पर्दा (Screen) रखा जाय तो यह प्रतिबिम्ब एक प्रकाश बिन्दु के रूप में दिखाई देगा। अतः वास्तविक प्रतिबिम्ब को पर्दे पर लिया जा सकता है।

(ii) आभासी प्रतिबिम्ब (Virtual Image)- यदि परावर्तन के पश्चात् किरणें अपसारी (divergent ) हों तो वे किसी बिन्दु पर वास्तव में अभिसरित नहीं होतीं। ऐसी किरणें दर्पण के पीछे स्थित किसी बिन्दु से अपसरित (di- verge) होती प्रतीत होती हैं। परावर्तन के पश्चात् प्रकाश की अपसारी किरणें, जिस बिन्दु से अपसरित होती प्रतीत होती हैं, उसे वस्तु का आभासी

प्रतिबिम्ब कहते हैं। चित्र (ख) में I, वस्तु O का आभासी प्रतिबिम्ब है। आभासी प्रतिबिम्ब को पर्दे पर नहीं देखा जा सकता क्योंकि इस बिन्दु’ प्रकाश वास्तव में नहीं पहुँचता।

प्रश्न 3.
केवल किरण आरेख बनाकर, अवतल दर्पण में प्रतिबिम्ब का बनना प्रदर्शित कीजिए जब
(i) वस्तु मुख्य फोकस पर हो
(ii) प्रतिबिम्ब मुख्य फोकस पर बने।
उत्तर:
(i) वस्तु मुख्य फोकस (F) पर

(ii) प्रतिबिम्ब मुख्य फोकस (F) पर

प्रश्न 4.
अवतल दर्पण आभासी प्रतिबिम्ब बनने का किरण आरेख बनाइए। इसके लिए वस्तु की स्थिति का उल्लेख कीजिए।
उत्तर:
वस्तु की स्थिति दर्पण के मुख्य फोकस (F) तथा ध्रुव (P) के बीच कहीं पर।

प्रश्न 5.
उत्तल दर्पण किस प्रकार का प्रतिबिम्ब बनता है? किरण आरेख बनाकर बताइए।
उत्तर:
प्रतिबिम्ब दर्पण के पीछे, आभासी सीधा तथा वस्तु के आकार से छोटा बनता है। प्रतिबिम्ब की स्थिति ध्रुव तथा मुख्य फोकस के बीच में ही होती है।

प्रश्न 6.
आरेख बनाकर स्पष्ट कीजिए कि उत्तल दर्पण की फोकस – दूरी धनात्मक तथा अवतल दर्पण की फोकस – दूरी ऋणात्मक क्यों मानी जाती है?
उत्तर:
दर्पणों में दूरियों के चिन्ह निर्देशांक ज्यामिति की प्रणाली के अनुसार निर्धारित किये जाते हैं। निर्देशांक ज्यामिति की चिन्ह प्रणाली के अनुसार, दर्पण के ध्रुव को मूल बिन्दु, मुख्य अक्ष को X- अक्ष तथा मुख्य अक्ष के लम्बवत् ध्रुव से गुजरने वाली रेखा को Y- अक्ष मानते हैं। इस प्रणाली के अनुसार-
(1) गोलीय दर्पणों (और लेंसों) पर प्रकाश किरणें सदैव बायीं ओर से डाली जाती हैं।

(2) समस्त दूरियाँ दर्पण के ध्रुव से मुख्य अक्ष के अनुदिश नापी जाती हैं।

(3) दर्पण के ध्रुव से दायीं नापी गयी दूरियाँ धनात्मक चिन्ह (+) के साथ ली जाती हैं या दूरियाँ जो आपतित किरण जाती हैं।

(4) दर्पण के ध्रुव से बायीं ओर नापी गयी दूरियाँ ऋणात्मक चिन्ह (-) के साथ ली जाती हैं या वे दूरियाँ जो आपतित किरण की विपरीत दिशा में नापी जाती हैं, ऋणात्मक चिन्ह (-) के साथ ली जाती हैं।

चित्र (क) तथा (ख) में अवतल दर्पण तथा उसल दर्पण के मुख्य फोक्सों की स्थितियाँ प्रदर्शित हैं। चित्रास स्पष्ट है कि उत्तल दर्पण में फोकस दूरी () धनात्मको तथा अवतल दर्पण में ऋणात्मक (-) होगी।

प्रश्न 7.
केवल प्रतिबिम्ब देखकर कैसे विभेद कीजिएगा?
(i) उत्तल दर्पण एवं अवतल दर्पण में,
(ii) उत्तल दर्पण एवं समतल दर्पण में,
(iii) समतल दर्पण एवं अवतल दर्पण में।
उत्तर:
(i) किसी दूरस्थ वस्तु का प्रतिबिम्ब उत्तल दर्पण में सीधा परन्तु अवतल दर्पण में उल्टा दिखाई देगा। किसी निकटस्थ वस्तु का प्रतिबिम्ब उत्तल दर्पण में छोटा परन्तु अवतल दर्पण में बड़ा दिखाई देगा।

(ii) उत्तल दर्पण में किसी भी वस्तु का प्रतिबिम्ब छोटा परन्तु समतल दर्पण में वस्तु के समान आकार का दिखाई देगा।

(iii) समतल दर्पण में दूरस्थ वस्तु का प्रतिबिम्ब सीधा परन्तु अवतल दर्पण में उल्टा दिखाई देगा। समतल दर्पण में निकटस्थ वस्तु का प्रतिबिम्ब समान आकार का परन्तु अवतल दर्पण में बड़ा दिखाई देगा।

प्रश्न 8.
अवतल दर्पण में वस्तु को किस स्थिति पर रखने से रेखीय आवर्धन 1 होता है? आवश्यक आरेख बनाकर बताइए।
उत्तर:
वस्तु को दर्पण के वक्रता केन्द्र पर रखने से प्रतिबिम्ब का आकार वस्तु के आकार के समान होता है अतः आवर्धन 1 होता है।

प्रश्न 9.
समान्तर आपाती किरणों के परावर्तन के आरेख बनाकर स्पष्ट कीजिए कि किस प्रकाश का दर्पण अभिसारी (convergent) होता है तथा किस प्रकार का दर्पण अपसारी (divergent )?
उत्तर:
चित्र (क) तथा (ख) में अवतल तथा उत्तल दर्पण से समान्तर आपाती किरणों का परावर्तन दिखाया गया है।

चित्र (क) से स्पष्ट है कि अवतल दर्पण समान्तर’ किरणों को एक बिन्दु (मुख्य फोकस) पर केन्द्रित अथवा अभिसरित कर देता है। अतः अवतल दर्पण अभिसारी होता है।

इसके विपरीत उत्तल दर्पण पर आपतित समान्तर किरणें परावर्तन के पश्चात् एक बिन्दु (फोकस) से अपसरित होती प्रतीत होती हैं। अतः उत्तल दर्पण अपसारी होता है।

प्रश्न 10.
किरण आरेख बनाकर, प्रदर्शित कीजिए कि-
(i) अवतल दर्पण
(ii) उत्तल दर्पण में परावर्तन के नियम का पालन किस प्रकार होता है?
उत्तर:
दोनों चित्रों में आपतित किरण AP दर्पण से परावर्तन के पश्चात् किरण PB के रूप में प्रदर्शित है। परावर्तन के बिन्दु P को दर्पण की वक्रता केन्द्र C से मिलाने पर अभिलम्ब CN प्राप्त होता है। दोनों ही दर्पणों में परावर्तन के नियम के अनुसार आपतन कोण (i) परावर्तन कोण (r) के बराबर होता है।

प्रश्न 11.
किरण- आरेख बनाकर समझाइए कि अवतल दर्पण द्वारा, किसी बिन्दु प्रकाश स्रोत से-
(i) समान्तर किरण पुंज,
(ii) अभिसारी किरण- पुंज।
(iii) अपसारी किरण पुंज कैसे प्राप्त किया जा सकता है?
उत्तर:
अवतल दर्पण –
(i) अवतल दर्पण के ध्रुव और फोकस के बीच रखी वस्तु का प्रतिबिम्ब वस्तु के सापेक्ष सीधा तथा बड़ा बनता है। अतः अवतल दर्पण का उपयोग दाढ़ी बनाने में किया जाता है।

(ii) कान, नाक व गले के आन्तरिक भागों की जाँच करने के लिए डॉक्टर अवतल दर्पण का उपयोग करते हैं।

(iii) अवतल दर्पण का उपयोग ऐसे परावर्तक के रूप में किया जाता है जो किसी प्रकाश स्रोत के प्रकाश को समान्तर पुंज (Parallel beam) के रूप में परिवर्तित करें. अथवा किसी अभिसारी किरण पुंज के रूप में किसी वस्तु पर केन्द्रित करें।

पहली दशा में प्रकाश स्रोत को दर्पण के मुख्य फोकस (F) पर रखना होगा [चित्र (क)]

दूसरी दशा में प्रकाश – स्रोत को दर्पण के मुख्य फोकस (F) तथा वक्रता केन्द्र (C) के बीच में इस प्रकार रखना होगा कि दर्पण से परावर्तित संपूर्ण प्रकाश, वस्तु पर केन्द्रित हो।

इस प्रकार का उपयोग बहुधा सूक्ष्मदर्शी, प्रोजेक्टर आदि में किया जाता है-

(iv) अवतल दर्पण का उपयोग परावर्तक दूरदर्शी (Re-flecting telescope) में किया जाता है।

प्रश्न 12.
रेखीय आवर्धन की परिभाषा लिखिए।
उत्तर:
रेखीय आवर्धन वस्तु के प्रतिबिम्ब की लम्बाई एवं वस्तु की लम्बाई के अनुपात को प्रतिबिम्ब का रेखीय आवर्धन कहते हैं। यदि प्रतिबिम्ब की लम्बाई तथा वस्तु की लम्बाई हो तो प्रतिबिम्ब का रेखीय आवर्धन M = \(\frac { I }{ O }\)।

प्रश्न 13.
अवतल दर्पण के ध्रुव एवं फोकस के मध्य रखी वस्तु के प्रतिबिम्ब बनने का किरण आरेख खींचिए।
उत्तर:
जब वस्तु अवतल दर्पण के सामने ध्रुव तथा फोकस के मध्य रखी हो तो प्रतिबिम्ब दर्पण के पीछे, आभासी, सीधा व वस्तु से बड़ा बनता है।

प्रश्न 14.
किसी लेंस के लिए फोकस दूरी एवं प्रकाशित केन्द्र को परिभाषित करो।
उत्तर:
(i) लेंस की फोकस दूरी-लेंस के प्रकाशिक केन्द्र से मुख्य फोकस की दूरी को लेंस की फोकस दूरी कहते हैं। इसे संकेत से दर्शाते हैं।

(ii) प्रकाशिक केन्द्र-लेंस के मुख्य अक्ष पर स्थित वह बिन्दु जिससे गुजरने वाली प्रकाश किरण अपवर्तन के पश्चात् आपतित किरण की दिशा में निर्गमित होती है, लेंस का प्रकाशिक केन्द्र कहलाता है।

प्रश्न 15.
क्या कारण है कि रात को तारे टिमटिमाते दिखाई देते हैं जबकि चन्द्रमा नहीं।
उत्तर:
तारों का टिमटिमाना प्रकाश के अपवर्तन के कारण होता है। तारों से आने वाला प्रकाश वायु की विभिन्न घनत्व की परतों से अपवर्तित होता है। वायुमण्डल की परिस्थिति (ताप) जल्दी-जल्दी बदलने के कारण वायुमण्डल की परतों का घनत्व तथा अपवर्तनांक भी बदलता रहता है जिससे तारों की आभासी स्थिति एक सी नहीं रहती है बल्कि प्रत्येक क्षण बदलती है। इसलिए तारे टिमटिमाते प्रतीत होते हैं।

चन्द्रमा पृथ्वी से तारों की अपेक्षा पास होने के कारण आकार में बड़ा दिखाई देता है तथा तारे की अपेक्षा कहीं अधिक किरणें भेजता है। अतः किरणों की संख्या अधिक होने के कारण किरणों पर होने वाले परिवर्तन का आँख पर कोई विशेष प्रभाव नहीं पड़ता है, इसलिए चन्द्रमा टिमटिमाता प्रतीत नहीं होता है।

प्रश्न 16.
तालाब में पड़ी मछली वहाँ नहीं होती है जहाँ वह दिखायी देती है। कारण बताइये।
उत्तर:
जब हम स्वच्छ जल से भरे किसी तालाब की तली को ऊर्ध्वाधर लम्बवत् देखते हैं तो हमें तली का प्रत्येक बिन्दु ऊपर उठा दिखाई देता है अर्थात् तालाब की आभासी गहराई, उसकी वास्तविक गहराई से कम प्रतीत होती है। चूँकि वायु के सापेक्ष जल का अपवर्तनांक

अर्थात् जल के भीतर स्थित किसी वस्तु को बाहर से देखने पर वह अपनी वास्तविक गहराई की \(\frac { 3 }{ 4 }\) गहराई पर ही दिखाई देगी। यही कारण है कि तालाब में पड़ी मछली वहाँ नहीं होती है जहाँ वह दिखाई देती है।

प्रश्न 17.
जल में डूबी हुई सीधी छड़ मुड़ी हुई दिखाई देती है? कारण स्पष्ट कीजिए।
उत्तर:
चित्रानुसार छड़ के बिन्दु C से चलने वाली प्रकाश की किरणें CP और CQ अपवर्तन के पश्चात् PR और QS मार्ग पर चलती हैं तथा नेत्र E को बिन्दु C से आती हुई प्रतीत होती हैं अर्थात् बिन्दु C का आभासी प्रतिबिम्ब है। इसी प्रकार BC के प्रत्येक बिन्दु के आभासी प्रतिबिम्ब BC के संगत बिन्दुओं पर बनते हैं अतः जल में डूबा हुआ भाग BC, BC’ पर प्रतीत होता है। इस प्रकार जल की सतह से सीधी छड़ मुड़ी प्रतीत होती है।

प्रश्न 18.
पानी में पड़ा सिक्का सतह दिखाई देता है? कारण दीजिए।
उत्तर:
चित्रानुसार पानी में पड़ा सिक्का P से आने वाली प्रकाश किरणें सघन माध्यम (जल) से विरल माध्यम (वायु) में आती हैं अतः अभिलम्ब से दूर हट जाती हैं जो P’ से आती हुई प्रतीत होती हैं जिससे सिक्के P का आभासी प्रतिबिम्ब P’ बनता है और सिक्का उठा हुआ P’ पर दिखाई देता है।

प्रश्न 19.
स्वच्छ जल के तालाब की गहराई, उसकी वास्तविक गहराई से कम क्यों प्रतीत होती है?
उत्तर:
माना तालाब के जल की सतह PR है। इसकी तली में स्थित वस्तुएँ O₁, O₂, O₃, इत्यादि से चलने वाली प्रकाश किरणें OP₁, OP₂, OP₃ …….. जल (सघन माध्यम) से चलकर वायु के अपवर्तक के पश्चात् E स्थान से देखी जाती हैं। नेत्र E को ये क्रमश: I₁, I₂, I₃……. बिन्दुओं से आती हुई प्रतीत होती हैं अतः स्पष्ट है कि दर्शक को दूर स्थित वस्तुएँ समीप स्थित वस्तुओं की अपेक्षा अधिक उठी हुई प्रतीत होती हैं। यही कारण है तालाब कम गहरा दिखाई देता है।

प्रश्न 20.
पानी को बीकर में रखी खाली परख नली चाँदी जैसी चमकदार दिखायी देती है। कारण समझाइए।
उत्तर:
पानी के बीकर में रखी खाली परखनली चाँदी जैसी चमकदार दिखायी देती है, इसका कारण यह है कि प्रकाश किरणें जल (सघन माध्यम) से परखनली की वायु (विरल माध्यम) में प्रवेश करती हैं। अतः वे किरणें जो जल के क्रान्तिक कोण (49° लगभग) से बड़ा कोण बनाते हुए परखनली पर आपतित होती हैं पूर्ण परावर्तित हो जाती हैं।

दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
अवतल दर्पण का उदाहरण लेकर सिद्ध कीजिए कि गोलीय दर्पण की फोकस दूरी, दर्पण की वक्रता त्रिज्या की आधी होती है।
उत्तर:
चित्र में एक किरण AM अवतल दर्पण के मुख्य अक्ष (CP) के समान्तर आपतित होती है। यह किरण दर्पण से MF दिशा में परावर्तित है जहाँ F दर्पण का फोकस है। दर्पण का ध्रुव P तथा वक्रता केन्द्र C है बिन्दु M पर अभिलम्ब MC है क्योंकि दर्पण के किसी बिन्दु से वक्रता केन्द्र को मिलाने वाली रेखा दर्पण पर अभिलम्ब होती है। ∠AMC आपतन कोण तथा ∠FMC परावर्तन कोण है।

परावर्तन के नियमानुसार,
∠AMC = ∠FMC
∵ AM || CP तथा CM इन्हें काटती है। … (1)
∴ ∠AMC = ∠FCM (एकान्तर कोण) … (2)
समीकरण (1) और समीकरण (2) से,
∠FMC = ∠FCM
अतः इन कोणों सम्मुख ∆FMC की भुजाएँ भी बराबर होंगी,
अर्थात् FC = MF … (3)
यदि अवतल दर्पण द्वारक बहुत अधिक छोटा है, तो बिन्दु M ध्रुव P के बहुत निकट होगा और तब इस दशा में,
FM = FP (लगभग) … (4)
समीकरण (3) में FM = FP रखने पर,
FC = FP
अर्थात् F, CP का मध्य बिन्दु होगा। …(5)
समीकरण (5) के दोनों पक्षों में FP जोड़ने पर,
FC + FP = FP + FP
या, PC = 2.FP
या, FP = \(\frac { PC }{ 2 }\)
अर्थात् फोकस दूरी = \(\frac { 1 }{ 2 }\) x वक्रता त्रिज्या
अथवा f = \(\frac { R }{ 2 }\)

प्रश्न 2.
उत्तल दर्पण के लिए सूत्र f = \(\frac { R { 2 \) प्राप्त कीजिए, जबकि तथा R अपने सामान्य अर्थों में प्रयुक्त हैं।
उत्तर:
उत्तल दर्पण के लिए फोकस दूरी एवं वक्रता त्रिज्या में सम्बन्ध (Relation between Radius of Curvature and Focal Length by Con- vex Mirror) – माना M₁ M₂ एक उत्तल दर्पण है, जिसके फोकस F. वक्रता केन्द्र C तथा ध्रुव P हैं। अतः इसकी वक्रता त्रिज्या PC = R तथा फोकस दूरी PF = f है। माना, एक प्रकाश किरण AB मुख्य अक्ष PC के समान्तर दर्पण पर आपतित होती है। परन्तु बिन्दु B पर CN अभिलम्ब है। अतः यह किरण BD दिशा में परावर्तित हो जाती है तथा फोकस F से आती हुई प्रतीत होती है। ∠ABN आपतन कोण तथा ∠DBN परावर्तन कोण है।
अत: ∠ABN = ∠DBN  (परावर्तन के नियम से)
तथा ∠ABN = ∠FBC (शीर्षाभिमुख कोण) … (1)
अब AB एवं PC परस्पर समान्तर हैं तथा NC इन्हें काटती है,
∴ ∠ABN = ∠FCB (संगत कोण) … (2)
समीकरण (1) व समीकरण (2) से ∠FBC = ∠FCB
अब ∆FBC में, ∠FBC = ∠FCB
FC = FB … (3)

यदि उत्तल दर्पण बहुत अधिक छोटे द्वारक का है, तो बिन्दु B ध्रुव P के बहुत निकट होगा और तब इस दशा में
PB = FP (लगभग) … (4)
समीकरण (3) में FB = FP रखने से,
FC = FP … (5)
समीकरण (5) के दोनों ओर PF जोड़ने पर,
PF + FC = PF + PF
PC = 2 PF
∴ PF = \(\frac { 1 }{ 2 }\) PC
अर्थात् फोकस-दूरी = \(\frac { 1 }{ 2 }\) x वक्रता त्रिज्या

प्रश्न 3.
अवतल दर्पण के समुख स्थित वस्तु के प्रतिबिम्ब का बनना किरण आरेख द्वारा प्रदर्शित कीजिए, जबकि वस्तु की स्थिति-
(i) वक्रता केन्द्र से अधिक दूरी पर हो,
(ii) वकता केन्द्र पर हो,
(iii) वक्रता केन्द्र तथा फोकस कें षीच हो,
(iv) फोकस तथा दर्पण के बीच में हो।
प्रत्येक प्रतिबिम्ब की प्रकृति बताइए।
उत्तर:
(i) वस्तु दर्पण के वक्रता केन्द्र से अधिक दूरी पर हो (u > R)
प्रतिबिम्ब की स्थिति-मुख्य फोकस तथा वक्रता केन्द्र के बीच में [R > v > F]
आकार-वस्तु से छोटा I = \(\frac { v { u \) x O [ [I] < O]
प्रकृति- वास्तविक तथा उल्टा।

(ii) वस्तु दर्पण के वक्रता केन्द्र पर हो – [u = R]
प्रतिबिम्ब की स्थिति वक्रता केन्द्र पर [v = R]
आकार-वस्तु के समान [I = O]
प्रकृति-वास्तविक तथा उल्टा।

(iii) वस्तु दर्पण के वक्रता केन्द्र तथा मुख्य फोकस के बीच में [R > u > f]
प्रतिविम्ब की स्थिति वक्रता केन्द्र के आगे (v > R)
आकार वस्तु से बड़ा (I > O)
प्रकृति – वास्तविक तथा उल्टा।

(iv) वस्तु फोकस तथा दर्पण के बीच हो [u < f]
प्रतिबिम्ब की स्थिति दर्पण के पीछे।
आकार – वस्तु से बड़ा (I > O)
प्रकृति – सीधा तथा आभासी।

प्रश्न 4.
अवतल दर्पण (गोलीय दर्पण) के लिए सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) का निगमन कीजिए।
उत्तर:
माना M₁M₂ एक अवतल दर्पण है जिसका P ध्रुव तथा C वक्रता केन्द्र है। दर्पण की मुख्य अक्ष पर AB एक वस्तु रखी है जिनका प्रतिबिम्ब A’B’ बनता है। दर्पण के ध्रुव (P) से वस्तु की दूरी PA माना है। ध्रुव से प्रतिबिम्ब की दूरी PA’ माना है तथा दर्पण की फोकस दूरी PF माना है। मुख्य अक्ष के समान्तर प्रकाश किरण है। D से मुख्य अक्ष पर लम्ब DN है।
अतः DN = AB
अब ∆ABC तथा ∆A’B’C में।
∠BAC = ∠B’A’C (प्रत्येक समकोण है)
∠BCA = ∠B’C’A ( शीर्षाभिमुख कोण)
अतः तीसरा कोण ABC = CB’A’
अर्थात् दोनों त्रिभुज समरूप हैं।
∴ \(\frac{A B}{A^{\prime} B^{\prime}}=\frac{C A}{C A^{\prime}}\) …(i)
अब ∆DNF तथा ∆ABF भी समरूप हैं
∴ \(\frac{D N}{A^{\prime} B^{\prime}}=\frac{F N}{F A^{\prime}}\) …(2)
समीकरण ( 1 ) व समीकरण (2) से,
∴ \(\frac{C A}{C A^{\prime}}=\frac{F N}{F A^{\prime}}\) …(3) (∵ DN = AB)
यदि अवतल दर्पण का द्वारक बहुत अधिक छोटा हो, तो बिन्दु N दर्पण के ध्रुव (P) के बहुत निकट होगा।
इस दशा में, FN = FP (लगभग)
तथा समीकरण (3) को निम्न प्रकार लिख सकते हैं

अवतल दर्पण में, चिन्हों के नियमानुसार-
वस्तु की दर्पण के ध्रुव से दूरी PA = – u
प्रतिबिम्ब की दर्पण के ध्रुव से दूरी PA’ = – v
दर्पण की फोकस दूरी PF = – f
दर्पण की वक्रता त्रिज्या PC = – r = 2f
समीकरण (4) में मान रखने पर,
\(\frac{-u-(-2 f)}{-2 f-(-v)}=\frac{-f}{-v-(-f)}\)
अथवा \(\frac{-u+2 f}{-2 f+v}=\frac{-f}{-v+f}\)
-f(-2f + v) = (- v + f)(-u + 2f)
अथवा 2f² – vf = uv – 2vf – uf + 2f²
अथवा uv = uf + vf
दोनों पक्षों में uvf से भाग देने पर,
अवतल दर्पण का सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}/latex]

प्रश्न 5.
उत्तल दर्पण के के सम्मुख किसी भी स्थिति में रखी वस्तु के दर्पण द्वारा बने प्रतिबिम्ब की स्थिति, आकार एवं प्रकृति किरण आरेख खींचकर बनाइए।
उत्तर:
नीचे चित्र में MPN एक उत्तल दर्पण है जिसके सम्मुख मुख अक्ष पर वस्तु AB स्थित है। बिन्दु A से चलने वाली प्रकाश किरणें AM और AQ दर्पण से परावर्तन के पश्चात् FM व CA में चली जाती है जो A से आती हुई प्रतीत होती है। इस प्रकार A’ आभासी प्रतिबिम्ब बनता है। इसी प्रकार AB के विभिन्न बिन्दुओं के अभासी A’B’ ‘के संगत बिन्दुओं पर बनते हैं। अतः उत्तल दर्पण में वस्तु का प्रतिबिम्ब सदैव वस्तु से छोटा, सीधा, आभासी एवं दर्पण के पीछे सदैव फोकस और ध्रुव के मध्य बनता है।

प्रश्न 6.
लेंस द्वारा प्रतिबिम्ब बनाने के नियम लिखो।
उत्तर:
लेंस द्वारा प्रतिबिम्ब बनाने के नियम निम्नलिखित है-

• वह प्रकाश किरण जो लेंस की मुख्य अक्ष के समान्तर आपतित होती है, उत्तल लेंस से अपवर्तन के पश्चात् मुख्य अक्ष की ओर झुककर फोकस F से गुजरती है तथा अवतल लेंस से अपवर्तन के पश्चात् मुख्य अक्ष से दूर हटकर लेंस के फोकस से आती प्रतीत होती है।
• लेंस के प्रकाशिक केन्द्र 0 से गुजरने वाली किरण अपने मार्ग पर ही बिना किसी विचलन के चली जाती है।
• लेंस के फोकस से गुजरने वाली आपतित किरण (उत्तल लेंस में) अथवा एकत्रित प्रतीत होने वाली आपतित किरण (अवतल लेंस में) लेन्स से परावर्तन के पश्चात् मुख्य अक्ष के समान्तर हो जाती है।

लेंस द्वारा किसी वस्तु का प्रतिबिम्ब बनाने के लिए उपर्युक्त तीनों में से किन्हीं दो किरणों का ही उपयोग किया जाता है।

प्रश्न 7.
किरण आरेख द्वारा पूर्ण परावर्तक प्रिज्म का (i) 90° विचलन, (ii) 180° विचलन प्रिज्म के लिए उपयोग बताइये।
उत्तर:
(i) 90° विचलन के लिए पूर्ण परावर्तक का उपयोग – चित्रानुसार जब वस्तु AB, प्रिज्म के सम्मुख रखते हैं तो उससे निकलने वाली प्रिज्म भुजा PQ किरणें AC व BD अत: तल PQ पर लम्बवत् आपतित होती हैं बिना मुड़े सीधे कर्ण PR के बिन्दु 0 एवं O’ पर पड़ती तथा अभिलम्ब के साथ 45° का कोण बनाती हैं। चूँकि 42° होता है अतः इनका पूर्ण परावर्तन हो जाता है।

पूर्ण हो जाता है। पूर्ण परावर्तित किरणें OE व OF क्रान्ति क्रमशः तल OR के बिन्दु E व F पर लम्ब लम्बवत् पड़ती हैं अत: सीधे EA’ व FB’ की दिशा में निकल जाती हैं। इस प्रकार AB का प्रतिबिम्ब A’B’ बनता है जो AB से 90° झुका हुआ है। स्पष्ट है इससे साधारण दर्पण की अपेक्षा अधिक चमकीला प्रतिबिम्ब बनता है। इसलिए इसका उपयोग पेरिस्कोप में समतल दर्पण के स्थान पर करते हैं।

(ii) 180° विचलन के लिए पूर्ण परावर्तक प्रिज्म का उपयोग – चित्रानुसार जब वस्तु AB, प्रिज्म के कर्ण भुजा PR के सम्मुख रखते हैं तो उससे निकलने वाली किरणें PR पर लम्बवत् आपतित होती हैं जिसका (काँच का क्रांतिक कोण 42° होने के कारण) प्रिज्म के फलक PQ व QR पर दो बार पूर्ण आन्तरिक परावर्तन हो जाता है। तथा निर्गत किरण का मार्ग आपतित किरण के विपरीत दिशा में अर्थात् 180° कोण से विचलित हो जाता है। इस प्रकार यह प्रिज्म उल्टे प्रतिबिम्ब को सीधा करने के लिए प्रयुक्त किया जाता है।

प्रश्न 8.
किसी अवतल लेंस द्वारा प्रतिबिम्ब बनने की प्रक्रिया का रेखाचित्र दीजिए।
उत्तर:
चित्रानुसार, अवतल लेंस के सम्मुख एक वस्तु AB रखी गयी है जिसका प्रतिबिम्ब A’B’ लेंस के सामने वस्तु की ओर लेंस व प्रकाशिक केन्द्र के बीच आभासी, सीधा तथा वस्तु छोटा बनता है। वस्तु को अवतल लेंस के सामने कितनी भी दूरी पर रखें, उसका प्रतिबिम्ब सदैव उसी ओर प्रकाशिक केन्द्र व फोकस के बीच वस्तु से छोटा, सीधा व आभासी बनता है।

प्रश्न 9.
काँच के गुटके का अपवर्तनांक ज्ञात करने की विधि को लिखकर आरेख खींचिए।
उत्तर:
काँच के गुटके का अपवर्तनांक ज्ञात करना – सर्वप्रथम ड्राइंग बोर्ड पर एक सफेद कागज ड्राइंग पिन की सहायता से लगा देते हैं। कागज पर काँच के गुटके की सीमा रेखा ABCD बना दें। अब पृष्ठ AB पर एक बिन्दु O लें तथा AB पर अभिलम्ब N₁ON₂ खींचे। अब रेखा N₁N₂ के साथ बिन्दु पर ∠N₁OX बनाते हुए रेखा OX खींचें। रेखा OX पर दो पिनें P₁ तथा P₂ ऊर्ध्वाधर ठोंक दो।

अब काँच के गुटके को ABCD पर रख दें तथा CD तल पर पिनों P₁ व P₂ के प्रतिबिम्बों को देखते हुए उनके ठीक सीध में दो अन्य पिनें P₃ तथा P₄ पुनः ठोंक दो। इस प्रकार चारों पिनें एक सीध में दिखेंगी। काँच का गुटका हटा देते हैं।

अब पिनों को हटाकर P₃ तथा P₄ को मिला देते हैं जो पृष्ठ CD को O₁ पर मिलती है। बिन्दु O व O₁ को मिला दो। अब O को केन्द्र मानकर किसी त्रिज्या का एक वृत्त बनाया जो OX को Q पर तथा OO₁ को S पर काटती है। अब Q व S से अभिलम्ब N₁ON₂ पर लम्ब OR व ST खींचें। OR व ST की लम्बाई माप लेते हैं।

प्रश्न 10.
प्रकाश के अपवर्तन से क्या तात्पर्य है? यह प्रकाश के परावर्तन से किस प्रकाश भिन्न होता है। जब किसी दूसरे पारदर्शी माध्यम (2) के तल पर आपतित
उत्तर:
किसी एक माध्यम (1) में चलता हुआ प्रकाश जब किसी दूसरे पारदर्शी माध्यम (2) के तल पर आपतित होता है तो आपतित प्रकाश का कुछ अंश, पहले माध्यम से ही वापस लौट जाता है-जिसे प्रकाश परावर्तन कहते हैं।

आपतित प्रकाश का अधिकांश भाग दूसरे माध्यम में अपने पहले मार्ग से कुछ विचलित होकर, गमन करता है। एक माध्यम से दूसरे माध्यम में, अपने पूर्व मार्ग से विचलित होकर गमन को प्रकाश का अपवर्तन कहते हैं।

प्रश्न 11.
प्रकाश के अपवर्तन के नियम लिखिए तथा उपयुक्त आरेख बनाकर स्नैल का नियम स्पष्ट कीजिए।
अथवा
प्रकाश के अपवर्तन सम्बन्धी स्नैल के नियम लिखिए तथा समझाइए।
उत्तर:
चित्र में प्रकाश के माध्यम-1 से माध्यम -2 में अपवर्तन को दर्शाया गया है। अपवर्तन बिन्दु 0 पर अभिलम्ब एवं आपतित किरण के बीच बने कोण (i) को आपतन कोण तथा अभिलम्ब एवं अपवर्तित किरण के बीच बने कोण (r) को आयतन कोण कहते हैं।

अपवर्तन के नियमों के अनुसार,
(1) आपतित किरण, अपवर्तित किरण तथा आपतन बिन्दु पर खींचा गया अभिलम्ब, तीनों एक ही समतल में होते हैं।

(2) आपतन कोण (sin i) तथा अपवर्तन कोण (sin r) की ज्या (sin e) परस्पर समानुपाती होते हैं।

इस नियम को स्नैल का नियम कहते हैं। इसके अनुसार,
sin i ∝ sin r
अथवा [latex]\frac { sin i }{ sin r }\) = n (n एक नियतांक है।)
नियतांक n को पहले माध्यम (1) के सापेक्ष दूसरे माध्यम (2) का अपवर्तनांक कहते हैं तथा प्रतीक ₁n₂ से व्यक्त करते हैं।

आंकिक प्रश्न

प्रश्न 1.
अवतल दर्पण से 2.0 सेमी दूर रखी वस्तु का वास्तविक प्रतिबिम्ब 30 सेमी दूर बनता है। दर्पण की फोकस – दूरी ज्ञात कीजिए।
उत्तर:
अवतल दर्पण के लिए,
वस्तु की दूरी (u) = – 20 सेमी
वास्तविक प्रतिबिम्ब की दूरी
(v) = – 30 सेमी
अवतल दर्पण के सूत्र
\(\frac{1}{f}=\frac{1}{-30}+\frac{1}{-20}\)
\(\frac{1}{f}=\frac{(-20)+(-30)}{600}=\frac{-50}{600}\)
= – \(\frac { 1 }{ 12 }\)
अतः अवतल दर्पण की फोकस-दूरी
(f) = – 12 सेमी।

प्रश्न 2.
एक अवतल दर्पण की फोकस-दूरी 25 सेमी है। दर्पण से $2.0$ सेमी दूरी पर रखी वस्तु के प्रतिबिम्ब की स्थिति एवं प्रकृति ज्ञात किजिए।
उत्तर:
अवतल दर्पण की फोकस-दूरी (f) = – 25 सेमी.
दर्पण से वस्तु की दूरी
(u) = – 20 सेमी.
अवतल दर्पण के सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) से,
\(\frac{1}{v}+\left(-\frac{1}{20}\right)=-\frac{1}{25}+\frac{1}{20}\)
\(\frac{1}{v}=-\frac{1}{25}+\frac{1}{20}\)
\(\frac{1}{v}=\frac{-4+5}{100}=\frac{1}{100}\)
∴ v = 100 सेमी, v का मान धनात्मक है। अत: प्रतिबिम्ब दर्पण के पीछे बनता है। प्रतिबिम्ब का आवर्धन
m = \(\frac { v }{ u }\)
= \(\frac { 100 }{ 20 }\) = 5
अर्थात् प्रतिबिम्ब सीधा तथा वस्तु से 5 गुना बड़ा बनता है। प्रतिबिम्ब दर्पण के पीछे बनने के कारण आभासी हैं। ]

प्रश्न 3.
एक अवतल दर्पण से 30 सेमी दूर रखी वस्तु का तीन गुना बड़ा
(i) आभासी,
(ii) वास्तविक प्रतिबिम्ब प्राप्त होता है। प्रत्येक दशा में अवतल दर्पण की फोकस-दूरी ज्ञात कीजिए।
उत्तर:
(i) आभासी प्रतिबिम्ब के लिए-
अवतल दर्पण से वस्तु की दूरी
(u) = – 30 सेमी
प्रतिबिम्ब का आवर्धन
(m) = + 3
प्रतिबिम्ब का आवर्धन
m = – \(\frac { v }{ u }\) से
v = – m u = – 3 x (- 30)
= – 90 सेमी
अवतल दर्पण के सूत्र \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) से,
\(\frac{1}{f}=\frac{1}{-30}+\frac{1}{90}\)
\(\frac{-3+1}{90}\)
= \(\frac { -2 }{ 90 }\) = – \(\frac { 1 }{ 45 }\)
अर्थात् अवतल दर्पण की फोकस दूरी
(f) = – 45 सेमी

(ii) वास्तविक प्रतिबिम्ब के लिए-
m = – 3
v = – m u
= – (- 3) x (- 30) = – 90 सेमी
अवतल दर्पण के सूत्र \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) से,
= – \(\frac{1}{30}-\frac{1}{90}\)
= \(\frac{-3-1}{90}=\frac{-4}{90}\)
f = – \(\frac{90}{4}\) = – 22.5 सेमी
अतः अवतल दर्पण की फोकस-दूरी
(f) = – 22.5 सेमी

प्रश्न 4.
उत्तल दर्पण से 40 सेमी दूर रखी वस्तु का प्रतिबिम्ब 10 सेमी दूर बनता है। उत्तल दर्पण की फोकस-दूरी ज्ञात कीजिए।
उत्तर:
उत्तल दर्पण में,
दर्पण से वस्तु की दूरी
(u) = – 40 सेमी
प्रतिबिम्ब की दूरी
(v) = + 10 सेमी
उत्तल दर्पण के सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) से,
\(\frac{1}{f} =-\frac{1}{40}+\frac{1}{10}\)
= \(\frac{-1+4}{40}=\frac{3}{40}\)
∴ उत्तल दर्पण की फोकस-दूरी
(f) = \(\frac{40}{3}\) = 13.33 सेमी
अतः उत्तल दर्पण की फोकस-दूरी = 13.33 सेमी।

प्रश्न 5.
एक उत्तल दर्पण की फोकस दूरी 20 सेमी है। इस दर्पण से 25 सेमी दूर रखी वस्तु के प्रतिबिम्ब की स्थिति ज्ञात कीजिए।
उत्तर:
उत्तल दर्पण में,
उत्तल दर्पण की फोकस दूरी
(f) = 20 सेमी
वस्तु की दूरी (u) = – 25 सेमी
उत्तल दर्पण के सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) से
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)
= \(\frac{1}{20}-\frac{1}{(-25)}\)
= \(\frac{1}{20}+\frac{1}{-25}\)
= \(\frac { 5+4 }{ 100 }\) = \(\frac { 9 }{ 100 }\)
अथवा प्रतिबिम्ब की स्थिति
v = \(\frac { 100 }{ 9 }\)
= 11.11 सेमी
अतः प्रतिबिम्ब उत्तल दर्पण के पीछे 11.11 सेमी पर बनता है।

प्रश्न 6.
एक उत्तल दर्पण से 25 सेमी दूर रखी वस्तु का प्रतिबिम्ब वस्तु की लम्बाई का आधा होता है। दर्पण की फोकस दूरी ज्ञात कीजिए।
उत्तर:
उत्तल दर्पण में,
वस्तु की दूरी (u) = -25 सेमी.
प्रतिबिम्ब की लम्बाई

अथवा दर्पण की फोकस दूरी f = 25 सेमी

प्रश्न 7.
6 सेमी लम्बाई की एक कील उत्तल दर्पण के सामने 20 सेमी दूर रखी है। यदि इस दर्पण की फोकस – दूरी 10 सेमी हो, तो कील के प्रतिबिम्ब की स्थिति एवं लम्बाई ज्ञात कीजिए।
उत्तर:
उत्तल दर्पण में,
कील की लम्बाई (O) = 6 सेमी
कील की दूरी (u) = – 20 + 10 सेमी
फोकस दूरी (f) = + 10 सेमी
∴ उत्तल दर्पण के सूत्र \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) से
\(\frac{1}{10}=-\frac{1}{20}+\frac{1}{v}\)
अथवा \(\frac{1}{v}=\frac{1}{10}+\frac{1}{20}\)
= \(\frac{2+1}{20}=\frac{3}{20}\)
अथवा v = \(\frac { 20 }{ 3 }\)
= 6.67 सेमी दर्पण के पीछे
अब आवर्धन m = \(\frac { v }{ u }\) = \(\frac { 20/3 }{ 20 }\)
= \(\frac { 1 }{ 3 }\) = तथा m = \(\frac { I }{ O }\)
∴ प्रतिबिम्ब की लम्बाई (I) = m x 0 = \(\frac { 1 }{ 3 }\) x 6 = 2 सेमी।
अतः प्रतिबिम्ब दर्पण के पीछे 6.67 सेमी दूरी पर 2 सेमी लम्बाई का बनता है।

प्रश्न 8.
एक वस्तु 30 सेमी वक्रता त्रिज्या वाले अवतल दर्पण से 25 सेमी की दूरी पर रखी है। प्रतिबिम्ब की स्थिति ज्ञात कीजिए।
उत्तर:
अवतल दर्पण में,
वस्तु की दूरी (u) = 25 सेमी
वक्रता त्रिज्या (R) = 30 सेमी
फोकस दूरी (f) = \(\frac { R }{ 2 }\) = \(\frac { -30 }{ 2 }\) = – 15 सेमी
∴ उत्तल दर्पण के सूत्र \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) से,
\(\frac{1}{v}=\frac{1}{-15}-\frac{1}{-25}\)
= \(-\frac{1}{15}+\frac{1}{25}\)
= \(\frac{-25+15}{375}=-\frac{10}{375}\)
v = – \(\frac { 375 }{ 10 }\)
= – 37.5
∴ प्रतिबिम्ब की दूरी (v) = 37.5 सेमी दर्पण के सामने।

प्रश्न 9.
एक 10 सेमी फोकस दूरी वाले अवतल दर्पण से कितनी दूरी पर कोई वस्तु रखी जाय कि उसका 5 गुना बड़ा प्रतिबिम्ब बने जबकि प्रतिबिम्ब (i) आभासी, (ii) वास्तविक हो।
उत्तर:
अवतल दर्पण के लिए,
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
(i) वास्तविक प्रतिबिम्ब वस्तु तथा फोकस सभी दर्पण के बायीं ओर होते हैं इसलिए निर्देशांक ज्यामिति की चिन्ह परिपाटी के अनुसार ७, ७ तथा सब ऋणात्मक होंगे। अतः आवर्धन
(m) = 5 = \(\frac { v }{ u }\)
∴ प्रतिबिम्ब की दूरी (v) = 5u
अव तल दर्पण की फोकस – दूरी
(f) = – 10 सेमी
∴ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
या, \(-\frac{1}{5 u}+\frac{1}{u}=-\frac{1}{10}\)
या, \(\frac{-1+5}{5 u}=-\frac{1}{10}\)
या, \(\frac{4}{5 u}=-\frac{1}{10}\)
5u = – 40
u = \(\frac { -40 }{ 5 }\)
∴ u = – 8 सेमी
अतः वस्तु से पाँच गुना बड़ा वास्तविक प्रतिबिम्ब प्राप्त करने के लिए वस्तु को अवतल दर्पण से 8 सेमी दूर रखना होगा।

(ii) यदि आभासी प्रतिबिम्ब बनेगा तो वह अवतल दर्पण के पीछे बनेगा। ऐसी दशा में धनात्मक होगा।
अवतल दर्पण के सूत्र से
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
∴ \(\frac{1}{5 u}+\frac{1}{u}=\frac{1}{-10}\)
\(\frac{1+5}{5 u}=\frac{1}{-10}\)
∴ \(\frac{6}{5u}=\frac{1}{-10}\)
5u = – 60
या u = – 12 सेमी।
∴ वस्तु का पाँच गुना बड़ा आभासी प्रतिबिम्ब अवतल दर्पण द्वारा प्राप्त करने के लिए वस्तु दर्पण के ध्रुव से 12 सेमी की दूरी पर रखना होगा।

प्रश्न 10.
एक उत्तल दर्पण 20 सेमी वक्रता त्रिज्या वाला है। इसके सामने कोई वस्तु रखी जाती है तो उसका प्रतिबिम्ब वस्तु के आधे आकार का बनता है। वस्तु तथा प्रतिबिम्ब की स्थिति ज्ञात कीजिए।
उत्तर:
उत्तल दर्पण की वक्रता-त्रिज्या = 20 सेमी
प्रश्नानुसार, आवर्धन
m = \(\frac { v }{ vu}\) = \(\frac { 1 }{ 2 }\)
अथवा u = 2v
फोकस – दूरी (f) = \(\frac { R }{ 2 }\) = + \(\frac { 20 }{ 2 }\) = 10 सेमी
चूँकि उत्तल दर्पण में प्रतिबिम्ब दर्पण के पीछे बनता है अतः v का चिन्ह धनात्मक एवं u का चिन्ह ऋणात्मक होगा।
उत्तल दर्पण के सूत्र \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ -2v }\) = \(\frac { 1 }{ 10 }\)
\(\frac { 1 }{ v }\) – \(\frac { 1 }{ v }\) = \(\frac { 1 }{ 10 }\)
\(\frac { 2-1 }{ 2v }\) = \(\frac { 1 }{ 10 }\)
अथवा \(\frac { 1 }{ 2v }\) = \(\frac { 1 }{ 10 }\)
∴ v = \(\frac { 10 }{ 2 }\) = 5 सेमी दर्पण के पीछे
तथा u = 2v = 5 x 2
= 10 सेमी दर्पण के सामने।

प्रश्न 11.
एक वस्तु अवतल दर्पण से 15 सेमी की दूरी पर रखी गयी है। इसका प्रतिबिम्ब वस्तु की ओर 30 सेमी की दूरी पर बनता है। यदि वस्तु 8 सेमी की दूरी पर रहे तो प्रतिबिम्ब का स्थान कहाँ होगा? उत्तर:
पहली दशा में वस्तु की दूरी (u) = 15 सेमी, प्रतिबिम्ब की दूरी (v) = – 30 सेमी
अतः अवतल दर्पण के सूत्र \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) से,
\(\frac{1}{f}=\frac{1}{-30}+\frac{1}{-15}\)
= \(\frac{-15-30}{450}=\frac{-45}{450}=\frac{1}{-10}\)
दूसरी दशा में वस्तु की दूरी (u) = 8 सेमी, प्रतिबिम्ब की दूरी (u) = ?
अतः \(\frac{1}{v}+\frac{1}{-8}=\frac{1}{-10}\)
अथवा \(\frac{1}{v}=\frac{1}{8} \frac{1}{-10}=\frac{8-10}{80}=\frac{1}{40}\)
v = + 40 सेमी
अतः प्रतिबिम्ब दर्पण के पीछे 40 सेमी दूरी पर बनेगा।

प्रश्न 12.
12 सेमी फोकस दूरी वाले अवतल दर्पण के सामने 4 सेमी लम्बी वस्तु रखी है। इसका 1 सेमी लम्बा प्रतिबिम्ब बनता है तो वस्तु की स्थिति ज्ञात कीजिए।
उत्तर:
अवतल दर्पण की फोकस दूरी f = 12 सेमी
वस्तु की लम्बाई (O) = 4 सेमी
आवर्धन (m) = \(\frac { I }{ O }\) = \(\frac { 1 }{ 4 }\) = \(\frac { v }{ u }\)
अथवा v = \(\frac { u }{ 4 }\); f = – 12 सेमी
अवतल दर्पण में छोटा प्रतिबिम्ब वस्तु की ही ओर बनता है अतः u तथा v दोनों के चिन्ह (-) होंगे। अवतल दर्पण के सूत्र
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) से,
\(\frac{1}{-(u / 4)}+\frac{1}{-u}=-\frac{1}{12}\)
अथवा \(\frac { 4 }{ u }\) – \(\frac { 1 }{ u }\) = – \(\frac { 1 }{ 12 }\) या \(\frac { 5 }{ u }\) = – \(\frac { 1 }{ 12 }\)
सरल करने पर u = 60 सेमी
अतः वस्तु दर्पण के सामने 60 सेमी दूर स्थित है।

प्रश्न 13.
50 सेमी फोकस दूरी वाले अवतल दर्पण के सामने 25 सेमी की दूरी पर रखी वस्तु के दर्पण से बने प्रतिविम्ब की स्थिति ज्ञात कीजिए।
उत्तर:
अवतल दर्पण में.
वस्तु की दूरी (u) = – 25 सेमी
फोकस – दूरी (f) = – 50 सेमी
प्रतिबिम्ब की दूरी (v) = ?
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
\(\frac{1}{-50}=\frac{1}{v}+\frac{1}{-25}\)
या, \(\frac{1}{v}=\frac{1}{25}-\frac{1}{50}=\frac{2-1}{50}=\frac{1}{50}\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ 50 }\)
∴ v = 50 सेमी, दर्पण के पीछे प्रतिबिम्ब बनेगा।

प्रश्न 14.
10 सेमी फोकस दूरी के उत्तल दर्पण एक वस्तु को 30 सेमी की दूरी पर रखने से प्रतिविम्ब कहाँ बनेगा? यदि वस्तु की लम्बाई 6 सेमी हो तो प्रतिबिम्ब की लम्बाई कितनी होगी?
उत्तर:
उत्तल दर्पण में,
f = + 10 सेमी
वस्तु की दूरी (u) = – 30 सेमी,
v = ?, O = 6 सेमी, I = ?

प्रश्न 15.
एक अवतल दर्पण की फोकस दूरी 5 सेमी है। इसके सामने 10 सेमी की दूरी पर रखी वस्तु का प्रतिबिम्ब कहाँ पर बनेगा? क्या यह वास्तविक होगा।
उत्तर:
अवतल दर्पण की फोकस दूरी (f) = – 5 सेमी
वस्तु की दूरी (u) = 10 सेमी
प्रतिबिम्ब की दूरी (v) = ?
∴ \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) से,
\(\frac{1}{v}-\frac{1}{10}=\frac{1}{-5}\)
\(\frac{1}{v}=-\frac{1}{5}+\frac{1}{10}=-\frac{1}{10}\)
v = – 10 सेमी
अर्थात् प्रतिबिम्ब दर्पण से 10 सेमी दूर वस्तु की ओर बनेगा। प्रतिबिम्ब वास्तविक होगा
दर्पण से 10 सेमी दूर वस्तु की ओर, वास्तविक।

प्रश्न 16.
यदि वायु के सापेक्ष जल का एवं शीशे का अपवर्तनांक क्रमशः 4/3 एवं 3/2 हो तो शीशे के सापेक्ष जल के अपवर्तनांक की गणना कीजिए।
उत्तर:
दिया है –

प्रश्न 17.
जल में प्रकाश की चाल 2.25 x 10⁸ मी/से है। यदि जल का अपवर्तनांक 4/3 हो तो निर्वात् में प्रकाश की चाल ज्ञात कीजिए।
उत्तर:

प्रश्न 18.
जल और काँच के अपवर्तनांक क्रमशः 1.35 एवं 1.50 हैं। यदि प्रकाश किरणें काँच से जल में जा रही हों, तो काँच के सापेक्ष जल का अपवर्तनांक ज्ञात कीजिए।
उत्तर:

प्रश्न 19.
यदि प्रिज्म का प्रिज्म कोण 60° तथा न्यूनतम विचलन कोण 36° हो तो प्रिज्म के पदार्थ का अपवर्तनांक ज्ञात कीजिए।
उत्तर:

प्रश्न 20.
वायु के सापेक्ष किसी द्रव का क्रान्तिक कोण 45° है। उस द्रव का अपवर्तनांक ज्ञात कीजिए।
उत्तर:
वायु के सापेक्ष द्रव का क्रान्तिक कोण (C) = 45°

प्रश्न 21.
यदि वायु के सापेक्ष किसी पारदर्शी द्रव का अपवर्तनांक 1.25 है तथा काँच का अपवर्तनांक 1.5 है, तो द्रव के सापेक्ष काँच के अपवर्तनांक की गणना कीजिए।
उत्तर:
वायु के सापेक्ष द्रव का अपवर्तनांक
= 1.25 (_an_w)
वायु के सापेक्ष काँच का अपवर्तनांक= 1.5 (_an_g)
तो द्रव के सापेक्ष काँच का अपवर्तनांक
= \({ }_w n_g=\frac{{ }_a n_g}{{ }_a n_w}=\frac{1.5}{1.25}\) = 1.2
= 1.2

प्रश्न 22.
संलग्न चित्र के अनुसार प्रकाश की किरण वायु से किसी माध्यम में प्रवेश करती है। वायु के सापेक्ष माध्यम का अपवर्तनांक ज्ञात कीजिए।

उत्तर:

बहुविकल्पीय प्रश्न

1. एक अवतल दर्पण की वक्रता त्रिज्या 20 सेमी है। इसकी फोकस दूरी होगी-
(a) 10 सेमी
(b) 15 सेमी
(c) 10 सेमी
(d) 20 सेमी
उत्तर:
(a) 10 सेमी

2. उस उत्तल दर्पण की वक्रता त्रिज्या क्या होगी जिसकी फोकस दूरी 10 सेमी है?
(a) 5 सेमी
(b) 10 सेमी
(c) 20 सेमी
(d) 30 सेमी
उत्तर:
(c) 20 सेमी

3. गोलीय दर्पण की फोकस दूरी और वक्रता त्रिज्या में सम्बन्ध होता है-
(a) f = \(\frac { R }{ 2 }\)
(b) f = 2R
(c) R = \(\frac { f }{ 2 }\)
(d) f = R
उत्तर:
(a) f = \(\frac { R }{ 2 }\)

4. 10 सेमी फोकस दूरी वाले अवतल दर्पण के सामने 20 सेमी पर रखी वस्तु का प्रतिबिम्ब बनेगा-
(a) दर्पण के पीछे
(b) फोकस पर
(c) दर्पण और फोकस के बीच
(d) दर्पण के वक्रता केन्द्र पर
उत्तर:
(d) दर्पण के वक्रता केन्द्र पर

5. एक अवतल दर्पण की वक्रता त्रिज्या 12 सेमी है, तो इसकी फोकस दूरी होगी-
(a) 12 सेमी
(b) + 6 सेमी
(c) 24 सेमी
(d) 6 सेमी
उत्तर:
(d) 6 सेमी

6. किसी अवतल दर्पण के सामने वस्तु कहाँ रखी जाय कि प्रतिविम्ब आभासी, सीधा बड़ा बने?
(a) फोकस तथा दर्पण के बीच
(b) f दूरी तथा 2f के बीच
(c) 2f दूरी पर
(d) कहीं भी
उत्तर:
(a) फोकस तथा दर्पण के बीच

7. उत्तल दर्पण के सामने रखी वस्तु का प्रतिबिम्ब बनता है-
(a) वस्तु की स्थिति पर ही
(b) दर्पण के सामने वस्तु की स्थिति से दुगुनी दूरी पर
(c) दर्पण के सामने वस्तु की स्थिति से आधी दूरी पर
(d) दर्पण के पीछे
उत्तर:
(d) दर्पण के पीछे

8. एक उत्तल दर्पण की फोकस दूरी 12 सेमी है दर्पण के उत्तल पृष्ठ की त्रिज्या होगी-
(a) 6 सेमी
(b) 12 सेमी
(c) 18 सेमी
(d) 24 सेमी
उत्तर:
(d) 24 सेमी

9. इनमें से कौन-सा दर्पण वस्तु छोटा व आभासी प्रतिविम्ब बनाता है?
(a) समतल
(b) अवतल
(c) उत्तल
(d) उपर्युक्त तीनों
उत्तर:
(c) उत्तल

10. एक व्यक्ति दर्पण में अपना सीधा और बड़ा प्रतिबिम्ब देखता है। यह दर्पण है-
(a) उत्तल
(b) अवतल
(c) समतल
(d) उपर्युक्त तीनों
उत्तर:
(b) अवतल

11. उत्तल दर्पण से बना प्रतिबिम्ब होता है सदैव-
(a) वस्तु से छोटा
(b) वस्तु से बड़ा
(c) समान आकार का
(d) वास्तविक
उत्तर:
(a) वस्तु से छोटा

12. किसका दृष्टि क्षेत्र सबसे अधिक होता है?
(a) समतल दर्पण
(b) उत्तल दर्पण
(c) अवतल दर्पण
(d) उत्तल लेंस
उत्तर:
(c) अवतल दर्पण

13. यदि किसी वस्तु को दर्पण के निकट रखने पर सीधा प्रतिबिम्ब बने तथा दूर रखने पर वास्तविक प्रतिबिम्ब बने तो वह दर्पण होगा-
(a) अवतल
(b) उत्तल
(c) समतल
(d) उत्तल अथवा समतल
उत्तर:
(a) अवतल

14. किसी अवतल दर्पण द्वारा आभासी, सीधा तथा आवर्धित प्रतिबिम्ब बनता है। वस्तु की स्थिति होगी-
(a) ध्रुव व फोकस के बीच
(b) फोकस तथा वक्रता केन्द्र
(c) वक्रता केन्द्र पर
(d) वक्रता केन्द्र से पीछे
उत्तर:
(b) फोकस तथा वक्रता केन्द्र

15. एक अवतल दर्पण की फोकस दूरी 10 सेमी है। दर्पण की वक्रता त्रिज्या होगी?
(a) 10 सेमी
(b) 20 सेमी
(c) 30 सेमी
(d) 40 सेमी
उत्तर:
(b) 20 सेमी

16. किसी अवतल दर्पण की फोकस दूरी 15 सेमी है। उसकी वक्रता त्रिज्या होगी-
(a) 15 सेमी
(b) 30 सेमी
(c) 45 सेमी
(d) 60 सेमी
उत्तर:
(b) 30 सेमी

17. समतल दर्पण की फोकस दूरी होती है-
(a) शून्य
(b) अनन्त
(c) 25 सेमी
(d) -25 सेमी
उत्तर:
(b) अनन्त

18. फोकस दूरी व वक्रता त्रिज्या के बीच सम्बन्ध है-
(a) f = r
(b) f = \(\frac { 1 }{ r }\)
(c) 2f = r
(d) f = 2r
उत्तर:
(c) 2f = r

19. एक अवतल दर्पण की वक्रता त्रिज्या 15 सेमी है। इसकी फोकस दूरी होगी-
(a) 15 सेमी
(b) 7.5 सेमी
(c) + 30 सेमी
(d) 7.5 सेमी
उत्तर:
(b) 7.5 सेमी

20. समतल दर्पण के सामने एक वस्तु दर्पण से 10 सेमी की दूरी पर रखी गयी है। दर्पण से प्रतिबिम्ब की दूरी होगी-
(a) 5 सेमी
(b) 10 सेमी
(c) 20 सेमी
(d) 0 सेमी
उत्तर:
(b) 10 सेमी

21. उत्तल दर्पण से प्रतिबिम्ब सदैव बनता है-
(a) वक्रता केन्द्र तथा फोकस के बीच
(b) वक्रता केन्द्र तथा अनन्त के बीच
(c) ध्रुव व फोकस के बीच
(d) कहीं भी बन सकता है।
उत्तर:
(a) वक्रता केन्द्र तथा फोकस के बीच

22. यदि आपतन कोण 42° तथा अपवर्तन कोण 30° हो तब अपवर्तित किरण विचलित होती है-
(a) 12°
(b) 72°
(c) 1.4°
(d) \(\frac { sin 45° }{ sin 30° }\)
उत्तर:
(a) 12°

23. वायु के सापेक्ष काँच का अपवर्तनांक 3/2 है। वायु प्रकाश की चाल 3 x 10⁸ मी/सेकण्ड है। काँच में प्रकाश की चाल होगी-
(a) 1.5 x 10⁸ मी/सेकण्ड
(b) 2 x 10⁸ मी/सेकण्ड
(c) 3 x 10⁸ मी/सेकण्ड
(d) 4.5 x 10⁸ मी/सेकण्ड
उत्तर:
(b) 2 x 10⁸ मी/सेकण्ड

24. जब एक श्वेत प्रकाश की किरण काँच से बने किसी प्रिज्म से गुजरती है, तो किस रंग की किरण के लिए सबसे अधिक विचलन होता है-
(a) पीला
(b) बैंगनी
(c) लाल
(d) हरा
उत्तर:
(b) बैंगनी

25. यदि वायु के सापेक्ष काँच का अपवर्तनांक 1.5 है तो काँच के सापेक्ष वायु का अपवर्तनांक होगा-
(a) 3.00
(b) 1.50
(c) 0.75
(d) 0.67
उत्तर:
(d) 0.67

रिक्त स्थानों की पूर्ति कीजिए

1. एक काँच की स्लैब की क्षमता …………………. होती है।
2. एक लेंस जो बीच में से मोटा और किनारों पर से पतला होता है …………………. लेंस कहलाता है।
3. एक मोटे लेंस की फोकस दूरी कम होती है। इसकी क्षमता …………………. होगी।
4. एक उत्तल लेंस की दक्षता का चिन्ह ………………….।
5. एक अवतल लेंस की क्षमता का चिन्ह …………………. होती है।

उत्तर:

1. शून्य
2. उत्तल
3. अधिक
4. धनात्मक होता है।
5. – 0.5 D
6. धनात्मक।