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Jharkhand Board JAC Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 11 Human Eye and Colourful World

Additional Questions and Answers

Question 1.
Give four points of difference between the following terms / quantities :
(1) Near-sightedness and Far-sightedness
Answer:

Near-sightedness	Far- sightedness
1. The eye lens does not become thin as required but remains thick.	1. The eye lens does not become thick as required but remains thin.
2. The light rays from objects at far distances are focused short of the retina. As a result, the distant objects cannot be seen clearly.	2. The light rays from objects nearby the eyes are focused behind the retina. As a result the nearby objects cannot be seen clearly.
3. The light rays from objects nearby eyes are focused on the retina. As a result, the nearby objects are seen clearly.	3. The light rays from distant objects are focused on the retina. As a result the distant objects are seen clearly.
4. This defect can be corrected by using concave lens of appropriate focal length.	4. This defect can be corrected by using convex lens of appropriate focal length.

(2) Near point and Far point
Answer:

Near point	Far point
1. The minimum distance at which the object can be seen clearly without contraction of eye lens is called the near point of an eye.	1. The farthest distance upto which the eye can see objects clearly is called far point of an eye.
2. For young adult with normal vision, this value is 25 cm.	2. For young adult with normal vision, this value is infinite.
3. For a person having defect of far-sightedness value of the near point is at a distance more than 25 cm.	3. For a person having defect of near-sightedness value of the far point is at a distance less than infinite distance.
4. For a person having defect of far-sightedness value of the near point is obtained at a distance of 25 cm using convex lens of appropriate focal length.	4. For a person having defect of near-sightedness value of the far point is obtained at distance of infinite distance using concave lens of appropriate focal length.

Question 2.
Give scientific reasons for the following statements:
(1) To rectify the defect of near-sightedness or myopia, concave lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of near-sightedness or myopia, does not become thin as per the requirement and so the rays after being refracted by the eye lens get focussed at a position short of retina. So distant objects cannot be seen clearly.

If a person having defect of near-sightedness, uses concave lens of suitable focal length, then light rays become slightly divergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the distant object may be seen clearly.

(2) To rectify the defect of far-sightedness or hypermetropia, convex lens of suitable focal length is used as corrective lens.
Answer:
The eye lens of a person having defect of far-sightedness as hypermetropia, does not become thick as per the requirement and so the rays after being refracted by the eye lens get focussed behind retina. So nearby objects cannot be seen clearly.

If a person having defect of far-sightedness, uses convex lens of suitable focal length, then light rays become slightly convergent. Hence, the image can be formed on retina.

Thus, if the image is formed on retina, then the nearby object may be seen clearly.

(3) A rainbow is visible in the sky only after rain shower.
Answer:
In rainy season there are many clouds in the sky having tiny water droplets. When the sunlight is incident on tiny water droplets, due to these tiny water droplets, refraction, dispersion, internal reflection and at the end again refraction of sunlight take place.

Due to which a band is created which contains seven colours in the sky which is known as rainbow.

In other seasons there are no clouds in the sky. Therefore there are no tiny water droplets s in the sky. Hence, no rainbow is formed in the sky in other seasons and so it is not visible in? the sky.

(4) The sunrise is experienced two minutes early and the sunset is experienced two minutes delayed.
Answer:
As the altitude (height) from the earth’s surface gradually increases, the earth’s atmosphere becomes rarer. Hence, the refractive index of the air decreases continuously.

So a ray of light coming from the Sun towards an observer continuously passes from an optically rarer to optically denser medium and bends towards the normal. Thus, its direction of propagation changes continuously.

The actual sunrise or sunset begins when the Sun reaches the horizon.
1. In figure 11.12, S₁ is the actual position of the Sun a little below the horizon.

2. In this case, the light rays coming from S₁, continuously get refracted in the earth’s atmosphere (atmospheric refraction) and reach the observer as shown in the figure.

3. The tangent drawn to the curved path of the ray at point P passes through S₂, above the horizon.

4. S₂ is the apparent position of the Sun.

5. Thus, during the sunrise, the Sun is seen even though it is little below the horizon. Similarly, during the sunset it is seen for sometime even though it is little below the horizon.

6. Taking the refractive index of air as 1.00029 the apparent angular shift in the position of the Sun is found to be approximately (\(\frac { 1 }{ 2 }\))°.
Now, for angular displacement of 180° of the Sun, time required is 12 hours. Hence for angular displacement of (\(\frac { 1 }{ 2 }\))° of the Sun, time required is
= \(\frac{\left(\frac{1}{2}\right)^{\circ} \times 12 \text { hours }}{180^{\circ}}\)
= 0.03333 hour
= 1.9998 minutes
≈ 2 minutes
Due to advance sunrise and delayed sunset the duration of a day increases by about 4 minutes.

(5) The clear sky appears blue in colour.
Answer:

The molecules of air and other fine particles in the atmosphere are smaller in size than the wavelength of light in the visible region.

• The wavelength of red light is about 1.8 times that of blue light.
• When the sunlight passes through the atmosphere, the fine particles in the air scatter blue light more strongly than red light.
• At this time if the observer looks upwards, he finds sky blue.
• If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark.

Important Note:
The sky appears dark to passengers flying at very high altitudes, as scattering of light is not prominent at such heights.

(6) The danger signal lights are red in colours.
Answer:
The red light gets scattered least by fog or smoke because of its longer wavelength relative to light of any other colour. So, it can be seen even from a long distance. Therefore it is used in signals showing danger.

(7) The sun appears reddish at sunrise and sunset.
Answer:

In figure 11.14, the situation at the sunrise is shown.
Here, the white light coming from the Sun near the horizon, passes through thick layers of air and covers larger distance in the earth’s atmosphere before reaching the observer. During this, more scattering of blue light and shorter wavelengths take place. Hence, the reddish light reaches the observer and the Sun appears reddish. The same thing occurs at the sunset.
[Note : Rising and setting of the full moon from the horizon appears reddish due to this reason.]

Objective Questions and Answers

Question 1.
Answer the following questions in one word / sentence :

Question 1.
What is dispersion of white light?
Answer:
The phenomenon of a splitting of white light into its constituent colours is called the dispersion of white light.

Question 2.
What happens to the image-distance in the normal eye, when we increase the distance of an object from the eye?
Answer:
The image distance always remains constant.

Question 3.
What can be said about the focal length of the eye lens if its curvature increases?
Answer:
decreases

Question 4.
What can be said about the curvature of the eye lens if it becomes thin?
Answer:
decreases

Question 5.
For normal eye vision what is the object-distance and image-distance when the object is placed at a near point? (Take the distance between the eye lens and the retina as 2.3cm.)
Answer:
u = – 25 cm, v = + 2.3 cm

Question 6.
For normal eye vision what is the object-distance and image-distance when the object? is placed at a far point? (Take the distance between the eye lens and the retina as 2.3 cm.)
Answer:
u = – 00, a = + 2.3 cm

Question 7.
State the type of image of an object formed s on the retina.
Answer:
real, inverted and diminished

Question 8.
Write the name of the most front part of human eye.
Answer:
cornea

Question 9.
State the function of the iris.
Answer:
Iris can control the amount of light entering into the eye as well as size of pupil can be controlled by it.

Question 10.
State the function of light sensitive cells present in retina.
Answer:
Light rays falling on the retina are converted into the electrical signals.

Question 11.
Write the function of optic nerves.
Answer:
The work of optic nerves is to sent electrical signals to the brain.

Question 12.
Write use of bifocal lens.
Answer:
To rectify/remove the eye-defect known as presbyopia.

Question 13.
How much duration in second increases per day due to early sunrise and delayed sunset?
Answer:
240s

Question 14.
Due to which effect does the smoke emitted by the combustion of the engine oil in a motorcycle sometimes appears blue in colour?
Answer:
The Tyndall effect

Question 15.
Which effect is developed commercially to determine the size and density of aerosol and other colloidal particles?
Answer:
The Tyndall effect

Question 16.
Wavelength of red colour is approximately how many times the wavelength of violet colour?
Answer:
1.8

Question 2.
Fill in the blanks :

1. The type of image formed by the eye lens is ………………. and ……………….
2. A triangular glass prism has ………………. triangular bases and ………………. rectangular surfaces.
3. Light enters our eye through the ……………….
4. A person suffering from far-sightedness or hypermetropia cannot see clearly ………………. objects.
5. A ………………. corrective lens is used to rectify near-sightedness.
6. An old person suffering from near-sightedness and a far-sightedness uses ………………. to rectify his vision.
7. While passing through the prism the light ray travelling from air to glass bends towards the ……………….
8. In a glass prism ………………. light propagates with maximum speed.
9. At night stars are seen slightly at a higher position than their actual position because of the ……………….
10. For a light ray passing through the prism, the angle between the incident ray and the emergent ray is known as the ……………….
11. At the time of sunrise the sun appears ………………. in colour.
12. The fine particles in air scatter ………………. light more strongly.
13. ………………. light, while passing through a prism, does not disperse.
14. Stars behave like ………………. sources of light and planets behave like ………………. sources of light.
15. Danger signals are red in colour because red light is ……………….
16. In the spectrum of white light, ………………. and ………………. colours seen at the two ends.
17. The diameter of the human eyeball is approximately ………………. cm.
18. The distance between the eye lens and the retina is known as the ……………….
19. In the normal situation in the relaxed position of the ciliary muscles, the eye lens is ……………….
20. At night as we move up in the atmosphere of the earth, the refractive index ………………. continuously.

Answer:

1. real, inverted
2. two, three
3. cornea
4. nearby
5. concave / diverging
6. spectacles with bi-focal lenses
7. normal
8. red
9. atmospheric refraction
10. angle of deviation
11. reddish
12. blue
13. Monochromatic (It means light of single frequency, i.e., of single colour.
14. point, extended
15. scattered the least
16. violet, red
17. 2.3
18. size of the eyeball
19. thin
20. decreases

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

1. The near point of every person is 25 cm.
2. The splitting of white light into its constituent colours is called the scattering of light.
3. Far-sightedness can be rectified by using a concave lens of suitable power.
4. In the eye of myopic person, the image of a distant object is formed behind the retina.
5. Near-sightedness arises due to more curvature of the cornea or due to the eye lens remaining thick permanently.
6. The speed of light decreases as it passes from an optically denser medium to an optically rarer medium.
7. A myopic person has the far point nearer than infinity.
8. A hypermetropic person has near point farther away from the normal near point (25 cm).
9. The construction of the human eye can be compared with that of a camera.
10. A rainbow is formed due to refraction taking place twice, one internal reflection and dispersion of sunlight by water droplets in the sky.
11. Planets twinkle.
12. When the sunlight passes through a canopy of dense forest, tiny water droplets in the mist, scatter the light. This effect is known as the Tyndall effect.

Answer:

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

Question 4.
Match the following:
(1)

Column I	Column II	Column III
1. Myopia	p. The focal length of the eye lens increase	a. Bifocal lens
2. Hypermetropia	q. The focal length of the eye lens decrease	b. Concave lens
3. Presbyopia	r. The power of accommodation of the eye decrease with ageing	c. Convex lens

Answer:
(1 – q – b), (2 – p – c), (3 – r – a).

(2)

Column I	Column II
1. TWinkling of stars	p. Tiny water droplets present (or suspended) in the atmosphere
2. Blue coloured sky	q. Band of colours
3. Rainbow	r. Scattering of light
4. Spectrum	s. Uneven atmosphere

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

(3)

Column I	Column II
1. Human eye or eyeball	a. It controls and regulates the amount of light entering the eye.
2. Self operated accommodation power of eye	b. Delicate membrane having large number of light sensitive cells.
3. Retina	c. Works as a photographic camera.
4. Ciliary muscles	d. Not able to see the nearby objects.
5. Myopia	e. Electrical signals related to image are sent to the brain.
6. Cataract	f. A circular muscular diaphragm which can control the size of pupil.
7. Presbyopia	g. Milky and cloudy layer is formed on the eye lens.
8. Iris	h. The capacity of eye to see objects clearly between 25 cm and infinite distance.
9. Pupil	i. Increases or decreases curvature of eye lens.
10. Optic nerves	j. Image of object at infinite distance is formed in front of retina.
11. Hypermetropia	k. Accommodation power of eye decreases with ageing.

Answer:
(1 – c), (2 – h), (3 – b), (4 – i), (5 – j), (6 – g), (7 – k), (8 – f), (9 – a), (10 – e), (11 – d).

Question 5.
Choose the correct option from those given below each question:
1. Splitting of white light into its seven constituent colours is called ……………
A. refraction
B. reflection
C. dispersion
D. interference
Answer:
C. dispersion

2. Which colour of light deviates maximum in the dispersion of white light by a prism?
A. Violet
B. Blue
C. Green
D. Red
Answer:
A. Violet

3. In the human eye, the image of an object is formed at the ……………….
A. iris
B. pupil
C. retina
D. cornea
Answer:
C. retina

4. The focal length of the eye lens is changed due to the action of the ……………….
A. pupil
B. retina
C. ciliary muscles
D. cornea
Answer:
C. ciliary muscles

5. A ………………. lens is used to correct presbyopia.
A. convex
B. concave
C. bi-focal
D. contact
Answer:
C. bi-focal

6. Out of the following, which phenomenon does not play a role in the formation of a rainbow?
A. Reflection
B. Refraction
C. Dispersion
D. Absorption
Answer:
D. Absorption

7. Where is the image formed in the eye of a person suffering from near-sightedness?
A. On the retina
B. Behind the retina
C. In front of the retina
D. On the pupil
Answer:
C. In front of the retina

8. Which phenomenon is responsible for the twinkling of stars?
A. Atmospheric reflection
B. Atmospheric refraction
C. Reflection
D. Total internal reflection
Answer:
B. Atmospheric refraction

9. The phenomenon of ………………. of light by the colloidal particles gives rise to the Tyndall effect.
A. reflection
B. refraction
C. scattering
D. dispersion
Answer:
C. scattering

10. What is the time difference between actual sunset and apparent sunset?
A. 2 seconds
B. 20 seconds
C. 2 minutes
D. 20 minutes
Answer:
C. 2 minutes

11. Which light gets scattered maximum due to atmosphere?
A. Blue
B. Yellow
C. Green
D. Red
Answer:
A. Blue

12. Which light has minimum speed in glass (prism)?
A. Red
B. Green
C. Blue
D. Violet
Answer:
D. Violet

13. When an eye is focussed on a distant object, the focal length of the eye lens is ……………….
A. maximum
B. minimum
C. half of its minimum
D. half of its maximum
Answer:
A. maximum

14. How many surfaces does a triangular prism have?
A. 3
B. 4
C. 5
D. 6
Answer:
C. 5

15. The wavelengths of violet, yellow and red light are λ_v, λ_y and λ_r respectively, then ……………………..
A. λ_v > λ_y > λ_r
B. λ_v < λ_y < λ_r
C. λ_y < λ_v < λ_r
D. λ_y < λ_r < λ_v
Answer:
B. λ_v < λ_y < λ_r

16. For normal vision, the far point is at ……………… distance.
A. 25 cm
B. 1 cm
C. 1 m
D. infinite
Answer:
D. infinite

17. For normal vision, the near point is ………………
A. 25 cm
B. 25 m
C. zero
D. infinite
Answer:

18. Which phenomenon can explain the advance sunrise and the delayed sunset?
A. Dispersion of light
B. Scattering of light
C. Tyndall effect
D. Atmospheric refraction
Answer:
D. Atmospheric refraction

19. Which of the following phenomena cannot be explained by scattering of light?
A. The red light used for signal lights for danger.
B. Blue colour of clear sky
C. White colour of clouds
D. Early sunrise
Answer:
D. Early sunrise

20. The base of an equilateral triangle ABC is BC. When it is arranged in four different situations and white light is incident on it, then in which of the following arrangements of the prism, the third colour from the top is the colour of clear sky in dispersion of light is produced?

A. (i)
B. (ii)
C. (iii)
D. (iv)
Answer:
B. (ii)
Hint: The dispersion of white light is shown in arrangement (ii) of prism

The third colour of light from the top is blue which is the colour of the clear sky.

21. The Sun appears white in afternoon. The reason is …
A. less scattering of light.
B. more scattering of all the colours of white light.
C. more scattering of blue colour.
D. more scattering of red colour.
Answer:
A. less scattering of light.
Hint: White light coming from the Sun has to travel less distance in the atmosphere before reaching the observer. So less scattering of light takes place and as a result the Sun appears white.

22. Sea water at more depth appears blue. The reason is …
A. presence of some plants in sea water.
B. the image of the sky appears in water.
C. scattering of light.
D. light is absorbed by sea water.
Answer:
C. scattering of light.

23. When the ciliary muscles are relaxed, the eye lens becomes …………….. and its focal length ……………… This enables us to see distant objects clearly.
A. thin, increases
B. thin, decreases
C. thick, increases
D. thick, decreases
Answer:
A. thin, increases

24. When the ciliary muscles contract, the eye lens becomes …………….. and its focal length ……………..
This enables us to see nearby objects clearly.
A. thick, decreases
B. thick, increases
C. thin, increases
D. thin, decreases
Answer:
A. thick, decreases

25. The rainbow on the moon ……………..
A. is not possible.
B. is rare.
C. is observed with the reverse order of colours.
D. is of two types.
Answer:
A. is not possible.

26. In dispersion of white light due to a triangular glass prism, the deviation of red colour is less compared to violet colour. The reason ………………
A. is n_v > n_r.
B. is n_r > n_v.
C. is n_v = n_r.
D. does not depend on n.
Answer:
A. is n_v > n_r.
Hint: In glass medium, speed of violet colour is less as compared to speed of red colour. So, n_v > n_r from n = \(\frac { c }{ v }\)

27. Which lens from the following, should a person suffering from near-sightedness use?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
B. A concave lens

28. Which lens is used by a person suffering from far-sightedness?
A. A convex lens
B. A concave lens
C. A cylindrical lens
D. A bi-focal lens
Answer:
A. A convex lens

29. Which of the following is true for near-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thick as required.
D. This defect can be rectified using spectacles of convex lenses.
Answer:
B. Distant objects cannot be seen clearly.

30. Which of the following is true for far-sightedness?
A. Nearby objects cannot be seen clearly.
B. Distant objects cannot be seen clearly.
C. The eye lens cannot become thin as required.
D. This defect can be rectified using spectacles of concave lenses.
Answer:
A. Nearby objects cannot be seen clearly.

31. Where is the image formed in the eye of a person suffering from far-sightedness?
A. On the retina
B. Behind the retina
C. On the pupil
D. In front of the retina
Answer:
B. Behind the retina

32. A person has a defect of eye vision. His near point is 40 cm. It means …
A. he cannot clearly see objects at a distance more than 40 cm from the eye.
B. he can clearly see objects at a distance of 40 cm only.
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
D. he can clearly see objects at a distance less than 40 cm e.g., 25 cm from the eye.
Answer:
C. he can clearly see objects at a distance equal to 40 cm or more from the eye.
Hint: Here the person suffers from far-sightedness.

33. A person has a defect of vision. His far point is 1.5m. It means …
A. he cannot clearly see objects at a distance more than 1.5 m from the eye.
B. he can clearly see objects at a distance more than 1.5 m from the eye.
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
D. he suffers from far-sightedness.
Answer:
C. he cannot clearly see objects at a distance less than 1.5 m from the eye.
Hint: Here the person suffers from near-sightedness.

34. Out of the following, which light is deviated minimum in the dispersion of white light through a glass prism?
A. Green
B. Violet
C. Yellow
D. Dispersion of the given three colours is the same.
Answer:
C. Yellow

35. Which light has maximum speed in glass?
A. Violet
B. Blue
C. Green
D. Red
Answer:
D. Red

36. Which ray of light is present exactly at the middle of the spectrum obtained from white light?
A. Green
B. Yellow
C. Red
D. Violet
Answer:
A. Green

37. The lens in human eye is a …………….
A. convex mirror
B. convex lens
C. concave mirror
D. concave lens
Answer:
B. convex lens

38. For persons suffering from near-sightedness the power of the lens used in spectacles is …………….
A. positive
B. zero
C. negative
D. infinite
Answer:
C. negative

39. For a person suffering from ……………. the power of the lens used in spectacles is positive.
A. far-sightedness
B. near-sightedness
C. presbyopia
D. cataract
Answer:
A. far-sightedness

40. Which phenomenon/phenomena of light is/ are involved in the formation of a rainbow?
A. Refraction
B. Dispersion
C. Internal reflection
D. All of the given above
Answer:
D. All of the given above

41. ……………. light from the following is least scattered by fog, dust and smoke.
A. Violet
B. Blue
C. Red
D. Yellow
Answer:
C. Red

42. Which of the following controls the amount of light entering into the human eye?
A. Ciliary muscles
B. Pupil
C. Cornea
D. Iris
Answer:
D. Iris

43. The refractive index of glass is maximum for ……………. light.
A. violet
B. green
C. blue
D. red
Answer:
A. violet

Question 6.
Answer the following questions in very short as directed (Miscellaneous) :
(1) What is the power of a lens that can be used to correct the eye defect of a person who cannot see the objects distinctly kept beyond 2m?
Answer:
0.5 D
(∵ For a myopic eye the focal length of the corrective lens is equal to the far point of the myopic person, i.e., f = – 2m).

(2) Why does the Sun appear white at noon?
Answer:
The Sun appears white at noon because white light (sunlight) is least scattered by the atmosphere.

(3) Why is the eye lens not perfectly solid?
Answer:
If it becomes solid, its focal length would be fixed. Then, we would not be able to focus the objects lying at different distances on the retina. In short, accommodation of the eye would be reduced to zero.

(4) What is the focal length of plain goggles?
Answer:
Infinity

(5) What happens when elasticity of the crystalline lens is reduced to zero?
Answer:
Power of accommodation would be almost zero.

(6) Which defect of the eye occurs due to distortion of cornea?
Answer:
Astigmatism occurs due to the distortion of cornea.

(7) How the defect of astigmatism can be corrected?
Answer:
Astigmatism is corrected by the use of a cylindrical lens.

(8) What is colour blindness? How can it be cured?
Answer:
It is a defect of the eye in which a person is unable to distinguish between certain colours due to insufficient or no cone shaped cells on retina. It cannot be cured.

(9) In hypermetropia, how does the size of eyeball change?
Answer:
In hypermetropia, the eyeball becomes too small (flat).

(10) What change occurs in the focal length, when our eye lens becomes thick?
Answer:
The focal length will decrease when the eye lens becomes thick.

(11) What are rods and cones?
Answer:
Sensitive portion of retina has large number of cells, rod shaped and cone shaped cells. Rod shaped cells are sensitive to the intensity or brightness of the light whereas cone shaped cells are sensitive to colours.

(12) What is cataract?
Answer:
Sometimes, the eye lens of a person becomes hazy or even opaque resulting in reduced or total loss of vision. This is called cataract.

(13) What would have been the colour of the sky, had there been no atmosphere?
Answer:
Black

(14) Due to which phenomenon is the colour of water in deep sea blue?
Answer:
Due to scattering of light.

(15) What is the cause for presbyopia?
Answer:
The eye lens becomes less elastic or non-elastic after 40 years as the, accommodation power of eye becomes less.

(16) Give the relationship between wavelength of light and its angle of deviation, when it is passed through a prism.
Answer:
Wavelength λ = \(\frac{1}{\text { Angles of deviation } \delta}\)

(17) For which colour has the glass larger refractive index – violet or green?
Answer:
Violet

(18) Which part of human eye is also known as ‘white of the eye’?
Answer:
Sclera

(19) Why is blind spot so called?
Answer:
Blind spot is a point at which the optic nerve leaves the eye. It contains no rods or cones. So an image formed at this point, is not sent to the brain.

(20) Which liquid is filled in the space between the eye lens and the retina?
Answer:
Vitreous humour (It is a transparent jelly.)

(21) What happens to the pupil of the eye when the light is (a) very bright and (b) very dim?
Answer:
(a) In very bright light, the size of the pupil becomes very small, so that less amount of light enters into the eye.
(b) In very dim light, the size of the pupil increases, so that more amount of light enters into the eye.

(22) A man is wearing spectacles of focal length +1 m. What can be the defect in the eye?
Answer:
As the focal length of spectacles is positive, the man is using convex lenses. So, he is suffering from hypermetropia.

(23) Which portion of a bi-focal lens is
(a) a concave lens
(b) a convex lens?
Answer:
(a) Upper portion
(b) Lower portion.

(24) When sunlight enters into a room filled with dark smoke, its path becomes visible. Name the phenomenon responsible for this.
Answer:
The Tyndall effect.

(25) What is the function of the iris?
Answer:
The iris controls the size of the pupil.

(26) What are light sensitive cells?
Answer:
Rods and cones

(27) What type of signals are generated and sent to the brain by light sensitive cells in the retina?
Answer:
Electrical signals.

(28) What holds the crystalline lens in the human eye?
Answer:
Ciliary muscles.

(29) Which part of the human eye helps in changing the thickness of the eye lens?
Answer:
Ciliary muscles.

(30) What is dispersion of white light?
Answer:
The splitting of white light into its various components (i.e., 7 colours) is called dispersion of white light.

(31) Give the main difference between the lens of the human eye and the lens of a camera.
Answer:
The lens of the human eye has flexible aperture, so that its focal length can be changed, while in a camera the focal length of the lens is fixed.

(32) The image formed on the retina is inverted, but we see the object erect. Why?
Answer:
The inverted image formed on the light sensitive cells (called rods and cones of retina), generates electrical signals. These signals reach the brain via the optic nerve. It is the brain? that interprets this image and while processing the image it helps in perceiving the objects as they are.

(33) The absolute refractive index of a medium is 2.0. The speed of light in vacuum/air is 3 x 10⁸ms^-1. Find the speed of light in the medium.
Answer:
Absolute refractive index of medium is,
n = \(\frac { c }{ v }\)
∴ v = \(\frac { c }{ n }\)
= \(\frac{3 \times 10^8}{2}\)
= 1.5 x 10⁸ms^-1

(34) Match the column properly :

Column I (eye defect)	Column II (correcting lens)
1. Myopia	p. Bi-focal lens
2. Hypermetropia	q. Concave lens
	r. Convex lens

Answer:
(1 – q), (2 – r).

(35) Match the following column :

Column I (eye defect)	Column II (correcting lens)
1. Astigmatism	p. Convex lens
2. Presbyopia	q. Cylindrical lens
	r. Concave lens

Answer:
(1 – q), (2 – p).

(36) The far point of a myopic eye is 100 cm. What is the focal length of the lens required to see very distant (normal far point) objects clearly?
Answer:
As the focal length of concave lens used for correcting the myopic eye is equal to distance of far point of the myopic eye, the focal length of correcting lens required is f = – x = – 100 cm = – 1 m.

(37) The near point of a hypermetropic eye is 75 cm. What is the focal length of the lens required to see clearly an object placed at 25 cm from the eye (normal near point)?
Answer:
The focal length of convex lens used for correcting the hypermetropic eye is given by
(where x’ = defective near point = 75 cm and d = normal near point = 25 cm)
f = \(\frac{x^{\prime} d}{x^{\prime}-d}\)
∴ f = \(\frac{75 \times 25}{75-25}\)
= 37.5 cm
= 0.375 m

(38) The eye lens of human eye is a double convex lens. Agree or Disagree?
Answer:
Agree.

(39) Cone-shaped retinal cells respond to the brightness or intensity of light. Agree or Disagree?
Answer:
Disagree.

(40) Which property of vision is used in cinematography?
Answer:
Property of persistence of vision is used in cinematography.

(41) What is aqueous humour?
Answer:
Aqueous humour is a transparent viscous liquid in the space between the cornea and the eye lens.

(42) What is the maximum power of accommodation of a normal eye?
Answer:
The maximum power of accommodation of a normal eye
= \(\frac{1}{\text { near point of the normal eye (in metre) }}\)
= \(\frac { 1 }{ 0.25 m }\)
= \(\frac { 100 }{ 25 m}\)
= 4 m^-1
= 4 D

(43) What is meant by scattering of light?
Answer:
It is the phenomenon of change in the direction of light on striking a scatterer.

(44) What is the basic cause of atmospheric refraction?
Answer:
The basic cause of atmospheric refraction is variation in the refractive index (i.e., optical density) of different layers of the earth’s atmosphere with altitude.

(45)

In the above figure a narrow beam of white light is shown to pass through a triangular glass prism. After passing through the prism, it produces a spectrum XY on a screen. State the colour seen at (i) X and (ii) Y.
Answer:
The colour seen at X is violet and that seen at Y is red.

(46)

In the above figure, which angles are correctly marked?
Answer:
∠A and ∠e are correctly marked.

(47)

In the above figure (ray diagram), state the angle of incidence and the angle of deviation.
Answer:
∠PQN = i = Angle of incidence
∠P’OR = D = Angle of deviation

(48)

In the above figure (ray diagram), state angle of incidence, angle of emergence and angle of deviation.
Answer:
∠p = Angle of incidence
∠y = Angle of emergence
∠z = Angle of deviation

(49) What is the principle of the working of s the human eye?
Answer:
It is like a camera having a lens system forming an inverted, real image on the light |> sensitive screen (retina) inside the eye.

(50) On which factor does the colour of the scattered white light depend?
Answer:
Size of the particles of the medium : through which it is passing.

(51) Give the scientific names of the following parts of the eye :
(a) Carrying signals from an eye to the brain.
(b) A small opening (hole) in the middle of the iris.
Answer:
(a) Optic nerve
(b) Pupil

(52) A near-sighted person has a near point 25 cm and a far point 50 cm. Can he see clearly an object at a distance of: (i) 5 cm, (ii) 25 cm, (iii) 60 cm. Write ‘Yes’ or ‘No’ only.
Answer:

1. No
2. Yes
3. No

(53) The near point of a far-sighted person is 50 cm.
Can the person see clearly an object at a distance of:
(i) 20 cm
(ii) ∞ (infinity)
Answer:
(i) No
(ii) Yes

(54) How much is our horizontal field of view (a) with one eye open
(b) with both eyes open?
Answer:
(a) about 150°
(b) about 180°.

(55) Which of the following have a wider s field of view?
(a) Animals having two eyes on the opposite 5 sides of their head or
(b) Animals having two eyes at the front of their head.
Answer:
(a) Animals having two eyes on the opposite 5 sides of their head or

Value Based Questions With Answers

Question 1.
Maull and Vlshva are best friends and they study in 4th grade. Recently, Mauli has been facing difficulty in reading the blackboard text from the last desk/bench. Vishva wonders why Mauli avoids sitting on the last desk/bench. On observation, Vishva found that Mauli often carries junk food in her lunch. Vishva has started sharing her lunch full of green vegetables and fruits with her. Mauli is now better and has also started taking a “balanced diet”.

1. Name the eye defect Mauli is suffering from.
2. What are two possible deformities related s to her eye defect?
3. What values do you learn from Vishva and Mauli?

Answer:

1. Myopia (near-sightedness)
2. Lens defect (increased thinness i.e., excessive curvature of the eye lens) and Eyeball defect (elongation of the eyeball).
3. Friendship, concern for each other, s importance of a balanced diet.

Question 2.
An eye camp was organised by the doctors in a village. They found that the eyes of imaged people in the village have the near point receded and the far point also gets reduced. Often aged people suffer from both myopia s and hypermetropia. Doctors (opthalmologists) provide these people spectacles of bi-focal lenses to correct the defects. The people were happy and grateful to the doctors.
(1) Name the eye defect from which the people were suffering.
(2) Give any two causes of this defect.
(3) What were benefits to organise such camps > in rural areas? Give two suggestions.
Answer:
(1) Presbyopia

(2) (a) Weakening of ciliary muscles
(b) Reducing ability of the lens to change the curvature.

(3) (a) To make people aware of eye diseases
(b) To tell people to take proper and balanced diet.

Question 3.
Four friends went to a picnic. The weather was pleasant. They played various games and then had snacks. Suddenly, Raju, one of them, observed seven colours in the sky. He said to others, “wow what a rainbow” !

Ram, one of them, asked him “What is a rainbow?” Raju then explained to all about its formation. After that everyone in the group thanked Raju for the knowledge, he had given to them.
(1) When Raju was facing the rainbow, where was the Sun?
(2) Which device can be used to obtain such a phenomenon?
(3) What moral value do you learn from Raju?
Answer:
(1) The Sun was behind Raju.
(2) A small prism. (Water droplets present in the atmosphere act like small prisms.)
(3) Knowledge increases by sharing, friendship, love and affection with nature.

Question 4.
In a beautiful valley, there was a village. When trains passed from the village, the whistle and the sound of train, mixed with the sound of waterfall, seemed to be very pleasant to everyone. Hence, children of that village used to play near the railway track. Once on a very light foggy day, a group of children found that a fish plate was missing from the track. As such, all the villagers were worried.

Prashant, one of the children, suddenly put his ear to the line and tried to know whether a train is coming or not. He knew a train is coming. He asked his friends to inform the railway cabin crew and he himself put off his red shirt and started running s towards the train, waving his red shirt. The driver and cabin man got the alert signal in time and thus a major accident was averted.?
(1) Name the two physical phenomena of science used by Prashant.
(2) Why did Prashant use his red shirt instead of any other coloured shirts?
(3 ) What moral values do you learn from Prashant?
Answer:
(1) (a) Sound travels through a medium.
(b) Scattering of light

(2) The red light is least scattered by fog or smoke so it can be seen from a large distance.

(3) (a) Proper knowledge and its application.
(b) Concern for others.

Question 5.
Millions of people of the developing countries of the world are suffering from corneal blindness. They can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact.
(1) State the objective of organising such campaigns.
(2) Write one argument which you would give s to motivate the people to donate their eyes after death.
(3) List two values which could be developed in the persons who actively participate and < contribute in such programme.
Answer:
(1) The objective of organising such campaign is to help those people who are suffering from corneal blindness and they can be cured by replacing their defective cornea with the cornea of a donated eye.

(2) Come forward to participate in this compaign because, if someone gets his vision through your eyes, it is an incredible help as the eye is one of the most valuable sense organs through which an individual can achieve so many things in his / her life.

(3) The persons who actively participate and contribute in such programme are

• strong hearted
• very much helpful for the people living in such situations.

Question 6.
Mr Bharat’s 65 year old mother was complaining about blurred vision in both the eyes due to which she could not see things clearly. Mr Bharat took his mother to an eye hospital. The doctor examined her eyes carefully and concluded that she has a medical condition which could not be corrected by using any type of spectacle lenses and it required surgery. Her eyes were operated upon and she could then see once again properly.
(1) What could be the defect in the eyes of Mr Bharat’s mother?
(2) What happens to the eye lens during this defect? What is done during surgical operation of the eyes to restore the correct vision?
(3) What values do you learn from Mr Bharat in this episode?
Answer:
(1) The defect in the eyes of Mr Bharat’s mother is known as cataract.

(2) During the development of cataract, a membrane is gradually formed over both the eye lenses making the eye lenses cloudy. This makes the vision blurred. During surgical operation, the cloudy eye lenses are removed from the eyes and suitable artificial lenses are inserted in their place.

(3)

• Awareness and knowledge about the eye defects that can be cured by eye specialist doctors.
• Desire to mitigate the suffering of others (here mother).
• Sense of responsibility.

Question 7.
Amit is a domestic help (or maid) working at ; Mr Dave’s house. One day Amit complained to Mr Dave that he had difficulty in reading the letter which he had received from his s parents. Mr Dave, realising that Amit had an eye defect, took him to an eye specialist doctor.

The doctor tested his eyes carefully and told Amit to wear spectacles containing certain type of lenses having specified power, Mr Dave bought the required spectacles for Amit. By wearing this spectacle, Amit could read and write easily. He was very happy and thanked Mr Dave.
(1) What could be the eye defect Amit was suffering from?
(2) What could be the two possible reasons responsible for his eye defect? What type of lenses do you think the doctor recommended for Amit’s spectacles?
(3) What values are displayed by Mr Dave in this episode??
Answer:
(1) Amit was suffering from an eye defect called hypermetropia (far-sightedness) in ; which a person cannot see the nearby objects clearly though he can see the distant objects clearly.

(2) (a) Low converging power of the eye lens (because of the eye lens being less convex or less thick)
(b) Eyeball being too small (flat) (because of which the distance of the retina from the eye lens is less than normal).

The doctor recommended convex lenses for the spectacles of Amit.

(3) Mr Dave displayed the values of –

• awareness, which means having s knowledge of a situation or facts.
• concern for others (to mitigate their suffering).
• kindness and generosity.

Question 8.
Rohit is a car driver working for Mr Joshi. One day Rohit complained that he had difficulty in driving the car because he could not see the distant traffic clearly though he could see the nearby objects clearly. Mr Joshi took Rohit to an eye hospital. The eye specialist doctor checked and tested his eyes with various machines and gave him the name and power of the lenses to be worn as spectacles.

Mr Joshi paid for the required spectacles for the driver. By wearing these spectacles, the driver could now see even the distant vehicles and people on the road clearly. He thanked Mr Joshi for this.
(1) Name the eye defect Rohit is suffering from.

(2) What could be the two possible reasons for his eye defect? What type of lenses do you think the doctor recommended for Rohit’s spectacles?

(3) What values (or qualities) do you learn from Mr Joshi in this episode?
Answer:
(1) Myopia (near-sightedness) in which a person cannot see distant objects clearly though he can see nearby objects clearly.

(2) (a) High converging power of the eye lens (because of the eye lens being too convex or too thick).
(b) Eyeball may be too long (elongated) (because of which the distance of the retina from the eye lens is more than normal).
The doctor recommended concave lenses for the spectacles of Rohit.

(3) (a) General awareness (that an eye defect can usually be corrected by wearing spectacles containing suitable lenses).
(b) Concern for others (because Mr Joshi wanted to mitigate or remove the suffering of driver).
(c) Kindness and generosity.

Practical Skill Based Questions With Answers

Question 1.
Dispersion is caused by refraction and not by reflection. Why?
Answer:
White light is composed of seven colours having different wavelengths. And speed of different colours is the same in vacuum / air while in different media it is different.

Now, for a given angle of incidence, the angle of reflection is the same for all the wavelengths of white light, while the angle of refraction is different for different wavelengths.

Question 2.
A beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the diagram.

(1) The colours at positions marked 3 and 5 are similar to the colour of the sky and the colour of gold (metal), respectively. Is the above statement made by a student correct or incorrect. Justify.
(2) Which of the positions shown above correspond approximately to the colour of (a) a brinjal, (b) danger signal, (c) neel (applied to clothes), (d) orange.
Answer:
(1) No, because 3 refers to yellow and 5 to blue colour of the spectrum.

(2) (a) 7 (b) 1 (c) 6 (d) 2.

Question 3.
When a beam of white light is passed through a triangular glass prism, it gets dispersed into its component colours. Why do we get these colours? In the given figure, the colours X and Y represent the extreme components of the spectrum. Identify X and Y.

Answer:
(1) Different colours of light while passing through a prism bend through different angles with respect to the incident ray, as they travel with different speeds, this leads to dispersion of light.

(2) X – violet, Y – red

Question 4.
A narrow beam PQ of white light passes through a glass prism ABC as shown in the diagram.

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.
(1) Write the name and cause of the phenomenon observed.
(2) Where else in nature is this phenomenon observed?
(3) Based on this observation, state the conclusion which can be drawn about the constituents of white light.
Answer:

(1) The phenomenon of splitting of white light into its constituent colours is called dispersion of light. It occurs because different constituent colours of light travel with different speeds in a material medium (other than air /vacuum) and hence bend through different angles.

(2) In nature, this phenomenon is observed in formation of a rainbow.

(3) Based on the phenomenon of dispersion, we can conclude that –

• White light consists of seven colours.
• Violet light suffers maximum deviation and red light suffers minimum deviation.

Question 5.
(a) A narrow beam of white light is incident on three glass objects as shown below. Comment on the nature of the behaviour of the emergent beam in all three cases.

(b) There is a similarity between two of the emergent beams. Identify the two.
Answer:
(a) (1) The incident beam of light after refraction through the glass slab emerges parallel to the incident beam but is laterally shifted. No dispersion takes place in this case.

(2) The incident beam of light after refraction through the prism splits into a band of seven colours which are violet, indigo, blue, green, yellow, orange and red. These coloured rays emerge out of the prism along different directions and become distinct. Thus, in this case, dispersion of white light takes place.

(3) When the incident beam passes through the first prism, the prism splits it into a band of seven colours. These coloured rays are then incident on an identical inverted prism.

Then the recombination of the coloured rays takes place. The emergent beam of light is parallel to the incident beam but slightly shifted outward.

(b) The emergent beam in the cases (1) and (3) are similar. In both the cases, the emergent beam is parallel to the incident beam and is laterally shifted.

Memory Map:

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 1 वास्तविक संख्याएँ Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 1 वास्तविक संख्याएँ

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
सिद्ध कीजिए कि प्रत्येक तीन क्रमागत धनात्मक पूर्णांकों में से एक 3 से विभाज्य है।
हल :
माना कि n, n + 1, n + 2 तीन क्रमागत धनात्मक पूर्णांक हैं।
∵ हम जानते हैं कि n, 3q या 3q + 1 या 3q + 2 के रूप का होता है।
∴ निम्नलिखित तीन स्थितियाँ सम्भव हैं।
स्थिति 1. जब n = 3q, जो कि 3 से विभाज्य है।
n + 1 = 3q + 1, 3 से विभाज्य नहीं है।
n + 2 = 3q + 2, 3 से विभाज्य नहीं है।
इस स्थिति में n, 3 से विभाज्य है परन्तु n + 1 और n + 2, 3 से विभाज्य नहीं हैं।
स्थिति 2. जब n = 3q + 1
इस स्थिति में n + 2 = 3q + 1 + 2 = 3(q + 1) जो कि 3 से विभाज्य है परन्तु तथा n + 1, 3 से विभाज्य नहीं है।
स्थिति 3. जब n = 3q + 2
इस स्थिति में n + 1 = 3q + 1 + 2 = 3(q + 1), 3 से
विभाज्य है परन्तु n या n + 2, 3 से विभाज्य नहीं है।
अत: n, n + 1 तथा n + 2 में से एक 3 से विभाज्य है।

प्रश्न 2.
सिद्ध कीजिए कि \(\sqrt{3}\) एक अपरिमेय संख्या
हल :
माना कि \(\sqrt{3}\) एक परिमेय संख्या है।
हम ऐसी सह अभाज्य (Co-prime) संख्याएँ a और b ज्ञात करते हैं कि
\(\sqrt{3}\) = \(\frac{a}{b}\) [जहाँ b ≠ 0]
दोनों पक्षों का वर्ग करने पर
3 = \(\frac{a^2}{b^2}\)
⇒ a² = 3b²
अतः 3, a² को विभाजित करता हैं
⇒ 3, a को विभाजित करेगा।
माना कि a = 3c (जहाँ c कोई पूर्णांक है)
⇒ a² = 9c²
⇒ 3b² = 9c² [∵ a² = 3b²]
⇒ b² = 3c²
अत: 3, b² को विभाजित करता है।
⇒ 3, b को विभाजित करेगा।
अतः a और b में कम से कम एक उभयनिष्ठ गुणनखण्ड 3 है।
परन्तु यह इस तथ्य का विरोध करता है कि a और b में 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है। अतः हमारी परिकल्पना गलत है।
अतः \(\sqrt{3}\) एक अपरिमेय संख्या है। इति सिद्धम् ।

प्रश्न 3.
संख्याओं 180, 72 व 252 का HCF और LCM ज्ञात कीजिए।
हल :
संख्याओं के गुणनखण्ड करने पर

प्रश्न 4.
अभाज्य गुणनखण्ड विधि द्वारा पूर्णांक 375 और 675 का HCF ज्ञात कीजिए।
हल :
अभाज्य गुणनखण्ड विधि द्वारा

375 = 3 × 5 × 5 × 5 = 3 × 5³
675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5²
अंतः म. स. 3 × 5² = 3 × 25 = 75

प्रश्न 5.
सिद्ध कीजिए \(\sqrt{6}\) एक अपरिमेय संख्या है।
हल :
माना \(\sqrt{6}\) एक परिमेय संख्या है।
∴ हम दो पूर्णांक r और s ऐसे ज्ञात कर सकते हैं कि हो तथा s ≠ 0 हो। यहाँ और सह अभाज्य हैं, अर्थात् इनमें 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है।
अब \(\sqrt{6}\) = \(\frac{r}{s}\)
दोनों पक्षों का वर्ग करने पर
6 = \(\frac{r^2}{s^2}\)
⇒ 6s² = r²
⇒ s² = \(\frac{r^2}{6}\)
⇒ 6, r² को विभाजित करता है।
⇒ इसलिए 6, r को भी विभाजित करता है।
⇒ 6, r का एक गुणनखण्ड है। ………..(A)
माना r = 6m
⇒ 6s² = (6m)²
⇒ 6s² = 36m²
⇒ s² = 6m²
⇒ m² = \(\frac{s^2}{6}\)
⇒ 6, s² को विभाजित करता है।
⇒ इसलिए 6, s को भी विभाजित करता है।
⇒ 6, s का एक गुणनखण्ड है। …(B)
समीकरण (A) और (B) अवलोकन करने पर हम पाते हैं कि r और s में 1 के अतिरिक्त एक अन्य उभयनिष्ठ गुणनखण्ड 6 है। परन्तु इससे इस तथ्य का विरोधाभास प्राप्त होता है कि r और s में 1 के अतिरिक्त कोई उभयनिष्ठ गुणनखण्ड नहीं है। इसका तात्पर्य है कि हमने एक त्रुटिपूर्ण कल्पना की थी।
अत: \(\sqrt{6}\) एक अपरिमेय संख्या है।

प्रश्न 6.
दो पूर्णांक संख्याओं का HCF व LCM क्रमशः 12 और 36 हैं, यदि एक पूर्णांक 48 है, तो दूसरा पूर्णांक ज्ञात कीजिए। (मा. शि. बो. राज. 2017 )
हल :
दिया है, HCF = 12, LCM = 336, एक पूर्णांक संख्या = 48
∴ HCF × LCM = पहली संख्या × दूसरी संख्या
⇒ 12 × 36 = 48 × दूसरी संख्या
अतः दूसरी संख्या = \(\frac{12 \times 36}{48}\) = 9

प्रश्न 7.
किसी परेड़ में 612 सदस्यों वाली एक सेना की टुकड़ी को 48 सदस्यों वाले एक बैंड के पीछे मार्च करना है। दोनों समूहों को समान संख्या वाले स्तम्भों में मार्च करना है। उन स्तम्भों की अधिकतम संख्या, जिसमें वे मार्च कर सकते हैं, क्या है?
हल :
स्तम्भों में अधिकतम संख्या ज्ञात करने के लिए में 612 और 48 का म.स. ज्ञात करना है।

∴ म. स. (612, 48) = 12
अतः अधिकतम स्तम्भों की संख्या 12 है।

प्रश्न 8.
वह सबसे छोटी संख्या लिखिए जो 306 तथा 657 से पूर्णतया विभाजित हो ।
हल :
सबसे छोटी संख्या जो 306 और 657 दोनों से पूर्णतया विभाजित होगी वह है :
ल. स. (657, 306)
657 = 3 × 3 × 73
306 = 3 × 3 × 2 × 17
अत: ल. स. (657, 306) = 3 × 3 × 73 × 2 × 17
= 22338

प्रश्न 9.
\(\sqrt{2}\) तथा \(\sqrt{3}\) के बीच में स्थित एक परिमेय संख्या ज्ञात कीजिए।
हल:
\(\sqrt{2}\) = 1.414
\(\sqrt{3}\) = 1.732
\(\sqrt{2}\) तथा \(\sqrt{3}\) के बीच परिमेय संख्या है,
= 1.5 (∵ 1.414 < 1.5 < 1.732)

प्रश्न 10.
दर्शाइए कि प्रत्येक विषम धनपूर्णांक (4q + 1) अथवा (4q + 3) के रूप का होता है, वहाँ q कोई पूर्णांक है।
हल :
मान लीजिए कि ‘a’ एक विषम धन पूर्णांक है तथा b = 4 है।
a तथा b में विभाजन एल्गोरिथ्म का प्रयोग करने पर,
a = bq + r, r < b
∴ a = 4q + r
∵ 0 ≤ r ≤ 4 इसलिए संभावित शेषफल 0, 1, 2, 3 हैं।
∵ a = 4q, 4q + 1, 4q + 2, 4q + 3; जहाँ q भागफल है।
∵ a एक विषम धन पूर्णांक 4q + 1, 4q + 3 के प्रकार का होगा।

प्रश्न 11.
दो धनपूर्णांक a तथा b जहाँ a = x³y² तथा b = xy³ के रूप में लिखे जा सकते हैं, जहाँ x, y अभाज्य संख्याएँ हैं, तो ल.स. (LCM) (a, b) का मान ज्ञात कीजिए ।
हल :
दिया है, a = x²y², b = xy³, और
म. स. (HCF) = xy²
∵ म. स. × ल. स. = a × b
xy² × ल. स. = x³y² × xy³

प्रश्न 12.
छोटी से छोटी अभाज्य संख्या तथा छोटी-से-छोटी भाज्य संख्या का म.स. (HCF) क्या है?
हल :
छोटी से छोटी अभाज्य संख्या = 2 तथा छोटी से
भाज्य संख्या = 4

अत: म. स. (HCF) = 2

प्रश्न 13.
404 तथा 96 का म. स. (HCF) तथा ल. स. (LCM) ज्ञात कीजिए तथा निम्न का सत्यापन कीजिए :
HCF × LCM = दोनों दी गई संख्याओं का गुणनफल
इल :
404 व 96 के अभाज्य गुणनखण्ड :

404 = 2 × 2 × 101 × 1
96 = 2 × 2 × 2 × 2 × 3 × 1
म. स. = 2 × 2 × 1 = 4
ल. स. = 2 × 2 × 101 × 2 × 2 × 2 × 3
= 9696
सत्यापन:
∵ HCF × LCM – दोनों दी गई संख्याओं का गुणनफल
⇒ 4 × 9696 = 404 × 96
⇒ 38784 = 38784
अतः दिया कथन सत्यापित हुआ। इति सिद्धम्

रिक्त स्थानों की पूर्ति कीजिए-

प्रश्न (क)

1. यूक्लिड विभाजन प्रमेयिका पूर्णांकों की ……………. के संबंध में है।
2. वह सबसे बड़ा धनात्मक पूर्णांक जो दिए हुए दो धनात्मक पूर्णांकों को पूर्णत: विभाजित कर देता है, संख्याओं का ………………… समापवर्तक कहलाता है।
3. सबसे छोटी भाज्य सम संख्या ………………… तथा सबसे छोटी भाग्य विषम संख्या …………………. है।
4. जिन संख्याओं का दशमलव प्रसार विभाजन प्रक्रिया के परिमित चरणों के बाद समाप्त हो जाता है ……………… दशमलव कहलाती है।
5. ऐसा दशमलव प्रसार जिसका अंत नहीं होता तथा भागफल में अंकों का एक पुनरावृत्ति खंड प्राप्त होता, …………… प्रसार कहलाता है।

हल :

1. विभाज्यता,
2. महत्तम,
3. 4, 9,
4. सांत,
5. असांत आवर्ती

निम्न में सत्य / असत्य कथन बताइए :

प्रश्न (ख)

1. प्रमेयिका एक सत्य कथन है, जिसका प्रयोग अन्य कथनों को सिद्ध करने के लिए किया जाता है।
2. संख्या 2 सबसे छोटी अभाज्य संख्या है।
3. दो या दो से अधिक संख्याओं का ल.स. वह छोटी से ‘छोटी संख्या होती है, जो दी गई संख्याओं से पूर्णतया विभाजित होती है।
4. असांत अनावर्ती दशमलव प्रसार वाली संख्याओं को परिमेय संख्या कहते हैं।
5. यदि किसी परिमेय संख्या को किसी अन्य शून्येत्तर परिमेय संख्या से भाग दिया जाए, तो प्राप्त संख्या हमेशा परिमेय होगी।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
दिया है, HCF (156, 78) = 78, तो LCM (156, 78) का मान है :
(A) 156
(B) 78
(C) 156 × 78
(D) 156 × 2
हल :
∵ HCF × LCM = संख्याओं का गुणनफल
⇒ 78 × LCM = 156 × 78
⇒ LCM = \(\frac{156 \times 78}{78}\) = 156
अत: सही विकल्प (A) है।

प्रश्न 2.
\(\frac{1095}{1168}\) का सरलतम रूप है :
(A) \(\frac{17}{26}\)
(B) \(\frac{25}{26}\)
(C) \(\frac{13}{26}\)
(D) \(\frac{15}{16}\)
हल :
1095 और 1168 का म. स. = 73

∴ सही विकल्प (D) है।

प्रश्न 3.
निम्न में से कौन-सी परिमेय संख्या को सांत दशमलव के रूप में व्यक्त किया जा सकता है?
(A) \(\frac{124}{165}\)
(B) \(\frac{131}{30}\)
(C) \(\frac{2027}{625}\)
(D) \(\frac{1625}{462}\)
हल :
\(\frac{124}{165}\) = \(\frac{124}{3^1 \times 5^1 \times 11^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
\(\frac{131}{30}\) = \(\frac{131}{2^1 \times 3^1 \times 5^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में है।
\(\frac{2027}{625}\) = \(\frac{2027}{2^0 \times 5^4}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
\(\frac{1625}{462}\) = \(\frac{1625}{2^1 \times 3^1 \times 7^1 \times 11^1}\)
गुणनखंड 2ⁿ × 5ⁿ के रूप में नहीं है।
अत: सही विकल्प (C) है।

प्रश्न 4.
वह बड़ी से बड़ी संख्या, जिससे 245 तथा 1029 को भाग देने पर क्रमशः 5 एवं शेष बचे है:
(A) 15
(B) 60
(C) 9
(D) 5
हल :
∵ प्रत्येक स्थिती में 5 शेष बचता है।
∴ 245 – 5 = 240 तथा 1029 – 9 = 1020
अभीष्ट संख्या 240 और 1020 का म. स. है ।

∴ म.स. = 5
अत: सही विकल्प (D) है।

प्रश्न 5.
225 को निम्न रूप में व्यक्त किया जा सकता है:
(A) 5 × 3².
(B) 5² × 3
(C) 5² × 3²
(D) 5 × 3
हल :
अभाज्य गुणनखण्ड विधि से

∴ 225 = 3 × 3 × 5 × 5 = 3² × 5²
अत: सही विकल्प (c) है।

प्रश्न 6.
\(2 . \overline{35}\) है एक :
(A) पूर्णांक
(B) परिमेय संख्या
(C) अपरिमेय संख्या
(D) प्राकृत संख्या
हल :
\(2 . \overline{35}\) का प्रसार असांत आवर्ती है।
∴ यह एक परिमेय संख्या है।
अतः विकल्प (B) सही है।

प्रश्न 7.
144 तथा 198 का महत्तम समापवर्तक है:
(A) 9
(B) 18
(C) 6
(D) 12
हल :

∴ म. स. = 18
अत: विकल्प (B) सही है।

प्रश्न 8.
दो संख्याओं का म. स. 27 है तथा उनका ल.स. 162 है। यदि एक संख्या 54 है, तो दूसरी संख्या है :
(A) 36
(B) 35
(C) 9
(D) 81
हल :
∵ म. स. × ल. स. = पहली संख्या × दूसरी संख्या
⇒ 27 × 162 = 54 × दूसरी संख्या
⇒ दूसरी संख्या = \(\frac{27 \times 162}{54}\) = 81
अत: सही विकल्प (D) है।

प्रश्न 9.
2\(\sqrt{3}\) एक :
(A) पूर्णांक है
(B) परिमेय संख्या है
(C) अपरिमेय संख्या है
(D) एक पूर्ण संख्या है.
हल :
एक परिमेय संख्या (2) और अपरिमेय संख्या (\(\sqrt{3}\)) का गुणनफल एक अपरिमेय संख्या होती है। अत: सही विकल्प (c) है।

प्रश्न 10.
संख्या \(\frac{3}{2^5 \times 5^2}\) का दशमलव प्रसार दशमलव के कितने स्थानों के बाद सांत होगा ?
(A) 2
(B) 4
(C) 5
(D) 1
हल :
यहाँ हर = 2⁵ × 5²
∴ दशमलव प्रसार 5 दशमलव स्थानों के बाद सांत होगा।
अतः सही विकल्प (c) है।

प्रश्न 11.
एक अभाज्य संख्या के कुल गुणनखण्डों की संख्या है :
(A) 1
(B) 0
(C) 2
(D) 3
हल :
दो 1 और स्वयं वही संख्या।
अत: सही विकल्प (c) है।

प्रश्न 12.
12, 21, 15 का म. स. और ल. स. है:
(A) 3, 140
(B) 12,420
(C) 3, 420
(D) 420, 3
हल :
यहाँ,
12 = 2 × 2 × 3
21 = 3 × 7
15 = 3 × 5
∴ म. स. = 3
ल. स. = 22 × 3 × 5 × 7
= 420
अत: सही विकल्प (C) है।

प्रश्न 13.
संख्या 196 के अभाज्य गुणनखण्ड में अभाज्य गुणनखण्डों की घातों का योग है :
(A) 3
(B) 4
(C) 5
(D) 2
हल :

196 = 2 × 2 × 7 × 7= 2² × 7²
घातों का योग = 2 + 2 = 4
अत: सही विकल्प (B) है।

प्रश्न 14.
यूक्लिड विभाजन प्रमेविका के अनुसार दो धनात्मक पूर्णांकों a और b के लिए ऐसी अद्वितीय पूर्ण संख्याएँ q और r विद्यमान है कि a = bq + r है तथा
(A) 0 < r < b
(B) 0 < r < b
(C) 0 ≤ r < b
(D) 0 ≤ r ≤ b
हल :
सही विकल्प (C) है।

प्रश्न 15.
संख्या n² – 1, 8 से विभाज्य होती है, यदि n है एक:
(A) पूर्णांक
(B) प्राकृत संख्या
(C) विषम संख्या
(D) सम संख्या
हल :
सही विकल्प (C) है।

प्रश्न 16.
यदि 65 और 117 के H.C.F. को 65m – 117 रूप में व्यक्त किया जा सके, तो m का मान है :
(A) 4
(B) 2
(C) 1
(D) 3

हल :
65 = 5 × 13
117 = 3 × 3 × 13
H.C.F. = 13
⇒ 65m – 117 = 13
[∵ H.C.F. = 65m – 117]
⇒ 65m = 13 + 117
⇒ 65m = 130
⇒ m = \(\frac{130}{65}\) = 2
अत: सही विकल्प (B) है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 रचनाएँ Ex 11.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 11 रचनाएँ Exercise 11.2

निम्नलिखित में से प्रत्येक के लिए रचना का औचित्य भी दीजिए:

प्रश्न 1.
6 सेमी त्रिज्या का एक वृत्त खींचिए । केन्द्र से 10 सेमी दूर एक बिन्दु से वृत्त पर स्पर्श रेखा -युग्म की रचना कीजिए और उनकी लम्बाइयाँ मापकर लिखिए।
हल :
दिया है : 6 सेमी त्रिज्या का एक वृत्त और उसके केन्द्र बिन्दु O से 10 सेमी की दूरी पर एक बिन्दु P है।

रचना के चरण :

1. सर्वप्रथम बिन्दु O को केन्द्र मानकर 6 सेमी त्रिज्या का एक वृत्त खींचा।
2. वृत्त के केन्द्र O से 10 सेमी की दूरी पर एक बिन्दु P लिया।
3. OP को मिलाया और OP की समद्विभाजक रेखा खींची जो OP को M बिन्दु पर प्रतिच्छेद करती है।
4. बिन्दु M को केन्द्र मानकर PM त्रिज्या का एक वृत्त खींचा जो O केन्द्र वाले वृत्त को T₁ तथा T₂ बिन्दुओं पर काटता है।
5. PT₁ और PT₂ को मिलाया, जो वृत्त की अभीष्ट स्पर्श रेखाएँ हैं।

औचित्य (उपपत्ति) : हम जानते हैं कि किसी बिन्दु पर स्पर्श रेखा, उस बिन्दु पर त्रिज्या पर लम्ब होती है।
∴ ∠OT₁P = ∠OT₂P = 90°
अब OT₁ और OT₂ को मिलाया, वृत्त OT₁PT₂ में OP व्यास है।
∴ ∠OT₁P अर्द्धवृत्त में बना कोण है।
∴ ∠OT₁P = 90°
इसी प्रकार ∠OT₂P = 90°
अत: PT₁ तथा PT₂ वृत्त की स्पर्श रेखाएँ हैं।
स्पर्श रेखा की लम्बाई नापने पर,
PT₁ = PT₂ = 8.0 सेमी

प्रश्न 2.
4 सेमी त्रिज्या के एक वृत्त पर 6 सेमी त्रिज्या के एक संकेन्द्रीय वृत्त के किसी बिन्दु से एक स्पर्श रेखा की रचना कीजिए और उसकी लम्बाई मापिए परिकलन से इस माप की जाँच भी कीजिए।
हल :

दिया है: 4 सेमी त्रिज्या का एक वृत्त और 6 सेमी त्रिज्या का एक संकेन्द्रीय वृत्त जिस पर एक बिन्दु P दिया है।
रचना के चरण :

1. 4 सेमी त्रिज्या लेकर केन्द्र O वाला एक वृत्त खींचा।
2. केन्द्र O से 6 सेमी त्रिज्या पर एक संकेन्द्रीय वृत्त खींचा और इस पर एक बिन्दु P लिया।
3.  रेखाखण्ड OP खींचा और इसका लम्ब समद्विभाजक खींचा जो OP को बिन्दु M पर काटता है।
4. बिन्दु M को केन्द्र मानकर MP त्रिज्या का एक वृत्त खींचा जो केन्द्र O के 4 सेमी त्रिज्या वाले वृत्त को T₁ और T₂ बिन्दुओं पर काटता है।
5. PT₁ और PT₂ को मिलाया जो वृत्त की अभीष्ट स्पर्श रेखाएँ है।

औचित्य (उपपत्ति) : ∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∴ ∠OT₁P = ∠OT₂P = 90°
OT₁ तथा OT₂ को मिलाया, OP वृत्त का व्यास है ।
∠OT₁P तथा ∠OT₂P अर्द्धवृत्त के कोण हैं।
∴ ∠OT₁P = 90° तथा ∠OT₂P = 90°
∴ OT₁ ⊥ PT₁ तथा OT₂ ⊥ PT₂
अत: PT₁ तथा PT₂ अभीष्ट स्पर्श रेखाएँ हैं।
स्पर्श रेखा की लम्बाई मापने पर,
PT₁ = PT₂ = 4.7 लगभग ।

प्रश्न 3.
3 सेमी त्रिज्या का एक वृत्त खींचिए। इसके किसी बढ़ाए गए व्यास पर केन्द्र से 7 सेमी की दूरी पर स्थित दो बिन्दु और 2 लीजिए। इन दोनों बिन्दुओं से वृत्त पर स्पर्श रेखाएं खींचिए ।
हल :
दिया है : 3 सेमी त्रिज्या का एक वृत्त है जिसका केन्द्र O है। AOB वृत का व्यास है जिसको इस प्रकार बिन्दुओं P व Q तक बढ़ाया गया है कि वृत्त के केन्द्र O से प्रत्येक बिन्दु P व Q की दूरियाँ OP व OQ, 7 सेमी हैं।
रचना के चरण :

1. O केन्द्र वाला 3 सेमी त्रिज्या का एक वृत्त खींचा।
2. वृत्त का व्यास AB खींचकर इसे दोनों ओर क्रमशः P व Q तक इस प्रकार बढ़ाया कि OP = OQ = 7 सेमी ।
3. OP और OQ के लम्ब समद्विभाजिक खींचे जो OP को M₁ तथा OQ को M₂ पर काटते हैं।
4. बिन्दु M₁ को केन्द्र मानकर M₁O त्रिज्या का वृत्त खींचा जो O केन्द्र वाले वृत्त को T₁ व T₂ पर काटता है।
5. PT₁ और PT₂ को मिलाया ।
6. बिन्दु M₂ को केन्द्र मानकर M₂O त्रिज्या का वृत्त को खींचा जो O केन्द्र वाले वृत्त को S₁ व S₂ पर काटता है।
7. QS₁ और QS₂ को मिलाया।

अत: PT₁, PT₂, QS₁ और QS₂ अभीष्ट स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) : केन्द्र O वाले वृत्त की त्रिज्याएँ OT₁, OT₂, OS₁ व OS₂ खींचीं।
∵ हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
अतः ∠OT₁P – ∠OT₂P = 90° तथा
∠OS₁Q = ∠OS₂Q = 90°
∵ केन्द्र M₁ वाले वृत्त में ∠OT₁P व ∠OT₂P अर्द्धवृत्तों में स्थित कोण है।
∴ ∠OT₁P व ∠OT₂P समकोण हैं जो क्रमश: त्रिज्याओं OT₁ OT₂ के सिरों T₁ व T₂ पर स्थित हैं।
∴ PT₁ व PT₂, केन्द्र O वाले वृत्त की स्पर्श रेखाएँ हैं। इसी प्रकार, QS₁ व QS₂ भी केन्द्र O वाले वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 4.
5 सेमी त्रिज्या के एक वृत्त पर ऐसी दो स्पर्श रेखाएँ खींचिए, जो परस्पर 60° के कोण पर झुकी हों।
हल :
दिया है एक वृत्त का केन्द्र O है तथा त्रिज्या 5 सेमी है।
रचना के चरण :
1. O को केन्द्र मानकर 5 सैमी त्रिज्या का एक वृत खींचा।
2. वृत्त का व्यास AB खींचा।
3. बिन्दु O पर OB त्रिज्या से 60° का कोण बनाती हुई एक OP रेखा खींची जो वृत्त को P बिन्दु पर काटती है।
4. बिन्दु A पर OA से 90° कोण बनाती हुई AX स्पर्श रेखा खींची।

5. इसी प्रकार बिन्दु P से स्पर्श रेखा PY खींची।
AX तथा PY एक दूसरे को T पर प्रतिच्छेद करती हैं।
अत: PT और AT वृत्त की दो अभीष्ट स्पर्श रेखाएँ हैं
जो एक-दूसरे के साथ 60° का कोण बनाती हैं।
औचित्य (उपपत्ति) : माना वृत्त का केन्द्र O है।
PT और AT वृत्त की दो स्पर्श रेखाएँ हैं जिनके बीच का कोण 60° है अर्थात्
∠PTA = 60°
परन्तु ∠POA + ∠PTA = 180° होता है।
∴ ∠POA = 180° – ∠PTA
= 180° – 60°
= 120°
∴ ∠POB = 180° – ∠POA
= 180° – 60°
= 120°
अत: PT तथा AT वृत्त की स्पर्श रेखाएँ हैं जो 60° के कोण पर झुकी हैं।

प्रश्न 5.
8 सेमी लम्बा एक रेखाखण्ड AB खींचिए । A को केन्द्र मानकर 4 सेमी त्रिज्या का एक वृत्त तथा B को केन्द्र मानकर 3 सेमी त्रिज्या का एक अन्य वृत्त खींचिए । प्रत्येक वृत्त पर दूसरे वृत्त के केन्द्र से स्पर्श रेखाओं की रचना कीजिए ।
हल :
दिया है रेखाखण्ड AB = 8.0 सेमी केन्द्र A से 4 सेमी त्रिज्या का एक वृत्त खींचा गया है तथा केन्द्र B से 3 सेमी त्रिज्या का एक अन्य वृत्त खींचा गया है।
रचना के चरण :

1. रेखाखण्ड AB = 8 सेमी खींचा।
2. केन्द्र A से 4 सेमी त्रिज्या का एक वृत्त खींचा और केन्द्र B से 3 सेमी त्रिज्या का एक वृत्त खींचा।
3. AB का लम्ब समद्विभाजक खींचा जो AB को M बिन्दु पर काटता है।
4. बिन्दु M को केन्द्र मानकर MB त्रिज्या का एक वृत्त खींचा जो A केन्द्र वाले वृत्त को S₁ व S₂ पर काटता है तथा B केन्द्र वाले वृत्त को T₁ व T₂ बिन्दुओं पर काटता है।
5. S₁B व S₂B और T₁A व T₂ को मिलाया।

अत: S₁B और S₂B केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं तथा T₁A और T₂A केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) :
∵ चूँकि हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
अतः ∠AS₁B = ∠AS₂B = 90°
तथा ∠BT₁A = ∠BT₂A = 90°
∵ केन्द्र बिन्दु M वाले वृत्त का व्यास AB है।
∴ ∠AS₁B = ∠AS₂B, ∠BT₁A व ∠BT₂A अर्द्धवृत्त में बने कोण हैं। अत: प्रत्येक कोण समकोण है।
∴ AS₁ ⊥BS₁; AS₂ ⊥ BS₂ और BT₁ ⊥ AT₁ व BT₂ ⊥ AT₂
∴ S₁B व S₂B केन्द्र A वाले वृत्त की स्पर्श रेखाएँ हैं और AT₁ व AT₂ केन्द्र B वाले वृत्त की स्पर्श रेखाएँ हैं।

प्रश्न 6.
माना ABC एक समकोण त्रिभुज है, जिसमें AB = 6 सेमी, BC = 8 सेमी तथा ∠B = 90° है। B से AC पर BD लम्ब है। बिन्दुओं B, C व D से होकर जाने वाला एक वृत्त खींचा गया है। A से इस वृत्त पर स्पर्श रेखा की रचना कीजिए।
हल :
दिया है : एक समकोण त्रिभुज ABC जिसमें ∠B = 90°, AB = 6 सेमी तथा BC = 8 सेमी है। शीर्ष B से भुजा AC पर BD लम्ब खींचा गया है।
रचना के चरण :
1. सर्वप्रथम रेखाखण्ड BC = 8 सेमी खींचा।
2. बिन्दु B पर 90° कोण बनाते हुए, AB रेखाखण्ड 6 सेमी खींचा।

3. AC को मिलाया। इस प्रकार समकोण ΔABC प्राप्त हुआ।
4. बिन्दु B से AC पर BD लम्ब खींचा जो AC को D बिन्दु पर काटता है।
5. अब ΔBCD की भुजाओं BD और CD के लम्ब समद्विभा जक किए जो परस्पर O बिन्दु पर काटते हैं।
6. O को केन्द्र मानकर OB त्रिज्या का एक वृत्त खींचा जो बिन्दुओं B, C D से होकर गुजरता है।
7. चूँकि AB स्वयं स्पर्श रेखा है। इसलिए A को केन्द्र मानकर AB त्रिज्या का चाप खींचा जो वृत्त को P बिन्दु पर काटता है। AP को मिलाया।
अतः AP अभीष्ट स्पर्श रेखा है।
औचित्य (उपपत्ति) :
∵ ∠ABC = 90° (रचना से)
स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∴ ∠ABC = 90°
अत: AB रेखा स्पर्श रेखा है।
∵ AP = PB
अत: रेखा AB बिन्दु A पर खींची गई दूसरी स्पर्श रेखा है।

प्रश्न 7.
किसी चूड़ी की सहायता से वृत्त खींचिए। वृत्त के बाहर एक बिन्दु लीजिए। इस बिन्दु से वृत्त पर स्पर्श रेखाओं की रचना कीजिए।
हल :
किसी चूड़ी की सहायता से एक वृत्त खींचने का अर्थ है कि वृत्त का केन्द्र अज्ञात है। सर्वप्रथम हम केन्द्र ज्ञात करेंगे।
रचना के चरण :
1. चूड़ी की सहायता से वृत्त खींचा। इस वृत्त का केन्द्रबिन्दु ज्ञात करने के लिए कोई दो जीवाएँ AB और CD खाँचते हैं।

2. जीवा AB और CD के लम्ब समद्विभाजक किए जो परस्पर बिन्दु पर काटते हैं।
बिन्दु O वृत्त का केन्द्र होगा।
[∵ OA = OB = OC = OD]
(वृत की त्रिज्याएँ हैं)
3. वृत्त के बाहर कोई बिन्दु P लिया।
4. OP को मिलाकर लम्ब समद्विभाजक किया जो OP को बिन्दु M पर काटता है।
5. बिन्दु M को केन्द्र मानकर MP त्रिज्या का वृत्त खींचते हैं जो O केन्द्र वाले वृत्त को T₁ व T₂ पर काटता है।
6. PT₁ और PT₂ को मिलाया।
अत: PT₁ व PT₂ अभीष्ट स्पर्श रेखाएँ हैं।
औचित्य (उपपत्ति) : OT₁ को मिलाया।
हम जानते हैं कि स्पर्श रेखा स्पर्श बिन्दु से होकर खींची गई त्रिज्या पर लम्ब होती है।
∵ ∠OT₁P = ∠OT₂P = 90°
∵ OP वृत्त का व्यास है, ∠OT₁P तथा ∠OT₂P अर्द्धवृत्त में बने कोण हैं।
∴ ∠OT₁P = 90° तथा ∠OT₂P = 90°
अत: PT₁ और PT₂ वृत्त की स्पर्श रेखाएँ हैं।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:

1. All circles are …………… (congruent, similar)
2. All squares are ……….. (similar, congruent)
3. All ……………. triangles are similar. (isosceles, equilateral)
4. Two polygons of the same of sides are similar, if (a) their corresponding angles are …… and (b) their corresponding sides are ……….. (equal, proportional)

Answer:

1. similar
2. similar
3. equilateral
4. equal, proportional

Question 2.
Give two different examples of pair of
1. similar figures.
2. non-similar figures.
Answer:
1. Similar figures: Any two squares are similar. Any two equilateral triangles are similar. Any two circles are similar.
2. Non-similar figures: An equilateral triangle and an obtuse angled triangle are non- similar. An equilateral triangle and a right angled triangle are non-similar.

Question 3.
State whether the following quadrilaterals are similar or not:
Answer:

No, because for any correspondence between their vertices, the corresponding sides are proportional but the corresponding angles are not equal.

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ

लयूत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
यदि किसी A. P. के प्रथम 12 पदों का योग 468 है तथा इसका सार्वअंतर 6 है, तो 10वाँ पद ज्ञात कीजिए।
हल :
दिया है,
S_n = 468, d = 6, n = 12
a₁₀ = ?
हम जानते हैं :
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
468 = \(\frac{12}{2}\)[2a + (12 – 1)6]
468 = 6(2a + 66)
= 12a + 396
a = \(\frac{468-396}{12}=\frac{72}{12}\)
a = 6
∴ a₁₀ = a + (n – 1)d
= 6 + (10 – 1)6
= 6 + 9 × 6 = 6 + 54
a₁₀ = 60.

प्रश्न 2.
एक समान्तर श्रेणी का 14वाँ पद उसके 8वें पद का दुगना है। यदि उसका छटा पद – 8 है, तो उसके प्रथम 20 पदों का योगफल ज्ञात कीजिए।
हल :
माना समान्तर श्रेणी का प्रथम पद = a
तथा सर्वाअंतर = d
∴ समान्तर श्रेढी के nth पद
a_n = a + (n – 1)d
इस प्रकार a₁₄ = a + (14 – 1)d
= a + 13d
a₈ = a + (8 – 1)d = a + 7d
a₆ = a + (6 – 1)d = a + 5d
प्रश्नानुसार-
a₁₄ = 2a₈
⇒ a + 13d = 2(a + 7d)
⇒ a + d = 0 ………..(1)
इसी प्रकार,
a₆ = a + 5d = – 8 ………..(2)
समी (1) व (2) को हल करने पर,
a = 2, d = – 2
∴ S₂₀ = \(\frac{20}{2}\) [2 × z + (20 – 1) (- z)]
S₂₀ = – 340
अतः प्रथम 20 पदों का योगफल = – 340.

प्रश्न 3.
1 से 100 तक के मध्य की 6 से विभाजित होने वाली संख्याओं का योगफल ज्ञात कीजिए।
हल :
1 से 100 तक के मध्य की 6 से विभाज्य संख्याएँ है….
6, 12, 18, 24, …, 96
स्पष्ट है कि उपर्युक्त संख्याएँ समान्तर श्रेढी में हैं।
यहाँ a = 6, d = 12 – 6 = 6 तथा a_n = 96
अत: a_n = 96
⇒ a + (n – 1)d = 96
⇒ 6 + (n – 1) × 6 = 96
⇒ (n – 1) × 6 = 96 – 6 = 90
⇒ n – 1 = \(\frac{90}{6}\) = 15
⇒ n = 15 + 1 = 16
∵ हम जानते हैं कि
S_n = \(\frac{n}{2}\)(a + l)
⇒ S₁₆ = \(\frac{16}{2}\)(6 + 96)
⇒ S₁₆ = 8 × 102
⇒ S₁₆ = 816
अतः S₁₆ = 816

प्रश्न 4.
समीकरण हल कीजिए :
1 + 4 + 7 + 10 +……+ x = 287.
हल :
दिया गया समीकरण है :
1 + 4 + 7 + 10 +………+ x = 287
स्पष्ट है कि यह एक समान्तर श्रेढी है।
यहाँ a = 1, d = 4 – 13, a_n = x तथा S_n = 287
∵ a_n = x
⇒ a + (n – 1)d = x
⇒ 1 + (n – 1) × 3 = x
⇒ 1 + 3n – 3 = x
⇒ x = 3n – 2 ……………(i)
और S_n = 287
⇒ \(\frac{n}{2}\)(a + l) = 287
⇒ \(\frac{n}{2}\)(1 + x) = 287
⇒ \(\frac{n}{2}\)(1 + 3n – 2) = 287
[समी. (i) का प्रयोग करने पर]
⇒ n(3n – 1) = 574
⇒ 3n² – n – 574 = 0
श्रीधराचार्य सूत्र से –
n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

∵ n एक प्राकृतिक धन पूर्णांक है। अतः n = \(\frac{-41}{3}\) को छोड़ देते हैं।
अत: n = 14
n का मान समीकरण (i) में रखने पर
x = 3 × 14 – 2
= 42 – 2 = 40
अतः x = 40.

प्रश्न 5.
A. P. 17, 15, 13 के कितने पद लिए जाएँ ताकि उनका योग 81 हो।
हल :
a = 17, d = 15 – 17 = – 2
S_n = 81
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
81 = \(\frac{n}{2}\)[2 × 17 + (n – 1) × (- 2)]
81 = \(\frac{n}{2}\)[34 – 2n + 2]
= n[17 – n + 1]
= n[18 – n]
81 = 18n – n²
n² – 18n + 81 = 0
n² – 9n – 9n + 81 = 0
n(n – 9) – 9 (n – 9) = 0
(n – 9) (n – 9) = 0
n = 9.

प्रश्न 6.
चार क्रमागत संख्याएँ जोकि समान्तर श्रेढी में हैं, का योग 32 है। प्रथम और अन्तिम संख्याओं का गुणनफल मध्य संख्याओं के गुणनफल से 7 : 15 के अनुपात में हो। वह संख्याएँ ज्ञात करो ।
हल :
माना चार क्रमागत संख्याएँ हैं-
a – 3d, a – d, a + d, a + 3d
प्रश्नानुसार,
संख्याओं का योग = 32
⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = \(\frac{32}{4}\) = 8
तथा \(\frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}\) = \(\frac{7}{15}\)
⇒ \(\frac{a^2-9 d^2}{a^2-d^2}\) = \(\frac{7}{15}\)
⇒ \(\frac{64-9 d^2}{64-d^2}\) = \(\frac{7}{15}\) [∵ a = 8]
⇒ 960 – 135d² = 448 – 7d²
⇒ 960 – 448 = 135d² – 7d²
⇒ 512 = 128d²
⇒ d² = \(\frac{512}{128}\) = 4
⇒ d = ± 2
अतः a – 3d = 8 – 3 × 2 = 2
a – d = 8 – 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3 × 2 = 14
अतः अभीष्ट संख्याएँ हैं-
2, 6, 10, 14.

प्रश्न 7.
यदि एक समान्तर श्रेढी के m पदों का योग ‘n’ तथा n पदों का योग ‘m’ हो, तो सिद्ध कीजिए कि (m + n) पदों का योगफल – (m + n) होगा।
हल :
माना दी गई श्रेढी का प्रथम पद a तथा सार्वअन्तर d हैं।
दिया है S_m = n
⇒ \(\frac{m}{2}\){2a + (m – 1)d} = n
⇒ 2am + m(m-1)d = 2n ……..(1)
तथा S_n = m
⇒ \(\frac{n}{2}\){2a + (n – 1)d} = m
⇒ 2am + n(n – 1)d = 2m …….(ii)
समीकरण (i) में से समीकरण (ii) को घटाने पर,
2a(m – n) + {m(m – 1) n (n-1)} d = 2n – 2m
⇒ 2a(m – n) + {(m² – n²) – (m – n)} d
= – 2(m – n)
⇒ 2a + (m + n – 1)d = – 2 ………….(iii)
[दोनों पक्षों में (m – n) से भाग देने पर]
अब S_m+n = \(\frac{m+n}{2}\){2a + (m + n – 1)d}
⇒ \(\frac{(m+n)}{2}\)(-2) [(iii) के प्रयोग से]
⇒ S_m+n = – (m + n)
अतः S_m+n = – (m + n) इति सिद्धम् ।

प्रश्न 8.
यदि एक समान्तर श्रेढी का m वाँ पद \(\frac{1}{n}\) तथा n वाँ पद \(\frac{1}{m}\) हो, तो दर्शाइए कि mn पदों का योगफल \(\frac{1}{2}\) (mn + 1) होगा।
हल :
माना दी गई समान्तर श्रेढी का प्रथम पद a तथा सार्वअन्तर d है।
दिया है : a_m = \(\frac{1}{n}\)
⇒ a + (m – 1)d = \(\frac{1}{n}\) ……..(i)
तथा a_n = \(\frac{1}{m}\)
⇒ a + (n – 1)d = \(\frac{1}{m}\) …………….(ii)
समीकरण (i) में से समीकरण (ii) को घटाने पर,
(m – n)d = \(\frac{1}{n}-\frac{1}{m}\)
⇒ (m – n)d = \(\frac{m-n}{m n}\)
⇒ d = \(\frac{1}{mn}\)
समीकरण (i) में d = \(\frac{1}{mn}\) रखने पर,
a + (m – 1)\(\frac{1}{mn}\) = \(\frac{1}{n}\)
⇒ a + \(\frac{1}{n}\) – \(\frac{1}{mn}\) = \(\frac{1}{n}\)
⇒ a = \(\frac{1}{mn}\)
∴ अभीष्ट योगफल S_mn = \(\frac{mn}{2}\){2a + (mn – 1)d}
⇒ S_mn = \(\frac{mn}{2}\){\(\frac{2}{mn}\) + (mn – 1) × \(\frac{1}{mn}\)}
⇒ S_mn = \(\frac{1}{2}\)(mn + 1)
अत: mn पदों का योगफल
= \(\frac{1}{2}\)(mn + 1) इति सिद्धम् ।

प्रश्न 9.
एक समान्तर श्रेढी के प्रथम n पदों केयोगफल को S_n द्वारा दर्शाया जाता है। इस श्रेढी में यदि S₅ + S₇ = 167 तथा S₁₀ = 235 है, तो समान्तर श्रेणी ज्ञात कीजिए ।
हल :
माना श्रेढी का प्रथम पद = a
और सार्वअंतर = d
n पदों का योगफल
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
दिया है,
S₅ + S₇ = 167
⇒ \(\frac{5}{2}\)(2a + 4d) + \(\frac{7}{2}\)(2a + 6d) = 167
⇒ 5a + 10d + 7a + 21d = 167
⇒ 12a + 31d = 167 ……….. (1)
इसी प्रकार,
S₁₀ = 235
⇒ \(\frac{10}{2}\)(2a + 9d) = 235
⇒ 2a + 9d = 47 ……….. (2)
समी. (1) व (2) से,
a = 1 और d = 5
अत: अभीष्ट A. P. 1, 6, 11… है ।

प्रश्न 10.
दर्शाइए कि उस A. P का योग, जिसका प्रथम पद a, द्वितीय पद b और अन्तिम पद c हो \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\) के बराबर है।
हल :
दिया है,
प्रथम पद = a
द्वितीय पद = b
∴ सार्वअन्तर (d) = ba
अन्तिम पद (l) = c
माना कि समान्तर श्रेढी में पद है।
∴ a_n = c
⇒ a + (n – 1) × d = c
⇒ a + (n – 1) × (b – a) = c
⇒ (n – 1) (b – a) = c – a
⇒ (n – 1) = \(\frac{c-a}{b-a}\)
⇒ n = \(\frac{c-a}{b-a}\) + 1
⇒ n = \(\frac{c-a+b-a}{b-a}\)
⇒ n = \(\frac{b+c-2 a}{b-a}\)
अब S_n = \(\frac{n}{2}\)[a + l]
⇒ S_n = \(\frac{(b+c-2 a)}{2(b-a)}\)[a + c]
⇒ S_n = \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\) इति सिद्धम् ।

प्रश्न 11.
एक मोटर कार A स्थान से B स्थान तक 175 किमी दूरी 70 किमी / घंटा समान गति से सभी 10 हरे यातायात्रा सिग्नलों को पार करती है। भारी यातायात के कारण यह प्रथम सिग्नल पर एक मिनट, दूसरे सिग्नल पर 3 मिनट, तीसरे सिग्नल पर 5 मिनट एवम् इसी प्रकार दसवें सिग्नल पर 19 मिनट रूकती हैं। स्थान B तक पहुँचने में इसे कुल कितना समय लगेना उपयुक्त गणितीय विधि से हल कीजिए ।
हल :
यहाँ, हम देखते हैं कि 10 सिग्नलों पर लगा समय समान्तर श्रेढी में हैं।

प्रथम पद a = 1, तथा सार्वअन्तर d = 3 – 1 = 2
∴ 10 सिग्नलों को पार करने के लगा समय = 1 + 3 + 5 + 5 + 9 + 11 + 13 + 15 + 17 + 19
= \(\frac{10}{2}\)[2 × 1 + (10 – 1) × 2]
= 5 [2 + 18] = 5 × 20 = 100 मिनट
= \(\frac{10}{60}\) घंटे = \(\frac{5}{3}\)घंटे
∴ A से B तक पहुँचाने में लगा कुल समय
= (\(\frac{175}{70}+\frac{5}{3}\))घंटे = (\(\frac{35}{14}+\frac{5}{3}\))घंटे
= (\(\frac{35 \times 3+5 \times 14}{42}\))घंटे = \(\frac{175}{42}\)घंटे
= 4\(\frac{1}{6}\) घंटे = 4 घंटे 10 मिनट

प्रश्न 12.
उस A. P. के प्रथम 15 पदों का योग ज्ञात कीजिए, जिसका पाँचवा और नवाँ पद क्रमश: 26 और 42 है।
हल :
माना सं. क्षे. का प्रथम पद a तथा सार्वअन्तर d है।
स.क्षे. का पाँचवाँ पद,
a₅ = a + (n – 1)d
⇒ 26 = a + (5 – 1)d
⇒ 26 = a + 4d ………..(i)
स.क्षे. का नव पद, a₉ = a + (9 – 1)d
42 = a + 8d ………….(ii)
समीकरण (ii) में से समीकरण (i) घटाने पर
a + 8d = 42
a + 4d = 26
4d = 16
d = 4
समीकरण (i) में d = 4 रखने पर
a + 4 × 4 = 26
a = 26 – 8 = 18
स.क्षे. के 15 पदों को योग
= [2 × 18 + (15 – 1) × 4]
= [36 + 56]
= 15 × 92
= 690

प्रश्न 13.
एक कार A स्थान से B स्थान पर 260 किमी दूरी 65 किमी / घंटा समान गति से सभी 13 हरे यातायात सिग्नलों को पार करती है। भारी यातायात के कारण यह प्रथम सिग्नल पर 4 मिनट, दूसरे सिग्नल पर 7 मिनट, तीसरे सिग्नल पर 10 मिनट एवम् इसी प्रकार तेरहवें सिग्नल पर 40 मिनट रूकती है। स्थान B तक पहुँचने इसे कुल कितना समय होगा? उपयुक्त गणितीय विधि से हल कीजिए।
हल :

A से B तक पहुँचने में लगा कुल समय
= \(\frac{260}{65}\)घंटे + \(\frac{13}{2}\)[2 × 4 + (13 – 1) × 3] मिनट
= \(\frac{20}{5}\)घंटे + \(\frac{13}{2}\) [8 + 36] मिनट
= 4 घंटे + \(\frac{13}{2}\) × 44 मिनट = 4 घंटे + 286 मिनट
= 4 घंटे + 46 मिनट = 8 घंटे 46 मिनट

प्रश्न 14.
दर्शाइए कि (a – b)², (a² + b²), (a + b)² एक समान्तर श्रेणी में हैं।
हल :
माना a₁ = (a – b)², a₂ = (a² + b²), a₃ = (a + b)²
a₂ – a₁ = (a² + b²) – (a – b)²
= (a² + b²) – (a² + b² – 2ab)
= a² + b² – a² – b² + 2ab = 2ab … (i)
तथा
a₃ – a₂ = (a + b)² – (a² + b²)
= a² + b² + 2ab – a² – b²
= 2ab ……………(ii)
समीकरण (i) व (ii) से,
∵(a² + b²) – (a – b)² = (a + b)² – (a² + b²)
a₂ – a₁ = a₃ – a₂
अत: (a – b)², (a² + b²), (a + b²) एक स.क्षे. मैं हैं।

प्रश्न 15.
एक स. क्षे. के प्रथम 7 पदों का योग 63 है और इसके अगले 7 पदों का योग 161 है। स. क्षे. ज्ञात कीजिए ।
हल :
माना स.क्षे. का प्रथम पद ‘a’ तथा सार्व अन्तर ‘d’ है।
स. क्षे. के प्रथम 7 पदों का योग,
S_n = \(\frac{n}{2}\)[2a+ (n – 1)d]
S₇ = \(\frac{7}{2}\)[2a + (7 – 1)d]
63 = \(\frac{7}{2}\)(2a + 6d)
⇒ \(\frac{1}{2}\)(2a + 6d) = 9
⇒ a + 3d = 9
a = 9 – 3d ……………(i)
प्रश्नानुसार,
S₁₄ – S₇ = 161
\(\frac{14}{2}\)[2a + (14 – 1)d] – 63 = 161
\(\frac{14}{2}\)[2a + (14 – 1)d] = 161 + 63
7[2(9 – 3d) + 13d] = 224
18 – 6d + 13d = \(\frac{224}{7}\)
7d = 32 – 18
d = \(\frac{14}{7}\) = 2
समीकरण (i) से, a = 9 – 3 × 2 = 3
अतः समान्तर श्रेढी हैं :
a, a + d, a + 2d, a + 3d, ……………..
= 3, 5, 7, 9, …………

प्रश्न 16.
एक समान्तर श्रेणी के सभी 11 पदों का योगफल ज्ञात कीजिए, जिसका मध्य पद 30 है।
हल :
माना स. क्षे. का प्रथम पद a तथा सार्वअन्तर d है।
∵ स. क्षे. में 11 पद हैं।
∴ मध्य पद = (\(\frac{n+1}{2}\)) वाँ पद = (\(\frac{11+1}{2}\))वाँ पद = 5वाँ पद
a₅ = a + (5 – 1)d
30 = a + 5d ……………(i)
स. क्षे. के 11 पदों का योग = \(\frac{11}{2}\)[2a + (11 – 1)d]
= \(\frac{11}{2}\)[2a + 10d]
= \(\frac{11}{2}\) × 2(a + 5d)
= 11 × 30 = 330

प्रश्न 17.
निम्न समीकरण को हल कीजिए: 1 + 5 + 9 + 13 + ……………. + x = 1326
हल :
यहाँ, प्रथम पद, a = 1
सार्वअन्तर, d = 5 – 1 = 4
प्रश्नानुसार,
S_n = 1326
⇒ \(\frac{n}{2}\)[2 × 1 + (n – 1)4] = 1326
⇒ \(\frac{n}{2}\)(2 + 4n – 4) = 1326
⇒ \(\frac{n}{2}\)(4n – 2) = 1326
⇒ n(2n – 1) = 1326
⇒ 2n² – n – 1326 = 0
⇒ 2n² – 52n + 51n – 1326 = 0
⇒ 2n(n – 26) + 51(n – 26) = 0
⇒ (n – 26) (2n + 51) = 0
यदि n – 26 = 0 तो n = 26
और यदि 2n + 51 = 0, तो n = \(\frac{-51}{2}\) असम्भव, क्योंकि ‘n’ एक प्राकृतिक संख्या है।
अर्थात् n = 14 है।
प्रश्नानुसार,
14वाँ पद = x
⇒ a + (14 – 1)d = x
⇒ 1 + 13 × 4 = x
अतः x = 53

प्रश्न 18.
यदि संख्याएँ a, 7, b, 23, c एक समान्तर श्रेढी में है, तो a, b तथा c के मान ज्ञात कीजिए।
हल :
माना स. क्षे. का सार्व अन्तर d है।
प्रश्नानुसार,
a₂ = 7
a + (2 – 1)d = 7
a + d = 7 ……………(i)
तथा a₄ = 23
a + (4 – 1)d = 23
a + 3d = 23 ……………(ii)
समीकरण (ii) में से समीकरण (i) घटाने पर

समीकरण (i) में, d = 8 रखने पर
a + 8 = 7
⇒ a = 7 – 8 = – 1
तीसरा पद, a₃ = a + (3 – 1)d
b = – 1 + 2 × 8 = 15
पाँचवाँ पद,
a₅ = a + (5 – 1)d
c = – 1 + 4 × 8 = 31.
अतः a = – 1, b = 15, c = 31

प्रश्न 19.
समानान्तर श्रेणी
20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), ………… का कौन-सा पद प्रथम ऋणात्मक पद है?
हल :
प्रश्नानुसार, समानान्तर श्रेणी
20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{3}{4}\), …….. = 20, \(\frac{77}{4}\), \(\frac{37}{2}\), \(\frac{71}{4}\), ………..
यहाँ, a = 20, d = \(\frac{77}{4}\) – 20 = \(\frac{77-80}{4}\) = \(\frac{-3}{4}\)
माना पहला ऋणात्मक पद a_n है।

अतः दी गयी समानान्तर श्रेणी का 28वाँ पद प्रथम ऋणात्मक पद होगा।

प्रश्न 20.
यदि दो समानान्तर श्रेणियों के प्रथम पदों के योगफलों का अनुपात (7+ 1) (4 + 27) है, तो उनके 9वें पदों का अनुपात ज्ञात कीजिए।
हल :
दिया है, समानान्तर श्रेणियों के प्रथम n पदों के योगफलों का अनुपात :

अतः दो समानान्तर श्रेणियों के 9वें पदों का अनुपात = 24 : 19

प्रश्न 21.
n के किस मान के लिए, दो समानान्तर श्रेणियों 63, 65, 67, ….. तथा 3, 10, 17 ……… के n वें पद समान होंगे?
हल :
पहल समानान्तर श्रेणी 63, 65, 67, …………..
a = 63, d = 65 – 63 = 2
a_n = a + (n – 1)d
= 63 + (n – 1)d
= 63 + 2n – 2 = 61 + 2n
दूसरी समानान्तर श्रेणी 3, 10, 17, ………..
a = 3, d = 10 – 3 = 7
a_n = a + (n – 1)d
= 3 + (n – 1) (7) = 3 + 7n – 7
= 7n – 4
प्रश्नानुसार
61 + 2n = 7n – 4
61 + 4 = 7n – 2n
65 = 5n
n = \(\frac{65}{5}\) = 13
अतः दोनों समानान्तर क्षेणियों का 13वाँ पद समान हैं।

प्रश्न 22.
एक समान्तर श्रेढी के प्रथम 6 पदों का योग 42 है। इसके 10वें पद तथा 30वें पद में अनुपात 1 : 3 का है। इस समान्तर श्रेढी का प्रथम पद तथा 13वाँ पद ज्ञात कीजिए।
हल :
माना स.क्षे. का प्रथम पद a तथा सार्व अन्तर d है।
स. क्षे. का 6 पदों का योग,
S₆ = \(\frac{6}{2}\)[2a + (6 – 1)d]
⇒ 42 = 3 (2a + 5d)
⇒ 2a + 5d = 14 ………(i)
10वें तथा 30वें पद में अनुपात 1 : 3

समीकरण (i) में d = 2 रखने पर
2a + 5 × 2 = 14
⇒ 2a = 14 – 10
⇒ a = \(\frac{4}{2}\) = 2
अतः स.क्षे. का 13वाँ पद = 2 + (13 – 1) × 2
= 2 + 24 = 26

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. संख्याओं के एक निश्चित नियमानुसार क्रम को ………………. कहते हैं।
2. वह श्रेणी जिसमें अगल पद, पहले पदे में एक निश्चित संख्या जोड़ने अथवा घटाने पर प्राप्त होती है, ………….. श्रेणी कहलाती है।
3. समान्तर श्रेणी का …………… धनात्मक, ऋणात्मक या शून्य हो सकता है।
4. यदि समान्तर श्रेणी में पदों की संख्या निश्चित है, तो उसे ……………… समान्तर श्रेणी कहते हैं।
5. यदि 3k – 2, 4k – 6 तथा k + 2 एक समान्तर श्रेणी के क्रमित पद हैं, तो k का मान ………….. हैं।

हल :

1. अनुक्रम
2. समान्तर
3. सार्व अन्तर,
4. परिमित,
5. 2.

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख)

1. यदि समान्तर श्रेणी में पदों की संख्या अपरिमित है, तो उसे परिमित समान्तर श्रेणी कहते हैं।
2. श्रेणी 21, 18, 15, …………… का 8वाँ पद शून्य है।
3. श्रेणी 5, 9, 13 …………. 185 में अन्तिम पद से पहले पद की और 9वीं पद 152 है।
4. दो अंकों की 30 संख्याएँ उसे भाग्य हैं।
5. समान्तर श्रेणी 3, 15, 27, 39, ….. का 31वाँ पद 21 वे पद से 120 अधिक है।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
एक समान्तर श्रेणी का प्रथम पद p है तथा सार्वअन्तर q है, तो उसका 10वाँ पद है :
(A) q + 9p
(B) p – 9q
(C) p + 9g
(D) 2p + 9q
हल :
a10 = a + (n – 1)d
⇒ a10 = p + (10 – 1) × q
⇒ a10 = p + 9q
अत: सही विकल्प (C) है।

प्रश्न 2.
x का मान जिसके लिए 2x, (x + 10) तथा (3x + 2) एक समान्तर श्रेणी के क्रमिक पद हैं, है:
(A) 6
(B) – 6
(C) 18
(D) – 18
हल :
∵ दिए गए पद स.क्षे. के क्रमिक पद हैं।
∴ (x + 10) – 2x = (3x + 2) – (x + 10)
⇒ – x + 10 – 2x = 3x + 2 – x – 10
⇒ – x + 10 = 2x – 8
⇒ – x – 2x = – 8 – 10
⇒ – 3x = – 18
⇒ x = 6
अत: सही विकल्प (A) है।

प्रश्न 3.
समांतर श्रेणी \(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\), ……………… का सार्वअंतर है :
(A) 1
(B) \(\frac{1}{p}\)
(C) – 1
(D) – \(\frac{1}{p}\)
हल :
सार्व अंतर = \(\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}\)
= \(\frac{-p}{p}\)
= – 1
अत: सही विकल्प (C) है।

प्रश्न 4.
समान्तर श्रेणी a, 3a, 5a, …………. का n वाँ पद है :
(A) na
(B) (2n – 1) a
(C) (2n + 1)a
(D) 2na
हल :
प्रथम पद, A = a
सार्वअन्तर
D = 3a – a = 2a
∴ n वाँ पद = A + (n – 1) D
= a + (n – 1) × 2a
= a + 2na – 2a
= 2na – a
= (2n – 1)a
अत: सही विकल्प (B) है।

प्रश्न 5.
समान्तर श्रेणी 5, 9, 13, ………., 185 में कितने पद हैं?
(A) 31
(B) 51
(C) 46
(D) 40
हल :
प्रथम पद, a = 5
सार्वअन्तर, d = 9 – 5 = 4
पदों की संख्या, n = ?
∵ an = a + (n – 1)d
⇒ 185 = 5 + (n – 1) × 4
⇒ n – 1 = \(\frac{180}{4}\) = 45
⇒ n = 45 + 1 = 46
अतः सही विकल्प (C) है।

प्रश्न 6.
उस समान्तर श्रेणी, जिसका n वाँ पद an = (3n + 7) है, का सार्वअंतर है :
(A) 3
(B) 7
(C) 10
(D) 6
हल :
n वाँ पद,
an = 3n + 7
प्रथम पद, a1 = 3 × 1 + 7 = 10
द्वितीय पद, a2 = 3 × 2 + 7 = 13
सार्व अंतर, = a2 – a1 = 13 – 10 = 3
अत: सही विकल्प (A) है।

प्रश्न 7.
एक समांतर श्रेणी का प्रथम पद 5 है तथा अंतिम पद 45 है। यदि सभी पदों को योगफल 400 हो, तो पदों की संख्या है :
(A) 20
(B) 8
(C) 10
(D) 16
हल :
दिया है, a = 5, l = 45, Sn = 400
∵ Sn = \(\frac{n}{2}\)(a + l)
⇒ 400 = \(\frac{n}{2}\) (5 + 45)
⇒ n = \(\frac{400 \times 2}{50}\) = 16
अत: सही विकल्प (D) है।

प्रश्न 8.
एक समान्तर श्रेणी – 15 – 11 – 7, …………… 49 का 9 वाँ पद है
(A) 32
(B) 0
(C) 17
(D) 13
हल :
दियाहै,
a = – 15, d = – 11 – (-15) = – 11 + 5 = 4 तथा n = 9
an = a + (n – 1)d
a9 = – 15 + (9 – 1) × 4
= – 15 + 32 = 17
अत: सही विकल्प (C) है।

प्रश्न 9.
निम्नलिखित में से कौन-सी समांतर श्रेढी नहीं है?
(A) – 1.2, 0.8, 2.8, ………….
(B) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),….
(C) \(\frac{4}{3}, \frac{7}{3}, \frac{9}{3}, \frac{12}{3}\), ………………..
(D) \(\frac{-1}{5}, \frac{-2}{5}, \frac{-3}{5}\), …………..
हल :
(C) में सार्वअन्तर समान नहीं है।
\(\frac{7}{3}-\frac{4}{3}=\frac{3}{3}\)
तथा \(\frac{9}{3}-\frac{7}{3}=\frac{2}{3}\)
अर्थात् \(\frac{7}{3}-\frac{4}{3} \neq \frac{9}{3}-\frac{7}{3}\)
अंत: सही विकल्प (C) है।

प्रश्न 10.
समान्तर श्रेणी 5, 8, 11, ………. 47 का अंतिम पद से (प्रथम पद की ओर) दूसरा पद है:
(A) 50
(B) 45
(C) 44
(D) 41
हल :
दिया है, a = 5, d = 8 – 5 = 3, l = 47 तथा n = 2
स. क्षे. का अंत से n वाँ पद = l – (n – 1)d
= 47 – (2 – 1) × 3
= 47 – 3 = 44
अत: सही विकल्प (C) है।

प्रश्न 11.
दी गई श्रेढी 2,\(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………….. A.P. में है, तो इसका 3 वाँ पद होगा : (A) – 3
(B) 3
(C) 2
(D) 4
हल :
a = 2, d = \(\frac{5}{2}\) – 2 = \(\frac{5-4}{2}=\frac{1}{2}\)
an = a + (n – 1)d
∴ a3 = 2 + (3 – 1) × \(\frac{1}{2}\)
= 2 + 2 × \(\frac{1}{2}\) = 2 + 1 = 3
अत: सही विकल्प (B) है।

प्रश्न 12.
समान्तर श्रेढी 21, 42, 63, 84, ………….. का कौन-सा पद 210 है ?
(A) 9th
(B) 10th
(C) 11th
(D) 12th
हल :
दी गयी समान्तर श्रेढी है :
21, 42, 63, 84, ………. 210
यहाँ a = 21, d = 42 – 21 = 21 तथा an = 210
∴ an = 210
a + (n – 1)d = 210
⇒ 21 + (n – 1) × 21 = 210
(n – 1) × 21 = 210 – 21 = 189
(n – 1) = \(\frac{189}{21}\) = 9
n = 9 + 1 = 10
अत: विकल्प (B) सही है।

प्रश्न 13.
3 के प्रथम पाँच गुणजों का योगफल है :
(A) 45
(B) 55
(C) 65
(D) 75
हल :
3 के प्रथम पाँच गुणज हैं :
3, 6, 9, 12, 15
यहाँ a = 3, n = 3 तथा l = 15
सूत्र Sn = \(\frac{n}{2}\)(a + l) से
S5 = \(\frac{5}{2}\)(3 + 15)
⇒ S5 = \(\frac{5}{2}\) × 18 = 45
अत: सही विकल्प (A) है।

प्रश्न 14.
m के किस मान के लिए 10, m, – 2 समान्तर श्रेढी में होंगे:
(A) m = 4
(B) m = 3
(C) m = 2
(D) m = 1
हल :
a2 – a1 = a2 – a2
⇒ m – 10 = – 2 – m
⇒ m + m = – 2 + 10
2m = 8
m = \(\frac{8}{2}\)
m = 4
अत: सही विकल्प (A) है।

प्रश्न 15.
यदि एक समान्तर श्रेढी का n वाँ पद (2n + 1) है, तो उसके प्रथम तीन पदों का योगफल है :
(A) 6n + 3
(B) 15
(C) 12
(D) 21
हलं :
दिया गया है:
an = 2n + 1
n = 1, 2, 3 रखने पर
a1 = 2 × 1 + 1 = 3
a2 = 2 × 2 + 1 = 5
a3 = 2 × 3 + 1 = 7
अतः समान्तर श्रेढी है-
3, 5, 7,…
सार्वअन्तर (d) = 5 – 3 = 2
या 7 – 5 = 2
प्रथम तीन पदों का योगफल (Sn)
= \(\frac{n}{2}\) [2a + (n – 1) × d)]
= \(\frac{3}{2}\) [2 × 3 + (3 – 1) × 2]
= \(\frac{3}{2}\) [6 + 4]
= \(\frac{3}{2}\) × 10 = 15
अत: विकल्प (B) सही है।

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 7 निर्देशांक ज्यामिति Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 7 निर्देशांक ज्यामिति

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
बिन्दु (- 5, 4) की x – अक्ष से दूरी लिखिये ।
हल :
बिन्दु (- 5, 4) की x-अक्ष से दूरी = 4 इकाई ।

प्रश्न 2.
बिन्दु (3, – 2) की -अक्ष से दूरी लिखिए।
हल :
बिन्दु (3, – 2) की y-अक्ष से दूरी 3 इकाई है।

प्रश्न 3.
k के मान ज्ञात कीजिए जिनसे (1, – 1), (- 4, 2k) तथा (- k, – 5) शीर्षों वाले त्रिभुज का क्षेत्रफल 24 वर्ग इकाई हो ।
हल :
माना A(1, – 1), B (- 4, 2k) तथा (- k, – 5) दिये गये ΔABC के शीर्ष हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[1(2k + 5) + (- 4) (- 5 + 1) + (-k) (- 1 – 2k)]
= \(\frac{1}{2}\)[2k + 5 + 16 + k + 2k²]
= \(\frac{1}{2}\)[2k² + 3k + 21] वर्ग इकाई
दिया है
ΔABC का क्षेत्रफल = 24 वर्ग इकाई
∴ \(\frac{1}{2}\)[2k² + 3k + 21] = 24
2k² + 3k + 21 = 48
2k² + 3k – 27 = 0
2k² + 9k bk – 27 = 0
k(2k + 9) – 3(2k + 9) (k – 3) (2k + 9) =0
तथा k = 3
अतः k के मान 3 और हैं।

प्रश्न 4.
x- अक्ष पर वह बिन्दु ज्ञात कीजिए जो बिन्दुओं 4 (6, 5) और B (- 4, 5) से समदूरस्थ है।
हल :
x- अक्ष पर किसी बिन्दु P के निर्देशांक (x, 0) हैं।
∴ बिन्दु P(x, 0) की बिन्दुओं A(6, 5) और B(- 4, 5) से दूरी समान होगी।
PA = PB
⇒ PA² = PB²
⇒ (x – 6)² + (0 – 5)² = (x + 4)² + (0 – 5)²
⇒ x² – 12x + 36 + 25 = x² + 8x + 16 + 25
⇒ – 12x + 36 = 8x + 16
⇒ – 12x – 8x = 16 – 36
⇒ – 20x = – 20
⇒ x = \(\frac{-20}{-20}\) = 1
अतः अभीष्ट बिन्दु के निर्देशांक = (1, 0).

प्रश्न 5.
यदि M(4, 5), रेखाखण्ड AB का मध्य बिन्दु है तथा 4 का निर्देशांक (3, 4) है, तो बिन्दु B के निर्देशांक ज्ञात कीजिए।
हल :
माना बिन्दु B के निर्देशांक (x, y) हैं।
दिया है (3, 4) तथा (x, y) के मध्य बिन्दु के निर्देशांक (4, 5) है।
4 = \(\frac{3+x}{2}\) ⇒ 3 + x = 8
x = 8 – 3 = 5
तथा 5 = \(\frac{y+4}{2}\) ⇒ 10 = y + 4
y = 10 – 4 = 6
अत: B के निर्देशांक B(5, 6) है।

प्रश्न 6.
यदि बिन्दु 4(x, y), B (- 5, 7) तथा C (- 4, 5) सरेखीय हों तो x तथा में सम्बन्ध ज्ञात कीजिए।
हल :
माना कि दिया गया बिन्दु (x, y), B(- 5, 7) और C(- 4, 5) है।
यहाँ x₁ = x, x₂ = – 5, x₃ = – 4
y₁ = y, y₂ = 7, y₃ = 5
तीन बिन्दु सरेखी होने के लिए ΔABC का क्षेत्रफल 0 होता है।
∴ \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\)[x(7 – 5) + (-5)(5 – y) + (-4) (y – 7)] = 0
⇒ \(\frac{1}{2}\)[2x – 5(5 – y) – 4(y – 7)] = 0
⇒ 2x – 25 + 5y – 4y + 28 = 0
⇒ 2x + y + 3 = 0
अत: 2x + y + 3 = 0 अभीष्ट सम्बन्ध हैं।

प्रश्न 7.
सिद्ध कीजिए कि बिन्दु A(2, – 1), B(3, 4), C (- 2, 3) तथा D(- 3, – 2) एक समचतुर्भुज ABCD के शीर्ष बिन्दु हैं। क्या ABCD एक वर्ग है?
हल :

= \(\sqrt{(-4)^2+(4)^2}\)
= \(\sqrt{16+16}\) = 4\(\sqrt{2}\)
∴ AB = BC = CD = AD
तथा विकर्ण AC ≠ विकर्ण BD
अत: ABCD वर्ग नहीं एक समचतुर्भुज है।
यही सिद्ध करना था ।

प्रश्न 8.
यदि बिन्दु A(1, – 2), B(2, 3), C(a, 2) और D (- 4, – 3) एक समान्तर चतुर्भुज बनाते हैं। a का मान तथा AB को आधार मानते हुए चतुर्भुज की ऊँचाई ज्ञात कीजिए।
हल :
बिन्दु 4 (1 , 2), B(2, 3), C(a, 2) तथा D(- 4, – 3) दिए गए समान्तर चतुर्भुज के शीर्ष हैं।
∵ समान्तर चतुर्भुज के विकर्ण परस्पर समद्विभाजित करते हैं।

∴ विकर्ण AC के मध्य बिन्दु O के निर्देशांक विकर्ण BD के मध्य बिन्दु O के निर्देशांक

अत: बिन्दु C के निर्देशांक = (- 3, 2)
आधार AB पर लम्ब DM खींचा। त्रिभुज ABD का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[1(3 – 3) + 2(3 + 2) + (- 4) (- 2 – 3)]
= \(\frac{1}{2}\)[0 + 10 + 20]
= \(\frac{1}{2}\) × 30
= 15 वर्ग इकाई
आधार (AB) की लम्बाई

प्रश्न 9.
बिन्दु A (4, 7), B (P, 3) तथा C (7, 3) एक समकोण त्रिभुज के शीर्ष हैं जिसमें B पर समकोण है। P का मान ज्ञात करो ।
हल :
समकोण ΔABC में,
AB² + BC² = AC² (पाइथागोरस प्रमेय से)

(3 – 7)² + (P – 4)² + (3 – 3)² + (7 – P)² = (3 – 7)² + (7 – 4)²
⇒ (P – 4)² + (7 – P)² = 9
⇒ P² + 16 – 8P + 49 + p² – 14P = 9
⇒ 2P² – 22P + 56 = 0
⇒ P² – 11P + 28 = 0
⇒ P² – 4P – 7P + 28 = 0
= P(P – 4) – 7(P – 4) = 0
= (P – 4) (P – 7) = 0
P = 4, 7
अत: P का मान 4, 7 है।

प्रश्न 10.
यदि बिन्दु 4 (6, 1), B(8, 2), C(9, 4) और D(x, y) क्रम में एक समान्तर चतुर्भुज के शीर्ष हैं, तो बिन्दु D(x, y) ज्ञात कीजिए।
हल :
बिन्दु 4 (6, 1), B(8, 2), C(9, 4) तथा D(x, y) दिए गए समान्तर चतुर्भुज के शीर्ष हैं।
∵ समान्तर चतुर्भुज के विकर्ण परस्पर समद्विभाजित करते हैं।

∴ विकर्ण AC के मध्य विन्दु के निर्देशांक = विकर्ण BD के मध्य बिन्दु के निर्देशांक

⇒ 8 + x = 15 और 2 + y = 5
⇒ x = 15 – 8 और y = 5 – 2
⇒ x = 7 और y = 3
अतः बिन्दु D के निर्देशांक = (7, 3).

प्रश्न 11.
बिन्दुओं P(- 3, 4) और Q(4, 5) को जोड़ने वाले रेखाखण्ड को समत्रिभाजित करने वाले बिन्दुओं के निर्देशांक ज्ञात कीजिए।
हल :
माना कि A(x₁, y₁) और B(x₂, y₂) अभीष्ट बिन्दु हैं जो बिन्दुओं P(- 3, 4) और Q( 4, 5) को जोड़ने वाले रेखा खण्ड को समत्रिभाजित करते हैं।

माना कि
PA = AB = QB = x
AQ = x + x = 2x
PB = x + x = 2x
\(\frac{PA}{AQ}=\frac{x}{2x}\) = \(\frac{1}{2}\) = 1 : 2
\(\frac{PB}{BQ}=\frac{2x}{x}\) = \(\frac{2}{1}\) = 2 : 1
अर्थात् बिन्दु A, PQ को 1 : 2 के अनुपात में तथा बिन्दु B, PQ को 2 : 1 के अनुपात में विभाजित करती है।
बिन्दु A के लिए:

अतः A और B के निर्देशांक क्रमश (\(\frac{5}{3}\), \(\frac{14}{3}\)) तथा (\(\frac{-2}{3}\), \(\frac{13}{3}\)) है।
यही सिद्ध करना था ।

प्रश्न 12.
यदि K(5, 4) रेखाखण्ड PQ का मध्यबिन्दु है तथा Q के निर्देशांक (2, 3) है तो P के निर्देशांक ज्ञात कीजिए ।
हल :
माना बिन्दु P के निर्देशांक (x, y) है।

तब बिन्दु P(x, y) तथा Q(2, 3) के मध्यबिन्दु K के निर्देशांक (5, 4).
5 = \(\frac{x+2}{2}\)
⇒ 10 = x + 2
⇒ x = 10 – 2 = 8
और 4 = \(\frac{y+3}{2}\)
⇒ 8 = y + 3
⇒ y = 8 – 3 = 5
P के निर्देशांक = (8, 5).

प्रश्न 13.
उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसके शीर्ष (- 3, – 2), (5, 2) और (5, 4) हैं। यह भी सिद्ध कीजिए कि यह समकोण त्रिभुज हैं।
हल :
माना ABC एक त्रिभुज है जिसके शीर्ष A(- 3, – 2), B (5, – 2) तथा C(5, 4) हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[- 3(- 2 – 4) + 5(4 + 2) + 5(- 2 + 2)]


∵ AB² + BC² = (8)² + (6)² = 100
= (10)² = CA²
⇒ AB² + BC² = CA²
अत: ΔABC एक समकोण त्रिभुज है।

प्रश्न 14.
त्रिभुज ABC, जिसमें A(1, – 4) तथा A से जाने वाली भुजाओं के मध्य बिन्दु (2, – 1) तथा (0, – 1) है, का क्षेत्रफल ज्ञात कीजिए।
हल
माना ABC एक त्रिभुज है, जहाँ B (x, y) तथा C(z, 1) है।
दिया है, P, AB का मध्य-बिन्दु है।
(2, – 1) = (\(\frac{1+x}{2}\), \(\frac{-4+y}{2}\))

∴ C के निर्देशांक = (- 1, 2)
∴ ΔABC के शीषों के निर्देशांक A(1, – 4), B(3, 2) तथा C(1, – 2) हैं।
ΔABC का क्षेत्रफल
= \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\)[1(2 – 2) + 3(2 + 4) – 1(- 4 – 2)]
= \(\frac{1}{2}\)[1(0) + 3(6) – 1(-6)]
= \(\frac{1}{2}\)[18 + 6] = 12 वर्ग इकाई

प्रश्न 15.
x – 3y = 0 बिन्दुओं (-2, -5) तथा (6, 3) को जोड़ने वाले रेखाखंड को किस अनुपात में विभाजित करती है? इस प्रतिच्छेद बिन्दु के निर्देशांक भी ज्ञात कीजिए।
हल :
माना रेखा x – 3y = 0, बिन्दुओं (- 2, – 5) तथा (6, 3) को जोड़ने वाले रेखाखंड को k : 1 अनुपात में विभाजित करती है।
यहाँ x₁ = – 2, x₂ = 6, y₁ = – 5, y₂ = 3, m = k तथा n = 1
∴ विभाजन सूत्र से,

अतः विभाजन बिन्दु (\(\frac{9}{2}\), \(\frac{3}{2}\)) तथा अभीष्ट अनुपात (\(\frac{13}{3}\) : 1) अथवा (13 : 3) है।

प्रश्न 16.
बिन्दु A, बिन्दुओं X(6, – 6) तथा Y(- 4, – 1) को मिलाने वाले रैखाखण्ड XY इस प्रकार स्थित है कि \(\frac{XA}{XY}=\frac{2}{5}\) है। यदि बिन्दु A रेखा 3x + k(y + 1) = 0 पर भी स्थित है, तो k का मान ज्ञात कीजिए।
हल :
दिया है, \(\frac{XA}{XY}=\frac{2}{5}\)

\(\frac{X A}{X A+A Y}\) = \(\frac{2}{5}\)
5XA = 2XA + 2AY
5XA = 2AY
\(\frac{XA}{XY}=\frac{2}{3}\)
अत: बिन्दु A रेखाखण्ड XY को 2 : 3 अनुपात में विभाजित करता है।
यहाँ m₁ = 2, m₂ = 3
अतः A बिन्दु के निर्देशांक

इसलिए A बिन्दु के निर्देशांक (2, – 4) है।
चूँकि बिन्दु A, रेखा 3x + k(y + 1) = 0 पर भी स्थित है, अतः यह रेखा के समीकरण की संतुष्ट करेगा।
∴ x = 2 तथा y = – 4 रखने पर,
3(2) + k (- 4 + 1) = 0
6 + k (- 3) = 0
– 3k = – 6
k = 2

प्रश्न 17.
यदि A(- 2, 1), B(a, 0), C(4, b) तथा D(1, 2) एक समानान्तर चतुर्भुज ABCD एके शीर्ष बिन्दु हैं, तो a तथा b के मान ज्ञात कीजिए। अतः इस चतुर्भुज की भुजाओं की लम्बाइयाँ ज्ञात कीजिए।
हल :
ज्ञात है, ABCD एक समानान्तर चतुर्भुज है।

विकर्ण AC तथा BD एक-दूसरे को बिन्दु O पर समद्विभाजित करते हैं।


अत: चतुर्भुज ABCD की भुजाओं की लम्बाई
AB = BC = CD = DA = \(\sqrt{10}\) इकाई

प्रश्न 18.
यदि बिन्दु A(k + 1, 2k), B(3k, 2k + 3) तथा C(5k – 1, 5k) सरेख हों, तो k का मान ज्ञात कीजिए।
हल :
बिन्दु A(k + 1, 2k), B (3k, 2k +3) और C(5k – 1, 5k) सरेख हैं। अतः त्रिभुज ABC का क्षेत्रफल = 0
Δ = \(\frac{1}{2}\)[(k + 1)(2k + 3) – 6k² + 15k² – (5k – 1)(2k + 3) + 2k(5k – 1) – (k + 1)(5k)]
0 = \(\frac{1}{2}\)[2k² + 5k + 3 – 6k² + 15k² – 10k² – 13k + 3 + 10k² – 2k – 5k² – 5k]
0 = \(\frac{1}{2}\)[6k² – 15k + 6]
⇒ 6k² – 15k + 6 = 0
⇒ 6k² – 12k – 3k + 6 = 0
⇒ 6k(k – 2) – 3 (k – 2) = 0
⇒ (k – 2) (6k – 3) = 0
⇒ k = 2 या k = \(\frac{1}{2}\)

प्रश्न 19.
दर्शाइए कि ΔABC जहाँ A(- 2, 0), B(2, 0) C(0, 2) तथा ΔPQR जहाँ P(- 4, 0), Q( 4, 0), R (0, 4) हैं, समरूप त्रिभुज है।
हल :
त्रिभुजों के शीर्षों के निर्देशांक है,
A (- 2, 0), B (2, 0), C(0, 2)
P(- 4, 0), Q(4, 0), R(0, 4)

यहाँ ΔPQR की भुजायें ΔABC को भुजाओं की दोगुनी हैं।
अत: दोनों त्रिभुज समरूप है। इति सिद्धम्

प्रश्न 20.
एक त्रिभुज का क्षेत्रफल 5 वर्ग इकाई है। इसके दो शीर्ष (2, 1) तथा (3, – 2) हैं। यदि तीसरा शीर्ष (\(\frac{7}{2}\), y), तो y का मान ज्ञात कीजिए।
हल :
दिया है, 4(2, 1), B(3, – 2) और C(\(\frac{7}{2}\), y)
अब, ΔMBC का क्षेत्रफल

प्रश्न 21.
निम्न आकृति में किसी कक्षा में रखे डेस्कों की व्यवस्था दर्शाइए गई है। आशिमा, भारती तथा आशा क्रमशः बिंदुओं A, B तथा C पर बैठी हैं। निम्न प्रश्नों के उत्तर दीजिए ।
(i) ज्ञात कीजिए कि क्या तीनों लड़कियाँ एक ही रेखा में बैठी हैं।
(ii) यदि A, B तथा C सरेख हैं तो ज्ञात कीजिए कि बिंदु B रेखाखंड AC को किस अनुपात में विभाजित करता है।

हल :
आलेख से बिंदु A, B और C के निर्देशांक हैं :
A = (3, 1), B (6, 4) तथा C = (8, 6)
A, B, C के सरेख के लिए प्रतिबंध
x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂) = 0
∴ x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)
= 3(4 – 6) + 6(6 – 1) + 8(1 – 4)
= 3(-2) + 6(5) + 8(-3)
= – 6+ 30 – 24
= 0
∴ A, B और C सरेख है।
अतः तीनों लड़कियाँ एक ही रेखा में बैठी है।

(ii) माना B, रेखाखंड AC को m : n अनुपात में विभाजित करता है।

⇒ 8m + 3n = 6m + 6n
⇒ 2m = 3n
⇒ \(\frac{m}{n}=\frac{3}{2}\)
⇒ m : n = 3 : 2
अत: B, AC को 3 : 2 के अनुपात में विभाजित करता है।

प्रश्न 22.
कृष्णा के पास एक सेवों का बाग है जिसके साथ एक 10 मी × 10 मी साइज का एक किचन गार्डन है। उसने उसे एक 10 × 10 ग्रिड के बाँटकर उसमें मिट्टी तथा खाद डाली है। उसने बिंदु पर एक नींबू का पौधा, बिंदु B पर धनिए का पौधा, बिंदु C पर प्याज का पौधा तथा बिंदु D पर एक टमाटर का पौधा लगाया है। उसका पति राम किचन गार्डन को देखकर तारीफ़ करता है तथा कहलाता है A, B, C तथा D को मिलाने पर शायद एक समांतर चतुर्भुज बन जाएं। नीचे दिए गए चित्र को ध्यानपूर्वक देखकर निम्नलिखित के उत्तर दीजिए :

(i) निर्देशांक अक्ष के रूप में 10 × 10 ग्रिड का उपयोग करते हुए बिंदुओं A, B, C तथा D के निर्देशांक ज्ञात कीजिए।
(ii) ज्ञात कीजिए कि क्या ABCD एक समांतर चतुर्भुज है या नहीं।
हल :
(i) चित्र से, A, B, C तथा D के निर्देशांक हैं :
A = (2, 2), B = (5, 4), C = (7, 7), D = (4, 5)

यहाँ AB = BC = CD = DA
लेकिन AC ≠ BD
अत: ABCD एक समांतर चतुर्भुज है।

प्रश्न 23.
यदि A (- 5, 7), B (- 4, – 5), C(- 1, – 6) तथा D(4, 5) एक समानान्तर चतुर्भुज ABCD के शीर्ष बिन्दु हैं, तो चतुर्भुज ABCD का क्षेत्रफल कीजिए ।
हल :
चतुर्भुज ABCD का एक विकर्ण BD को मिलाया।

ΔABD के लिए शीर्ष A(- 5, 7), B(-4, -5), D(4, 5)
ΔABD का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₃)]
= \(\frac{1}{2}\)[- 5(- 5 – 5) + (- 4)(5 – 7) + 4(7 + 5)]
= \(\frac{1}{2}\)[-5(-10) – 4(-2) + 4(12)]
= \(\frac{1}{2}\)[50 + 8 + 48]
= \(\frac{1}{2}\)[106]
= \(\frac{106}{2}\) = 53 वर्ग इकाई
ΔBCD के लिए शीर्ष
B(- 4, – 5), C(-1, – 6), D(4, 5)
ΔBCD का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₃)]
= \(\frac{1}{2}\) [-4(- 6 – 5) + (-1)(5 + 5) + 4(-5 + 6)]
= \(\frac{1}{2}\) [-4(- 11) – 1(10) + 4(1)]
= \(\frac{1}{2}\) [44 – 10 + 4]
= \(\frac{1}{2}\) [38] = \(\frac{38}{2}\)
= 19 वर्ग इकाई
चतुर्भुज ABCD का क्षेत्रफल = 53 + 19
= 72 वर्ग इकाई

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. x निर्देशांक को ……….. भी कहते हैं।
2. y- निर्देशांक को ………….. भी कहते हैं।
3. AOBC एक आयत है जिसके तीन शीर्ष- बिंदु A(0, – 3), O(0, 0) एवं B (4, 0) हैं इसके विकर्ण की लंबाई …………… है।
4. बिन्दुओं (- 3, – 3) तथा (- 3, 3) को जोड़ने वाले रेखाखंड का मध्य बिंदु ………………. है।
5. बिंदुओं (- a, a) तथा (- a, – a) के बीच की दूरी ………………. है।

हल :

1. भुज,
2. कोटि,
3. 5 इकाई,
4. (- 3, 0),
5. 2a इकाई ।

निम्न में सत्य / असत्य ज्ञात कीजिए :

प्रश्न (ख)

1. यदि बिंदु (3, – 6) बिंदुओं (0, 0) तथा (x, y) को जोड़ने वाले रेखाखंड का मध्य-बिंदु है, तो बिन्दु (x, y) का मान (6, – 12) होगा।
2. x – अक्ष पर स्थित वह बिंदु जो (2, 3) तथा (6, – 9) को जोड़ने वाले रेखाखंड को 1 : 3 के अनुपात में विभाजित करता है, के निर्देशांक (0, 3) है।
3. यदि बिंदुओं A(-3, b) तथा B (1, b + 4) को मिलाने वाले रेखाखंड का मध्यबिंदु P(- 1, 1) है, तो b का मान 2 है।
4. यदि एक वृत्त का केंद्र (3, 5) है तथा एक व्यास के अंत बिंदु (4, 7) तथा (2, y) हैं, तो y का मान 3 है।
5. बिंदुओं A(2, – 3) तथा B (5, 6) को मिलाने वाले रेखाखंड x-अक्ष को 1 : 2 के अनुपात में बाँटता है।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
बिंदुओं (a cos θ + b sin θ, 0) तथा (0, a sin θ – b sin θ) के बीच की दूरी है
(A) a² + b²
(B) a² – b²
(C) \(\sqrt{a^2+b^2}\)
(D) \(\sqrt{a^2-b^2}\)
हल :

अत: सही विकल्प (C) है।

प्रश्न 2.
यदि बिंदु P(k, 0), बिंदुओं A(2, – 2) तथा B(- 7, 4) को मिलाने वाले रेखाखंड को 1 : 2 के अनुपात में विभाजित करता है, तो k का मान है:
(A) 1
(B) 2
(C) – 2
(D) 1
हल :
बिंदु P के निर्देशांक

⇒ P (k, 0) = (- 1, 0)
∴ k = – 1
अत: सही विकल्प (D) है।

प्रश्न 3.
p का वह मान जिसके लिए बिंदु 4 (3, 1), B (5, p) तथा C(7, – 5) सरेख हैं, है :
(A) – 2
(B) 2
(C) – 1
(D) 1
हल :
बिंदु A(3, 1), B(5, p) तथा C(7, – 5) सरेख होंगे यदि,
x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) = 0
⇒ 3 (p + 5) + 5 (- 5 – 1) + 7 (1 – p) = 0
⇒ 3p + 15 – 30 + 7 – 7p = 0
⇒ – 4p – 8 = 0
⇒ – 8 = 4p
⇒ p = \(\frac{-8}{4}\) = – 2
अत: सही विकल्प (A) हैं।

प्रश्न 4.
x- अक्ष पर स्थित बिंदु P जो बिंदुओं B(5, 0) से समदूरस्थ है, हैं
(A) (2, 0)
(B) (0, 2)
(C) (3, 0)
(D) (2, 2)
हल :
माना x- अक्ष पद P(x, 0) कोई बिंदु है।
दिया है, PA = PB
⇒ PA² = PB²
(दोनों पक्षों का वर्ग करने पर)
⇒ (x + 1)² + (0 – 0)² = (x + 5)² + (0 – 0)²
⇒ x² + 1 + 2x = x² + 25 – 10x
⇒ 12x = 24
⇒ x = 2
∴ बिंदु (2, 0)
अत: सही विकल्प (A) हैं।

प्रश्न 5.
उस बिंदु के निर्देशांक जो बिंदु (- 3, 5) का x- अक्ष में प्रक्षेप है, है:
(A) (3, 5)
(B) (3, – 5)
(C) (- 3, – 5)
(D) (- 3, 5)
हल :
P(-3, 5) का प्रेक्षप बिंदु P'(- 3, – 5) है ।

अत: सही विकल्प (C) है।

प्रश्न 6.
यदि बिंदुओं A(10, – 6) तथा B (k, 4) को मिलाने वाले रेखाखंड का मध्य-बिन्दु (a, b) है, तथा a – 2b = 18 है, तो k का मान है:
(A) 30
(B) 22
(C) 4
(D) 40
हल :
प्रश्नानुसार,

प्रश्न 7.
उस त्रिभुज जिसके शीर्ष बिंदु (0, 4), (0, 0) तथा (3, 0) है का परिमाप है।
(A) 7 + \(\sqrt{5}\)
(B) 5
(C) 10
(D) 12
हल :
माना ABC एक त्रिभुज है।

त्रिभुज ABC की परिमाप = AB + BC + CA
= 4 + 3 + 5 = 12 इकाई ।
अत: सही विकल्प (D) है।

प्रश्न 8.
बिंदु P(3, 4) की x-अक्ष से दूरी है :
(A) 3 इकाई
(C) 5 इकाई
(B) 4 इकाई
(D) 1 इकाई
हल :
बिंदु P(3, 4) की x अक्ष से दूरी इस बिंदु के y-निर्देशांक के बराबर होती है।
अत: सही विकल्प (B) है।

प्रश्न 9.
यदि बिंदुओं A(4, p) तथा B (1, 0) के बीच की दूरी 5 इकाई है, तो का मान है :
(A) केवल 4
(B) केवल – 4
(c) ± 4
(D) 0
हल :
∵ AB = 5
⇒ \(\sqrt{(4-1)^2+(p-0)^2}\) = 5
⇒ \(\sqrt{3^2+p^2}\) = 5
⇒ 3² + p² = 25
(दोनों पक्षों का वर्ग करने पर)
⇒ p² = 25 – 9
⇒ p² = 16
⇒ p = ± 4
अत: सही विकल्प (C) है।

प्रश्न 10.
आयत AOBC के तीन शीर्ष A(0, 3), O(0, 0) तथा C(5, 0) है। इसके विकर्ण की लम्बाई है:
(A) 5
(B) 3
(C) \(\sqrt{34}\)
(D) 6
हल :
आयत AOBC के तीन शीर्ष 4 (0, 3), O(0, 0) तथा C(5, 0) है।
आयत के विकर्ण (AC) की लम्बाई
= बिन्दु A(0, 3) तथा C(5, 0) के बीच की दूरी (AC)

अत: विकल्प (C) सही है।

प्रश्न 11.
यदि (a, b + c), (b, c + a) और (c, a + b) त्रिभुज के शीर्ष बिन्दु हैं तो त्रिभुज का क्षेत्रफल है :
(A) (a + b + c)²
(B) 0
(C) a + b + c
(D) abc
हल :
(a, b + c), (b, c + a) और (c, a + b) दिए गए त्रिभुज के शीर्ष बिन्दु हैं।
यहाँ
x₁ = a,
x₂ = b,
x₃ = c
y₁ = b + c
y₂ = c + a
y₃ = a + b
त्रिभुज का क्षेत्रफल
= \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂))]
= \(\frac{1}{2}\)[a(c + a – a – b) + b(a + b – b – c) + c(b + c – c – a)]
= \(\frac{1}{2}\)[a(c – b) + b(a – c) + c(b – a)]
= \(\frac{1}{2}\)[ac – ab + ab – bc + bc – ac]
= \(\frac{1}{2}\) × 0 = 0
अत: विकल्प (B) सही है।

प्रश्न 12.
यदि एक चतुर्भुज के शीर्ष (1, 4), (- 5, 4), (- 5, – 3) और (1, – 3) हों, तो चतुर्भुज का प्रकार है:
(A) वर्ग
(B) आयत
(C) समान्तर चतुर्भुज
(D) समचतुर्भुज
हल :
माना कि चतुर्भुज के शीर्ष 4(1, 4), B(- 5, 4) तथा D(1, – 3) है तब

अतः AB = CD और BC = DA और
विकर्ण AC = विकर्ण BD

अत: दिये गये बिन्दु आयत के शीर्ष हैं।
अत: सही विकल्प (B) है।

प्रश्न 13.
यदि बिन्दु (1, 2), O(0, 0) तथा C (a, b) संरेखी हैं, तब :
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = – b
हल :
यदि दिए गए बिन्दु A(1, 2), O(0, 0) तथा C(a, b) सरेखी हैं, तो इनसे बने त्रिभुज का क्षेत्रफल शून्य होगा ।
त्रिभुज ABC का क्षेत्रफल = 0
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac{1}{2}\)[1(0 – b) + 0(b – 2) + a(2 – 0)] = 0
⇒ \(\frac{1}{2}\)[- b + 0 + 2a] = 0
⇒ 2a – b = 0
⇒ 2a = b
अत: विकल्प (C) सही है।

प्रश्न 14.
यदि A(- 2, – 1), B (a, 0), C(4, b) और D (1, 2) समान्तर चतुर्भुज के शीर्ष हों, तो a और b का मान होगा :
(A) 1, 3
(B) 2, 4
(C) 2, 3
(D) 1, 4
हल :
समान्तर चतुर्भुज के विकर्ण परस्पर सम- द्विभाजित करते हैं।
∴ विकर्ण AC का मध्य-बिन्दु = विकर्ण BD का मध्य बिन्दु

अत: सही विकल्प (A) है।

प्रश्न 15.
आकृति में बिन्दु P(5, – 3) तथा Q(3, y) बिन्दुओं A (7, – 2) तथा B (1, – 5) को मिलाने वाले रेखाखण्ड को समनिभाजित करते हैं, तो y बराबर है-

(A) 2
(B) 4
(C) – 4
(D) \(\frac{-5}{2}\)
हल :
∵ बिन्दु P तथा Q, AB को समत्रिभाजित करते हैं।

∴ AP = PQ = BQ
AQ = AP + PQ = 2AP
∴ \(\frac{AQ}{BQ}=\frac{2AP}{AP}\) = \(\frac{2}{1}\) = 2 : 1
बिन्दु Q का निर्देशांक है।

अतः विकल्प (C) सही है।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 4 Carbon and Its Compounds Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 4 Carbon and Its Compounds

Additional Questions and Answers

Question 1.
(1) Write the IUPAC names of the following compounds using their structural formulae:
1. CH₃ – CH₂ – CH₂ – Cl

7. CH₃ – CH₂ – CH₂ – CHO
8. CH₃ – CHO
9. CH₃ – CH₂ – CH₂ – CH₂ – CHO
10. CH₃ – CO – CH₃
11. CH₃ – CH₂ – CO – CH₃
12. CH₃ – CH₂ – CH₂ – CO – CH₃
13. CH₃ – CH₂ – COOH
14. CH₃ – CH₂ – CH₂ – COOH
15. CH₃ – COOH
16. CH₃ – CH = CH₂
17. CH₃ – CH₂ – CH = CH₂
18. CH₃ – C ≡ CH
19. CH₃ – CH₂ – C ≡ CH
20. CH ≡ CH
Answer:

Structural formula	IUPAC name
1. CH3 – CH2 – CH2 – Cl	1-Chloropropane
	1-Bromo-2-methylpropane
	3-Bromopentane
4. CH3 – CH2 – CH2 – CH2 – OH	Butan-l-ol(Butanol)
	2-Methylpropan-2-ol
	Butan-2-ol
7. CH3 – CH2 – CH2 – CHO	Butanal
8. CH3 – CHO	Ethanal
9. CH3 – CH2 – CH2 – CH2 – CHO	Pentanal
10. CH3 -CO-CH3	Propanone
11. CH3 – CH2 – CO – CH3	Butanone
12. CH3 – CH2 – CH2 – CO – CH3	2-Pentanone
13. CH3 – CH2 – COOH	Propanoic acid
14. CH3 – CH2 – CH2 – COOH	Butanoic acid
15. CH3 – COOH	Ethanoic acid
16. CH3 – CH = CH2	Propene
17. CH3 – CH2 – CH = CH2	1-Butene (or But-l-ene)
18. CH3 – C = CH	1-Propyne
19. CH3 – CH2 – C = CH	1-Butyne (or But-l-yne)
20. CH = CH	Ethyne

(2) Give structural formulae of the following compounds using their IUPAC names:
1. 2-Chloropropane
2. 1-Bromopentane
3. 2-Bromo-2-methylpropane
4. 1-Pentanol
5. Ethanol
6. Methanol
7. Methanal
8. Ethanal
9. 3-Pentanone
10. Propanone
11. Methanoic acid
12. Pentanoic acid
13. Ethene
14. 2-Butene
15. 1-Pentyne
Answer:

IUPAC name	Structural formula
1. 2-Chloro-propane
2. 1-Bromo-pentane	CH3 – CH2 – CH2 – CH2 – CH2 – Br
3. 2-Bromo-2-methyl-propane
4. 1-Pentanol	CH3 – CH2 – CH2 – CH2 – CH2 – OH
5. Ethanol	CH3 – CH2 – OH
6. Methanol	CH3 – OH
7. Methanal	H – CHO
8. Ethanal	CH3 – CHO
9. 3-Pentanone	CH3 – CH2 – CO – CH2 – CH3
10. Propanone	CH3 – CO – CH3
11. Methanoic acid	HCOOH
12. Pentanoic acid	CH3 – CH2 – CH2 – CH2 – COOH
13. Ethene	CH2 = CH2
14. 2-Butene	CH3 – CH = CH – CH3
15. 1-Pentyne	CH3 – CH2 – CH2 – C ≡ CH

Question 2.
Distinguish between:
(1) Ionic compounds and Covalent compounds
Answer:

Ionic compounds	Covalent compounds
1. The compounds formed by transfer of electrons between two atoms are known as ionic compounds.	1. The compounds formed by sharing of electrons between two atoms are known as covalent compounds.
2. They are mostly solids.	2. They may be solid, liquid or gases.
3. They have comparatively high melting point and boiling point.	3. They have comparatively low melting point and boiling point.
4. They conduct electricity through a solution or in molten state.	4. They are generally non-conductors of electricity.

(2) Diamond and Graphite
Answer:

Diamond	Graphite
1. In diamond, each carbon atom is covalently bonded to four other carbon atoms forming a hard and three-dimensional tetrahedral structure.	1. In graphite, each carbon atom is covalently bonded to three other carbon atoms forming two-dimensional hexagonal structure.
2. It is the hardest natural substance known.	2. It is soft and greasy.
3. It is non-conductor of electricity.	3. It is good conductor of electricity.
4. Chemically, diamond is unreactive.	4. Chemically, graphite is reactive.

(3) Saturated carbon compounds and Unsaturated carbon compounds
Answer:

Saturated carbon compounds	Unsaturated carbon compounds
1. Compounds of carbon which have only single covalent bonds between carbon atoms are called saturated carbon compounds.	1. Compounds of carbon which have double or triple bonds between carbon atoms are called unsaturated carbon compounds.
2. They are less reactive.	2. They are more reactive.
3. For example, compounds of alkanes.	3. For example, compounds of alkenes and alkynes.
4. They burn with blue flame.	4. They burns with yellow sooty flame and produce black smoke.

Question 3.
Give scientific reasons for the following statements:
(1) Candle burns with a yellow flame.
Answer:
Candle is made-up of wax, which consists of saturated hydrocarbons containing 18 to 20 carbon atoms.

• When a candle is lighted (ignited), the wax melts s which rises up the wick and gets covered by vapour. As a result, there is no proper mixing of O₂ of air and wax vapours.
• Therefore, here incomplete combustion of wax takes place, and complete combustion of wax does not take place.
• Furthermore, incomplete combustion of wax s produces unburnt carbon particles, which rise up in the flame, get heated and imparts yellow colour to the flame.

(2) Burning substances (fuels) burn with or without a flame.
Answer:
When the gaseous substances are heated, a luminous flame is seen and they start to glow.

• Some liquid and solid fuels burn with a flame because on heating, their atoms easily get converted into vapours; and burn with a luminous flame starting with glow.
• When volatile substances present in this type of fuel vapourises, then the fuel just glows red and gives out heat without flame.

(3) Diamond has high melting point, in spite of having covalent bonds.
Answer:
In diamond, each carbon atom is attached to four other carbon atoms by strong covalent bonds and forms three-dimensional tetrahedral structure.

• A large amount of energy is needed to break the network of large number of covalent bonds.
• Hence, diamond has high melting point.

Objective Questions and Answers

Question 1.
Answer the following questions in one word or in one sentence:

1. State the proportion of carbon in the earth’s crust in the form of minerals.
2. Mention the percentage proportion of carbon dioxide in the atmosphere.
3. Write the number of electrons in shell of carbon, oxygen and nitrogen respectively.
4. Mention the major component of biogas and CNG.
5. Are the boiling points and melting points of covalent compounds high or low in comparison to ionic compounds?
6. Write the property of electrical conductivity of carbon compounds.
7. Which allotrope of carbon is very hard?
8. Name the element which is next to carbon according to catenation property.
9. What is the number of carbon compounds estimated approximately?
10. State the general formula of saturated alkane compounds.
11. State the general formula of unsaturated alkene compounds.
12. Mention the general formula of unsaturated alkync compounds.
13. Compare the chemical reactivity of saturated and unsaturated compounds.
14. State the minimum number of carbon atoms required to form the ring structure.
15. How many isomers are possible for hexane?
16. Write the structural formula of functional groups (i) aldehyde, (ii) ketone and (iii) carboxylic acid respectively.
17. What is the difference of molecular masses between two successive compounds of homologous series?
18. State the elements used as catalyst in hydrogenation reaction.
19. Which ions are present in hard water?
20. Identify the functional groups in butanone and butanal respectively.

Answer:

1. 0.02 %
2. 0.03 %
3. 4, 6 and 5
4. Methane
5. Melting points and boiling points of covalent compounds are comparatively lower than the melting points and boiling points of ionic compounds.
6. Carbon compounds are non-conductor of electricity.
7. Diamond
8. Sulphur
9. Three million
10. C_nH_2n+2
11. C_nH_2n
12. C_nH_2n-2
13. Unsaturated compounds are more reactive chemically than the saturated compounds.
14. Three
15. Five
16. 
17. 14u
18. Nickel, palladium
19. Calcium (Ca²⁺), magnesium (Mg²⁺)
20. Ketone, aldehyde

Question 2.
Define:
(1) Covalent bond
Answer:
A chemical bond formed between two or more atoms by mutual sharing of valence electrons is known as a covalent bond.

(2) Catenation
Answer:
Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to a large number of molecules. This property of carbon is called catenation.

(3) Saturated carbon compounds
Answer:
Compounds of carbon which are linked by only single covalent bonds between carbon atoms are called saturated carbon compounds.

(4) Unsaturated carbon compounds
Answer:
Compounds of carbon which have double or triple bonds between carbon atoms are called unsaturated carbon compounds.

(5) Hydrocarbons
Answer:
The compounds containing carbon and hydrogen only are called hydrocarbons.

(6) Functional group
Answer:
An atom or a group of atoms which imparts specific properties to the compound is called a functional group.

(7) Homologous series
Answer:
A series of organic compounds in succession which differ by a definite group (like – CH₂ -) is called homologous series.

(8) Combustion reaction
Answer:
Carbon (or most of the organic compounds), in all its allotropic forms, burns in air to give carbon dioxide along with the release of heat and light. This is called combustion reaction.

(9) Oxidising agent
Answer:
Some substances are capable of adding oxygen to other substances. These substances are known as oxidising agents.

(10) Addition reaction
Answer:
The reaction in which unsaturated hydrocarbons add hydrogen in the presence of catalysts such as palladium or nickel to form saturated hydrocarbons is called an addition reaction.

(11) Substitution reaction
Answer:
The reaction in which H-atoms of saturated hydrocarbons are displaced by another atom or functional group is called a substitution reaction.

(12) Esterification reaction
Answer:
A reaction in which a carboxylic acid and an alcohol react in the presence of acid catalyst forming esters and water is known as an esterification reaction.

(13) Saponification reaction
Answer:
The reaction of forming alcohol and sodium salt of carboxylic acid from ester is known as saponification.

Question 3.
Fill in the blanks :

1. All living structures are based on …………………
2. The earth’s crust has ………………… carbon in the form of minerals.
3. Carbon possesses ………………… electrons in its outermost shell.
4. The atomic number of chlorine is
5. The molecular formula of ammonia is …………………
6. ………………… is a major component of biogas and CNG.
7. Carbon atoms are linked together with a ………………… in saturated carbon compounds.
8. Fuels such as coal and petroleum contain some amount of ………………… and ………………… in them.
9. The molecular formula of propane is …………………
10. The general formula of alkene is …………………
11. Most of the carbon compounds release a large amount of ………………… and ………………… on burning.
12. Unsaturated hydrocarbons gives a ………………… flame with lots of ………………… smoke.
13. Alcohols are oxidised to …………………
14. ………………… are substances that cause a reaction to proceed at a higher rate without the reaction itself being affected.
15. ………………… catalyst is used in hydrogenation of vegetable oils.
16. Animals fats contain esters of …………………
17. Substitution of hydrogen of methane by chlorine in the presence of sunlight forms …………………
18. ………………… is used in the preparation of tincture iodine.
19. Reaction of alcohol with sodium forms …………………
20. Concentrated H₂SO₄ is a …………………
21. Reaction of ethanoic acid with ethanol in presence of acid catalyst forms …………………
22. The molecules of soap are ………………… salts of long chain carboxylic acids.
23. Micelles forms an ………………… in water.
24. Unsaturated hydrocarbons add hydrogen in the presence of catalyst such as ………………… to give saturated hydrocarbons.
25. Compound made-up of only carbon and hydrogen is called …………………

Answer:

1. carbon
2. 0.02 %
3. four
4. 17
5. NH₃
6. Methane
7. single bond
8. nitrogen, sulphur
9. C₃H₈
10. C_nH_2n
11. heat, light
12. yellow, black
13. carboxylic acids
14. Catalysts
15. Nickel (Ni)
16. saturated fatty acids
17. CH₃Cl
18. Ethanol
19. hydrogen gas (H₂)
20. dehydrating agent
21. ester (ethyl acetate) and water
22. sodium/ potassium
23. emulsion
24. palladium, nickel (Pd/Ni)
25. hydrocarbon

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

1. Propene is a saturated hydrocarbon.
2. Alkenes or alkynes are more reactive than their corresponding alkane.
3. The difference in number of atoms or molecular formula between two successive members in homologous series is – CH₂.
4. The solution of ethanol containing 5 % water is called as an absolute alcohol.
5. The molecular formula of formaldehyde is HCHO.
6. IUPAC name of CH₃CH₂COOH is butanoic acid.
7. Detergents are used to prepare shampoos and products for cleaning clothes.
8. Non-polar end of soap is hydrophilic while the polar end (head) is hydrophobic.
9. Soap solution appears cloudy because micelles scatter the light.
10. Reaction of carboxylic acid with sodium carbonate produces carbon dioxide.
11. Dyes are added to colour the alcohol blue. This is called denatured alcohol.
12. Dehydration of ethanol gives propene.
13. Coal and petroleum have some amount of nitrogen and sulphur in them, hence, their combustion causes pollution in environment.

Answer:

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

Question 5.
Match the following:
(1)

Answer:
(1 → S), (2 → P), (3 → Q), (4 → R).

(2)

Column I (Molecule)	Column II (Number of bonds)
1. Molecule of hydrogen	P. Double bond and single bond
2. Molecule of nitrogen	Q. Only single bond
3. Molecule of oxygen	R. Triple bond
4. Benzene	S. Only double bond

Answer:
(1 → Q), (2 → R), (3 → S), (4 → P).

(3)

Answer:
(1 → S), (2 → P), (3 → Q), (4 → R).

Question 6.
Mention the formulae, and names of the products in the following reactions:
( 1 ) C + O₂ →
( 2 ) CH₄ + O₂ →
( 3 ) CH₃CH₂OH + O₂ →

( 14 ) CH₃COOH + KOH →
( 15 ) CH₃CH₂COOH + KOH →
( 16 ) CH₃COOH + Na₂O₃ →
( 17 ) CH₃CH₂COOH + Na₂CO₃ →
( 18 ) CH₃COOH + NaHCO₃ →
( 19 ) CH₃CH₂COOH + NaHCO₃ →
( 20 ) CH₃CH₂CH₂COOH + NaHCO₃ →
Answer:

Question 7.
Choose the correct option from those given below each question:
1. By which name the compounds containing functional group – CHO are known?
A. Amide
B. Aldehyde
C. Ketone
D. Alcohol
Answer:
B. Aldehyde

2. Which functional group is present in carboxylic acid?
A.
B. – COOH
C. – CHO
D. – OH
Answer:
B. – COOH

3. Which functional group is to be given suffix -ol in the nomenclature?
A. – CHO
B.
C. – OH
D. – X
Answer:
C. – OH

4. Which functional group is present in meyhyl ethanoate?
A. Alcohol
B. Halide
C. Ketone
D. Ester
Answer:
D. Ester

5. Which of the following is obtained by the reduction of methanal?
A. Ethanol
B. CO₂ and O₂
C. Methanol
D. All of the given
Answer:
C. Methanol

6. Which of the following reaction takes place between the reaction of alcohol and carboxylic acid in the presence of concentrated H₂SO₄?
A. Hydrolysis
B. Beta elimination
C. Saponification
D. Esterification
Answer:
D. Esterification

7. In order to form a compound, which of the following functional group possess minimum three carbon atoms?
A. – COOH
B. – CHO
C.
D. – C – O –
Answer:
C.

8. Which functional group is present in ketone?
A.
B. – COOH
C. – CHO
D. – OH
Answer:
A.

9. Which functional group is present in aldehyde?
A.
B. – COOH
C. – CHO
D. – OH
Answer:
C. – CHO

10. Name the substance having functional group – OH.
A. Alcohol
B. Ketone
C. Ester
D. Carboxylic acid
Answer:
A. Alcohol

11. How many carbon atom/s are present in formic acid?
A. 1
B. 2
C. 3
D. 4
Answer:
A. 1

12. Which of the following is used as food preservator?
A. CH₃OH
B. CH₃COOH
C. CH₃CHO
D. CH₃COCH₃
Answer:
B. CH₃COOH

13. In soap and detergent, non-polar tail is attracted towards ………………. and an anionic head is attracted towards ……………….
A. stain, glycerol
B. water molecules, stain
C. stain, water molecules
D. water molecules, glycerol
Answer:
C. stain, water molecules

14. Write the common name of ethanoic acid.
A. Formic acid
B. Acetic acid
C. Propanoic acid
D. Butanoic acid
Answer:
B. Acetic acid

15. A molecule of NH₃ (ammonia) has …
A. only single bonds.
B. only double bonds.
C. only triple bonds.
D. two double bonds and one single bond.
Answer:
A. only single bonds.

16. Fullerene is an allotropic form of…
A. phosphorus.
B. sulphur.
C. carbon.
D. tin.
Answer:
C. carbon.

17. Which of the following are correct structural isomers of butane?

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

18.
In the above given reaction, alkaline KMnO₄ acts as …
A. reducing agent.
B. oxidising agent.
C. catalyst.
D. dehydrating agent.
Answer:
B. oxidising agent.

19. Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats. This is an example of…
A. addition reaction.
B. substitution reaction.
C. rearrangement reaction.
D. oxidation reaction.
Answer:
A. addition reaction.

20. Structural formula of ethyne is …
A. H – C ≡ C – H
B. H₃C – C ≡ CH

Answer:
A. H – C ≡ C – H

21. Which of the following are unsaturated compounds ?
(i)Propane
(ii) Propene
(iii) Propyne
(iv) Chloropropane
A. (i) and (ii)
B. (ii) and (iv)
C. (iii) and (iv)
D. (ii) and (iii)
Answer:
D. (ii) and (iii)

22. In which condition, chlorine reacts with saturated hydrocarbons at room temperature?
A. In the absence of sunlight
B. In the presence of sunlight
C. In the presence of water
D. In the presence of hydrochloric acid
Answer:
B. In the presence of sunlight

23. Which products are formed by the reaction between ethanol and sodium?
A. Sodium ethanoate and hydrogen
B. Sodium ethanoate and oxygen
C. Sodium ethoxide and hydrogen
D. Sodium ethoxide and oxygen
Answer:
C. Sodium ethoxide and hydrogen

24. The correct structural formula of butanoic acid is …

Answer:

25. Vinegar is a solution of…
A. 50%-60% acetic acid in alcohol.
B. 5%-8% acetic acid in alcohol.
C. 5%-8% acetic acid in water.
D. 50%-60% acetic acid in water.
Answer:
C. 5%-8% acetic acid in water.

26. Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, for example, hydrogen. After the formation of four bonds, carbon attains the electronic configuration of….
A. helium
B. neon
C. argon
D. krypton
Answer:
B. neon

27. Which of the following compounds does not belong to the homologous series?
A. CH₄
B. C₂H₆
C. C₃H₆
D. C₄H₈
Answer:
D. C₄H₈

28. The first member of alkyne homologous series is … .
A. ethyne
B. ethene
C. propyne
D. methane
Answer:
A. ethyne

29. Which of the following represents saponification reaction?

Answer:
CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH

30. The number of covalent bonds present in pentane is –
A. 5
B. 12
C. 16
D. 17
Answer:
C. 16

Question 8.
Choose more than one correct options from those given below each question:
1. Which of the following are not correct for carbon compounds?
A. Three hydrocarbon compounds can be represented by a general formula.
B. Three hydrocarbon compounds are isomers of each other.
C. Three hydrocarbon compounds are unsaturated hydrocarbons.
D. Chemical reactions of three hydrocarbon compounds are similar.
Answer:
B, D.

2. Which of the following are unsaturated compounds?
A. Ethene, ethyne, propene, propyne
B. Ethyne, butyne, propanol, butanol
C. Ethene, propene, butene, pentene
D. Ethyne, propyne, butyne, pentyne
Answer:
A, C, D

3. Which of the following statements are correct for the given hydrocarbon compounds?

A. Three hydrocarbon compounds can be represented by a general formula.
B. Three hydrocarbon compounds are isomers of each other.
C. Three hydrocarbon compounds are unsaturated hydrocarbons.
D. Chemical reactions of three hydrocarbon compounds are similar.
Answer:
A, B, D

4. Which of the following are isomers of butane?

Answer:
A, B

5. Which of the following reactions are correct?

Answer:
B, C, D

Question 9.
In each of the following questions, two statements are given. A statement of Assertion (A) is followed by another statement of Reason (R). Study the statements carefully and choose the correct option:
A. Both assertion (A) and reason (R) are true, and reason (R) is the true explanation of the assertion (A).
B. Both assertion (A) and reason (R) are true, but reason (R) is not the true explanation of the assertion (A).
C. Assertion (A) is true and reason (R) is false.
D. Both assertion (A) and reason (R) are false.
1. (A) A mixture of oxygen and ethyne is burnt for welding.
(R) Burning of the mixture releases a large amount of heat.
Answer:
A. Both assertion (A) and reason (R) are true, and reason (R) is the true explanation of the assertion (A).

2. (A) Acidified K₂Cr₂O₇ is used in oxidation of ethanol.
(R) Acidic K₂Cr₂O₇ is a reducing agent.
Answer:
C. Assertion (A) is true and reason (R) is false.

3. (A) Animal fats should be chosen for cooking.
(R) Animal fats are not harmful for health.
Answer:
D. Both assertion (A) and reason (R) are false.

4. (A) Carbon could not form C^4- anion by gaining four electrons.
(R) It would be difficult for carbon atom with six protons in its nucleus to accomodate ten electrons.
Answer:
A. Both assertion (A) and reason (R) are true, and reason (R) is the true explanation of the assertion (A).

5. (A) C₂H₅OH, C₃H₇OH and C₄H₉OH are members of homologous series.
(R) These compounds do not have same functional group.
Answer:
C. Assertion (A) is true and reason (R) is false.

Value Based Questions With Answers

Question 1.
One day Mudra was talking to her mother who was cooking vegetables in a stainless steel? utensils. Mudra observed that the bottom of cooking utensils was getting blackened from outside. She showed this to her mother. The mother told Mudra that the bottoms of all the cooking utensils kept on the gas stove were getting blackened for the last few days and she had tough time cleaning these utensils.

Being a science student of class X, Mudra S checked the gas stove thoroughly and could understand the reason for this problem. She explained everything to her mother. As Mudra S was getting late for school, she asked her mother to take a particular step to stop the blackening of cooking utensils. Mudra’s mother S did the same. The mother was glad that the? bottoms of cooking utensils kept on gas burner were no longer being blackened.

Questions:

1. Why is the bottom of cooking utensils getting blackened? Explain.
2. Apart from blackening the utensils, state two other disadvantages in this condition.
3. What did Mudra find on checking the gas stove thoroughly which was causing this problem?
4. What step was taken by Mudra to get rid of this problem?
5. What type of flame was produced by the gas stove burner after the required step was taken by Mudra’s mother?
6. What values are shown by Mudra in this act?

Answer:

1. The fuel LPG is burning incompletely in the gas stove producing yellow, sooty flame. The unburnt particles present in sooty flame stick to the bottom of the cooking utensils and blacken it.
2. Since the fuel is not burning completely, there is wastage of fuel. and The incomplete combustion of LPG puts unburnt carbon particles into the air and pollutes the environment.
3. Mudra had found that the air holes of the gas stove were partially blocked.
4. Mudra had suggested her mother to clean the holes of the gas burner properly to bring in free flow of air sufficient for the complete combustion of fuel.
5. After the air holes of the gas stove were opened fully by proper cleaning, then sufficient air was made available for the complete combustion of fuel to produce smokeless, blue flame.’
6. The values shown by Mudra are : (1) Good observation skills, (2) Understanding of combustion of fuels, (3) Application of knowledge in everyday life and (4) Concern for the environment.

Question 2.
Yug studies in tenth standard. One day his science teacher was discussing about oils and fats in the class. During this discussion, Yug came to know many facts about oils and fats which he did not know earlier. When Yug came back home from school, he asked his mother about oil she used to prepare food for the family.

His mother replied that she was using vegetable ghee for cooking food. Yug requested his mother not to use vegetable ghee because it is harmful for health. He asked her to use vegetable oil for cooking the food because vegetable oil is good for health. Yug’s mother agreed to do the same; and she started the use of vegetable oil.

Questions :

1. Write the physical states of vegetable ghee and vegetable oil.
2. How do vegetable oils and fats differ chemically?
3. By which method, the vegetable oil is converted into vegetable ghee?
4. Why is vegetable ghee not considered good for health?
5. Name the most common animal fat consumed by people.
6. Why is vegetable oil better for health?

Answer:

1. Vegetable ghee is a solid and vegetable oil is a liquid at room temperature.
2. Vegetable oils are unsaturated organic compounds while fats are saturated organic compounds.
3. Vegetable oil is converted into vegetable ghee by hydrogenation of vegetable oil.
4. Vegetable ghee consists of saturated fatty acids, which raise the level of bad cholesterol in blood after a long time use and increase the risk of heart disorder. Hence vegetable ghee is not good for health.
5. Butter is the most common animal fat consumed by people.
6. Vegetable oil contains unsaturated fatty acids, which does not increase the level of bad cholesterol in blood and reduces the risk of heart disorder. Therefore, vegetable oil is preferable and good for health.

Practical Skill Based Questions With Answers

Question 1.
An organic compound A’ is widely used to preserve the pickles and has a molecular formula C₂H₄O₂. This compound reacts with ethanol to form a sweet smelling compound ‘B
( 1 ) Identify the compound A.
( 2 ) Write the chemical equation for the reaction of A with ethanol in the presence of acid.
(3 ) Which gas is produced when compound A reacts with washing soda? Write the chemical equation.
(4) How can we obtain compound A from ‘B’?
Answer:
( 1 ) Compound A is acetic acid (CH₃COOH).

( 3 ) Carbon dioxide gas is produced :

( 4 ) Saponification :

Question 2.
Identify the compounds A, B, C, D, E and F in the following reactions:

Answer:
( 1 ) A = Alkaline KMnO₄ or Acidic K₂Cr₂O₇
( 2 ) B = CH₃COOC₂H₅
( 3 ) C = CH₃COONa, B = CH₃COOC₂H₅
( 4 ) D = CH₃COOH, E = CO₂
( 5 ) E = CO₂, F = CaCO₃

Question 3.
An organic compound P when heated with excess concentrated H₂SO₄ at 443 K produced compound Question The number of carbon atoms in P and Q are same. Compound Q when treated with hydrogen in presence of nickel or palladium formed compound R. One mole of ‘R’ when reacts with sufficient O₂ gives two moles of carbon dioxide and three moles of water. Identify the compounds P, Q and R. Write chemical equations for the reactions.
Answer:
P : C₂H₅OH (Ethanol)
Q : H₂C = CH₂ (Ethene)
R : H₃C – CH₃ (Ethane)
Chemical reactions :

Memory Map

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 2 बहुपद Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 2 बहुपद

लयूत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
बहुपद ax² + bx + c के आलेखों की आकृतियाँ, चित्र में दी गयी हैं। प्रत्येक आकृति में a, b और c के चिह्न ज्ञात कीजिए:

हल :
(i) बहुपद y = ax² + bx + c का परवलय नीचे की ओर खुलता है।
∴ a < 0 होगा तथा परवलय का शीर्ष प्रथम चतुर्थांश में है। –\(\frac{b}{2a}\) > 0 ⇒ – b < 0 ⇒ b > 0
परलवय y-अक्ष को बिन्दु P पर प्रतिच्छेद करता है। y-अक्ष पर x = 0 होता हैं। x = 0 का मान समीकरण y = ax² + bx + c में रखने पर हम y = c प्राप्त करते हैं। अतः P के निर्देशांक (o, c) हैं। ∵ P, y- अक्ष की धनात्मक दिशा में है । अतः
∴ c > 0.
∴ a < 0, b > 0 तथा c > 0

(ii) दिशा है : y = ax² + bx + c परवलय ऊपर की ओर खुलता है। अत: a > 0 हैं। परवलय के शीर्ष चौथे चतुर्थाश में हैं।
∴ – \(\frac{b}{2a}\) > 0 ⇒ b < 0 ⇒ b < 0
परवलय y-अक्ष को बिन्दु p पर प्रतिच्छेद करता है। x = 0, अत: y = ax 2 + bx + c में x = 0 रखने पर y = c प्राप्त होता हैं।
∴ p के निर्देशांक (0, c) हैं।
∵ बिन्दु P, OY’ पर स्थित है। अतः c < 0 होगा। a > 0, b < 0 ओर c <0

(iii) बहुपद y = ax² + bx + c का परवलय ऊपर की ओर खुलता है। अतः a > 0 है।
परवलय का शीर्ष OX पर स्थित है।
∴ – \(\frac{b}{2a}\) > 0 ⇒ – b > 0 ⇒ b < 0 परवलय y = ax² + bx + c Y-अक्ष को बिन्दु P पर प्रतिच्छेद करता है जो OY पर स्थित है। समीकरण y = ax² + bx + c में x = 0 रखने पर, y = c प्राप्त होता है। अत: बिन्दु P के निर्देशांक (0, c) हैं। ∵ P, OY अक्ष पर स्थित है। ∴ c > 0 होगा।
∴ a > 0, b < 0 और c > 0.

(iv) परवलय y = ax² + bx + c नीचे की ओर खुलता है। अत: a < 0 है।
परवलय का शीर्ष (\(\frac{-b}{2a}\), \(\frac{-D}{4a}\))OX’ पर स्थित है।
∴ – \(\frac{b}{2a}\) < 0 ⇒ b > 0
परवलय y = ax² + bx + c, Y-अक्ष को P(0, c) पर प्रतिच्छेद करता है जो OY’ पर है। अतः c < 0 होगा।
∴ a < 0, b > 0 तथा c < 0.

प्रश्न 2.
यदि बहुपद p(x) = 2x² + 5x + k के शून्यक α तथा β, सम्बन्ध α² + β² + αβ = \(\frac{21}{4}\) को सन्तुष्ट करते हैं, तो k का मान ज्ञात कीजिए।
हल :
∵ αβ बहुपद p(x) = 2x² + 5x + k के शून्यक हैं।
∴ α + β = \(\frac{-5}{2}\)
तथा α × β = \(\frac{k}{2}\) ………(1)
α² + β² + αβ = \(\frac{21}{4}\) (दिया है)
⇒ α² + β² + 2αβ – αβ = \(\frac{21}{4}\) (αβ जोड़ने तथा घटाने पर)
⇒ (α + β)² – αβ = \(\frac{21}{4}\)
⇒ (\(\frac{-5}{2}\))² – \(\frac{k}{2}\) = \(\frac{21}{4}\)
[समी. (1) का प्रयोग करने पर]
⇒ \(\frac{25}{4}-\frac{k}{2}\) = \(\frac{21}{4}\)
\(\frac{25}{4}-\frac{21}{4}\) = \(\frac{k}{2}\)
\(\frac{4}{4}\) = \(\frac{k}{2}\)
\(\frac{k}{2}\) = 1
⇒ k = 2
अतः k = 2

प्रश्न 3.
यदि α और β द्विघात बहुपद f(x) = x² – px + g के शून्यक हैं तो निम्न के मान ज्ञात कीजिए :
(i) α² + β²
(ii) \(\frac{1}{α}=\frac{1}{β}\)
हल :
∵ α और β बहुपद f(x) = x² – px + q के शून्यक हैं।
∴ शून्यकों का योग (α + β) = – (\(\frac{-p}{1}\)) = p
शून्यकों का गुणनफल (αβ) = \(\frac{q}{1}\) = q

(i) α² + β² = (α + β)² – 2αβ
= p² – 2q

(ii) \(\frac{1}{α}=\frac{1}{β}\) = \(\frac{\alpha+\beta}{\alpha \beta}\) = \(\frac{p}{q}\)

प्रश्न 4.
यदि α, β द्विघात बहुपद f(x)= 2x² – 5x + 7 के शून्यक हों तो एक बहुपद ज्ञात कीजिए जिसके शून्यक 2α + 3β और 3α + 2β हों।
हल
∵ α और β द्विघात बहुपद f(x) = 2x² – 5x + 7 के शून्यक हैं।
∴ α + β = -(\(\frac{-5}{2}\)) = \(\frac{5}{2}\) और αβ = \(\frac{7}{2}\)
माना अभीष्ट बहुपद के शून्यकों के योग व गुणनफल को S व P से व्यक्त करें तो :
S = (2α + 3β) + (3α + 2β)
= 5(α + β) = 5 × \(\frac{5}{2}=\frac{25}{2}\)
या P = (2α + 3β) × (3α + 2β)
या P = 6(α² + β²) + 13αβ
या P = 6α² + 6β² + 12αβ + αβ
या P = 6 (α + β)² + αβ
या P = 6(\(\frac{5}{2}\))² + \(\frac{7}{2}\)
या P = 6 × \(\frac{25}{4}+\frac{7}{2}=\frac{75}{2}+\frac{7}{2}\)
∴ P = \(\frac{75+7}{2}=\frac{82}{2}\) = 41
अत: अभीष्ट बहुपद g(x) निम्न होगा :
g(x) = k(x² – Sx + P)
⇒ g (x) = k(x² – \(\frac{25}{2}\)x + 41)
जहाँ k एक अशून्य वास्तविक संख्या है।

प्रश्न 5.
बहुपद f(x) = 3x³ + ax² + 4x + b का एक गुणनखण्ड (x + 2) है। यदि इसमें (x – 3) का भाग दिया जाये तो शेषफल – 5 बचता है। a तथा b के मान ज्ञात कीजिए।
हल :
यहाँ f(x) = 3x³ + ax² + 4x + b का एक गुणनखण्ड (x + 2) है।
अत: x + 2 = 0
⇒ x = – 2 रखने पर
f(-2) = 0 होगा।
यहाँ f(x) = 3x³ + ax² + 4x + b
∴ f(-2) = 3(-2)³ + a(-2)² + 4(-2) + b
⇒ 0 = 3(-8) + a(4) – 8 + b
⇒ 0 = – 24 + 4a – 8 + b
⇒ 0 = 4a + b – 32
⇒ 4a + b = 32 ……..(i)
पुन: (x – 3) से भाग देने पर शेषफल – 5 बचता है।
अत: f(3) = – 5.
∵ f(x) = 3x³ + ax² + 4x + b
∴ f(3) = 3(3)³ + a(3)² + 4(3) + b
⇒ – 5 = 3 × 27 + a(9) + 12 + b
⇒ – 5 = 81 + 9a + 12 + b
⇒ – 5 = 93 + 9a + b
⇒ 9a + b = – 5 – 93
⇒ 9a + b = – 98 ….(ii)
समीकरण (ii) में से समीकरण (i) को घटाने पर,

∴ a = \(\frac{-130}{5}\) = – 26
a का मान समीकरण (i) में रखने पर,
⇒ 4(-26) + b = 32
⇒ – 104 + b = 32
∴ b = 32 + 104 = 136
अतः a = – 26 और b = 136 होगा।

प्रश्न 6.
3x³ + x² + 2x + 5 को 1 + 2x + x² से भाग दीजिए।
हल :
3x³ + x² + 2x + 5 ÷ 1 + 2x + x²
1 + 2x + x² को x² + 2x + 1 लिख सकते है

जाँच-
(3x – 5) (1 + 2x + x²) + 9x + 10
= 3x + 6x² + 3x² – 5 – 10x – 5x² + 9x + 10
= 3x³ + x² – 7x – 5 + 9x + 10
= 3x³ + x² + 2x + 5

प्रश्न 7.
x³ – 3x² + 3x – 5 को x – 1 – x² से भाग दीजिए और विभाजन एल्गोरिथ्म की सत्यता की जाँच कीजिए।
हल :
माना कि
f(x) = x³ – 3x² + 3x – 5
तथा g(x) = x – 1 – x² = – x² + x + 1
अब f(x) को g(x) से विभाजित करेंगे।

अत: भागफल q(x) = – x + 2
तथा शेषफल (x) = – 3
विभाजन एल्गोरिथ्म के प्रयोग से भाग्य = भाजक × भागफल + शेषफल
⇒ x³ – 3x² + 3x – 5 = (- x² + x – 1) (- x + 2) – 3
⇒ x³ – 3x² + 3x – 5 = (x³ – 2x² – x² + 2x + x – 2) – 3
⇒ x³ – 3x² + 3x – 5 = x³ – 3x² + 3x – 2 – 3
⇒ x³ – 3x² + 3x – 5 = x³ – 3x² + 3x – 5
चूँकि बायाँ पक्ष = दायाँ पक्ष
अतः विभाजन एल्गोरिथ्म सत्यापित होता है।

प्रश्न 8.
वह प्रतिबन्ध ज्ञात कीजिए कि बहुपद p(x) = ax² + bx + c के शून्यक एक दूसरे के व्युत्क्रम हैं।
हल :
माना कि α बहुपद p(x) = ax² + bx + c का प्रथम शून्यक है।
प्रश्नानुसार बहुपद p(x) का दूसरा शून्यक (β) = \(\frac{1}{α}\)
α × β = \(\frac{c}{a}\)
⇒ α × \(\frac{1}{α}\) = \(\frac{c}{a}\)
⇒ 1 = \(\frac{c}{a}\)
⇒ c = a
अतः प्रतिबन्ध c = a है ।

प्रश्न 9.
x³ – 6x² + 11x – 6 को x – 2 से भाग दीजिए और विभाजन एल्गोरिथ्म की सत्यता की जाँच कीजिए ।
हल :

अतः भागफल = x² – 4x + 3 तथा शेषफल = 0
विभाजन एल्गोरिथ्य की सत्यता
भागफल भाजक + शेषफल
= (x² – 4x + 3) × (x – 2) + 0
= x³ – 4x² + 3x – 2x² + 8x – 6
= x³ – 6x² + 11x – 6
= भाज्य
अतः भागफल × भाजक + शेपफल = भाज्य
इति सिद्धम्

प्रश्न 10.
यदि त्रिघातीय बहुपद x³ – 3x² – 10x + 24 का एक शून्यक 4 है, तो इसके अन्य दो शून्यक ज्ञात कीजिए।
हल :
माना p(x) = x³ – 3x² – 10x + 24
दिया है, 4, p(x) का एक शून्यक है।
∴ (x – 4), p(x) का एक गुणनखंड होगा।

भागफल = x² + x – 6 तथा शेषफल = 0
∴ x³ – 3x² – 10x + 24 = (x – 4)(x² + x – 6).
अब p(x) के अन्य शून्यक ज्ञात करने के लिए भागफल (x² + x – 6) = 0 रखकर गुणनखंड करने पर,
x² + x – 6 = 0
⇒ x² + 3x – 2x – 6 = 0
⇒ x(x + 3) – 2(x + 3) = 0
⇒ (x + 3) (x – 2) = 0
यदि x + 3 = 0, तो x = – 3
और यदि x – 2 = 0, तो x = 2
अत: – 3 और 2 बहुपद x³ – 3x² – 10x + 24 के अन्य दो गुणनखंड है।

प्रश्न 11.
एक अध्यापक ने अपने 10 विद्यार्थियों में से प्रत्येक को एक कागज पर एक चर वाला एक बहुपद लिखकर देने को कहा। विद्यार्थियों के उत्तर निम्न थे :
2x + 3, 3x² + 7x + 2, 4x² + 3x² + 2,
x² + \(\sqrt{3}\)x + 7, 5x³ – 7x + 2,
2x² + 3 – \(\frac{5}{x}\), 5x – \(\frac{1}{2}\),
ax³ + bx³ + cx + d, x + \(\frac{1}{x}\), x³ – 3
निम्न प्रश्नों के उत्तर दीजिए:
(i) उपरोक्त दस में कितने बहुपद नहीं हैं?
(ii) उपरोक्त दस में कितने द्विघात बहुपद हैं?
हल :
(i) तीन, x² + \(\sqrt{3x}\) + 7, 2x² + 3 – \(\frac{5}{x}\), x + \(\frac{1}{x}\) बहुपद नहीं है।
(ii) एक 3x² + 7x + 2 एक द्विघातीय बहुपद है।

प्रश्न 12.
यदि बहुपद f(x) = x² – 8x + k के शून्यकों के वर्गों का योग 40 है, तो का मान ज्ञात कीजिए।
हल :
दिया है, f(x) = x² – 8x + k
माना f(x) के शून्यक α और β हैं।
∴ शून्यकों का योग = – \(\frac{b}{a}\)
α + β = \(\frac{-(-8)}{1}\)
और शून्यकों का गुणन = \(\frac{c}{a}\)
αβ = \(\frac{k}{1}\)
प्रश्नानुसार, α² + β² = 40
हम जानते हैं कि
(α + β)² = α² + β² + 2αβ
⇒ (8)² = 40 + 2k
⇒ 64 – 40 = 2k
⇒ 24 = 2k
⇒ k = \(\frac{24}{2}\) = 12

प्रश्न 13.
बहुपद p(x) = 2x⁴ – x³ – 11x² + 5x + 5 के दो शून्यक \(\sqrt{5}\) और – \(\sqrt{5}\) है। इस बहुपद के अन्य दो शून्यक ज्ञात कीजिए।
हल :
\(\sqrt{5}\) और – \(\sqrt{5}\) दिए गए बहुपद के शून्यक है।
∴ (x – \(\sqrt{5}\))(x + \(\sqrt{5}\)) = x² – 5, दिए गए बहुपद का एक गुणनखंड होगा।

विभाजन एल्गोरिथ्य प्रमेय से
2x⁴ – x³ – 11x² + 5x + 5 = (2x² – x – 1)(x² – 5)
अब p(x) के अन्य दो शून्यक ज्ञात करने के लिए भागफल (2x² – x – 1) = 0 रखकर गुणनखंड करने पर
2x² – x – 1 = 0
⇒ 2x² – 2x + x – 1 = 0
⇒ 2x(x – 1) + 1 (x – 1) = 0
⇒ (x – 1) (2x + 1) = 0
यदि x – 1 = 0, तो x = 1
और यदि 2x + 1 = 0, तो x = – \(\frac{1}{2}\)
अतः 1 और – \(\frac{1}{2}\) व्यंजक p(x) के अन्य दो गुणनखंड है।

प्रश्न 14.
बहुपद 2x³ – 3x² + 6x + 7 में कम-से-कम क्या जोड़ा जाए कि प्राप्त बहुपद x² – 4x + 8 से पूर्णतया विभाजित हो जाए?
हल :
माना p(x) = 2x³ – 3x² + 6x + 7
और q(x) = x² – 4x + 8
अब p(x) को q(x) से भाग देने पर

यहाँ शेषफल 10x – 33 प्राप्त होता है। यदि 2x³ – 3x² + 6x +7 में 10x – 33 घटा दिया जाए तो जो बहुपद प्राप्त होगा वह q(x) से पूर्णतया विभाजित हो जाएगा।

प्रश्न 15.
यदि α तथा β बहुपद f(x) = 5x² – 7x + 1 के शून्यक है, तो (\(\frac{α}{β}=\frac{β}{α}\)) का मान ज्ञात कीजिए।
हल :
दिया, f (x)= 5x² – 7x + 1

प्रश्न 16.
k के किस मान के लिए, बहुपद
f(x) = 3x⁴ – 9x³ + x² + 15x + k,
माना, g(x) = 3x² – 5
g(x) से f(x) को भाग करने पर,

∵ f(x), g(x) से पूर्णतया विभाजित होता है, इसलिए शेषफल को शून्य के बराबर रखने पर,
k + 10 = 0
k = – 10

प्रश्न 17.
3x³ + 10x² – 9x – 4 के सभी शून्यक ज्ञात कीजिए यदि इसका एक शून्य 1 है।
हल :
दिया गया बहुपद है –
3x³ + 10x² – 9x – 4
माना, इस बहुपद के शून्यक α, β तथा γ है तथा α = 1 है।
दिए गए समीकरण को, ax³ + bx² + cx + d = 0 से तुलना करने पर प्राप्त होता है,
a = 3 b = 10, c = – 9 तथा d = – 4

⇒ 3β² + 4 = – 13β
⇒ 3β² + 13β + 4 = 0
⇒ 3β² + 12β + β + 4 = 0
⇒ 3β(β + 4) + 1(β + 4) = 0
⇒ (β + 4) (3β + 1) = 0
β = – 4 या β = \(\frac{-1}{3}\)
तब समीकरण (ii) से,
βγ = \(\frac{4}{3}\)
– 4 × γ = \(\frac{4}{3}\) या – \(\frac{1}{3}\) × γ = \(\frac{4}{3}\)
γ = –\(\frac{1}{3}\)
γ = – 4
अतः इसके शून्यक 1, – 4, – \(\frac{1}{3}\) या 1, – \(\frac{1}{3}\), – 4 होंगे।

प्रश्न 18.
p का यह मान ज्ञात कीजिए जिसके लिए बहुपद px² – 14x + 8 = 0 का एक शून्यक दूसरे का 6 गुना है।
हल :
दिया गया समीकरण है
px² – 14x + 8 = 0
माना, एक मूल = α
दूसरा मूल = 6α
मूलों का योग = – \(\frac{b}{a}\)
α + 6α = \(\frac{-(-14)}{p}\)
7α = \(\frac{14}{p}\)
α = \(\frac{14}{p \times 7}\)
α = \(\frac{2}{p}\) ……(i)
मूलों का गुणनफल = \(\frac{c}{a}\)
(α) (6α) = \(\frac{8}{p}\)
6α² = \(\frac{8}{p}\) ……(ii)
α का मान समीकरण (ii) में रखने पर,
6(\(\frac{2}{p}\))² = \(\frac{8}{p}\)
⇒ 6 × \(\frac{4}{p^2}\) = \(\frac{8}{p}\)
⇒ 24p = 8p²
⇒ 8p² – 24p = 0
⇒ 8p (p – 3) = 0
⇒ 8p = 0 ⇒ p = 0
या p – 3 = 0 ⇒ p = 3
p = 0 दिए द्विघात समीकरण px² – 14x + 8 = 0 की सन्तुष्ट नहीं करती है।
अत: p = 3

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क)

1. बहुपद में चर राशि की उच्चतम घात वाले पद के घातांक को बहुपद की …………… कहते हैं।
2. …………… बहुपद की कोई घात नहीं होती है।
3. यदि ग्राफ x-अक्ष के दो भिन्न बिन्दुओं पर काटता है, तो बहुपद के ……………. शून्यक होते हैं।
4. 1 घात के बहुपद के अधिकतम …………………. वास्तविक शून्यक हो सकते हैं।
5. ऐसे बहुपद जिसमें चर राशि की उच्चतम घात 1 हो, …………….. बहुपद कहलाते हैं।

हल:

1. घात,
2. शून्य,
3. दो,
4. 1,
5. रैखिक ।

निम्न में सत्य / असत्य बताइए :

प्रश्न (ख)

1. ऐसे बीजीय व्यजंक जिनमें चर राशियों की घात धनात्मक, ऋणात्मक अथवा शून्य हो, बहुपद कहलाते हैं।
2. बहुपद 5y³ + 2y² + 4y⁵ + y + 7 की घात 3 है।
3. बीजीय व्यंजक \(\frac{3}{4}\)x³ – \(\frac{5}{4}\)x² + 2x – 5 चर x का परिमेय संख्याओं पर बहुपद है।
4. ऐसे बहुपद को, जिसमें चर राशि की उच्चतम घात 2 हो विघातीय बहुपद कहलाता है।
5. यदि कोई व्यजंक f(x), x में बहुपद है, तब एक वास्तविक संख्या, बहुपद (x) का शून्यक कहलाती है, यदि और केवल यदि (k) = 0 हो ।

हल :

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

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
यदि द्विघात बहुपद x² + 3x + k का एक शून्यक 2 है, तो k का मान है:
(A) 10
(B) – 10
(C) – 7
(D) – 2
हल :
माना p(x) = x² + 3x + k
∵ 2, p(x) का एक शून्यक है।
∴ p(2) = 0
⇒ (2)² + 3(2) + k = 0
⇒ 4 + 6 + k = 0
⇒ k = – 10
अत: सही विकल्प (B) है।

प्रश्न 2.
वह द्विघात बहुपद जिसके तथा गुणनफल 6 है, है :
(A) x² + 5x + 6
(B) x² – 5x + 6
(C) x² – 5x – 6
(D) – x² + 5x + 6
हल :
बहुपद = x² – (शून्यकों का योग)x + शून्यकों का गुणनफल
= x² – (- 5)x + 6
= x² + 5x + 6
अत: सही विकल्प (A) है।

प्रश्न 3.
निम्न आकृति में, बहुपद p(x) का आलेख दिया है। बहुपद के शून्यकों की संख्या है।

(A) 1
(B) 2
(C) 3
(D) 0
हल :
∵ बहुपद का आलेख x अक्ष को दो बिन्दुओं पर काटता है।
∴ बहुपद के दो शून्यक हैं।
अतः सही विकल्प (B) है।

प्रश्न 4.
बहुपद p(x) को x² – 4 से विभाजित करने पर भागफल तथा शेषफल क्रमशः x तथा 3 पाए गए। बहुपद p(x) है:
(A) 3x² + x – 12
(B) x³ – 4x + 3
(C) x² + 3x – 4
(D) x³ – 4x – 3
हल :
∵ भाज्य = भागफल × भाजक + शेषफल
p(x) = x × (x² – 4) + 3
= x³ – 4x + 3
अत: सही विकल्प (D) है।

प्रश्न 5.
ऐसे बहुपद जिनके शून्यक केवल 3 तथा 4 हैं, की घात है :
(A) 2
(B) 1
(C) 3 से अधिक
(D) 3
हल :
ऐसा बहुपद जिसके दो शून्यक होते हैं, द्विघातीय बहुपद होता है।
अत: सही विकल्प (A) है।

प्रश्न 6.
यदि α और β बहुपद x² + 2x + 1 के शून्यक हैं, तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) बराबर है।
(A) – 2
(B) 2
(C) 0
(D) 1
हल :
दिया है, बहुपद = x² + 2x + 1
∴ α + β = \(\frac{-b}{a}=\frac{-2}{1}\) = – 2
और αβ = \(\frac{c}{a}=\frac{c}{a}=\frac{1}{1}\) = 1
अब \(\frac{1}{\alpha}+\frac{1}{\beta}\) = \(\frac{\beta+\alpha}{\alpha \beta}\) = \(\frac{-2}{1}\) = – 2
अत: सही विकल्प (A) है।

प्रश्न 7.
यदि बहुपद (3x² + 8x + k) का एक शून्यक दूसरे का व्युत्क्रम है, तो k का मान है :
(A) 3
(B) – 3
(C) \(\frac{1}{3}\)
(D) \(\frac{-1}{3}\)
हल :
दिया है,
बहुपद = 3x² + 8x + k
माना एक शून्यक α है, तब दूसरा शून्यक α होगा।
शून्यकों का गुणनफल = \(\frac{c}{a}\)
⇒ α × \(\frac{1}{α}\) = \(\frac{k}{3}\)
⇒ 1 = \(\frac{k}{3}\)
⇒ k = 3
अत: सही विकल्प (A) है।

प्रश्न 8.
एक तीन घात वाले बहुपद के शून्यकों की अधिकतम संख्या है :
(A) 1
(B) 4
(C) 2
(D) 3
हल :
किसी बहुपद के शून्यकों की अधिकतम संख्या उसके घात के बराबर होती है।
अत: सही विकल्प (D) है।

प्रश्न 9.
यदि बहुपद kx² + 2x + 3 के शून्यकों का योग उनके गुणनफल के बराबर है, तो k बराबर है:
(A) \(\frac{1}{3}\)
(B) \(\frac{-1}{3}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{-2}{3}\)
हल :
शून्यकों का योग = \(\frac{-b}{a}=\frac{-2}{k}\)
शून्यकों का गुणनफल = \(\frac{c}{a}=\frac{3 k}{k}\) = 3
प्रश्नानुसार,
शून्यकों का योग = शून्यकों का गुणनफल
⇒ \(\frac{-2}{k}\) = 3
⇒ k = \(\frac{-2}{3}\)
अत: सही विकल्प (D) है।

प्रश्न 10.
द्विघात समीकरण x² + 99x + 127 के शून्यक होंगे:
(A) दोनों धनात्मक
(B) दोनों ऋणात्मक
(C) एक धनात्मक और एक ऋणात्मक
(D) दोनों समान।
हल :
∵ a > 0, b > 0 तथा c > 0 तब दोनों शून्यक ऋणात्मक होते हैं। अतः विकल्प (B) सही है।

प्रश्न 11.
चित्र में बहुपद f(x) = ax² + bx + c का आलेख दर्शाया गया हो तो :

(A) a > 0, b > 0 तथा c > 0
(B) a > 0 b < 0 तथा c < 0
(C) a > 0, b < 0 तथा c > 0
(D) a > 0, b > 0 तथा c < 0.
हल :
विकल्प (B) सही है।

प्रश्न 12.
चित्र में बहुपद f (x) = ax² + bx + c का आलेख दर्शाया गया हो तो :

(A) a < 0, b < 0 तथा c > 0
(B) a < 0, b < 0 तथा c < 0
(C) a < 0, b > 0 तथा c > 0
(D) a < 0, b > 0 तथा c < 0.
हल :
विकल्प (A) सही है।

प्रश्न 13.
त्रिघात बहुपद ax³ + bx² + cx+ d के दो शून्यक 0 दिए हैं। तीसरा शून्यक है:
(A) \(\frac{-b}{a}\)
(B) \(\frac{b}{a}\)
(C) \(\frac{c}{a}\)
(D) \(\frac{-d}{a}\)
हल :
बहुपद ax³ + bx² + cx + d के दो शून्यक 0 हैं। माना कि तीसरा शून्यक α है।
अतः शून्यकों का योग = α + 0 + 0 = \(\frac{-b}{a}\)
= α = \(\frac{-b}{a}\)
अंत: विकल्प (A) सही है।

प्रश्न 14.
यदि a – b, a तथा a + b बहुपद x³ – 3x² + x + 1 के शून्यक हैं तो (a + b) का मान होगा :
(A) 1 ± \(\sqrt{2}\)
(B) – 1 ± \(\sqrt{2}\)
(C) – 1 – \(\sqrt{2}\)
(D) 3
हल :
शून्यकों का योगफल = – \(\frac{b}{a}\)
⇒ a – b + a + a + b = \(\frac{-(-3)}{1}\)
⇒ 3a = 3
⇒ a = 1
तथा (a – b)a + a(a + b) + (a + b) (a – b) = \(\frac{c}{a}\)
⇒ (1 – b) × 1 + 1(1 + b) + (1 + b) (1 – b) = \(\frac{1}{1}\)
⇒ 1 – b + 1 + b + 1 – b² = 1
⇒ 3 – b² = 1
⇒ b² = 2
⇒ b = ± \(\sqrt{2}\)
अब a + b = 1 ± \(\sqrt{2}\)
अतः सही विकल्प (A) है।

प्रश्न 15.
यदि बहुपद f(x) = x² – 5x + k के शून्यक α तथा β इस प्रकार हों कि α – β = 1 तो k की मान होगा:
(A) 6
(B) 4
(C) 3
(D) शून्य।
हल :
बहुपद x² – 5x + k के शून्यक α, β हों तो
α + β = – (\(\frac{-5}{1}\)) = 5
αβ = \(\frac{k}{1}\) = k
अब α – β = 1
या (α – β)² = 1
या (α + β)² – 4αβ = 1
या 25 – 4k = 1
या 24 = 4k
∴ k = 6
अत: विकल्प (A) सही है।

Jharkhand Board JAC Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

Jharkhand Board Class 10 Science Magnetic Effects of Electric Current Textbook Questions and Answers

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.
Hint: This is an experimental fact. Which is only exhibited by the right-hand thumb rule.

Question 2.
The phenomenon of electromagnetic induction is…
(a) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coll.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called a…
(a) generator.
(b) galvanometer.
(c) ammeter.
(d) motor
Answer:
(a) generator.
Hint: A generator converts mechanical energy into electrical energy.

Question 4.
The essential difference between an AC generator and a DC generator is that…
(a) an AC generator has an electromagnet while a DC generator has permanent magnet.
(b) a DC generator will generate a higher voltage.
(c) an AC generator will generate a higher voltage.
(d) an AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) an AC generator has slip rings while the DC generator has a commutator.
[Due to the slip rings, the direction of the current produced by an AC generator is reversed at regular interval of time, while the current produced by a DC generator always flows in one direction.]

Question 5.
At the time of short circuit, the current in the circuit…
(a) reduces substantially.
(b) does not change.
(c) increases heavily.
(d) varies continuously.
Answer:
(c) increases heavily.

Question 6.
State whether the following statements are true or false :
(a) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The magnetic field lines at the centre of a long circular coil carrying current are parallel straight lines.
(d) The wire with green insulation is usually the live wire of an electric supply.
Answer:
(a) False
[An electric motor converts electrical energy into mechanical energy.]
(b) True
(c) True
(d) False
[The wire with green insulation is the earth wire, whereas the wire with red insulation is the live wire of an electric supply.]

Question 7.
List two methods of producing magnetic fields.
Answer:
Magnetic fields can be produced by passing an electric current through:

• a straight conductor
• a circular loop and
• a solenoid.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:

• A current-carrying solenoid behaves like a bar magnet and the polarities of its ends depend upon the direction of the current through it.
• Yes, we can use a bar magnet to determine the north and south poles of a current-carrying solenoid.
• In order to determine the magnetic poles of a current-carrying solenoid, place it in a brass hook and suspend it with a long thread so that it can moves freely in a horizontal plane.

Bring the north pole of a bar magnet near one of its ends. If that end of the solenoid moves towards the bar magnet, then it must be its south pole and the other end its north pole, On the contrary if that end of the solenoid moves away from the magnet, then it must be its north pole and the other end its south pole.

Question 9.
When is the force experienced by a current carrying conductor placed in a magnetic field largest?
Answer:
The force experienced by a current-carrying conductor placed in a magnetic field is the largest, when the current in the conductor is perpendicular to the magnetic field.

For reference see the following figure :

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:

• For reference, see figure (a) and (b).
• The direction of the conventional current is opposite to the direction of motion of the electron.
• Applying Fleming’s left-hand rule, we find that when the middle finger points in the direction of the current and the thumb points in the direction of the force, the forefinger points downward giving the direction of the magnetic field.

[Note : IMg indicates a vector (arrow) going into the plane of figure, while IMG indicates a vector (arrow) coming out of the plane of the figure.]

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of the split ring in an electric motor?
Answer:
Construction: The construction of an electric motor is as shown in figure 13.24.

• Electric motor consists of a rectangular coil ABCD of insulated copper wire.
• The coil is placed between the two poles of a permanent magnet such that the arms AB and CD are perpendicular to the direction of the magnetic field.
• The ends of the coil are connected to two halves P and Q of a split ring.
• The inner sides of these halves are insulated and attached to an axle such that they can rotate easily on it.
• The external conducting edges of P and Q touch two conducting stationary brushes (i.e., carbon strips) X and Y respectively.
• These brushes are connected to a plug key and battery as shown in figure 13.24.

Working:
(1) Current in coil ABCD enters from the source – battery through conducting brush X and flows back to the battery through brush Y.

(2) Current in arm AB of the coil flows from A to B while in arm CD it flows from C to D, i.e., opposite to the direction of the current through AB. Both the currents flowing in AB and CD are perpendicular to the magnetic field.

(3) On applying Fleming’s left hand rule for finding the direction of the force on a current carrying conductor in a magnetic field (see figure 13.24), we find that the force acting or arm AB pushes it downwards while the force acting on arm CD pushes it upwards. These forces are also equal in magnitude and perpendicular to the respective lengths of arms AB and CD.

(4) So, the coil and the axle, mounted free to turn about an axis, rotate anticlockwise.

(5) At half rotation, Q makes contact with brush X and P with brush Y. Therefore, the current in the coil gets reversed and flows along the path DCBA.

(6) A device that reverses the direction of flow of current through a circuit is called a commutator.
In an electric motor the split rings act as a commutator.

(7) Now, the reversal of current also reverses the direction of the force acting on the arms AB and CD. Thus arm AB that was earlier pushed down, is now pushed up and arm CD previously pushed up, is now pushed down.

(8) Therefore, the coil and the axle rotate half a turn more in the same direction.

(9) The reversing of the current is repeated at each half rotation, giving rise to a continuous rotation of the coil and the axle.

Important note : When currents through arms BC and DA are either parallel or antiparallel to the magnetic field, magnetic force does not act on them.

When a rectangular coil carrying a current is placed in a magnetic field, then forces, equal in magnitude and opposite in direction, act on its two parallel sides, perpendicular to their lengths, due to which the coil rotates continuously.

The function of the split ring : Due to the split ring commutator in an electric motor, the direction of the current through the coil is reversed after every half rotation of the coil. This reverses the direction of the force acting on the two arms of the coil. The arm that was earlier pushed downward is now pushed upward, while the other arm is now pushed downward. Therefore, the coil continues to rotate in the same direction.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motors are used in electric fans, refrigerators, mixers and grinders, washing machines, water-pumps, coolers, etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil, (ii) withdrawn from inside the coil, (iii) held stationary inside the coil?
Answer:
(i) The galvanometer will show a momentary deflection in one direction. It means a current is induced in the coil in one direction due to the relative motion between the coil and the magnet.

(ii) The galvanometer will show a momentary deflection in the opposite direction. It means a current is induced in the coil in the opposite direction due to the relative motion between the coil and the magnet.

(iii) There will be no deflection in the galvanometer. It means no current is induced in the coil, as there is no relative motion between the coil and the magnet.

[Note: The greater the speed of the magnet, the greater is the deflection of the pointer in the galvanometer.]

Question 14.
Two circular coils A and B are placed close to each other. If the current in coil A is changed, will some current be induced in coil B? Give reason.
Answer:
Yes. Current will be induced in coil B. Reason : When the current in coil A is changed, the magnetic field around it changes.

As the two circular coils (with their planes parallel to each other) are placed close to each other, the magnetic field lines linked with coil B also change. Therefore, some current is induced in coil B.

Question 15.
State the rule to determine the direction of a (i) magnetic field produced around a straight conductor-carrying current, (ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and (iii) current induced in a coil due to its rotation in a magnetic field.
Answer:
(i) Right-hand thumb rule : This rule is also called ‘Maxwell’s corkscrew rule.
Rule : Imagine that you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers of right hand wrap around the conductor in the direction of the field lines of the magnetic field.

This is known as the Right-hand thumb rule.

(ii) Fleming’s left-hand rule : In this case, Fleming’s left hand rule is used to find the direction of the force acting on the conductor.
Fleming left-hand rule:

Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular (figure 13.22). If the first (fore) finger points in the direction of magnetic field and second (middle) finger points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Note : Experimentally it is found that, when direction of current through the rod and direction of the magnetic field are perpendicular to each other, the force exerted on the rod is perpendicular to both of them.

(iii) Fleming’s right-hand rule : Direction of induced current in a conductor can be known with the help of Fleming’s right-hand rule.

Stretch the thumb, forefinger and middle finger of right hand so that they are perpendicular to each other. If forefinger indicates the direction of the magnetic field and thumb shows the direction of motion of conductor, then the middle finger will show the direction of induced current.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of the brushes?
Answer:
Principle : When a coil is rotated properly in a magnetic field, electromotive force is induced in it. As a result, a current flows in the circuit containing the coil.
OR
When a coil is rotated in a magnetic field, electric current is induced in the circuit containing the coil.

Labelled diagram and working : An electric generator is a device that converts mechanical energy into electrical energy. Its working is based on electromagnetic induction.

Construction :

• Figure 13.30 (a) shows the construction of an AC generator. It consists of a rectangular coil ABCD placed between the two poles of a permanent magnet.
• Two ends of this coil are connected to the two metal (copper) rings R₁ and R₂. The inner side of these rings are insulated.
• The two conducting stationary brushes R₁ and R₂ are kept pressed separately on the rings R₁ and R₂ respectively.
• The two rings R₁ and R₂ are internally attached to an axle. The axle may be mechanically rotated from outside to rotate the coil inside the magnetic field.
  
• Outer ends of the two brushes are s connected to the galvanometer to show the flow of current in the given external circuit.

Working:
(1) Suppose, the axle attached to the two rings is rotated such that the arm AB moves up and the arm CD moves down in the magnetic field produced by the permanent magnet. Then the coil ABCD rotates clockwise in the arrangement shown in the figure. By applying Fleming’s right- hand rule, we find that the induced currents are set up in these arms along the directions AB and CD. This means that the current in external circuit flows from brush B₂ to B₁.

[If the coil is made of a larger number of turns, the current generated in each turn adds up to give a large current through the coil.]

(2) After half a rotation, arm CD starts moving up and AB moving down. As a result, the directions of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. This means that the current in the external circuit now flows from brush B₁ to B₂.

(3) Thus, after every half rotation the direction of the current in the respective arms changes. Such a current, which changes direction after equal intervals of time, is called an alternating current (abbreviated as AC).

The function of the brushes in an electric generator : Brushes are used to transmit current induced externally from coil ABCD to the external circuit.

Question 17.
When does an electric short circuit occur?
Answer:
An electric short circuit takes place when the live wire and neutral wire of the electric supply line touch each other directly or indirectly via a conducting wire. This occurs when the plastic insulation of the wires gets torn or there is a fault in the electrical appliance.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
Function of an earth wire : The function of an earth wire is to provide a low-resistance conducting path to the current leaking from an electrical appliance to the earth and thereby prevent electric shock to the user of the electrical appliance.

Therefore, it is necessary to earth metallic appliances such as an electric press, toaster, table fan, refrigerator, etc.

Sometimes accidently, the insulation of the wires connected to an electrical appliances may melt resulting in a current through the metallic casing of the appliance. A person touching such an appliance may get a severe electric shock which can be fatal. To avoid this, the current should be diverted to the earth.

Moreover, due to very low resistance (almost nil) offered by the earth wire, the current in the circuit rises to a very high value, thereby melting fuse in that circuit and cutting off its electric supply.

Thus, earthing the electrical appliances properly, saves us from possible electric shock.

Jharkhand Board Class 10 Science Magnetic Effects of Electric Current InText Questions and Answers

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
A compass needle is a small bar magnet with one north pole and the other south pole. When a compass is brought near a bar magnet, compass needle gets deflected due to the forces acting on its pole due to the magnetic field of the bar magnet.

Note: When compass needle is brought near a bar magnet it is acted upon by the force due to the magnetic field of the earth as well as that due to the bar magnet. Therefore, it gets deflected and finally comes to rest in the direction of the net force.

Question 2.
Draw magnetic field lines around a bar magnet.
Answer:

Question 3.
List the properties of magnetic field lines.
Answer:
(1) The magnetic field lines emerge from north pole and merge at the south pole outside the magnet, while inside the magnet the direction of field lines is from its south pole to its north pole.
Thus, the magnetic field lines are closed and continuous curve.

(2) The magnetic field lines are crowded near the pole where the magnetic field is strong and are far apart near the middle of the magnet and far from the magnet where the magnetic field is weak.

(3) The magnetic field lines never intersect each other because if they do so, there would be two directions of magnetic field at that point which is absurd.

(4) In case the field lines are parallel and equidistant, these represent a uniform magnetic field.

Important note : The relative strength of the magnetic field is shown by the degree of closeness of the field lines.

Question 4.
Why don’t two magnetic field lines intersect each other?
Answer:
The direction of the magnetic field B at a point is obtained by drawing a tangent to the magnetic field line at that point.

If two magnetic field lines intersect each other, it would mean that there are two directions of the magnetic field at the point of intersection, which is not possible because magnetic field is a vector, at a given point in space, it can have only one direction, (or the resultant force on a pole (north / south) at any point can only be in one direction.)

Question 5.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the Right-hand thumb rule to And out the direction of the magnetic field inside and outside the loop.
Answer:
Using the Right-hand thumb rule, the direction of the magnetic field inside and outside the circular loop of wire carrying an electric current can be found. This is shown in the following figure:

The dotted magnetic field lines are perpendicular to the plane of the paper.
The front face of the loop behaves as the south pole and the back face, i.e., the face touching the plane of the table behaves as the north pole.

Question 6.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
A uniform magnetic field in a given region can be represented by straight, parallel equally spaced field lines.

Question 7.
Choose the correct option :
The magnetic field inside a long straight solenoid carrying current…
(a) is zero.
(b) decreases as we move towards its end.
(c) increases as we move towards its end.
(d) is the same at all points.
Answer:
(d) is the same at all points.
Hint: The magnetic field inside a long straight current-carrying solenoid is uniform which is represented by straight parallel equally spaced lines. Thus, it is the same at all the points.

Question 8.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
(c) velocity and
(d) momentum
Hint: The proton is a charged particle. Whenever a charged particle moves in a magnetic field except in the direction of the magnetic field and opposite to the direction of the magnetic field, a magnetic force exerts on it. As a result its velocity and momentum (mass x velocity) both change.

Note: The magnetic force is perpendicular to the velocity of the charged particle (here, the proton). Therefore, only the direction of the velocity of the particle changes, i.e., there is no change in magnitude of the velocity (speed) of the particle. The mass does not change.

Question 9.
In Activity 13.7. how do we think the displacement of the rod AB will be affected if (i) the current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) the length of the rod AB is increased?
Answer:
Experiments have shown that, when a current-carrying conductor is placed in a magnetic field such that its length is perpendicular to the magnetic field, the force acting on the conductor and hence its displacement (as the conductor is initially at rest) is proportional to…

• the current in the conductor
• the strength of the magnetic field
• the length of the conductor

(i) When the current in the rod AB is increased, then more force will act on the rod and hence the displacement of the rod will also be more (in the same proportion).

(ii) If a stronger horse-shoe magnet is used, then the strength of magnetic field will increase leading to a greater force on the rod. Due to the greater force, the displacement of the rod will also be more (in the same proportion.).

(iii) If the length of the rod AB is increased, then more force will act on the rod and hence the displacement of the rod will also be more (in the same proportion).

Question 10.
A positively charged particle (alpha particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is…
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) upward
Hint:

• Here, the positively charged particle (alpha particle) is initially moving towards the west, so the direction of the corresponding conventional current (current opposite to electron-current) is towards the west.
• The deflection of alpha particle is towards north. This shows that the magnetic force acts on it towards the north (direction).
• Hence, here (a) direction of current I is towards west and (b) direction of force F is towards north
  
• Let us now hold the first (fore) finger, middle finger and thumb of your left hand at right angles to one another.
• Adjust the left hand in such a way that your middle finger points towards the west (in the direction of the current) and the thumb points towards the north (in the direction of the force.)

In this case, your first (fore) finger points upwards [shown by the symbol G], Since the direction of the forefinger gives the direction of the magnetic field, the magnetic field would be in the upward direction, [see figure 13.23 (a) or 13.23 (b)].

Question 11.
State Fleming’s left-hand rule.
Answer:
In this case, Fleming’s left hand rule is used to find the direction of the force acting on the conductor.

Fleming left-hand rule:

Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular (figure 13.22). If the first (fore) finger points in the direction of magnetic field and second (middle) finger points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Note : Experimentally it is found that, when direction of current through the rod and direction of the magnetic field are perpendicular to each other, the force exerted on the rod is perpendicular to both of them.

Question 12.
What is the principle of an electric motor?
Answer:
When a rectangular coil carrying a current is placed in a magnetic field, then forces, equal in magnitude and opposite in direction, act on its two parallel sides, perpendicular to their lengths, due to which the coil rotates continuously.

Question 13.
What is the role of the split rings s in an electric motor?
Answer:
The split rings act as a commutator and its function is to reverse the direction of the current flowing through the coil after every half rotation of the coil.

Due to reversing of the current, the direction of rotating couple (i.e., torque) remains unchanged and the coil continues to rotate in the same direction.

Question 14.
Explain different ways to induce a current in a coil.
Answer:

• A current can be induced in a coil by moving a magnet towards or away from it or by moving the coil towards or away from the magnet.
• A current can be induced in a coil by changing the current in the coil placed near it.
• A current can be induced in a coil by moving a coil properly in a non-uniform magnetic field or by changing a magnetic field around steady coil by some means.
• A current can be induced in a coil by rotating it properly in a magnetic field or by rotating a magnet properly placed near the coil.

[Note : The magnet used may be a bar magnet or a current-carrying conductor.]

Question 15.
State the principle of an electric generator.
Answer:
The working of an electric generator is based upon the principle of electromagnetic induction.
Principle : An electric current produced in a closed circuit or coil by a changing magnetic field is called an induced current. This phenomenon is called electromagnetic induction.

Question 16.
Name some sources of direct current.
Answer:
Electrochemical dry cells, batteries, DC generators, solar cells, etc. are sources of direct current (DC).

Question 17.
Which sources produce alternating current?
Answer:
AC generators (or powerhouse generators), car alternators, bicycle dynamos are sources of alternating current (AC).

Question 18.
Choose the correct option :
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each…
(a) two rotations
(b) one rotation
(c) half rotation
(d) one-fourth rotation
Answer:
(c) half rotation
Hint: After each half rotation, the direction of motion of the two parallel sides of the coil (AB and CD) is reversed and so the direction of the induced current changes once in each half rotation.

Question 19.
Name two safety measures commonly used in electric circuits and appliances.
Answer:

1. A safety fuse of proper rating : It prevents damage to the appliances and circuit due to overloading.
2. An earth wire : It prevents possible electric shock when the live wire accidentally touches the metallic body of an appliance.

Question 20.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
The current drawn by this electric oven,
I = \(\frac { P }{ V }\)
= \(\frac { 2000 W }{ 220 V }\)
= 9.09 A
Since the current rating of the circuit is 5 A. the fuse used in this circuit is only of 5 A capacity.

Hence, when the oven is switch ON, the fuse wire (rated 5 A) will get heated too much, melt and break the circuit. This will save the electric oven from getting damaged.

[If fuse of current rating more than 9.09 A is used in the circuit or if no fuse has been put in the circuit, there may be a fire.]

Question 21.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:
To avoid overloading of domestic electric circuits, the following precautions should be taken :

• Wires used for carrying current should be of proper current rating.
• Two separate circuits should be used, one of 5 A current for lighting electric bulbs, tubes, TV, etc. and another of 15 A for higher current rating appliances such as heating appliances, AC, etc.
• Parallel circuits should be used and each circuit should have a fuse of proper rating.
• Too many higher power rating electrical appliances such as electric iron, geyser, air-conditioner, etc. should not be switched on at the same time.
• Too many electrical appliances should not be operated on a single socket simultaneously.
• Wires should be replaced by new wires of proper rating and good insulation after every 5-6 years.
• PVC of good quality should be used.

Activity 13.1 [T. B. Pg. 223]

To show that the magnetic field is produced s due to electric current.

Procedure:
1. Take a straight thick copper wire and place it between points X and Y in an electric circuit, as shown in figure 13.1. Wire XY is kept perpendicular to the plane of the paper.

2. Horizontally place a small compass near this copper wire.
See the position of its needle.

3. Pass the current through the circuit by inserting the key into the plug.

• Observe the change in the position of the compass needle.
• What does it indicate? (or What does it mean?)

Observation:
1. When no current flows in the straight thick copper wire (i.e., when plug key K is open), the compass needle (i.e., magnetic needle) remains stationary in the geographical north- south direction of the earth.

2. On passing the current through the copper wire XY (i.e., conducting wire), by inserting the key into the plug, compass needle is deflected.
[Any change in the direction of current through copper wire will show a variation in deflection.]

3. This indicates (shows) that the electric current through copper wire (the conductor) has produced a magnetic effect. It means that magnetic field is produced around the (copper) wire.

Conclusion :
Magnetic field is produced due to electric current. Thus, we can say that +electricity and ++magnetism are linked to each other.

Note: The ‘magnetic effect of electric current’ means that an electric current flowing in a conducting wire produces a magnetic field around it.

Activity 13.2 [T. B. Pg. 224]

To understand the magnetic field produced by a bar magnet.
OR
To obtain the magnetic field lines around a bar magnet.

Procedure :
1. Fix a sheet of white paper on a drawing board using some adhesive material.

2. Place a bar magnet in the centre of it.

3. Sprinkle some iron filings uniformly around a bar magnet as shown in figure 13.3. A salt- sprinkler may be used for this purpose.

4. Now tap the board gently.

• What do you observe?
• Why do the iron filings arrange in such a pattern?
• What does this pattern demonstrate?

Observation :
1. The iron filings arrange themselves in a pattern as shown in figure 13.3.

2. A bar magnet exerts its influence in the region surrounding it. Therefore, the iron filings experience a magnetic force. This force makes iron filings arrange in a specific (particular) pattern.

3. The region surrounding a magnet, in which the force of the magnet can be detected, is said to have a magnetic field.
The lines along which the iron filings align themselves represent magnetic field lines.

Conclusion:
A bar manget possesses a magnetic field which can be detected by sprinkling iron filings around it.

Activity 13.3 [T. B. Pg. 224]

To draw magnetic field lines of a bar magnet using a compass.

Procedure:
1. Take a small compass and a bar magnet.

2. Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.

3. Mark the boundary of the magnet.

4. Place a compass near the north pole of the magnet. How does it behave?

5. Mark the positions of two ends of the needle.

6. Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole.
In this way, proceed step by step till you reach the south pole of the magnet as shown in figure 13.4.

7. Join the points marked on the paper by a smooth curve. This curve represents a magnetic field line.

8. Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in figure 13.5.
These lines represent the magnetic field around the magnet.
These are known as magnetic field lines.

• Observe the deflection in the compass needle as you move it along a field line. The deflection increases as the needle is moved towards the poles.

Observation:
1. The south pole of the needle points towards the north pole of the magnet. The north pole s of the compass needle is directed away from the north pole of the magnet.

2. Magnetic field is a quantity that has both direction and magnitude. The direction of the magnetic field is taken to be direction in which a north pole of the compass needle moves inside it.

Therefore, the field lines emerge from the north pole and merge at the south pole of a magnet. Inside the magnet, the direction of the field lines is from its south pole to its north pole. Thus, the magnetic field lines are closed curves.

Conclusion:

1. The region surrounding a magnet has a magnetic field, i.e., a bar magnet possesses magnetic field.
2. The direction of magnetic field is the direction in which the north pole of the compass needle move / deflect.
3. The field is stronger where the field lines are crowded and weaker where the field lines are farther away from each other.
4. The deflection of the needle increases as the needle is moved towards the poles.

Activity 13.4 [T. B. Pg. 226]

To show that the direction of the magnetic field due to a current depends on the direction of the current.

Procedure:
1. Take a long straight copper wire, two or three cells of 1.5 V each, and a plug key. Connect all of them in series as shown in figure 13.6 (a).

2. Place the straight wire parallel to and over a compass needle.

3. Plug the key in the circuit.
Observe the direction of deflection of the north pole of the needle.

4. Replace the cell connections in the circuit as shown in figure 13.6 (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.
Observe the change in the direction of deflection of the needle.

Observation:
1. If the current flows from north to south as shown in figure 13.6 (a), the north pole of the compass needle moves towards the east.

2. Now, when the current flowing through the copper wire is reversed, the compass needle aligns / moves in the opposite direction i.e., towards the west as shown in figure 13.6 (b).

It means that the direction of the magnetic field due to a current is reversed, when the direction of the current is reversed.

Conclusion :
The direction of the magnetic field due to a current depends on the direction of the current.

Activity 13.5 [T. B. Pg. 226]

To study the pattern of the magnetic field (lines) due to a current through a straight conductor.

Procedure:
1. Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0-5 A), a plug key, connecting wires and a long straight thick copper wire.

2. Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.

3. Connect the copper wire vertically between the points X and Y, as shown in figure 13.7 (a), in series with the battery and a plug key.

4. Sprinkle some iron filings uniformly on the cardboard. (You may use a salt-sprinkler for this purpose.)

5. Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.

6. Keep the variable of the variable resistance (or rheostat) at a fixed position and note the current through the ammeter.

• Gently tap the cardboard a few times. Observe the pattern of the iron filings.
• What do these concentric circles represent?
• How can the direction of the magnetic field be found?
  (Place a compass at a point (say P) over a circle. Observe the direction of the needle.)
• Does the direction of the magnetic field lines get reversed if the direction of the current through the straight copper wire is reversed? Check it.
  
• What happens to the deflection of the compass needle placed at a given point if the current in the copper wire is changed?
• (To see this, vary the current in the wire using the variable resistance.)
• What happens to the deflection of the needle if the compass is moved away from the copper wire keeping the current through the wire remains tire same?
• (To see this, now place the compass needle at a farther point from the conducting wire [say at point Q (figure 13.7 (b)]).
• What change do you observe?

Observation:

• It is found that the iron filings align themselves forming a pattern of concentric circles around the copper wire (figure 13.7 (c)).
• These concentric circles represent the (pattern of) magnetic field lines.
• The direction shown by the north pole of the compass needle gives the direction of the magnetic field lines at point P. In figure 13.7 (a), it is shown by an arrow.
• Yes. The direction of the magnetic field lines get reversed if the direction of the current through the straight copper wire is reversed (see figure 13.7 (b)).
• We find that the deflection of the needle changes, when the current is changed.

If the current is increased, the deflection increases and if the current is decreased, the deflection decreases.

• When the compass is moved away from the wire, the deflection of the needle decreases.
• When the compass is moved away from the wire, it is observed that the concentric circles representing the magnetic field around a current-carrying straight wire become larger and larger.
• In other words, as we move away from the wire, the radius of a circle representing the magnetic field line increases.

Conclusion :
(1) Pattern of the magnetic field produced due to a straight conductor carrying current is concentric circles, which are nothing but the magnetic field lines of the produced magnetic field.

(2) Direction of magnetic field at a point is in the direction of tangent drawn at that point on particular circle, which is nothing but the direction of north pole of compass needle at that point.

(3) Direction of magnetic field lines get reversed if direction of current through the straight copper wire is reversed, i.e., it depends on the direction of the current.

(4) Magnitude of the magnetic field produced at a given point increases as the current through the wire increases.
i.e., Magnetic field B ∝ Current I.

(5) Magnitude of the magnetic field produced by a given current in the conductor decreases as the distance from wire increases.
i.e., Magnetic field B ∝ \(\frac{1}{\text { Distance from the wire }}\)

Activity 13.6 [T. B. Pg. 229]

To study the magnetic field produced near a circular coil carrying current.

Procedure:

• Take a rectangular cardboard having two holes. Insert a circular coil having a large number of turns through them, normal to the plane of the cardboard.
• Connect the ends of the coil in series with a battery, a plug key and a rheostat, as shown in figure 13.13.
• Sprinkle iron filings uniformly on the cardboard.
• Plug the key.
• Tap the cardboard gently a few times.
• Note the pattern of the iron filings that emerges on the cardboard.
  

Observation :
The iron filings arrange themselves in the pattern as shown in figure 13.13.

Conclusion:

• The magnetic field at the centre of the coil is nearly uniform.
• The magnetic field lines near the coil are circular and concentric.
• At the centre, as the field lines are crowded, the strength of the magnetic field is maximum.

Activity 13.7 [T. B. Pg. 230]

To understand the force experienced by a current-carrying small aluminium (conducting) rod placed in a magnetic field.

Procedure:
1. Take a small aluminium rod AB (of length about 5 cm.) Using two connecting wires suspend it horizontally from a stand, as shown in figure 13.20.

2. Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and the south pole vertically above the aluminium rod (see figure 13.20).

3. Connect the aluminium rod in series with a battery, a plug key and a rheostat.

4. Now pass a current through the aluminium rod from end B to end A.

• What do you observe?
• Reverse the direction of current flowing through the rod and observe the direction of its displacement.
  
• Why does the rod get displaced?
• Change the direction of magnetic field by interchanging the two poles of the magnet to vertically downwards and observe the direction of the force / displacement of the current-? carrying rod.

Observation :
1. It is observed that the rod is displaced to the left.

2. The direction of the displacement of the rod is reversed, i.e., now it is to the right, on reversing the direction of the current flowing through the rod.

It suggests that the direction of the force is also reversed when the direction of the current through the conductor (rod) is reversed.

3. The displacement of the rod is due to the force exerted on the current- carrying aluminium rod when it is placed in a magnetic field.

4. The direction of the force acting on the current-carrying rod and hence that of the displacement of the rod (as the rod is initially ; at rest) gets reversed on changing the direction of the magnetic field to vertically downwards.

Conclusion :

• A current-carrying conductor (here a rod) experiences a force when placed in a magnetic field.
• The direction of the force and hence that of the displacement of the conductor (rod) depends upon the direction of the current and the direction of the magnetic field.

Notes :

• Here, the displacement of the rod has the same direction as that of the force acting on the rod because the rod is initially at rest. Recall the kinematical equations.
• The magnetic force also depends upon the length of the rod in the magnetic field.

Activity 13.8 [T. B. Pg. 233]

To study the phenomenon of electromagnetic induction.

Procedure:
1. Take a coil of wire AB having a large number of turns.

2. Connect the ends of the coil to a galvanometer as shown in figure 13.25.

3. Take a strong bar magnet and move its north pole towards the end B of the coil.
Do you find any change in the galvanometer needle?

4. Now withdraw the north pole of the magnet away from the coil.
What do you observe?

5. Place the magnet stationary at a point near the coil, keeping its north pole towards the end B of the coil.

6. Move the coil to right and later to left.
What do you observe?

7. When the coil is kept stationary with respect to the magnet.
What do you observe?

8. Now, move the south pole of the magnet towards the end B of the coil.

• What do you observe?
• When the coil and the magnet both are stationary what do you observe?
• What do you conclude from this activity?
  

Observation :
1. There Is a momentary deflection in the needle of the galvanometer, say to the right.
This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.

2. When the north pole of the magnet is moved away from the coil, the galvanometer needle is deflected to the left, showing that the current is now set up in the direction opposite to the first.

3. We see that the galvanometer needle deflects to the right when the coil is moved towards the north pole of the magnet. Similarly, the needle moves to the left when the coil is moved away.

4. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer needle drops to zero.

5. If you move the south pole of the magnet ‘c towards the end B of the coil, the deflections in the galvanometer would just be opposite > to those in the previous case.

6. When the coil and magnet both are stationary, there is no deflection in the galvanometer.

Conclusion :
Whenever there is a relative motion between s a magnet and coil, potential difference is produced between two ends of the coil, which sets up an electric current in the circuit.

This potential difference is called the induced potential difference and the corresponding current is called the induced current.

Activity 13.9 [T. B. Pg. 235]

To study the phenomenon of electromagnetic induction.

Procedure:
1. Take two different coils of copper wire having large number of turns (say 50 and loo turns respectively). Insert them over a non-conducting cylindrical roll, as shown in figure 13.27 (You may use a thick paper roll for this purpose.)

2. Connect the coil 1. having larger number of turns, in series with a battery and a plug key. Also connect the other coil 2 wIth a galvanometer as shown.

3. Plug in the key. Observe the galvanometer.
Is there a deflection in its needle?

4. Disconnect coil 1 from the battery.
What do you observe?

Observation :
1. When the key is plugged, the needle of the galvanometer is instantly deflected to one side and then quickly returns to zero, indicating a momentary current in coil 2.

2. When coil 1 is disconnected from the battery, the needle of the galvanometer momentarily moves, but to the opposite side, i.e., now the current flows in the opposite direction in coil 2.

In short in this activity, we observe that as soon as the current in coil 1 reaches either a steady value or becomes zero, the galvanometer connected to coil 2 shows no deflection.

Conclusion :
(a) A potential difference is induced in coil 2 whenever the electric current through the coil 1 changes with time.

(b) As the current in coil 1 (which is called Primary coil) changes, the magnetic field associated with it also changes. Then the magnetic field lines around coil 2 (which s is called Secondary coil) also change.

Thus, the change in magnetic field lines associated with the secondary coil is the cause of induced electric current in it. s

This process, by which a changing magnetic field in a conductor induces a current in another conductor, is called electromagnetic induction.

Students must go through these JAC Class 10 Science Notes Chapter 12 Electricity to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 12 Electricity

→ Electric charge: An electric charge is an intrinsic property of the proton, electron and many other particles. There are two types of electric charges :

• Positive electric charge
• Negative electric charge.

→ Electric current: The electric current is the net amount of electric charge that passes through any cross-sectional area of the conductor in unit time (I = \(\frac { Q }{ t }\)).
Its SI unit is the coulomb/second (C/s) or the ampere (A).

The direction of the conventional current is taken as opposite to the direction of the flow of electrons.

ampere: If 1 coulomb charge flows through any cross-section of a conductor in 1 second, then the electric current flowing through the conductor is said to be 1 ampere.
1 mA = 10^-3 A, 1 μA = 10^-6A

→ Electric potential and Electric potential difference: The work done in bringing a unit positive charge from infinity to a particular point in the electric field against the electrostatic force due to the electric field is called the electric potential at that point.

(The charge is kept in equilibrium.)

The electric potential difference (p.d.) between any two points A and B in an electric field is the work done to move a unit positive charge ( + 1C) from one point A to the another point B against the electric force due to the electric field.
The SI unit of potential and potential difference is the joule /coulomb or volt (V).

→ volt: The potential difference between two points in an electric field is said to be 1 volt if 1 J of work is done to move a charge of 1 C from one point to another point.
1 V = \(\frac { 1 J }{ 1 C }\)

→ Ohm’s law: The current flowing through a conductor, such as a metallic wire, is directly proportional to the potential difference across its ends, provided its temperature and other physical conditions remains the same.

→ Resistance: Whenever a current flows through a conductor, the free electrons moving in one direction in the conductor collide with the ions, atoms and molecules of the conductor. Due to these collisions, the motion of electrons is opposed. The resulting opposition to the current is called resistance of the conductor.
Resistance R = \(\frac { V }{ I }\)
where, V = voltage applied between the two ends of the conductor
I = electric current flowing through the conductor
Unit: The SI unit of resistance is Ohm. It is denoted by Ω (omega).

Ohm : If the potential difference across the two ends of a conductor is 1V and the current through it is 1 A, then the resistance R of the conductor is 1Ω.
1 Ω = \(\frac { 1 V }{ 1 A }\)

→ Resistivity: The electrical resistivity of a material is the resistance of a conductor (of that material) having unit length and unit area of cross-section. Its SI unit is Ωm.

→ Series combination of resistors: When two (or more) resistances are connected end to end consecutively with one source such that the close path is formed, then they are said to be connected in series.
OR
When two or more than two resistors are connected between two points in a circuit in such a way that the current has only one way to flow and current flowing through every resistor in the circuit remains the same, such a combination of the resistors is called series combination of resistors.

The total resistance of the circuit increases in a series combination of resistors. This results in a reduction in the current.

→ Parallel combination of resistors: When two (or more) resistances and one source are connected between the same two points, they are said to be connected in parallel.
OR
When two or more than two resistors are connected between two points in a circuit in such a way that more than one paths are available for the current to flow and the voltage drop across two ends of each resistor remains the same, then the resistors are said to be connected in parallel between these two points and the combination of resistors is called parallel combination.

The total resistance in the circuit is reduced in a parallel combination of the resistors. This results in an increase in the electric current.

→ Heating effect of electric current: The production of heat in a metallic conductor by the electric current flowing through it is callled the heating effect of electric current.

The heat generated due to flow of electric current depends on the following factors:

• Electric current
• Resistance and
• Time taken to pass electric current.

The electrical energy consumed (or heat energy produced) when an electric current (7) flows in a conductor for time t is
H = I²Rt = \(\frac{V^2}{R}\) t = \(\frac { V }{ R }\) x Vt = IVt (joule)
The unit of electric energy is joule.

→ Electric power: Electric energy consumed (or heat energy produced) per unit time is termed as electric power.
OR
Electric power is the rate of consumption of electric energy with time. The SI unit of electric power is watt.

watt: If a device carries a current of 1 A when operated at a potential difference of 1 V, the power consumed is 1W.
1 watt = 1 volt x 1 ampere.

Students must go through these JAC Class 10 Science Notes Chapter 16 Management of Natural Resources to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 16 Management of Natural Resources

→ Natural resources : The naturally occurring resources which are used by human beings for various purposes are called natural resources. Soil, water, air, forests, wildlife, coal, petroleum, etc. are the examples of natural resources.

→ In 1985, Ganga Action Plan was started for cleaning and maintaining the quality of water in the river Ganga.
The presence of faecal coliform bacteria in water indicates contamination by disease causing microorganisms.

→ There are some measurable factors which are used to quantify pollution or the quality of water.
The pH of water can be easily checked using universal indicator. Water with a pH level between 6.5-8.5 is safe for drinking.

→ Conservation is controlled utilization of natural resources for the benefit of all life forms.

→ The five R’s to save the environment are:

• Refuse : It means not to buy product that can harm the environment.
• Reduce : It means use less and save natural resources.
• Reuse : It means use things again and again instead of throwing away.
• Repurpose : It means a product can be used for other purpose, when it is no more useful for its primary and main purpose.

→ Recycle : It means the processing of things that are considered waste and turning them into useful products.

→ The concept of sustainable development encourages eco-friendly approach that meets current basic human needs. Economic development is linked to environmental conservation.

→ The natural resources need to be managed due to following reasons :

• They should last for future generation
• They should be equally distributed
• The resource are used in such a way that the environment should not be harmed.

→ Forests are important renewable natural resources. A wide range of different life forms are found there. They are biodiversity hot spots.

→ The different stakeholders in a forest are :

• People who live in or around forests
• The Forest Department of the Government
• The industrialists
• The wildlife and nature enthusiasts.

→ The Government of India has instituted an ‘Amrita Devi Bishnoi National Award for Wildlife Conservation’, in the memory of Amrita Devi Bishnoi. who in 1731 sacrificed her life along with 363 others for the protection of ‘Khejri’ trees.

→ Chipko Andolan (‘Hug the Trees Movement’) was started in early 1970 from Reni village, in Garhwal.
Chipko Andolan was the result of a grass root level effort to end the alienation of people from forests.

→ Sal forests in the Arabari forest range of Midnapore district of West Bengal were saved by the involvement of the local villagers.

→ Monoculture of pine, teak and eucalyptus in the forest area are an important source of revenue for the Forest Department, but such practise destroys biodiversity.

→ Water is a basic necessity for all terrestrial forms of life.

• Rain is a very important source of fresh water.
• Large dams can ensure the storage of adequate water mainly for irrigation and for generating electricity.
• Large dams cause social, economic and environmental problems.

→ Water harvesting means storing rainwater where it falls or storing run off water in a local areas. Dr Rajendra Singh won the Stockholm Water Prize in 2015. He is known as India’s ‘Waterman’.

→ Fossil fuels, i.e., coal and petroleum were formed from the degradation of biomass millions of years ago. Energy needs have been largely met by the reserves of coal and petroleum.

We would need to look for alternative sources of energy because coal and petroleum are exhaustible energy resources.