Class 10

Haryana State Board HBSE 10th Class Maths Notes Chapter 2 बहुपद Notes.

Haryana Board 10th Class Maths Notes Chapter 2 बहुपद

→ बहुपद की पात- घर x के बहुपद p (x) में x को उच्चतम घात (power) यहुपद की घात कहलाती है; जैसे बहुपद 5x³ – 4x² + x – \(\sqrt{2}\) चर x में घात 3 का बहुपद है।

→ घातों एक, दो और तीन के बहुपद क्रमशः रखिक बहुपद, द्विघात बहुपद एवं त्रिवात बहुपद कहलाते हैं।

→ एक द्वियात बहुपद ax² + bx + c जहाँ a, b, c वास्तविक संख्याएँ हैं और a ≠ 0 है के रूप का होता है।

→ यदि p (x), x में कोई बहुपद है और कोई वास्तधिक संख्या है, तो p(x) में x को k से प्रतिस्थापित करने पर प्राप्त वास्तविक संख्या p(x) का x = k पर मान कहलाती है और इसे P (k) से निरूपित करते हैं।

→ कोई वास्तविक संख्या k, बहुपद p(x) का शून्यक कहलाती है यदि p(k) = 0 हो।

→ किसी रैखिक बहुपद ax + b का शून्यक \(\frac{-b}{a}\) होता है।

→ एक बहुपद p (x) के शून्यक उन बिंदुओं के x-निर्देशांक होते हैं जहाँ y = p(x) का ग्राफ x-अक्ष को प्रतिच्छेद करता है।

→ एक द्विघात बहुपद के अधिक-से-अधिक दो शून्यक और एक त्रिघात बहुपद के अधिक-से-अधिकं तीन शून्यक हो सकते हैं।

→ यदि द्विघात बहुपद ax² + bx + c के शून्यक α और β हो, तो उनका योगफल व गुणनफल निग्न होगा-
α + β = \(-\frac{b}{a}\), αβ = \(\frac{c}{a}\)

→ यदि किसी द्विधात बहुपद के दो शून्यक दिए गए हों तो बहुपद होगा-
बहुपद = x² – (शून्यकों का योग) x + शून्यकों का गुणनफल

→ यदि α, β, γ त्रिपात बहुपद ax³ + bx² + cx + d = 0 के शून्यक हों, तो उनका योगफल व गुणनफल निम्न होगा-
α + β + γ = \(\frac{-b}{a}\)
αβ + βγ + γα = \(\frac{c}{a}\)
और αβγ = \(\frac{-d}{a}\)

→ यदि α, β व γ एक विधात बहुपद के शून्यक हों तो बहुपद = x³ – (α + β + γ) x² + (αβ + βγ + γα) x – αβγ

→ विभाजन एल्गोरिथ्म के अनुसार दिए गए बहुपद p (x) और शून्येतर बहुपद g(x) के लिए दो ऐसे बहुपदों q(x) तथा r(x) का अस्तित्व है कि P(x) = g (x) q(x) + r(x),
जहाँ r(x) = 0 है या बात r(x) < घात g(x) है।

Haryana State Board HBSE 10th Class Maths Notes Chapter 1 वास्तविक संख्याएँ Notes.

Haryana Board 10th Class Maths Notes Chapter 1 वास्तविक संख्याएँ

→ यूक्लिड विभाजन प्रमेयिका- दो धनात्मक पूर्णाक a और b दिए होने पर, ऐसी अद्वितीय पूर्ण संख्याएँ q और r विद्यमान होती हैं कि a = bq + r, 0 ≤ r < b है।

→ यूक्लिड विभाजन एल्गोरिम द्वारा- दो धनात्मक पूर्णाकों, c और d(c > d) का HCF ज्ञात करने के लिए नीचे दिए हुएं चरणों का अनुसरण किया जाता है-
चरण I-c और के लिए यूक्लिड विभाजन प्रमेयिका से हम ऐसे q और r ज्ञात करते हैं कि c = dq + r, 0 ≤ r < d हो।
चरण II-यदि r = 0 है, तो d पूर्णाको c और d का HCF है। यदि r ≠ 0 है, तो और के लिए, यूक्लिड विभाजन प्रमेविका का प्रयोग पुनः कीजिए।
चरण III-इस प्रक्रिया को तब तक जारी रखिए, जब तक शेषफल 0न प्राप्त हो जाए। इसी स्थिति में, प्राप्त भाजक ही वांछित HCF है।

→ अंकगणित की आधारभूत प्रमेय- प्रत्येक भाज्य संख्या को अभाज्य संख्याओं के एक गुणनफल के रूप में व्यक्त (गुणनखडित) किया जा सकता है तथा यह गुणनखंडन अभाज्य गुणनखंडों के आने वाले क्रम के बिना अद्वितीय होता है।

→ यदि p कोई अभाज्य संख्या है और p, a² को विभाजित करता है तो p, a को भी विभाजित करेगा, जहाँ a एक धनात्मक पूर्णाक है।

→ किन्हीं दो धनात्मक पूर्णांकों a और b के लिए HCF(a, b) × LCM(a, b) = a × b होता है।

→ परिमेय और अपरिमेय संख्याएँ मिलकर वास्तविक संख्याएँ बनाती हैं।

→ \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) तथा व्यापक रूप में \(\sqrt{p}\) अपरिमेय संख्याएँ हैं, जहाँ पर p एक अभाज्य संख्या है।

→ यदि x एक परिमेय संख्या हो जिसका दशमलब प्रसार सांत हो, तो हम x को \(\frac{p}{q}\) के रूप में व्यक्त कर सकते हैं, जहाँ p औरq सहअभाज्य होते हैं तथा q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का होता है, जहाँ n, m प्रणेतर पूर्णाक होते हैं।

→ माना x = \(\frac{p}{q}\) एक ऐसी परिमेय संख्या है कि q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का है, (जहाँ n, m ऋणेतर पूर्णाक है) तो x का दशमलव प्रसार सांत होगा।

→ माना x = \(\frac{p}{q}\) एक ऐसी परिमेय संख्या है कि q का अभाज्य गुणनखंडन 2ⁿ5^m के रूप का नहीं है, (जहाँ n, m ऋणेतर पूर्णाक है) तो x का दशमलव प्रसार असांत आवर्ती होगा।

→ किन्हीं तीन संख्याओं p, q तथा r के लिए LCM होगा-

→ किन्हीं तीन संख्याओं P, q तथा r के लिए HCF होगा-

→ एक परिमेय संख्या और एक अपरिमेय संख्या का योग या अंतर एक अपरिमेय संख्या होती है।

→ एक शून्येतर परिमेय संख्या और एक अपरिमेय संख्या का गुणनफल या भागफल एक अपरिमेय संख्या होती है।

Haryana Board HBSE 10th Class Maths Notes

HBSE 10th Class Maths Notes in English Medium

• Chapter 1 Real Numbers Notes
• Chapter 2 Polynomials Notes
• Chapter 3 Pair of Linear Equations in Two Variables Notes
• Chapter 4 Quadratic Equations Notes
• Chapter 5 Arithmetic Progressions Notes
• Chapter 6 Triangles Notes
• Chapter 7 Coordinate Geometry Notes
• Chapter 8 Introduction to Trigonometry Notes
• Chapter 9 Some Applications of Trigonometry Notes
• Chapter 10 Circles Notes
• Chapter 11 Constructions Notes
• Chapter 12 Areas Related to Circles Notes
• Chapter 13 Surface Areas and Volumes Notes
• Chapter 14 Statistics Notes
• Chapter 15 Probability Notes

HBSE 10th Class Maths Notes in Hindi Medium

• Chapter 1 वास्तविक संख्याएँ Notes
• Chapter 2 बहुपद Notes
• Chapter 3 दो चरों वाले रखिक समीकरण का युग्म Notes
• Chapter 4 द्विघात समीकरण Notes
• Chapter 5 समांतर श्रेढ़ियाँ Notes
• Chapter 6 त्रिभुज Notes
• Chapter 7 निर्देशांक ज्यामिति Notes
• Chapter 8 त्रिकोणमिति का परिचय Notes
• Chapter 9 त्रिकोणमिति का अनुप्रयोग Notes
• Chapter 10 वृत्त Notes
• Chapter 12 वृतों से संबंधित क्षेत्रफल Notes
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Notes
• Chapter 14 सांख्यिकी Notes
• Chapter 15 प्रायिकता Notes

Haryana State Board HBSE 10th Class Science Solutions Chapter 10 Light Reflection and Refraction Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 10 Light Reflection and Refraction

HBSE 10th Class Science Light Reflection and Refraction Textbook Questions and Answers

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

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

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

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

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

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

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:
1. When the object is placed between the pole and the principal focus of the concave mirror, we get an erect, virtual and enlarged image.
2. We are given that the focal length of concave mirror = 15 cm
3. So, the object should be placed in front of the concave mirror at a distance less than 15 cm.

Question 8.
Name the type of mirror used in the following situations.
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Answer:
(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.

(b) A convex mirror is used as a side/rear-view mirror of a vehicle because of the following reasons:

• A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front of it.
• A convex mirror has a wider field of view than a plane mirror of the same size.

(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces. Concave mirrors are able to raise the temperature drastically.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
As shown in figure, when the lower half of the convex lens is covered with a black paper, it still forms the complete image of the object, but the intensity of the image is reduced.

The rays coming from the object get refracted by the upper half of the lens and meet at the other side of the lens to form the image.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the Image formed.
Answer:
The final position, nature and size of the image A’B’ are:
Here, Object size h = +5 cm
Object distance u = – 25 cm
Focal length of lens f = + 10 cm
( ∵ Converging, i.e., Convex lens)
Image distance y =?
Image height h’ =?

Image distance y is positive. This shows that the image formed is real and on the other side of the lens, at 16.67 cm from the lens as shown in the figure. AB : Object, A’B’: Image
Now, Magnification m = \(\frac{h^{\prime}}{h}=\frac{v}{u}\)

The negative (minus) sign shows that the image is inverted, real, diminished (3.3 cm). Figure 10.60 show the positive, size and nature of the image formed.

Question 11.
A concave lens of focal length 15 cm forms an Image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Scale: 1 cm = 5 cm
Focal length, f = – 15 cm [f is – ve for a concave lens]
Image distance, v = – 10 cm
[Concave lens forms virtual image on same side as the object, so v is – ve]

Object distance, u = – 30 cm.
The negative sign shows that the object is placed on the left side of the lens.

Question 12.
An object Is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Object distance, u = – 10 cm
Focal length, f = + 15 cm [f is +ve for a convex mirror]

Image distance, v = + 6 cm.
As v is +ve, so a virtual, erect image is formed at a distance 6 cm behind the mirror.

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification of a mirror \(m=\frac{h^{\prime}}{h}=\frac{-v}{u}\)
Here, for a plane mirror, m = + 1, So, h’ = h and y = – u

• m = 1 indicates that both ¡mage and object are of same size.
• Positive sign of m indicates that a virtual image is formed behind the mirror.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Object size, h = + 5cm
Object distance, u = – 20 cm
Radius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]
Focal length, \(f=\frac{R}{2}\) = + 15 cm
From mirror formula,

A virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.

Question 15.
An object of size 7.0 cm Is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharply focused image can be obtained? Find the size and the nature of the Image.
Answer:
Object size, h = + 7.0 cm
Focal length, t = – 18 cm
Image size, h’ = ?

Object distance, u = – 27 cm
image distance,v = ?
The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain sharp image.

The image is real, inverted and enlarged in size.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Here, P = – 2.0 D f =\(\frac{1}{P}=\frac{1}{-2.0 D}\) = – 0.5 m.
Since the power of lens is negative, the lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Here P = +1.5D
\(f=\frac{1}{P}=\frac{1}{+1.5 D}=+\frac{10}{15} m\) = + 66.67cm.

• As the focal length is positive, the prescribed lens is converging.
• Opposite side at F, real and inverted, highly diminished (point sized)

HBSE 10th Class Science Light Reflection and Refraction InText Activity Questions and Answers

Textbook Page no – 168

Question 1.
Define the principal focus of a concave mirror.
Answer:
The point (F) on the principle axis where the beam of light parallel to the principle axis either actually converges to or diverges from, after reflection from a mirror is called the principle focus (F).

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
R = 20cm.
Focal length (f) = \(\frac{\mathrm{R}}{2}=\frac{20}{2}\) = 10 cm

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

Question 4.
Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror is preferred as a rear-view mirror because —

• Convex mirror always forms a virtual, erect and a diminished (point-like) image of an object placed anywhere in front of it. This means the driver can see the image of person even at a very distant position.
• Convex mirror has a wider field of view. So it covers a large rear distance. This helps the driver as he can see more.

Textbook Page no – 171

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

Question 2.
A concave mirror produces three times magnified (enlarged) real Image of an object placed at 10 cm in front of It. Where is the image located?
Answer:
As the image is real, so magnification m must be negative.
\(m=\frac{h^{\prime}}{h}=-\frac{v}{u}=-3\)
∴ v = 3 x (-10) = – 30 cm.
Thus the image is real and it is located at a distance of 30 cm from the mirror in the front of mirror.

Textbook Page no – 176

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
1. Light travels faster in rarer medium air and slower in denser medium water.
2. Since the ray of light travelling in air enters obliquely into water, it slows down and bends towards the normal.

Question 2.
Light enters from air to glass having refractive Index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 x 108 mr1.
Answer:
1. Refractive index of glass, ng = 1.50
Speed of light in vacuum, c = 3 x 10⁸ ms^-1

2. Absolute refractive index of glass, ng = ((c/v))
∴ Speed of light in glass \(v=\frac{c}{n_g}=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{1.5}=2 \times 10^8 \mathrm{~ms}^{-1}\)

Question 3.
Find out, from table 10.3 of the text book, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
1. From table 10.3, diamond has highest refractive index (= 2.42), so it has largest optical density.
2. Air has lowest refractive index (= 1.0003), so it has lowest optical density.

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given In table 10.3 of the text book.
Answer:
For kerosene, n = 1. 44,   For turpentine oil, n = 1. 47,   For water, n = 1. 33
As water has lowest refractive index, so light travels fastest in this optically rarer medium compared to kerosene and turpentine oil.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It indicates that the ratio of speed of light in air to that in diamond is 2.42.

Textbook Page no – 184.

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre (1 D) is the power of a lens whose focal length is 1 meter.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed ¡n front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
Answer:
1. Here, image distance y = + 50 cm (∴ y is positive for real images)
2. As mentioned, the real image is of the same size as the object.

Thus, object (needle) is placed at 50 cm from the lens.

Power of lens:

∴ Focal length (f) = 25 cm = 0.25 m
Now, \(P=\frac{1}{f}=\frac{1}{+25}=+4 D\)
Thus, power of lens = + 4D

Question 3.
Find the power of a concave lens of focal length 2 m.
Answer:
Since the focal length of a concave lens is negative; f = – 2m.
Power \(P=\frac{1}{f}=\frac{1}{-2 m}=-0.5 \mathrm{D}\)

Activities

Activity 1.

Aim: To study the formation of images by both sides of a large shining spoon.

Apparatus required: A large shining spoon.

Procedure:

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

Observation and Conclusion:

• Initially, we can see our tace inside the curved portion of the spoon. The image of our face looks erect and enlarged. As we move the spoon at a farther distance, the image goes on becoming smaller and then gets inverted. The inner portion of the spoon behaves as a concave mirror.
• When we see the spoon from outer surface, it behaves as a convex mirror. The image formed is erect and small Irrespective of the distance of face from the spoon.

Activity 2.

Aim: To find the approximate focal length of a concave mirror.

Apparatus required: Concave mirror, a sheet of paper and meter scale

Caution: Do not look at the sun directly or even Into a mirror reflecting sunlight. It may damage your eyes.

Procedure:

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

Question 1.
Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?
Answer:
Observation
Initially, the paper turns blackish and emits smoke while burning slowly. The paper then catches fire.

Conclusion:

• The point at which the sun-rays get converged to produce tire is the principal focus (F) of the concave mirror.
• The distance of this image of sun at point F gives the approximate value of focal length (t) of the concave mirror.

Activity 3.

Aim: To locate the image formed by a concave mirror for different positions of the object.

Procedure:
1. Take a concave mirror. Find out its approximate focal length. Let f = 10 cm.

2. Mark a line on a table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.

3. Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will correspond to the position of the points P, F and C, respectively.

4. Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the
mirror till you obtain a sharp bright image of the candle flame on it.

5. Observe the image carefully. Note down its nature, position arid relative size with respect to the object size.

6. Repeat the experiment by placing the candle —
(a) just beyond C, (b) at C, (C) between F and C, (d) at F, and (e) between P and F

7. In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.

8. Note down and tabulate your observations.

Observation:
Table 1. Nature, size and position of Images formed by a concave mirror for different positions of the object

Activity 4.

Students should do this activity on their own. Make use of activity 3 for drawing the diagrams.

Activity 5.

Aim: To locate the image formed by a convex mirror for different positions of the object.

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

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

Question 2.
Move the pencil away from the mirror slowly. Does the image become smaller or larger?
Answer:
The image formed is smaller and erect.

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

Observation:

Activity 6.

Aim: To demonstrate that convex mirror has a wide field of view.

Question 1.
Observe the Image of a distant object, say a distant tree, in a plane mirror. Could you see a full-length Image?
Answer:
No, we cannot see the full-length image of the object in plane mirror.

Question 2.
Try with plane mirrors of different sizes. Did you see the entire object In the Image?
Answer:
When we try with plane mirrors of different sizes, we find that the entire object in the image is seen only when the size of the plane mirror is at least half the size of the object.

Question 3.
Repeat this activity with a concave mirror. Did the mirror show full length Image of the object?
Answer:
The full length image is not formed even in concave mirror.

Question 4.
Now try using a convex mirror. Did you succeed? Explain your observations with reason.
Answer:
A full length image of the object is obtained in a small convex mirror.

Reasons for these observations:

• In a plane mirror, size of image is always equal to the size of the object.
• In case of a concave mirror, when the object is between P and F only then we get an enlarged, virtual
  and erect image behind the mirror. But here, since the object is far away, its full length image cannot be seen.
• In a convex mirror, the image is always virtual, erect and smaller than the object.

Activity 7.

Aim: To demonstrate refraction of light.

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

Observation and conclusion:

• Due to refraction of light from water, the coin appears to be slightly above its actual position.
• This visual illusion makes it difficult to identify exact position of the coin and pick it. However, after trying a while we can pick it up.

Activity 8.

1. Place a large shallow bowl on a table and put a coin in it.
2. Move away slowly from the bowl. Stop when the coin just disappears from your sight.
3. Ask a friend to pour water gently into the bowl without disturbing the coin.
4. Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?

Observation and Conclusion: On pouring water into the bowl, the coin becomes visible again. Moreover, due to refraction of light, the coin appears slightly raised above its actual position. This happens due to refraction of light.

Activity 9.

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

Question 1.
Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
Answer:
We see the line first in the air medium and then in the glass. So, the line under the glass slab appears bent at the edges-due to refraction.

Question 2.
Next, place the glass slab such that it ¡s normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
Answer:
Now, the line does not appear bent. The reason for this is that the light rays incident normally at the glass and air do not undergo refraction.

Question 3.
Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
Answer:
Yes, part of the line under the glass slab appears to be raised. This is due to the refraction of light rays when they travel from glass to air.

Activity 10.

1. Fix a sheet of white paper on a drawing board using drawing pins.
2. Place a rectangular glåss slab over the sheet in the middle.
3. Draw the outline of the slab with a pencil. Let us name the outline as ABCO.
4. Take four identical pins.
5. Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
6. Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
7. Remove the pins and the slab.
8. Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O’.
9. Join O and O’. Also produce EF up to P, as shown by a dotted line in Figure.

Activity 11.

Aim: To show the converging nature of convex lens.
Caution: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.

Procedure:

• Hold a convex lens in your hand. Direct it towards the Sun.
• Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
• Hold the paper and the lens in the same position for a while. Keep observing the paper. What happens? Why? Recall your experience in Activity 2.

Observation and conclusion:

• The paper begins to burn producing smoke. Later, it catches fire.
• We conclude that the light rays from the sun gets converged when they pass through the lens and form a bright spot on the paper. This spot shows the focus of the lens.

Activity 12.

Aim: To locate the position and examine the nature of the image formed by a convex lens for different positions of the object.

1. Take a convex lens. Find its approximate focal length in the way described in Activity 11.

2. Draw five parallel straight lines, using chalk on a long table such that the distance between the successive lines is equal to the focal length of the lens.

3. Place the lens on a lens stand. Place it on the central line such that the optical centre of the lens lies just over the line.

4. The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2F₁, F₁, F2 and 2F₂, respectively.

5. Place a burning candle, far beyond 2F₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.

6. Note down the ‘nature, position and relative size of the image.

7. Repeat this activity by placing the object just behind 2F₁, between F₁ and 2F₁ at F₁ at F₁, between F₁ and O. Note down and tabulate your observations.

Observation and conclusion:

The nature, position and size of the images formed by a convex lens depend on the position of the object in relation, to the points O, F₁ and 2F₁. After nothing the various characteristics of the image, the observations are mentioned in the table below.

Activity 13.

Aim: To locate the position and examine the nature of the image formed by a concave lens for different positions of the object.

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

Observation and Conclusion:

• On observing the image directly through the lens, we can see that the image of the candle is virtual, erect, smaller than the candle and closer to the lens.
• On moving the candle away from the lens, the size of the image decreases.
• When the candle is placed very far from the lens, the image becomes highly diminished and appears point-like.

Haryana State Board HBSE 10th Class Science Solutions Chapter 9 Heredity and Evolution Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 9 Heredity and Evolution

HBSE 10th Class Science Heredity and Evolution Textbook Questions and Answers

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

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

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

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light – coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:

Reason:
1. No, the given information is not enough to consider the dominance or recessiveness of eye colour.

2. However, we can observe that generally in a population, the proportion of the people having light coloured eyes is much less than the people having dark coloured eyes. Based on this observation, we may consider that the light coloured eyes are a recessive trait and dark coloured eyes may be dominant.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
1. In classification, the groups of organisms are based on the similarities and differences of characteristics of the organisms.

2. The organisms that have more similarities belong to nearer groups while the organisms having lesser similarities i.e. having more differences belong to distant groups.

3. In evolution, a study of identifying hierarchies of characteristics between the species is done. So, in such studies small groups of species with recent common ancestors are formed. Thus, the areas of study – evolution and classification are interlinked.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
1. In dogs, the coat colour is either black or white. The character of black coloured coat is indicated by ‘B’ and white coloured coat is indicated by ‘b’.

2. First of all the homozygous/pure parents with opposite characters are fertilized. Thus, applying Mendelian’s law, in the above case, the black coloured coat is a dominant character. So, black coat is dominant trait in dog.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
1. It is believed that life must have developed from simple inorganic molecules which were present on earth soon after it was formed.
2. It was also speculated that the conditions on earth at that time could have given rise to more complex organic molecules that were necessary for life.
3. It is believed that the first primitive organisms would have arisen from further chemical synthesis.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
(1) First, in sexual reproduction, there occurs a combination of genetic material of two different parents i.e. a male and a female. Hence, each new generation is a combination of two different parental DNAs.

(2) Second, during the formation of gametes i.e. formation of sperms and ovum, the cell undergoes cell division phase called meiosis.

• During meiosis, the genetic material gets distributed randomly. Moreover, all the gametes are not always similar and so there are more chances of variation during fertilization, in other words, there are so many sperm cells but only one fertilizes with the ovum. Naturally, this will create variation.
• Owing to these two reasons, there occur more variations in sexual reproduction as compared to asexual.
• The combination of different types of gametes which takes place is the root of variations. This also gives rise to new type of characters too. This also favours evolution.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
1. During gametogenesis (production of male and female sex cells), the meiosis type of cells division occurs.
2. During, meiosis the number of chromosomes becomes half. The DNA gets distributed to half in each gamete.
3. During fertilization, both the types of sex cells get fused and the quantity of DNA remains maintained. Thus, equal genetic contribution of male and female parents is ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, this is true.

Reason:

• If the variation achieved by an organism is helpful to adapt to the surrounding environment then it increases the probability of the survival of that organism.
• The survived organisms then produces more offspring with such genetic variations and thus the species survives.

HBSE 10th Class Science Heredity and Evolution InText Activity Questions and Answers

Textbook Page no – 143

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B which exists in 60% of population of an asexually reproducing species must have arisen earlier than trait A.

Reason:

• In asexually reproducing species, DNA copy error takes place, in very small amount and so new traits do not occur very fast.
• Since, the percentage of trait B is very large as compared to trait A. This suggests that trait B must have arisen quite before to trait A.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
1. Variations in a species are created through two reasons,

• Due to inaccuracies in DNA copying or
• During sexual reproduction.

2. Different individuals gain different types of advantages with these variations. Several variations enable the individual to adapt to the environmental conditions. This increases the chances of their survival.

3. The individuals that survive then reproduce offspring and the existence of species remains continued.

Textbook Page no – 147

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
1. Mendel conducted cross fertilization between pure tall plants (TT) and pure short (tt) plant. This resulted in all (Tt) plants in F₁ generation.

2. This shows that single copy of T is enough to make the plant tall. The trait that gets expressed over the other is called the dominant trait and the other recessive.

3. In Mendel’s trial, the trait which gets expressed in 75% individuals in F₂ generation after self-fertilization is dominant whereas the trait which appears in 25% individual is recessive.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
1. Mendel performed the experiments of heredity on pea plant. He studied the heredity of two characters simultaneously and performed the dihybrid experiment.

2. On the basis of this experiment we can say that even though the F₁ progeny plants were dominant in both the characters, the F₂ progeny showed the new combinations. In F₂ progeny plants some were tall and wrinkled, while some were short and round.

3. The experiment clearly shows that the factors that control the shape, size and height of the seeds are inherited independently and that is the main reason why new combinations could be seen in F₂ generation.

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
No, the given information is not enough to determine which trait is dominant.

Reason:

• The daughter (or any child) receives two copies of genes, one from each parent.
• The trait of blood group depends on the genotype i.e. the types of genes which get combined. Hence, the provided information is insufficient and so we cannot tell which one is a dominant blood group.

Question 4.
How is the sex of the child determined in human beings?
Answer:
1. In human beings, sex of the child to be born is determined by the sex chromosome of the father.

2. The father or say male contains XY chromosomes and so during spermatogenesis, the male produces two types of sperms. 50% of these sperms have ‘X’ sex chromosome and 50% sperms have ‘Y’ sex chromosome.

3. If the sperm having ‘X’ chromosome fertilizes the ovum, a female child (= a girl) is born and if the sperm having ‘Y’ chromosome fertilizes the ovum, the male child (= a boy) is born.

Textbook Page no – 150

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A particular trait may increase in a population through two main ways. They are –

• Natural selection – It directs evolution with a survival advantage.
• Genetic drift – It leads to accumulation of different changes.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Change in non-reproductive tissues cannot be passed on to the DNA of the germ cells. Hence the experiences of an individual during its lifetime can neither be passed on to its progeny nor can such experiences result in evolution.

Example:

• If we breed a group of mice, all their progeny will have tails.
• Suppose we remove the tails of each generation of these mice by surgery then it does not mean that the progeny is tailless.
• Artificially removing the tail does not bring any change in the genes of the germ cells of the mice.
• So, the cut-tail that a generation of mice experienced cannot become a trait to be passed on to the next generation.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
1. A small population generally does not show much variation from generation to generation.
2. A very small population of surviving tigers indicates that very less amount of variation will occur in the genetic characteristics. This means they will not be able to adapt to the major environmental changes if any, in the future.
3. The tigers will not survive and the species will become extinct.

Textbook Page no – 151

Question 1.
What factors could lead to the rise of a new species?
Answer:
Factors that can lead to the rise of a new species are:

• Accumulated variations favourable to natural environment,
• Geographic isolation of a population,
• Gene flow,
• Genetic drift and
• Natural selection

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
No, geographical isolation will not be a major factor in the speciation of a self-pollinating plant species.

Reason:

• Only a single parent is involved in self-pollination. So, there is no gene flow between two geographically isolated populations.
• If it would be a cross-pollination species, the geographical isolation would be a major factor as it would lead to faster accumulation of variation (genetic drift) in the two geographically separated populations.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
No, geographical isolation will not be a major factor to the speciation that reproduces asexually.

Reason:
Asexually reproducing organisms show very little variation over generations. These little variations are not sufficient enough to raise a new species having such variations.

Textbook Page no – 156

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
1. One of the ways to determine how close two species are with respect to evolution is to obtain evidence from their homologous organs.

2. For example, the basic design of internal structure of bones of forelimbs of a frog, lizard, bird, bat and man is same, even though these organs perform different functions.

3. This indicates that all these forelimbs have evolved from a common ancestral animal, which had a same basic internal structure.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
No, the wings of a butterfly and the wings of a bat cannot be considered homologous organs.

Reason:
Although the wings of a butterfly and the wings of a bat perform similar function, but the design, structure and components of their wings are very different. Hence, the wings of these two can be considered as analogous organs and not homologous.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
1. The remains of dead organisms buried under the earth for millions of years are known as fossils.
2. Fossils are impressions of dead plants or animals that lived in the past.
3. When plants or animals die, the micro-organisms get decomposed in the presence of moisture and oxygen.
4. However, sometimes due to environmental conditions, their bodies do not decompose completely.
5. Such body parts of the plants and animals become fossil and can be available on digging the earth.
6. Fossils can be in the form of imprints, burrow of a worm, or even an actual bone.

Example:

• If the dead leaf gets caught in the mud, leaf will not decompose completely.
• The mud around the leaf will set around it as a mould which will then slowly become hard to form a rock and retain the impression of the leaf. Thus fossil of a leaf is formed.
• By studying these fossils, scientists learn how organisms evolved over time.

Textbook Page no – 156

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
1. Across the earth, there is a great diversity in humans and so we look so different from each other in terms of size, colour and looks.

2. Initially, human race was identified on the basis of skin colour and the humans were known as yellow, black, white or brown. However, in the recent years, it has been proved that all the human beings have evolved from a single species called the Homo sapiens.

3. Hence, in spite of having different skin colour, looks and size, we all belong to the same species.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
1. In evolutionary terms, we can say that chimpanzees have a better developed body design.
2. The reason for this is that chimpanzees are highly complex organisms compared to remaining three.

Activities:

Activity -1

Aim: To study inheritance in the type of earlobes.

Background:

• The lowest part of the ear pinna (external ear) is known as an earlobe.
• The earlobe may be either closely attached to the (lateral) side of the head, or not-attached i.e. might be free.

Procedure:

• Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and
• Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.
• Students can observe the earlobes of other students. They can also ask about the type of earlobes of their parents.
• Make a chart as given below. Write 1F’ for the free earlobes and A’ for the attached earlobes.

Observation:
We can observe that more number of students have inherited the trait of ‘free earlobes’ i.e. trait F as compared to attached earlobes i.e. A.

Conclusion:
From studies it has been proved that free earlobe is a dominant trait whereas an attached earlobe is a recessive trait.

Activity 2.

In Fig. 9.3 of textbook, what experiment would we do to confirm that the F₂ generation did in fact have a 1:2:1 ratio of TT, it and tt trait combinations?

Observation:
Although the experiment phenotypically shows the ratio of 3 : 1 ratio between tall and dwarf, genetically the ratio is 1 : 2 : 1.
This can be proved by the following experiment. We can raise F₃ generation by allowing self-pollination of F₃ generation plants.

Then we may calculate ratio for —

• Tall plants which produced only tall plants. These plants will be pure dominant (TT).
• Tall plants which produced both tall and dwarf pea plants. These plants are hybrid (Tt).
• Dwarf plants produced only dwarf plants. This means they are pure recessive (tt).

Observing the data carefully, one can find the appearance of 1 : 2 : 1 ratio of TT Tt and tt trait combination in F₃ generation (also known as genotypic ratio).

Haryana State Board HBSE 10th Class Science Solutions Chapter 8 How do Organisms Reproduce? Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 8 How do Organisms Reproduce?

HBSE 10th Class Science How do Organisms Reproduce? Textbook Questions and Answers

Question 1.
Asexual reproduction takes place through budding In
(a) amoeba
(b) yeast
(c) plasmodium
(d) leis mania
Answer:
(b) yeast

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

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

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:
The advantages of sexual reproduction over asexual reproduction are as follows:
1. Sexual reproduction involves fusion of male and female gametes from male and female parents to form zygote. Thus, new generation shows variation as DNA copies from two different individuals are received.

2. Sexual reproduction involves gamete formation. Meiosis takes place during male and female gamete formation and new combination of genes occur during meiosis. This also leads to variation. Combining variation from two individuals would create new combination of variants.

3. Sexual reproduction create variation generation after generation, which is useful for survival of species over time.

4. The inbuilt tendency for variation during sexual reproduction is the basis for evolution.

Question 5.
What are the functions performed by the testes ¡n human beings?
Answer:
Testes does two functions.

• To produce sperms (male sex cells) and
• To produce hormone called testosterone.

Question 6.
Why does menstruation occur?
Answer:
When egg is not fertilized, the thick-spongy and highly vascular inner inning of uterus breaks. This causes menstruation in woman.

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

Flowering plants belong to the group of angiosperms. The reproductive parts of angiosperms are located in the flower. So, we can say that flower is the reproductive organ of the plant. The four main reproductive parts of a flower are:

• Stamen
• Pistil
• Petal and
• Sepal

Stamen and pistil contain germ-cells (gametes).
1. Unisexual flower: If the flower contains only one part out of stamen or pistil then such a flower is called unisexual. For example, papaya and watermelon.

2. Bisexual flower: If the flower contains both stamen as well as pistil the flower is called bisexual For example. hibiscus and mustard.

3. Male reproductive part: Stamen is the male reproductive part. It produces yellowish-coloured pollen grains. Stamen is made up of two parts namely

• anther and
• filament.

4. Female reproductive part:
Pistil (or carpel) is the female reproductive part. It is present at the centre of a flower. It is made of three parts namely,

• ovary,
• style and
• stigma.

The swollen bottom part is the ovary, middle elongated part is the style and the terminal part which may be sticky is the stigma. The ovary contains ovules and each ovule contains an egg cell.

Question 8.
What are the different methods of contraception?
Answer:
Contraception :
The method to prevent pregnancy in women is called contraception. Methods adopted to prevent pregnancy are called contraceptive methods.

Methods:
(1) Birth control tools:

• Under this method, a mechanical barrier is created which prevents the entry of sperm into the genital tract. As a result, fertilization does not occur.
• The tools include condoms to be worn on penis and/or diaphragm worn in the vagina by female.
• Another tool is placing lntra-Uterine Contraceptive Devices (IUCDs) like Copper-T into the uterus of females to prevent pregnancy.

(2) Birth control pills:
There are oral pills that a women can take. These pills change the hormonal balance of the body so that the eggs are not released by the ovaries and fertilization is prevented.

(3) Surgical methods:

• In males, a small portion of the vas deferens is surgically removed or tied.
• This process is known as vasectomy and it prevents the sperms from entering urethra.
  (Note: ectomy means to remove by surgery)
• In females, the fallopian tube is surgically cut and tied. This process is known as tubectomy. It will not allow the sperm to reach the uterus.

(4) Abortion:
Another method is to surgically remove the foetus from the body of pregnant women. However, this is not a method to prevent pregnancy but to prevent child-birth after pregnancy.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:
1. In unicellular organisms, reproduction occurs by simple cell division. For example, through fission and budding.
2. Multi-cellular organisms have complex body organization. So, they have a complex method of reproduction. They undergo sexual reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
1. Every organism that takes birth has to die. if organisms do not reproduce, their species will become extinct.
2. Species can reproduce and gives rise to new offspring. This gives stability to its population.

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

• To prevent unwanted pregnancies –
• To control human population
• To prevent the transmission of sexually transmitted disease.

HBSE 10th Class Science How do Organisms Reproduce InText Activity Questions and Answers

Textbook Page no – 128

Question 1.
What Is the importance of DNA copying in reproduction?
Answer:
Importance of DNA copying:

• For continuity of species
• For transmitting hereditary information of parents to off-springs.
• Occurrence of variation in DNA copying, leads to evolution of a species and its better adaptability to its surroundings.

Question 2.
Why is variation beneficial to the species but not necessarily for the individuals?
Answer:
1. DNA copying occurs at the time of reproduction. DNA copying shows some variations.

2. The variations may not benefit an individual all the time, but when such variations keep on taking place over generations within a large population, they may bring some changes in the species. These changes may help the generation to be better adaptive to the changing surroundings.

3. Thus, variation in individual may not seem beneficial but when studied from the point of view of entire species, it may be beneficial.

Textbook Page no – 133

Question 1.
How does binary fission differ from multiple fission?
Answer:

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
1. Spore can survive under unfavourable conditions as they are covered by a hard protective coat.
2. Spores can also be carried away easily by agents like wind, water, etc.

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
More complex organisms cannot give rise to new individuals through regeneration because

• In complex organisms, there are specialized cells for performing different functions as well as specialized functions.
• Complex organisms have tissues and organs which cannot be produced through regeneration. Moreover, the regenerative ability of complex organism is lost due to specialization of tissues and organs.
• Specialized cells needed to conduct regeneration are not present in complex animals.

Question 4.
Why Is vegetative propagation practiced for growing some types of plants?
Answer:
Advantages of vegetative propagation:

• Plants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds.
• Vegetative propagation is also helpful in growing plants such as banana, orange, rose and jasmine that have lost the capacity to produce seeds.
• The plants produced are genetically similar to the parent plant and so have similar characteristics as the parent plant.

Question 5.
Why is DNA copying an essential part of the process of reproduction?
Answer:
The first and foremost task of the reproduction process is to make copies of blueprints of body design’. This is done by creating copies of DNA.

• Through DNA copying the genetic information of parents is passed to next generation.
• DNA present in the nucleus of the cell is the source of information for making proteins which helps in designing the body.

Textbook Page no – 140

Question 1.
How is the process of pollination different from fertilization?
Answer:
1. In pollination, there occurs the transfer of pollen grains from anther of the stamen to the stigma of the carpel. So, the main process is transfer of pollen grains.
2. In fertilization, the male gamete fuses with female gamete to form a zygote. So. fertilization is the process of fusion of germ cells.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
The seminal vesicles and prostate gland secrete fluids which make transport of sperms easier. Moreover, the fluid also provide nutrition to the sperms.

Question 3.
What are the changes seen girls at the time of puberty?
Answer:
1. The size of the breasts starts increasing. The nipples become dark.
2. The ovary starts secereting female sex hormones. Hence, girls begin to menstruate at around this time.
3. Hair start growing in pubic area and in arm-pits.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
1. The embryo gets nourishment from the mother’s blood with the help of special tissue called placenta.
2. The placenta contains villi on embryo’s side. Glucose, oxygen and important nutrients pass via. villi from the mother to embryo. This is how the embryo gets the nutrition.

Question 5.
If a woman is using a copper-T, will it help In protecting her from sexually transmitted diseases?
Answer:
No, copper-T does not restrict the flow of fluid that enters the vagina. Hence, if the women involves in sex with a person affected with a sexual disease, the fluid from the affected person’s body can enter into the women’s body and cause her sexual disease.

Activities

Activity 1.

Aim: To observe a sexual reproduction method of budding in yeast.

Procedure:

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

Observation and conclusion:

• We can see budding occurring in yeast cells. The nucleus of the cell divides.
• Bud is an outgrowth on yeast cell surface which develops due to cell division. Here, budding has shown the asexual method of reproduction.

Activity 2.

Aim: To study growth of fungus called rhizopus on a piece of bread.

Procedure:

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

Observation:
A white cotton-like mass appears on the moist bread which is known as bread mould. Mould is a type of fungus. Later, in most cases it becomes black to produce sporangia which produces spores.

Conclusion:

• We will be able to see a scattered white cotton type structure on the moist bread. This is called bread mould or fungus.
• The fungus grown is rhizopus. It becomes black to produce sporangia which produce spores.

Activity 3.

Aim: To observe amoeba and its binary fission under a microscope

Procedure:

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

Observation and conclusion:
It can be seen in the slide that amoeba is a unicellular microscopic organism. It does not have an shape i.e. its shape Is irregular.

Binary fission in amoeba:

• While studying the binary fission in amoeba, we can see that first its nucleus elongates and the divides into two parts.
• After that the cytoplasm divides into two parts. The cytoplasm gets accumulated around each nucleus. This results in two daughter cells. Each daughter cell grows into an adult organism.

Activity 4.

Aim: To observe filamentous algae under a microscope

Procedure:

• Collect water from a lake or pond that appears dark green and contains filamentous structures.
• Put one or two filaments on a slide.
• Put a drop of glycerine on these filaments and gently cover it with a coverslip.
• Observe the slide under a microscope.
• Can you identify different tissues in the spirogyra filaments?

Observation and conclusion:

• Spirogyra has a filament type structure. It is a multicellular organism which is in the form of green coloured algae.
• We can observe that the different cells of the spirogyra filament are alike and there is no tissue differentiation.

Activity 5.

Aim: To study vegetative propagation in potato

Procedure:

• Take a potato and observe its surface. Can notches be seen?
• Cut the potato into small pieces such that some pieces contain a notch or bud and some do not.
• Spread some cotton on a tray and wet it. Place the potato pieces on this cotton. Note where the pieces with the buds are placed.
• Observe changes taking place in these potato pieces over the next few days. Make sure that the cotton is kept moistened.
• Which potato pieces give rise to fresh green shoots and roots?

Observation and conclusion:

New plant grew from that part of potato which had notches (or say buds) whereas the other piece of potato that did not have notches (buds) did not show growth of a new plant.

Activity 6.

Aim: To study vegetative propagation by stem cutting.

Procedure:

• Select a money plant.
• Cut some pieces such that they contain at least one leaf.
• Cut out some other portions between two leaves.
• Dip one end of all the pieces in water and observe over the next few days.
• Which ones grow and give rise to fresh leaves?
• What can you conclude from your observations?

Observation and conclusion:

• The piece which has at least one leaf develops into fresh leaves and branch. The reason for this is that the money plant contains an auxiliary bud between leaf and stem. This bud develops into new i.e. fresh leave.
• The other piece i.e. the inter-node piece does not bear buds and.So fresh leaves do not develop on this piece.

Activity 7.

Aim: To observe parts of a seed

Procedure:

• Soak a few seeds of Bengal gram (chana) and keep them overnight.
• Drain the excess water and cover the seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry.
• Cut open the seeds carefully and observe the different parts.
• Compare your observations with the figure and see if you can identify all the parts.

Observation and conclusion:

• The seed consists of two cotyledon i.e. it is dicotyledon.
• When the seed starts growing, it shows the growth of shoot called pumule’. In future, the root will develop which will be called ‘radicle’.
• Radicle comes out of the seed during first germination.