Class 9

Haryana State Board HBSE 9th Class Science Notes Chapter 9 बल तथा गति के नियम Notes.

Haryana Board 9th Class Science Notes Chapter 9 बल तथा गति के नियम

→ बल एक सदिश राशि है।

→ गैलीलियो का जन्म 15 फरवरी, 1564 में इटली के पीसा नामक शहर में हुआ।

→ गति के पहले नियम को जड़त्व का नियम भी कहते हैं।

→ संवेग की SI इकाई किलोग्राम-मीटर प्रति सेकंड (kgm/s) होती है।

→ संवेग परिवर्तन की दर, वस्तु पर लगने वाले बल के समानुपाती होती है।

→ किसी वस्तु में उत्पन्न त्वरण, उस पर लगे बल के समानुपाती होता है।

→ संवेग एक सदिश राशि है।

→ गति के द्वितीय नियम से हमें किसी वस्तु पर लगने वाले बल को मापने की विधि मिलती है।

→ बल को वस्तु में उत्पन्न त्वरण तथा वस्तु के द्रव्यमान के गुणनफल से प्राप्त किया जाता है।

→ किसी विलगित निकाय का कुल संवेग संरक्षित रहता है।

→ घर्षण बल सदैव वस्तु की गति का प्रतिरोध करता है।

→ सभी ठोस सतहें, उनके संपर्क में गतिशील वस्तुओं पर घर्षण बल आरोपित करती हैं।

→ सभी द्रव व गैसीय सतहें, उन पर या उनसे होकर जाने वाली वस्तुओं पर घर्षण बल आरोपित करती हैं।

→ जब कोई वस्तु किसी दूसरी वस्तु के ऊपर खिसकती है तो उन दोनों के बीच के घर्षण को सी घर्षण कहते हैं।

→ रोलरों पर गति करने वाली वस्तुओं की स्थिति में घर्षण को लोटनिक घर्षण कहते हैं।

→ घर्षण उन दो सतहों के चिकनेपन अथवा खुरदरेपन पर निर्भर करता है, जो परस्पर संपर्क में हैं।

→ किसी दी हुई वस्तु के लिए सी घर्षण सदैव लोटनिक घर्षण से अधिक होता है।

→ घर्षण के अवांछनीय प्रभावों को कुछ सीमा तक नियंत्रित किया जा सकता है।

→ जो उल्का पृथ्वी की सतह तक पहुँच जाती हैं, उसे उल्का पिंड कहते हैं।

→ घर्षण कम करने के लिए मशीनों में प्रायः बॉल-बेयरिंगों का उपयोग किया जाता है।

→ बल-बल वह बाह्य कारक है, जो किसी वस्तु की स्थिति में परिवर्तन करता है या करने की चेष्टा करता है।

→ गति का प्रथम नियम-वस्तु अपनी विरामावस्था अथवा सरल रेखा के अनुरूप एकसमान गति की अवस्था में तब तक बनी रहती है, जब तक कि उस पर कोई असंतुलित बल कार्य न करे।

→ जड़त्व-वस्तुओं द्वारा अपनी गति की अवस्था में परिवर्तन का प्रतिरोध करने की प्रवृत्ति को जड़त्व कहते हैं।

→ द्रव्यमान-किसी वस्तु का द्रव्यमान उसके जड़त्व की माप है। इसका SI मात्रक किलोग्राम (kg) है।

→ गति का द्वितीय नियम-वस्तु का संवेग उसके द्रव्यमान व वेग का गुणनफल होता है और उसकी दिशा वही होती है, जो वस्तु के वेग की है अर्थात् p = mv

→ एक न्यूटन-एक न्यूटन (N) वह बल है, जो एक किलोग्राम द्रव्यमान वाली वस्तु में एक मीटर प्रति वर्ग सेकंड का त्वरण उत्पन्न करता है।

→ गति का तृतीय नियम प्रत्येक क्रिया के लिए उसके बराबर व विपरीत प्रतिक्रिया होती है और यह दो भिन्न-भिन्न वस्तुओं पर कार्य करती है।

→ संतुलित बल यदि किसी वस्तु पर क्रिया कर रहे विभिन्न बलों का परिणाम शून्य हो तो ऐसे बलों को संतुलित बल कहते हैं।

→ असंतुलित बल-यदि वस्तु पर क्रिया कर रहे विभिन्न बलों का परिणामी बल शून्य न हो तो, ऐसे बलों को असंतुलित बल कहा जाता है।

Haryana State Board HBSE 9th Class Science Notes Chapter 8 गति Notes.

Haryana Board 9th Class Science Notes Chapter 8 गति

→ स्थिति में परिवर्तन एक गति है।

→ गति की व्याख्या तय की गई दूरी या विस्थापन के रूप में की जा सकती है।

→ सभी सजीव वस्तुएँ चाहे वे पौधे हों या जंतु, किसी-न-किसी प्रकार की गति करती हैं।

→ पौधों की तुलना में जंतुओं की गति का प्रेक्षण करना आसान होता है।

→ निर्जीव वस्तुओं को गति में लाने के लिए किसी बाहरी कारक की आवश्यकता होती है।

→ दूरी एक अदिश राशि है।

→ विस्थापन एक सदिश राशि है।

→ चाल का SI मात्रक मीटर प्रति सेकंड (ms^-1) है। अन्य मात्रक सेंटीमीटर प्रति सेकंड (cms^-1) व किलोमीटर प्रति घंटा (kmh^-1) है।

→ यदि कोई वस्तु । समय में s दूरी तय करती है तो उसकी चाल (v) निम्न होगी

→ वायु में ध्वनि की चाल 346ms^-1 है।

→ यदि त्वरण वेग की दिशा में हो तो उसे धनात्मक लिया जाता है तथा यदि त्वरण वेग की विपरीत दिशा में हो तो उसे ऋणात्मक लिया जाता है।

→ त्वरण का SI मात्रक ms^-2 है।

→ स्वतंत्र रूप से गिर रही वस्तु की गति एकसमान त्वरित गति का उदाहरण है।

→ दूरी-समय ग्राफ में समय को X-अक्ष और दूरी को Y-अक्ष पर प्रदर्शित किया जाता है।

→ एकसमान चाल के लिए, समय के साथ तय की गई दूरी का ग्राफ एक सरल रेखा होता है।

→ एक सरल रेखा में चल रही वस्तु के वेग में समय के साथ परिवर्तन को वेग-समय ग्राफ द्वारा दर्शाया जाता है।

→ वेग-समय ग्राफ में समय को X-अक्ष पर और वेग को Y-अक्ष पर दर्शाया जाता है।

→ एकसमान त्वरण से चल रही एक वस्तु की गति की व्याख्या तीन समीकरणों के माध्यम से की जा सकती है। वे हैं

• v = u + at,
• s = ut + \(\frac{1}{2}\)at²
• 2as = v² – u²

→ यदि कोई एथलीट। त्रिज्या वाले वृत्तीय पथ का एक चक्कर लगाने में 1 सेकंड का समय लेता है तो वेग (v) = \(\frac{2 \pi \mathrm{r}}{\mathrm{t}}\)

→ चाल-प्रति इकाई समय में वस्तु के द्वारा तय की गई दूरी वस्तु की चाल कहलाती है।

→ वेग–प्रति इकाई समय में हुआ विस्थापन वस्तु का वेग कहलाता है।

→ दूरी-किसी वस्तु द्वारा तय किया गया वास्तविक पथ दूरी कहलाता है।

→ विस्थापन-किसी वस्तु की प्रारंभिक व अंतिम स्थिति के बीच की न्यूनतम दूरी को वस्तु का विस्थापन कहते हैं।

→ मूल बिंदु-किसी वस्तु की स्थिति को बताने के लिए हमें एक निर्देश बिंदु की आवश्यकता होती है, जिसे मूल बिंदु कहा जाता है।

→ एकसमान गति-यदि कोई वस्तु समान समयांतराल में समान दूरी तय करे, तो उसकी गति को एकसमान गति कहते हैं।

→ असमान गति-यदि कोई वस्तु समान समयांतराल में असमान दूरी तय करे, तो उसकी गति को असमान गति कहते हैं।

→ औसत चाल-किसी वस्तु द्वारा तय की गई कुल दूरी और कुल समय के अनुपात को औसत चाल कहते हैं।
अर्थात्

→ औसत वेग-यदि किसी वस्तु का वेग समान रूप से परिवर्तित होता हो तो प्रारंभिक वेग और अंतिम वेग के अंकगणितीय माध्य को औसत वेग कहते हैं। अर्थात्

→ त्वरण-किसी वस्तु का त्वरण प्रति इकाई समय में उसके वेग में होने वाला परिवर्तन है। अर्थात्

→ एकसमान त्वरण-यदि बराबर समयांतराल में किसी वस्तु के वेग में बराबर परिवर्तन हो, तो वस्तु एकसमान त्वरण से गतिशील कहलाती है।

→ असमान त्वरण-यदि किसी वस्तु का वेग असमान रूप से बदलता हो तो उसे असमान त्वरण कहते हैं।

→ एकसमान वृत्तीय गति-यदि एक वस्तु वृत्तीय रास्ते पर एकसमान चाल से चलती है तो उसकी गति को एकसमान वृत्तीय गति कहा जाता है।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 Number Systems Ex 1.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 1 Number Systems Exercise 1.1

Question 1.
Is zero a rational number? Can you write it in the form \(\frac {p}{q}\) where p and q are integers and q ≠ 0?
Solution :
Yes, zero is a rational number.
Yes, it can be written in the form \(\frac {p}{q}\) as follows:
\(\frac{0}{1}=\frac{0}{2}=\frac{0}{3}\) etc., denominator q can also be taken as negative integer.

Question 2.
Find six rational numbers between 3 and 4.
Solution:
To find six rational numbers between 3 and 4, we write rational numbers 3 and 4 with denominator 7(6 + 1) as follows:
3 = \(\frac{3 \times 7}{7}=\frac{21}{7}\)
4 = \(\frac{4 \times 7}{7}=\frac{28}{7}\)
Six rational numbers between 3 and 4 are
\(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}\) and \(\frac {27}{7 }\)
Hence, six rational numbers between 3 and 4 are
\(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}\) and \(\frac {27}{7 }\)

Question 3.
Find five rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\)
Solution:
To find 5 rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\), we multiply the numerator and denominator of \(\frac {3}{5}\) and \(\frac {4}{5}\) by 6(5 + 1).
\(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\)
\(\frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\)
Five rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\) are :
\(\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}\) and \(\frac {23}{30}\)
Hence, five rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\) are :
\(\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}\) and \(\frac {23}{30}\)

Question 4.
State whether the following statements are true or false. Give reasons for your answers:
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Solution :
(i) True, since the collection of the whole numbers contains all the natural numbers.
(ii) False, since negative integers are not whole numbers.
(iii) False, since \(\frac {2}{5}\) is a rational number but not a whole number.

Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Solution:
The frequency distribution table is given below :
Frequency distribution of blood groups

Most common blood group is O and rarest blood group is AB.

Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0 – 5 (5 is not included). What main features do you observe from this tabular representation?
Solution :
Minimum and maximum observations are 2 and 32. And class size is 5, first class interval is 0 – 5.
So, class intervals of same size are 0 – 5, 5 – 10, 10 – 15, 15 – 20, 20 – 25, 25 – 30 and 30 – 35. The frequency distribution table is given below
Frequency distribution of distance

We observe that:
(i) The residence of 36 engineers are within 5 km to 20 km.
(ii) The residence of 4 engineers are within 20 km to 35 km.

Question 3.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84 – 86, 86 – 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution :
Minimum and maixmum observations are 84.9 and 99.2 respectively.
∴ Class intervals of same size are 84 – 86, 86 – 88, 88 – 90, 90 – 92, 92 – 94, 96 – 98 and 98 – 100.
Frequency distribution table of relative humidity is given below :
(i) Frequency distribution of relative humidity :

(ii) The data appears to be taken in the rainy season as the relative humidity is high.
(iii) Range = 99.2 – 84.9 = 14.3.

Question 4.
The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:

(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 – 165, 165 – 170 etc.
(ii) What can you conclude about their heights from the table?
Solution :
Minimum and maximum observations are 150 and 173 respectively. So, class intervals of same size are
150 – 155, 155 – 160, 160 – 165, 165 – 170 and 170 – 175
The frequency distribution table of heights of 50 students is given below.
(i) Frequency distribution of heights.

(ii) One conclusion that we can draw from the above table is that more than 50% of students are shorter than 165 cm. Ans.

Question 5.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

(i) Make a ground frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08 and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
Solution :
Minimum and maximum observations are 0.01 and 0.22.
Class intervals of same size are 0.00 – 0.04, 0.04 – 0.08, 0.08 – 0.12, 0.12 – 0.16, 0.16 – 0.20 and 0.20 – 0.24.
Frequency distribution table of concentration of sulphur dioxide in the air is given below :
(i) FREQUENCY DISTRIBUTION OF CONCENTRATION OF SULPHUR DIOXIDE

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

Question 6.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows :

Prepare a frequency distribution table for the data given above.
Solution :
Frequency distribution table for the given data is given below :
Frequency distribution table of coins

Question 7.
The value of π up to 50 decimal places is given below :
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) Frequency distribution table of digits from 0 to 9 after the decimal point is given below:
Frequency distribution of digit from 0 to 9

(ii) The most frequently occurring digits are 3 and 9. The least occurring is 0.

Question 8.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

(i) Make a grouped frequency distribution table for this data, using class width 5 and one of the class intervals as 5 – 10.
(ii) How many children watched television for 15 or more hours a week?
Solution :
Minimum and maximum observations are 1 and 17 respectively.
Class width = 5 and one of the class interval is 5 – 10.
So, class intervals of the following data are 0 – 5, 5 – 10, 10 – 15 and 15 – 20.
(i) Frequency distribution table of hours of T.V. programme is given below :
Frequency distribution of hours of T.V. programme

(ii) Number of children watched television for 15 or more hours a week = 2.

Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 – 2.5.
Solution :
The minimum and maximum observations of the data are 2.2 and 4.6 respectively.
Class size = 0.5 and one of the class interval is 2 – 2.5.
So, class intervals of the following data are :
2 – 2.5, 2.5 – 3.0, 3.0 – 3.5, 3.5 – 4.0, 4.0 – 45 and 4.5 – 5.0.
The frequency distribution table of car batteries is given below :
Frequency distribution of car batteries

Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Ex 6.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 6 रेखाएँ और कोण Exercise 6.1

प्रश्न 1.
आकृति में, रेखाएँ AB और CD बिंदु O पर प्रतिच्छेद करती हैं। यदि ∠AOC + ∠BOE = 70° है और ∠BOD = 40° है, तो ∠BOE और प्रतिवर्ती ∠COE ज्ञात कीजिए।

हल :
क्योंकि आकृति अनुसार OA तथा OB विपरीत किरणें हैं। इसलिए AB एक रेखा है। अब किरण OC, AB पर स्थित है। [रैखिक युग्म]
⇒ ∠AOC + ∠COB = 180°
या ∠AOC + ∠COE + ∠BOE = 180°
या (∠AOC + ∠BOE) + ∠COE = 180°
∠AOC + ∠BOE = 70°
⇒ 70° + ∠COE = 180°
या ∠COE = 180° – 70° = 110°
इस प्रकार प्रतिवर्ती
∠COE = 360° – 110° = 250° उत्तर
हम जानते हैं कि OC तथा OD विपरीत किरणें हैं इसलिए
∠COE + ∠EOD = 180° [रैखिक युग्म]
या ∠COE + ∠BOE + ∠BOD = 180°
या 110° + ∠BOE + 40° = 180° [∵ ∠COE = 110° व ∠BOD = 40°]
या ∠BOE = 180° – 110° = 40°
या ∠BOE = 180° – 150° = 30° उत्तर

प्रश्न 2.
आकृति में, रेखाएँ XY और MN बिंदू O पर प्रतिच्छेद करती हैं। यदि ∠POY = 90° और a : b = 2 : 3 है, तो c ज्ञात कीजिए।
हल :

क्योंकि a : b = 2 : 3 और
a + b = ∠POX = ∠POY = 90°
अब अनुपाती योग = 2 + 3 = 5
⇒ a = \(\frac {2}{5}\) × 90° = 36°
तथा b = \(\frac {3}{5}\) × 90° = 54°
आकृति अनुसार OM और ON विपरीत किरणें हैं। इसलिए MN एक रेखा है।
क्योंकि किरण OX, MN पर स्थित है, इसलिए
∠MOX + ∠XON = 180° [रैखिक युग्म]
या b + c = 180° या c + 54° = 180° [∵ ∠b = 54°]
या c = 180° – 54° = 126°
या c = 126° उत्तर

प्रश्न 3.
आकृति में, यदि ∠PQR = ∠PRQ है, तो सिद्ध कीजिए कि ∠PQS = ∠PRT है।

हल :
क्योंकि QS और QR विपरीत किरणें हैं, इसलिए SR एक रेखा है।
∵ QP, SR पर स्थित है।
∴ ∠PQS + ∠PQR = 180° [रैखिक युग्म] … (i)
अब RQ और RT विपरीत किरणें हैं। इसलिए QT एक रेखा है।
इसी प्रकार PR, QT पर स्थित है।
∴ ∠PRQ + ∠PRT = 180° [रैखिक युग्म] … (ii)
समीकरण (i) व (ii) की तुलना से
∠PQS + ∠PQR = ∠PRQ + ∠PRT [∵ प्रत्येक युग्म = 180°] …. (iii)
या ∠PQR = ∠PRO [दिया है]… (iv)
समीकरण (iv) को (iii) में से घटाने पर प्राप्त होगा
∠PQS = ∠PRT [इति सिद्धम]

प्रश्न 4.
आकृति में, यदि x + y = w + z है, तो सिद्ध कीजिए कि AOB एक रेखा है।
हल :

क्योंकि एक बिंदु के चारों ओर के कोणों का योग 360° होता है।
∴ (∠BOC + ∠COA) + (∠BOD + ∠AOD) = 360°
⇒ (x + y) + (w + z) = 360°
परंतु x + y = w + z (दिया है)
अतः (x + y) + (x + y) = 360°
या 2 (x + y) = 360°
या x + y = \(\frac {360°}{2}\) = 180°
अतः ∠BOC और ∠COA रैखिक युग्म बनाते हैं।
अर्थात OA और OB दो विपरीत किरणें हैं।
इसलिए AOB एक रेखा है। [इति सिद्धम]

प्रश्न 5.
आकृति में, POQ एक रेखा है। किरण OR रेखा PQ पर लंब है। किरणों OP और OR के बीच में OS एक अन्य किरण है। सिद्ध कीजिए
∠ROS = \(\frac {1}{2}\) (∠QOS – ∠POS)

हल :
क्योंकि \(\overline{O R}\) रेखा PQ पर लंब है
इसलिए
∠POR = ∠QOR
या ∠POS + ∠ROS = ∠QOS – ∠ROS
या ∠ROS + ∠ROS = ∠QOS – ∠POS
या 2∠ROS = (∠QOS – ∠POS)
या ∠ROS = \(\frac {1}{2}\)(∠QOS – ∠POS) [इति सिद्धम]

प्रश्न 6.
यह दिया है कि ∠XYZ = 64° है और XY को बिंद्र P तक बढ़ाया गया है। दी हुई सूचना से एक आकृति खींचिए। यदि किरण YQ, ∠ZYP को समद्विभाजित करती है, तो ∠XYQ और प्रतिवर्ती ∠QYP के मान ज्ञात कीजिए।
हल :

क्योंकि रेखाखंड XY को बिंदु P तक बढ़ाया गया है।
इसलिए XP एक रेखा है।
क्योंकि YZ, रेखा XP पर स्थित है।
∴ ∠XYZ + ∠ZYP = 180° [रैखिक युग्म]
⇒ 64° + ∠ZYP = 180° [∵ ∠XYZ = 64°]
⇒ ∠ZYP = 180° – 64° = 116°
क्योंकि किरण YQ, ∠ZYP को समद्विभाजित करती है।
इसलिए
∠QYP = ∠ZYQ
= \(\frac {116°}{2}\) = 58°
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58° = 122°
तथा
प्रतिवर्ती ∠QYP = 360° – ∠QYP = 360° – 58° = 302°
अतः ∠XYQ = 122° व प्रतिवर्ती ∠QYP = 302° उत्तर

Haryana State Board HBSE 9th Class Science Solutions Chapter 4 परमाणु की संरचना Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 4 परमाणु की संरचना

HBSE 9th Class Science परमाणु की संरचना Intext Questions and Answers
(पृष्ठ संख्या-53)
प्रश्न 1.
केनाल किरणें क्या हैं?
उत्तर:
ई० गोल्डस्टीन ने 1886 ई० में धनावेशित विकिरण की खोज की, जिन्हें केनाल किरणें कहा जाता है।

प्रश्न 2.
यदि किसी परमाणु में एक इलेक्ट्रॉन और एक प्रोटॉन है, तो इसमें कोई आवेश होगा या नहीं?
उत्तर:
यदि किसी परमाणु में एक इलेक्ट्रॉन व एक प्रोटॉन हो तो उस पर कोई आवेश नहीं होगा क्योंकि प्रोटॉन और इलेक्ट्रॉन परस्पर आवेशों को संतुलित करते हैं।

(पृष्ठ संख्या -56)

प्रश्न 1.
परमाणु उदासीन है, इस तथ्य को टॉमसन के मॉडल के आधार पर स्पष्ट कीजिए।
उत्तर:
टॉमसन के मॉडल अनुसार परमाणु धन आवेशित गोले का बना होता है तथा इलेक्ट्रॉन उसमें धंसे होते हैं। इस प्रकार ऋणात्मक और धनात्मक आवेश परिमाण समान होने के कारण परमाणु वैद्युतीय रूप से उदासीन होता है।

प्रश्न 2.
रदरफोर्ड के परमाणु मॉडल के अनुसार, परमाणु के नाभिक में कौन-सा अवपरमाणुक कण विद्यमान है?
उत्तर:
रदरफोर्ड के परमाणु मॉडल के अनुसार, परमाणु के नाभिक में प्रोटॉन अवपरमाणुक कण विद्यमान है क्योंकि वह अल्फा कण को विक्षेपित करता है।

प्रश्न 3.
तीन कक्षाओं वाले बोर के परमाणु मॉडल का चित्र बनाइए।
उत्तर

प्रश्न 4.
क्या अल्फा कणों का प्रकीर्णन प्रयोग सोने के अतिरिक्त दूसरी धातु की पन्नी से संभव होगा?
उत्तर:
हाँ, अल्फा कणों का प्रकीर्णन प्रयोग सोने के अतिरिक्त किसी भी दूसरी धातु की पन्नी से संभव होगा।

(पृष्ठ संख्या – 56)

प्रश्न 1.
परमाणु के तीन अवपरमाणुक कणों के नाम लिखें।
उत्तर:
परमाणु के तीन अवपरमाणुक कण इलेक्ट्रॉन, प्रोटॉन व न्यूटॉन हैं।

प्रश्न 2.
हीलियम परमाणु का परमाणु द्रव्यमान 4u है और उसके नाभिक में दो प्रोटॉन होते हैं। इसमें कितने न्यूट्रॉन होंगे?
उत्तर:
हीलियम परमाणु का परमाणु द्रव्यमान = 4u
हीलियम के परमाणु नाभिक में उपस्थित प्रोटॉन = 2
हीलियम के परमाणु नाभिक में उपस्थित न्यूट्रॉन = परमाणु द्रव्यमान – प्रोटॉन
= 4 – 2 = 2

(पृष्ठ संख्या-57)

प्रश्न 1.
कार्बन और सोडियम के परमाणुओं के लिए इलेक्ट्रॉन-वितरण लिखिए।
उत्तर:
1. कार्बन
द्रव्यमान संख्या = 12
परमाणु संख्या = 6
इसलिए प्रोटॉनों की संख्या = 6
इलेक्ट्रॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 12 – 6 = 6
इलेक्ट्रॉन वितरण = K = 2, L=4

2. सोडियम
द्रव्यमान संख्या = 23
परमाणु संख्या = 11
इसलिए प्रोटॉनों की संख्या = 11
इलेक्ट्रॉनों की संख्या = 11
न्यूट्रॉनों की संख्या = 23 – 11 = 12
इलेक्ट्रॉन वितरण = K = 2
L = 8
M = 1

प्रश्न 2.
अगर किसी परमाणु का K और L कोश भरा है, तो उस परमाणु में इलेक्ट्रॉनों की संख्या क्या होगी? उत्तर
कोश K = 2 इलेक्ट्रॉन
कोश L = 8 इलेक्ट्रॉन
अतः परमाणु में कुल इलेक्ट्रॉन = 2 + 8 = 10 इलेक्ट्रॉन .

(पृष्ठ संख्या-58)

प्रश्न 1.
क्लोरीन, सल्फर और मैग्नीशियम की परमाणु संख्या से आप इनकी संयोजकता कैसे प्राप्त करेंगे?
उत्तर:
(1) क्लोरीन की परमाणु संख्या 17 है इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 7 इलेक्ट्रॉन
इस प्रकार क्लोरीन को अपना अष्टक पूरा करने के लिए (8-7) = 1 इलेक्ट्रॉन की आवश्यकता है। अतः क्लोरीन की संयोजकता 1 है।

(2) सल्फर की परमाणु संख्या 16 है इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 6 इलेक्ट्रॉन
इस प्रकार सल्फर को अपना अष्टक पूरा करने के लिए (8 – 6) = 2 इलेक्ट्रॉनों की आवश्यकता है। अतः सल्फर की संयोजकता 2 है।

(3) मैग्नीशियम की परमाणु संख्या 12 है। इसलिए इसका इलेक्ट्रॉन वितरण निम्न होगा
K = 2 इलेक्ट्रॉन
L = 8 इलेक्ट्रॉन
M = 2 इलेक्ट्रॉन
इस प्रकार मैग्नीशियम को अपना बाह्य अष्टक पूरा करने के लिए 2 इलेक्ट्रॉनों को त्यागना आसान है। अतः मैग्नीशियम की संयोजकता 2 है।

(पृष्ठ संख्या – 59)

प्रश्न 1.
यदि किसी परमाणु में इलेक्ट्रॉनों की संख्या 8 है और प्रोटॉनों की संख्या भी 8 है तब,
(a) परमाणु की परमाणुक संख्या क्या है?
(b) परमाणु का क्या आवेश है?
उत्तर:
परमाणु में इलेक्ट्रॉनों की संख्या = 8
परमाणु में प्रोटॉनों की संख्या = 8
परमाणु में प्रोटॉनों की संख्या = इलेक्ट्रॉनों की संख्या
= 8
(a) परमाणु की परमाणुक संख्या = परमाणु में प्रोटॉनों तथा इलेक्ट्रॉन की संख्या
= 8 + 8 = 16

(b) इलेक्ट्रॉन वितरण = K = 2, L = 6
अतः परमाणु के बाह्यतम कोश को पूरा करने के लिए 2 इलेक्ट्रॉनों की आवश्यकता है। इसलिए आवेश -2 है।

प्रश्न 2.
पाठ्यपुस्तक की सारणी 4.1 की सहायता से ऑक्सीजन और सल्फर-परमाणु की द्रव्यमान संख्या ज्ञात कीजिए।
उत्तर:
दी गई सारणी अनुसार
(1) ऑक्सीजन में परमाणु संख्या = 8
ऑक्सीजन में प्रोटॉनों की संख्या = 8
ऑक्सीजन में न्यूट्रॉनों की संख्या = 8
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 8 + 8 = 16

(2) सल्फर में परमाणु संख्या = 16
सल्फर में प्रोटॉनों की संख्या = 16
सल्फर में न्यूट्रॉनों की संख्या = 16
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 16 + 16 = 32

(पृष्ठ संख्या – 60)

प्रश्न 1.
चिह्न H, D और T के लिए प्रत्येक में पाए जाने वाले तीन अवपरमाणुक कणों को सारणीबद्ध कीजिए।
उत्तर:
(1) चिह्न H प्रोटियम का प्रतीक है अर्थात ,\({ }_1 \mathrm{H}^1\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 1
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 1 – 1 = 0

(2) चिह्न D ड्यूटीरियम का प्रतीक है अर्थात \({ }_1 \mathrm{H}^2\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 2
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 2 – 1 = 1

(3) चिह्न T ट्राइटियम का प्रतीक है अर्थात \({ }_1 \mathrm{H}^3\)
परमाणु संख्या = 1
द्रव्यमान संख्या = 3
इलेक्ट्रॉनों की संख्या = 1
प्रोट्रॉनों की संख्या = 1
न्यूट्रॉनों की संख्या = 3 – 1 = 2

प्रश्न 2.
समस्थानिक और समभारिक के किसी एक युग्म का इलेक्ट्रॉनिक विन्यास लिखिए।
उत्तर:
(i) समस्थानिक युग्म क्लोरीन \({ }_{17} \mathrm{Cl}^{35}\) व \({ }_{17} \mathrm{Cl}^{37}\) का इलेक्ट्रॉनिक विन्यास निम्नलिखित होगा

(ii) समभारिक युग्म कैल्शियम तथा आर्गन का इलेक्ट्रॉनिक विन्यास निम्न होगा
(1) कैल्शियम \({ }_{20} \mathrm{Cl}^{40}\)
e = 20
P = 20
N = 40 – 20 – 20

(2) आर्गन ₁₈Ar⁴⁰
e = 18
P = 18
N = 40 – 18 = 22

HBSE 9th Class Science परमाणु की संरचना Textbook Questions and Answers

प्रश्न 1.
इलेक्ट्रॉन, प्रोटॉन और न्यूट्रॉन के गुणों की तुलना कीजिए।
उत्तर:

प्रश्न 2.
जे.जे. टॉमसन के परमाणु मॉडल की क्या सीमाएँ हैं?
उत्तर:
जे.जे. टॉमसन के परमाणु मॉडल ने परमाणु के उदासीन होने की व्याख्या की जिसके अनुसार परमाणु धन आवेशित गोले का बना होता है और इलेक्ट्रॉन उसमें धंसे होते हैं, परंतु यह मॉडल न्यूट्रॉनों के बारे में तथा इलेक्ट्रॉनों की स्थिति के बारे में कुछ भी नहीं बताता जो दूसरे वैज्ञानिकों द्वारा बाद में स्पष्ट किया गया।

प्रश्न 3.
रदरफोर्ड के परमाणु मॉडल की क्या सीमाएँ हैं?
उत्तर:
अरनेस्ट रदरफोर्ड के परमाणु मॉडल की मुख्य सीमा यह थी कि यह परमाणु की स्थिरता का वर्णन नहीं करता। उसके मॉडल अनुसार, इलेक्ट्रॉन नाभिक के चारों ओर वृत्तीय पथ में चक्कर काटते हैं जिसका परिणाम उनकी त्वरित गति होता है। अतः यह ऊर्जा को विकरित करके वृत्तीय पथ के नाभिक आकार को छोटा कर देगा। परिणामस्वरूप इलेक्ट्रॉन नाभिक में गिर जाएँगे और परमाणु को अस्थिर कर देंगे, परंतु वास्तव में परमाणु अस्थिर नहीं है। चित्र 4.8 : रदरफोर्ड के परमाणु मॉडल की कमी

प्रश्न 4.
बोर के परमाणु मॉडल की व्याख्या कीजिए।
उत्तर:
नील्स बोर के परमाणु मॉडल के अनुसार
(1) इलेक्ट्रॉन केवल कुछ निश्चित कक्षाओं में ही चक्कर लगा सकते हैं, जिन्हें इलेक्ट्रॉन की विविक्त कक्षा कहते हैं।
(2) जब इलेक्ट्रॉन इस विविक्त कक्षा में चक्कर लगाते हैं, तो उनकी ऊर्जा का विकिरण नहीं होता। इन कक्षाओं (या कोशों) को ऊर्जा-स्तर कहते हैं। चित्र में एक परमाणु के ऊर्जा स्तर को दिखाया गया है।

प्रश्न 5.
इस अध्याय में दिए गए सभी परमाणु मॉडलों की तुलना कीजिए।
उत्तर:
इस अध्याय में दिए गए विभिन्न परमाणु मॉडलों की तुलना निम्नलिखित प्रकार से हैं-
1. टॉमसन के परमाणु मॉडल अनुसार
(1) परमाणु धन आवेशित गोले का बना होता है और इलेक्ट्रॉन उसमें धंसे होते हैं।
(2) ऋणात्मक और धनात्मक आवेश परिणाम में समान होते हैं। इसलिए परमाणु वैद्युतीय रूप से उदासीन होते हैं।

2. रदरफोर्ड के परमाणु मॉडल अनुसार
(1) परमाणु का केंद्र धनावेशित होता है जिसे नाभिक कहा जाता है। एक परमाणु का लगभग संपूर्ण द्रव्यमान नाभिक में होता है।
(2) इलेक्ट्रॉन नाभिक के चारों ओर निश्चित कक्षाओं में चक्कर लगाते हैं।
(3) नाभिक का आकार परमाणु के आकार की तुलना में काफी कम होता है।

3. बोर मॉडल के अनुसार
(1) इलेक्ट्रॉन केवल कुछ निश्चित कक्षाओं में ही चक्कर लगा सकते हैं, जिन्हें इलेक्ट्रॉन की विविक्त कक्षा कहते हैं।
(2) जब इलेक्ट्रॉन इस विविक्त कक्षा में चक्कर लगाते हैं, तो उनकी ऊर्जा का विकिरण नहीं होता है। इन कक्षाओं को ऊर्जा स्तर कहते हैं।

प्रश्न 6.
पहले अठारह तत्त्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम को लिखिए।
उत्तर:
पहले अठारह तत्त्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के लिए बोर और बरी द्वारा निम्न नियम प्रस्तुत किए गए
(1) किसी कक्षा में उपस्थित अधिकतम इलेक्ट्रॉनों की संख्या को सूत्र 2n² से दर्शाया जाता है, जहाँ ‘n’ कक्ष की संख्या या ऊर्जा स्तर है। इसलिए इलेक्ट्रॉनों की अधिकतम संख्या पहले कक्ष या K कोश में (2 x 1²) = 2 होगी, दूसरे कक्ष या L सेल में (2 x 2²) = 8 होगी, तीसरे कक्ष या M सेल में होगी = (2 x 3²)= 18, चौथे कक्ष या N सेल में होगी = (2 x 4²)= 32.

(2) सबसे बाहरी कोश में इलेक्ट्रॉनों की अधिकतम संख्या 8 हो सकती है।

(3) किसी परमाणु के अगले कोश में इलेक्ट्रॉन तब तक स्थान नहीं लेते हैं जब तक कि उससे पहले वाले भीतरी कोश पूर्ण रूप से भर नहीं जाता। इससे स्पष्ट होता है कि कक्षाएँ क्रमानुसार भरती हैं।

प्रश्न 7.
सिलिकॉन और ऑक्सीजन का उदाहरण देते हुए संयोजकता की परिभाषा दीजिए।
उत्तर:
परमाणु के बाह्यतम कक्ष में इलेक्ट्रॉनों के अष्टक बनाने के लिए जितनी संख्या में इलेक्ट्रॉनों की साझेदारी या स्थानांतरण होता है, वह उस तत्त्व की संयोजकता कहलाती है। जैसे

(1) सिलिकॉन \(({ }_{14} \mathrm{Si}^{28})\) में इलेक्ट्रॉनों का वितरण निम्न होगा
K = 2
L = 8
M = 4
अतः यह अपना अष्टक पूरा करने के लिए 4 इलेक्ट्रॉन स्थानांतरित कर सकता है। इसलिए इसकी संयोजकता 4 है।

(2) ऑक्सीजन (\(({ }_{8} \mathrm{Si}^{16})\)) में इलेक्ट्रॉनों का वितरण निम्न होगा
K = 2
L = 6
अतः इसे अपना अष्टक पूरा करने के लिए (8 – 6) = 2 इलेक्ट्रॉनों की साझेदारी करनी होगी। इसलिए इसकी संयोजकता 2 है।

प्रश्न 8.
उदाहरण के साथ व्याख्या कीजिए-परमाणु संख्या, द्रव्यमान संख्या, समस्थानिक और समभारिक समस्थानिकों के कोई दो उपयोग लिखिए।
उत्तर:
परमाणु संख्या-परमाणु के नाभिक में उपस्थित कुल प्रोटॉनों की संख्या उसकी परमाणु संख्या (Z) कहलाती है; जैसे ऑक्सीजन की परमाणु संख्या 8 व कार्बन की परमाणु संख्या 6 है।

द्रव्यमान संख्या – परमाणु के नाभिक में उपस्थित प्रोटॉनों और न्यूट्रॉनों की कुल संख्या के योग को द्रव्यमान संख्या (A) कहते हैं। जैसे ऑक्सीजन की द्रव्यमान संख्या 16u तथा कार्बन की द्रव्यमान संख्या 12u है।

समस्थानिक – एक ही तत्त्व के ऐसे परमाणु जिनकी परमाणु संख्या समान हो, परंतु द्रव्यमान संख्या भिन्न हो तत्त्व के समस्थानिक कहलाते हैं। जैसे हाइड्रोजन के तीन समस्थानिक प्रोटियम \(({ }_{1} \mathrm{H}^{1})\), ड्यूटीरियम \(({ }_{1} \mathrm{H}^{2})\) व ट्राइटियम \(({ }_{1} \mathrm{H}^{3})\) तथा कार्बन के दो समस्थानिक \(({ }_{6} \mathrm{C}^{12})\) व \(({ }_{6} \mathrm{C}^{14})\) हैं।

समभारिक – समभारिक वे परमाणु होते हैं जिनकी द्रव्यमान संख्या समान लेकिन परमाणु संख्या भिन्न-भिन्न होती है; जैसे कैल्शियम \(({ }_{20} \mathrm{Ca}^{40})\) तथा आर्गन \(({ }_{18} \mathrm{Ar}^{40})\) समभारिक हैं।

समस्थानिकों के उपयोग
(1) यूरेनियम के एक समस्थानिक (U-235) का उपयोग परमाणु भट्टी (reactor) में ईंधन के रूप में होता है।
(2) कैंसर के उपचार में कोबाल्ट के समस्थानिक का उपयोग होता है।
(3) पेंघा रोग के इलाज में आयोडीन के समस्थानिक का उपयोग होता है।

प्रश्न 9.
Na⁺ के पूरी तरह से भरे हुए Ka L कोश होते हैं व्याख्या कीजिए। उत्तर-सोडियम परमाणु की परमाणु संख्या 11 होती है। इसलिए सोडियम परमाणु में इलेक्ट्रॉनों की संख्या = 11 अतः सोडियम परमाणु का इलेक्ट्रॉनिक वितरण होगा
K = 2
L = 8
M = 1
यह सोडियम परमाणु स्थाई संरचना प्राप्त करने के लिए एक इलेक्ट्रॉन त्याग कर Na (सोडियम आयन) बन जाएगा जिस कारण इसका इलेक्ट्रॉनिक वितरण K = 2 तथा L = 8 होगा अर्थात K और L कोश पूर्ण भरे होंगे।

प्रश्न 10.
अगर ब्रोमीन परमाणु दो समस्थानिकों \({ }_{35}^{79} \mathrm{Br}\)(49.7%) तथा \({ }_{35}^{81} \mathrm{Br}\) (50.3%)] के रूप में है, तो ब्रोमीन परमाणु के औसत परमाणु द्रव्यमान की गणना कीजिए।
उत्तर:
ब्रोमीन परमाणु दो समस्थानिकों \({ }_{35} \mathrm{Br}^{79}\) और \({ }_{35} \mathrm{Br}^{81}\) के रूपों में पाया जाता है जिनका द्रव्यमान क्रमशः 49.7% व 50.3% है।
इसलिए ब्रोमीन का औसत परमाणु द्रव्यमान होगा
\(\left(79 \times \frac{49.7}{100}+81 \times \frac{50.3}{100}\right)\)
\(\left(\frac{79 \times 497}{1000}+\frac{81 \times 503}{1000}\right)\)
\(\left(\frac{39263}{1000}+\frac{40743}{1000}\right)\) = 39.263 + 40.743 = 80.006 = 80u

प्रश्न 11.
एक तत्त्व X का परमाणु द्रव्यमान 16.2u है तो इसके किसी एक नमूने में समस्थानिक \({ }_{8}^{16} \mathrm{X}\) और \({ }_{8}^{18} \mathrm{X}\) का प्रतिशत क्या होगा?
हल
माना समस्थानिक \({ }_{8} \mathrm{X}^{16}\) का प्रतिशत = x
तो \({ }_{8} \mathrm{X}^{18}\) का प्रतिशत = 100 – x
तत्त्व X का औसत परमाणु द्रव्यमान = 16.2u
\(\frac{16 \times x}{100}+\frac{18 \times(100-x)}{100}\) = 16.2
दोनों ओर 100 से गुणा करने पर
16x + 1800 – 18x = 1620
या -2x = 1620 – 1800
या -2x = -180
या x = \(\frac{-180}{-2}\) = 90
अतः नमूने (सैंपल) में \({ }_8 \mathrm{X}^{16}\) का प्रतिशत = 90%
तथा \({ }_8 \mathrm{X}^{18}\) का प्रतिशत = 100 — 90 = 10%

प्रश्न 12.
यदि तत्त्व का Z= 3 हो तो तत्त्व की संयोजकता क्या होगी? तत्त्व का नाम भी लिखिए।
हल
तत्त्व की परमाणु संख्या (Z) = 3
तत्त्व में इलेक्ट्रॉनों की संख्या = 3
इलेक्ट्रॉन वितरण
K = 2
L = 1
अतः तत्त्व के बाह्यतम कोश में 1 इलेक्ट्रॉन है। यह स्थाई विन्यास प्राप्त करने के लिए एक इलेक्ट्रॉन का त्याग कर सकता है। इसलिए तत्त्व की संयोजकता 1 है। अतः तत्त्व लीथियम (Li) है।

प्रश्न 13.
दो परमाणु स्पीशीज़ के केंद्रकों का संघटन नीचे दिया गया है

X और Y की द्रव्यमान संख्या ज्ञात कीजिए। इन दोनों स्पीशीज़ में क्या संबंध है?
हल–परमाणु x के लिए
प्रोटॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 6
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 6+6
= 12
परमाणु Y के लिए
प्रोटॉनों की संख्या = 6
न्यूट्रॉनों की संख्या = 8
द्रव्यमान संख्या = प्रोटॉनों की संख्या + न्यूट्रॉनों की संख्या
= 6+ 8
= 14
ये दोनों स्पीशीज \({ }_6 X^{12}\) और \({ }_6 Y^{14}\) समस्थानिक हैं।

प्रश्न 14.
निम्नलिखित वक्तव्यों में गलत के लिए F और सही के लिए T लिखें।
(a) जे.जे. टॉमसन ने यह प्रस्तावित किया था कि परमाणु के केंद्रक में केवल न्यूक्लीयॉन्स होते हैं।
(b) एक इलेक्ट्रॉन और प्रोटॉन मिलकर न्यूट्रॉन का निर्माण करते हैं इसलिए यह अनावेशित होता है।
(c) इलेक्ट्रॉन का द्रव्यमान प्रोटॉन से लगभग \(\frac{1}{2000}\) गुना होता है।
(d) आयोडीन के समस्थानिक का इस्तेमाल टिंक्चर आयोडीन बनाने में होता है। इसका उपयोग दवा के रूप में होता है।
उत्तर:
(a) F,
(b) F
(c) T,
(d) F

नोट-प्रश्न संख्या 15, 16 और 17 में सही के सामने (√ ) का चिह्न और गलत के सामने (X) का चिह्न लगाइए।

प्रश्न 15.
रदरफोर्ड का अल्फा कण प्रकीर्णन प्रयोग किसकी खोज के लिए उत्तरदायी था?
(a) परमाणु केंद्रक
(b) इलेक्ट्रॉन
(c) प्रोटॉन
(d) न्यूट्रॉन
उत्तर:
(a) परमाणु केंद्रक (√),
(b) इलेक्ट्रॉन (x),
(c) प्रोटॉन (x),
(d) न्यूट्रॉन (x)

प्रश्न 16.
एक तत्त्व के समस्थानिक में होते हैं
(a) समान भौतिक गुण
(b) भिन्न रासायनिक गुण
(c) न्यूट्रॉनों की अलग-अलग संख्या
(d) भिन्न परमाणु संख्या
उत्तर:
(a) समान भौतिक गुण (x),
(b) भिन्न रासायनिक गुण (x),
(c) न्यूट्रॉनों की अलग-अलग संख्या (√),
(d) भिन्न परमाणु संख्या (x)।

प्रश्न 17.
Cl^– आयन में संयोजक इलेक्ट्रॉनों की संख्या है
(a) 16
(b) 8
(c) 17
(d) 18
उत्तर:
(a) 16 (x)।
(b) 8 (√)।
(c) 17 (x)।
(d) 18 (x)।

प्रश्न 18.
सोडियम का सही इलेक्ट्रॉनिक विन्यास निम्न में कौन-सा है?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
उत्तर:
सोडियम का सही इलेक्ट्रॉनिक विन्यास
(d) 2, 8, 1 है।

प्रश्न 19.
निम्नलिखित सारणी को पूरा कीजिए

उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 – 44 (in years) worldwide, found the following figures (in %):

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death world-wide ?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution :
(i) We draw the bar graph of this data in the following steps :
(a) We draw a horizontal and vertical line.
(b) We represent the causes of death on horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one head be represented by one unit.
(c) We represent the female fatality rate on the vertical axis. Let 1 unit = 4%.
(d) To represent first head i.e., reproductive health conditions, we draw a rectangular bar with width 1 unit and height 7.95 units.
(e) Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) Reproductive health conditions is the major cause of women’s ill health and death world wide.
(iii) Two factors which play a major role is :
(a) There are unqualified doctors in a maternity hospitals, where the pregnant women did not get better treatment.
(b) Unwanted female child.

Question 2.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below :

(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution :
(i) We draw the bar graph of this data in the following steps :
(a) We draw a horizontal and vertical line.
(b) We represent section of society on horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one head be represented by one unit.
(c) We represent the number of girls (to nearest ten) per thousand boys on the vertical axis. Let 1 unit = 10 girls.
(d) To represent the first head i.e., scheduled caste (8.c.), we draw a rectangular bar with width 1 unit and height 4 units.
(e) Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.

(ii) From the graph, we observe that the number of girls to the nearest ten per thousand boys are maximum in scheduled tribe but minimum in urban.

Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections :

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i) We draw the bar graph of the given data in the following steps:
(a) We draw a horizontal and vertical line.
(b) We represent political party on the horizontal axis, choosing any scale. We take equal widths for all bars and maintain equal gaps in between. Let one political party represented by 1 unit.

(c) We represent the number of seats won on the vertical axis. Let 1 unit = 10 seats.
(d) To represent the political party A, we draw a rectangular bar with width 1 unit and height 7.5 units.
(e) Similarly, other political parties we represented leaving a gap of 1 unit in between two consecutive bars.
(ii) Party A won the maximum number of seats.

Question 4.
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table :

(i) Draw a histogram to represent the given data.
[Hint : First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long ? Why?
Solution :
(i) The given frequency distribution is in inclusive form. So, first we convert it into exclusive form,
The difference between the lower limit of a class and upper limit of preceding class (h)= 1.
To convert the given frequency distribution into exclusive frequency distribution, we subtract \(\frac{h}{2}=\frac{1}{2}\) =0.5 from each of lower limit and add 0-5 to each upper limit. The exclusive frequency distribution table obtained as follows:

We represent the length (in mm) along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale.
We construct rectangles with class intervals as bases and the corresponding frequencies as heights.
Thus, we obtain a histogram as shown below :

(ii) Frequency polygon is the another method of representing this data graphically.
(iii) No, it is not correct to conclude that maximum number of leaves are 153 mm long because number of leaves having length 153 mm or less than 153 mm = 3 + 5 + 9 + 12 = 29 and number of leaves having length more than 153 mm = 5 + 4 + 2 = 11. But there might be no leave of length 153 mm.

Question 5.
The following table gives the life times of 400 neon lamps :

(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 hours?
Solution:
(i) The given frequency distribution is in the exclusive form. So we represent the lifetime (in hours) along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights. Thus, we obtain a histogram as shown below :

(ii) Number of lamps having lifetime more than 700 hrs = 74 + 62 + 48 = 184.

Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution :
To represent the frequency polygon, we need the class marks of given classes. For class interval (0 – 10), the upper limit is 10 and lower limit is 0.
So, the class mark = \(\frac{10+0}{2}\) = 5
Continuing in the same manner, we find the class marks of the other classes as well. So, we obtained the following table :

We represent class marks along the x-axis on a suitable scale and the corresponding frequency along the y-axis on a suitable scale. To obtain the frequency polygon of section A, we plot the points (5, 3), (15, 9), (25, 17),(35, 12), (45, 9) and join thuse points by line segments.
To obtain the frequency polygon of section B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10) and (45, 1) on the same scale and join these points by dotted line segments.
The two frequency polygons are shown given figure.

Question 7.
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

Represent the data of both the teams on the same graph by frequency polygons.
(Hint First make the class intervals continuous.)
Solution:
Here the class intervals are discontinuous. So, we convert it into continuous class intervals. For this, we subtract \(\frac{h}{2}=\frac{1}{2}\) = 0.5 from each of lower limit and add 0.5 to each upper limit. So, continuous class intervals obtained as follows:

To represent the frequency polygon, we need the class marks of the given classes. For class interval (0-5 -6.5), the upper limit is 65 and lower limit is 0.5.
So, the class mark = \(\frac{6.5+0.5}{2}\) = 3.5
Continuing in the same manner, we find the class marks of the other classes as well. So, we obtained the following table :

We represent the class marks along x-axis on a suitable scale and the corresponding frequencies along y-axis on a suitable scale.
To obtain the frequency polygon of team A, we plot the points (3.5, 2), (9.5, 1), (15.5, 8), (21.5, 9), (27.5, 4), (33.5, 5), (39.5, 6), (45.5, 10), (51.5, 6) and (57.5, 2) and join these points by line segments.
To obtain the frequency polygon of team B, we plot the points (3.5, 5), (9.5, 6), (15.5, 2), (21.5, 10), (27.5, 5), (33.5, 6), (39.5, 3), (45.5, 4), (51.5, 8) and (57.5, 10) on the same scale and join these points by dotted line segments.
The two frequency polygons are shown given figure.

Question 8.
A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above.
Solution :
Here the class intervals are not same size. Here, minimum class size is 1, then lengths of the rectangles are modified to be proportionate to the class size 1. Lengths of rectangles are computed as follows in the given table :

We represent the ages of children along x-axis on a suitable scale and number of children along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights.

Question 9.
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Here the class iņtervals are not same size and minimum class size is 2, then lengths of the rectangles are modified to be proportionate to the class size 2.
Lengths of the rectangles are computed as follow in the given table :

We represent the number of letters along x-axis on a suitable scale and number of surnames along y-axis on a suitable scale. We construct rectangles with class intervals as bases and corresponding frequencies as heights.

(ii) The maximum surnames lies in the class interval 6 – 8.

Haryana State Board HBSE 9th Class Science Important Questions Chapter 6 Tissues Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 6 Tissues

Very Short-Answer Type Questions

Question 1.
Are there same type of tissues in plants and animals?
Answer:
No, plants and animals have different types of tissues.

Question 2.
How are newly cells formed?
Answer:
Through cell division process.

Question 3.
What is called as unicellular organisms?
Answer:
All those organisms in which all the biological activities are fulfilled through single cell are called as unicellular organisms.

Question 4.
Which organisms undergo labour division?
Answer:
Multicellular organisms.

Question 5.
How many types of tissues are there in plants?
Answer:
Two types:
(i) Meristematic tissues
(ii) Permanent tissues.

Question 6.
Where are Meristematic cells found?
Answer:
Dividing cells (meristematic tissues) found at the tip of the shoot and at the tip of the root.

Question 7.
What is the function of meristematic tissue?
Answer:
Building up of new cells.

Question 8.
Do the permanent tissues have the ability of divide?
Answer:
No.

Question 9.
How many types of permanent plant tissues are there?
Answer:
Two types:
(i) Simple tissues
(ii) Complex tissues.

Question 10.
What are Parenchymatous tissues ?
Answer:
Parenchymatous tissues can be living and oval, rounded elongated or polygonal in shape. They have thin cell wall and the cells are isodiametric.

Question 11.
What are collenchymatous tissues ?
Answer:
Due to deposition of cellulose and Pectin at the angles of these tissues they get thinken in size. These tissues do not have intercellular spaces.

Question 12.
What are sclerenchymatous tissues ?
Answer:
These are dead cells. These cells do not have any intercellular space in them. They provide strength to the plant parts.

Question 13.
What are simple permanent tissues ?
Answer:
These tissues are formed of parenchymatous cells. Their origin, structural formation and function are same.

Question 14.
Which part of the plant consists of meristematic tissues ?
Answer:
Green part of the plant.

Question 15.
What are protective tissues made up of?
Answer:
By sclerenchymatous tissues.

Question 16.
Which parts of the plant consists of protective tissues.
Answer:
Leaves, flowers, stem and roots.

Question 17.
What is the other name of sclereids ?
Answer:
Brachysclereids or stone cells.

Question 18.
What is called as complex tissue ?
Answer:
Cells of different shape and size together form a group which works out a special function, is called as complex tissue.

Question 19.
How many types of complex tissues are there ?
Answer:
Two type:
(i) Phloem
(ii) Xylem.

Question 20.
Name the tissue that transports food in plants. [H.B.S.E. March, 2019]
Answer:
Sieve elements (phloem).

Question 21.
How many types of animal tissues are there ? Write their names.
Answer:
Four types:
(i) Epithelial tissues
(ii) Connective tissues
(iii) Muscular tissues
(iv) Nervous tissues.

Question 22.
Where are epithelial tissues found ?
Answer:
In the external and internal layers of animals.

Question 23.
Which tissue is called a protective covering in animals ?
Answer:
Epithelial tissue.

Question 24.
Which tissue is responsible for producing sperm and egg in animals ?
Answer:
Epithelial tissue.

Question 25.
What are the types of muscular tissues ?
Answer:
Three types:
(i) striated
(ii) unstriated
(iii) cardiac muscles.

Question 26.
Which muscles function according to owr will ?
Answer:
Voluntary muscles (Striated muscles).

Question 27.
Which type of muscles present in heart ?
Answer:
Cardiac muscles.

Question 28.
How many types of connective tissues are there ?
Answer:
Three.

Question 29.
Write two main functions of connective tissues ?
Answer:
Joining different organs and provide protection.

Question 30.
What are the types of skeletal connective tissues ?
Answer:
Two types:
(i) Bone
(ii) Cartilage.

Question 31.
Which tissue functions to connect bone to another bone?
Answer:
Ligaments.

Question 32.
What do tendons interconnect with?
Answer:
Muscle with bone.

Question 33.
Write down the name of the fluid tissue found in the body.
Answer:
Blood.

Question 34.
How many types of blood cells or corpuscles are found in blood?
Answer:
Three types of blood cells or corpuscles.
They are:
(i) Erythrocytes-Red Blood Cells (RBC)
(ii) Leucocytes: White Blood Cells (WBC) and
(iii) Blood Platelets.

Question 35.
What is the fluid part in blood called as ?
Answer:
Plasma.

Question 36.
What percentage of water is there in blood ?
Answer:
90%

Question 37.
Which tissues are brain, spinal cord and nerves formed of ?
Answer:
Nervous tissues.

Question 38.
What is nervous tissue called as ?
Answer:
Neuron.

Question 39.
What is the main function of neuron ?
Answer:
Sending impulses of messages.

Short-Answer Type Questions

Question 1.
Multicellular organisms are more superior in comparison of unicellular organisms. Why ?
Answer:
The unicellular organism can accomplish but a few functions and in doing that they are comparatively less expert to that of multicellular organisms. Among multicellular organisms by forming groups of lacs of cells not only those functions, rather they can perform more functions with complete skill, since they have the quality of labour-division in them.

Question 2.
Is there some difference between plants and animals tissues ?
Answer:
The plant tissues perform the function of supporting. Most of the tissues of plants are dead. The dead tissues too provide mechanical strength like the living tissues. They need little care of maintenance, where as the animals roam about here and there in search of companion, food and shelter. They consume more energy than plants. They possess maximum number of living tissues. Growth in some parts of the plants remains constrained, whereas in animals the case is different.

Question 3.
What are the different types of tissues found in plants ?
Answer:
Two types of tissues are found in plants:
(1) Meristematic tissues
(2) Permanent tissues

Question 4.
Write down the features of meristematic tissues.
Answer:
The features of meristematic tissues are as follows :
(1) Meristematic cells have same structure and their cell wall is thin.
(2) The shape of the cells is round, oval or polygonal.
(3) They are compactly arranged with each other and they don’t have any intercellular space in them.
(4) These cells have dense or sufficient cytoplasm (cell liquid) and a big nucleus.
(5) They have comparatively few vacuoles or no vacuole.

Question 5.
Where are meristematic tissues found in plants ?
Answer:
Meristematic tissues are found in the growing up parts of the plants only; Like-apical meristem and cambium (the region that increases thickness)

Question 6.
Write down the characteristics of permanent tissues.
Answer:
The characteristics of permanent tissues are as follows:
(1) Permanent tissues are formed of meristematic tissues.
(2) Their cells have no ability of division.
(3) They have a definite shapes.
(4) Their wall can be dead, thin or thick.
(5) They have enlarged cells and vacuolic cytoplasm.

Question 7.
Differentiate between meristematic tissue and permanent tissue.
Answer:
The difference between meristematic tissue and permanent tissue is as follows:
Meristematic Tissue:
1. They do not have intercellular space.
2. They have the ability of divide.
3. The are responsible for growth.
4. They are thin-walled and isodiametric.
5. They have an enlarged nucleus.

Permanent Tissue:
1. It can be or cannot be there in them.
2. They lack the ability of division.
3. They are no responsible for growth.
4. They are thin or thick-walled but not isodiametric.
5. They have small nucleus.

Question 8.
What are permanent tissues?
Answer:
Permanent tissues:
Permanent tissues are formed of meristematic tissues. Their tissues have no ability of division. Their size and volume is definite.

Permanent tissues are of two types:
(1) Simple tissues
(2) Complex or compound tissues.

Question 9.
Differentiate between parenchyma and collenchyma tissues.
Answer:
Following are the differences between parenchyma and collenchyma:
Parenchyma Tissue:
1. They are found in roots, stem and leaves.
2. Their walls are not thick.
3. They are formed of round and fine walled cells.
4. Their wall is made up of pectin and cellulose.
5. They helps in storing food and provide mechanical support.
6. They have intercellular space in them.

Collenchyma Tissue:
1. They are found in the middle vein of the stem and leaves.
2. Their walls are thick.
3. They have thick and elongated wall.
4. Their cell wall possesses collens of pectin.
5. They just provide mechanical support.
6. It is absent in them.

Question 10.
Enunciate different types of simple permanent tissues and write down the function of each. Answer: Simple permanent tissues are of three types :
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma

(a) Functions of Parenchyma:
(i) Storing of food and assimilation.
(ii) To provide mechanical strength to the parts of plant.
(iii) Storing of resin, tanin, gum particles, and resin of morganie waste material.
(iv) Due to presence of chloroplast or chlorenchyma, they prepare food themselves.

(b) Functions of Collenchyma:
(i) They provide flexibility and strength to the plants.
(ii) Due to presence of chloroplast or chlorenchyma, they prepare glucose and starch.

(c) Functions of Sclerenchyma:
(i) They provide more strength to cartex and hard seeds.
(ii) They get transformed into protective tissues and protect the interior parts.

Question 11.
Differentiate among parenchyma, collenchyma and sclerenehyma.
Answer:
Parenchyma:
1. Their cell wall is constructed ot pectin and cellulose.
2. These are living cells.
3. Their function is food-storage and providing support.
4. They are found in root, stem and leaves.
5. They are formed of rounded thin walled cells.

Collenchyma :
1. Their cell wall is also formed of pectin and cellulose.
2. They too have living cells.
3. Their function is to provide mechanical support.
4. They are found in the middle vein of leaf and stem.
5. They are formed of multiarmed cells.

Sclerenchyma:
1. Their cell wall is formed of lignin.
2. Their cells are dead.
3. Their function is only to provide support
4. They are found in the stem and pulp of the fruits.
5. They are formed of thick-walled cells.

Question 12.
Describe Sclerenchvmatous tissue.
Answer:
Sclerenchyma:
These ceils are elongated in size and thin but are equipped with lignin. These are dead cells and are attached together. These cells are pointed at both ends. There is a clear mid lamella between adjacent cells. Cells consist of skewed regions that are called as pit canals. The cells provide strength to the parts of the plant.

Question 13.
Name different components found in Xylem (xylem tissues) components.
Answer:
The four components found in Xylem (xylem tissues) are as follows:
(1) Tracheids,
(2) Vessels,
(3) Xylem or wood Parenchyma
(4) Wood or Xylem fibers.

Question 14.
Give the importance of phloem in plants.
Answer:
Importance of Phloem: It is planted conductive tissue. It transports the starch (food) prepared by leaves to different parts of the plant.

Question 15.
Write functions of the Xylem (xylem tissues).
Answer:
Functions of Xylem:
It is too plant conductive tissue. It’s function is to transport the absorbed water and dissolved mineral salts by the roots to the leaves of the
plants and the apex. It also provide mechanical power to the plant.

Question 16.
Write short note on protective tissue.
Answer:
Protective Tissue:
Sclerenchyma found in the outermost parts of plants gets transformed into protective tissues. These tissues remain in the external layer of leaves, stems and roots. These tissues protect the interior tissues of plants. These tissues are covered with cutin. The cells of cork are dead. Cork is a lighter and protective tissue, it is more compressible and does not catch fire quickly.

Question 17.
What is stomata? Write its functions.
Answer:
Stomata:
They are found in the epidermal cells of the lower surface of leaves of plants which are called as stomatal pore. Stomata get protected by two guard cells.

Functions:
(1) Stomata are the nostrils of plants.
(2) Stomata plays a vital role in the exchange of gases and through them water loss takes place in plants.

Question 18.
Write names of different tissues found in animals.
Answer:
Animal tissues:
There are four types of tissues found in animals.
(1) Epithelial tissue
(2) Connective tissue
(3) Muscular tissue
(4) Nervous tissue.

Question 19.
Give the names and important functions of animal tissues.
Answer:
Animal tissue:
(a) Epithelial tissues:
Functions: They participate in absorption, secretion, protection, excretion, etc.

(b) Connective tissue:
Functions: This tissue participates in interlink the various parts of the body, provide support, turgidity, storage, protection and transportation.

(c) Muscular tissue :
Functions: Muscular tissue participates in motion, conduction, contraction, etc.

(d) Nervous tissue:
Functions: These do the main work of controlling and management.

Question 20.
Write a typical note on epithelial tissue.
Answer:
This is a covering or outer protective tissue that cover the body of an animal. Also, it protects a number of organs and cavities inside the body. Skin, alimentary can be capillary tube’s coat, food pipe, hollow organs of lungs all are formed of epithelial tissues. There is very’ narrow space. In these tissues, the cells are compactly arranged.

Question 21.
Classify Epithelial tissues.
Answer:
On the basis of shape and functions of the cells, epithelial tissues are classified into following types:
1. Squamous epithelium: These flat and disc like cells help in covering the organs of the body providing protection to them.
2. Cubical epithelium: These cubical or isodiametric cells are found in ducts of salivary glands and kidney tubules.
3. Columnar epithelium: The pillar or column like wide cells found in the layer of stomach and intestine.
4. Ciliated epithelium: Columnar and cubical shaped ceils germinate hair like structures at their ends that are called as cilia.
5. Glandular epithelium: These cells are found in secretory glands.

Question 22.
Write down the functions of Epithelial tissue.
Answer:
The functions of Epithelial tissue are the following :
(1) It performs the function to protect and cover the body organs
(2) It performs the functions of secretion, (in the Stomach and salivary glands)
(3) It produces sensitivity. (Eye and retinal cells)
(4) It produces sperm and egg.
(5) These absorb water and other salts.
(6) They also play role in excretion.

Question 23.
Write down the functions of connective tissue.
Answer:
The functions of connective tissues are as follows :
(1) These tissues join different organs of body with one another.
(2) They perform the function to tieup up organs of the body.
(3) These tissues give support to the organs of the body.
(4) Bone and cartilage build up the skeleton of the body.
(5) Adipose tissues store fat.
(6) These tissue perform the function of blood circulation.
(7) White blood corpuscles (WBC) fight against harmful bacteria.

Question 24.
What is Blood?
Answer:
Blood:
Blood is a liquid connective tissue. It tissue drift in liquid plasma (matrix). Blood cells are callei
(i) Red blood corpuscles (RBC)-Erythrocytes
(ii) White blood corpuscles (WBC)-Leucocytes
(iii) Blood platelets
Blood runs in each part of the body and thus, it connects all the parts of the body.

Question 25.
What are the functions of blood ? Explain.
Answer:
Functions of Blood:
(1) Red blood corpuscles receive oxygen from lungs and send it to different parts of the body.
(2) Blood collects hormones from the hormone glands and sends it to different parts.
(3) It helps in excreting excretory products from the body.
(4) White blood corpuscles or leucocytes protects from harmful bacteria.
(5) Blood checks the body temperature.
(6) Blood platelets clots the blood and avoid the more discharge of blood out of the body.

Question 26.
What is Bone? What are its types ?
Answer:
Bone: Skeletal tissue is connective tissue. In this tissue, ’/3 part organic and 2/3 part inorganic material is found. Its matrix consists of a peculiar type of protein ‘ossein’ which is in the form of collagen fibres. Bone possesses matrix osteoblasts, cytus, fibers and matrix have calcium phosphate, calcium carbonate and calcium fluoride. It give strength to the bone.

Question 27.
What is compact bone ?
Answer:
Compact bone:
This bone used to be strong and hard. Inside the matrix of bone, calcium salts provide hardness to the bone. Compact bone is formed of round and irregular dense layers of bones. Compact bone is found in the long shaft of bones like-femur. It involves Haversian canals.

Question 28.
Write down two differences each between bone and cartilage.
Answer:
Following are the differences between bone and cartilage :
Bone Cartilage:
1. It’s matrix is hard and flexible.
2. It’s matrix is formed of condin.
3. It is covered with pericondrium.
4. Haversian system is absent in it.

bone:
1. It’s matrix is too rigid and strong.
2. It’s matrix is formed of ossein.
3. It is covered with peraosteium.
4. Haversian system is present in it.

Question 29.
Write down the difference between tendon and ligament. Answer: Following are the difference between tendon and ligament:
Tendon:
1. It is flexible and moulding.
2. It connects muscle with bone.
3. It is made up of yellow fibre.

Ligament:
1. it is inflexible and hard.
2. It connects bone with another bone.
3. It is made up of white fibred connective tissues.

Question 30.
Write down the functions of bone, cartilage tendon and ligament.
Answer:
(i) Bone : It constructs the structure of body and protects the soft orgAnswer:
(ii) Cartilage: It constructs the skeletons of cartilaginous fishes. In our body the tip of nose and external ear (pinna), is formed of it.
(iii) Tendon : It connects muscle with bone.
(iv) Ligament: It joins bone with another bone.

Question 31.
Write down the functions of skeletal tissue.
Answer:
Skeletal tissue: Skeletal tissue is rigid, there is sufficient quantity of calcium salts in bone-matrix. Following are the significant functions of skeletal tissue:
(1) Skeletal tissue enshapes the body.
(2) Skeletal tissue protect the sensitive organs of the body.
(3) Skeletal tissue helps in joining muscles together.
(4) This tissue provides movement and locomotion to the body.
(5) In the middle of the bones erythrocytes (RBC) are produced.
(6) It provides surface for attachment of muscles (tendons and ligaments).

Question 32.
What is areolar tissue?
Answer:
Areolar tissue:
This tissue is found below the skin, between the muscles, blood vessels and around the nervous tissue. These cells possess jelly types transparent and sticky matrix. This is connective tissue. It interlinks body cavity and other parts of the body.

Question 33.
What is muscular tissue ? What are its types ?
Answer:
Muscular tissue:
The cells of these tissues are elongated, spindle shaped and unbranched. Cardiac muscles are branched. Muscular tissues are found in different sizes. Muscular tissues provide locomotion and movement to different parts of body.

Muscular tissues are of three types :
(a) Striated muscles. (Skeletal or voluntary muscles)
(b) Unstriated muscles. (Smooth muscles or involuntary muscles)
(c) Cardiac muscles.

Question 34.
Make difference between striated muscles arid cardiac muscies.
Answer:
The differences between striated muscies and cardiac muscles are as follows:
Striated muscle:
1. These muscles work accordingly to our will i.e., they are voluntary muscles.
2. These muscles are jointed with bones.
3. They are unbranched.
4. They have a number of nuclei.
5. They have light or dark bands on them.

Cardiac muscle:
1. These muscles are involuntary, they do not work accordingly to our command.
2. These muscles are found in the heart.
3. They are branched.
4. They have one or two nuclei.
5. They have transverse indistinct light bands which show faint cross striations.

Question 35.
Describe the structure of skeletal and cardiac muscles.
Answer:
Skeletal Muscles:
We can move some muscles by the conscious will. Muscles present in our limbs move when we want them to, and stop when we so decide. Such muscles are called voluntary muscles. These muscles are also called skeletal muscles as they are mostly attached to bones and help in body movement. Under the microscope, these muscles show alternative light and dark bands or striations when stained appropriately. As a result, they are also called striated muscles. The cells of this tissue are long, cylindrical, unbranched and multinucleate (having many nuclei).

Cardiac Muscles: The muscles of the heart show rhythmic contraction and relaxation throughout life. These involuntary muscles are called cardiac muscles. Heart muscle cells are cylindrical, branched and uninucleate.

Essay Type Questions

Question 1.
What is called a tissue ? How many types of plant cells are there ? Describe meristematic tissues.
Answer:
Tissue:
A group of several cells that have common origin, structure and function is called a tissue. In plants there are two types of tissues.
(i) Meristematic tissues
(ii) Permanent tissues.

Meristematic Tissue:
In these tissues, division of cells takes places continuously and these cells that have the ability to divide, they keep on reproducing newly cells. Meristematic tissues are found only in those parts that participate in the growth process like tip of the shoot-root, and cambium. Due to these tissues, the length and thickness of the plants get increased.

On the basis of origination, meristematic tissues are of two types:
(i) Primary meristematic tissues.
(ii) Secondary meristematic tissues.

On the basis of position meristematic tissues are of three types:
(a) Apical meristem: They are found at the tip of the root, stem and branches; They are also of three types:
(1) Epidermis: These tissues buiid up the outer skin of the plant.
(2) Cortex: These tissues buiid up the filling up system.
(3) Hypodermis: These tissues make transport tissues, phioem and xylem,
(b) Intercalary meristem: This tissue is found in the stem of monocotyledones. Due to activity of this tissue plant grow up in height.
(c) Lateral meristem: This tissue is found in the stems and root of dicotyledones.

Question 2.
How many types of simple tissue of plant are there ? Explain them.
Answer:
Simple tissue When the meristematic tissues totally lose the ability of division, then they transform into permanent tissues. Permanent tissues are of two types: Simple tissues and complex tissues. Simple tissues are formed of only one type of same structural cells. Their function is also same. On the basis of structure these tissues are of three types:

(i) Parenchyma:
These tissues are found in the shoot system (stem, leaves and flower) and root in plants. In the green part of the plant their number used to be more. These cells can be round, elongated or polygonal shape. These are living cells. Their cell wall is thin and is made up of cellulose. These cells contain dense protoplasm. The main functions of these tissues are storage of food, assimilation, provide strength, storing of resin tanin and gum and many times preparing food too.

(ii) Collenchyma:
These cells too are living. These are multiarmed, oval and rounded shaped. Their cell wall is thin, but at the corners of the cell due to deposition of cellulose and pectin they get hard or stout. In these cells generally there are some chloroplasts. This tissue brings flexibility and strength in plants and prepare glucose and starch due to presence of chloroplast in it.

(iii) Sclerenchyma:
These cells are elongated in size and thin but are equipped with lignin. These are dead cells and are attached together. These cells are pointed at both ends. There is a clear mid lamella between adjacent ceils. Cells consist of askewed regions that are called as pit canals. The cells provide strength to the parts of plant.

Question 3.
What is called as complex tissues ? What are their types in plants ? Explain each of them.
Answer:
Complex Tissue:
Complex tissue is a group of more than one type of cells which perform a special function. Complex tissue is of two types :
(1) Xylem: Xylem is a transport tissue. It transports water and dissolved mineral salts from root to the shoot-system. All the cells of this tissue are dead, thick walled and lignified. This tissue is formed of with the combination of four cells.
(i) Trachieds
(ii) vessels
(iii) xylem or wood-parenchyma
(iv) xylem fibres.

(2) Phloem:
It is a living transport tissue. Starch prepared due to photosynthesis process is transported to different parts of the plant by it. This tissue is also formed of with the combination of four components :
(i) Sieve tubes
(ii) companion cells
(iii) phloem parenchyma
(iv) phloem fibres.
Out of these sieve tubes are significant. These sieve tubes consist of porous wall that send food from the leaves to different parts of the plant.

Question 4.
What is called as nervous tissue ? How many types does it have ?
Or
Define the neuron with diagram.
Answer:
Nerve cells (neurons): It is a specialised elongated cell which is the basic unit of conduction of impulses Tn this cell there a nucleus surrounded with cytoplasm. Many small branches project out of cytoplasm, which are called as dendrites. One of these branches is the longest one which is called as Axon. Axon carries the impulses away from cells. One nerve cell does not directly join with the another neuron. They have some empty space between them in which there is a very closed transportation. It is called as synapses.

If we have pain in our hand, this information gets collected by the dentrites of the sensory neuron located in the hand. The sensory neuron converts it into electric signal. This signal is transmitted by the motor neurons and passing through synapse reaches the brain. The brain on receiving the impulse reacts to that. The motor nerve carries this reaction to the muscles of the hand and thus, the muscles of the hand make suitable reaction.

Thus, nerve cells (neurons) are of three types:
1. Sensory nerve cells (neurons): They carry sensations in different parts of the body towards brain.
2. Motor neurons: These receive message from the brain and further send it to the muscles.
3. Multipolar neurons: They does both the functions they carry sensations towards brain and further carry the commands of brain towards the muscles.

Question 5.
What is blood (human blood) ? Explain its constituents.
Answer:
Blood: Blood is liquid connective tissue which is red in colour. Blood circulates in the organs of body, thus it can be said that it connects the entire parts of the body together.

Human blood has mainly two constituents:
(A) Plasma: This is a substance in the blood which is light yellow in colour. It contains 90% of water. The remaining salts, glucose, amino acid, protein, hormone, oxygen, carbon dioxide gas and certain digested undigested food materials are also found in it.

Function: It consists of fibrin named protein, that helps in coagulation or clotting of blood. It also consists anti bodies which destroy foreign bodies harmful bacteria etc.

(B) Blood corpuscles (Blood cells): There are three types of corpuscles found in blood which are described as ahead:

(a) Red Blood Corpuscles or Erythrocytes: These are round in shape. Each red corpuscle has the life time of nearly four months. They have a substance namely haemoglobin in them which provides red colour to blood. They are round in shape, flat and small in size. They have no nucleus in them. They have the largest number (50 lakh oer cubic milliletre).

Functions:
(i) Red blood corpuscles provide red colour to the blood.
(ii) Their haemoglobin distributes oxygen throughout the body. On receiving oxygen these corpuscles i become oxy-haemoglobin.

(b) White Blood corpuscles or Leucocytes : The number of white blood corpuscles is lesser than red blood corpuscles, but they are bigger in size than them. They are white in colour, since they have no haemoglobin in them. Therefore they do not have a definite shape. They also have nucleus. They are of five types.
Function: They kill the harmful bacteria in the body and gobble them up. Thus, they protect us from diseases.

(c) Thrombocytes or Platelets: They are thick in the centre and thin at the ends. They are very small. They also have nucleus.
Function: They help in coagulation of blood. When in case of injury, blood oozes out from the wound, these platelets gather around the injury and by a chemical action clot there, so that there might not be any more bleeding.

Practical Work

Experiment-1:
Observing the structure of stem by cutting it in a transverse section (T.S.).

Procedure:
(i) Peel off a thin slice or transverse section of a thin and soft stem of a plant, keep it into the safranine solution.
(ii) Take a slice of stem section and keep it on a glass slide carefully leave a drop of glycerine on it and cover it with a coverslip.
(iii) Observe the prepared slide under the compound microscope and draw a sketch diagram of the setting up of the cells that you observe.

Quick Review of the Chapter

1. Blood transports
(A) oxygen
(B) food
(C) harmones and waste substances
(D) all of the above
Answer:
(D) all of the above

2. Nearly all tissues of are dead.
(A) plants
(B) animals
(C) both (A) and (B)
(D) saprophytes
Answer:
(A) plants

3. New cells in plants are formed by tissues.
(A) parenchyma
(B) collenchyma
(C) meristematic
(D) sclerenchyma
Answer:
(C) meristematic

4. Which of the follow ing is not the characteristic of meristematic?
(A) highly active
(B) more cytoplasm
(C) vacuole present
(D) distinct nucleus
Answer:
(C) vacuole present

5 tissues are found in the growing up parts of root and stem.
(A) Parenchyma
(B) Meristematic
(C) Sclerenchyma
(D) Collenchyma
Answer:
(B) Meristematic

6. The helpful tissue in photosynthesis is :
(A) parenchyma
(B)
(C) sclerenchyma
(D)
Answer:
(A) parenchyma

7 tissues carries the function of storage.
(A) collenchyma
(B)
(C) sclerenchyma
(D)
Answer:
(B) parenchyma

8. The tissues which makes a plant hard and strong is :
(A) collenchyma
(C) parenchyma
Answer:
(B) sclerenchyma

9. Which tissues make up the husk of coconut ?
(A) sclerenchyma
(C) parenchyma
Answer:
(A) sclerenchyma

10. Sclerenchyma is found in the of a plant.
(A) stem
(B) near vascular bundle
(C) buds of leaves and seeds
(D) all of the above
Answer:
(D) all of the above

11. Holes/pores are found :
(A) on the surface of leaf
(B) in vessels
(C) in xylem
(D) in phloem
Answer:
(A) on the surface of leaf

12. Which of the following is not a function of epidermis ?
(A) reduces the loss of water
(B) protects from mechanical shocks against water cell
(C) helpful in entry of parasite fungi
(D) all of the above
Answer:
(C) helpful in entry of parasite fungi

13. The shape of protective cells/tissues in stomata is
(A) like liver
(B) like kidney
(C) like the seed of bean
(D) like the seed of gram
Answer:
(B) like kidney

14. The function of stomata is:
(A) evaporation
(B) intake and release of oxygen
(C) intake and release of carbon dioxide
(D) all of the above
Answer:
(D) all of the above

15. The capacity of water absorption in roots increases by:
(A) increase in the size of root
(B) increase in the length of root
(C) due to hair roots present on roots
(D) all of the above
Answer:
(C) due to hair roots present on roots

16. What makes bark impermeable for air and water ?
(A) waxy cover
(B) thick layer of bark
(C) chemical named subrin
(D) all of the above
Answer:
(C) chemical named subrin

17. The example of complex tissue is:
(A) xylem
(B) meristematic
(C) phloem
(D) both (A) and (C)
Answer:
(D) both (A) and (C)

18. ………….. is Conductive tissue.
(A) xylem
(B) phloem
(C) both (A) and (B)
(D) neither (A) nor (B)
Answer: (C) both (A) and (B)

19. The parts of xylem are:
(A) tracheids
(B) vessels
(C) xylem parenchyma and fibre
(D) all of the above
Answer:
(D) all of the above

20. The parts of phloem are:
(A) sieve tubes
(B) companion cells
(C) phloem parenchyma and fibres
(D) all of the above
Answer:
(D) all of the above

21. Minerals and water are transported by:
(A) phloem
(B) xylem
(C) sieve tube
(D) companion cells
Answer:
(B) xylem

22. Phloem conducts:
(A) synthesised food
(B) mineral salts
(C) water
(D) all of the above
Answer:
(A) synthesised food

23. Which tissue is not a simple tissue?
(A) sclerenchyma
(B) phloem
(C) parenchyma
(D) collenchyma
Answer:
(B) phloem

24. Which tissue is not an animal tissue?
(A) epithelial
(B) epidermal
(C) connective
(D) muscular
Answer:
(B) epidermal

25. Which tissue carries the function of covering the body and making covers for different organs inside the body?
(A) epithelium
(B) connective
(C) nervous
(D) muscular
Answer:
(A) epithelium

26. Which one is not a type of epithelium tissues?
(A) squamous
(B) cubical
(C) rounded
(D) columnar and ciliated
Answer:
(C) rounded

27. Which tissue forms the difference of alimentary canal and mouth?
(A) glandular
(B) squamous
(C) columnar
(D) cubical
Answer:
(B) squamous

28. What kind of tissue is blood?
(A) epithelium
(B) connective
(C) muscular
(D) nervous
Answer:
(B) connective

29. Which blood cells carries the function of intake of oxygen?
(A) RBC
(B) WBC
(C) platelets
(D) all of the above
Answer:
(A) RBC

30. Bone is an example of what type of tissue?
(A) epithelium
(B) connective
(C) muscular
(D) nervous
Answer:
(B) connective

31. The tissue which provides motion to body is:
(A) epithelium
(B) connective
(C) muscular
(D) all of the above
Answer:
(C) muscular

32. The feature of voluntary muscle is:
(A) long in size
(B) cylindrical
(C) unstriated
(D) all of the above
Answer:
(D) all of the above

33. Involuntary muscles are not found in:
(A) bones
(B) eyelids
(C) urinal
(D) trachea of lungs
Answer:
(A) bones

34. Which cell can be longest?
(A) epithelium
(B) nervous
(C) connective
(D) muscular
Answer:
(B) nervous

35. Which tissue rs found in heart?
(A) muscular
(B) nervous
(C) epithelium
(D) connective
Answer:
(A) muscular

Haryana State Board HBSE 9th Class Science Important Questions Chapter 8 Motion Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 8 Motion

Very Short-Answer Type Questions

Question 1.
What do you mean by rest ?
Answer:
When a body does not change its position with respect to its surroundings, then the body is said to be at rest.

Question 2.
What is meant by motion ?
Answer:
When the position of the body continuously changes with respect to its surroundings, then the body is said to be in motion.

Question 3.
What is meant by relative motion ?
Answer:
The increase or decrease in distance between two moving bodies is known as relative motion.

Question 4.
How the motion of air can be measured ?
Answer:
The motion of air can be measured by its effect.

Question 5.
Passengers are sitting in a moving train. Tell, about which objects the passengers are in rest position and about which objects the passengers are in motion ?
Answer:
A passenger sitting in a moving train is in the state of rest in comparison of other passengers and the luggage kept in train. But the same person is in motion as compared to outsider trees and buildings.

Question 6.
Give one difference betw een the motion of living and non-living things.
Answer:
Living organisms move by themselves but external factor is required for the motion of non-living things.

Question 7.
What is distance ?
Answer:
The total path travelled by a body between initial and final points is called distance.

Question 8.
What is displacement?
Answer:
The shortest distance between the initial and final position of the body is called displacement.

Question 9.
What is the SI unit of distance and displacement ?
Answer:
The SI unit of distance and displacement is metre.

Question 10.
What is meant by reference point or origin point ?
Answer:
A fixed point with respect of which a body is at rest or in motion is known as reference point or origin point.

Question 11.
What those quantities are called, which require magnitude and direction both for their expression ?
Answer:
Those quantities which require magnitude and direction both for their expression are called vector quantities.

Question 12.
Which quantities are expressed by magnitude only ?
Answer:
Scalar quantities are expressed by magnitude only.

Question 13.
Write two examples each of vector and scalar quantities.
Answer:
Vector quantities – Velocity and displacement.
Scalar quantities – Distance and speed.

Question 14.
When an object covers equal distance in equal intervals of time, then what its speed is called ?
Answer:
When an object covers equal distance in equal intervals of time, its speed is known as uniform speed.

Question 15.
Give two examples of uniform speed.
Answer:
(i) Motion of earth
(ii) Motion of pendulum of a watch.

Question 16.
When an object covers unequal distance in equal intervals of time, then what its speed is called ?
Answer:
When an object covers unequal distance in equal intervals of time, its speed is known as non¬uniform speed.

Question 17.
Write two examples of non-uniform speed.
Answer:
(1) Train starting from a station.
(2) Birds flying in the sky.

Question 18.
Write down the formula to find out the average speed of a moving object.
Answer:
The distance travelled by an object per unit time is known as average speed.

Question 19.
What is meant by velocity ?
Answer:
The distance covered by a moving body in one second in a certain direction is called velocity.

Question 20.
What is the unit of velocity ?
Answer:
The unit of velocity is metre per second (ms’1).

Question 21.
What is the relation among distance, speed and time ?
Answer:
Distance = Speed x Time

Question 22.
What will be the velocity of an athlete, if he takes 10 seconds to cover a distance of 100 metres ?
Answer:

Question 23.
What is meant by acceleration ?
Answer:
The rate of change of velocity of a moving body is called acceleration.
a = \(\frac {v – u}{t}\)

Question 24.
What is meant by retardation ?
Answer:
The rate of decrease in velocity in a certain direction is called retardation i.e. negative acceleration s called retardation.

Question 25.
What is SI unit of acceleration ?
Answer:
SI unit of acceleration is metre Der second sauare

Question 26.
Write example ot accelerated motion.
Answer:
A body moving in a circle with uniform speed is an example of accelerated motion.

Question 27.
Choose scalar and vector from following :
(i) Delhi is at a distance of 1450 km from Mumbai.
(ii) Delhi is at a distance of 1450 km in north from Mumbai.
Answer:
(i) It is an example of scalar quantity.
(ii) It is an example of vector quantity.

Question 28.
When, a body is said to be moving with uniform acceleration ?
Answer:
If there is equal change in velocity in equal intervals of time, then the body is said to be moving with uniform acceleration.

Question 29.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is known as uniform circular motion.

Question 30.
Which type of motion is shown, if the speed-time graph is a straight line ?
Answer:
It shows uniformly accelerated motion.

Question 31.
What will be the acceleration of the body, if the speed-time graph is parallel to time axis ?
Answer:
The speed of the body will uniform, therefore acceleration will be zero.

Question 32.
What shows the slope of speed-time graph of an object ?
Answer:
Acceleration of the object.

Question 33.
What the area under and between interval of speed-time graph of moving body shows ?
Answer:
The distance covered by the body in that time interval.

Question 34.
If the distance-time graph, parallel to time axis is a straight line, then average speed in different intervals will be equal or different.
Answer:
Equal.

Question 35.
If the distance-time graph of a body is a straight line, what you will say about the motion of the body ?
Answer:
The body will be in constant position.

Question 36.
If the distance-time graph of a body is a curve, then its average speed in different intervals will be different or uniform.
Answer:
The average speed will be different in different intervals.

Question 37.
If the speed of a moving body becomes 30 km/h from 0 in 2 hours, then what will be its acceleration?
Answer:
Acceleration =img= 15 kmh^-2

Question 38.
If a body is moving on circular track with uniform-speed, then which type of its motion will be ?
Answer:
Non-uniform and accelerate motion.

Question 39.
Speed of light is 3 x 10⁸ ms^-1. What will be its speed in kmh^-1 ?
Answer:
speed of light = 3 x 10⁸ms^-1 = \(\frac{3 \times 10^8 \times 3600}{1000}\)kmh^-1
= 1.08 x 10⁹kmh^-1

Question 40.
Rajdhani Express covers a distance of 1384 km in 17 hours. What will be its average speed?
Answer:

Question 41.
What will be the average speed of a car, if it covers a distance of 80 km in two hours ?
Answer:
Average speed = \(\frac {80}{2}\) = 40 kmh^-1

Question 42.
What is meant by circular motion ?
Answer:
When a body moves on circular track, its motion is known as circular motion.

Question 43.
Give two examples of circular motion.
Answer:
(1) The motion of the wheel of a vehicle.
(2) The motion of a stone tied on the end of a rope in plane circle.

Question 44.
What is meant by graph ?
Answer:
The figure representation of any information is called graph.

Question 45.
Which type of motion is shown by a freely falling object ?
Answer:
It shows a uniformly accelerated motion.

Question 46.
A stone is thrown in vertical direction, then its velocity is continuously decreasing. Why ?
Answer:
When a stone is thrown in vertical direction, its velocity continuously decreases because the acceleration due to gravity is always directed towards the centre of earth.

Question 47.
You have gone to school on 8 o’clock in morning and come back 2 o’clock in afternoon. What will be your displacement?
Answer:
Zero.

Question 48.
What will be the velocity of that object which is thrown vertically upward from the surface of earth and attains maximum height ?
Answer:
Zero.

Question 49.
What is the relation among s, u, a and t ?
Answer:
s = ut + \(\frac {1}{2}\) at².

Question 50.
What is the relation among v, u, a and t ?
Answer:
v = u + at

Question 51.
What is the relation among v, u, a and s ?
Answer:
v² – u² = 2as

Question 52.
What will be the speed of an athlete, when he completes a round in t seconds on a circular track of radius r ?
Answer:
Speed (v) = \(\frac {2πr}{t}\)

Question 53.
When there are clouds in the sky, the lightning and thundering happen in same time. First, you see the eightning After some time, the sound of thundering reaches to you. Can you tell, why it is so ?
Answer:
Because the velocity of light (3 x 10⁸ ms^-1) is much more than the velocity of sound (346 m/s).

Short Answer Type Questions

Question 1.
What is the difference between rest and motion ?
Answer:
When a body does not change its position with respect to its surroundings, it is called rest, e.g. a book lying on the table, a stone lying on the floor, etc. When the position of the body continuously changes with respect to its surroundings, it is called motion, e.g. a moving car, a running horse, etc.

Question 2.
What is meant by displacement ? When it is positive and negative ?
Answer:
The change in the position of body in a certain direction is called displacement.
The position of a body changes from x₁ to x₂, then the displacement (d) can be written as:
d = (x₂ – x₁)
The +ve and -ve signs of displacement shows the direction of change in position of body. If displacement is +ve, then body moves from left to right. If the displacement is -ve, then body moves from right to left. It is a vector quantity.

Question 3.
What is meant by speed ? Write its unit.
Answer:
Speed is the distance travelled by the body in unit time interval in any direction.

Suppose, a body covers a distance ‘s’ in time ‘ f, then its speed v = –
Unit of speed is metre per second (ms^-1). 1

Question 4.
What is meant by non-uniform motion ? Draw a graph for non-uniform motion.
Answer:
Motion of a body is said to be non-uniform, if it covers unequal distance in equal intervals of time, e.g. motion of a car on road, because its speed keeps on changing. Similarly, the motion of a train is an example of non-uniform motion, because train covers different distance in every second. Distance-time graph is shown to clear the motion of train. It is clear from the figure that speed changes with time and distance-time graph is not a straight line. It is an example of

Question 5.
How vector is shown ?
Answer:
Vector is represented by a straight line with an arrow head over it. The length of this line is proportional to the magnitude of vector and arrow head shows the direction, which is known as the ‘head’ of vector and other end is known as the ‘tail’ of vector. For example, a velocity of 30 ms^-1, 30° in south-west direction can be shown in the form of a line carrying arrow head, as shown in figure.

Question 6.
Two cars moving in opposite directions, cover an equal distance d in an hour. What will be displacement of each after an hour, if they are moving in north and south directions respectively ?
Answer:
Let O is the reference point of both cars.
The distance covered by car A from reference point O in hour in north direction = d
∴Displacement OA = d
The distance covered by car B from reference point O in hour in south direction = d
∴Displacement OB = d
Total displacement of both cars, AB = d + d = 2d

Question 7.
What is meant by graph ? What are its uses ?
Answer:
To draw the change in one physical quantity7 with respect to another physical quantity on two different axes is called graph.

Uses: The uses of graph are as follows:
(l) The relation among different quantities can be studied easily.
(2) The table of two or more than two quantities can be compared easily.
(3) It is more simple to explain quantity in graph than table.
(4) Graph represents data in less space in comparison to table which takes larger space to represent data.
(5) Quantity related to slope of graph can be found.

Question 8.
What is distance-time graph ? The distance covered by a bus in different time intervals is given in the table. Draw a distance-time graph with the help of it.

Answer:
A simple and successive method, which shows the different distances at different time of a moving body, is called distance-time graph. The distance covered by the bus in different time-intervals is shown in distance-time graph.

Question 9.
Draw distance-time graph for tw o bodies moving with uniform speed of 4 ms^-1 and 7 ms^-1 and compare it.
Answer:
We know that,
Distance = Speed x Time
(i) Distance-time graph for a body moving with uniform speed of 4 ms^-1.


(ii) Distance-time graph for a body moving with uniform speed of 7 ms^-1.

On drawing these values on graph paper, we see that distance-time graph of given body will be as the adjacent figure.

Comparison of Graphs:
The slope of the body moving with uniform speed of 7 ms^-1 will be more than that of the body moving with uniform speed of 4 ms^-1.

Question 10.
Distance-time graph for a body is shown in figure. Is it the graph of real situation ? Give reason.
Answer:
No, this is not the graph of real situation. Reasons are as follows:
(1) According to distance-time graph, the time taken to reach that point from where it started its journey is less than the time taken by it to complete that journey.
(2) Time (t) is such a quantity which is continuously increasing. This will be larger after t = 14.

Question 11.
Distance-time table of a moving car is as follows:

Draw distance-time graph for car on the graph paper. Answer: Distance-time graph for car:

Question 12.
The distances covered by Amitabh and Archna by their bicycles in different times:

Show both by distance-time graph.
Answer:
The distance-time graph for both is given below:

Question 13.
Draw velocity-time graph for a body moving with a constant velocity of 40 kmh^-1.
Answer:
To draw a velocity-time graph of a car moving with constant velocity’ of 40 kmh^-1. It means, car covers a distance of 40 km in an hour, 80 km in two hours, 120 km in 3 hours and so on, it can be seen from figure, that velocity-time graph is a straight line and parallel to time axis. It is true for all velocity-time graphs when motion is uniform.

Question 14.
How we can calculate the distance covered by a body in a given time by using velocity¬time graph ?
Answer:
Suppose the velocity-time graph of a car moving with speed of 40 kmh^-1 is as the given figure. Let us consider, we have to calculate the distance covered by the car between time and t₂. Draw perpendiculars from points t₁ and t₂ on time axis as shown in figure.

By this, we obtain a rectangle ABCD between graph and X-axis. In this rectangle side AD is equal to BC, which is equal to (t₂ -t₁), and side AB is equal to CD, which is similar to 40 kmh^-1. We know that, if a body moving with velocity (v) covers a distance (.s’) in time (?) then,
s = vt
∴ Distance covered by the car in time (t₂ – t₁) = [(40 kmh^-1) (t₂ – t₁)h]
= 40(t₂ – t₁) km
= Area of rectangle ABCD or the area under velocity-time graph shows the distance covered by the car. This is true for any velocity-time graph, whether velocity is uniform or non-uniform,

Question 15.
How velocity is calculated by straight-lined distance-time graph ? Explain.
Answer:
We can calculate the velocity of a car from its distance-time graph. In fig., think about the smaller part AB of distance-time graph of motion of car. We draw a straight-line from point A parallel to X-axis and from point B parallel to Y-axis to find the velocity of car. These two straight lines meet at a point C and make a AABC. Now, on graph, AC represents time-interval (t₂ – t₁), and BC is equal to distance (s₂ – s₁). It can be seen from the graph that when car move from point A to B, it covers a distance (s₂ – s₁) in time (t₂ – t₁)
Velocity of car = \(\frac{s_2-s_1}{t_2-t_1}\)

Question 16.
Show uniformly accelerated motion, non-uniformly accelerated motion and uniform retarded motion by graph.
Answer:
1. Uniformly Accelerated Motion: In uniformly-accelerated motion, speed-time graph is a straight a line.

2. Non-uniformly Accelerated Motion: In non-uniformly accelerated motion, speed-time graph is a

3. Uniformly Retarded Motion: In uniformly retardation, the acceleration is negative (-ve) and velocity time graph is a straight line.

Question 17.
Draw speed-time graph for a body, moving in a straight line. What the slope of this straight line show ?
Answer:
If there is equal change in the velocity of a body in equal intervals of time, then we can say that the body is moving with uniform acceleration. The speed-time graph for a body moving in a straight-line is shown in figure. If the speed of the body is v₁ at time t₁ and v₂ at time t₂, then
acceleration of the body, a = \(\frac{v_2-v_1}{t_2-t_1}\)

The time intervals (t₂ – t₁) and (t₄ -t₃) are equal as shown in figure, because the change in heights of speed are also equal. Therefore, this graph shows the motion of such a body, which is moving with uniform acceleration.
The speed-time graph of a body moving with uniform acceleration is a straight line. The acceleration of such body is the slope of this straight line.

Question 18.
How you will prove that the motion is relative ?
Answer:
In order to prove that the motion is relative, we can compare the passengers sitting in a moving bus or train with road or trees standing near the track, vehicle or persons, e.g. the passengers sitting inside the bus . or train are in rest as compared to the luggage lying inside, but are in motion as compared to outsided objects.

Question 19.
Write the difference between-
(i) vector and scalar.
(ii) uniform and non-uniform motion.
(iii) distance and displacement.
Answer:
The difference between vector and scalar:
Vector:
1. Those physical quantities which contains both direction and magnitude are known as vector quantities, e.g. velocity, displacement, force, etc.
2. Vector quantities are combined according to rule of triangle or parallelogram.
3. Vector quantities are represented by a straight line with an arrow head over it.

Scalar:
1. Those physical quantities which contain magnitude only is known as scalar quantities, e.g. distance, volume, mass, etc.
2. Scalar quantities can be combined algebraically.
3. There is no special method to represent scalar quantities.

(ii) The difference between uniform and non-uniform motion :
Uniform Motion:
When a body covers equal distance in equal intervals of time, its motion is called uniform motion, e.g. motion of earth, motion of pendulum of watch, the motion of moon around the earth.

Non-uniform Motion:
When a body covers unequal distance in equal intervals of time, its motion is called non¬uniform motion, e.g. motion of train starting from station, the motion of bus starting from bus stop, motion of cycle.

(iii) The difference between distance and displacement:

Distance:
1. The length of the path actually covered by the body is called distance.
2. It is a scalar quantity.
3. The distance between two points through different paths can be different.

Displacement:
1. The distance between the initial and final positions of the body is called displacement.
2. It is a vector quantity.
3. The value of the displacement between any two points is always same.

Question 20.
What is meant by uniform circular motion ? Justify your answer with example.
Answer:
When a body moves on circular track, it changes its direction at every point. Actually, circle is a polygon figure which has infinite sides. In figure any circular shaped track can be called a polygon of inner-sides. Therefore, a body moving in a circle with uniform speed is an example of accelerated motion. Although, there is no change in the speed of the body, but the direction of the motion of body is continuously changes. Example – Take a piece of thread and tie one stone on its one end and rotate it, it will be in a circular motion.

Numerical Questions

Question 1.
The distance between the home of a boy and school is 1.8 km, if he takes 5 minutes to reach school by cycle, then find his speed In metre/second.
Solution:
Distance (s) = 1.8 km = 1.8 x 1000 metres = 1800 metres
Time taken (t) = 5 minutes = 5 x 60 seconds = 300 seconds

= 6 ms^-1

Question 2.
A person takes 10 minutes to reach office from his home. If the office is at a distance of 3.6 km, then find the average speed of person.
Solution:
Total distance 3.6 km = 3.6 x 1000 m = 3600 metres
Total time = 10 minutes = l0 x 60seconds = 600 seconds
Total distance 3600 m

= 6 ms^-1

Question 3.
Trivendrum express moves from Guwahati at 23 : 30 hrs on Monday and after covering a distance of 3574 km, It reaches Trivendrum at 22 : 30 hrs on Thursday. Find out the average speed of
this train.
Solution:
The total distance covered by the train 3574 km
Total time between 23:30 hrs (Monday) to 22:30 hrs (Thursday) = 24 + 24 + 23 – 71h

= 50.34 kmh^-1

Question 4.
A car, starting from rest, acquire a velocity of 36 kmh’ in 10 seconds. Find out acceleration of the car.
Solution:
The car started from rest
∴ Initial velocity (u) = O
Final velocity (v) = 36 kmh^-1 = 36 x \(\frac {5}{18}\) = 10 ms^-1
Time (t) = 10 s
We know that. v = u + at
10 = 0 + a x 10
10 = 10 a
a = \(\frac {10}{10}\) = 1 ms^-1

Question 5.
A cycle moving with a velocity of 3 metre per second increase its motion with acceleration of 0.5 ms² . What will be Its velocity after 5 seconds and how much distance it will cover ¡n this time?
Solution:
Here, Initial velocity (u) = 3 ms
Acceleration (a) = 0.5 ms²
Time (t) = 5 s
Final velocity (v) = ?
Distance (s) = ?
We know that,
v = u + at = 3 + 0.5 x 5 = 3 + 2.5 = 5.5 ms^-1

(2) s = ut+ \(\frac {1}{2}\) at²
= 3 x 5 + \(\frac {1}{2}\) x 0.5 x 5 x 5
15 + 6.25 = 21.25 m

Question 6.
The uniform acceleration of a scooter is 4 ms2. How much distance will scooter cover after 10 seconds from its movement ? What will be its velocity after 10 s?
Solution:
Here,
Initial velocity of scooter (u) = 0
Acceleration (a) = 4 ms
Time (t) = 10 s
Distance (s) = ?
Final velocity (y) = ?
We know that
(1) s = ut + \(\frac {1}{2}\) at²
s = 0 x 10 + \(\frac {1}{2}\) x 4 x (10)²
s = 0 + \(\frac {1}{2}\) x 4 x 100 = 200m
(2) v = u + at = 0 + 4 x 10 = 40 ms^-1

Question 7.
A bicycle is travelling at a speed of 10 ms. The brakes are applied so as to produce a retardation of 0.5 ms2. Find, how far the bicycle travel before it is brought to rest?
Solution:
Initial velocity of bicycle (u) = 10 ms^-1
Final velocity (v) = 0
Retardation (a) = 0.5 ms^-2
We know that, v² – u² = 2as
(0)² – (10)² = 2(- 0.5) x s
– 100 = – 1s
s = 100 m

Question 8.
A car travels with uniform velocity 20 ms’ for 5 s. On applying brakes, the car comes to rest in 8s with uniform retardation. Draw, velocity-time graph. How far, the car will go after applying brakes?
Solution:
The velocity-time graph is shown in figure.
The distance covered by the car, after applying brakes
= area of ∆BDC = \(\frac {1}{2}\) x DC x DB
[ ∴ area of triangle = \(\frac {1}{2}\) x base x perpendicular]
= \(\frac {1}{2}\) x (13 – 5) x 20
= \(\frac {1}{2}\) x 8 x 20 = 80 m.

Question 9.
In a long running, the athletes have to move 4 rounds of running track The starting and end of race is at the same place. If the length of one round is 200 m then –
(a) how much distance is to be covered by athletes?
(b) what will be the total displacement when the athletes has completed the race?
(c) is the motion of athletes uniform or non-uniform?
(d) are the distance covered by the athletes and displacement equal, at the end of race?
Answer:
(a) The total distance to be covered by the athletes = 200 x 4 = 800 metres.
(b) Because the initial and final places of running track are at same place, therefore the total displacement of athletes = zero.
(c) The motion of athletes is non-uniform.
(a) The distance covered by the athletes at the end of race is not equal to displacement. Because –
Distance = 800 metres
Displacement = 0

Question 10.
Write the speed of following bodies in descending order –
(i) Bicycle running with a speed of 18 kmh^-1.
(ii) An athlete running with a speed of 7 ms^-1.
(iii)A car running with a speed of 2000 m min^-1
Solution:
We know that,
(i) 18 kmh^-1 = \(\frac {18000}{3600}\) ms^-1 = 5 ms^-1
(ii) 7 ms^-1
(iii) 2000 m min^-1 = \(\frac {2000}{60}\) ms^-1 = 33.33 ms^-1
∴ The descending order of speed of bodies is as follows:
(i) A car running with a speed of 2000 m min^-1.
(ii) An athlete running with a speed of 7 ms^-1.
(iii) A bicycle running with a speed of 18 kmh^-1.

Question 11.
A body moves with velocity of 2 ms^-1 for 5 seconds, in next 5 seconds, its velocity increases to 10 ms^-1 due to uniform acceleration. After that, its velocity decreases uniformly and comes in rest in 10 seconds, then
(a) draw velocity-time and distance-time graphs for the motion of the body.
(b) Show that portion in graph, where the motion is uniform and non-uniform.
(c) From graph, find the distance covered by the body after 2s and 12s and in the last lOs.
Answer:
(a) Velocity-time graph: The velocity of the body is 2 ms for first 5 seconds, which is marked
by AB in the graph. After that, it becomes 10 mst in another 5 seconds (between 5 and 10 s), which is marked
by BC in the graph. After that the body comes in rest in next 10 seconds, which is marked by CD in the graph.

(b) Distance-time graph: Because the speed of the body is not uniform. Therefore, distance-time graph is not a straight-line

From velocity-time graph, we find that the motion of the body is uniform in AB part, whereas the motion is non-uniform in BC and CD parts.
(c) (1)The distance covered by the body in 2s = 2 x 2 = 4 m.
(2) The distance covered by the body in 12 s = area of [OB’BA + BBC’C + CC’FF’]
Now, the area of rectangle OBBA = 2 x 5 = 10
Area of trapezium BB’C’C = \(\frac {1}{2}\) x (BB’ + CC’) x B’C
= \(\frac {1}{2}\) (2 + 10) x 5
= \(\frac {1}{2}\) x 12 x 5 = 30
Area of trapezium CC’FF’ = 4- (CC’ + FF’) x CF’
= \(\frac {1}{2}\) x (10 + 8) x 2 = \(\frac {1}{2}\) x 18 x 2 = 18
∴ The distance covered by the body in 12 sec 10 + 30 + 18 = 58 metres

(3) Distance covered in last lO seconds area of (trap. CC’ F’F + right-angle ∆FF’D
Now, area of trapezium CC’F’F = x (CC + FF’) x C’F’
= \(\frac {1}{2}\) x (10 + 8) x 2
= \(\frac {1}{2}\) x 18 x 2 = 18

Area of right-angle ∆FF’D = \(\frac {1}{2}\) x F’D x F’F = \(\frac {1}{2}\) x 8 x 8 = 32
∴ Distance covered in last 10 seconds = 18 + 32 = 50 metres.

Do Yourself

1. Rita takes 20 minutes to cover a distance of 3.2 km on a bicycle. Calculate her velocity in units of a kilometre minute and kilometre/hour.
Answer:
0.16 km/minute, 9.6 km/hour

2. Ahmed is driving his car with a velocity of 45 kmh^-1. How much distance will be covered in
(a) one minute and
(b) one second?
Answer:
(a) 750 metres
(b) 12.5 metres

3. Starting from rest, Saurabh paddles his bicycle to attain a velocity of 6 ms^-1 in 30 seconds. Then, he applies brakes after that the velocity of the bicycle comes down to 4 ms^-1 in next 5 seconds. Calculate the acceleration of the bicycle in both the cases.
Answer:
(a) 0.2 ms^-2
(b) -0.4 ms^-2

4. A train starting from rest attains a velocity of 108 kmh^-1 in 10 minutes. Calculate (I) acceleration (ii) distance covered by the train.
Answer:
(1) 0.5 ms²
(ii) 900 m

5. A train starting from rest attains a velocity of 72 kmh^-1 in 5 minutes. Assuming that the acceleration is uniform, find (j) acceleration of train and (ii) the distance travelled by the train while it attained this velocity.
Answer:
(i) ms^-2
(ii) 3 km

6. A car accelerates uniformly from 18 kmh^-1 to 36 kmh^-1 in 5 seconds. Calculate : (1) the acceleration and (ii) the distance covered by the car in that time.
Answer:
(i) 1 ms^-2
(ii) 37.5 m

7. The brakes applied to a car produces a negative acceleration of 6 ms2. if the car takes 2 seconds to stop after applying the brakes, calculate the distance it travels, during this time.
Answer:
12 m

8. A train covers a distance of first 30 km out of 60 km lengthy path with a uniform speed of 45 kmh^-1. By how much speed, the train will cover the remaining 30 km path so that its average speed will be 60 kmh^-1 ?
Answer:
90 kmh^-1

9. Choose right answer out of the following questions:
(a) How much distance will a car cover in 12 minutes moving with a speed of 35 kmh^-1’?
(i) 7 km
(ii) 35 km
(iii) 14 km
(iv) 28 km
Answer:
(i) 7 km

(b) If, the acceleration of the body moving with a speed of 20 ms^-1 in a straight line ¡s 4 ms^-2, then
how much speed of it will be after 2 seconds?
(1) 8 ms^-1
(ii) 12 ms^-1
(iii) 16 ms^-1
(iv) 28 ms^-1
Answer:
(iv) 28 mst

(C) The speed of a car becomes 50 kmh^-1 from 20 kmh^-1 in 10 seconds. The acceleration of this car is –
(i) 30 ms²
(ii) 3 ms²
(iii) 1 8 ms²
(iv) 0.83 ms²
Answer:
(iv) 0.83 mc²

10. A piece of wood slides down from rest on plane of 10m tong with an acceleration of 5 ms² . What will be the speed of piece of wood on reaching at the base of inclined plane?
Answer:
10 m /s

11.. A train is travelling at a speed of 90 kmh^-1 On applying brakes, a retardation of 0.5 mc² s produced. Find how far the train goes before it is brought to rest?
Answer:
625 m

12. A body covers a distance of 16 metres in 4 seconds and again covers a distance of 16 metres in 2 seconds. What will be the average speed of body?
Answer:
5.33 ms^-1

13. The odometer of a car represents 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 hours then, calculate the average speed of the car in kmh’ and ms.
Answer:
50kmh^-1, 13.9 ms^-1

14. Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Answer:
Speed 3 ms^-1, Velocity = O ms^-1

15. Starting from a rest position, Rahul paddles his bicycle to attain a velocity of 6 ms1 in 30s. Then he applies brakes such that the velocity of bicycle comes down to 4 ms^-1 in the next 5s. Calculate the acceleration of the bicycle ¡n both the cases.
Answer:
0.2 m² 0.4 ms

Essay Type Question

Question 1
Prove the following equations:
(i) v = u + at²
(ii) s = ut + ut²
(iii) v² – u² = 2as
Solution:
(i) Let the initial velocity of the body = u
Time = t
Final velocity = v
Change in velocity = v – u
Rate of change in velocity (acceleration), a = \(\frac {v – u}{t}\)
at = v – u
at+u = v
v = u + at
Let the initial velocity of the body = u
Acceleration = a
and after time t, final velocity = v
1f after t sec, the distance is s, then distance (s) = average velocity x time
But, average velocity = \(\frac {v – u}{t}\)
∴Distance (s) = \(\frac {v – u}{t}\)
s = \(\frac {v – u}{2}\) x t
s = \(\frac {u + u + at}{2}\) x t [∴v = u + at]
s = \(\frac {2u + 2t}{2}\) x t
s = \(\left(\frac{2 u t+a t^2}{2}\right)=\frac{2 u t}{2}+\frac{a t^2}{2}\)
s = ut + \(\frac {1}{2}\) at²

(iii) We know that, v = u + at
Squaring both sides
(v)² = (u + at)²
v² = u² + a²t² + 2uat
v² – u² = a²t² + 2uaI
v²– u² = 2uat + a²t²

v²– u² = 2a(ut + \(\frac {1}{2}\) at²)
v²– u² = 2as

Practical Work

Experiment 1:
Describe one activity for the difference between distance and displacement.

Procedure:
Take a metre scale and a long rope. Walk from one corner of a basket-ball court to its opposite corner way and measure with the help of scale. This will be the distance covered by you, whereas you will get displacement by putting a straight-rope from one corner to another. Suppose ABCD is basket-ball court. Distance to go from A to C AB + BC, whereas displacement = AC.

Experiment 2:
Clarify with an activity that direction of motion changes at every point on circular track

Procedure:
Take a piece of thread and tie a small piece of stone at one of its ends, move the stone to describe a circular path with constant speed by holding the thread at the other end as shown in figure. Now, let the stone go by releasing with thread and measure the direction of stone.

By repeating the activity for a few times and releasing the stone at different positions of circular path, check whether the direction in which the stone moves remains the same or not If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is’ released, it continues to move along the direction it has been moving at that instant. This shows that the direction of mot ion changes at every point when the stone was moving along the circular path.

Quick Review of the Chapter

1. A train is moving. The passengers sitting inside it, are ¡n rest compared to :
(A) other passengers
(B) the luggage in the train
(C) outsider objects
(D) other passengers and luggage in train
Answer:
(D) other passengers and luggage ¡n train

2. A train is moving. The passengers sitting inside it, are in motion as compared to :
(A) other passengers
(B) the luggage in the train
(C) outsider buildings and trees
(D) other passengers and luggage in train
Answer:
(C) outsider buildings and trees

3. If any object changes its position continuously with respect to other object, then objett is in state of
(A) motion
(B) rest
(C) constant
(D) none of these
Answer:
(A) motion

4. If any object does not change its position with respect to its surroundings, then the object is in state of
(A) motion
(B) rest
(C) circular motion
(D) movement
Answer:
(B) rest

5. Which of the following moves by themselves?
(A) living things
(B) non-living things.
(C) both living and non-living things
(D) neither living nor non-living
Answer:
(A) living things

6. SI unit of distance is:
(A) km
(B) m
(C) cm
(D) mm
Answer:
(B) m

7. WhIch of the following is a scalar quantity?
(A) momentum
(B) velocity
(C) displacement
(D) distance
Answer:
(D) distance

8. Rate of change of velocity is known as:
(A) Speed
(B) Acceleration
(C) Momentum
(D) Force
Answer:
(B) Acceleration

9. When a body covers equal distance in equal intervals of time, it is said to be
(A) uniform speed
(B) non-uniform speed
(C) average speed
(D) accelerated motion
Answer:
(A) uniform speed

10. If acceleration is in the direction of velocity, then it is called:
(A) Negative acceleration
(B) zero acceleration
(C) Positive acceleration
(D) none of these
Answer:
(C) Positive acceleration

11. If acceleration is ¡n the opposite direction of velocity then it is called:
(A) Negative acceleration
(B) zero acceleration
(C) positive acceleration
(D) none of these
Answer:
(A) Negative acceleration

12. If an object moves ¡n a circular path with uniform speed, Its motion is called: [H.B.S.E. 2020]
(A) Uniform circular motion
(B) Non-uniform circular motion
(C) Uniform curved motion
(D) Non-uniform curved motion
Answer:
(A) Uniform circular motion

13. SI unit of speed Is: [H.B.S.E. March, 2019]
(A) ms^-2
(B) ms^-1
(C) kmh^-1
(D) kms^-1
Answer:
(B) ms^-1

14. The distance covered by a moving body in one second in a certain direction is called:
(A) velocity
(B) acceleration
(C) non-uniform speed
(D) uniform speed
Answer:
(A) velocity

15. What is the relation among distance, speed and time?
(A) distance x speed = time
(B) time x distance = speed
(C) distance speed x time
(D) none of these
Answer:
(C) distance = speed x time

16. What will be the velocity of athlete if he takes 10 s to cover a distance of 100m?
(A) 100 ms^-1
(B) 10ms^-1
(C) 10 ms^-2
(D) 10 ms^-2
Answer:
(B) 10 ms^-1

17. The rate of decrease In velocity in a certain direction is:
(A) +ve acceleration
(B) – ve acceleration
(C) +ve and – ve acceleration
(D) none of these
Answer:
(B) – ve acceleration

18. The downward slope in velocity-time graph shows:
(A) accelerated motion
(B) equal motion
(C) retarded motion
(D) none of these
Answer:
(C) retarded motion

19. Rajdhanl Express covers a distance of 1384 km in 17 hours. Its average speed will be:
(A) 81.4 ms^-1
(B) 81.4 kms^-1
(C) 81.4 mh^-1
(D) 81.4 kmh^-1
Answer:
(D) 81.4 kmlf1

20. If a car covers a distance of 80 km in 2 hours, then its average speed will be:
(A) 40 kmh^-1
(B) 40 kms^-1
(C) 40 ms^-1
(D) 40 mh^-1
Answer:
(A) 40 kmh^-1

21. Cheetah is the fastest land animal and can achieve a peak velocity of 100 kmh upto a distance less than 500 m. H the cheetah spots his prey at a distance of 100 m, what ¡s the minimum time it will take to get the prey, if the average velocity attained by ¡t ¡s 90 kmh?
(A) 3s
(B) 4s
(C) 5s
(D) 6s
Answer:
(B) 4 s

22. The distance-time graph of a body is parallel to X-axis. It shows that:
(A) body is moving with non-uniform motion
(B) body is moving with accelerated motion
(C) the body is at rest
(D) body is moving with uniform motion
Answer:
(C) the body is at rest

23. By converting the speed of 2 km/h in m/s, we will get:
(A) \(\frac {5}{36}\) m/s
(B) \(\frac {5}{18}\) m/s
(C) \(\frac {5}{9}\)m/s
(D) \(\frac {50}{9}\) m/s
Answer:
\(\frac {5}{9}\)m/s

24. What is the speed of light in:
(A) 3 x 10⁸ m/s
(B) 3 x 10⁷ m/s
(C) 3 x 10¹⁰
(D) 3 x 10⁵ m/S
Answer:
(A) 3 x 10⁸ m/s

25. Which of the following is right for displacement?
(A) it cannot be zero
(B) its magnitude is more than the distance covered by the body
(C) it can be zero
(D) none of these
Answer:
(C) it can be zero

26. The odometer of any vehicle shows
(A) speed
(B) velocity
(C) distance
(D) time
Answer:
(A) speed

27. When an object Is In uniform motion, its path seems to be
(A) straight line
(B) curve
(C) irregular
(D) circular
Answer:
(A) straight line

28. During an experiment, a wireless signal from a spaceship reached ground station in 5 seconds. What Is the distance of spaceship from the ground station?
(A) 9 x 10¹⁰kjn
(B) 9 x 10¹⁰m
(C) 9 x 10¹⁰cm
(D) 9 x 10¹⁰ mm
Answer:
(B) 9 x 10¹⁰ m

29. A train is moving with a speed of 90 kmh^-1, on applying brakes, it produces a retardation of 0.5 ms². How much distance will it cover before coming to rest?
(A) 62.5 m
(B) 6.25 m
(C) 625 m
(D) 625 km
Answer:
(C) 625 m

30. A trolley while going down an inclined plane has an acceleration of 0.2 ms^-2. What will be its velocity In 3s after the starts?
(A) 0.6 ms^-1
(B) 0.6 cms^-1
(C) 0.6 mms^-1
(D) 0.6 ms^-2
Answer:
(A) 0.6 ms^-1

31. A racing car has a uniform acceleration of 4 ms^-2. What distance will it cover in lOs after the
start from rest?
(A) 20m
(B) 200m
(C) 2000m
(D) 2m
Answer:
(B) 200 m

32. A stone is thrown in vertically upward direction with a velocity of S ms⁴. If the acceleration of the stone during its motion is 10 ms², in downward direction, what will be the height aftalned by the stone?
(A) 12.5m
(B) 125m
(C) 1.25m
(D) 0.125m
Answer:
(C) 1.25 m

33. A stone Is thrown in vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 10 mr2, in downward direction, how much time will ¡t take to reach at maximum height?
(A) 0.5s
(B) 5s
(C) 5 min
(D) 0.5 mm
Answer:
(A) 0.5 s

34. An artificial satellite Is moving in a circular orbit of radius of 42,250 km. What will be Its speed, if it takes 24 hours to revolve round the earth?
(A) 3.07 ms^-1
(B) 3.07 kmh^-1
(C) 3.07 kms^-1
(D) 3.07 cms^-1
Answer:
(C) 307 kms^-1

35. An athlete completes one round of a circular track of radius r in time t, then its speed will be:
(A) 2πrt
(B) \(\frac{2 \pi r}{t}\)
(C) \(\frac{t}{2 \pi r}\)
(D) none of these
Answer:
(B) \(\frac{2 \pi r}{t}\)

36. The proper relation among s, u, t land a is:
(A) s = \(\frac{2 \pi r}{t}\)
(B) s = u + at
(C) s = ut + \(\frac {1}{2}\) at²
(D) s = uv + at
Answer:
(C) s = ut + \(\frac {1}{2}\) at²

37. The relation among y, u, a and t ¡s:
(A) v = u – at
(B) v = u + at
(C) v² = u² – at
(D) v² = u² + at
Answer:
(B) v = u + at

38. The proper relation between y, u, a and s is:
(A) v²= u² – 2as
(B) v² = 2as – u²
(C) v² – u² = 2as
(D) v² + u² = as
Answer:
(C) v² – u² = 2as

Haryana State Board HBSE 9th Class Maths Solutions Chapter 10 वृत्त Ex 10.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 10 वृत्त Exercise 10.5

प्रश्न 1.
आकृति में, केंद्र O वाले एक वृत्त पर तीन बिंदु A, B और C इस प्रकार हैं कि ∠BOC = 30° तथा ∠AOB = 60° है। यदि चाप ABC के अतिरिक्त वृत्त पर D एक बिंदु है, तो ∠ADC ज्ञात कीजिए।

हल :
प्रश्नानुसार,
∠BOC = 30° …..(i)
∠AOB = 60° ….(ii)
समीकरण (i) तथा (ii) को जोड़ने पर,
∠AOB + ∠BOC = 60° + 30°
∠AOC = 90°
चाप AC केंद्र पर ∠AOC तथा शेष वृत्त पर ∠ADC बनाती है।
∴ ∠AOC = 2∠ADC
90° = 2∠ADC
∠ADC = \(\frac{90^{\circ}}{2}\)
∠ADC = 45° उत्तर

प्रश्न 2.
किसी वृत्त की एक जीवा वृत्त की त्रिज्या के बराबर है। जीवा द्वारा लघु चाप के किसी बिंदु पर अंतरित कोण ज्ञात कीजिए तथा दीर्घ चाप के किसी बिंदु पर भी अंतरित कोण ज्ञात कीजिए।
हल :

माना PQ, एक जीवा है। OP तथा OQ को मिलाया।
दिया है : PQ = OP = OQ [∵ जीवा = त्रिज्या]
∴ ΔOPQ एक समबाहु त्रिभुज है।
∠POQ = 60°
∴ क्योंकि चाप PBQ वृत्त के केंद्र पर परावर्तन ∠POQ = 360° – 60° = 300° बनाती है तथा ∠PBQ वृत्त के लघु चाप पर बनाती है।
∠PBQ = \(\frac{1}{2}\) × परावर्तन ∠POQ
= \(\frac{1}{2}\) × 300° = 150°
इसी प्रकार, ∠PAQ = \(\frac{1}{2}\)(∠POQ) [∵ वृत्त के शेष भाग पर बना कोण केंद्र पर बने कोण का आधा होता है]
∴ = \(\frac{1}{2}\) × 60° = 30°
अतः जीवा द्वारा लघु चाप पर बनाया गया कोण 150° तथा दीर्घ चाप पर बनाया गया कोण = 30°

प्रश्न 3.
आकृति में, ∠PQR = 100° है, जहां P,Qतथा R, केंद्र O वाले एक वृत्त पर स्थित बिंदु हैं। ∠OPR ज्ञात कीजिए।

हल :
क्योंकि एक चाप द्वारा केंद्र पर बनाया गया कोण, उसी चाप द्वारा बनाई गई शेष परिधि के बिंदु पर कोण का दोगुना होता है।
∴ परावर्तन ∠POR = 2∠PQR
या परावर्तन ∠POR = 2 × 100° = 200°
∠POR = 360° – 200° = 160°
ΔOPR में,
या OP = OR [एक ही वृत्त की त्रिज्याएं]
∠OPR = ∠ORP [बराबर भुजाओं के सम्मुख कोण बराबर होते हैं]
तथा ∠POR = 160° [प्रमाणित]
अतः ∠OPR = ∠ORP = \(\frac{1}{2}\)(180° – 160°)
= \(\frac{1}{2}\) × 20° = 10°
∠OPR = 10° उत्तर

प्रश्न 4.
आकृति में, ∠ABC = 69° और ∠ACB = 31° हो, तो ∠BDC ज्ञात कीजिए। [B.S.E.H. March, 2020]

हल :
ΔABC में,
∠A + ∠ABC + ∠ACB = 180°
∠A + 69° + 31° = 180° (∵ ∠ABC = 69° व ∠ACB = 319)
या ∠A = 180° – 69° – 31° = 180° – 100° = 80°
परंतु ∠A = ∠BDC [एक ही वृत्तखंड में बने कोण]
अतः
∠BDC = 80° उत्तर

प्रश्न 5.
आकृति में, एक वृत्त पर A, B, C और D चार बिंदु हैं। AC और BD एक बिंद्र E पर इस प्रकार प्रतिच्छेद करते हैं कि ∠BEC = 130° तथा ∠ECD = 20° है। ∠BAC ज्ञात कीजिए।

हल:
∠BEC + ∠DEC = 180° (रैखिक युग्म]
⇒ 130° + ∠DEC = 180° [∵ ∠BEC = 130°]
या ∠DEC = 180° – 130°
या ∠DEC = 50°
अब ΔDEC में,
∠DEC + ∠DCE + ∠D = 180°
या 50° + 20° + ∠D = 180° [त्रिभुज के कोणों का योग]
या ∠D = 180° – 50° – 200
∠D = 180° – 70° = 110°
परंतु ∠BAC = ∠D [एक ही वृत्तखंड में बने कोण]
अतः ∠BAC = 110° उत्तर

प्रश्न 6.
ABCD एक चक्रीय चतुर्भुज है जिसके विकर्ण एक बिंदु E पर प्रतिच्छेद करते हैं। यदि ∠DBC = 70° और ∠BAC = 30° हो, तो ∠BCD ज्ञात कीजिए। पुनः यदि AB = BC हो, तो ∠ECD ज्ञात कीजिए।
हल :

∠BDC = ∠BAC [एक ही वृत्तखंड में बने कोण]
परंतु ∠BAC = 30°
∠BDC = 300
अब ΔBCD में,
या ∠BDC + ∠DBC + ∠BCD = 180° [त्रिभुज के कोणों का योग]
या 30° + 70° + ∠BCD = 180° [∵ ∠DBC = 70°, ∠BDC = 30°]
∠BCD = 180° – 30° – 70° = 180° – 100° = 80°
यदि AB = BC, तब ∠BCA = ∠BAC = 30° [बराबर भुजाओं के सम्मुख कोण]
अब ∠ECD = ∠BCD – ∠BCE
= 80° – 30° = 50°
अतः ∠BCD = 80° व ∠ECD = 50° उत्तर

प्रश्न 7.
यदि एक चक्रीय चतुर्भुज के विकर्ण उसके शीर्षों से होकर जाने वाले वृत्त के व्यास हों, तो सिद्ध कीजिए कि वह एक आयत है।
हल :

दिया है : एक चक्रीय चतुर्भुज ABCD जिसमें विकर्ण AC और BD एक ही वृत्त के व्यास हैं।
सिद्ध करना है : चतुर्भुज ABCD एक आयत है।
प्रमाण : ∵ एक वृत्त की सभी त्रिज्याएं बराबर होती हैं।
∴ OA = OB = OC = OD
या OA = OC = \(\frac{1}{2}\)AC
OB = OD = \(\frac{1}{2}\)BD
या AC = BD
∴ चक्रीय चतुर्भुज ABCD के विकर्ण बराबर हैं तथा एक-दूसरे को समद्विभाजित करते हैं।
अतः चक्रीय चतुर्भुज ABCD एक आयत है। [इति सिद्धम]

प्रश्न 8.
यदि एक समलंब की असमांतर भुजाएं बराबर हों, तो सिद्ध कीजिए कि वह चक्रीय है।
हल :

दिया है : एक समलंब ABCD में AB || CD तथा AD = BC.
सिद्ध करना है : समलंब ABCD चक्रीय है।
रचना : DL ⊥ AB व CM ⊥ AB खींचो।
प्रमाण : ΔADL व ABCM में,
AD = BC [दिया है]
DL = CM [दिया है]
∠DLA = ∠CMB [प्रत्येक 90°]
∴ ΔADL ≅ ΔBCM [समकोण-कर्ण-भुजा सर्वांगसमता]
⇒ ∠A = ∠B [सर्वांगसम त्रिभुजों के संगत भाग]
∵ AB || CD [दिया है]
∠ADC + ∠A = 180° [तियर्क रेखा के एक ओर बने कोण]
या ∠ADC + ∠B = 180° [∵ ∠A = ∠B]
अतः ABCD एक चक्रीय समलंब है।
[इति सिद्धम]

प्रश्न 9.
दो वृत्त दो बिंदुओं B और C पर प्रतिच्छेद करते हैं। B से जाने वाले दो रेखाखंड ABD और PBQ वृत्तों को A, D और P, Q पर क्रमशः प्रतिच्छेद करते हुए खींचे गए हैं। (आकृति अनुसार)। सिद्ध कीजिए कि ∠ACP = ∠QCD है।

हल :
दिया है : दो वृत्त दो बिंदुओं B व C पर प्रतिच्छेद करते हैं। B से जाने वाले दो रेखाखंड ABD व PBQ वृत्तों को क्रमशः A,D और P, Q पर प्रतिच्छेद करते हैं।
सिद्ध करना है : ∠ACP = ∠QCD
प्रमाण : क्योंकि एक ही वृत्तखंड के कोण बराबर होते हैं।
∴ ∠ACP = ∠ABP ……(i)
∠QCD = ∠QBD ……(ii)
∠ABP = ∠QBD ……(iii)
[शीर्षाभिमुख कोण]
∴ समीकरण (i), (ii) एवं (iii) से,
∠ACP = ∠QCD [इति सिद्धम]

प्रश्न 10.
यदि किसी त्रिभुज की दो भुजाओं को व्यास मानकर वृत्त खींचे जाएं, तो सिद्ध कीजिए कि इन वृत्तों का प्रतिच्छेद बिंदु तीसरी भुजा पर स्थित है।
हल :

दिया है : ΔABC की भुजाओं AB तथा AC को व्यास लेते हुए दो वृत्त बनाए गए हैं। वृत्त एक-दूसरे को A तथा D बिंदुओं पर प्रतिच्छेदित करते हैं।
सिद्ध करना है : D, BC पर स्थित है।
रचना : A तथा D को मिलाओ।
प्रमाण : ∵ AB तथा AC दो वृत्तों के व्यास हैं। [दिया है]
∴ ∠ADB = 90° [अर्द्धवृत्त में बना कोण]
तथा ∠ADC = 90° [अर्द्धवृत्त में बना कोण]
जोड़ने पर,
∠ADB + ∠ADC = 90° + 90° = 180° या.
या BDC एक सीधी रेखा है।
अतः D, BC पर स्थित है। [इति सिद्धम]

प्रश्न 11.
उभयनिष्ठ कर्ण AC वाले दो समकोण त्रिभुज ABC और ADC हैं। सिद्ध कीजिए कि ∠CAD = ∠CBD है।
हल :

दिया है : कर्ण AC वाले दो समकोण त्रिभुज ABC और ADC हैं अर्थात ∠B = ∠D = 90°
सिद्ध करना है : ∠CAD = ∠CBD.
रचना : बिंदु B और D को मिलाओ। प्रमाण:
∠ABC = 90° …(i) [दिया है]
∠ADC = 90° …(ii) [दिया है]
समीकरण (i) तथा (ii) को जोड़ने पर
∠ABC + ∠ADC = 180°
अतः ABCD एक चक्रीय चतुर्भुज है।
∴ ∠CAD = ∠CBD [एक ही वृत्तखंड में बने कोण]
[इति सिद्धम]

प्रश्न 12.
सिद्ध कीजिए कि चक्रीय समांतर चतुर्भुज आयत होता है।
हल :

दिया है : एक चक्रीय समांतर चतुर्भुज ABCD है।
सिद्ध करना है : ABCD एक आयत है।
प्रमाण : क्योंकि ABCD एक चक्रीय समांतर चतुर्भुज है।
∴ ∠A + ∠C = 180° [चक्रीय चतुर्भुज के सम्मुख कोण]
∠A = ∠C (समांतर चतुर्भुज के सम्मुख कोण]
∴ ∠A = ∠C = 90°
यदि किसी समांतर चतुर्भुज का एक कोण 90° हो जाए तो वह आयत होता है।
∴ चक्रीय समांतर चतुर्भुज ABCD एक आयत है। [इति सिद्धम]

Haryana State Board HBSE 9th Class Science Important Questions Chapter 14 Natural Resources Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 14 Natural Resources

Very Short-Answer Type Questions

Question 1.
What are natural resources?
Answer:
Useful substances available in nature are called natural resources.

Question 2.
Give examples of natural resources.
Answer:
Air, water, soil, minerals, forests, etc.

Question 3.
What are essential factors for life ?
Answer:
Ambient temperature, water and food.

Question 4.
What is called lithosphere ?
Answer:
The outermost layer of earth is called lithosphere.

Question 5.
How much area of earth is covered with water ?
Answer:
75%.

Question 6.
What is called hydrosphere?
Answer:
The part of earth where water is available is called hydrosphere.

Question 7.
Write the definition of atmosphere.
Answer:
Earth is covered with an envelope of air, this envelope of air is called atmosphere.

Question 8.
What is called biosphere ?
Answer:
Integeration of atmosphere, lithosphere and hydrosphere which makes the life possible, is called biosphere.

Question 9.
Name the inorganic components of biosphere.
Answer:
Air, water, soil.

Question 10.
What is called air ?
Answer:
The mixture of various gases, water vapours, dust particles is called air.

Question 11.
How much percentage of C02 is found on Venus and Mars ?
Answer:
95-97%.

Question 12.
is oxygen is available at Venus and Mars ?
Answer:
No.

Question 13.
How much percentage of C02 is in earth atmosphere ?
Answer:
0.033%.

Question 14.
Is atmosphere there on the surface of moon ?
Answer:
No.

Question 15.
What are called winds ?
Answer:
Blowing air is called wind or wave.

Question 16.
What is called air pollution ?
Answer:
Addition of undesirable substances in air is called air pollution.

Question 17.
Write the names of waste air pollutants.
Answer:
Waste gases (CO₂, nitrogen and oxides of sulphur, hydrocarbon, etc.) water vapours, dust particles, etc.

Question 18.
What is called fog ?
Answer:
In winter season, when water is also condensed with air, then it is called fog.

Question 19.
Write the names of the diseases caused by air pollution.
Answer:
Respiratory diseases, cancer, heart diseases, allergy.

Question 20.
Mainly how many kinds of water are ?
Answer:
Two, saline and fresh water (non-saline).

Question 21.
What is called the outermost layer of earth ?
Answer:
Earth-crust.

Question 22.
How is soil formed ?
Answer:
By the breaking of rocks.

Question 23.
What is called humus ?
Answer:
Substance prepared after the decomposition of dead decayed animals is called humus.

Question 24.
What is called soil pollution ?
Answer:
Addition of undesirable substances in soil is called soil pollution.

Question 25.
Who is most helpful in preventing soil erosion ?
Answer:
Trees and plants.

Question 26.
How is underground water obtained ?
Answer:
By digging wells.

Question 27.
How much percentage of nitrogen gas is in the earth atmosphere ?
Answer:
78%

Question 28.
Mainly which gas is responsible for the greenhouse effect ?
Answer:
CO₂.

Question 29.
What is the main reason of depletion of Ozone Layer?
Answer:
Chloro Fluoro Carbon (CFC)

Short-Answer Type Questions

Question 1.
What are called natural resources ?
Answer:
Many useful substances are found in nature which are called natural resources, e.g., air, water, soil, forests, minerals and living organisms.

Question 2.
What is called atmosphere ? What is the importance of atmosphere for organisms ?
Answer:
An unseen envelope is there around the earth which is called atmosphere. Atmosphere is extended upto 40 km height. 99% of the total air is found in this region. A little volume of air is found at the height of 1000 km from the earth. Water vapours and dust particles are found only near the surface of earth. Above the height of 10-12 km water vapours are not found. Oxygen, the breath of life is present in atmosphere. Ozone layer protects from ultraviolet radiations of the sun.

Question 3.
Write two uses of oxygen present in atmosphere.
Answer:
Following are the uses of oxygen:
(i) Oxygen is used to get energy by breaking the glucose molecules.
(ii) Oxygen is used in combustion and carbon dioxide is released.

Question 4.
Write the names of methods for the fixation of carbon dioxide.
Answer:
Following are the two methods for the fixation of carbon dioxide:
(i) Formation of glucose by green plants in photosynthesis.
(ii) To form the shells from dissolved carbonates in the water by aquatic animals.

Question 5.
On which factor blowing of winds is dependent in coastal areas ?
Answer:
We know that land gets heated much factor and cool down than water. During daytime the air above the land is heated in sunlight and rises up and there creates a low pressure. The air over sea being at a higher pressure moves to area of low pressure. Thus, during daytime winds blow from the sea to land while during night position becomes reverse.

Question 6.
Besides unequal heating of the winds which are other factors which influence these winds ?
Answer:
Unequally heated winds blow on the earth to unequal heating of the atmosphere, but following are the two more factors :
(i) The rotation of earth around the sun and around its own axis.
(ii) Location of mountains in the path of these winds.

Question 7.
How does rain occur ? What is called hailing ?
Answer:
During daytime, water of different water-bodies and flora and fauna evaporates and goes into air. Heated air carry water vapours with it where due to cooling of air, water vapours condense and become droplets. These droplets become bigger and heavier. These droplets come on to the earth in the form of rain. Sometimes, droplets by cooling down become solid ice which are called hail and the rain of hail is called hailing.

Question 8.
What is called air pollution ? Describe its factors.
Answer:
Air Pollution:
Addition of undesirable particles in air makes it harmful to organisms is called air pollution. Following are the natural factors of air pollution:
(i) Gases and toxic particles released oy tne volcano activities.
(ii) Storms.
(iii) Rising of smoke and gases from the forest fires.
(iv) Releasing of methane gas from marshy lands.
(v) Decomposition of plants and animals and pollen grains etc.

Following are the artificial factors of air pollution:
(i) Smoke of vehicles.
(ii) Burning of fuels.
(iii) Industrial smokes and vaporised substances.
(iv) Metallurgical activities
(v) Extraction of minerals
(vi) Pesticides
(vii) Released radio active radiations at the time of originating atomic energy.

Question 9.
What is smog ? Explain its ill effects.
Answer:
Smog:
Dust and smoke particles on combining with water vapours from smog. Toxic gases from houses, industries and toxic gases with the burning of fuels of vehicles are also added in it. Smoke and dust + Fog and toxic gases → Smog Poisonous gases such as sulphur dioxide (S02) or peroxiacetil nitrate (PAN) are added in smog. The smog does not rise high in the atmosphere but on the ocean it makes a form of cloud. It forms commonly in winter season.

Ill Effects:
Following are the ill effects of smog:
(i) It hampers in transportation because long distance things are not visible.
(ii) Creates burning sensation in eye, nose and throat.
(iii) It is harmful to all these organisms, in whom it can easily enter by respiration. Increases the incidence of asthma and heart disorders.
(iv) It affects the growth and development of plants.

Question 10.
What is called acid rain ?
Answer:
Acidic gases in the atmosphere, such as, sulphur dioxide, nitrogen dioxide, hydrogen chloride, etc. adding with droplets made acids, rain of these acids is called acid rain. Acid rain is harmful to monuments, rocks, as well as to fauna and flora.

Question 11.
What is water pollution ? Give reasons of water pollution.
Answer:
Water pollution:
Addition of undesirable substances in water is called water pollution. Following are the reasons of water pollution :
(i) Release of industrial waste and chemical substances etc.
(ii) Agriculture waste such as fertilizers, pesticides, insecticides, weedicides, etc.
(iii) Domestic waste and sewage excreta, such as, garbage, rubbish, polythene, etc.
(iv) Thermal waste such as release of hot water from thermal plants.
(v) Radioscopy wastes such as waste of radioactive substances.
(vi) Social-religious wastes such as festival wastes, by throwing dead bodies of men and animals in rivers.
(vii) The ejected waste, excreta, dead animals from natural resources such as mining, storms, wind stroms due to such type of disasters.

Question 12.
Soil is a natural resources how ?
Answer:
Soil is the uppermost layer of the earth’s crust which is formed by the breaking up of the rocks. Soil is an important natural resources which is essential for the development of life. Our prime requirements of life such as food, clothing and shelter all are obtained from plants that grow in soil or from animals which are dependent upon these plants. Therefore, soil is a precious natural resources for us.

Question 13.
Write the usefulness of humus.
Answer:
Following is the usefulness of humus:
(i) Mixing with humus soil becomes porous which helps air and water to enter in the soil.
(ii) Required nutrients for plants are found in humus.
(iii) Several organisms get nutrients from humus such as earthworm.
(iv) By mixing humus in sandy soil, water bearing capacity of soil increases.
(v) By mixing humus in sandy soil, soil erosion decreases.

Question 14.
Write four common sources of soil pollution.
Answer:
Following are the four common sources of soil pollution:
(i) Excess use of fertilizers in soil.
(ii) Excess spray of insecticides and pesticides on the soil to protect crops.
(iii) To mix the kitchen waste and other garbage of house.
(iv) Mixing of industrial undesired wastes in soil.

Question 15.
What do you know about water cycle ?
Answer:
Ponds, lakes, rivers and oceans are the different water bodies of water on land. Due to water cycle, water of these water bodies evaporates and make water vapours. These water vapours make clouds. On cooling, condensed clouds make droplets. In the form of rain water falls down again on earth surface. This cycling of water is called ‘water cycle’.

Question 16.
In which substances nitrogen is found ?
Answer:
Nitrogen is mainly found in protein, nucleic acid, DNA, RNA, Vitamins, etc. Vitamins are also found in alkaloids and urea.

Question 17.
How does nitrogen deficiency create in soil ? What measure should be adopted for this kind of soil?
Answer:
Content of nitrogen decreases when the soil is utilized again and again to obtain crop yield. Nitrogen enriched fertilizers are mixed in soil to eradicate this deficiency. Most prevailing fertilizer is ammonium nitrate which is prepared from ammonia gas and nitric acid.

Question 18.
How do green plants prepare starch ?
Answer:
Plants get carbon from the atmosphere in the form of carbon dioxide. By getting energy from the sunlight, plants combine this carbon

These are the carbohydrates and fats which provide energy to the living organisms by which all activities of living beings are completed.

Question 19.
Write two benefits of greenhouse effect.
Answer:
The two benefits of greenhouse effect are following:
(i) The temperature of atmosphere increases by absorption of infrared radiations by greenhouse effect and the life remain smooth in cold climate also.
(ii) The snow covered mountains melt due to greenhouse effect and there is water in rivers throughout the year by this reason.

Question 20.
In which form, the oxygen is found on earth ?
Answer:
21 % oxygen is found in our environment. The oxygen is present in the compound form in carbon dioxide. The oxygen is also found in oxides of metals and silicons, carbonates, sulphates, nitrate and in other minerals. The oxygen is also present in biomolecules like carbohydrates, proteins, nucleic acid and fats.

Question 21.
What will be happen if the ozone layer disappear from atmosphere ?
Answer:
At a height of 15 km above the earth’s surface ozone (O₃) layer exists. It grows thicker with the increase in height. It is thickest at the height of 23 km. This layer absorbs the ultraviolet radiation that is present in the sunlight. This ultraviolet light develops skin cancer in the living beings. Therefore, if ozone layer gets disappeared from the atmosphere, the protection envelop will come to an end. The harmful ultraviolet radiation after, reaching to us will produce skin cancer to us and other organisms. Thus, the stratosphere will run with a huge loss. Along with, ultraviolet radiation will become the cause of diseases like glucoma and heat immolation. Also, the division of cells in plants will stop.

Essay Type Questions

Question 1.
How does water-cycle In nature accomplish?
Answer:

Water is continuously flow in environment. This is called hydrological cycle. In water, hydrogen is the basic element that circulates in the form of water compound water constantly. Evaporation from water bodies (ocean, sea, lakes and rivers) and from the skin of the living organisms and it goes on mingling into the air. Alongwith the water absorbed by the plants from the underground too transpires through stomata of leaves and gathers into the atmosphere. Thus, by different in sources the water mingled in atmosphere in form of water vapours. Water vapours present in atmosphere changes in rain drops by condensation at low temperature. These causes raining or hailing and in this way water again comes on earth (in soil and ponds). Thus water cycle in continues in biosphere.

Question 2.
Describe briefly the nitrogen cycle.
Answer:
Proteins are the compounds of nitrogen. They are the important physical part of plants and animals. Due to the deficiency of protein growth of plants and animals stops because protein is responsible for physical growth ¡n living beings.

Animals consume their food from plants. Also, it contains protein. When the plants and animals becomes dead and decay then some saprophytic bacteria and fungi turns them into the ammonia salt. These ammonia-salts are turned into nitrates by nitrifying bacteria. Some of its part is absorbed by the root rains of roots tips of plants which is further converted into protein. Remaining part of the nitrate gets transformed into nitrogen by the micro-organisms (nitrifying bacteria). Thus, nitrogen once again appears in the atmosphere.

The bacteria that stabilise cosmic nitrogen are found in the soil and other root-glands of the legume plants. These bacteria convert the nitrogen in free state into its compounds, which are absorbed by the plants. Some part of atmospheric nitrogen remains fixed into the earth by thunder lightning. Fertilizers, manufactured in the factories are also a fixed form of nitrogen. This nitrogen gets absorbed by the plants and passes through the nitrogen food network. Thus, this nitrogen cycle continues in nature.

Question 3.
How does Carbon cycle gets completed in the biosphere ? Give explanation.
Answer:
Carbon is an important physical element for all living beings. In living beings it is found in the form of carbohydrates, fats, protein and nucleic acids. Its transfer takes place through food chain. In atmosphere, it is found in the form of CO₂. In oceans it is found in the form of biocarbon and carbonates. The green plants through photosynthesis process consume CO₂and prepare carbohydrates. These carbohydrates then get transferred to animals. Decomposes again set them free into the environment. All living beings produce CO₂ during respiration. Burning of fuels consisting ef carbon to increase the quantity of CO₂ in the atmosphere. Eruption of volcanoes too is the main source of receiving CO₂.

Question 4.
Draw a labelled diagram of Oxygen-cycle in nature.
And.
Oxygen in nature, is nearly the 21 % part of the gaseous constituents in atmosphere. It gets dissolved into the water stores, which is utilised by the aquatic animals in breathing process. In bio-bodies it enters and exits in the form of compounds of CO₂ and H₂O. In biosphere the main source of providing oxygen are the green plants. Two stages of oxygen cycle are as under:

1. Intrusion of oxygen in the atmosphere: Carbon dioxide reached from different sources is utilised by the green plants to undergo photosynthesis process. In photosynthesis, oxygen is produced as a coproduct.

2. Expulsion of oxygen from the atmosphere: Atmospheric oxygen is utilised by all animals in respiration and they release in form of CO₂ in the atmosphere. In same way the oxygen dissolved in the water bodies is utilised by the aquatic creatures and they release CO₂ as by product underwater. Atmospheric oxygen is utilised in burning of coal, wood and carbon other fuels and is released in the form of CO₂ in the atmosphere. In this way, a balance is created in the atmosphere for intrusion and expulsion of oxygen.

Practical Work

Experiment 1.
To study the conduction current waves.

Procedure:
Fix a candle in a beaker and light it up, not take a glowing incense stick near to the mouth of the beaker and observe it:
(1) On bringing the glowing incense stick near to the mouth of beaker the smoke is pushed back, since hot air escapes out.
(2) On keeping the glowing incense stick above the candle flame the smoke directly rises up because the flame of the candle heats up the upper air quickly and thus, let is escape right above.
(3) While keeping the glowing incense stick around the burning candle at differeht positions the smoke is repelled because the entire air from inside the beaker escapes out on heating up.

Experiment 2.
To study the process of formation of fog.

Procedure:
(1) Take 5-10 ml of water in a plastic bottle and tight-up the cap.
(2) Now, shake it well and let escape little quantity of smoke form a glowing incense stick into the bottle by unscrewing the cap. Now place the cap once again to the mouth of the bottle.
(3) Take the bottle in between both palms and squeeze the bottle hard. After a short while, again squeeze the bottle and make your observation. On the water vapours saturated and with an increase in the pressure, fog gets formed inside the bottle. The fog is disappeared during keeping the bottle for a short time still. The pressure increases by squeezing and keep it in motion. During the collection of water vapour around the smoke particles, the fog is formed.

Experiment 3.
Making of a hygrometer and note down the time of maximum rainfall in your city and state.

Procedure:
Take a glass funnel with its mouth equal to that of the size of the mouth of the beaker and fit it to the mouth of the beaker, mark the wall of the beaker and draw a scale.

Hygrometry:
As soon as the rain starts keep the hygrometer at such a place, where water may fall down into the funnel without any obstacle. Note down the reading as soon as the rain stops. With the help of this device, note down the quantity of rainfall for a fixed period (months) of everyday. Thus, we can find out the day which received the maximum rainfall. To note down the maximum rainfall of a particular state the quantity of rainfall is noted down of metropolitan cities or big cities and thus average rainfall is known.

Quick Review of the Chapter

1. The gas found in our atmosphere different to the atmosphere of Venus and Mars is :
(A) CO₂
(B) O₂
(C) Ar
(D) N₂
Answer:
(B) O₂

2. The cause of air blow s:
(A) pressure difference
(B) temperature difference
(C) density difference
(D) surface difference
Answer:
(A) pressure difference

3. Which of the following pollute the air ?
(A) burning of fossil fuels
(B) movement of vehicles
(C) smoke of factories
(D) all of the above
Answer:
(D) all of the above

4. The main cause (s) of soil erosion is (are)
(A) floods
(B) heavy winds
(C) deforestation
(D) all of the above
Answer:
(D) all of the above

5. Earthworms are helpful ¡n:
(A) making nitrogen
(B) making humas
(C) decreasing the fertily
(D) all of the above
Answer:
(B) making humas

6. Pattern of rainfall decided on which factor?
(A) wind pattern
(B) forest pattern
(C) evaparation pattern
(D) people’s life style pattern
Answer:
(A) wind pattern

7. Soil erosion can be controlled by:
(A) growing grass
(B) anti air
(C) levelling fields
(D) all of the above
Answer:
(D) all of th above

8. The protective cover ¡n our environment (atmosphere) is of:
(A) O₂
(B) CO₂
(C) O₃
(D) N₂
Answer:
(C) O₃

9. The water vapours when passes through cold regions, condense in the form of water drops which is called:
(A) snow
(B) water
(C) sky
(D) cloud
Answer:
(D) cloud

10. The human activity/activities which increase the quantity of CO₂ in atmosphere is/are:
(A) burning of fuels
(B) use of petrol in vehicles
(C) respiration process
(D) all of the above
Answer:
(D) all of the above

11. The increase in atmospheric temperature by the absorption of infrared radiations by CO₂ reflected from the surface of earth is called:
(A) ozone hole
(B) greenhouse effect
(C) ultraviolet effect
(D) absorption effect
Answer:
(B) greenhouse effect

12. The outermost layer of earth is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) ozonosphere
Answer:
(B) lithosphere

13. The cover of air spread around the earth is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) ozonosphere
Answer:
(C) atmosphere

14. The part of earth which contains water is called:
(A) hydrosphere
(B) lithosphere
(C) atmosphere
(D) biosphere
Answer:
(A) hydrosphere

15. The non-living factors of biosphere are:
(A) air
(B) water
(C) soil
(D) all of the above
Answer:
(D) all of the above

16. The air pollutant (s) ¡s (are):
(A) CO
(B) NO₂
(C) SO₂
(D) all of the above
Answer:
(D) all of the above

17. Gas produced in combustion is:
(A) oxygen
(B) nitrogen
(C) carbon dioxide
(D) helium
Answer:
(C) carbon dioxide

18. What is the formula of ozone?
(A) O₂
(B) O₃
(C) O₄
(D) O₅
Answer:
(B) O₃

19. Necessary for photo-synthesis:
(A) sunlight
(B) chlorophil
(C) water and carbon dioxide
(D) all of the above
Answer:
(D) all of the above

20. The substance made of decomposition of dead animals ¡s called:
(A) humus
(B) fertiliser
(C) soil
(D) coal
Answer:
(A) humus

21. The mixing of undesirable materials in soil is called:
(A) water pollution
(B) air pollution
(C) soil pollution
(D) noise pollution
Answer:
(C) soil pollution

22. The percentage of nitrogen in our atmosphere is:
(A) 21 %
(B) 78 %
(C) 0.033 %
(D) 0.1 %
Answer:
(B) 78%

23. The change of nitrogen of atmosphere into nitrates and nitrites by bacteria is called:
(A) oxygen fixation
(B) denitrification
(C) nitrogen fixation
(D) ammonification
Answer:
(C) nitrogen fixation

24. Soil ¡s a natural resource which:
(A) provides necessary nutrients to plants
(B) is compulsory for life and development
(C) provides eating materials
(D) all of the above
Answer:
(D) all of the above

25. CO₂ absorbs:
(A) visible light
(B) infrared radiations
(C) ultraviolet radiations
(D) none of above
Answer:
(B) infrared radiations

26. Oxides of nitrogen and sulphur are present in:
(A) acid rain
(B) basis rain
(C) normal rain
(D) all of the above
Answer:
(A) acid rain

27. Acid rain results due to:
(A) Nitrogen and sulphur oxides in air
(B) Increase in suspended practices in air
(C) Increase in hydrocarbons
(D) Excess C0 in air
Answer:
(A) Nitrogen and sulphur oxides in air

28. Which of the following ¡s wrong regarding causes of water pollution?
(A) Presence of unwanted substances in water bodies
(B) Decrease in amount of dissolved oxygen in water:
(C) Change in temperature of water
(D) Increased no. of aerobic bacteria
Answer:
(D) Increased no. of aerobic bacteria

29. Green House Effect is caused by:
(A) Green plants
(B) mfra-red rays
(C) UV rays
(D) X-rays
Answer:
(C) UV rays

30. Cause of Ozone depletion is:
(A) CO₂
(B) O₂
(C) CFC
(D) N₂
Answer:
(C) CFC