Prasanna

MP Board Class 7th Sanskrit Model Question Paper (आदर्श प्रश्नपत्रम्)

Sanskrit Gadyansh Class 7 प्रश्न 1.
(अ) सही विकल्प चुनकर लिखिए
(क) बदरीनाथधामास्ति
(i) गुजरातराज्ये
(ii) उड़ीसाराज्ये
(iii) कर्नाटकराज्ये
(iv) उत्तराखण्डराज्ये।
उत्तर:
(iv) उत्तराखण्डराज्ये

(ख) सिंह पीडितः आसीत्
(i) पिपासया
(ii) क्षुधया
(iii) ज्वरेण
(iv) शत्रुणा।
उत्तर:
(ii) क्षुधया

(ग) अपदं दूरगामी अस्ति
(i) पत्रम्
(ii) पक्षी
(iii) पशुः
(iv) मनुष्यः।
उत्तर:
(i) पत्रम्

(घ) पृथिव्याः उपग्रहः अस्ति
(i) बुधः
(ii) शनिः
(iii) चन्द्रः
(iv) शुक्रः।
उत्तर:
(iii) चन्द्रः

(ङ) लोकमान्यतिलकेन आरब्धः उत्सवः अस्ति
(i) दीपोत्सवः
(ii) होलिकोत्सवः
(iii) गणेशोत्सवः
(iv) स्वतन्त्रतादिवसोत्सवः।
उत्तर:
(iii) गणेशात्सवः

(ब) दिये गये शब्दों से रिक्त स्थानों की पूर्ति कीजिए
(खड्गं, मूलाधारः, पूर्णिमा, तक्षशिला, शूलपाणिः)
(क) धर्म एव भारतस्य एकतायाः ………….. अस्ति।
(ख) ………. विश्वविख्यातम् अध्ययनकेन्द्रमासीत्।
(ग) त्रिनेत्रधारी न च ……….. ।
(घ) …………. गृहीत्वा युद्धं कुरु।
(ङ) शुक्लपक्षे ………… तिथि भवति।
उत्तर:
(क) मूलाधारः
(ख) तक्षशिला
(ग) शूलपाणिः
(घ) खड्गं
(ङ) पूर्णिमा।

Class 7 Sanskrit Abhyas Prashn Patra प्रश्न 2.
अधोलिखित गद्यांश को पढ़कर प्रश्नों के उत्तर लिखो-

एकस्मिन् पर्वते दुर्मुखः नाम महौजस्वी सिंहः वसति स्म। सः च सदैव बहूनां पशूनां वधं करोति स्म। एकदा सर्वे पशवः सिंहस्य समीपम् अगच्छन् अवदन् च मृगेन्द्र! त्वं सदैव पशूनां वधं कथं करोषि? प्रसीद वयं स्वयं तव भोजनाय प्रतिदिनम् एकैकं पशुं प्रेषयिष्यामः।

(क) पर्वते किं नाम सिंहः प्रतिवसति स्म? (पर्वत पर किस नाम का शेर रहता था?)
उत्तर:
पर्वते दुर्मुखः नाम सिंहः वसति स्म। (पर्वत पर दुर्मुख नाम का शेर रहता था।)

(ख) सः केषां वधं करोति स्म? (वह किनका करता था?)
उत्तर:
स: बहूनां पशूनां वधं करोति स्म। (वह बहुत से पशुओं का वध किया करता था।)

(ग) के सिंहस्य समीपम् अगच्छन्? (कौन शेर के पास आये थे?)
उत्तर:
सर्वे पशवः सिंहस्य समीपम् अगच्छन् स्म। (सभी पशु शेर के पास गये थे।)

(घ) वयं प्रतिदिन किं प्रेषयिष्यामः? (हम सब प्रतिदिन किसको भेजेंगे?)
उत्तर:
वयं प्रतिदिनं एकैकं पशु प्रेषयिष्यामः। (हम सब प्रतिदिन एक-एक पशु को भेजेंगे।)
अथवा

योगेन शरीरं चित्तम् अपि स्वस्थं भवति। “शरीरमाद्यं खलु धर्मसाधनम्” इति प्रसिद्धं वचनम्। स्वस्थ शरीर अध्ययनं सुकरं भवति। तेन चित्तस्य एकाग्रता भवति। कार्ये कौशलं जायते। योगस्य अभ्यासेन अनेके लाभाः सम्भवन्ति।

(क) शरीरं चित्तं न केन स्वस्थं भवति? (शरीर और चित्त किससे स्वस्थ होता है?)
उत्तर:
योगेन शरीरं चित्तं न स्वस्थं भवति। (योग से शरीर और चित्त स्वस्थ होता है।)

(ख) धर्मस्य आद्यं साधनं किम् अस्ति? (धर्म का आदि साधन क्या है?)
उत्तर:
धर्मस्य आद्यं साधनं शरीरं अस्ति। (धर्म का आदि साधन शरीर है।)

(ग) स्वस्थे शरीरे किं सुकरं भवति? (स्वस्थ शरीर में क्या आसान होता है?)
उत्तर:
स्वस्थ शरीरे अध्ययनं सुकरं भवति। (स्वस्थ शरीर में अध्ययन आसान होता है।)

(घ) कार्ये किं जायते? (कार्य करने पर क्या पैदा होता है?)
उत्तर:
कार्ये कौशलं जायते। (कार्य करने पर कुशलता पैदा होती है।)

Class 7 Sanskrit Apathit Gadyansh Mcq प्रश्न 3.
अधोलिखित पद्यांश को पढ़कर प्रश्नों के उत्तर लिखो-
माता शत्रुः पिता वैरी, येन बालो न पाठितः।
न शोभते सभामध्ये, हंसमध्ये बको यथा।।

(क) यया बालो न पाठितः सा माता कीदृशी? (जिस माता के द्वारा बालक को शिक्षित नहीं कराया जाता बह माता कैसी होती है?)
उत्तर:
सा माता शत्रुः अस्ति। (वह माता शत्रु होती है)

(ख) येन बालो न पाठितः स पिता कीदृशः? (जिस पिता के द्वारा बालक को शिक्षित नहीं कराया जाता वह पिता कैसा होता है?)
उत्तर:
सः पिता वैरी अस्ति। (वह पिता बैरी होता है।)

(ग) यः न पठितवान् स: कुत्र न शोभते? (जो अशिक्षित है वह कहाँ शोभा नहीं देता?)
उत्तर:
सः सभामध्ये न शोभते। (वह सभा के बीच शोभा नहीं देता।)

(घ) यः न पठितवान् सः कथं न शोभते? (जो अशिक्षित है वह कैसे शोभा नहीं देता?)
उत्तर:
सः हंसमध्ये बको यथा न शोभते। (वह हंसों के बीच बगुले की तरह शोभा नहीं देता।)
अथवा

वरमेको गुणी पुत्रो न मूर्ख-शतान्यपि।
एकश्चन्द्रस्तमो हन्ति, न च तारागणा अपि॥

(क) कः पुत्रः वरम् अस्ति? (कौन-सा पुत्र श्रेष्ठ होता है?)
उत्तर:
गुणी पुत्रः वरम् अस्ति। (गुणवान पुत्र श्रेष्ठ होता है।)

(ख) कति मूर्खपुत्राः न वराणि? (कितने मूर्ख पुत्र अच्छे नहीं होते हैं?)
उत्तर:
शतानि मूर्खपुत्राः न वराणि। (सौ मूर्ख पुत्र अच्छे नहीं होते हैं।)

(ग) कः तमो हन्ति? (कौन अन्धकार को नष्ट कर देता है?)
उत्तर:
चन्द्रः तमो हन्ति। (चन्द्रमा अन्धकार को नष्ट कर देता है।)

(घ) क तमो न घ्नन्ति? (कौन अन्धकार को नष्ट नहीं कर पाते हैं?)
उत्तर:
ताराः तमो न घ्नन्ति। (तारे अन्धकार को नष्ट नहीं कर पाते हैं।)

Class 7th Sanskrit Paper प्रश्न 4.
(अ) पाठ्य पुस्तक से कण्ठस्थ किया हुआ एक श्लोक लिखो जो इस प्रश्न-पत्र में न हो।
उत्तर:
विद्या विवादाय धनं मदाय, शक्तिः परेषां परिपीडनाय।
खलस्य साधोः विपरीतमेतत्, ज्ञानाय दानाय च रक्षणाय।।

(ब) श्लोक को पूरा करो
जलबिन्दुं …………… क्रमशः …………. घटः।
स ………… सर्वविद्यानां ……….. च धनस्य च ॥
उत्तर:
निपातेन, पूर्यते, हेतु, धर्मस्य।

(स) पाठ्य पुस्तक से कण्ठस्थ की हुई एक सूक्ति लिखो।
उत्तर:
आचार: परमो धर्मः।

Class 7 Sanskrit Paper प्रश्न 5.
(अ) अधोलिखित प्रश्नों के उत्तर एक शब्द में लिखो
(क) प्रयत्नेन के विश्वप्रियाः? (प्रयत्न करने से कौन विश्वप्रिय बन गये?)
उत्तर:
भारतीयाः

(ख) के उत्सवप्रियाः भवन्ति? (कौन उत्सवप्रिय होते हैं?)
उत्तर:
जनाः

(ग) सिक्खानां दशमः गुरुः क आसीत्? (सिक्खों के दसवें गुरु कौन थे?)
उत्तर:
गुरुगोविन्दसिंहः

(घ) सुप्तोऽपि नेत्रे कः न निमीलयति? (सोने पर भी दोनों नेत्रों को कौन बन्द नहीं करती है?)
उत्तर:
मत्स्यः

(ङ) कस्य सहायतां प्रभुः करोति? (प्रभु किसकी सहायता करते हैं?)
उत्तर:
श्रमशीलस्य

(च) केन कार्याणि सिद्धयन्ति? (किससे कार्य सिद्ध हो जाते हैं)
उत्तर:
उद्यमेन

(ब) अधोलिखित प्रश्नों के उत्तर एक वाक्य में लिखो
(क) मठानि किमर्थ स्थापितानि? (मठों की स्थापना किसलिए की गई?)
उत्तर:
धर्मरक्षार्थं वेदान्ततत्त्वानां प्रचारार्थम् च मठानि स्थापितानि। (धर्म की रक्षा और वेदान्त तत्वों के प्रचार के लिए मठों की स्थापना की गई।)

(ख) कौ द्वौ पक्षौ भवतः? (कौन से दो पक्ष होते हैं?)
उत्तर:
शुक्लपक्षः कृष्णपक्ष: च इति दौ पक्षौ भवतः। (शुक्ल पक्ष और कृष्ण पक्ष नामक दो पक्ष होते हैं।)

(ग) भास्कराचार्यः किं प्रतिपादितवान्? (भास्कराचार्य ने क्या प्रतिपादित किया?)
उत्तर:
भास्कराचार्यः गुरुत्वाकर्षणसिद्धांत π (पै) इति गणितचिह्नस्य मानं त्रैराशिकनियमादीन् प्रतिपादितवान्। (भास्कराचार्य ने गुरुत्वाकर्षण सिद्धान्त, गणित चिह्न π (पाई) का मान, त्रैराशिक नियम आदि का प्रतिपादन किया।)

(घ) परोपकारः किमर्थ भवति? (परोपकार किसके लिए होता है)
उत्तर:
परोपकारः पुण्याय भवति। (परोपकार पुण्य के लिए होता है।)

(ङ) बालचरस्य प्रथमा प्रतिज्ञा का अस्ति? (बालचरस्य की पहली प्रतिज्ञा क्या है?)
उत्तर:
‘ईश्वरं स्वदेशं प्रति च कर्त्तव्यपालनं’ बालचरस्य प्रथमा प्रतिज्ञा अस्ति। (‘ईश्वर और अपने देश के प्रति कर्त्तव्य का पालन करना’ बालचर की पहली प्रतिज्ञा है।)

(च) सुलभा कस्य मूर्तिम् अपश्यत्? (सुलभा ने किसकी मूर्ति को देखा?)
उत्तर:
सुलभा मुनेः पतञ्जले: मूर्तिम् अपश्यत्। (सुलभा मे मुनि पतञ्जलि की मूर्ति को देखा।)

MP Board Sanskrit Class 7th प्रश्न 6.
(अ) अधोलिखित शब्दों के रूप तीनों वचनों में लिखो-
(क) लेखनी-पञ्चमी विभक्ति
(ख) सर्व-तृतीया विभक्ति (पुल्लिङ्ग)
(ग) मधु-चतुर्थी विभक्ति।
उत्तर:

(ब) अधोलिखित के धातुरूप निर्देशानुसार तीनों वचनों में लिखो-
(क) पठ्-लोट्लकारः (आज्ञार्थकः), उत्तमपुरुषः।
(ख) गम् (गच्छ्)-विधिलिङ्लकारः, प्रथमपुरुषः।
(ग) वन्द-(आत्मनेपद) लट्लकारः, मध्यमपुरुषः।
उत्तर:

(स) अधोलिखित में रेखांकित शब्दों के कारक नाम लिखो-
(क) खगः वृक्षे निवसति।
(ख) रामः पठति।
(ग) हिमालयात् गङ्गा प्रभवति।
(घ) राजा ब्राह्मणाय धनं ददाति।
उत्तर:
(क) अधिकरणकारकम् (सप्तमी विभक्तिः)
(ख) कर्तृकारकम् (प्रथमा विभक्तिः)
(ग) अपादानकारकम् (पञ्चमी विभक्तिः)
(घ) सम्प्रदानकारकम् (चतुर्थी विभक्तिः)

Kaksha Satvin Sanskrit प्रश्न 7.
(अ) अधोलिखित शब्दों के धातु और प्रत्यय अलग करो-
(क) विलिख्य
उत्तर:
विलिख्या = वि (उपसर्गः) + लिख् (धातुः) + य (ल्यप्)

(ख) कृतवान्
उत्तर:
कृतवान् = कृ धातुः + क्तवतु प्रत्ययः

(ग) लिखित्वा
उत्तर:
लिखित्वा = लिख (धातुः) + त्वा (क्त्वा प्रत्ययः)

(घ) क्रीडितः।
उत्तर:
क्रीडितः = क्रीड् धातुः + क्त प्रत्ययः

(ब) अधोलिखित के उपसर्ग अलग करो-
(क) उपकरोति
(ख) अनुधावति
(ग) पराजयते
(घ) उद्भवति।
उत्तर:
(क) उप
(ख) अनु
(ग) परा,
(घ) उत्।

(स) अधोलिखित में से अव्यय चुनकर लिखो-
(क) धेनु
(ख) अतः
(ग) नगरम्
(घ) पुरतः
(ङ) मा।
उत्तर:
(ख) अतः
(घ) पुरतः
(ङ) मा।

Class 7 Sanskrit Sample Paper 2023 With Answers प्रश्न 8.
(अ) अधोलिखित शब्दों की सन्धि विच्छेद करके सन्धि का नाम लिखो-
(क) देवर्षिः
(ख) अजन्तः
(ग) पावकः
(घ) सुबन्तः।
उत्तर:
(क) देवर्षिः = देव + ऋषिः (स्वरसन्धिः)
(ख) अजन्तः = अच् + अनतः (व्यञ्जनसन्धिः)
(ग) पावकः = पौ + अकः (स्वरसन्धिः)
(घ) सुबन्तः = सुप् + अन्तः (व्यञ्जनसन्धिः)

(ब) अधोलिखित शब्दों के समास विग्रह करके समास का नाम लिखो-
(क) चौरभयम्
(ख) पंचवटी
(ग) उपकृष्णम्।
उत्तर:
(क) चौरभयम्-चौराद् भयम् (तत्पुरुषसमासः)
(ख) पंचवटी-पञ्चानां वटानां समाहारः (द्विगुसमास:)
(ग) उपकृष्णम्-कृष्णस्य समीपम् (अव्ययीभावसमासः)

(स) अधोलिखित संख्याओं को संस्कृत में लिखो-
(क) 14
(ख) 18
(ग) 16
उत्तर:
(क) 14-चतुर्दश
(ख) 18-अष्टादश
(ग) 16-षोडश।

Sanskrit Ka Paper 7th Class 9.
अधोलिखित शब्दों से पत्र को पूरा करो-
(भ्रमणार्थं, स्वास्थ्यम्, प्रणामाः, परीक्षा, कुशलिनी)

खजूरीपन्थतः
3 जनवरी, 20……….

पूज्यमातः! …………..
अहम् ईश्वरस्य कृपया ………….. अस्मि।
भवत्याः ………….. कथम् अस्ति?
अहं अस्मिन् मासे ………. गमिष्यामि। आगामिमासे मम ……….. अस्ति। पितृचरणौ वन्दे।

भवत्याः पुत्री
शैलजा

उत्तर:
प्रणामाः, कुशलिनी, स्वास्थ्यम्, भ्रमणार्थम्, परीक्षा।

Sanskrit Prashn Patra प्रश्न 10.
अधोलिखित में से किसी एक विषय पर पाँच वाक्यों में संस्कृत में निबन्ध लिखो
(क) मम विद्यालयः
उत्तर:

1. मम विद्यालयः ‘खाईखेड़ा’ ग्रामे स्थितः अस्ति।
2. विद्यालयस्य भवनम् अतीवसुन्दरम् अस्ति।
3. अहं विद्यालयं गत्वा गुरून् प्रणमामि।
4. विद्यालये एकम् उद्यानम् अपि अस्ति।
5. विद्यालये एक विशालं क्रीडाक्षेत्रम् अस्ति।

(ख) पुस्तकम्
उत्तर:

1. पुस्तकानि मह्यम् अतीव रोचन्ते।
2. पुस्तकानि ज्ञानस्य भण्डारः भवन्ति।
3. पुस्तकानि अस्माकं मित्राणि सन्ति।
4. पुस्तकानां सङ्गति लाभप्रदा भवति।
5. अस्माभिः पुस्तकानि रक्षणीयानि।

(ग) उद्यानम्
उत्तर:

1. उद्यानम् अत्यन्तं रमणीयं भवति।
2. बालकाः उद्यानं क्रीडन्ति।
3. उद्याने तडागः अपि अस्ति।
4. जनाः उद्यानं भ्रमणार्थं गच्छन्ति।
5. खगाः वृक्षेषु निवसन्ति।

(घ) धेनुः।
उत्तर:

1. धेनुः अस्माकं माता अस्ति।
2. धेनूनां विविधाः वर्णाः भवन्ति।
3. धेनुः तृणानि भक्षयति।
4. धेनुः जनेभ्यः मधुरं पयः प्रयच्छति।
5. वयं धेनुं मातृरूपेण पूजयामः।

अथवा
अधोलिखित शब्दों की सहायता से पाठ्य पुस्तक में चित्र देखकर संस्कृत में पाँच वाक्य लिखो-
(धेनुः, कृषकः, गृहाणि, वृक्षौ, क्षेत्रम्)
उत्तर:
(1) गोपालः धेनुम् दुहति।
(2) कृषकः तापं शीतं वृष्टिं सहित्वा कृषिकर्म करोति।
(3) ग्रामे बहूनि गृहाणि सन्ति।
(4) ग्रामे दौ वृक्षौ स्तः।
(5) प्रात:काले कृषकाः स्व-स्व क्षेत्रम् गत्वा कृषिकर्माणि कुर्वन्ति।

MP Board Class 7th Sanskrit Solutions

In this article, we will share MP Board Class 10th Science Solutions Chapter 6 Life Processes Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 6 Life Processes

MP Board Class 10th Science Chapter 6 Intext Questions

Class 10th Science Chapter 6 Intext Questions Page No. 95

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multi cellular organisms like humans?
Answer:
Diffusion is insufficient to meet the oxygen requirement of multicellular organisms like humans because multicellular organisms have complex body designs. Moreover, all their cells may not be in direct contact with the surrounding environment.

MP Board Class 10 Science Chapter 6 Question 2.
What criteria do we use to decide whether something is alive?
Answer:
We tend to think of some sort of movement, either growth-related or not, as common evidence for being alive.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Food, water and oxygen are outside raw materials mostly used by an organism. Depending on the complexity of the organism its requirement varies from organism to organism.

MP Board Class 10th Science Chapter 6 Question 4.
What processes would you consider essential for maintaining life?
Answer:
Nutrition, Respiration, Transportation of materials into the body and Excretion are the processes we consider essential for maintaining life.

Class 10th Science Chapter 6 Intext Questions Page No. 101

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic Nutrition	Heterotrophic Nutrition
It is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This mate­rial is taken in the form of carbon dioxide and water which is converted into carbohydrates in the presence of sunlight and chlorophyll	This involves the intake of complex material prepared by other convert organisms
Chlorophyll is necessary.	Chlorophyll is absent.

Class 10th Science Chapter 6 Question Answer Question 2.
Where do plants get each of the raw materials required for photosynthesis?
Answer:
The following raw materials are required for photosynthesis:

• The CO₂ enters from the atmosphere through stomata.
• Water is absorbed from the soil by the plant roots.
• Sunlight, an important component for manufacture food, is absorbed by the chlorophyll and other green parts of the plants.

Question 3.
What is the role of the acid in our stomach?
Answer:

1. The hydrochloric acid reacts an acidic medium which facilitates the action of the enzyme pepsin.
2. Acid in the stomach kills micro organisms.

Ch 6 Science Class 10 Question 4.
What is the function of digestive enzymes?
Answer:
Digestive enzymes such as amylase, lipase, pepsin, trypsin etc. help in the breaking down of complex food particles into simpler ones. These simple particles can be easily absorbed by the blood and thus, transported to all the cells of the body.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
The small intestine is designed in such a way that the digested end products are easily absorbed into the body. The innermost lining of the small intestine has many finger-like foldings called villi which increase the surface area for absorption. The villi are richly supplied with blood capillaries and a large lymph vessel which takes the absorbed food to all the cells of the body.

Class 10th Science Chapter 6 Intext Questions Page No. 105

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
Terrestrial organisms take up oxygen from the atmosphere whereas aquatic animals need to utilize oxygen present in the water. Air contains more O₂ as compared to water. Since, the content of O₂ in air is high, the terrestrial animals do not have to breathe faster to get more oxygen. Therefore, unlike aquatic animals, terrestrial animals do not show various adaptations for better gaseous exchange.

10th Science 6th Lesson Question Answer Question 2.
What are the different ways in which glucose is oxidised to provide energy in various organisms?
Answer:
Glucose is first broken down in the cell cytoplasm into a three carbon molecules called pyruvate. Pyruvate is further broken down by different ways to provide energy. The breakdown of glucose by different pathways can be illustrated as follows,

Break-down of glucose by various pathways.
In yeast and human muscle cells, the breakdown of pyruvate occurs in the absence of oxygen whereas in mitochondria, the breakdown of pyruvate occurs in the presence of oxygen.

Science Class 10 Chapter 6 Question 3.
How is oxygen and carbon dioxide transported in human beings?
Answer:
In human beings, respiratory pigments take up oxygen from the air in the lungs and carry it to tissues which are deficient in oxygen before releasing it. The respiratory pigment is haemoglobin which has a very high affinity for oxygen. This pigment is present in the red blood corpuscles. Carbon dioxide is more soluble in water than oxygen is and hence is mostly transport the dissolved from in our blood.

Question 4.
How are the lungs designed in human beings to maximize the area for exchange of gases?
Answer:
Within the lungs, the passage divides into smaller and smaller tubes which finally terminate in ballon like structures which are called alveoli. The alveoli provide a surface where the exchange of gases can take place.

Class 10th Science Chapter 6 Intext Questions Page No. 110

Question 1.
What are the components of the transport system in human beings? What are the functions of these components?
Answer:
The components of the transport system in human beings are the heart, blood and blood vessels and lymph:

• Heart pumps oxygenated blood throughout the body. It takes deoxygenated blood from the various body parts and sends this impure blood to the lungs for oxygenation.
• Being a fluid connective tissue, blood helps in the transport of oxygen, nutrients, CO₂ and nitrogenous wastes.
• The blood vessels (arteries, veins and capillaries) carry blood either away from the heart to various organs or from various organs, back to the heart.

Class 10 Science Chapter 6 Exercise Question Answer Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
Because mammals and birds require more energy. Hence there must be separation of oxygenated blood and deoxygenated blood. By this these organisms get sufficient oxygen and helps to maintain their body temperature.

Question 3.
What are the components of the transport system in highly organised plants?
Answer:
Xylem and Phloem are the two main types of conducting tissues. In highly organised plants, xylem conducts water and minerals obtained from the soil (via roots) to the rest of the plant. Phloem transports food materials from the leaves to different parts of the plant body.

Class 10th Chapter 6 Science Question 4.
How are water and minerals transported in plants?
Answer:
In xylem tissue, vessels and tracheids of the roots, stems and leaves are inter connected to form a continuous system of water conducting channels reaching all parts of the plant. At the roots, cells in contact with the soil actively take up ions. This creates a difference in the concentration of these ions between the root and the soil. Water therefore moves into the root from the soil to eliminate this difference. This means that there is steady movement of water into root xylem, creating a column of water that is steadily pushed upwards.

Question 5.
How is food transported in plants?
Answer:
The transport of soluble products of photosynthesis is called translocation and it occurs in the part of the vascular tissue known as phloem. Besides the products of photosynthesis, the phloem transports amino acids and other substances. These substances are specially delivered to the storage organs of roots, fruits and seeds and to growing organs. The translocation of food and other substances takes place in the sieve tubes with the help of adjacent companion cells both in upward and downward directions.

Class 10th Science Chapter 6 Intext Questions Page No. 112

Question 1.
Describe the structure and functioning of nephrons.
Answer:
Nephrons are the basic units of kidneys. Each kidney possesses large number of nephrons, approximately 1-1.5 million. The main components of the nephron are glomerulus, Bowman’s capsule and a long renal tubule.

Structure of a nephron:

Structure of a nephron

Functioning of a nephron:

• The blood with metabolic waste enters the kidney through the renal artery, which branches into many capillaries associated with glomerulus.
• The water and solute are drained to the nephron at Bowman’s capsule.
• In the proximal tubule, some substances such as amino acids, glucose and salts are selectively reabsorbed and unwanted molecules are added in the urine.
• The filtrate then moves down into the loop of Henle, where more water is absorbed. The amount of water reabsorbed depends on how much excess water is present in the body and on, how much of dissolved waste is to be excreted.
• From here, the filtrate moves into the distal tubule and finally reach to the collecting duct Collecting duct collects urine from many nephrons.
• The urine formed in each kidney enters a long tube called ureter. From ureter, it gets transported to the urinary bladder and then into the urethra.

MP Board Solution Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Plants can get rid of excess water by transpiration. For other wastes, plants use the fact that many of their tissues consist of dead cells, and that they can even lose some parts such as leaves. Many plant waste products are stored incellular vacuoles waste products may be stored in leaves that fall off. Other waste products are stored as resins and gums especially in old xylem. Plants also excrete some waste substances into the soil around them.

Question 3.
How is the amount of urine produced regulated?
Answer:
The amount of urine produced depends on the amount of excess water and dissolved wastes present in the body. Some other factors such as habitat of an organism and hormone such as Antidiuretic hormone (ADH) also regulates the amount of urine produced.

MP Board Class 10th Science Chapter 6 NCERT Textbook Exercises

Question 1.
The kidneys in human beings are a part of the system for:
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) In human beings, the kidneys are a part of the system for excretion.

Class X Science Chapter 6 Question Answer Question 2.
The xylem in plants are responsible for:
(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen
Answer:
(a) In a plant, the xylem is responsible for transport of water.

Question 3.
The autotrophic mode of nutrition requires:
(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above.
Answer:
(d) The autotrophic mode of nutrition requires carbon dioxide, water, chlorophyll and sunlight, all the four components.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in:
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus
Answer:
(b) The breakdown of pyruvate to give carbon dioxide, water and energy takes place in mitochondria.

Transportation In Human Beings Class 10 Questions And Answers Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
Fats are present in the intestine in the form of large globules which makes it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The enzymes present in the wall of the small intestine converts fats into fatty acids and glycerol.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
Saliva is secreted by the salivary Glands, located around tongue. It moistens the food for easy swallowing. It contains a digestive enzyme – amyjase, which breaks down bulky starch into sugar. So, sometimes it is advised to consume less water during a meal.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by-products?
Answer:
Autotrophic nutrition is accomplished by the process of photosynthesis. Carbon dioxide, water, chlorophyll pigment and sunlight are the necessary conditions required for autotrophiq nutrition. Carbohydrates (food) and 02 are the by-products of photosynthesis.

MP Board Class 10th Solution Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:

Aerobic respiration	Anaerobic respiration
i) This take place in presence of oxygen	No oxygen is
ii) It takes place in cytoplasm and mitochondria.	it takes place only in cytoplasm
iii) By this process more energy is released.	Less energy is released by this process.

Question 9.

How are the alveoli designed to maximise the exchange of gases?
Answer:
The alveoli are the small hollow structures present in the lungs. The walls of the alveoli consist of extensive network of blood vessels. Each lung contains 300-350 million alveoli, making it a total of approximately 700 millions in both the lungs. The alveolar surface when spread out covers about 80m² area. This large surface area makes the gaseous exchange more efficient in alveoli and capillaries.

Question 10.
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Haemoglobin is the respiratory pigment that transports oxygen to the body cells for cellular respiration. Therefore, deficiency of haemoglobin in blood can affect the oxygen supplying capacity of blood. This can lead to deficiency of oxygen in the body cells. It can also lead to a disease called anemia.

Class 10th Science Chapter Number 6 Question 11.
Describe double circulation in human beings. Why is it necessary?
Answer:
Oxygen rich-blood from the lungs comes to the thin walled upper chamber of the heart on the left, the left atrium. The left atrium relaxes when it is collecting this blood. It then contracts while the next chamber, the left ventricle, relaxes, so that the blood is transferred to it. When the muscular left ventricle contracts in its turn, the blood is pumped out to th body. De-oxygenated blood comes from the body to the upper chamber on the right, the right atrium, as it relaxes. As the right atrium contracts, the corresponding lower chamber, the right ventricle, dilates. This transfer blood to the right ventricle, which in turn pumps it to the lungs for oxygenation.

Blood goes through the heart twice during each cycle in other vertebrates. This is known as double circulation. The separation of the right side and the left side of the heart is useful to keep oxygenated and deoxygenated blood from mixing.

Importance of double circulation:

The separation of oxygenated and de-oxygenated blood allows a more efficient supply of oxygen to every single cells. This efficient system of oxygen supply is very useful in warm-blooded animals such as human beings. As we know, warm-blooded animals have to maintain a constant body temperature. Thus, the circulatory system of humans becomes more efficient because of the double circulation.

Schematic sectional view of the human heart.

Question 12.
What are the differences between the transport of materials in xylem and phloem?
Answer:
(i) Xylem tissue helps in the transport of water and minerals.

• Phloem tissue helps in the transport of food.

(ii) Water is transported upwards from roots to all other plant parts.

• Food is transported in both upward and downward directions. Transport in xylem occurs with the help of simple physical forces such as transpiration pull.

(iii) Transport of water and minerals do not require energy in the form of ATP.

• Transport of food in phloem requires energy in the form of ATP.

Ncert Class 10 Science Ch 6 Extra Questions Question 13.
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:
Structure of Alveoli

1. Alveoli are tiny balloon-like structures present inside the lungs.
2. The walls of the alveoli are one cell thick and it contains an extensive network of blood capillaries.

Functions:

1. The exchange of O₂ and CO₂ takes place between the blood of the capillaries that surround the alveoli and the gases present in the alveoli.
2. Alveoli are the site of gaseous exchange.

Structure Nephrons

1. Nephrons are tubular structures present inside the kidneys.
2. Nephrons are made of glomerulus, bowman’s capsule, and a long renal tube. It also contains a cluster of thin walled capillaries.

Functions:

(i) The blood enters the kidneys through the renal artery which branches into many capillaries in the glomerulus. The water and solute are transferred to the nephron at Bowman’s capsule. Then, the filtrate moves through the proximal tubule and then down into the loop of henle. From henle’s loop, filtrate passes into the distal tubule and then to the collecting duct. The collecting duct collects the urine from many nephrons and passes it to the ureter. During the ’ flow of filtrate some substances such as glucose, amino acid and water are selectively re-absorbed.

MP Board Class 10th Science Chapter 6 Additional Questions

MP Board Class 10th Science Chapter 6 Multiple Choice Questions

Question 1.
Which out of the following can prepare their own food:
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(d) Autotrophs

Question 2.
Which of the following type includes plant as majority of its population?
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(d) Autotrophs

Class 10 Chapter 6 Science Solutions Question 3.
Which of the given organisms can feed on plants and animals as well?
(a) Carnivores
(b) Omnivores
(c) Herbivores
(d) Autotrophs
Answer:
(b) Omnivores

Question 4.
Amylase is secreated by:
(a) Pancreas
(b) Stomach
(c) Small intestine
(d) Oesophagus
Answer:
(a) Pancreas

Question 5.
of the following metal is associated with chlorophyll?
(a) Aluminium
(b) Iron
(c) Potassium
(d) Calcium
Answer:
(b) Iron

Class 10 Science Chapter 5 Life Processes Question Answer Question 6.
Which of the following metals are helpful for cellular level transport?
(a) AI – Zn
(b) Fe
(c) Na – K
(d) Calcium
Answer:
(c) Na – K

Question 7.
Where does water get absorbed in body?
(a) Stomach
(b) Food canal
(c) Large intestine
(d) Small intestine
Answer:
(c) Large intestine

Question 8.
The lungs in human beings helps in:
(a) Excretion
(b) Nutrition
(c) Respiration
(d) Transportation
Answer:
(c) Respiration

Question 9.
The liver in human beings helps in:
(a) Excretion
(b) Digestion
(c) Respiration
(d) Transportation
Answer:
(b) Digestion

Class 10 Science Chapter 6 Explanation Question 10.
Bile is originated from in human digestive system.
(a) Pancreas
(b) Liver
(c) Kidney
(d) Stomach
Answer:
(b) Liver

Question 11.
Villi are find inside
(a) Brain
(b) Stomach
(c) Small intestine
(d) Oesophagus
Ans.
(c) Small intestine

Question 12.
The decomposition of carbohydrate is a process of:
(a) Esterification
(b) Hydrogenation
(c) Oxidation
(d) Emulsification
Answer:
(c) Oxidation

Life Processes Class 10 Question 13.
Which one of the following organism use air dissolved in water for respiration?
(a) Amoeba
(b) Sheep
(c) Lion
(d) Leech
Answer:
(a) Amoeba

Question 14.
Energy is released and stored in the form of ATP during:
(a) Excretion
(b) Nutrition
(c) Respiration
(d) Transportation
Answer:
(c) Respiration

Question 15.
Nephron are unit of kidney, it works for:
(a) Nutrition
(b) Respiration
(c) Excretion
(d) Transportation
Answer:
(c) Excretion

Question 16.
The phloem in plants are responsible for:
(a) Transport of water
(b) Transport of food
(c) Transport of amino acids
(d) Transport of oxygen
Answer:
(b) Transport of food

Breakdown Of Glucose By Various Pathways Question 17.
The autotrophic mode of nutrition requires:
(а) Carbon dioxide and water
(b) Chlorophyll and sunlight
(c) Carbohydrate
(d) (a) and (b)
Answer:
(d) (a) and (b)

Question 18.
Energy generated during cellular level of metabolism is stored in:
(a) Cytoplasm
(b) Mitochondria
(c) Chloroplast
(d) Nucleus
Answer:
(b) Mitochondria

Question 19.
During expiration, the lungs are:
(a) Arched
(b) Flattened s
(c) Perforated
(d) None of these
Answer:
(a) Arched

Question 20.
The correct pathway of blood in double circulatory system is:
(a) atria → ventricles → arteries → veins
(b) atria → veins → arteries
(c) veins → arteries → atria
(d) veins → ventricles → atria → arteries
Answer:
(a) atria → ventricles → arteries → veins

MP Board Class 10th Science Chapter 6 Very Short Answer Type Questions

Question 1.
Name one process which cannot be seen in non living things in comparison to living beings.
Answer:
Reproduction.

Activity 6.1 Class 10 Science Question 2.
Najne two pigments which absorb sunlight.
Answer:
Carotenoids and chlorophyll.

Question 3.
Which life process synthesise chemical energy and turn it into heat energy?
Answer:
Nutrition and respiration.

Question 4.
Name some carbohydrate rich food.
Answer:
Rice and wheat.

Question 5.
Name raw material required for photosynthesis.
Answer:
Carbon dioxide, water, chlorophyll rich living cell, sunlight.

10th Science Drawing Question 6.
In which form, the food is stored in leaves?
Answer:
Starch.

Question 7.
What is the colour of iodine-starch complex?
Answer:
Blue-Purple.

Question 8.
During blood circulation in humans, in one cycle how many times blood moves to heart?
Answer:
Two times.

Question 9.
How many chambers do a human heart have?
Answer:
Four.

Question 10.
What is the term for process of taking food in the body?
Answer:
Ingestion.

Question 11.
Which gland secretes amylase enzyme?
Answer:
Salivary gland.

Question 12.
What is the function of large intestine during digestion?
Answer:
Absorption of water.

Question 13.
How do aquatic plants and animal get oxygen for photosynthesis?
Answer:
Aquatic plants and animals obtain oxygen through the process of diffusion.

Question 14.
Where does light reaction and dark reaction of photosynthesis occur?
Answer:
In the grana thylakoids of chloroplasts, light reaction occurs while dark reaction occur in the stroma of chloroplasts.

Activity 6.3 Class 10 Science Question 15.
Which wavelength of light is best absorbed by chlorophyll?
Answer:
Red colour wavelength.

Question 16.
Which products formed during light reaction in photosynthesis process are used by dark reaction?
Answer:
NADPH and ATP.

Question 17.
What is the function of thylakoid membranes in chloroplast?
Answer:
It provides large surface area for light absorption.

Question 18.
Write down the full form of the following:

(a) ATP
(b) NADP

Answer:

(a) ATP : Adenosine triphosphate.
(b) NADP : Nicotinamide adenine dinucleotide phosphate.

Question 19.
Define excretion.
Answer:
Excretion is a biological process by which an organism gets rid of metabolic toxic wastes from its body.

Question 20.
Name the toxic wastes obtained as by products of metabolism.
Answer:
These wastes are nitrogenous materials i.e., ammonia, urea and uric acid, carbon dioxide, inorganic salts, excess of water.

Question 21.
What is osmoregulation?
Answer:
It is a process that maintains the amount of water and ionic balance in the body.

Question 22.
Where urea is formed in humans?
Answer:
Urea is made in the liver via the ornithine cycle by combining the ammonia made by deamination with carbon dioxide made from respiration.

Question 23.
What are the major excretory products of plants?
Answer:
Oxygen, water, carbon dioxide, latex, gums, resins, excessive salts, calcium oxalate and other toxic substances are excretory organs of plants.

Question 24.
Name the excretory organs of earthworm.
Answer:
Nephridia.

MP Board Class 10th Science Chapter 6 Short Answer Type Questions

Question 1.
What is gout? How it affects life?
Answer:
In humans, the high concentrations of uric acid in the blood cause uric acid crystals to precipitate in the kidneys and joints. This can cause huge pain and swelling in the joints, particularly in the big toes and disturb the movement of body. .

Question 2.
List the excretory system organs in human beings.
Answer:
The excretory system in humans consists of:

• a pair of kidneys
• a pair of ureter
• a urinary bladder
• a urethra

Question 3.
Explain unit of kidney.
Answer:
Nephron is the structural and functional unit of kidney. Each kidney of the pair contains millions of nephrons.

Question 4.
Which organ in human beings is related to homeostasis?
Answer:
Kidney is also concerned with homeostasis since, it carries out osmoregulation of body fluids and controls the pH of the blood.

Question 5.
How does excretion takes place in segmented worms, like earthworms?
Answer:
Segmented worms, such as earthworms, produce urea that is excreted through long tubules called nephridia.

Question 6.
What are the excretory organs of insects like grasshoppers?
Answer:
Malpighian tubules.

Question 7.
Why does uric acid is the best nitrogenous waste product for insects, reptiles and birds?
Answer:
Uric acid has low solubility and does conserve water in insects, reptiles and birds.

Question 8.
Name the two possible treatments during chronic renal (kidney) failure.
Answer:
Hemodialysis and kidney transplant.

Question 9.
What is the main excretory product in Amoeba and jellyfish?
Answer:
Ammonia (NH₃).

Question 10.
What are the two main functions of kidneys?
Answer:
The two main functions of kidneys are:

1. To remove toxic metabolic waste products (for example, urea, uric acid, ammonia, salts etc.) from the blood.
2. To regulate the blood pH, blood water and salt content, blood osmotic pressure, blood pressure (homeostasis).

Question 11.
Why is photosynthesis important to the global world?
Answer:

(a) In photosynthesis, solar energy (sunlight) is converted to chemical energy and is stored in plants as starch.
(b) Plants also store excess sugar by synthesis of starch.
(c) Heterotrophs, including humans, may completely or partially consume plants for fuel and raw materials.
(d) Photosynthesis is responsible for the presence of oxygen in our atmosphere.
(e) Each year, photosynthesis synthesizes approx. 160 billion metric tons of carbohydrate.

MP Board Class 10th Science Chapter 6 Long Answer Type Questions

Question 1.
What is photosynthesis? Explain in detail.
Answer:
The process by which green plants make their own food from carbon- dioxide and water in the presence of sunlight and chlorophyll is called photosynthesis.

During this process, oxygen gas is released. This process can be represented as:

The green plants convert energy of sunlight into chemical energy by making glucose.

The extra glucose formed changes into starch which is stored in leaves. The oxygen released comes from the water.

Question 2.
(a) Define translocation.
Answer:
Transportation of food from photosynthetic parts of the plant to the non-green part of the plant through phloem is known as translocation.

(b) Name the correct substrates for the following enzymes.
(i) Trypsin
(ii) Amylase
(iii) Pepsin
(iv) Lipase
Answer:

S.No.	Enzyme	Substrate
(i)	Trypsin	Proteins (Peptides)
(ii)	Amylase	Starch
(iii)	Pepsin	Proteins
(iv)	Lipase	Emulsified fats

Question 3.
What are the important events that occur during photosynthesis process?
Answer:
The following events occur during this process:

(a) Absorption of light energy by chlorophyll.
(b) Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen. (Light reaction)
(c) Reduction of carbon dioxide to carbohydrates. (Dark reaction by Calvin cycle)

Question 4.
Mention the conditions necessary for photosynthesis. Also, mention the process involved in each of these steps.
Answer:

1. Sunlight: Chlorophyll of the leaves of the plant trap sunlight and converts them into chemical energy during photosynthesis. Plant utilizes visible light only which is made up of 7 colours and green . colour is least absorbed (it reflects green the most – that’s why the leaves appear green).
2. Chlorophyll: Chlorophyll is a green pigment mainly present in the leaves of the plant. There are many types of chlorophyll named a, b, c, d, e and bacteria – chlorophyll. Chlorophyll ‘a’ and chlorophyll ‘b’ are most abundantly present in the nature.
3. Carbon Dioxide: Plants take carbon dioxide through stomata to make glucose (food for the plants).
4. Water: Water is absorbed by the roots of the plants through osmosis. Through xylem, water is transported to all parts of the plant.

MP Board Class 10th Science Chapter 6 Textbook Activities

Class 10 Science Activity 6.1 Page No. 96

1. Take a potted plant with variegated leaves – for example, money plant or crotons.
2. Keep the plant in a dark room for three days so that all the starch gets used up.
3. Now keep the plant in sunlight for about six hours.
4. Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
5. Dip the leaf in boiling water for a few minutes.
6. After this, immerse it in a beaker containing alcohol.
7. Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
8. What happens to the colour of the leaf? What is the colour of the solution?
9. Now dip the leaf in a dilute solution of iodine for a few minutes,
10. Take out the leaf and rinse off the iodine solution.
11. Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning Fig. 6.4.
12. What can you conclude about the presence of starch in various areas of the leaf?

Observations:

1. The green coloured leaf become colourless. The solution of alcohol becomes green in colour as chlorophyll of leaf gets dissolved it upon immersing.
2. leaf On dipping in iodine solution, green areas of leaf turns dark blue whereas colourless part of leaf show no formation of starch.

Class 10 Science Activity 6.2 Page No. 97

1. Take two healthy potted plants which are nearly the same size.
2. Keep them in a dark room for three days.
3. Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants. The potassium hydroxide is used to absorb carbon dioxide.
4. Cover both plants with separate bell-jars as shown in Fig. 6.5.

1. Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
2. Keep the plants in sunlight for about two hours.
3. Pluck a leaf from each plant and check for the presence of starch as in the above activity.
4. Do both the leaves show the presence of the same amount of starch?
5. What can you conclude from this activity?

Observations:

1. No. both leaves do not show the presence of the same amount of starch as starch is produced during the process of photosynthesis utilizing sunlight, chlorophyll and CO₂ In first set up, availability of CO₂, will be less as potassium hydroxide present absorbs CO₂, In the second set up, leaves will have more amount of starch.
2. From this activity, we can conclude that amount of carbon dioxide affects the process and outcome of photosynthesis.

Class 10 Science Activity 6.3 Page No. 99

1. Take 1 mL starch solution (1%) in two test tubes (A and B).
2. Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
3. Now add a few drops of dilute iodine solution to the test tubes.
4. In which test tube do you observe a colour change?
5. What does this indicate about the presence or absence of starch in the two test tubes?
6. What does this tell us about the action of saliva on starch.

Observations:

1. The colour change is observed in test tube B, showing presence of starch. Whereas test tube A will show no colour change as saliva present converts sugar into starch.
2. This tells us about the action of salivary amylase enzymes present in starch. The salivary amylase acts on starch and breakdown into sugar.
   

Class 10 Science Activity 6.4 Page No. 101

1. Take some freshly prepared lime water in a test tube.
2. Blow air through this lime water.
3. Note how long it takes for the lime water to turn milky.
4. Use a syringe or pichkari to pass air through some fresh lime water taken in another test tube Fig. 6.6.

1. Note how long it takes for this lime water to turn milky.
2. What does this tell us about the amount of carbon dioxide in the air that we breathe out?

Observations:

1. Immediately, on blowing air in turns milky,
2. On using syringe / pichkari, much time is taken as amount of CO₂ entering now is restricted.
3. It Shows, the air we breathe out contains higher amount of CO₂.

Class 10 Science Activity 6.5 Page No. 101

1. Take some fruit juice or sugar solution and add some yeast to this. Take this mixture in a test tube fitted with a one-holed cork.
2. Fit the cork with a bent glass tube. Dip the free end of the glass tube into a test tube containing freshly prepared lime water.
3. What change is observed in the lime water arid how long does it take, for this change to occur?
4. What’does this tell us about the pradabts of fermentation?

Observations:

1. Lime water turns milky as CO₂ produced gets mixed with yeast, sugar and afeohol,
2. The products of fermentation.are-CO₂ and alcohol.

Class 10 Science Activity 6.6 Page No. 103

1. Observe fish in an aquarium. They open and close their mouths and the gill-slits (or the operculum which covers the gill-slits) behind their eyes also open and close. Are the timings of the opening and closing of the mouth and gill-slits coordinated In some mariner?
2. Count the number of times the fish opens and closes its mouth in a minute.
3. Compare this to the number of times you breathe in and out in a minute.

Observations:

1. Yes, the timings of opening and closing of mouth and gill slits are coordinated. They open and close alternatively.
2. In dissolved water, availability of oxygen is less so fish breathes at higher rate.

Class 10 Science Activity 6.7 Page No. 105

1. Visit a health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
2. Is it the same for children and adults?
3. Is there any difference In the haemoglobin levels for men and women?
4. Visit a veterinary clinic in your locality. Find out what is the normal range of haemoglobin content in an animal like the buffalo or cow.
5. Is this content different in calves, male and female animals?
6. Compare the difference seen in male and female human beings and animals.
7. How would the difference. If any, be explained?

Observations:

1. The normal haemoglobin content for human male is 13.8-17.2 g/ decilitre and for female is 12.1 – 15.1 g/dl.
2. No. it is not same for children and adults.
3. Male has higher level of Hb level as compared to females.
4. Normal buffalo or low Hb level is 10.4-16.4 g/dl.
5. Yes, Hb content in calves is higher than male and female animals.

Class 10 Science Activity 6.8 Page No. 109

1. Take two small pots of approximately the same size and having the same amount of soil. One should have a plant in it. Place a stick of the same height as the plant in the other pot.
2. Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
3. Cover both sets, one with the plant and the other with the stick, with plastic sheets and place in bright sunlight for half an hour.
4. Do you observe any difference in the two cases?

Observations:

1. Yes, water droplets can be seen on the plastic sheet covering the pot with plant. This is due to condensation of water vapours released during transpiration.

MP Board Class 10th Science Solutions

MP Board Class 8th Science Solutions Chapter 10 Reaching the Age of Adolescence

MP Board Class 8 Science Chapter 10 Question 1.
What is the term used for secretions of endocrine glands responsible for changes taking place in the body?
Answer:
Hormones.

Class 8 Science Chapter 10 MP Board Question 2.
Define adolescence.
Answer:
Adolescence is defined as the period of life (11 to 18 or 19 years of age) when the body undergoes changes leading to reproductive maturity.

MP Boar d Class 8 Science Chapter 7 Reaching The Age Of Adolescence Question 3.
What is menstruation? Explain.
Answer:
Menstruation. In females, the ova begin to mature with the onset of puberty. One ovum matures and is released by the ovaries once in 28 to 30 days. In case, the fertilization does not take place, the released egg and the thickened lining of the uterus along with its blood vessels are shed off. This causes bleeding in woman and is called, menstruation: It occurs once in about 28 to 30 days.

MP Board Class 8th Science Chapter 10 Question 4.
List changes in the body that take place at puberty.
Answer:
The following changes take place in the body at puberty.

1. Sudden increase in height. The bones of the arms and legs elongate and make a person tall.
2. Change in the body shape. In boys, the shoulders become broader and the chests become wider. In girls, the region below the waist becomes wider.
3. Voice change. In boys, the voice becomes deep or hoarse. Generally the girls have high pitched voice.
4. Increased activity of sweat and sebaceous glands glands. Pimples may develop on the face.
5. Development of sex organs. In boys, testes begin to produce sperms. In girls, ovaries enlarge and start releasing matured eggs.
6. Reaching mental, intellectual and emotional maturity. These changes are natural part of growing up.

Class 8th Science Chapter 10 Question Answer Question 5.
Prepare a table having two columns depicting names of endocrine glands and hormones secreted by them.
Answer:
Hormones secreted by Endocrine organs.

Class 8 Science Chapter 10 Reaching The Age Of Adolescence Question 6.
What are sex hormones? Why are they named so? State their function.
Answer:
The hormones which are secreted by the testes and ovaries are called sex hormones. They are so called because they are secreted by the two human sexes of male and female.
Testosterone which is a male hormone causes changes like growth of facial hair in boys.
Oestrogen which is a female hormone makes the breasts develop in girls.

Class 8 Science Chapter 10 Question 7.
Choose the correct option:
(a) Adolescents should be careful about what they eat, because
(i) proper diet develops their brains.
(ii) proper diet is needed for the rapid growth taking place in their body.
(iii) adolescents feel hungry all the time.
(iv) taste buds are well developed in teenagers.
Answer:
(iii) adolescents feel hungry all the time.

(b) Reproductive age in women starts when their
(i) menstruation starts.
(ii) breasts start developing
(iii) body weight increases.
(iv) height increases.
Answer:
(i) menstruation starts.

(c) The right meal for adolescents consists of
(i) Chips, noodles, coke.
(ii) Chapati, dal, vegetables.
(iii) Rice, noodles and burger
(iv) Vegetable cutlets, chips and lemon drink.
Answer:
(ii) Chapati, dal, vegetables.

Class 8 Science Chapter 10 Question Answer Question 8.
Write notes on:
(a) Adam’s apple.
(b) Secondary sexual characters.
(c) Sex determination in the unborn baby.
Answer:
(a) At puberty, the voice box or the larynx begins to grow. The growing voice box in boys can be seen as a protruding part of the throat called Adam’s apple.

(b) In girls, breasts begin to develop at puberty and boys begin to grow facial hair, that is mustaches and beard. As these features help to distinguish the male from the female, they are called secondary sexual characters.

(c) All human beings have 23 pairs of chromosomes (fertilized egg) in the nuclei of their cells. Two chromosomes out of these are the sex chromosomes, named X and Y.

A female has two X chromosomes while a male has one X and one Y chromosome, The unfertilized egg always has one X chromosome. But, testes produce two kinds of sperms. One kind has an X chromosome and the other kind has a Y chromosome. When a sperm containing X chromosome fertilizes the egg, the zygote would have two X chromosomes and develop into a female child. If the sperm contributes a Y chromosome to the egg (ovum) at fertilization, the zygote would develop into a male child.

This is how sex determination in the unborn baby occurs.

Class 8 Science Ch 10 Question 9.
Word game: Use the clues to work out the words.
Across
3. Protruding voice box in boys
4. Glands without ducts
7. Endocrine gland attached to brain
8. Secretion of endocrine glands
9. Pancreatic hormone
10. Female hormone
Down
1. Male hormone
2. Secretes thyroxine
3. Another term for teenage
5. Hormone reaches here through bloodstream
6. Voice box
7. Term for change at adolescence

Answer:

Class 8 Science Reaching The Age Of Adolescence Question Answer Question 10.
The table given shows the data on the likely heights of boys and girls as they grow in age. Draw graphs showing height and age for both boys and girls on the same graph paper. What conclusions can be drawn from these graphs?

Answer:
We conclude that girls are taller than boys at the age of 12 years generally and their height is same at the age of 16 years but some boys gain height and generally become taller than girls.

Reaching the Age of Adolescence NCERT Extended Learning – Activities and Projects

Class 8 Science Chapter 10 Question Answer Pdf Question 1.
Find out from your elder relatives about their awareness of the legal status of early marriage. You yourself may get information on it from your teacher, parents, a doctor or the Internet. Write a two-minute speech explaining why early marriage is not good for the couple.
Answer:
The legal marriageable age for girls is 18 years and for boys it is 21 years. Before this age marriage is considered illegal. Early marriage is not good for couples because they are not prepared physically and mentally for the responsibilities of married life. A girl attains sexual maturity at age of about eighteen years. Before that her reproductive system is not fully prepared to bear a baby. Similarly boys are not capable of running the household as most of the boys at this age are pursuing their studies. So, the boys and girls should attain physical and mental maturity before getting married.

Reaching The Age Of Adolescence Class 8 Questions Answers Question 2.
Collect newspaper cuttings and information in magazines about HIV/AIDS. Write a one page article of 15 to 20 sentences on HIV/AIDS.
Answer:
HIV/AIDS is a sexually transmitted diseased (STD) that means this diseases mostly spreads by sexual contacts with an HIV/AIDS infected persons. AIDS stands for Acquired Immune Deficiency Syndrome. It is created by HIV virus. This disease is a fatal disease for which no permanent treatment has been deviced by The scientists. AJDS is actually a disease, which destroys the patients immune’ system slowly. The damage is most of the Permanent and beyond repair. Due to weak immunity the patient easily falls prey to many diseases which eventually kill the person. Prevention is the only cure for HIV/ AIDS. It is advised that one should maintain sexual hygiene and practice safe sexual relationship with a single partner.

Class 8 Science Chapter Number 10 Question Answer Question 3.
In our country, according to a census, there are 882 adolescent females for every 1000 males. Find out:
(a) The concerns of the community regarding this low ratio. Remember that the chance of having a boy or a girl is equal.
(b) What amniocentesis is and how useful this technique is. Why is its use for identification of sex of the unborn child banned in India?
Answer:
(a) The society is quite concerned about this issue. This is going to create a problem for them in future as it will become difficult for the boys to find girls for marriage.

(b) It is used to see, if all the organs of a foetus are developing well. It is, banned in India to use this technique to determine the sex of the unborn child because they kill the female foetus to have baby boys in future. This is called female foeticide. The female foeticide has disturbed the ratio of boys and girls in India.

Science Ch 10 Class 8 Question 4.
Put your ideas together and write a short note on the importance of knowing facts about reproduction.
Answer:
Do yourself.

MP Board Class 8th Science Reaching the Age of Adolescence NCERT Intext Activities and Projects

Activity 10.3
Colled information from magazines or from doctors and prepare a note on the importance of consuming iodised salt. You can also look for this information on the internet.
Answer:
Iodine in the diet prevents the humans from getting goitre. This is due to the fact that the production of thyroxine requires the presence of iodine in water or any food. Thyroxine is produced by thyroid. The lack of iodine in the food can cause goitre.

Activity 10.4
Make a group with your friends. Write down the items of food in your breakfast, launch and dinner you had on the previous day. Identify the items responsible for proper growth. Also identified the junk food items that you consumed the previous day.
Answer:
I have three friends, Shyam Chatterjee a Bengali, Maninder Singh a Punjabi Sikh, and Kunju Ratnam a South Indian, This is the group I am in. I give below the items of food in our breakfast, lunch and dinner.

(i) Myself
Breakfast: Bread (chapati), vegetable and a glass of milk.
Lunch: Chapati, rice, vegetable, dal, curd, an orange or an apple.
Dinner: Chapati, dal, vegetable, kheer, a fruit and a glass of milk.
There is a little change in the food items of my friends. Over and above, my meal is a balanced food and I think this will give me proper growth.

(ii) Shyam Banerjee
Breakfast: Bread, butter and tea.
Lunch: Rice, fish curry and vegetable (fried).
Dinner: Chapati, rice, dal, vegetable and curd.
Up to some extent, the diet is balanced. So, proper growth of the body and mind is possible up to an extent.

(iii) Maninder Singh
Breakfast: Chhole-bhature or puri-vegetable with tea.
Lunch: Chapati, dal, breakfast or meat/fish etc. + curd
Dinner: Chapati, fish/meat/egg curry + curd.
To some extent this meal is a balanced one. Proper growth is possible.

(iv) Kunju Ratnam
Breakfast: Sambhar, bada/idli and tea
Lunch: Rice, sambhar, chatni and curd.
Dinner: Same as Lunch + chapati.
This meal is not balanced one. So, proper growth may not be possible.

Activity 10.6
Collect data on the number of children in your class who exercise regularly and who do not exercise regularly. Did you notice any difference in their fitness and health? Prepare a report on the benefits of regular exercise.
Answer:
Exercise of various kinds keeps our body and mind fit, alert and strong. As it is said that healthy mind live in a healthy body, so exercise is a boon for human life. The children who do regular exercise remain very healthy and fit not only by body but also by mind. And those who do not do any exercise remain sickly and lack in studies also.

Reaching the Age of Adolescence NCERT Additional important Questions

A. Short Answer Type Questions

Science Class 8 Chapter 10 Reaching The Age Of Adolescence Question 1.
What is adolescence?
Answer:
The period of life, when the body undergoes changes, leading to reproductive maturity is called adolescence.

MP Board 8th Class Science Book Solutions Question 2.
Why can humans reproduce only after a certain age?
Answer:
Human reproduce only when they get reproductive maturity. The period of life when the body undergoes changes, leading to reproductive maturity is called adolescence. Adolescence begins around the age of 11 and last up to 18 or 19 years of age.

What Is Menstruation Class 8 Question 3.
Which hormone is responsible for the worried. maturity of ova or sperms?
Answer:
Follicle stimulating hormone is responsible for the maturity of ova or sperms.

MP Board Solution Class 8 Science Question 4.
What is the function of adrenalin
Answer:
It helps the body to adjust to stress age. when one is very angry, embarrased or worried.

Reaching The Age Of Adolescence Question Answer Question 5.
What is a balanced diet?
Answer:
A balanced diet means that the meals responsible for the maturity of ova or include proteins, carbohydrates, fats and sperms. vitamins in requisite proportions.

Class 8 Science Chapter Reaching The Age Of Adolescence Question 6.
Which virus causes AIDS? hormone?
Answer:
Virus named HIV causes AIDS.

B. Long Answer Type Questions

Define Adolescence Short Answer Class 8 Question 7.
Explain the reproductive health 30 days. It is the time when egg is matured
Answer:
The physical and mental well being and released by ovaries. If fertilisation does of an individual is regarded as an not occur, the released egg and the thickened individual s health. To keep the body lining of uterus along with its blood vessels healthy/every human being, at any age, are shed off This causes bieeding in women needs to have a balanced diet. The person which is called menstruation. must also observe personal hygiene and Q10 Write a short note on personal undertake adequate physical exercise. hygiene. During adolescence, however, these, become Answer: Everyone should have a bath at least even more essential as the body is growing.

Class 8 Chapter 10 Science Question 8.
Why some young boys and girls get teenagers because the increased activity of acnes and pimples on their face? sweat glands sometimes makes the body
Ans:
During adolescence the sweat glands smeUy An parts Gf the body should be and oil glands become more active. The Washed and cleaned everyday. If cleanliness sebaceous glands secrete more oil which is not maintained there are chances of blocks the skin and more secretions from catching bacterial infection. Girls should take sweat glands cause infections. Thus over special care of cleanliness during the time secretions from these glands cause acnes and 0f menstrual flow. They should keep track pimples on the faces of young boys and girls. of their menstrual cycle and be prepared

Science Chapter 10 Class 8 Question 9.
What is menstruation? for onset of menstruation.
Answer:
Monthly bleeding in women is called menstruation. It occurs once in about 28 to

Chapter 10 Class 8 Science Question 10.
Write a short note on personal hygiene.
Answer:
Everyone should have a bath at least once everyday. It is more necessary for teenagers because the increased activity of sweat glands sometimes makes the body smelly. All parts of the body should be washed and cleaned everyday. If cleanliness is not maintained there are chances of catching bacterial infection. Girls should take special care of cleanliness during the time of menstrual flow. They should keep track of their menstrual cycle and be prepared for the onset of menstruation.

MP Board Class 8th Science Solutions

In this article, we will share MP Board Class 6th Sanskrit Solutions विविधप्रश्नावलिः 1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 6th Sanskrit Solutions Surbhi विविधप्रश्नावलिः 1

Class 6 Sanskrit Vividh Prashnavali 1 प्रश्न 1.
एक पदेन उत्तरं लिखत (एक शब्द में उत्तर लिखो)
(क) रुचिकरा भाषा का अस्ति? (रुचिकर भाषा कौन-सी है?)
उत्तर:
संस्कृतभाषा

(ख) वयं नेत्राभ्याम् किं कुर्मः? (हम सब दोनों नेत्रों से क्या करते हैं?)
उत्तर:
पश्यामः

(ग) मोहनः रात्रौ कदा शयनं करोति? (मोहन रात को कब सोता है?)
उत्तर:
नववादने

(घ) तृणानां मेलनेन किं भवति? (तिनकों के मेल से क्या होता है?)
उत्तर:
रज्जुनिर्माणं।

Class 6 Sanskrit Chapter 1 MP Board प्रश्न 2.
एकवाक्येन उत्तरं लिखत (एक वाक्य में उत्तर लिंखो)
(क) स्वदेशे कः पूज्यते? (अपने देश में किसकी पूजा होती है?)
उत्तर:
स्वदेशे राजा पूज्यते। (स्वदेश में राजा की पूजा होती है।)

(ख) मोहनः कदा क्रीडति? (मोहन कब खेलता है?)
उत्तर:
मोहनः सायंकाले क्रीडति। (मोहन सायंकाल खेलता है।)

(ग) वयं कया जिघ्रामः? (हम किससे सूंघते हैं?)
उत्तर:
वयं नासिकया जिघ्रामः। (हम नाक से सूंघते हैं।)

(घ) वृक्षः कैः शोभते? (वृक्ष किससे शोभा पाता है?)
उत्तर:
पर्णैः, पुष्पैः, फलैः, शाखाभिः च वृक्ष शोभते। (पत्तों, फूलों, फलों और शाखाओं से वृक्ष शोभा पाता है।)

Ek Paden Uttar In Sanskrit प्रश्न 3.
कोष्ठकशब्दैः अधोलिखितप्रश्नानां उत्तरं लिखत (कोष्ठक के शब्दों से निम्नलिखित प्रश्नों के उत्तर लिखो)
(एतौ, ते, चटके, सः, ताः, नदी)
(क) कः पठति? – (कौन पढ़ता है?)
(ख) कौ धावतः? –  (कौन दो दौड़ते हैं?)
(ग) के रक्षन्ति? – (कौन रक्षा करते हैं?)
(घ) के कूजतः? – (कौन दो कूजती हैं?)
(ङ) का प्रवहति? – (कौन बहती है?)
(च) काः भ्रमन्ति? – (कौन घूमते हैं?)
उत्तर:
(क) सः पठति। – (वह पढ़ता है।)
(ख) एतौ धावतः। – (ये दोनों दौड़ते हैं।)
(ग) ताः रक्षन्ति। – (ये सब रक्षा करती हैं।)
(घ) चटके कूजतः। – (दो चिड़ियाँ कूजती हैं।)
(ङ) नदी प्रवहति। – (नदी बहती है।)
(च) ते भ्रमन्ति। – (वे सब घूमते हैं।)

Class 6th Sanskrit Surbhi प्रश्न 4.
श्लोकान् पूरयत (श्लोकों की पूर्ति करो)
(क) विदेशेषु धनं विद्या, …………….।
………….. , शीलं सर्वत्र वै धनम्॥
(ख) विद्वत्वं च नृपत्वं च, ………..।
………… , विद्वान सर्वत्र पूज्यते।
(ग) विद्या ददाति विनयं …………….।
पात्रत्वाद् धनमाप्नोति …………….॥
(घ) अलसस्य कुतो विद्या, …………….।
अधनस्य कुतो मित्रम्, …………
उत्तर:
(क) विदेशेषु धनं विद्या, व्यसनेषु धनं मतिः।
परलोके धनं धर्मः, शीलं सर्वत्र वै धनम्॥
(ख) विद्वत्वं च नृपत्वं च, नैव तुल्यं कदाचन।
स्वदेशे पूज्यते राजा, विद्वान् सर्वत्र पूज्यते ॥
(ग) विद्या ददाति विनयं, विनयाद् याति पात्रताम्।
पात्रत्वाद् धनमाप्नोति, धनाद् धर्मः ततः सुखम्॥
(घ) अलसस्य कुतो विद्या, अविद्यस्य कुतो धनम्।
अधनस्य कुतो मित्रम्, अमित्रस्य कुतो सुखम्॥

Sanskrit Class 6 Surbhi प्रश्न 5.
उचितशब्देनं रिक्त स्थानं पूरयत(उचित शब्दों से रिक्त स्थानों को पूरा करो)
(क) चतस्त्रः …………….। (बालिकाः/बालकाः/गृहाणि)
(ख) त्रीणि ……………। (छात्राः/फलानि/मयूराः)
(ग) अष्ट ……………। (पत्रम्/लते/भवनानि)
उत्तर:
(क) बालिकाः,
(ख) फलानि,
(ग) भवनानि।

MP Board Class 6 Sanskrit Solutions प्रश्न 6.
उचितं मेलयत (उचित मिलान कराओ)

उत्तर:
(क) → 4
(ख) → 1
(ग) → 2
(घ) → 3

Sanskrit Surbhi Class 6 प्रश्न 7.
निम्नलिखितशब्दानां तृतीयाविभक्तेः रूपाणि लिखत (निम्नलिखित शब्दों के तृतीया विभक्ति के रूप लिखो)
(क) घट
(ख) हस्त
(ग) नासिका
(घ) शाखा
(ङ) पुष्प
(च) चक्र।
उत्तर:
तृतीया विभक्ति-

Class 6th Surbhi Sanskrit प्रश्न 8.
उचितं मेलयत (उचित का मिलान करो)(अ)

उत्तर:
(क) → 5
(ख) → 4
(ग) → 3
(घ) → 7
(ङ) → 1
(च) → 9
(छ) → 6
(ज) → 1
(झ) → 8

Class 6 Surbhi Sanskrit प्रश्न 9.
उचितं मेलयत (उचित का मिलान करो)

उत्तर:
(क) → 9
(ख) → 5
(ग) → 7
(घ) → 6
(ङ) → 8
(च) → 4
(छ) → 2
(ज) → 3
(झ) →1

Class 6 Sanskrit Solutions MP Board प्रश्न 10.
अन्वयपूर्तिम् कुरुत (अन्वय की पूर्ति करो)
(क) कुरूपाणां…………. विद्या, ………….. धनं (विद्या) तथा, निर्बलानाम् बलं ………… , अतः………… साधनीया।
(ख) विदेशेषु…….. धनं, व्यसनेषु मतिः………….. । परलोके………… धनं………… सर्वत्र वै धनम्।
उत्तर:
(क) रूपं, निर्धनानां, विद्या, प्रयत्नतः।
(ख) विद्या, धनं, धर्मः, शीलं।

MP Board Class 6th Sanskrit Solutions प्रश्न 11.
वर्णमालानुसारं क्रमेण स्थापयत (वर्णमाला के अनुसार क्रम स्थापित कीजिए)
ऊर्णाः, बकः, औषधम्, हलम्, सरः, मयूरः, रथः, आम्रम्, घटः, कमलम्।
उत्तर:
आम्रम्, ऊर्णाः, औषधम्, कमलम्, घटः, बकः, मयूरः, रथः, सरः, हलम्।

MP Board Class 6th Sanskrit Solutions Surbhi Chapter 2 कर्त्तृक्रिर्त्तृयासम्बन्धः

MP Board Class 6th Sanskrit Chapter 2 अभ्यासः

MP Board Class 6 Sanskrit Chapter 2 प्रश्न 1.
उचितं मेलनं कुरुत (उचित मेल करो)

उत्तर:
(क) → 5
(ख) → 3
(ग) → 1
(घ) → 2
(ङ) → 4

Class 6 Sanskrit Chapter 2 MP Board प्रश्न 2.
उचितं क्रियापदं चित्वा रिक्तस्थानानि पूरयत
(उचित क्रिया पद चुनकर खाली स्थानों को पूरा करो)
(गच्छति, खादतः, पठन्ति, पततः, भ्रमन्ति)
(क) छात्राः ……….।
(ख) चक्राणि …………
(ग) यानं …………।
(घ) बालिके …………।
(ङ) पत्रे………..।
उत्तर:
(क) पठन्ति
(ख) भ्रमन्ति
(ग) गच्छति
(घ) खादतः
(ङ) पततः।

Class 6 Sanskrit Surbhi Chapter 2 प्रश्न 3.
चित्रं दृष्ट्वा उत्तरत (चित्र देखकर उत्तर दो)

(क) किं पतति?
(ख) के क्रीडन्ति?
(ग) काः धावन्ति?
(घ) कानि विकसन्ति?
(ङ) कः हसति?
(च) बालिका किं करोति?
उत्तर:
(क) पत्रं पतति।
(ख) बालकाः क्रीडन्ति।
(ग) बालिकाः धावन्ति।
(घ) पुष्पाणि विकसन्ति।
(ङ) बालकः हसति।
(च) बालिका भोजनं करोति।

MP Board Class 6th Sanskrit Chapter 2 प्रश्न 4.
शुद्धम् (✓)/अशुद्धं (✗) वा इति सूचयत (शुद्ध (✓) अथवा अशुद्ध (✗) है, की सूचना दो)
(क) मयूरा: रक्षति।
(ख) छात्रा लिखति।
(ग) बालिका कूजति।
(घ) कलिका विकसति।
(ङ) चटका पठति।
उत्तर:
(क) ✗
(ख) ✓
(ग) ✗
(घ) ✓
(ङ) ✗

Class 6 Sanskrit Chapter 2 प्रश्न 5.
बहुवचने परिवर्तनं कुरुत(बहुवचन में बदलिए)
(क) बालिका पठति।
(ख) चटके कूजतः।
(ग) फलं पतति।
(घ) गृहे स्तः।
(ङ) गजौ गच्छतः।
उत्तर:
(क) बालिकाः पठन्ति।
(ख) चटकाः कूजन्ति।
(ग) फलानि पतन्ति।
(घ) गृहाणि सन्ति।
(ङ) गजा: गच्छन्ति।

कर्त्तृक्रिर्त्तृयासम्बन्धः हिन्दी अनुवाद

अकारान्त पुल्लिङ्ग शब्दप्रयोगः

अन्यानि वाक्यानि (अन्य वाक्य)

आकारान्त स्त्रीलिङ्ग शब्दप्रयोगः

अन्यानि वाक्यानि (अन्य वाक्य) :

अकारान्त नपुंसकलिङ्ग शब्दप्रयोग :

अन्यानि वाक्यानिः (अन्य वाक्य):

MP Board Class 6th Sanskrit Solutions

MP Board Class 8th Maths Solutions Chapter 8 राशियों की तुलना Ex 8.3

Class 8 Maths Chapter 8 Exercise 8.3 MP Board प्रश्न 1. निम्नलिखित के लिए कुल राशि एवं चक्रवृद्धि ब्याज ज्ञात कीजिए

1. ₹ 10,800 पर 3 वर्ष के लिए 12\(\frac{1}{2}\)% वार्षिक दर से वार्षिक रूप से संयोजित करने पर।
2. ₹ 18,000 पर 2\(\frac{1}{2}\) वर्ष के लिए 10% वार्षिक दर से वार्षिक रूप से संयोजित करने पर।
3. ₹ 62,500 पर 1\(\frac{1}{2}\) वर्ष के लिए 8% वार्षिक दर से अर्धवार्षिक रूप से संयोजित करने पर।
4. ₹ 8000 पर 1 वर्ष के लिए 9% वार्षिक दर से अर्धवार्षिक रूप से संयोजित करने पर।
   (आप सत्यापन करने के लिए साधारण ब्याज के सूत्र का उपयोग करते हुए एक के बाद दूसरे वर्ष के लिए परिकलन कर सकते हैं।)
5. ₹ 10,000 पर 1 वर्ष के लिए 8% वार्षिक दर से अर्धवार्षिक रूप से संयोजित करने पर।

हल:
1. यहाँ, P = ₹ 10,800
R = 12\(\frac{1}{2}\)% = 25%
वार्षिक, n= 3 वर्ष

= ₹ 15,377.34
∴ चक्रवृद्धि ब्याज = मिश्रधन – मूलधन
= ₹ 15,377.34 – ₹ 10,800
= ₹ 45,77.34
अतः 3 वर्ष बाद कुल राशि = ₹ 15,377.34 तथा चक्रवृद्धि ब्याज = ₹ 45,77.34

2. यहाँ, P= ₹ 18,000,
R = 10%
वार्षिक, n = 2\(\frac{1}{2}\) वर्ष

= ₹ 22,869
चक्रवृद्धि ब्याज = ₹ 22,869 – ₹ 18,000
= ₹ 4,869
अतः मिश्रधन = ₹ 22,869 तथा चक्रवृद्धि ब्याज = ₹ 4,869

3. यहाँ, P = ₹ 62500,
R = 8%
वार्षिक = 4% प्रति छमाही,
n = 1\(\frac{1}{2}\) वर्ष = 3 अर्धवर्ष (3 छमाही)

= ₹ 70,304
चक्रवृद्धि ब्याज = ₹ 70,304 – ₹ 62,500
= ₹ 7,804
अतः मिश्रधन = ₹ 70,304 तथा चक्रवृद्धि ब्याज = ₹ 7,804

4. यहाँ, P = ₹ 8000
R = 9%
वार्षिक = \(\frac{9}{2}\)%
प्रति अर्धवार्षिक, n = 1 वर्ष = 2 छमाही

= ₹ 8,736.20
चक्रवृद्धि ब्याज = ₹ 8,736.20 – ₹ 8,000
= ₹ 736.20
अतः मिश्रधन = ₹ 8,736-20 तथा चक्रवृद्धि ब्याज = ₹ 736.20

5. यहाँ, P = ₹ 10,000
R = 8%
वार्षिक = 4%
प्रति छमाही, n = 1 वर्ष = 2 छमाही

= ₹ 10,816
चक्रवृद्धि ब्याज = ₹ 10,816 – ₹ 10000
= ₹ 816
अत: मिश्रधन = ₹ 10,816 तथा चक्रवृद्धि ब्याज = ₹ 816

MP Board Class 8 Maths Chapter 8 Exercise 8.3 प्रश्न 2.
कमला ने एक स्कूटर खरीदने के लिए किसी बैंक से ₹ 26,400, 15% वार्षिक दर से उधार लिए जबकि ब्याज वार्षिक संयोजित होना है। 2 वर्ष और 4 महीने के अन्त में उधार चुकता करने के लिए उसे कितनी राशि का भुगतान करना पड़ेगा?
(संकेत: ब्याज को वार्षिक संयोजित करते हुए पहले 2 वर्ष के लिए A ज्ञात कीजिए और दूसरे वर्ष की कुल राशि पर + वर्ष का साधारण ब्याज ज्ञात कीजिए।)
हल:
यहाँ, P = ₹ 26,400
समय n = 2 वर्ष 4 महीने
R= 15%

= ₹ 1,745.70
कुल मिश्रधन = ₹ 34,914 + ₹ 1,745.70
= ₹ 36,659.70
अतः कमला को उधार चुकता करने के लिए ₹ 36,659.70 का भुगतान करना पड़ेगा।

Rashiyon Ki Tulna Class 8 प्रश्न 3.
फैबिना ने ₹ 12,500, 3 वर्ष के लिए 12% वार्षिक दर से साधारण ब्याज पर उधार लिए और राधा ने उतनी ही राशि उतने ही समय के लिए 10% वार्षिक दर से चक्रवृद्धि ब्याज पर उधार ली जबकि ब्याज वार्षिक रूप से संयोजित होना है। किसे अधिक ब्याज का भुगतान करना है और कितना अधिक करना है?
हल:
फैबिना के लिए
मूलधन P = ₹ 12,500
समय T = 3 वर्ष
दर = 12% वार्षिक

= ₹ 4,500
राधा के लिए
मूलधन P = ₹ 12,500
दर R = 10%
वार्षिक, समय, n = 3 वर्ष

= ₹ 16,637.50
∴ चक्रवृद्धि ब्याज = ₹ 16,637.50 – ₹ 1,2500
= ₹ 4,137.50
अतः फैबिना को ₹ 4,500 – ₹ 4,137.50 = ₹ 362.50 अधिक भुगतान करना है।

Class 8 Maths Ex 8.3 Solutions In Hindi प्रश्न 4.
मैंने जमशेद से ₹ 12,000, 2 वर्ष के लिए 6% वार्षिक दर से साधारण ब्याज पर लिए। यदि मैंने यह राशि 6% वार्षिक दर से चक्रवृद्धि ब्याज पर ली हुई होती तो मुझे कितनी अतिरिक्त राशि का भुगतान करना पड़ता?
हल:
यहाँ, P = ₹ 12,000
समय = 2 वर्ष
दर = 6% वार्षिक

= ₹ 13,483.20
∴ चक्रवृद्धि ब्याज = ₹ 13,483.20 – ₹ 12,000
= ₹ 1,483.20
अतः मुझे ₹ 1,483.20 – ₹1,440 = ₹ 43.20 का अतिरिक्त भुगतान करना पड़ता।

Rashiyo Ki Tulna Class 8 प्रश्न 5.
वासुदेवन ने 12% वार्षिक दर पर ₹ 60,000 का निवेश किया। यदि ब्याज अर्धवार्षिक संयोजित होता है तो ज्ञात कीजिए कि वह –

1. 6 महीने के अन्त में
2. एक वर्ष के अन्त में, कुल कितनी राशि प्राप्त करेगा?

हल:
यहाँ, P = ₹ 60,000
दर = 12%
वार्षिक = 6% प्रति अर्धवार्षिक,

1. 6 महीने = 1 अर्धवर्ष
∴ 1 अर्धवर्ष के बाद मिश्रधन = ₹ 60,000 (1 + \(\frac{6}{100}\))¹
= ₹ 60,000 (1 + \(\frac{3}{50}\))
= ₹ 60,000 x \(\frac{53}{50}\)
= ₹ 63,600
अत: वासुदेवन 6 महीने के अन्त में ₹ 63,600 प्राप्त करेगा।

2. समय = 1 वर्ष = 2 अर्धवर्ष

= ₹ 67,416
अतः वासुदेवन 1 वर्ष के अन्त में ₹ 67,416 प्राप्त करेगा।

Class 8th Maths Chapter 8 Exercise 8.3 In Hindi प्रश्न 6.
आरिफ ने एक बैंक से ₹ 80,000 का कर्ज लिया। यदि ब्याज की दर 10% वार्षिक है, तो 1\(\frac{1}{2}\) वर्ष पश्चात् उसके द्वारा भुगतान की जाने वाली राशियों में अन्तर ज्ञात कीजिए यदि ब्याज –

1. वार्षिक संयोजित होता है
2. अर्धवार्षिक संयोजित होता है।

हल:
1. यहाँ, P = ₹ 80,000
दर = 10%
वार्षिक, समय = 1\(\frac{1}{2}\) वर्ष

= ₹ 92,400

2. जब कि ब्याज अर्धवार्षिक संयोजित होता है।
समय = 1\(\frac{1}{2}\) वर्ष = \(\frac{3}{2}\) x 2 = 3 अर्धवर्ष,
दर = 10% वार्षिक = 5% प्रति अर्धवार्षिक

= ₹ 92,610
∴ मिश्रधन में अन्तर = ₹ 92,610 – ₹ 92,400 = ₹ 210

MP Board Class 8 Maths Chapter 8 प्रश्न 7.
मारिया ने किसी व्यापार में ₹ 8000 का निवेश किया। उसे 5% वार्षिक दर से चक्रवृद्धि ब्याज का भुगतान किया जाएगा। यदि ब्याज वार्षिक रूप से संयोजित होता है, तो –

1. दो वर्ष के अन्त में उसके नाम से जमा की गई राशि ज्ञात कीजिए।
2. तीसरे वर्ष का ब्याज ज्ञात कीजिए।

हल:
1. यहाँ, P = ₹ 8,000
दर = 5%
वार्षिक, समय = 2 वर्ष

= ₹ 8,820
∴ 2 वर्ष के अन्त में उसके नाम से जमा की गई राशि = ₹ 8,820

2. यहाँ तीसरे वर्ष के लिए मूलधन = ₹ 8,820

= ₹ 441
∴ तीसरे वर्ष का ब्याज = ₹ 441

Class 8 Maths Chapter 8.3 In Hindi प्रश्न 8.
₹ 10,000 पर 19 वर्ष के लिए 10% वार्षिक दर से चक्रवृद्धि ब्याज और कुल राशि ज्ञात कीजिए जबकि ब्याज अर्धवार्षिक संयोजित होता है। क्या यह ब्याज उस ब्याज से अधिक होगा जो उसे वार्षिक रूप से संयोजित करने पर प्राप्त होगा?
हल:
यहाँ, P = ₹ 10,000
समय = 1\(\frac{1}{2}\) वर्ष = 3 अर्धवर्ष
दर = 10%
वार्षिक = 5% प्रति अर्धवार्षिक

= ₹ 11,576.25
चक्रवृद्धि ब्याज = ₹ 11,576.25 – ₹10,000
= ₹1576-25
अतः कुल राशि = ₹ 11,576.25 तथा चक्रवृद्धि ब्याज = ₹ 1,576.25
हाँ, यह ब्याज उस ब्याज से अधिक होगा जो उसे वार्षिक रूप से संयोजित करने पर प्राप्त होगा।

Class 8 Maths Exercise 8.3 Solutions In Hindi प्रश्न 9.
यदि राम ₹ 4,096, 18 महीने के लिए 125% वार्षिक दर पर उधार देता है और ब्याज अर्धवार्षिक संयोजित होता है तो ज्ञात कीजिए कि राम कुल कितनी राशि प्राप्त करेगा?
हल:
यहाँ, P = ₹ 4,096
समय = 18
महीने = 1\(\frac{1}{2}\) वर्ष = 3 अर्धवर्ष
दर = 12\(\frac{1}{2}\)%
वार्षिक = 25% प्रति अर्धवार्षिक

= ₹ 4,913
राम कुल ₹ 4,913 प्राप्त करेगा।

MP Board Class 8 Maths Solutions English Medium प्रश्न 10.
5% वार्षिक दर से बढ़ते हुए वर्ष 2003 के अन्त में एक स्थान की जनसंख्या 54,000 हो गई निम्नलिखित को ज्ञात
कीजिए –

1. वर्ष 2001 में जनसंख्या
2. वर्ष 2005 में कितनी जनसंख्या होगी?

हल:
1. वर्तमान जनसंख्या = 54,000
दर = 5%
माना कि 2001 में जनसंख्या = P थी (दो वर्ष पूर्व)

= 48,979.59 (लगभग 48,980)
अतः 2001 में जनसंख्या लगभग 48,980 थी।

2. माना कि वर्तमान जनसंख्या = 54000 (2003 में जनसंख्या)

= 59,535
अतः वर्ष 2005 में जनसंख्या 59,535 होगी।

Class 8 Maths 8.3 In Hindi प्रश्न 11.
एक प्रयोगशाला में किसी निश्चित प्रयोग में बैक्टीरिया की संख्या 2.5% प्रति घण्टे की दर से बढ़ रही है। यदि प्रयोग के शुरू में बैक्टीरिया की संख्या 5,06,000 थी तो 2 घण्टे के अन्त में बैक्टीरिया की संख्या ज्ञात कीजिए।
हल:
यहाँ, वर्तमान बैक्टीरिया की संख्या P = 5,06,000,
दर = 2.5%
प्रति घण्टा, समय = 2 घण्टे

= 5,31,616.25
= 5,31,616 (लगभग)
अत: 2 घण्टे बाद बैक्टीरिया की संख्या = 5,31,616

Math Class 8 Chapter 8.3 प्रश्न 12.
एक स्कूटर ₹ 42,000 में खरीदा गया। 8% वार्षिक दर से इसके मूल्य का अवमूल्यन हो गया। 1 वर्ष बाद स्कूटर का मूल्य ज्ञात कीजिए।
हल:
स्कूटर का वर्तमान मूल्य = ₹ 42,000
अवमूल्यन दर = 8% वार्षिक

= ₹ 38,640
अतः 1 वर्ष बाद स्कूटर का मूल्य = ₹ 38,640

MP Board Class 8th Maths Solutions

In this article, we will share MP Board Class 11th Biology Solutions Chapter 1 जीव जगत Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 11th Biology Solutions Chapter 1 जीव जगत

जीव जगत NCERT प्रश्नोत्तर

Sajeev Jivo Ko Kyon Vargikrit Kiya Gaya Hai प्रश्न 1.
जीवों को वर्गीकृत क्यों करते हैं ?
उत्तर:
जैव-वर्गीकरण का उद्देश्य बड़ी संख्या में ज्ञात पादपों और जन्तुओं को ऐसे वर्गों में व्यवस्थित करना है, कि उन्हें नाम प्रदान किया जा सके एवं अध्ययन में सरलता हो। जीवों को वर्गीकृत करने से निम्नलिखित लाभ हैं –

1. इसके कारण विश्व के विविध प्रकार, असंख्य जीवों के अध्ययन में सुविधा होती है।
2. इसके कारण जन्तु पादपों के सम्बन्धों का पता चलता है।
3. इसके कारण जीवों को पहचानने में सरलता होती है।
4. जीवों की उत्पत्ति तथा दूसरे जीवों से सम्बन्ध का पता चलता है।
5. इसके कारण जीवों के विकास के क्रम एवं प्रमाण का पता लगता है।

Sajeev Jivo Ko Kyon Vargikrit Kiya Jata Hai प्रश्न 2.
वर्गीकरण प्रणाली को बार-बार क्यों बदलते हैं ?
उत्तर:
वर्गीकरण प्रणाली को बार-बार बदलने का प्रमुख कारण जैव – विकास है। जीवों में सतत् चलने वाली विकास प्रक्रिया के कारण नई-नई विभिन्न प्रजातियों के पादप एवं जन्तु पहले से मौजूद जीव-विविधता (Bio – diversity) से जुड़ते जाते हैं। इन नवीन जीवों को पहचानकर वर्गीकरण प्रणाली से संबद्ध किया जाता है। विकास के कारण पादप एवं जन्तुओं की जातियाँ (Species) बदलती रहती हैं। अत: इसके कारण प्रचलित वर्गीकरण प्रणाली को परिवर्तित कर जीवों को उनके क्रम में रखना पड़ता है।

जीव जगत के प्रश्न उत्तर प्रश्न 3.
जिन लोगों से प्रायः आप मिलते रहते हैं, आप उन्हें किस आधार पर वर्गीकृत करना पसंद करेंगे?(संकेत-ड्रेस, मातृभाषा, प्रदेश जिसमें वे रहते हैं, आर्थिक स्तर आदि)।
उत्तर:
सबसे पहले हम मातृभाषा के आधार पर अपने से मिलने वाले लोगों का वर्गीकरण करते हैं, उसके पश्चात् प्रदेश जहाँ वह रहता है तथा अन्त में उसकी वेशभूषा, धर्म, जाति, शारीरिक रंगरूप की बनावट, आर्थिक स्थिति के आधार पर हम वर्गीकृत करना पसंद करेंगे।

जीव जगत के प्रश्न उत्तर Pdf प्रश्न 4.
व्यष्टि तथा समष्टि की पहचान से हमें क्या शिक्षा मिलती है ?
उत्तर:
व्यष्टि (Individual) तथा समष्टि (Population) की पहचान से वर्तमान के सभी जीवों के परस्पर संबंधों के बारे में जानकारी मिलती है। इसमें हमें समान प्रकार के जीवों तथा अन्य प्रकार के जीवों में समानता तथा विभिन्नता को पहचानने में मदद मिलती है। उदाहरण-आलू की दो विभिन्न जातियाँ (Species) हैं –

Sajeev Jivo Ko Kyon Vargikrit Kiya Gaya प्रश्न 5.
आम का वैज्ञानिक नाम निम्नलिखित है-इसमें से कौन-सा सही है? मैगिफेरा इंडिका (Mangifera indica) या मैंगिफेरा इंडिका (Mangifera indica)
उत्तर:
आम के वैज्ञानिक नाम को लिखने की सही विधि है-मैगिफेरा इंडिका (Mangifera indica)।

जीवों को वर्गीकृत क्यों करते हैं क्लास 11 प्रश्न 6.
टैक्सॉन की परिभाषा दीजिए। विभिन्न पदानुक्रम स्तर पर टैक्सा के कुछ उदाहरण दीजिए।
उत्तर:
टैक्सॉन (Taxon) जीवों का एक समूह है जो कि किसी भी स्तर की वर्गिकी पदानुक्रम (Hierarchical classification) में पाया जाता है। टैक्सॉन मुख्यत: जीवों के समान लक्षणों पर आधारित होता है।

उदाहरण:
कीट, संघ ऑर्थोपोडा के एक वर्ग का प्रतिनिधित्व करता है। सभी कीटों में एक समान लक्षण तीन जोड़ी युग्मित उपांग पाये जाते हैं। टैक्सा के प्रमुख उदाहरण हैं – जगत, संघ, वर्ग, गण, कुल, जाति एवं वंश। ये सभी टैक्सा (संवर्ग) मिलकर वर्गिकी पदानुक्रम बनाते हैं।

उदाहरण – मनुष्य का टैक्सा (संवर्ग) है –

1. संघ – कॉर्डेटा
2. वर्ग – स्तनधारी
3. गण – प्राइमेट्स
4. कुल – होमिनिडी
5. वंश – होमो
6. जाति – सैपियन्स।

Sajeev Jio Ko Kyon Vargikrit Kiya Gaya Hai प्रश्न 7.
क्या आप वर्गिकी संवर्ग का सही क्रम पहचान सकते हैं –
(अ) जाति (स्पीशीज) → गण ( ऑर्डर) → संघ (फाइलम) → जगत (किंगडम)
(ब) वंश (जीनस) → जाति → गण → जगत
(स) जाति → वंश → गण → संघ।
उत्तर:
वर्गिकी संवर्ग (Taxonomical categories) का सही क्रम है –
(स) जाति → वंश → गण → संघ।

वर्गीकरण प्रणाली को बार-बार क्यों बदलते हैं प्रश्न 8.
जाति शब्द के सभी मानवीय वर्तमान कालिक अर्थों को एकत्र कीजिए। क्या आप अपने शिक्षक से उच्च कोटि के पौधों तथा प्राणियों तथा बैक्टीरिया की स्पीशीज का अर्थ जानने के लिए चर्चा कर सकते हैं ?
उत्तर:
जाति वर्गीकरण की सबसे छोटी इकाई है। मेयर (1942) के अनुसार, “जाति आपस में संकरण या संयोग करने वाले एक समान जीवों का समूह है।” आधुनिक विचारधारा के अनुसार, निम्नलिखित लक्षण वाले जीवों को जाति (Species) कहते हैं –

1. इनके सदस्यों में अन्तरा प्रजनन की क्षमता पायी जाती है।
2. इस समूह या आबादी में एक समान जीन पूल (जीन समूह) पाया जाता है।
3. प्रत्येक जाति में वातावरण के साथ अनुकूलन एवं प्राकृतिक चयन चलता रहता है।
4. प्रत्येक जाति में जैव विकास के द्वारा नयी जाति पैदा करने की क्षमता पायी जाती है।
5. प्रत्येक जाति में पृथक्करण के कुछ ऐसे कारक पाये जाते हैं, जो निकट सम्बन्धी जाति के सदस्यों से प्रजनन करने में रुकावट पैदा करते हैं।

मैंगिफेरा इंडिका (आम), सोलेनम ट्यूबरोसम (आलू), तथा पैंथेरा लियो (शेर) उच्चकोटि के पौधे तथा प्राणी के उदाहरण हैं। इन सभी नामों में ” इंडिका, ट्यूबरोसम'” तथा “लियों” जाति संकेत के पद हैं। जबकि पहले शब्द ” मैंगिफेरा’, “सोलेनम” तथा ” पैंथेरा” वंश के नाम हैं और यह टैक्सा अथवा संवर्ग का भी निरुपण करते हैं। प्रत्येक वंश में एक अथवा एक से अधिक जाति के संकेत पद हो सकते हैं जो विभिन्न जीवों जिनमें आकारिकी गुण समान हों को प्रदर्शित करते हैं। उदाहरणार्थ “पैंथेरा” में एक अन्य जाति संकेत पद है जिसे टिग्रिस कहते हैं। सोलेनेम’ वंश में नाइग्रम, मैलान्जेना भी आते हैं। लेकिन बैक्टीरिया को उनके आकार के आधार पर चार वर्ग समूहों में रखा जाता है-गोलाकार, छड़नुमा, कॉमा एवं तर्कुरूपी। अतः इस प्रकार का अर्थ उच्च जीवों के लिए है, बैक्टीरिया के लिए अलग-अलग है।

Sajeev Jivon Ko Kyon Vargikrit Kiya Gaya Hai प्रश्न 9.
निम्नलिखित शब्दों को समझाइए तथा परिभाषित कीजिए –

1. संघ
2. वर्ग
3. कुल
4. गण
5. वंश।

उत्तर:
वर्गीकरण की पदानुक्रमी स्तरें:
जीवों का वर्गीकरण करते समय अध्ययन की सुविधा के लिए इन्हें कई छोटे-बड़े समूहों में रखा गया है, जिन्हें संवर्ग या श्रेणी (Category) कहा जाता है। इन संवर्गों का वर्गीकरण में एक निश्चित स्थान होता है और इन्हें इनके गुणों के आधार पर बढ़ते हुए क्रम में रखा जाता है। इसी क्रम को वर्गीकरण का पदानुक्रम (Hierarchy) कहते हैं। इस पदानुक्रम की सबसे छोटी इकाई जाति तथा बड़ी इकाई जगत है। वर्गीकरण में संवर्गों का क्रम निम्नानुसार होता है –

1. जाति:
यह वर्गीकरण की मूल तथा बहुत छोटी इकाई है। वर्गीकरण की इस इकाई में आपस में संकरण करने वाले जीवों के समूह को रखा जाता है जैसे-सभी प्रकार के मानवों को होमो सैपियन्स में रखा जाता है, जबकि वे बाह्य आकार में विविधता प्रदर्शित करते हैं।

2. वंश:
कुछ एक समान गुणों को प्रदर्शित करने वाली जातियों को एक वंश में रखा जाता है, जैसेशेर, बाघ, चीता को एक वंश पैंथेरा में रखा गया है।

3. कुल:
कुछ एक समान गुणों वाले वंशों को एक कुल में रखते हैं। जैसे-सभी दालों का प्रतिनिधित्व करने वाले पादपों को एक कुल पैपीलियोनेसी में रखा जाता है।

4. गण:
एक या कई मिलते-जुलते गुणों वाले कुलों को एक गण में रखा जाता है। जैसे-जन्तुओं के फेलिडी तथा कैनिडी कुल को कार्निवोरा गण में रखा जाता है।

5. वर्ग:
कुछ सर्वश्रेष्ठ गुणों वाले गणों को एक वर्ग में रखा जाता है।

6. संघ:
कुछ सर्वनिष्ठ गुणों वाले वर्गों को एक संघ में रखा जाता है। जैसे-नोटोकॉड, नर्वकॉर्ड तथा गिल की उपस्थिति के कारण मत्स्य, ऐम्फिबिया, सरीसृप, पक्षी एवं स्तनी वर्ग को संघ कॉर्डेटा में रखा जाता है।

7. जगत:
कई सर्वनिष्ठ गुणों वाले संघों को एक जगत में रखा जाता है। उपर्युक्त वर्णित संवर्ग मुख्य संवर्ग हैं, इन्हें आवश्यकतानुसार कई छोटे संवर्गों में भी बाँटा जाता है।

जीवों को वर्गीकृत क्यों करते हैं प्रश्न 10.
जीव के वर्गीकरण तथा पहचान में कुंजी किस प्रकार सहायक है ?
उत्तर:
पादपों एवं जन्तुओं को पहचानने की रूपरेखा कुंजी (Key) है। वर्गिकी की कुंजियाँ विपरीत लक्षणों पर आधारित होती हैं। कुल, वंश और जाति जैसी वर्गिकी की प्रत्येक श्रेणी के लिए अलग अलग वर्गिक की कुंजियों की आवश्यकता होती है। यह अज्ञात जीवों की पहचान के लिए अधिक उपयोगी होती है। कुंजियाँ दो प्रकार की होती हैं –

1. दोहरे प्रलेखधारी अथवा द्विशाखित (Yolked)
2. कोष्ठधारी कुंजी (Bracketed key)।

1. द्विशाखित कुंजी:
यह एक अन्य साधन सामग्री है, जिसका प्रयोग समानताओं और असमानताओं पर आधारित होकर पौधों तथा प्राणियों की पहचान में किया जाता है।

2. कोष्ठधारी कुंजी:
यह कुंजी विपर्यायी लक्षणों (Contrasting:characters), जो प्रायः युग्मों के आधार पर होती है। कुंजी दो विपरीत विकल्पों को चुनने को दिखाती है। इसके परिणामस्वरूप एक को मान्यता तथा दूसरे को अमान्य किया जाता है।

कुंजी में प्रत्येक कथन मार्गदर्शन का कार्य करता है। पहचानने के लिए प्रत्येक वर्गिकी संवर्ग जैसे-कुल, वंश, तथा जाति के लिए अलग-अलग वर्गिकी कुंजी की आवश्यकता होती है।

Jeev Jagat प्रश्न 11.
पौधों तथा प्राणियों के उचित उदाहरण देते हुए वर्गिकी पदानुक्रम का चित्रण कीजिए।
उत्तर:
वर्गीकरण (Classification) एकल सोपान प्रक्रम नहीं है बल्कि इसमें पदानुक्रम (Hierarchy) सोपान होते हैं जिसमें प्रत्येक सोपान पद (Rank) अथवा संवर्ग (Category) को प्रदर्शित करता है। चूंकि संवर्ग समस्त वर्गिकी व्यवस्था है इसलिए इसे वर्गीकरण संवर्ग (Taxonomic categories) कहते हैं और सभी संवर्ग मिलकर वर्गिकी पदानुक्रम (Taxonomic hierarchy) बनाते हैं। प्रत्येक संवर्ग वर्गीकरण की एक इकाई को प्रदर्शित करता है, इसे प्रायः वर्गक या टैक्सॉन (Taxon) कहते हैं।

सभी ज्ञात जीवों के वर्गीकीय अध्ययन से सामान्य संवर्ग जैसे-जगत (Kingdom), संघ (Phylum), वर्ग (Class), गण (Order), कुल (Family), वंश (Genus) तथा जाति (Species) का विकास हुआ। पौधे तथा प्राणियों, दोनों में जाति (Species) सबसे निचले संवर्ग में आती है। किसी भी जीव को विभिन्न संवर्गों में रखने के लिए उनके वर्ग के गुणों का ज्ञान होना आवश्यक है। जाति से लेकर जगत तक विभिन्न वर्गिकी संवर्ग को आरोही क्रम में निम्नानुसार दर्शाया जा सकता है –

उपरोक्त पदानुक्रम अनुसार, जैसे – जैसे हम जाति से जगत की ओर ऊपर जाते हैं वैसे ही समान गुणों में कमी आती जाती है। सबसे नीचे जो टैक्सा होगा, उसके सदस्यों में सबसे अधिक समान गुण होंगे।

तालिका – वर्गिकी संवर्ग सहित कुछ जीव
(Organisms with their Taxanomic Categories)

जीव जगत अन्य महत्वपूर्ण प्रश्नोत्तर

जीव जगत वस्तुनिष्ठ प्रश्न

Jivo Ko Vargikrit Kyon Karte Hain प्रश्न 1.
सही विकल्प चुनकर लिखिए –

1. सजीव जब अपने शत्रुओं से बचने के लिए स्वयं को उस परिस्थिति के अनुसार ढाल लेता है, उसे कहते हैं –
(a) अनुकूलन
(b) मिमिक्री
(c) हीमोलॉजी
(d) एनोलॉजी।
उत्तर:
(a) अनुकूलन

2. जीवों में ऊर्जा प्रवाह तथा ऊर्जा के रूपान्तरण में किस नियम का पालन होता है –
(a) सीमित कारक का नियम
(b) ऊष्मागतिक का नियम
(c) लिबिंग्स के न्यूनतता का नियम
(d) बायोजेनिक के नियम।
उत्तर:
(b) ऊष्मागतिक का नियम

3. बाहरी वातावरण में बदलाव के बाद भी जीवों में उचित आंतरिक अवस्था को बनाए रखना कहलाता है –
(a) एन्थैल्पी
(b) साम्यावस्था
(c) एन्ट्रॉपी
(d) स्थायी अवस्था।
उत्तर:
(b) साम्यावस्था

4. ग्लाइकोजन बहुलक है –
(a) ग्लैक्टोज
(b) ग्लूकोज
(c) फ्रक्टोज
(d) सुक्रोज।
उत्तर:
(b) ग्लूकोज

5. एक ऑर्किड का पुष्य किसी कीट के मादा के समान दिखाई देता है ताकि उसमें परागण क्रिया आसानी से हो सके, यह घटना कहलाती है-
(a) मिमिक्रीम
(b) स्यूडोकॉपुलेशन
(c) स्यूडोपॉलीनेशन
(d) स्यूडोकार्पोनोकाी।
उत्तर:
(a) मिमिक्रीम

6. जीवन की सबसे छोटी इकाई है –
(a) DNA
(b)RNA
(c) कोशिका
(d) प्रोटीन।
उत्तर:
(a) DNA

7. निम्न में से किसे होमियोस्टेसिस या साम्यावस्था कहते हैं –
(a) वातावरण के साथ बदलाव लाना
(b) नियंत्रण में बदलाव
(c) बदलाव का प्रतिरोध करना
(d) पादप एवं जन्तु रस का होम्योपैथी उपयोग।
उत्तर:
(b) नियंत्रण में बदलाव

8. शरीर का तापक्रम किसके द्वारा नियंत्रित होता है –
(a) फेफड़ा, पेशी तथा त्वचा
(b) केवल त्वचा
(c) परिसंचरण तंत्र
(d) कंकाल तंत्र।
उत्तर:
(a) फेफड़ा, पेशी तथा त्वचा

9. पसीना बहने का उद्देश्य होता है –
(a) त्वचा पर उपस्थित जीवाणुओं को मारना
(b) शरीर के ताप का नियमन
(c) अधिक लवण का निष्कासन
(d) अधिक जल का निष्कासन।
उत्तर:
(b) शरीर के ताप का नियमन

10. आयोडीन किसका घटक होता है –
(a) नाइट्रेट रिडक्टेज
(b) थायरॉक्सीन हॉर्मोन्स
(c) TSH हॉर्मोन्स
(d) नाइट्रोजिनेज।
उत्तर:
(b) थायरॉक्सीन हॉर्मोन्स

11. जीवित कोशिकाओं के लिए कौन-सा सही है –
(a) पहले ऊर्जा का स्थानांतरण तब ऊर्जा का रूपान्तरण
(b) पहले ऊर्जा का रूपान्तरण तब स्थानांतरण
(c) दोनों का साथ-साथ होना
(d) दोनों का लगातार होना।
उत्तर:
(d) दोनों का लगातार होना।

12. जीवों के सामान्य लक्षण होते हैं –
(a) कोशिकीय संरचना
(b) उपापचय
(c) श्वसन
(d) उपर्युक्त सभी।
उत्तर:
(d) उपर्युक्त सभी।

13. जीवों की आधारभूत आवश्यकता होती है –
(a) विकास
(b) क्रम
(c) ऊर्जा
(d) वृद्धि।
उत्तर:
(c) ऊर्जा

14. निम्न में से स्टोरेज पॉलीसैकेराइड है –
(a) सुक्रोज
(b) सेल्युलोज
(c) स्टार्च
(d) स्टार्च एवं ग्लाइकोजन।
उत्तर:
(d) स्टार्च एवं ग्लाइकोजन।

15. जहाँ पर पौधों के नमूने एकत्रित करके रखे जाते हैं, उस स्थान को कहा जाता है –
(a) हरियम
(b) संग्रहालय
(c) वनस्पति उद्यान
(d) उपरोक्त में से कोई नहीं।
उत्तर:
(a) हरियम

16. जिस स्थान पर जीवित पौधों का संग्रहण किया जाता है, उसे कहा जाता है –
(a) वनस्पति उद्यान
(b) संग्रहालय
(c) हर्बेरियम
(d) जूलॉजिकल पार्क।
उत्तर:
(a) वनस्पति उद्यान

17. जिन स्थानों पर जीवित प्राणी रखे जाते हैं, उसे कहा जाता है –
(a) संग्रहालय
(b) जूलॉजिकल पार्क
(c) वनस्पति उद्यान
(d) हर्बेरियम।
उत्तर:
(b) जूलॉजिकल पार्क

18. जीवधारियों को वैज्ञानिक नाम दिये जाते हैं, क्योंकि –
(a) प्रत्येक तकनीकी ज्ञान की शाखा की अपनी शब्दावली होती है
(b) वैज्ञानिक लोगों पर अपना अमिट प्रभाव डालना चाहते थे
(c) वैज्ञानिक नहीं चाहते थे कि आम आदमी जीव विज्ञान पढ़ सके
(d) बिना किसी संशय के वैज्ञानिकों में विचार-विनिमय हो सके।
उत्तर:
(d) बिना किसी संशय के वैज्ञानिकों में विचार-विनिमय हो सके।

वर्गीकरण प्रणाली को बार बार क्यों बदलते हैं प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए –

1. त्वचा का धूप में काला होना ………….. अनुकूलन है।
2. जैव संगठन का सूक्ष्मतम जैविक स्तर …………. है।
3. जीवन का भौतिक आधार …………… है।
4. समष्टि के कुल जीन समूह को …………… कहते हैं।
5. वह तंत्र जिसमें पदार्थों का विनिमय वातावरणीय परिवेश से नहीं होता ………….. कहलाता है।
6. ………….. जीवधारियों के शरीर में होने वाली समस्त क्रियाओं के लिए ऊर्जा उपलब्ध कराता है।
7. जीवधारियों में वातावरण के अनुसार परिवर्तित होने की क्रिया …………… कहलाती है।
8. ………….. पदार्थ की सबसे छोटी इकाई है, जो सबसे छोटी जैविक इकाई का निर्माण करता है।
9. जीवों के शरीर में पाये जाने वाली समस्त क्रियाओं को सामूहिक रूप से …………… कहते हैं।
10. जीवों में पाये जाने वाले स्थिर अवस्था को …………… कहते हैं।

उत्तर:

1. अल्पकालिक
2. आण्विक स्तर
3. जीवद्रव्य
4. जीन पूल
5. बंद तंत्र
6. ATP
7. अनुकूलन
8. परमाण
9. उपापचय
10. साम्यावस्था।

Jeev Jagat Question Answer प्रश्न 3.
एक शब्द में उत्तर दीजिए –

1. ऊतकों व अंग तंत्रों की कार्यक्षमता में बढ़ती उम्र के कारण होने वाले ह्रास को क्या कहते हैं ?
2. जैव-संगठन के उच्चतम स्तर का नाम लिखिए।
3. जीवों में आनुवंशिक अणु किसे कहते हैं ?
4. वातावरण के परिवर्तन को जीवों द्वारा अनुभव करने की क्षमता को क्या कहते हैं ?
5. जैव-मण्डल के अपघटक जीवों का नाम दीजिए।

उत्तर:

1. वयता या जरण
2. जीव जगत (समष्टि)
3. D.N.A.
4. संवेदनशीलता
5. विषाणु, जीवाणु, कवक (सूक्ष्मजीव)।

Vargikaran Pranali Ko Bar Bar Kyon Badalte Hain प्रश्न 4.
उचित संबंध जोडिए –

उत्तर:

1. (c) जनन क्रिया न होने से
2. (d) वृद्धि
3. (a) स्टार्च व मंड के रूप में संग्रहण
4. (b) जल

उत्तर:

1. (c) वंश
2. (e) कुल
3. (d) गण
4. (a) जगत
5. (b) संघ

जीवों को वर्गीकृत क्यों करते प्रश्न 5.
सत्य / असत्य बताइये –

1. पौधे तथा प्राणियों दोनों में जाति (Species) सबसे निचले संवर्ग में आती है।
2. जीवों के वर्ग जिसमें मौलिक समानता होती है, उसे जाति कहते हैं।
3. मानव का वैज्ञानिक नाम “मैंगिफेरा इंडिका’ है।
4. वंश (Genus) समीपस्थ संबंधित जातियों का समूह है।
5. कुल (Family) के वर्गीकरण का आधार पौधों के कायिक तथा जनन गुण हैं।
6. वर्ग (Class) संवर्ग में प्राइमेट गण जिसमें बंदर, गोरिल्ला तथा गिब्बन आते हैं।
7. मानव का वर्ग इंसेक्टा है।
8. कुंजी में प्रत्येक कथन मार्गदर्शन का कार्य करता है। पहचानने के लिए प्रत्येक वर्गिकी संवर्ग जैसे – कुल, वंश, तथा जाति के लिए अलग, वर्गिकी कुंजी की आवश्यकता होती है।

उत्तर:

1. सत्य
2. सत्य
3. असत्य
4. सत्य
5. सत्य
6. सत्य
7. असत्य
8. सत्य।

जीव जगत अति लघु उत्तरीय प्रश्न

Sajeev Jio Ko Kyon Vargikrit Kiya Jata Hai प्रश्न 1.
वर्गीकरण विज्ञान का जनक किसे कहा जाता है?
उत्तर:
कैरोलस लिनीयस (1707 – 1778) को वर्गीकरण विज्ञान का जनक कहा जाता है।

1. जीवों को वर्गीकृत क्यों करते हैं? प्रश्न 2.
स्वयंपोषी से क्या तात्पर्य है ?
उत्तर:
वे जीव, जो अपना भोजन स्वयं बनाते हैं, उन्हें स्वयंपोषी जीव कहते हैं। उदाहरण – समस्त हरे पौधे।

अध्याय 1 जीव जगत प्रश्न 3.
पाँच जगत वर्गीकरण के प्रणेता कौन हैं ?
उत्तर:
पाँच जगत वर्गीकरण के प्रणेता आर. एच. ह्विटैकर (1969) हैं।

कक्षा 11 वीं जीव विज्ञान अध्याय 1 Question Answer प्रश्न 4.
जाति किसे कहते हैं ?
उत्तर:
मेयर (1942) के अनुसार, “आपस में संकरण करने वाले जीवों के समूह को जाति (Species) कहते हैं।”

जीव जगत प्रश्न 5.
वर्गीकरण की प्राकृतिक पद्धति क्या है ?
उत्तर:
वर्गीकरण वह पद्धति है, जिसमें गुणों के एक विस्तृत समूह को वर्गीकरण का आधार बनाया जाता है।

Sajeev Jio Ko Kyon Vargikrit Kiya Gaya प्रश्न 6.
वर्गीकरण की कृत्रिम पद्धति क्या है ?
उत्तर:
वर्गीकरण की वह पद्धति जिसमें एक या कुछ गुणों को वर्गीकरण का आधार बनाया जाता है कृत्रिम पद्धति कहलाती है। यह वर्गीकरण की अप्राकृतिक पद्धति है।

Jivo Ko Vargikrat Kyo Karte Hai प्रश्न 7.
लिनीयस द्वारा लिखित दो पुस्तकों तथा उनके दो योगदानों को लिखिए।
उत्तर:
लिनीयस द्वारा लिखित पुस्तकें –

• जेनेरा प्लाण्टेरम
• सिस्टेमा नैचुरी।

लिनीयस के दो प्रमुख योगदा –

• द्वि – जगत वर्गीकरण को प्रस्तुत करना।
• जीवों के लिए द्वि – नामकरण पद्धति को प्रस्तुत करना।

Sajeev Jivon Ko Kyon Vargikrit Kiya Jata Hai प्रश्न 8.
वर्गक एवं संवर्ग को परिभाषित कीजिए।
उत्तर:
वर्गक (Texa):
जीवों के वर्गीकरण में प्रयुक्त विभिन्न समूहों को वर्गक कहते हैं, चाहे वर्गीकरण में इनका स्थान कुछ भी हो, जैसे-शैवाल, कीट, मछली आदि।

संवर्ग या श्रेणी (Category):
वर्गीकरण में प्रयुक्त समूहों की विभिन्न स्तरों को संवर्ग या श्रेणी या वर्गीकरण की इकाई कहते हैं। वर्गीकरण का सबसे छोटा संवर्ग जाति तथा बड़ा संवर्ग जगत है।

जीव जगत लघु उत्तरीय प्रश्न

Sajeev Ko Kyon Vargikrit Kiya Jata Hai प्रश्न 1.
निम्नलिखित जीवों के वैज्ञानिक नाम लिखिए –

1. प्याज
2.  मनुष्य
3. हाथी
4. फीताकृमि
5. मेढक
6. गेहूँ
7. चावल
8. सरसों
9. मटर
10. आम।

उत्तर:
सामान्य नाम – वैज्ञानिक नाम

1. प्याज – एलियम सेपा
2. मनुष्य – होमो सैपियन्स
3. हाथी – एलिफस इन्डीकस
4. फीताकृमि – टीनिया सोलियम
5. मेढक – राना टिग्रिना
6. गेहूँ – ट्रिटिकम एस्टीवम
7. चावल – ओराइजा सटाइवा
8. सरसों – ब्रेसिका कम्पेस्ट्रीस
9. मटर – पाइसम सटाइवम
10. ‘आम – मैंगिफेरा इंडिका।

प्रश्न 2.
द्विपद नामकरण पद्धति से आप क्या समझते हैं ? उदाहरण देकर समझाइए।
उत्तर:
द्विपद नामकरण पद्धति जीवों के नामकरण की एक ऐसी पद्धति है, जिसमें प्रत्येक जीव का नाम दो शब्दों का होता है। इसका प्रथम शब्द जीव के वंश तथा दूसरा शब्द उसकी जाति को व्यक्त करता है। इस पद्धति का आविष्कार कैरोलस लिनीयस ने किया था। इसमें नाम के अक्षर इटैलिक में छापे जाते हैं या लिखते समय इनके नीचे रेखाएँ खींचते हैं।

प्रथम शब्द का पहला अक्षर कैपिटल लेटर तथा शेष सभी छोटे अक्षरों में ही होते हैं। दूसरे शब्द के सभी अक्षर छोटे अक्षरों में होते हैं। जैसे-मेढक का इस पद्धति में नाम Rana tigrina होता है। इसमें राना वंश तथा टिग्रिनाजाति को प्रदर्शित कर रहा है। यह नाम रखने का अधिकार उस जीव के आविष्कारकर्ता को होता है।

दो जीवों के वैज्ञानिक नाम –
सामान्य नाम – वैज्ञानिक नाम

1. मेढक – राना टिग्रिना
2. मनुष्य – होमो सेपियन्स।

प्रश्न 3.
जातिवृत्तीय रेखा को समझाइए।
उत्तर:
किसी एक जाति के विकासात्मक इतिहास को जातिवृत्ति कहते हैं। जातिवृत्ति की विभिन्न जातियों के क्रम को जातिवृत्ति रेखा (Phylogenic – line) कहते हैं। जातिवृत्तीय रेखा किसी जाति विकास के क्रम को प्रदर्शित करती है। जिस प्रकार विकसित जीवों का जीवन एक कोशिका से शुरू होता है, धीरे – धीरे इस कोशिका में परिवर्तन होता रहता है और कुछ समय बाद इसी एक कोशिका से विशालकाय जीव बन जाता है।

ठीक इसी तरहं इस पृथ्वी पर सबसे पहले एककोशिकीय जीव बना। इसके बाद वातावरण के अनुसार, इसमें परिवर्तन होता गया और विविध प्रकार की जातियाँ बनीं। इस प्रकार जातिवृत्तीय रेखा में सबसे पहले मोनेरा जगत के जीव बनें, जिन्होंने विकसित होकर प्रोटिस्टा जगत के जीव बनाए। प्रोटिस्टा जगत के जीवों ने कई दिशाओं में विकसित होकर पादप कवक एवं जन्तु जगतों का निर्माण किया।

जीव जगत दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
वानस्पतिक उद्यान को संक्षेप में समझाइए।
उत्तर:
वानस्पतिक उद्यान मानव द्वारा स्थापित प्राकृतिक स्थल होते हैं जहाँ पर पौधों को जीवित अवस्था में संरक्षित रखा जाता है। यहाँ पर पौधों के विभिन्न प्रजातियों का समुचित प्रबंध किया जाता है। एक आधुनिक वानस्पतिक उद्योग में निम्न प्रकार के पौधे लगाये जाते हैं –

1. कृषि में उपयोगी पौधे की विभिन्न प्रजातियाँ
2. औषधीय एवं अन्य महत्व वाले पौधे
3. विभिन्न भौगोलिक क्षेत्रों में पाये जाने वाले पौधे
4. धर्म ग्रंथों एवं साहित्यों में उल्लेखित पौधे।

वानस्पतिक उद्यान मूलतः जीवित पौधों का एक खुला संग्रह होता है जिससे हमें विभिन्न प्रकार के पौधों के बारे में मूलभूत जानकारियाँ प्राप्त होती हैं।

प्रमुख वानस्पतिक उद्यान:

1. रॉयल बॉटनिकल गार्डन, किव (इंग्लैंड)
2. इंडियन बॉटनिकल गार्डन, शिबपुर (कोलकाता)
3. लॉयड बॉटनिकल गार्डन, दार्जिलिंग
4. नेशनल बॉटनिकल गार्डन, लखनऊ
5. वन अनुसंधान संस्थान, वानस्पतिक उद्यान, देहरादून।

MP Board Class 11th Biology Solutions

MP Board Class 6th General English Solutions Practice Test Paper 1

Class 6 English Practice Test Paper 1 Question 1.
Read the paragraph from your book.
(अपनी पुस्तक से एक पैराग्राफ पढ़ें।)
Answer:
Students should do themselves.

Practice Test Paper 1 Class 6 Question 2.
Recite a poem that you know.
(एक कविता का पाठ करे जो तुम जानते हो।)
Answer:
Students should do themselves.

MP Board Class 6 English Paper Question 3.
Fill in the blanks and complete the poem.
(रिक्त स्थानों को भरें एवं कविता पूर्ण करें।)
Answer:
How do you like to go up in a swing.
Up in the air so blue?
Oh, I do think it’s the pleasantest thing.
Ever a child can do!

Practice Test Paper 1 Question 4.
(1) Find the words rhyming with the following words
(तुकान्त शब्दों का ढूँढ़ों।)
Answer:
Wall      tall
Wide     Side
brown   town
blue      glue
swing    thing

(2) Find out the hidden words from the given words.
(निम्न शब्दों में से छिपे हुए शब्द ढूँढ़ो।)
Answer:
Flower – flow, lower.
Search – sea, ear.
Swing – wing, in.

Class 6th English Practice Test Paper 1 Question 5.
(1) Make new words by joining others words to the given words.
(दिये शब्दों में अतिरिक्त शब्द जोड़कर नये शब्द बनाओ।)
Answer:

1. rain – rain coat.
2. hand – hand bag.
3. bath – bath tub.

(2) Write opposites of the following words.
(निम्न शब्दों के विपरीत अर्थ के शब्द लिखो।)
Answer:

1. win – lose.
2. sad – happy.
3. weakness – strength.
4. Pass – fail.
5. friend – enemy.

(3) Choose correct words to fill in the blanks.
(खाली स्थान भरने के लिए सही शब्द चुनें।)
(lie, reality, power, copied, remember.)
Answer:
Till today Gandhiji is a living reality for Indians. He is an example to be copied. He did not remember ever having told a lie during his school days. He taught us the power of non-violence and non-cooperation.

MP Board Class 6 English Practice Test Paper 1 Question 6.
(1) Use these words to complete the conversation.
(बात-चीत को पूरा करने के लिये इन शब्दों को प्रयोग
Yes, I do.
No, I don’t
Answer:
Mona : Do you like trees?
Shalu : Yes, I do.
Shalu : Do you like bananas?
Mona : No, I don’t.

(2) Complete the following phrases.
(निम्न वाक्यांशों को पूश करें।)
Answer:

1. a bunch of bananas.
2. a team of players.
3. a cluster of stars.
4. a herd to cattle.
5. an army of soldiers.
6. a fleet of ships.

(3) Fill up the blanks with the correct form of the words.
(शब्दों की उचित form से रिक्त स्थान भरो।)
Answer:
wide      wider       widest
deep     deeper     deepest
sweet    sweeter    sweetest

Class 6 English Reader Question 7.
(1) Tick ✓ or ✗
(हाँ के लिये (✓) या ना के लिये (✗) चिन्हित करें।)

1. The spider did not succeed in climbing.
2. It is a pleasant thing to sway in a swing.
3. It was a fifty year old mango tree.
4. Gandhiji left Porbander at the age of eight.
5. Gandhiji misspelt the word Kettle.

Answer:

1. ✗
2. ✓
3. ✓
4. ✗
5. ✓

(2) Answer these questions:
(इन प्रश्नों के उत्तर दीजिए:)

Class 6th English Reader Question 1.
What do the village children do when they get tired?
Answer:
The village children sleep under the tree shade when they get tired.

Class 6 English Test Paper 2.
What did the King see in the cave?
Answer:
King saw a spider trying to climb up a wall in the cave and slipping down again and again. It tried it until it succeeded to climb the wall.

Class 6th English Question Paper Question 3.
What left a lasting impression on Gandhiji’s mind?
Answer:
The plays ‘Harish Chandra’ and ‘Shravana Kumar’ left a lasting impression on his mind.

Practice Test Paper-1 Class 5 Question 4.
What can you see when going up in a swing?
Answer:
While going up in the swing we can see rivers, trees, cattle and countryside.

English Reader Class 6 Question 5.
What does the child see in the country side?
Answer:
The child sees rivers, trees and cattle in the countryside.

Class 6 English MP Board Question 6.
Who is the writer of the poem ‘The Swing’
Answer:
Robert Louis Stevensen.

MP Board Class 6 English Book Pdf Question 8.
Write a paragraph on the ‘Uses of Trees’.
(“पेड़ों के प्रयोग” पर एक गद्यांश लिखे।)
Answer:
Trees are very useful to us. They make the air pure by taking in carbon dioxide and giving out oxygen. The oxygen given by them is used by living beings in breathing. They bring rainfall and prevent floods. They also prevent soil erosion or washing away of soil. The remove pollution. Their wood is used in making furniture, pencils; and paper is made from its bark. Some trees like neem, tulsi and cinchona are medicinally important. They also add to the beauty of nature. Life without trees cannot be imagined.

MP Board Class 6 English Solutions

MP Board Class 12th Physics Important Questions Chapter 1 Electric Charges and Fields

Electric Charges and Fields Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Class 12 Physics Chapter 1 Notes MP Board Question 1.
The least charge is:
(a) 1 coulomb
(b) 1 stat coulomb
(c) 1 micro coulomb
(d) Electronic charge.
Answer:
(d) Electronic charge.

Class 12th Physics Chapter 1 Important Questions MP Board Question 2.
The number of electrons in one coulomb charge is:
(a) 5.46 × 10²⁹
(b) 6.25 × l0¹⁸
(c) 1.6 × l0¹⁹
(d) 9.0 × l0²⁹
Answer:
(b) 6.25 × l0¹⁸

MP Board Class 12th Physics Chapter 1 Notes Question 3.
An object has – 80 µ C. What is the excess number of electrons:
(a) 5 × 10⁸C
(b) 16 × 10¹⁴C
(c) 5 × 10¹⁴ C
(d) 16 × 10⁸C.
Answer:
(c) 5 × 10¹⁴ C

Class 12 Physics Chapter 1 Notes MP Board English Medium Question 4.
The intensity of electric field E due to charge Q at distance r as compared to its normal state:
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

MP Board Class 12th Physics Important Questions Question 5.
The value of permitivity E₀ of vacuum (or free space) is :
(a) 8.85 × 10^-12 C
(b) 8.85 × 10^-12C²N^-1m^-2
(c) 9 × 10⁸ Nm²C^-2
(d) 9 × 10 C²N^-1m^-2.
Answer:
(b) 8.85 × 10^-12C²N^-1m^-2

Physics Important Questions Class 12 MP Board 2023 Question 6.
The work done in rotating a dipole of dipole moment p through an angle 180° from direction of uniform electric field E is:
(a) 2pE
(b) pE
(c) \(\frac { 1 }{ p}\)pE
(d) Zero.
Answer:
(a) 2pE

Point estimate calculator helps to calculate point estimate and find best guess with step by step solution.

MP Board 12th Physics Important Questions 2023 Pdf Question 7.
The intensity of electric field E due to a dipole of dipole moment p at a point a distance r from center of dipole depends on r as :
(a) E ∝ r
(b) E ∝ \(\frac { 1 }{ { r }^{ 2 } } \)
(c) E ∝ \(\frac { 1 }{ r }\)
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)
Answer:
(d) E ∝ \(\frac { 1 }{ { r }^{ 3 } } \)

Imp Questions Of Physics Class 12 MP Board Question 8.
A surface S is placed in an electric field E parallel to the field. The flux linked with the surface be:
(a) ES
(b) \(\frac{ E }{S}\)
(c) Zero
(d) Infinite.
Answer:
(c) Zero

Physics Important Questions Class 12 MP Board Question 9.
The total electric flux emerges out from unit positive charge in air is:
(a) E₀
(b) \(\frac { 1 }{ { E }_{ 0 } } \)
(d) \(\frac { 1 }{ { 4\pi E }_{ 0 } } \)
(d) 4πE₀
Answer:
(b) \(\frac { 1 }{ { E }_{ 0 } } \)

MP Board Class 12 Physics Notes In English Question 10.
The dielectric constant of metal is:
(a) Infinite
(b) Zero
(c) One
(d) None of these.
Answer:
(a) Infinite

Class 12 Physics Important Questions MP Board Question 11.
The dielectric constant of air is:
(a) 8.85 × l0^-12 C² N^-1m^-2
(b) 1
(c) Infinite
(d) None of these
Answer:
(c) Infinite

Physics Notes Class 12th MP Board Question 2.
Fill in the blanks:

1. The dimensional formula of charge is …………….
2. The two like charges …………… each other.
3. Two unlike charges ……………. each other.
4. The SI unit of electric field is …………….
5. The dimensional formula of electric field is …………….
6. The size of an ideal dipole is …………….
7. Two charges – q and + q are situated at a distance / its dipole moment will be …………….
8. The size of an ideal dipole is …………….
9. Two charges – q and + q are situated at a distance lits dipole moment will be …………….
10. The maximum torque acting on a dipole of dipole moments p in a uniform electric field E is …………….

Answers:

1. [AT]
2. Repel
3. Attract
4. NC^-1
5. [ML^-3A^-1’]
6. Point size
7. q.I
8. pE sin G.

MP Board 12th Physics Notes Pdf English Medium Question 3.
Match the Column:

Answers:
(b) newton / coulomb
(e) coulomb
(a) coulomb x metre
(d) N – m²/C
(c) coulomb²/newton x meter ²

answers:

1. (e) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)
2. (d) \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { Kr }^{ 2 } } \)
3. (b) \(\frac { q }{ { E }_{ 0 } } \)
4. (c) pE(1 – cosθ)
5. (a) – pE cos θ

MP Board Class 12th Physics Imp Questions 2023 Question 4.
Write the answer in one word / sentence:

1. What type of charge get accumulated in a glass rod rubbed with silk?
2. What type of charge get accumulated in rubber rod when rubber with fur fabric?
3. What is the cause of quanti zation of charge?
4. What is dielectric constant of a metal ?
5. What is the direction of electric dipole moment?

Answers:

1. Positive charge
2. Negative charge
3. Transfer of electrons takes place in integral multiple. This is the cause of quantization
4. Infinity
5. Direction of dipole moment is from negative charge to positive charge

Electric Charges and Fields Important Questions Very Short Answer Type Questions

MP Board 12th Physics Objective Pdf Question 1.
What does q₁ + q₂ = 0 signify in electrostatics?
Answer:
The equation signifies that the electric charges are algebraically additive and here q₁ and q₂ are equal and opposite.

Electric Charges And Fields Class 12 Important Questions Question 2.
Why electric lines of force do not form closed loop? (AH India 2014)
Answer:
Because electric lines of force start from the positive charge and end on die negative charge.

Physics Class 12 Chapter 1 Important Questions Question 3.
Why must electrostatic field at the surface of charged conductor be perpendicular to every point on it ? (Foreign 2014)
Answer:
As, electric field inside a conductor is always zero. So electric lines of force exert lateral pressure on each other and hence repulsion between like charges take place. Thus, in order to stable the spacing, the lines of force are normal to surface.

Chapter 1 Important Questions Physics Class 12 Question 4.
What quantity is electric flux? Write its SI unit.
Answer:
It is a scalar quantity. S.I. unit of electric flux is Nm²/C 

Important Questions Chapter 1 Physics Class 12 Question 5.
Charge of on QNC is placed as centre of a cube. What will be the electric flux passing through a face of it?
Answer:

MP Board Class 12 Physics Solutions English Medium Question 6.
Charges +q₁ -5q₁, +2q and +2q is at Gaussian surface. What will be the magnitude of electric flux passing through Gaussian surface.
Answer:
Zero, because total charges is zero as surface of Gaussian surface.

Class 12 Physics Chapter 1 Most Important Questions Question 7.
Fig. shows three charges + 2q, – q and+ 3q. Two charges T 2q and – q are enclosed within a surface S. What is the electric flux through the surface S?
Answer:

Physics Class 12 Chapter 1 Important Questions Cbse Board Question 8.
Does charge get changes with the velocity of charged conductor?
Answer:
No? charge does not get change with velocity of charged conductor.

Question 9.
Which scientist first discovered the charge as positive and negative?
Answer:
Benjamin Franklin.

Question 10.
Is static electric force is Newtonian between two charges?
Answer:
Yes, static electric force Newtonian between two charges.

Question 11.
What is the relation between dielectric constant in vacuum and medium?
Answer:
£ = £₀ × £_r

Question 12.
What is the electric force acting upon charge q placed in electric field E?
Answer:
qE

Electric Charges and Fields Important Questions Short Answer type Questions

2 Mark Questions For Physics Class 12 Chapter 1 Question 1.
Write the limits of inverse square law.
Answer:

• It is true only for point charge.
• It is only for that charges which is at rest.

Question 2.
Write the basic properties of electric charge.
Answer:

• Quantization of charge i.e. q = ±ne.
• Charge cannot be created nor be destroyed.

Question 3.
What is dielectric constant of a metal?
Answer:
The dielectric constant of a metal is ∞

Question 4.
Define intensity of electric field and write its unit.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge qo placed at a point in the electric field of intensity\(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) = \(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its S.I. unit is newton/coulomb.

Question 5.
What is an electric dipole?
Answer:
A system of two equal and opposite charges kept very close to each other is • called an electric dipole.

Question 6.
What is an electric flux? Give its SI Units.
Answer:
Electric flux associated with electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. Its SI unit is newton meter per coulomb.

Question 7.
What do you mean by quantization of electric charge?
Answer:
Quantization of electric charge means that the total charge (q) of a body is always ah integral multiple of basic charge (e) which is the charge on the electron.
.’. Charge on any body, q = ± ne
Where, e = 1.6 x 10^-19 coulomb and n = 0,1,2,3 In above relation, +ve sign denotes the loss of electrons and – ve sign denotes gain of electrons. certain particles are known having these charges which are called quarks.
Charge on any body will never be a fraction of e Like \(\frac { 3 }{ 2 }\)e, \(\frac { 5 }{ 2 }\)e, \(\frac { 7 }{ 2 }\)e. But, now days, certain particles are known having these charges which are called quarks.

Question 8.
Ordinary rubber are insulators but the types of an aeroplane are made a little bit conducting. Why?
Answer:
When the aeroplane lands and takes off, the friction between types and the run-way may cause the electrification of types. Due to conducting nature of the types, the charge so produced is conducted to the earth and electrical sparking is avoided.

Question 9.
State Gauss’ theorem in electrostatics.
Answer:
The net electric flux passing through any closed surface is \(\frac { 1 }{ { E }_{ 0 } } \) times the net charge present inside it. Mathematically,

Question 10.
The mass of an isolated conductor decreases when some positive charge is given to it. Why?
Answer:
Mass will reduce because when electrons escape out from the conductor then only it becomes positively charged.

Question 11.
Write the nature of the force acting between q₁ and q₂ if –

1. q₁ q₂ > 0
2. q₁q₂ <0
3. q₁q₂ = 0.

Answer:

1. If q₁q₂ > 0, then both charges are similar and repulsive force acts in between them.
2. If q₁ q₂ < 0, then both charges are of opposite nature and attractive force acts in between them.
3. If q₁q₂ = 0 then one of the two charges is neutral and no force acts in between them.

Question 12.
State principle of conservation of charges. Give examples.
Answer:
According to principle of conservation of charge “Net charge in any isolated system always remains constant”. In the other words “Charge can neither be created nor be , destroyed although it may be transferred from one body to another body”.

Example:
1. Pair annihilation:
When an electron and a positron come near to each other, then they get destroyed and form a y ray photon. Thus the total charge becomes zero.

2. Pair production:
A gamma ray photon splitted into a positron and an electron

Question 13.
Define electric dipole moment. Is it scalar or vector? Write its SI unit and dimensional formula.
Answer:
Electric dipole moment is the product of magnitude of any one charge of dipole and the distance between both the charges. If + q and – q are two charges placed at a distance 2l apart, then electric dipole moment will be
\(\vec{p}\) = q. 2l

It is a vector quantity whose direction is from negative charge to positive charge.
Unit: Its SI unit is coulomb – meter.
Dimension: [p] = [q][2l] = [ATL]
|p] = [M⁰L¹T¹A¹].

Question 14.
Metallic ropes are attached with the vehicles carrying inflammable substances, connected to earth. Why? (NCERT)
Answer:
Moving vehicles get charged due to friction. The inflammable material may catch fire due to spark produced by charged vehicle. When metallic ropes or chain is used, the charge developed on the vehicle is transferred to the ground and so the fire is prevented.

Question 15.
When the electric dipole becomes in equilibrium state when kept in electric field?
Answer:
We know that U = – pE cosθ
When the angle between electric dipole and electric field is zero then the electric dipole will be in equilibrium.

Question 16.
State Coulomb’s inverse square law of electrostatic charges. On this basis define unit charge. What is the condition for the law to be applicable?
Answer:
Coulomb’s inverse square law:
According to it, force of attraction or repulsion between two point charges is directly proportional to the product of their magnitude and inversely proportional to the square of distance between them. Force of attraction or repulsion is the central force which acts along the line joining the two charges. Let the charges q₁ and q₂ be placed at a distance r apart from each other. So, the force acting between them according to Coulomb’s law will be
F ∝ q₁q₂ and F ∝ \(\frac { 1 }{ { r }^{ 2 } } \)

Combining the above two relations, we get

To define unit charge : Let q₁ = q₂ = q, r = lm and F = 9 x 10⁹ newton.
Putting these values in eqn. (1), we get
9 x 10⁹N = 9 × 10⁹ × \(\frac { { q }^{ 2 } }{ { 1m }^{ 2 } } \)
q = ±1 coulomb
So, if two charges of same type and equal magnitude are placed at 1 m apart in vacuum and if force of repulsion between them is 9 × 10⁹ newton, then the two charges are said to be unit charges of magnitude 1 coulomb.

Condition for the applicability of the law:

• It is applicable for the point charges which are stationary’, not for moving charges.
• It is not valid for large distances and distances less than 10^-15 meter because at distances less than 10^-15 m, nuclear forces are dominant.

Question 17.
Define intensity of electric field and write its unit Four properties of electric lines of force.
Answer:
The electric field intensity at a point due to a source charge is defined as the force experienced by a unit positive test charge placed at that point. If \(\overrightarrow{\mathrm{F}}\) is the force acting on test charge q placed at a point in the electric field of intensity \(\overrightarrow{\mathrm{E}}\) then \(\overrightarrow{\mathrm{E}}\) =\(\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
Unit: Its SI unit is newton / coulomb.

Properties:
(a) Electric lines of force start from positive charge and end on negative charge.
(b) Tangent drawn at any point of the lines of force gives the direction of electric field intensity.
(c) Two lines of force will never intersect each other.
(d) Lines of force try to contract longitudinally. This demonstrates why unlike charges attract each other.
(e) They are emitted normal to the surface of conductor and end normally.

Question 18.
Why the two lines of force do not cross or intersect each other?
Answer:
If two lines of force intersect, then there would be two tangents and hence two directions of electric field at the point of intersection, which is not possible.

Question 19.
Prove that electrostatic force is much stronger than the gravitational force.
Or
Compare the electrostatic force and gravitational force.(NCERT)
Answer:
Let a proton and an electron be placed in air or vacuum at lm distance apart, mass of proton and electron are mp and me respectively.
m_p = l.67 × 10^-27 kg
m_e = 9.1 × 10^-31 kg
Charge on proton is q_p = 1.6 ×10^-19C. and charge on electron is q_e = 1.6 x 10^-19C The electrostatic force acting between them is –
\(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ p }{ q }_{ e } }{ { Kr }^{ 2 } }\)

The gravitational force between proton and electron is –

Dividing eqn. (1) by eqn. (2), we get

F_e = 2.27 × l0³⁹ × F_G
Thus, the electrostatic force of attraction is 1039 times more than the gravitational force for the electron – proton system. Obviously the electrostatic force acting between two charged particles is very much greater than the gravitational force acting between them.

Question 20.
(a) Define electric flux. Write its SI unit and dimensional formula.
Answer:
Electric flux associated with the electric field is a measure of total lines of force passing normally to the surface, when held in the electric field. It is denoted by Φ_E If the normal drawn to the surface dS makes an angle 0 with the electric field E, then the flux through this area is –
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\)
Unit: Its SI unit is Nm²/ C.

Dimension:

Question 20. (b)
Under what conditions flux is said to be positive and negative?
Answer;
Electric flux:
The electric flux associated with a surface is a measure of total lines of force passing normally to the surface when held in the electric field. It is denoted by ∅_E There are two types of electric flux:

• Positive electric flux – When electric lines of force leaves any body through its surface, it is considered as positive flux.
• Negative electric flux – When electric lines of force enters into body through any surface, it is considered as negative flux.

Question 20. (c)
Show that the electric flux passing through a surface parallel to electric field is zero.
Answer:
Let \(\overrightarrow{\mathrm{E}}\) be the intensity of electric field and \(\overrightarrow{\mathrm{dS}}\) be the normal to the arbitary surface. If the normal to the surface (\(\overrightarrow{\mathrm{dS}}\)) marks an angle θ with E, then,
The electric flux can be given by
d∅_E = E. dS cos θ
If surface is parallel to the electric field
i.e., θ = 90°
d∅_E = E dS cos 90°
= E.dS × 0
d∅_E = 0

Thus flux passing through a parallel surface to an electric field is zero.

Question 21.
State the principle of electrostatic shielding.
Answer:
The electric field inside the charged hollow conductor (spherical shell) is zero. Therefore, when an object is placed inside a hollow conductor then there will be no effect of electric field on the object i.e., the hollow conductor shields the object from external electric field.

The concept of electrostatic shielding is used to shield wires carrying audio signal from external fields like atmospheric electricity or due to electric sparks. If such wires are not shielded, the audio signal will give rise to noisy sound or reproduction.

Question 22.
Derive an expression of electric field intensity on a point in axial position (end on position) of an electric dipole.
Answer:
Consider an electric dipole made up of charges + q and – q separated by a distance 2/apart and placed in vacuum. We have to find out the electric field at point P situated at a distance r from the center of dipole system. To find out the electric field intensity, imagine a unit positive test charge situated at P.

Electric field intensity due to charge (+q) situated at A will be

Electric field intensity due to charge (-q) situated at B will be 1

As AP < PB, hence repulsive effect due to charge +q will be more effect then the attractive effect of charge – q. Therefore, E₁> E₂ but their directions are opposite. Hence, the net electric field will be

This is the expression for the electric field which is directed from A to P. For small and strong dipoles,
r>>2l ⇒ r>>l ⇒ r² >>>l²
∴ r² – l² ≈ r²
Hence, eqn. (1) becomes

This is the required expression.

Question 23.
Derive the expression for the intensity of electric field on the equatorial position of the dipole.
Answer:
Equatorial position is a point on the line which lies on the perpendicular bisector of the axial line, then their position is called equatorial position or broadside on position. Let AB be an electric dipole formed by charges +q and -q. The distance between the charges is 21. P is a point on the equatorial position of the dipole, situated at a distance r from the center of the dipole. Sup- pose unit positive test charge is situated at P. Electric field at P due to +q charge will be

Electric field at P due to -q charge will be


Resolving E and E₂ into its components, we get the component of E₁ are E₁cosθ parallel to AB and E₁ sin θ along OP. Also, components of E₂ are E₂ cosθ parallel to AB and E₂ sin O along PO.
From eqns. (1) and (2), we find |E₁| = |E₂|, hence E₁ sinθ = E₂ sinθ , but their directions are opposite. So, they cancel each other. The only resultant electric field will be
E = E₁ cos θ + E₂ cos θ
= E₁ cosθ + E₁ cos θ
E = 2E₁ cos θ  …………(3)

If the charges are kept in any other medium other than vacuum, then the electric field becomes

The direction of electric field is along AB i.e., it is parallel to axial line which acts from positive charge to negative charge. For small and strong dipoles, r >> 2l ⇒ r >> l
r2 + l2 ≈ r2
Hence, eqn. (4) becomes

Question 24.
Obtain an expression for the intensity of electric field due to a charge q at a distance r from it.
Answer:
Consider a point charge q which is placed at the origin O of the generalized coordinate system. We have to find out the electric field at point P at a distance r from it. For this, imagine a test charge q₀ placed at point P. Both the charges q and q₀ have been placed in vacuum. According to Coulomb’s law, the force on charge q₀ will be

Where, r is the unit vector in the direction from q to q₀. But intensity of electric field is given by the formula,

If any other medium is present having dielectric constant then

Where, £0 = k = dielectric constant of the medium. In magnitude,

Question 25.
Derive the expression for the torque acting on the dipole when kept in a uniform electric field and hence define dipole moment.
Answer:
Consider an electric dipole made up of charges +q and -q and length 2l which is placed in a uniform electric field of strength E. The dipole moment P makes an angle Q with respect to the electric field at any instant of time. As a result, a pair of forces +qE (along the direction of electric field) and -qE (opposite to the direction of electric field) acts on the dipole. These two opposite forces are of equal magnitude. So, they apply a couple on electric dipole, which tries to bring it along the direction of electric field. This couple is called restoring couple.

So, moment of couple or torque (τ)
= Magnitude of any force x Perpendicular distance between the forces
= qE × BC …….(1)
In A ABC, sinθ = \(\frac { BC }{ AB }\)
⇒ BC = AB sinθ = 2l sinθ
Putting the value of BC in eqn. (1), we get,
τ= qE × 2l sin θ
⇒ τ = pEsin θ,
(∴q 2l = p, where p is the dipole moment) ……..(2)
In vector form,
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{p}}\) × \(\overrightarrow{\mathrm{E}}\)
From eqn. (2), if E = 1 and 0= 90°.Then,
τ = p sin 90° = p
Thus, dipole moment of an electric dipole is numerically equal to the torque acting on the dipole when it is placed perpendicular to a unit electric field.

Question 26.
Derive an expression for the potential energy of a dipole in an electric field.
Answer:
Work done in rotating any electric dipole from standard position to any angle with respect to the electric field is called potential energy. The standard position is considered to θ = i.e., perpendicular to the field. Let p be the dipole moment of the dipole, placed in an electric field of intensity E and it makes an angle θ with respect to the electric field, then
Torque, τ = p E sinθ
Also, work done in displacing the dipole through an angle dθ will be
dW = τdθ = pE sinθ dθ
Now, the work done in displacing from θ = \(\frac { π }{ 2 }\) to θ = θ is

This work done is stored in the form of potential energy,
U = – p E cosθ
In vector form, U = \(\overrightarrow{\mathrm{-p}}\).\(\overrightarrow{\mathrm{E}}\)
This is the required expression.

Question 27.
Derive the expression for the amount of work done in rotating a dipole in a uniform electric field.
Answer:
If a dipole placed in a uniform electric field is rotated from the equilibrium position, then work has to be done. Consider an electric dipole of dipole moment p(= 2 ql) which has been rotated in an electric field E by an angle 0 from the equilibrium position, then torque acting on it will be –
\(\overrightarrow{\mathrm{τ}}\) = \(\overrightarrow{\mathrm{-p}}\) × \(\overrightarrow{\mathrm{E}}\) = pE sin θ, (in magnitude form)
Work done in rotating the dipole through an angle dθ is
dw = τ.dθ
Therefore, work done in rotating it by an angle θ can be obtained from the above equation under proper limits.

Above equation is the total work done in rotating any electric dipole through an angle θ

Question 28.
Write and prove Gauss’ theorem.
Or
Prove that net electric flux through any closed surface is 1 / £₀ times of total charge present inside it, where s £₀ is
permittivity of free space.
Answer:
Gauss’ law:
According to Gauss’ law, the net electric flux through any closed surface is 1 / £₀ times of the total charge present inside it.

Where, £₀ is permittivity of vacuum or air.
Proof:
Let at any point O, inside any closed surface, a charge +q is present. A point P lies at a distance r from O on surface dS. Normal to the surface dS can be represented by \(\overrightarrow{\mathrm{dS}}\) (say). Therefore, electric field at point P due to charge +q can be given by

.’. Electric flux passing through area \(\overrightarrow{\mathrm{dS}}\)will be
dΦ_E = \(\overrightarrow{\mathrm{E}}\).\(\overrightarrow{\mathrm{dS}}\) = E dS cos0

Where, 0 is an angle between electric field f and the normal to the surface.
Net flux through the closed surface is.

But \(\frac { dscos\theta }{ { r }^{ 2 } } \) = dw is the solid angle subtended by the small surface dS at O.

As total solid angle subtended by any closed surface at any point inside it is 4π hence –

Above equation is mathematical form of Gauss’ law.

Electric Charges and Fields Important Questions Long Answer Type Questions

Question 1.
Obtain an expression for the intensity of electric field due to linear charge distribution.
Answer:
Consider a thin wire of length ‘l’ Let the charge given to it be q. So, the linear charge density will be λ = \(\frac { q }{ l }\) ⇒ q = λl
If a test charge ¡s brought near to the wire, then it will be repelled. Hence, the electric field (\(\overrightarrow{\mathrm{E}}\)) will be radially out ward. Draw a normal \(\overrightarrow{\mathrm{dS}}\) on the surface at P. The direction of \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) are the same. The electric flux passing through the curved surface of the cylinder is

= E 2πrl (∴\(\overrightarrow{\mathrm{dS}}\) = Total surface areaofcurved surface 2,πrl …(2)
But according to Gauss’ theorem,
Φ = \(\frac { 1 }{ { E }_{ 0 } } \) x (Total charge present inside the Gaussian surface) …(3)
Φ = \(\frac { q }{ { E }_{ 0 } } \)
From eqns. (2) and (3), we have
E 2πrl = \(\frac { q}{ { E }_{ 0 } } \)
E 2πrl = \(\frac { λl}{ { E }_{ 0 } } \)
[∴q = λl from eqn. (1)]

This is the expression for the electric field due to a linear charge distribution having a linear charge density λ.

Question 2.
Derive an expression for the intensity of electric field due to a uniform infinite plane sheet of charge.
Answer:
Consider an infinite plane sheet on which charge given is q. If S is the total surface area of the sheet, then the surface charge density will be σ = \(\frac {q }{ S}\) = \(\frac { charge }{ area }\) the charge will be distributed uniformly on the surface of the sheet. Hence, surface charge density (= σ) remains constant. Any test charge held near to the surface will be repelled on both the sides i.e., either to paper and the other will go inside, hence the Gaussian surface formed will be cylindrical; which is having three surfaces S₁, S₂ and S₃.

The electric field \(\overrightarrow{\mathrm{dS}}\) is directed radially outwards away from the sheet. Moreover, \(\overrightarrow{\mathrm{dS}}\) is the outward drawn normal on the Gaussian surface. Hence, the total electric flux is

The angle between normal to the curved surface S3 in the direction of electric field \(\overrightarrow{\mathrm{E}}\)] is 90°, hence

For the surfaces Sj and S2, \(\overrightarrow{\mathrm{E}}\) and\(\overrightarrow{\mathrm{dS}}\) are along the same direction, hence the angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°, so

By Guss’ theorem,
Φ_E = \(\frac { 1}{ { E }_{ 0 } } \) q ……(2)
from eqns. (1) and (2) we get
2ES = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac{q}{2 \varepsilon_{0} S}\)
Hence, E = \(\frac { σ}{ { 2E }_{ 0 } } \)
(∴ σ = \(\frac { q }{ S }\))
This is the expression for the intensity of electric field due to a uniformly charged Sheet

Question 3.
Prove that intensity of field is zero inside a spherical shell, while in outside it is such that all the charges are concentrated at the center.
Or
Determine the intensity of electric field by Gauss’ law, due to a uniformly charged spherical shell at a point

1. Outside the shell
2. On the surface of shell
3. Inside the shell.

Draw a graph showing the variation of electric field with the distance from the center.
Answer:
Intensity of electric field due to uniformly charged spherical shell:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity at any point due to this charged sphere depends on its position relative to the sphere.

(i) When point P lies outside the spherical shell:
Consider a point P which lies outside the spherical shell of radius R. Now, imagine a sphere of radius r which passes through P. This closed surface behaves as Gaussian surface. Since, sphere is uniformly charged, so that electric field at each point on the surface is equal and points radially outward.

Net flux through the Gaussian sphere of radius r will be

But, charge enclosed inside the Gaussian surface is q. Therefore, from Gauss’ law
Φ_E = \(\frac { q }{ { E }_{ 0 } } \) …….(2)
Equating eqns. (1) and (2), we have
E. 4πr² = \(\frac { q }{ { E }_{ 0 } } \)
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \) ……..(3)

It is clear from above equation that, the electric field outside any uniformly charged spherical shell is exactly to that electric field when we take the same charge at the center of sphere.

(ii) Point P lies on the surface of spherical shell:
In this case, r = R
From eqn. (3),
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { R }^{ 2 } } \)

(iii) When point P lies inside the spherical shell:
As we know, the charge given to spherical shell spreads equally all over the surface, there is no charge present inside the sphere. Hence, by Gauss’ law,


From equs. (4) and (5),
E.4πr² =  0
⇒ E = 0
Therefore, electric field inside a spherical shell is always zero. The adjacent figure shows the variation of electric field with distance.

Question 4.
Obtain an expression for the intensity of electric field due to a solid sphere of charge using Gauss’ theorem at the following points:

1. Outside the sphere
2. On the surface of sphere
3. Inside the sphere.

Also prove that the intensity of the electric field at the center will be zero.
Answer:
Suppose that a sphere of radius R is uniformly charged with +q charge. Intensity of electric field at any point depends on its position with respect to sphere. The volume charge density of sphere is
p = \(\frac { Charge }{ volume }\)

(i) When the point lies outside the sphere:
Consider a point P which lies outside the spherical shell of radius R. Now imagine a sphere of radius r which passes through P.This closed surface behaves as a Gaussian surface. The electric field at each point outside the sphere points radially outwards.

.’. Total flux through the Gaussian sphere of radius r will be –

But charge enclosed inside the Gaussian surface is q. Hence, from Gauss’ law
Φ = \(\frac { 1}{ { E }_{ 0 } } \) q …..(3)
From eqns. (2) and (3),
E 4πr² = \(\frac { q}{ { E }_{ 0 } } \)
E = \(\frac { q }{ { 4\pi { E }_{ 0 }r }^{ 2 } }\)
Hence, the electric field outside any uniform charged sphere is same as if the entire charge is present at the centre.
Putting the value of q from eqn. (1) in eqn. (4), we get

(ii) When the point lies on the surface of sphere:
r = R
From eqn. (4), we have E =

and from eqn. (5), we have E =

(iii) When the point lies inside the sphere:
Suppose the point P lies at a distance r from the center of the sphere such that r < R. At radius r draw a Gaussian surface which is a sphere. Then electric field \(\overrightarrow{\mathrm{E}}\) is radially outwards. Moreover, normal to surface \(\overrightarrow{\mathrm{dS}}\) is along the direction of \(\overrightarrow{\mathrm{E}}\). Hence, angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{dS}}\) is 0°
The flux coming out of the Gaussian surface is –

But by Gauss’ law, the flux Φ = \(\frac { 1}{ { E }_{ 0 } } \) x (Total charge inside the Gaussian surface)
Φ = \(\frac { q}{ { E }_{ 0 } } \) q ……(2) (where q’ is the total charge inside the sphere of radius r)
On volume \(\frac { 4 }{ 3 }\) πR³charge given is q
On volume 1, charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \)
On volume\(\frac { 4 }{ 3 }\) πR³ nr charge given is = \(\frac { q }{ \frac { 4 }{ 3 } \pi R^{ 3 } } \) x \(\frac { 4 }{ 3 }\) πR³ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
q’ = \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \)
‘Putting the value of q’ in.eqn. (2), we get
Φ = \(\frac { q}{ { E }_{ 0 } } \) \(\frac { q{ r }^{ 3 } }{ { R }^{ 3 } } \) …….(3)
from equs. (1) and (3) we get,

At the center of sphere, r = 0, hence E = 0.

Question 5.
Neeta return home from school. When she was taking out the sweater, she heard twinkling sound. She asked her mother but she was unable to answer. Next day she went to school and asked the science teacher, she was satisfied with her answer. Write down the answer of the question given below on the basis of above statement:

1. What values does she show?
Answer:
She shown the value of answer, questioning.

2. What was the reason for the sound when she was taking out her sweater?
Answer:
Due to the friction between her school dress and sweater, its acquire negative charge which get discharge. Therefore the sound was been listen by her.

3. Presence of charge can be determined by which apparatus?
Answer:
Gold leaf electroscope.

4. Does any new charges get created on rubbing? If yes or not than why?
Answer:
No, new charges get created, only transfer of electron occur form one body to another. There one become positive charge and other become negative charge.

Electric Charges and Fields Important Questions Numerical Questions

Question 1.
Two positive ions which have same charges having a force of repulsion of
3.7 × 10^-9N. The distance between them is 5 Å. Find the deficiency of electron on each ions.
Solution Given:
F = 3.7 × 10^-9N, r = 5Å = 5 × 10^-10m
q₁ = q₂ = q

= 10.28 × 10³⁸
q = 3.2 × 10^-19
n = \(\frac { q }{ e }\)

There will be deficiency of two electron in each ions.

Question 2.
Calculate force of repulsion between two protons when they are separated with 4.0×10¹⁵m.
Solution:
We know that, \(\frac { 1 }{ 4\pi { E }_{ 0 } } \) = 9 × 10⁹Nm²C^-2
Given: q₁= q₂ = +1.6 × 10^-19 C,r = 4.0 × 10^-15m.
From formula, F = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { Kr }^{ 2 } }\)

or F = 14.4 newton.

Question 3.
Can 1 coulomb of charge be gives to a metallic sphere of radius 1cm? Give reason.
Solution:
Here, q = 1 C, r = lcm = 10^-2m
Electric field on the surface of sphere is
E = \(\frac { 1 }{ { 4\pi E }_{ 0 } }\) \(\frac { q }{ { r }^{ 2 } } \)
= 9 × 10⁹ × \(\frac { 1 }{ { \left( { 10 }^{ 2 } \right) }^{ 2 } } \) = 9 × 10³ N /C
But when intensity of electric field exceeds 3 × 10⁶ N / C air gets ionized and hence, electric discharge takes place. So l Cm of charge cannot be given to a sphere of radius 1 cm.

Question 4.
Two charge of one coulomb are separated by a distance of 1 meter from each other, find:

1. Force between the charge when was in air
2. If it is in dielectric medium, calculate its force.

Solution:
Given:
q₁= q₂ = 1 coulomb, r = l m

Question 5.
Two point charges + 9e and + e are placed 8 m apart. Where should the third charge q be placed on the line joining the two charges so that q should be in equilibrium?
Solution:
Let the charge q be placed at distance x from +9 e charge, then its distance from e will be 8 – x.
In equilibrium,


Taking + ve sign \(\frac { 3 }{ x }\) = \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = x
24 = 4x or x = 6m
Taking – ve sign, \(\frac { 3 }{ x }\) = – \(\frac {1 }{ 8 – x}\)
⇒ 24 – 3x = -x
24 – 2x or x = 12 m
Since, +9e and+e are similar charges, hence q will be in equilibrium when x = 6 m i.e., q should be placed 6 m apart from +9e between the charges.

Question 6.
What is the force between two small sphere having charge of 2 × 10^-7 C and 3 × 10^-7C? (NCERT)
Solution:
Given: q₁ = 2 × 10^-7C,
q₂ = 3 × 10^-7C,
and r = 30 cm = 30 × 10^-2
m = 0.3 m

The force will be repulsion force a both the charge are same.

Question 7.
The electrostatic force on a small sphere of charge 0.4µC due to another small sphere of charge 0.8µC in air is 2 N.

1. What is the distance between two sphere?
2. What is the force on the second sphere due to the first?

Solution:
Given:
q₁ = 0.4µC = 0.4 × 10^-6 C
q₂ = -0.8µC = -0.8 × 10^-6 C
F = 0.2N

= 9 × 2 × 8 × 10^-4
r = 3 × 4 × 10^-2
= 0.12 m = 12 cm.

(ii) Both the charges apply same force on each other. There force of first sphere due to second sphere will be 0.2 N. The force between them will be force of attraction.

Question 8.
A polythene piece is rubbed with woollen cloth as a result 3 x 10.7 coulomb charge is produced in it, then: .

1. How many electrons are transferred from which to which substance?
2. Is mass transferred from woollen to polythene?

Solution:

1. Polythene gets negative charge, so that electron is transferred from woollen to polythene.
We know that, q = ne

or n = 1.875 × 10¹² electrons.

2. As electrons are transferred from woolen to polythene, mass is also transferred in the same way.

Question 9.
A conductor has 14.4 × 10¹⁹ C positive charge. How many electrons are lacking or in excess?
Solution:
We know that, q = ne
⇒ n =\(\frac { q }{ e}\)

Therefore, body is lacking by 9 electrons (as it has positive charge).

Question 10.
What is the net flux of uniform electric flux there through a cube of field oriented so that its faces are parallel to the coordinate plane?
Solution:
All the faces of a cube are parallel to the coordinate axis. Therefore the Number of hold lines entering the cube is equal to the number of field lines piercing out of the cube. As a result net flux through the cube in zero.

MP Board Class 12th Physics Important Questions

In this article, we will share MP Board Class 7th Sanskrit Solutions Chapter 13 सौरमण्डलम् 2 Pdf, Sanskrit Class 7 Chapter 13 Pdf With Answers, These solutions are solved subject experts from the latest edition books.

MP Board Class 7th Sanskrit Solutions Surbhi Chapter 13 सौरमण्डलम्

MP Board Class 7th Sanskrit Chapter 13 अभ्यासः

Class 7 Sanskrit Chapter 13 MP Board प्रश्न 1.
एक शब्द में उत्तर लिखो
(क) स्वप्रकाशरहिताः के? [अपने प्रकाश से रहित कौन होते हैं?]
उत्तर:
ग्रहाः

(ख) सर्वे ग्रहाः कं परिभ्रमन्ति? [सभी ग्रह किसके चारों ओर घूमते हैं?]
उत्तर:
सूर्यम्.

(ग) पृथिव्याः उपग्रह कः? [पृथ्वी का उपग्रह कौन है?]
उत्तर:
चन्द्रः

(घ) कदा चन्द्रस्य दर्शनं न भवति? [चन्द्र का दर्शन कब नहीं होता है?]
उत्तर:
अमावस्यायां रात्रौ।

MP Board Class 7th Sanskrit Chapter 13 प्रश्न 2.
एक वाक्य में उत्तर लिखो
(क) आकाशपिण्डेषु के के दृश्यन्ते? [आकाशपिण्डों में कौन-कौन दिखते हैं?]
उत्तर:
आकाशपिण्डेषु तारागणा: महाः उपग्रहाः च दृश्यन्ते। [आकाशपिण्डों में तारागण, ग्रह और उपग्रह दिखाई देते हैं।]

(ख) पृथिव्याः सूर्य परिक्रमण कालः कः? [पृथ्वी का सूर्य की परिक्रमा करने का समय कौन-सा है?]
उत्तर:
पृथिव्याः सूर्यपरिक्रमणकालः २३.५६ होरासु भ्रमन्ती ३६५.२५ दिवसेषु सूर्य परिक्रमणकालः वर्तते।। [पृथ्वी सूर्य की परिक्रमा करते समय २३.५६ घंटों में घूमती है तथा ३६५.२५ दिनों में सूर्य की परिक्रमा करने का समय है।]

(ग) सूर्यम् परितः के ग्रहा: परिभ्रमन्ति? [सूर्य के चारों ओर कौन-से ग्रह घूमते हैं?]
उत्तर:
सूर्यम् परितः क्रमेण बुधः, शुक्रः, पृथिवी, मङ्गल, गुरुः, शनिः, अरुणः, वरुणः, यम इति ग्रहाः परिभ्रमन्ति। [सूर्य के चारों ओर क्रमशः बुध, शुक्र, पृथ्वी, मंगल, गुरु, शनि, अरुण, वरुण, यम इत्यादि ग्रह घूमते हैं।]

(घ) चन्द्रस्य पृथिवीं परितः भ्रमणकालः कः? [पृथ्वी के चारों ओर घूमने का चन्द्रमा का समय क्या है?]
उत्तर:
चन्द्रस्य पृथिवीं परितः भ्रमणकालः २७.३२ दिवसाः। [चन्द्रमा का पृथ्वी के चारों ओर घूमने का समय २७.३२ दिन है।]

(ङ) सौरमण्डलमिति किमर्थम् उच्यते? [सौरमण्डल किस अर्थ से कहा जाता है?]
उत्तर:
सूर्यस्य मण्डले एव एतेषां सर्वेषां ग्रहाणां उपग्रहाणां च स्थिति सौरमण्डलम् उच्यते। [सूर्यमण्डल में ही इन सभी ग्रहों और उपग्रहों की स्थिति सौरमण्डल कही जाती है।]

Class 7 Sanskrit Chapter 13 प्रश्न 3.
उचित मेल करो

उत्तर:
(क) → (7)
(ख) → (4)
(ग) → (5)
(घ) → (1)
(ङ) → (8)
(च) → (2)
(छ) → (6)
(ज) → (3)

MP Board Class 7 Sanskrit Chapter 13 प्रश्न 4.
रेखांकित शब्दों के आधार पर प्रश्न बनाओ
(क) पूर्णिमायां रात्रौ चन्द्रस्यं दर्शनं भवति।
(ख) वयं प्रतिदनं सूर्यं चन्द्रं नक्षत्राणि च पश्यामः।
उत्तर:
(क) कस्याम् रात्रौ चन्द्रस्य दर्शनं भवति?
(ख) वयं प्रतिदिनम् कानि कानि पश्याम:?

Class 7th Sanskrit Chapter 13 प्रश्न 5.
आत्मनेपद धातु रूपों को लिखो
(क) याच
(ख) सेव्
(ग) शुभ् (शोभ)।
उत्तर:

Class 7 Sanskrit Chapter 13 Question Answer प्रश्न 6.
रिक्त स्थानों को भरो
(क) सूर्य ………… प्रकाशयति।
(ख) सर्वे ग्रहाः ………… परिक्रमन्ति।
(ग) ………….. ग्रहं परितः भ्रमन्ति।
(घ) …………. रात्रौ पूर्णचन्द्रस्य दर्शनम् भवति।
(ङ) प्रतिदिनं चन्द्रस्य …………. दृश्यते।
उत्तर:
(क) स्वप्रकाशेन
(ख) सूर्य
(ग) उपग्रहाः
(घ) पूर्णिमायाम्
(ङ) आकारभेदः।।

Sanskrit Class 7 Chapter 13 प्रश्न 7.
भिन्न प्रकृति के क्रियारूपों का वर्गीकरण करो-
मोदते, जानाति, पठथ, याचे, गच्छामि, सेवामहे, करोषि, शोभेते।
उत्तर:
जानाति, पठथ, गच्छामि, करोषि।

सौरमण्डलम् हिन्दी अनुवाद

वयं प्रतिदिनं सूर्यं चन्द्रं नक्षत्राणि च पश्यामः, चकिताः आह्लादिताश्च भवामः। ब्रह्माण्डस्य आकाशगङ्गायाम् आकाशपिण्डानां समूहाः वर्तन्ते। तेषु आकाशपिण्डेषु तारागणाः ग्रहाः उपग्रहाः च दृश्यन्ते।

ये स्वप्रकाशयुक्ताः भवन्ति ते ताराः इति कथ्यन्ते। तेषु सूर्यः प्रमुखः। स स्वप्रकाशेन ग्रहान् उपग्रहान् च प्रकाशयति।

स्वप्रकाशरहिताः पिण्डाः ग्रहाः इति कथ्यन्ते। सर्वे ग्रहाः सूर्य परिक्रमन्ति। पृथिवी अपि सूर्यात् सुदूरे स्थित्वा स्वयं स्वधुरि २३.५६ होरासु भ्रमन्ती ३६५.२५ द्विवसेषु सूर्य परिक्रामति। पृथिव्या सह अन्ये ग्रहाः अपि सूर्यप्रकाशेनैव प्रकाशिताः भवन्ति। सूर्यः केन्द्रस्थाने तिष्ठति। सूर्यं परितः क्रमेण बुधः,शुक्रः, पृथिवी, मङ्गलः, गुरुः, शनिः, अरुणः, वरुणः, यम’ इति ग्रहाः परिभ्रमन्ति। यद्यपि अब इसे ग्रह नहीं माना जाता है।

अनुवाद :
हम प्रतिदिन सूर्य, चन्द्रमा और नक्षत्रों को देखते हैं, चकित और प्रसन्न होते हैं। ब्रह्माण्ड की आकाश गंगा में आकाश पिण्डों का समूह विद्यमान है। उन आकाशपिण्डों में तारों के समूह, ग्रह और उपग्रह दिखायी पड़ते हैं।

ये सभी अपने प्रकाश से युक्त होते हैं, उन सब को तारे कहते हैं। उनमें सूर्य प्रमुख है। वह अपने प्रकाश से ग्रहों और उपग्रहों को प्रकाशित करता है।

अपने प्रकाश से रहित पिण्ड ग्रह कहे जाते हैं। सभी ग्रह सूर्य की परिक्रमा करते हैं। पृथ्वी भी सूर्य से बहुत दूर रहकर स्वयं अपनी धुरी पर २३.५६ घण्टों में घूमती हुई ३६५.२५ दिनों में सूर्य का चक्कर लगाती है। पृथ्वी के साथ अन्य ग्रह भी सूर्य के प्रकाश से ही प्रकाशित होते हैं। सूर्य केन्द्र स्थान में होता है। सूर्य के चारों ओर क्रमशः बुध, शुक्र, पृथ्वी, मंगल, गुरु, शनि, अरुण, वरुण, यम आदि नवग्रह (नौ ग्रह) घूमा करते हैं।

उपग्रहाः ग्रहं परितः भ्रमन्ति। प्रत्येकस्य ग्रहस्य उपग्रहाः सन्ति। चन्द्रः पृथिव्याः उपग्रहः। चन्द्रस्य पृथिवीं परितः भ्रमण कालः २७.३२ दिवसाः। प्रतिदिनं चन्द्रस्य आकारभेदः दृश्यते। पौर्णिमायाः रात्रौ पूर्णचन्दस्य दर्शनं भवति। अमावस्यायां रात्रौ तु चन्द्रस्य दर्शनं न भवति।

सूर्यस्य मण्डले एव एतेषां सर्वेषां ग्रहाणां उपग्रहाणां च स्थितिः अस्ति। अतः एतदेव सौरमण्डलम् इति उच्यते। एतान् अतिरिच्य आकाशगङ्गायाम् ईदृशानां तारागणनां अनेके समूहाः वर्तन्ते। ब्रह्माण्डस्य विशालतायाः कल्पना दुष्करा एव अस्ति।

अनुवाद :
उपग्रह ग्रह के चारों ओर घूमते हैं। प्रत्येक ग्रह के उपग्रह होते हैं। चन्द्रमा पृथ्वी का उपग्रह है। चन्द्रमा का पृथ्वी के चारों ओर घूमने का समय २७.३२ दिन होता है। प्रत्येक दिन चन्द्रमा के आकार में भेद दिखाई देता है। पूर्णिमा की रात्रि को पूर्णचन्द्र का दर्शन होता है। अमावस्या की रात्रि को तो चन्द्रमा के दर्शन ही नहीं होते हैं।

सूर्यमण्डल में ही इन सभी ग्रहों और उपग्रहों की स्थिति होती है। इसलिए इसे ही सौरमण्डल कहा जाता है। इनके अतिरिक्त आकाशगंगा में ऐसे तारागणों के अनेक समूह होते हैं। ब्रह्माण्ड की विशालता की कल्पना अति दुष्कर ही है।

सौरमण्डलम् शब्दार्थाः

आह्लादिताः = प्रसन्न (बहुवचन)। अतिरिच्य = छोड़कर, अतिरिक्त। दृश्यन्ते = दिखाई देते हैं। कथ्यन्ते = कहे जाते हैं। होरासु = घण्टों में। परितः = चारों ओर। दुष्कराः = कठिन। ईदृशानां = इस प्रकार से।।

MP Board Class 7th Sanskrit Solutions