Class 10

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory:

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Solution:

752 + 20f = 792 + 18.1
20f – 18f = 792 – 752
2f = 40
f = \(\frac{40}{2}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO2 (in ppm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO₂ in the air.
Solution:

Mean concentration of SO₂ in air = 0.099 ppm

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:


Mean number of days a student was absent is 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Jharkhand Board JAC Class 10 Sanskrit Solutions Shemushi Chapter 1 शुचिपर्यावरणम् Textbook Exercise Questions and Answers.

JAC Board Class 10th Sanskrit Solutions Shemushi Chapter 1 शुचिपर्यावरणम्

JAC Class 10th Sanskrit शुचिपर्यावरणम् Textbook Questions and Answers

प्रश्न 1.
एकपदेन उत्तरं लिखत। (एक शब्द में उत्तर लिखिए।)
(क) अत्र जीवितं कीदृशं जातम्? (यहाँ जीवन कैसा हो गया है?)
(ख) अनिशं महानगरमध्ये किं प्रचलति? (दिन-रात महानगर में क्या चलता है?)
(ग) कुत्सितवस्तुमिश्रितं किमस्ति? (खराब वस्तु मिलावट वाला क्या है?)।
(घ) अहं कस्मै जीवनं कामये? (मैं किसके लिये जीवन की कामना करता हूँ?)
(ङ) केषां माला रमणीया? (किसकी माला सुन्दर है?)
उत्तरम् :
(क) दुर्वहम् (दूभर)।
(ख) कालायसचक्रम् (लोहे का पहिया)।
(ग) भक्ष्यम् (भोजन)।
(घ) मानवाय (मनुष्य के लिये)।
(ङ) हरिततरूणां ललितलतानाम् (हरे वृक्षों और सुन्दर बेलों की।)

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत (निम्नलिखित प्रश्नों के उत्तर संस्कृत भाषा में लिखिए)
(क) कविः किमर्थं प्रकृतेः शरणम् इच्छति ?
(कवि किसलिए प्रकृति की शरण चाहता है ?)
उत्तरम् :
कवि महानगरस्य दुर्वहं जीवनं दृष्ट्वा तस्मात् भीत: शुद्धपर्यावरणाय प्रकृतेः शरणम् इच्छति।
(कवि महानगर के दुष्कर जीवन को देखकर उससे डरा हुआ शुद्ध पर्यावरण के लिए प्रकृति की शरण में जाना चाहता है।)।

(ख) कस्मात् कारणात् महानगरेषु संसरणं कठिनं वर्तते ?
(किस कारण से महानगरों में विचरण कठिन हो गया है ?)
उत्तरम् :
अहर्निशं लौहचक्रस्य सञ्चरणात् यानानां बाहुल्यात् च महानगरेषु संसरणं कठिनं वर्तते।
(दिन-रात लौहचक्र के घूमने से तथा वाहनों की बहुलता के कारण महानगरों में चलना कठिन है।)

(ग) अस्माकं पर्यावरणे किं किं दूषितम् अस्ति ?
(हमारे पर्यावरण में क्या क्या दूषित है ?)
उत्तरम् :
अस्माकं पर्यावरणे वायुमण्डलं, जलं, भक्ष्यं, धरातलं च दूषितम् अस्ति।
(हमारे पर्यावरण में वायुमण्डल, जल, खाद्य पदार्थ और पृथ्वी दूषित हैं।)

(घ) कविः कुत्र संचरणं कर्तुम् इच्छति ?
(कवि कहाँ विचरण (भ्रमण) करना चाहता है ?)
उत्तरम् :
कविः ग्रामान्ते एकान्ते कान्तारे सञ्चरणं कर्तुम् इच्छति।
(कवि गाँव के बाहर एकान्त वन में भ्रमण करना चाहता है।)

(ङ) स्वस्थजीवनाय कीदृशे वातावरणे भ्रमणीयम् ?
(स्वस्थ जीवन के लिए कैसे वातावरण में घूमना चाहिए ?)
उत्तरम् :
स्वस्थजीवनाय शुचि-वातावरणे (पर्यावरणे) भ्रमणीयम्।
(स्वस्थ जीवन के लिए शुद्ध वातावरण में घूमना चाहिए।)

(च) अन्तिमे पद्यांशे कवेः का कामना अस्ति ?
(अन्तिम पद्यांश में कवि की कामना क्या है ?)
उत्तरम् :
अन्तिमे पद्यांशे कवेः कामना अस्ति यत् निसर्गे पाषाणी सभ्यता समाविष्टा न स्यात्।।
(अन्तिम पद्यांश में कवि की कामना है कि प्रकृति में पाषाण युग की सभ्यता का समावेश न हो।)

प्रश्न 3.
सन्धिं/सन्धिविच्छेदं कुरुत –
(सन्धि/सन्धिविच्छेद कीजिए)
(क) प्रकृतिः + …………… = प्रकृतिरेव।
(ख) स्यात् + …….. + …………….. = स्यान्नैव।
(ग) …….. + अनन्ताः = ह्यनन्ताः।
(घ) बहिः + अन्तः + जगति = ……..।
(ङ) …….. + नगरात् = अस्मान्नगरात्।
(च) सम् + चरणम् = ……..।
(छ) धूमम् + मुञ्चति = ……..।
उत्तरम् :
(क) प्रकृतिः + एव = प्रकृतिरेव।
(ख) स्यात् + न + एव = स्यान्नैव।
(ग) हि + अनन्ताः = ह्यनन्ताः।
(घ) बहिः + अन्तः + जगति = बहिरन्तर्जगति।
(ङ) अस्मात् + नगरात् = अस्मान्नगरात्।
(च) सम् + चरणम् = सञ्चरणम्।
(छ) धूमम् + मुञ्चति = धूमं मुञ्चति।

प्रश्न 4.
अधोलिखितानाम् अव्ययानां सहायतया रिक्तस्थानानि पूरयत –
(निम्नलिखित अव्ययों की सहायता से रिक्त स्थानों की पूर्ति कीजिए)
भृशम्, यत्र, तत्र, अत्र, अपि, एव, सदा, बहिः
(क) इदानी वायुमण्डलं ………….. प्रदूषितमस्ति।
(ख) ………… जीवनं दुर्वहम् अस्ति।
(ग) प्राकृतिक वातावरणे क्षणं सञ्चरणम् …….. लाभदायकं भवति।
(घ) पर्यावरणस्य संरक्षणम् ………. प्रकृतेः आराधना।
(ङ) ………. समयस्य सदुपयोगः करणीयः।
(च) भूकम्पित-समये ………….. गमनमेव उचितं भवति।
(छ) ……………… हरीतिमा …………. शुचि पर्यावरणम्।
उत्तरम् :
(क) इदानी वायुमण्डलं भृशं प्रदूषितमस्ति। (अब वायुमण्डल अत्यधिक प्रदूषित है।)
(ख) अत्र जीवनं दुर्वहम् अस्ति। (यहाँ जीवन दूभर है।)
(ग) प्राकृतिकवातावरणे क्षणं सञ्चरणम् अपि लाभदायकं भवति। (प्राकृतिक वातावरण में क्षणभर घूमना भी लाभदायक होता है।)
(घ) पर्यावरणस्य संरक्षणम् एव प्रकृते: आराधना। (पर्यावरण का संरक्षण ही प्रकृति की आराधना है।)
(ङ) सदा समयस्य सदुपयोगः करणीयः। (सदा समय का सदुपयोग करना चाहिए।)
(च) भूकम्पित-समये बहिः गमनमेव उचितं भवति। (भूकम्प के समय में बाहर जाना ही उचित है।)
(छ) यत्र हरीतिमा तत्र शुचि पर्यावरणम्। (जहाँ हरियाली है वहाँ शुद्ध पर्यावरण है।)

प्रश्न 5.
(अ) अधोलिखितानां पदानां पर्यायपदं लिखत –
(निम्नलिखित शब्दों के पर्याय शब्द लिखिए)
(क) सलिलम्
(ख) आम्रम्
(ग) वनम्
(घ) शारीरम्
(ङ) कुटिलम्
(च) पाषाणम्।
उत्तरम् :
(क) सलिलम् = जलम्
(ख) आम्रम् = रसालम्
(ग) वनम् = काननम् (कान्तारम्)
(घ) शरीरम् = तनुः (देहम्)
(ङ) कुटिलम् = वक्रम्
(च) पाषाणाम् = प्रस्तरम्।

(आ) अधोलिखितापदानां विलोमपदानि पाठात् चित्वा लिखत।
(निम्नलिखित पदों के विलोम पद पाठ से चुनकर लिखिये।)
पद – विलोमपद
(क) सुकरम् – दुष्करम्
(ख) दूषितम् – शुचि
(ग) गृह्णन्ती – वितरन्ती
(घ) निर्मलम् – समलम्
(ङ) दानवाय – मानवाय
(च) सान्ताः – अनन्ताः

प्रश्न 6.
उदाहरणमनुसृत्य पाठत् चित्वा समस्तपदानि समासनाम च लिखत –
(उदाहरण के अनुसार पाठ से चुनकर समस्त पद तथा समास का नाम लिखिए)

उत्तरम् :
(ख) हरिततरूणाम् (कर्मधारय)
(ग) ललितलतानाम् (कर्मधारय)
(घ) नवमालिका (कर्मधारय)
(ङ) धृतसुखसन्देशम् (बहुव्रीहि)
(च) कज्जलमलिनम् (कर्मधारय)
(छ) दुर्दान्तदशनैः (कर्मधारय)।

प्रश्न 7.
रेखाङ्कितपदमाधृत्य प्रश्ननिर्माणं कुरुत –
(रेखांकित शब्द के आधार पर प्रश्न बनाइए)
(क) शकटीयानम् कज्जलमलिनं धूमं मुञ्चति।
(मोटरगाड़ी काजल की तरह मलिन धुआँ छोड़ती है।)
(ख) उद्याने पक्षिणां कलरवं चेतः प्रसादयति।
(उद्यान में पक्षियों का कलरव चित्त को प्रसन्न करता है।)
(ग) पाषाणीसभ्यतायां लतातरुगुल्माः प्रस्तरतले पिष्टाः सन्ति।
(पाषाणकालीन सभ्यता में बेल, वक्ष, झाडियाँ पत्थरों के नीचे पिसी हैं।)
(घ) महानगरेषु वाहनानाम् अनन्ताः पङ्क्तयः धावन्ति।
(महानगरों में वाहनों की असीम पंक्तियाँ दौड़ती हैं।)
(ङ) प्रकृत्याः सन्निधौ वास्तविकं सुखं विद्यते।
(प्रकृति के सामीप्य में वास्तविक सुख है।)
उत्तरम् :
(क) शकटीयानं कीदृशं धूमं मुञ्चति ?
(मोटरगाड़ी कैसा धुआँ छोड़ती है ?)
(ख) उद्याने केषां कलरवं चेतः प्रसादयति ?
(उद्यान में किनका कलरव चित्त को प्रसन्न करता है?)
(ग) पाषाणीसभ्यतायां के प्रस्तरतले पिष्टाः सन्ति ?
(पाषाणकालीन सभ्यता में कौन पत्थर के नीचे पिसे हैं?)
(घ) कुत्र वाहनानाम् अनन्ताः पङ्क्तयः धावन्ति ?
(कहाँ वाहनों की असीम पंक्तियाँ दौड़ती हैं ?)
(ङ) कस्याः सन्निधौ वास्तविकं सुखं विद्यते ?
(किसके सामीप्य में वास्तविक सुख है ?)

योग्यताविस्तार :

समास – समसनं समासः

समास का शाब्दिक अर्थ होता है – संक्षेप। दो या दो से अधिक शब्दों के मिलने से जो नया और संक्षिप्त रूप बनता है, वह समास कहलाता है। समास के मुख्यतः चार भेद हैं –

1. अव्ययीभाव
2. तत्पुरुष
3. बहुव्रीहि
4. द्वन्द्व

1. अव्ययीभाव – 

इस समास में पहला पद अव्यय होता है और वही प्रधान होता है और समस्तपद अव्यय बन जाता है।
यथा – निर्मक्षिकम्-मक्षिकाणाम् अभावः।
यहाँ प्रथमपद निर् है और द्वितीयपद मक्षिकम् है। यहाँ मक्षिका की प्रधानता न होकर मक्षिका का अभाव प्रधान है, अतः यहाँ अव्ययीभाव समास है। कुछ अन्य उदाहरण देखें –

1. उपग्रामम् – ग्रामस्य समीपे – (समीपता की प्रधानता)
2. निर्जनम् – जनानाम् अभावः – (अभाव की प्रधानता)
3. अनुरथम् – रथस्य पश्चात् – (पश्चात् की प्रधानता)
4. प्रतिगृहम् – गृहं गृहं प्रति – (प्रत्येक की प्रधानता)
5. यथाशक्ति – शक्तिम् अनतिक्रम्य – (सीमा की प्रधानता)
6. सचक्रम् – चक्रेण सहितम् – (सहित की प्रधानता)

2. तत्पुरुष –

‘प्रायेण उत्तरपदार्थप्रधानः तत्पुरुषः’ इस समास में प्रायः उत्तरपद की प्रधानता होती है और पूर्व पद उत्तरपद के विशेषण का कार्य करता है। समस्तपद में पूर्वपद की विभक्ति का लोप हो जाता है।
यथा – राजपुरुषः अर्थात् राजा का पुरुष। यहाँ राजा की प्रधानता न होकर पुरुष की प्रधानता है।

1. ग्रामगतः – ग्रामं गतः।
2. शरणागतः – शरणम् आगतः।
3. देशभक्तः, – देशस्य भक्तः।
4. सिंहभीतः – सिंहात् भीतः।
5. भयापनः – भयम् आपन्नः।
6. हरित्रातः – हरिणा त्रातः। तत्पुरुष समास के दो प्रमुख भेद हैं-कर्मधारय और द्विगु।

(ii) कर्मधारय – इस समास में एक पद विशेष्य तथा दूसरा पद पहले पद का विशेषण होता है। विशेषण विशेष्य भाव के अतिरिक्त उपमान उपमेय भाव भी कर्मधारय समास का लक्षण है।
यथा – पीताम्बरम् – पीतं च तत् अम्बरम्।
महापुरुषः – महान् च असौ पुरुषः।
कज्जलमलिनम् – कज्जलम् इव मलिनम्।
नीलकमलम् – नीलं च तत् कमलम्।
मीननयनम् – मीन इव नयनम्।
मुखकमलम् – कमलम् इव मुखम्।

(ii) द्विगु- ‘संख्यापूर्वो द्विगुः’ इस समास में पहला पद संख्यावाची होता है और समाहार (एकत्रीकरण या समूह) अर्थ की प्रधानता होती है।
यथा – त्रिभुजम् – त्रयाणां भुजानां समाहारः। इसमें पूर्वपद ‘त्रि’ संख्यावाची है।
पंचपात्रम् – पंचानां पात्राणां समाहारः।
पंचवटी – पंचानां वटानां समाहारः।
सप्तर्षिः . – सप्तानां ऋषीणां समाहारः।
चतुर्युगम् – चतुर्णा युगानां समाहारः।

3. बहुव्रीहि –

‘अन्यपदार्थप्रधानः बहुब्रीहिः’ इस समास में पूर्व तथा उत्तर पदों की प्रधानता न होकर किसी अन्य पद की प्रधानता होती है।

यथा –
पीताम्बरः – पीतम् अम्बरम् यस्य सः (विष्णुः)। यहाँ न तो पीतम् शब्द की प्रधानता है और न अम्बरम्
शब्द की अपितु पीताम्बरधारी किसी अन्य व्यक्ति (विष्णु) की प्रधानता है।
नीलकण्ठः – नीलः कण्ठः यस्य सः (शिवः)।
दशाननः – दश आननानि यस्य सः (रावण:)।
अनेककोटिसारः – अनेककोटिः सारः (धनम्) यस्य सः।
विगलितसमृद्धिम् – विगलिता समृद्धिः यस्य तम् (पुरुषम्)।
प्रक्षालितपादम् – प्रक्षालितौ पादौ यस्य तम् (जनम्)।

4. द्वन्द्व –

‘उभयपदार्थप्रधानः द्वन्द्वः’ इस समास में पूर्वपद और उत्तरपद दोनों की समान रूप से प्रधानता होती है। पदों के बीच में ‘च’ का प्रयोग विग्रह में होता है।
यथा –
रामलक्ष्मणौ – रामश्च लक्ष्मणश्च।
पितरौ – माता च पिता च।
धर्मार्थकाममोक्षाः – धर्मश्च, अर्थश्च, कामश्च, मोक्षश्च।
वसन्तग्रीष्मशिशिरा: – वसन्तश्च ग्रीष्मश्च शिशिरश्च।

भावविस्तार –

पर्यावरण-परिभाषा – ‘आवियते परितः लोकोऽनेनेति पर्यावरणम्’ [चूँकि इसके द्वारा लोक को ढका (घेरा या आच्छादित किया) जाता है अतः पर्यावरण कहलाता है।
पृथ्वी, जल, तेज, वायु और आकाश ये पाँच महाभूत प्रकृति के प्रमुख तत्त्व हैं। इन पाँच तत्त्वों से ही पर्यावरण की रचना होती है। जो चारों ओर से लोक को घेरता या आच्छादित करता (ढकता) है, वह पर्यावरण है।
शुद्ध और प्रदूषणरहित पर्यावरण हमें सब प्रकार के जीवन के सुख देता है। हमारे द्वारा सदैव ऐसे प्रयत्न किए जाने चाहिए जिससे जल, स्थल और आकाश निर्मल हों। पर्यावरण सम्बन्धी कुछ श्लोक नीचे लिखे हुए हैं –

यथा – पृथिवीं परितो व्याप्य तामाच्छाद्य स्थितं च यत्।
जगदाधाररूपेण, पर्यावरणमुच्यते।।

अनुवाद – जगत् के आधार के रूप में पृथ्वी के चारों ओर व्याप्त और उसको घेरकर जो स्थित है, पर्यावरण कहलाता है।

प्रदूषण के विषय में –

सृष्टौ स्थितौ विनाशे च नृविज्ञैर्बहुनाशकम्।
पञ्चतत्त्वविरुद्धं यत्साधितं तत्प्रदूषणम्।।

अनुवाद – सृष्टि की स्थिति और विनाश में बुद्धिमान् मनुष्यों द्वारा पंचतत्त्वों के विरुद्ध बहुत अधिक विनाशकारी जो साधा जाता है अर्थात् किया जाता है, वह प्रदूषण है।

वायुप्रदूषण के विषय में –

प्रक्षिप्तो वाहनधूमः कृष्णे बह्वपकारकः।
दुष्टैरसायनैर्युक्तो घातकः श्वासरुग्वहः।।

अनुवाद – वाहनों द्वारा फेंके गए (निकाले गए) काले धुएँ में घातक हानिकारक रसायनों से युक्त बहुत-से अपशिष्ट श्वास रोगों के वाहक होते हैं।

जल-प्रदूषण के विषय में –

यन्त्रशाला परित्यक्तैर्नगरे दूषितद्रवैः।
नदीनदौ समुदाश्च प्रक्षिप्तैर्दूषणं गताः।।

अनुवाद – कारखानों द्वारा छोड़े गए और नगरों के दूषित द्रवों (जलों) के नदी, नालों और समुद्रों में डाले जाने के द्वारा (फेंकने से) इनके जल दूषित किए जाते हैं।

प्रदूषण-निवारण एवं संरक्षण के लिए –

शोधनं रोपणं रक्षावर्धनं वायुवारिणः।
वनानां वन्यवस्तूनां भूमेः संरक्षणं वरम्।।

अनुवाद – वायु, जल की शुद्धि, वन्यवस्तुओं तथा भूमि के संरक्षण के लिए वनों का लगाना तथा रक्षापूर्वक उनकी वृद्धि करना श्रेष्ठ उपाय है।
ये पाँचों श्लोक पर्यावरण काव्य से संकलित हैं।
तत्सम-तद्भव शब्दों का अध्ययन

निम्नलिखित तत्सम और उनसे बने तद्भव शब्दों का परिचय करने योग्य है –

छन्द-परिचय – इस गीत में शुद्ध पर्यावरण स्थायी तत्त्व है। इसके अतिरिक्त सब स्थानों पर प्रत्येक पंक्ति में 26 मात्राएँ हैं। यह गीतिका छन्द का रूप है।

JAC Class 10th Sanskrit शुचिपर्यावरणम् Important Questions and Answers

शब्दार्थ चयनम् –

अधोलिखित वाक्येषु रेखांकित पदानां प्रसङ्गानुकूलम् उचितार्थ चित्वा लिखत –

प्रश्न 1.
दुर्वहमत्र जीवितं जातं प्रकृतिरेव शरणम्।
(अ) जातम्
(ब) दुष्करम्
(स) कालायसचक्रम्
(द) दुर्दान्तैः
उत्तरम् :
(ब) दुष्करम्

प्रश्न 2.
मनः शोषयत् तनुः पेषयद् भ्रमति सदा वक्रम –
(अ) कुटिलम्
(ब) अमुना
(स) वक्रम्
(द) पेषयद
उत्तरम् :
(अ) कुटिलम्

प्रश्न 3.
कज्जलमलिनं धूमं मुञ्चति शतशकटीयानम् –
(अ) ध्वानम्
(ब) वाष्पयानमाला
(स) त्यजति
(द) संधावति
उत्तरम् :
(स) त्यजति

प्रश्न 4.
यानानां पङ्क्तयो ह्यनन्ताः, कठिनं संसरणम् –
(अ) यानानाम्
(ब) अनन्ताः
(स) शरणं
(द) संचलनम्
उत्तरम् :
(द) संचलनम्

प्रश्न 5.
कुत्सितवस्तुमिश्रितं भक्ष्यं समलं धरातलम्।
(अ) अशुद्धम्
(ब) भृशम्
(स) मिश्रितम्
(द) शुद्धीकरणम्
उत्तरम् :
(अ) अशुद्धम्

प्रश्न 6.
प्रपश्यामि ग्रामान्ते निर्झर-नदी-पयःपूरम् –
(अ) कान्तारे
(ब) ग्रामान्ते
(स) प्रपात
(द) नगरात
उत्तरम् :
(स) प्रपात

प्रश्न 7.
कुसुमावलिः समीरचालिता स्यान्मे वरणीया –
(अ) पुष्पाणां पंक्तिः
(ब) ललितलतानाम्
(स) कुसुमावलिः
(द) नवमालिका
उत्तरम् :
(अ) पुष्पाणां पंक्तिः

प्रश्न 8.
अयि चल बन्धो ! खगकुलकलरव गुञ्जितवनदेशम् –
(अ) जनेभ्यः
(ब) गुञ्जित
(स) कान्तार प्रदेश
(द) खगकलकलरव
उत्तरम् :
(स) कान्तार प्रदेश

प्रश्न 9.
प्रस्तरतले लतातरुगुल्मा नो भवन्तु पिष्टाः –
(अ) लतातरुगुल्माः
(ब) दमिता
(स) निसर्गे
(द) समाविष्टाः
उत्तरम् :
(ब) दमिता

प्रश्न 10.
पाषाणी सभ्यता निसर्गे स्यान्न समाविष्टा –
(अ) कामये
(ब) मानवाय
(स) पर्यावरणम्
(द) प्रकृती
उत्तरम् :
(स) पर्यावरणम्

II. संस्कृतमाध्यमेन प्रश्नोत्तराणि –

एकपदेन उत्तरत –
(एक शब्द में उत्तर दीजिए)

प्रश्न 1.
कालायसचक्रं किं शोषयति ?
(लौहचक्र किसका शोषण करता है ?)
उत्तरम् :
मनः (मन का)।

प्रश्न 2.
दुर्दान्तैः दशनैः केन जनग्रसनं न स्यात् ?
(किसके विकराल दाँतों द्वारा लोगों का विनाश न हो?)
उत्तरम् :
कालायसचक्रेण
(लौहचक्र द्वारा)।

प्रश्न 3.
धावन्ती वाष्पयानमाला किं वितरति ?
(दौड़ती रेलगाड़ियों की माला क्या वितरण करती है ?)
उत्तरम् :
ध्वानम्
(शोर)।

प्रश्न 4.
कज्जलमलिनं धूमं किं मुञ्चति ?
(काजल-सा काला धुआँ कौन त्यागती है ?)
उत्तरम् :
शतशकटीयानम्
(सैकड़ों मोटरगाड़ियाँ)।

प्रश्न 5.
‘अनन्ताः’ कस्य पदस्य विशेषणम् ?
(‘अनन्ताः’ किस शब्द का विशेषण है ?)
उत्तरम् :
पङ्क्तयः
(पंक्तियाँ)।

प्रश्न 6.
कुत्सितवस्तु मिश्रितं किम् ?
(गंदी वस्तुओं से मिश्रित क्या है ?)
उत्तरम् :
भक्ष्यम्
(खाद्य)।

प्रश्न 7.
बहिरन्तर्जगति किं बहुकरणीयम् ?
(बाह्य और आन्तरिक जगत में क्या बहुत करना है ?)
उत्तरम् :
शुद्धीकरणम्
(शुद्धीकरण)।

प्रश्न 8.
कविः कुतः बहुदूरं गन्तुम् इच्छति ?
(कवि कहाँ से बहुत दूर जाना चाहता है ?)
उत्तरम् :
नगरात्
(नगर से)।

प्रश्न 9.
‘माम् नगरात् बहुदूरं नय’ इति वाक्ये सर्वनामपदं लिखत।
(‘माम् नगरात् बहुदूरं नय’ इस वाक्य में सर्वनाम पद लिखिए।)
उत्तरम् :
माम् (मुझे)।

प्रश्न 10.
नवमालिका के मिलिता रुचिरं प्रतिभाति ?
(नवमल्लिका किससे मिली सुन्दर लगती है ?)
उत्तरम् :
रसालम् (आम से)।

प्रश्न 11.
कुसुमावलिः केन चालिता मे वरणीया?
(किसके द्वारा संचालित फूलों की पंक्ति मेरे द्वारा वरण करने योग्य
उत्तरम् :
समीरेण (हवा द्वारा)।

प्रश्न 12.
जनाः केन सम्भ्रमिता: ?
(लोग किससे भ्रमित हैं ?)
उत्तरम् :
पुरकलरवेण (नगर के कोलाहल से)।

प्रश्न 13.
किं जीवितरसहरणम् ?
(जीवन के आनन्द रस का हरण करने वाला क्या है ?)
उत्तरम् :
चाकचिक्यजालम् (चकाचौंध करने वाला जगत्)।

प्रश्न 14.
का निसर्गे न समाविष्टा स्यात् ?
(प्रकृति में किसका समावेश न हो ?)
उत्तरम् :
पाषाणी सभ्यता (पाषाणकालीन सभ्यता)।

प्रश्न 15.
‘पाषाणी’ पदस्य विशेष्यं लिखत।
(‘पाषाणी’ पद का विशेष्य लिखिए।)
उत्तरम् :
सभ्यता।

प्रश्न 16.
‘मानवाय जीवनं कामये’ अत्र क्रियापदं किम् ?
(‘मानवाय जीवनं कामये’ यहाँ क्रियापद क्या है ?)
उत्तरम् :
कामये (चाहता हूँ)।

प्रश्न 17.
कुत्र दुर्वहम् जीवितम् ?
(जीना दूभर कहाँ है ?)
उत्तरम् :
महानगरमध्ये (महानगर के बीच में)।

प्रश्न 18.
अनिशं किं चलति?
(दिन-रात क्या चलता है?)
उत्तरम् :
कालायसचक्रम् (लोहे का पहिया)।

प्रश्न 19.
धूमं कीदृशम् अस्ति ?
(धुओँ कैसा है ?)
उत्तरम् :
कज्जलमलिनम् (काजल-सा काला)।

प्रश्न 20.
ध्वानं वितरन्ती का धावति ?
(शोर करती क्या दौड़ती है ?)
उत्तरम् :
वाष्पयानमाला (रेलगाड़ियों की माला)।

प्रश्न 21.
वायुमण्डलं कीदृशम् ?
(वायुमण्डल कैसा है?)
उत्तरम् :
दूषितम् (दूषित)।

प्रश्न 22.
धरातलं कीदृशम् अस्ति ?
(धरातल कैसा है ?)
उत्तरम् :
समलम् (मलिन)।

प्रश्न 23.
कविः निर्झर नदी-पयः पूरं कुत्र पश्यति ?
(कवि जल से भरे नदी-झरना कहाँ देखता है ?)
उत्तरम् :
ग्रामान्ते (गाँव की सीमा पर)।

प्रश्न 24.
कविः कीदृशे कान्तारे सञ्चरणम् इच्छति ?
(कवि कैसे वनप्रदेश में घूमना चाहता है ?)
उत्तरम् :
एकान्ते (एकान्त में)।

प्रश्न 25.
हरिततरूणां माला कीदृशी ?
(हरे वृक्षों की माला कैसी है?)
उत्तरम् :
रमणीया (सुन्दर)।

प्रश्न 26.
का समीरचालिता मे वरणीया स्यात् ?
(वायु से हिलाई गई मेरे वरण करने योग्य क्या हो ?)
उत्तरम् :
कुसुमावलिः (फूलों की पंक्ति)।

प्रश्न 27.
हरिततरूणां माला कीदृशी ?
(हरे वृक्षों की माला कैसी है?)
उत्तरम् :
रमणीया (सुन्दर)।

प्रश्न 28.
का समीरचालिता मे वरणीया स्यात् ?
(वायु से हिलाई गई मेरे वरण करने योग्य क्या हो ?)
उत्तरम् :
कुसुमावलिः (फूलों की पंक्ति)।

प्रश्न 29.
वनदेशं केन गुञ्जितम् ?
(वनप्रदेश किससे गुंजित है ?)
उत्तरम् :
खगकलरवेण (पक्षियों के कलरव से)।

प्रश्न 30.
अत्र केभ्यः सुख सन्देशम् ?
(यहाँ सुख का सन्देश किनके लिए है ?)
उत्तरम् :
सम्भ्रमितजनेभ्यः (भ्रमित लोगों के लिए)।

प्रश्न 31.
कविः कस्मै जीवनं कामयते ?
(कवि किसके जीवन की कामना करता है ?)
उत्तरम् :
मानवाय (मानव के)।

प्रश्न 32.
लतातरुगुल्माः कस्य तले न पिष्टाः भवन्तु ?
(बेलों और पेड़ों के समूह किसके नीचे नहीं पिसें ?)
उत्तरम् :
प्रस्तरतले (पत्थर के नीचे)।
पूर्णवाक्येन उत्तरत (पूरे वाक्य में उत्तर दीजिए)

प्रश्न 33.
कालायसचक्रं किं किं करोति. ?
(लौहचक्र क्या-क्या करता है ?)
उत्तरम् :
कालायसचक्रं मनः शोषयति तनुः च पेषयति।
(लौहचक्र मन का शोषण करता है और शरीर को पीसता है।)

प्रश्न 34.
नगरेषु किं किं प्रदूषितम् भवति ?
(नगरों में क्या-क्या प्रदूषित होता है ?)
उत्तरम् :
नगरेषु वायुमण्डलं, जलं, भोजनं धरातलं च दूषितं भवति।
(नगरों में वायुमण्डल, जल, भोजन और धरती दूषित होती है।)

प्रश्न 35.
ग्रामान्ते कविः किं पश्यति?
(गाँव की सीमा पर कवि क्या देखता है ?)
उत्तरम् :
ग्रामान्ते कविः निर्झर-नदी-पयः पूरं पश्यति।
(गाँव की सीमा पर कवि नदी, झरने जल से परिपूर्ण देखता है।)

प्रश्न 36.
केषां माला रमणीया ?
(किनकी माला सुन्दर है ?)
उत्तरम :
हरिततरूणां ललितलतानां माला रमणीया।
(हरे वृक्षों और सुन्दर बेलों की पंक्ति रमणीय है।)

प्रश्न 37.
अत्र जीवितं कीदृशं जातम् ?
(यहाँ जीवन कैसा हो गया है?)
उत्तरम् :
अत्र जीवितं दुर्वहं जातम्।
(यहाँ जीवन दूभर हो गया है।)

प्रश्न 38.
शतशकटीपान कि मुञ्चति ?
(सैकड़ों मोटरगाड़ियाँ क्या छोड़ती है ?)
उत्तरम् :
शतशकटीयानं कज्जलमलिनं धूम मुन्धति।
(सैकड़ों मोटरगाड़ियाँ काजल के समान काला धुआँ छोड़ती हैं।)

प्रश्न 39.
नगरेषुकीदर्श भक्ष्य मिलति ?
(नगरों में कैसा खाना मिलता है?)
उत्तरम् :
नगरेषु कत्सितवस्तु-निर्मितं भोजनं मिलति।
(नगरों में दूषित वस्तुओं से बना भोजन मिलता है।)

प्रश्न 40.
कविः कुत्र सञ्चरणम् इच्छति ?
(कवि कहाँ विचरण करना चाहता है ?)
उत्तरम् :
कवि: एकान्ते कान्तारे सञ्चरणम् इच्छति।
(कवि एकान्त वनप्रदेश में घूमना चाहता है।)

प्रश्न 41.
कीदृशी कुसुमावलिः कवये वरणीया स्यात् ?
(कैसी फूलों की पंक्ति कवि के लिए वरण करने योग्य है ?)
उत्तरम् :
समीरचालिता कुसुमावलिः कवये वरणीया स्यात्।
(वायु द्वारा हिलाई गई फूलों की पंक्ति कवि के लिए वरण करने योग्य है।)

प्रश्न 42.
कीदृशी कुसुमावलिः कवये वरणीया स्यात् ?
(कैसी फूलों की पंक्ति कवि के लिए वरण करने योग्य है ?)
उत्तरम् :
समीरचालिता कुसुमावलिः कवये वरणीया स्यात्।
(वायु द्वारा हिलाई गई फूलों की पंक्ति कवि के लिए वरण करने योग्य है।)

प्रश्न 43.
कविः कीदृशं देशं गन्तुम् इच्छति ?
(कवि कैसे देश को जाना चाहता है ?)
उत्तरम् :
कविः खगकुलकलरव गुञ्जित वनदेशं गन्तुम् इच्छति।
(कवि पक्षियों के कलरव से गुञ्जित वनप्रदेश को जाना चाहता है।)

प्रश्न 44.
प्रस्तरतले के न पिष्टाः भवन्तु ?
(पत्थरों के नीचे क्या नहीं कुचले?)।
उत्तरम् :
प्रस्तरतले लतातरुगुल्माः पिष्टाः न भवन्तु।
(पत्थरों के नीचे बेल, वृक्ष और झाड़ियाँ न कुचलें।)

प्रश्न 45.
‘शुचिपर्यावरणम्’ इत्यस्य पाठस्य भावबोधनं सरलसंस्कृतभाषया लिखत।
(‘शुचिपर्यावरणम्’ पाठ का भाव सरल संस्कृत भाषा में लिखिए।)
उत्तरम् :
अद्य महानगरेषु यन्त्राणि चलानि। वाष्पयानानि शकटीयानानि चलन्ति। एते पर्यावरणं प्रदूषयन्ति। भोजनम् अपि कुत्सितवस्तु मिश्रितम् अस्ति। ध्वनिः अपि कौँ स्फोटयति। अतः अत्र जीवितं दुर्वहं जातम्। अधुना प्रकृतिः एव मात्र शरणम् अस्ति। कवि प्रकृतेः शरणं गन्तुम् इच्छति। सः मानवाय जीवनं कामयते।

(आज महानगरों में मशीनें चलती हैं। रेलगाड़ियाँ और मोटरगाड़ियाँ चलती हैं। ये सब पर्यावरण को प्रदूषित करती हैं। भोजन भी गन्दी वस्तुओं से मिश्रित है। ध्वनि भी कानों को फोड़ती है। अत: यह जीवन दूभर हो गया है। अब प्रकृति ही एकमात्र आश्रय है। कवि प्रकृति की शरण में जाना चाहता है। वह मानव के लिए जीवन की कामना करता है।)

III. अन्वय-लेखनम् –

अधोलिखितश्लोकस्यान्वयमाश्रित्य रिक्तस्थानानि मञ्जूषातः समुचितपदानि चित्वा पूरयत।
(नीचे लिखे श्लोक के अन्वय के आधार पर रिक्तस्थानों की पूर्ति मंजूषा से उचित पद चुनकर कीजिए।)

1. महानगरमध्ये ……………………………………. जनग्रसनम्।।

मञ्जूषा – दशनैः, महानगरमध्ये, अनिशं, पेषयत्।

कालायसचक्रम् (i) ………………. (ii)……………….. चलत्, मनः शोषयत्, तनुः (iii) ……………. सदा वक्र भ्रमति, अमुना दुर्दान्तैः (iv) ……………. जनग्रसनम् एव स्यात्।
उत्तरम् :
(i) अनिशं
(i) महानगरमध्ये
(iii) पेषयत्
(iv) दशनैः।

2. कज्जलमलिनं ……………………………………. संसरणम्।।

मञ्जूषा – यानानाम्, वाष्पयानमाला, कज्जलमलिनम, संसरणम्।

शतशकटीयानं (i) ……………. धूमं मुञ्चति। ध्वानं वितरन्ती (ii) …………… संधावति (iii) …………. पतयः अनन्ताः हि (iv) …………… कठिनम्।
उत्तरम् :
(i) कज्जलमलिनम्
(ii) वाष्पयानमाला
(iii) यामामाम्
(iv) संसरणम्।

3. वायुमण्डल …………………………….. शुद्धीकरणम्।।

मञ्जूषा – कुत्सितबस्तु, निर्मलम्, पूषितम्, शुद्धीकरणम्।

वायुमण्डलं भृशं (i)……. हि जलं (ii) …….. न, भक्ष्यं (iii) …….. मिश्रितं, धरातलं, समलं, तु बहि-अन्तः जगति बहु (iv)…… करणीयं स्यात्।
उत्तरम् :
(i) दूषितम्
(ii) निर्मलम्
(iii) कुत्सितवस्तु
(iv) शुद्धीकरणम्।

4. कश्चित्कालं ……………………………………. सञ्चरणम्।।

मञ्जूषा – अस्मात्, कान्तारे, प्रपश्यामि, नय।

कश्चित् कालात् माम् (i) ……… नगरात् बहुदूरं (ii) …………..। ग्रामान्ते निर्झर-नदी-पयः पूरं (iii) ………….. एकान्ते (iv) …………….. क्षणमपि मे सञ्चरणं स्यात्।
उत्तरम् :
(i) अस्मात्
(ii) नय
(iii) प्रपश्यामि
(iv) कान्तारे।

5. हरिततरूणां …………………………. संगमनम्।

मञ्जूषा – नवमालिका, समीरचालिता, ललितलतानां, रुचिरं।

हरिततरूणां (i) ……… रमणीया माला, (ii) ……… कुसुमावलिः मे वरणीया स्यात्। (iii) ……… मिलिता रसालं (iv) ……… सङ्गमनम्।
उत्तरम् :
(i) ललितलतानां
(ii) समीरचालिता
(iii) नवमालिका
(iv) रुचिरं।

6. अयि चल …………………… कुर्याज्जीवितरसहरणम्।

मञ्जूषा – हरणम्, सम्भ्रमित, खगकुल, चाकचिक्यजालं।

अयि बन्धो ! (i) ……… कलरव गुञ्जित वनदेशं चल। पुर-कलरव (ii) ……… जनेभ्यः सुख सन्देशं धृत, (iii) ……… नो जीवित रस (iv) ……… कुर्यात्।
उत्तरम् :
(i) खगकुल
(ii) सम्भ्रमित
(iii) चाकचिक्यजालं
(iv) हरणम्।

7. प्रस्तरतले ……………………………… जीवन्मरणम्।

मञ्जूषा – निसर्गे, प्रस्तरतले, जीवन्, मानवाय।

लतातरुगुल्मा (i)……… पिष्टाः न भवन्तु। पाषाणीसभ्यता (ii)……… समाविष्टा न स्यात्। (iii)……… जीवनं कामये नो (iv)……… मरणम्।
उत्तरम् :
(i) प्रस्तरतले
(ii) निसर्गे
(iii) मानवाय
(iv) जीवन्।

IV. प्रश्ननिर्माणम –

अधोलिखित वाक्येषु स्थूलपदमाभृत्य प्रश्न निर्माणं कुरुत –

1. दुर्वहम् अत्र जीवितं जातम्। (यहाँ जीवन दूभर हो गया है।)
2. महानगरमध्ये चलदनिशं कालायसचक्रम्। (महानगर में दिन-रात लोहे का पहिया चलता है।)
3. कज्जलमलिनं धूमं मुञ्चति शतशकटीयानम्। (सैकड़ों मोटरगाड़ियाँ काजल-सा मलिन धुऔं त्यागती हैं।)
4. वाष्पयानमाला संधावति वितरन्ती ध्वानम्। (रेलगाड़ियों की पंक्ति शोर करती दौड़ रही है।)
5. वाष्पयानमाला ध्वानं वितरन्ती संधावति। (रेलगाड़ियों की पंक्ति शोर करती दौड़ रही है।)
6. यानानां पङ्क्तयो ह्यनन्ता। (वाहनों की पंक्तियाँ अनन्त हैं।)
7. वायुमण्डलं भृशं दूषितम्। (वायुमण्डल बहुत दूषित है।)।
8. बहिरन्तर्जगति तु बहु शुद्धीकरणं करणीयम्। (बहिर् और अन्तर्जगत में बहुत शुद्धीकरण करना चाहिए।)
9. ग्रामान्ते निर्झर-नदी-पयःपूरं प्रपश्यामि। (गाँव के अंत में जल से भरपूर झरने और नदी देखता हूँ।)
10. प्रकृत्याः सन्निधौ वास्तविकं सुखं विद्यते। (प्रकृति के सान्निध्य में असली सुख विद्यमान है।)

उत्तराणि :

1. कीदृशम् अत्र जीवितं जातम् ?
2. महानगरमध्ये किं चलदनिशम् ?
3. कीदृशं धूमं मुञ्चति शतशकटीयानम्?
4. का सन्धावति वितरन्ती ध्वानम् ?
5. वाष्पयानमाला किं वितरन्ती सन्धावति ?
6. केषां पङ्क्तयो ह्यनन्ताः ?
7. किं भृशं दूषितम्।
8. बहिरन्तर्जगति तु किं करणीयम् ?
9. ग्रामान्ते किं प्रपश्यामि ?
10. केषां सन्निधौ वास्तविक सुखं विद्यते ?

V. भावार्थ लेखनम् –

अधोलिखितपद्यानां संस्कृते भावार्थं लिखत –

1. दुर्वहमत्र जीवितं ……………………… स्यान्नैव जनग्रसनम् ॥

भावार्थ – अस्मिन् स्थाने जीवनमपि दुष्करमभवत्। अत प्राकृतिक स्थलानां शुद्ध पर्यावरणमेव (एकमात्र) आश्रयः। लौहचक्रम दिवारात्रम् महत्स नगरेष गतिमानस्ति। मनसः शोषणं कुर्वन् शरीरं च पिष्टीकर्वन चूर्णयन वा सदैव कटिलं (चक्रवत्) घूर्णति। अनेन विकरालै दन्तैः मानवानां भक्षणं (पीडनं) न भवेत्। अतः शुद्धं पर्यावरणमेव (प्राणिनाम्) आश्रयमस्ति।

2. कज्जलमलिनं धूमं ………………………. कठिनं संसरणम्।

भावार्थ – शतसंख्यकानि वाहनानि कज्जल सदृशं मलिनं धूमजालं वमन्ति त्यजन्ति वा ध्वनिं कुर्वन्ती लौहपथ गामिनीनां पंक्ति निरन्तरं धावति। वाहनानां पङ्क्तयः असीमिता: अन्तहीनाः वा पंक्तयः सन्ति अतः सञ्चलनमपि दुष्करं जातम् अतः शुद्ध वातावरणम् (इदानी) एकमात्रं आश्रयम् अस्ति।

3. वायुमण्डलं भृशं ………………………. बहु शुद्धीकरणम्।

भावार्थ – समीर मण्डलं प्रभूतं दोषपूर्णं (जातम्)। यतः न वारि शुद्धं न भोजनं (शुद्धम्) पदार्थाः दूषिताः मिश्रिताः च पृथ्वीतलमपि अशुद्धमस्ति। तर्हि आन्तरिके बाह्ये च जगति भृशं पवित्रीकरणं कर्त्तव्यम्। शुद्धं वातावरणमिदानीं एकमात्र आश्रयम् अस्ति।

4. कञ्चित् कालं नय ……………………… मे स्यात् सञ्चरणम्।

भावार्थ – किञ्चित्समयं माम् एतस्मात् पत्तनात् भृशं दूरं गमय, ग्रामस्य सीम्नि अहं जलपूर्ण प्रपातं सरितां चावलोकयामि, निर्जने वने क्षणमपि मम विचरणं भवेत् यतः शुद्धं पर्यावरणमेव एकमात्रं आश्रय।

5. हरिततरूणां ललितलतानां …………………. रुचिरं संगमनम्।

भावार्थ – हरितवर्ण वृक्षाणां, रम्य वल्लरीणां रम्या पङ्क्तिः पवनेन वेपिता पुष्पाणां पङ्क्तिः मह्यं वरणाय भवेत्। नूतना मल्लिका सहकारेण मिलित्वा सुन्दरं मेलनं प्राप्नोति। शुद्ध पर्यावरणमेवाद्य (मानव जीवनस्य) आश्रय।

6. अयि चल बन्धो ! …………………. नो कुर्याज्जीवितरसहरणम्।

भावार्थ – हे भ्रातः! पक्षिणां समवेत स्वरेण गुञ्जायमानं कान्तारप्रदेशं चल यत् नगरस्य ध्वानेन सम्यक् भ्रान्तेभ्यः मानवेभ्यः सुखस्य समाचारं धारयति। अप्राकृतं जीवनरस हर्तारं इदं जगत् जीवन सुखं न हरेत्। (अद्य मानवाय) शुद्ध वातावरणमेव मात्राश्रयः।

7. प्रस्तरतले लतातरुगुल्मा …………………………….. नो जीवन्मरणम्।

संस्कृत व्याख्याः – वल्लर्यः वृक्षाः कण्टस्तवा इत्यादयः प्राकृतिक सम्पदः शिलातले दमिता न भवन्तु। प्रकृतौ पाषाणकालीन सभ्यतायाः समावेशः न भवेत्। अहं मनुष्याय जीवनं इच्छामि न तु जीवनं मरणं च। शुद्ध वातावरणमेवाश्रय अस्ति।

शुचिपर्यावरणम् Summary and Translation in Hindi

पाठ-परिचय – निरन्तर बढते हए पर्यावरण-प्रदषण से आज सम्पूर्ण विश्व पीडित है। पथ्वी, जल, व तेज सभी तो प्रदूषित हो गए हैं। मानव मन को शान्ति कहाँ मिले। विश्वव्यापी इसी समस्या से अनुप्रेरित होकर आधुनिक संस्कृत-कवि हरिदत्त शर्मा ने यह कविता लिखी है। ‘शुचिपर्यावरणम्’ उनके ‘लसल्लतिका’ गीत-संग्रह में संकलित है। इसमें कवि महानगरों की यान्त्रिक बहुलता से बढ़ते प्रदूषण पर चिन्ता व्यक्त करते हुए कहता है कि यह लौहचक्र तन-मन का शोषक है, जिससे वायु-मण्डल और भूमण्डल दोनों मलिन हो रहे हैं। प्रस्तुत कविता के प्रथम तीन श्लोकों में कवि ने यान्त्रिक वृद्धि के कारण प्रदूषित पर्यावरण तथा वायुमण्डल प्रदूषण का वर्णन किया है, चौथे से सातवें श्लोकों में कवि शुद्ध जलवायु-युक्त हरे-भरे वृक्षों से घिरे पक्षियों की संगीत लहरी से गुंजित शुद्ध-पर्यावरण में जाने की अभिलाषा करता है।

मूलपाठः, अन्वयः,शब्दार्थाः, सप्रसंग हिन्दी-अनुवादः

1 दुर्वहमत्र जीवितं जातं प्रकृतिरेव शरणम्।
शुचि-पर्यावरणम्॥
महानगरमध्ये चलदनिशं कालायसचक्रम्।
मनः शोषयत् तनुः पेषयद् भ्रमति सदा वक्रम्॥
दुर्दान्तैर्दशनैरमुना स्यान्नैव जनग्रसनम् ॥ शुचि…. ॥1॥

अन्वयः – अत्र जीवितं दुर्वहं जातं, (अत:) प्रकृतिः, शुचि-पर्यावरणम् एव शरणम्। कालायसचक्रम् अनिशं महानगरमध्ये चलत्, मनः शोषयत्, तनुः पेषयत् सदा वक्रम् भ्रमति, अमुना दुर्दान्तैः दशनैः जनग्रसनम् एव स्यात्। शुचिपर्यावरणमेव शरणम्।

शब्दार्थाः – अत्र = अस्मिन् संसारे (यहाँ, इस संसार में), जीवितम् = जीवनम् (जीवन), दुर्वहम् = दुष्करम् (कठिन, दूभर), जातम् = भूतं, उत्पन्नम् (हो गया है), अतः प्रकृतिः = अतः प्राकृतिकस्थलम् (अतः प्रकृति), शुचि-पर्यावरणम् एव = शुद्ध-पर्यावरणम् एव (स्वच्छ, शुद्ध वातावरण ही), शरणम् = आश्रयः (आश्रय योग्य है) अर्थात् शुद्ध पर्यावरण की शरण में ही जाना चाहिए। कालायसचक्रम् = लौहचक्रम् (लोहे का पहिया), अनिशम् = अहर्निशम् (दिन-रात), महानगरमध्ये = विशाल नगरेषु, महत्सु नगरेषु (महानगरों में), चलत् = गतिमत्, प्रवृत्तमानम् (चलता हुआ), मनः शोषयत् = मनसः शोषणं कुर्वत् (मन का शोषण करता हुआ), तनुः = शरीरम् (शरीर को), पेषयद् = पिष्टीकुर्वत् (पीसता हुआ), सदा = सर्वदा, सदैव (हमेशा, सदैव), वक्रम् = कुटिलम् (टेढ़ा-मेढ़ा, वक्रगति से), भ्रमति = घूर्णति (घूमता है), अमुना = अनेन (इसके द्वारा), दुर्दान्तः = विकरालैः, भयङ्करैः (विकराल), दशनैः = दन्तैः (दाँतों से), जन-ग्रसनम् एव = मानवानां भक्षणम्, नीगीर्णनिम् एव (मानव का विनाश ही), स्यात् = भवेत् (होना चाहिए, होगा), अतः शुचि-पर्यावरणम् एव = शुद्धं वातावरणमेव (स्वच्छ जलवायु ही), शरणम् = आश्रयः (एकमात्र सहारा है।)

सन्दर्भ – प्रसङ्गश्च – यह गीतांश हमारी ‘शेमुषी’ पाठ्यपुस्तक के ‘शुचि पर्यावरणम्’ पाठ से लिया गया है। यह पाठ प्रो. हरिदत्त शर्मा-रचित ‘लसल्लतिका’ गीतिसंग्रह से सङ्कलित है। इस पाठ में कवि पर्यावरण प्रदूषण के आधार पर लिखता

हिन्दी-अनुवादः – इस संसार में जीवन दूभर हो गया है, अतः प्रकृति का शुद्ध-पर्यावरण ही (एकमात्र) सहारा है अर्थात् शुद्ध पर्यावरण की शरण में ही जाना चाहिए। (यह) लोहे का पहिया (यन्त्रजाल) दिन-रात महानगरों में चलता हुआ, मन का शोषण करता हुआ, शरीर को पीसता हुआ हमेशा टेढ़ा-मेढ़ा वक्रगति से घूमता है। इसके विकराल दाँतों द्वारा मानव का विनाश ही (किया जा रहा है) होगा। अतः शुद्ध पर्यावरण की शरण में ही जाना चाहिए।

2. कज्जलमलिनं धूमं मुञ्चति शतशकटीयानम्।
वाष्पयानमाला सन्धावति वितरन्ती ध्वानम् ॥
यानानां पङ्क्तयो ह्यनन्ताः, कठिनं संसरणम्। शुचि……॥2॥

अन्वयः – शतशकटीयानम् कज्जलमलिनं धूमं मुञ्चति, ध्वानं वितरन्ती वाष्पयानमाला संधावति, यानानां पङ्क्तयः अनन्ताः हि संसरणं कठिनम्। शुचिपर्यावरणं शरणम्।।

शब्दार्थाः – शतशकटीयानम् = शकटीयानानां शतम् (सैकड़ों मोटर-गाड़ियाँ), कज्जलमलिनं धूमम् = कज्जलेन मलिनम् वाष्पः (काजल-सा मलिन अर्थात् काला धुआँ), मुञ्चति = त्यजति, वमति (त्यागती, छोड़ती या उगलती हैं)। ध्वानम् = (शोर), वितरन्ती = ददाति (देती, करती हुई), वाष्पयानमाला = वाष्पयानानां, लौहपथगामिनीनां पङ्क्तिः (रेलगाड़ियों की पंक्ति), संधावति = निरन्तरं धावति (निरन्तर दौड़ रही हैं), यानानाम् = वाहनानाम् (वाहनों की), पङ्क्तयः = मालाः, परम्पराः (पङ्क्तियाँ), अनन्ताः = असीमिताः, अन्तहीनाः (असीमित हैं), हि = अतः (इसलिए), संसरणम् = गमनम्, सचलनम् (चलना), कठिनम् = दुष्करम् (दूभर हो गया है), शुचि पर्यावरण = शुद्धवातावरण (शुद्ध पर्यावरण), शरणं = आश्रयः (आश्रय है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी ‘शेमुषी’ पाठ्यपुस्तक के ‘शुचि पर्यावरणम्’ पाठ से लिया गया है। यह पाठ प्रो. हरिदत्त शर्मा रचित ‘लसल्लतिका’ गीति-संग्रह से संकलित है।

हिन्दी-अनुवादः – सैकड़ों मोटरगाड़ियाँ काजल-सा मलिन धुआँ अर्थात् काला-काला धुआँ छोड़ती हैं। शोर करती हुई रेलगाड़ियों की पंक्तिः निरन्तर दौड़ रही है। वाहनों की पंक्तियाँ असीमित (अन्त न होने वाली) हैं इसलिए (रास्ते में) चलना भी दूभर हो गया है। अतः शुद्ध पर्यावरण की शरण में चलना चाहिए।

3. वायुमण्डलं भृशं दूषितं न हि निर्मलं जलम्।
कुत्सितवस्तुमिश्रितं भक्ष्यं समलं धरातलम् ॥
करणीयं बहिरन्तर्जगति तु बहु शुद्धीकरणम्। शुचि……॥3॥

अन्वयः – वायुमण्डलं भृशं दूषितं हि जलं निर्मलं न, भक्ष्यं कुत्सित वस्तु-मिश्रितम्, धरातलम् समलम् तु बहिः अन्तः जगति बहु शुद्धीकरणं करणीयम्। शुचि-पर्यावरणं शरणम्।

शब्दार्थाः – वायुमण्डलम् = वातावरणम्, समीरमण्डलम् (वायुमण्डल), भृशम् = अत्यधिकं, बहुः, प्रभूतम् (अत्यधिक, अत्यन्त), दूषितम् = दोषपूर्णम्, अशुद्धं, दूषणमयम् जातम् (प्रदूषित, दूषित हो गया है), हि = यतः (क्योंकि), जलम् = तोयः, वारि, पानीयम् (पानी भी), निर्मलम् = शुद्धम्, मलहीनम् (शुद्ध, स्वच्छ), न = नहि (नहीं है), भक्ष्यम् = खाद्यम्, भोजनम् (खाने की वस्तुएँ), कुत्सितवस्तु = दूषित पदार्थ, गर्हित वस्तु (बुरी या दूषित वस्तु), मिश्रितम् = युक्तम् (युक्त, मिलावट वाला है), धरातलम् = पृथ्वीतलम्, भूतलम् (धरती), समलम् = मलयुक्तम्, अशुद्धम्, अपावनम् (मैली अर्थात् गन्दगी युक्त है), तु = तर्हिः, तदा (तो, अतएव), बहिः अन्तः जगति = बाह्ये आन्तरिके च जगति (बाह्य एवं आन्तरिक जगत् में), बहुः = भृशम्, प्रभूतम् (अत्यधिक, अनेक, बहुत), शुद्धीकरणम् = पवित्रीकरणम् (शुद्धीकरण), करणीयम् = कर्त्तव्यम्, कार्यम् (करना चाहिए), शुचि = शुद्धम् (शुद्ध, स्वच्छ), पर्यावरणम् = वातावरणम् (पर्यावरण ही), शरणम् = मात्र आश्रयम् जातम् (एकमात्र शरण रह गया है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी ‘शेमुषी’ पाठ्यपुस्तक के ‘शुचिपर्यावरणम्’ पाठ से लिया गया है। यह गीत प्रो. हरिदत्त शर्मा लिखित ‘लसल्लतिका’ गीति-संग्रह से संकलित है। इस गीतांश में कवि वाहनों के धुएँ से प्रदूषित पर्यावरण के विषय में कहता है।

हिन्दी-अनुवादः – वायुमण्डल अत्यधिक दूषित हो गया है, क्योंकि पानी भी स्वच्छ नहीं है। खाने की वस्तुएँ भी दूषित पदार्थों से युक्त हैं। धरती मैली अर्थात् गन्दगीयुक्त है। अतएव बाह्य एवं आन्तरिक जगत् में अत्यधिक शुद्धीकरण करना चाहिए। शुद्ध पर्यावरण ही एकमात्र आश्रय है।

4. कञ्चित् कालं नय मामस्मान्नगराद् बहुदूरम्।
प्रपश्यामि ग्रामान्ते निर्झर-नदी-पयःपूरम् ॥
एकान्ते कान्तारे क्षणमपि मे स्यात् सञ्चरणम्। शुचि…… ॥4॥

अन्वयः – कञ्चित् कालम् माम् अस्मात् नगरात् बहुदूरं नय। (अहं) ग्रामान्ते पयः पूरं निर्झरं, नदी (च) प्रपश्यामि!, एकान्ते कान्तारे क्षणमपि मे सञ्चरणं स्यात्। शुचिपर्यावरणं शरणम्।

शब्दार्थाः – कञ्चित् कालम् = किञ्चित् समयम् (कुछ समय के लिए), माम् = मा, कविम् (मुझ कवि को), अस्मात् = एतस्मात् (इस), नगरात् = पुरात्, पत्तनात् (शहर से), बहुदूरम् = भृशंदूरम्, अत्यधिक दूरम् (बहुत दूर), नय गमय, प्रापय (ले चल), ग्रामान्ते = ग्रामस्य सीमायाम्, सीम्नि (गाँव की सीमा पर), निर्झरं – प्रपात, स्रोत (झरना), नदी सरिता (नदी), पयःपूरम् = जलेनपूर्णम् जलाशयम् (जलाशय को), प्रपश्यामि = अवलोकयामि, ईक्षे (देखता हूँ), एकान्ते = निर्जने (निर्जन, शान्त), कान्तारे = वने (जंगल में), क्षणमपि = क्षणमात्रम् (क्षणभर), मे = मम (मेरा), सञ्चरणम् = विचरण, भ्रमणम् (घूमना), स्यात् = भवेत् (होना चाहिए), अतः शुचि = शुद्धं (शुद्ध), पर्यावरणम् = वातावरणम् (पर्यावरण ही), शरणम् = आश्रयः (सहारा है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी ‘शेमुषी’ पाठ्यपुस्तक के ‘शुचिपर्यावरणम्’ पाठ से लिया गया है। यह पाठ प्रो. हरिदत्त शर्मा रचित ‘लसल्लतिका’ गीति-संग्रह से संकलित है। इस गीतांश में कवि पावन पर्यावरण का चित्रण करता है।

हिन्दी-अनुवादः – कुछ समय के लिए मुझे इस शहर से बहुत दूर ले चलो। गाँव की सीमा पर (मैं) जल से पूर्ण झरने और नदी को देख रहा हूँ। (उस) निर्जन शान्त वन में मेरा एक क्षण भी घूमना हो जाए अर्थात् मैं वहाँ क्षणभर घूमना चाहता हूँ। अतः शुद्ध पर्यावरण ही एकमात्र आश्रय है।

5. हरिततरूणां ललितलतानां माला रमणीया।
कुसुमावलिः समीरचालिता स्यान्मे वरणीया ॥
नवमालिका रसालं मिलिता रुचिरं संगमनम्। शुचि…..॥5॥

अन्वयः – हरिततरूणां ललितलतानां रमणीया माला, समीरचालिता कुसुमावलिः मे वरणीया स्यात्। नवमालिका मिलिता रसालं रुचिरं सङ्गमनम्। शुचिपर्यावरणं शरणम्।

शब्दार्थाः – हरिततरूणाम् = हरिद्वर्णानां वृक्षाणाम् (हरे-भरे वृक्षों की), ललितलतानाम् = रम्याणां बल्लरीणाम् (सुन्दर बेलों की), रमणीया – रम्या, मनोहरा (सुन्दर), माला = पंक्तिः, श्रेणिः, हारः (माला, पंक्ति), समीर = पवनेन, वायुना (वायु द्वारा), चालिता = वेपिता, सञ्चालिता (हिलाई हुई), कुसुमावलिः = पुष्पाणां पंक्तिः (फूलों की पंक्ति), मे = मह्यम् (मेरे लिए), वरणीया = वरणाय, वरप्रदानाय (वरण करने योग्य, प्रसन्नता देने वाली), स्यात् = भवेत् (होनी चाहिए), नवमालिका = नूतना मल्लिका (नवमल्लिका), मिलिता रसालम् = आम्रवृक्षेण, सहकारेण मिलिता, बलयिता (आम के पेड़ से लिपटी हुई), रुचिरं = सुन्दरम् (सुन्दर), सङ्गमनम् = मिलनं प्राप्नोति (संगम प्राप्त कर रही है), शुचि = स्वच्छं, शुद्धं (स्वच्छ), पर्यावरणम् = वातावरणम् (जलवायु, पर्यावरण), शरणम् = आश्रयः (आश्रय है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी शेमुषी’ पाठ्य-पुस्तक के ‘शुचिपर्यावरणम्’ पाठ (गीत) से लिया गया है। यह पाठ प्रो. हरिदत्त शर्मा रचित ‘लसल्लतिका’ गीति-संग्रह से संकलित है। इस गीत में कवि प्रकृति की रमणीयता का वर्णन करता है।

हिन्दी-अनुवादः – हरे-भरे वृक्षों और सुन्दर बेलों की पंक्ति (और) वायु द्वारा हिलाई जाती हुई फूलों की पंक्ति मेरे लिए वरण करने योग्य होनी चाहिए। नवमल्लिका आम के वृक्ष के साथ मिलकर सुन्दर संगम प्राप्त कर रही है। अतः स्वच्छ वातावरण ही एकमात्र आश्रय है।

6. अयि चल बन्धो ! खगकुलकलरव गुञ्जितवनदेशम्।
पुर-कलरव सम्भ्रमितजनेभ्यो धृतसुखसन्देशम् ॥
चाकचिक्यजालं नो कुर्याज्जीवितरसहरणम्। शुचि……॥6॥

अन्वयः – अयि बन्धो ! खगकुल कलरव गुञ्जित वनदेशं चल। पुर-कलरव सम्भ्रमित जनेभ्यः सुख सन्देशं धृत, चाकचिक्यजालं नो जीवित रस हरणम् कुर्यात्। शुचि पर्यावरणं शरणम्।

शब्दार्थाः – अयि बन्धो ! = हे भ्रातः ! (हे भाई !), खगकुलकलरव = पक्षिणां सामूहिक ध्वनिः, ध्वानेन (पक्षियों की सामूहिक ध्वनि से), गुञ्जित = गुञ्जायमानम् (गूंजते हुए), वनदेशम् चल = कान्तार प्रदेशं चल (वनप्रदेश को चलो), पुर-कलरव = नगरस्य ध्वानेन (नगर के शोरगुल से), सम्भ्रमित = सम्यक् भ्रमित, भयभीत (भ्रमित हुए या भयभीत), जनेभ्यः = मानवेभ्यः (लोगों के लिए), धृतसुखसन्देशम् = (सुख का सन्देश धारण किया हुआ है), चाकचिक्यजालम् = कृत्रिमं प्रभावपूर्ण जगत् (बनावटी आकर्षण पैदा करने वाला, चकाचौंध करने वाला जगत्), जीवित रस हरणम् = जीवनस्य सुखस्यापहरणम् (जीवन के सुखरूपी रस का हरण), नो = (नहीं), कुर्यात् = कुर्वीत (करना चाहिए), शुचि = शुद्धम् (शुद्ध), पर्यावरणम् = वातावरणम् (पर्यावरण ही), शरणम् = आश्रयः (आश्रय है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी ‘शेमुषी’ पाठ्य-पुस्तक के ‘शुचिपर्यावरणम्’ पाठ से लिया गया है। यह गीत प्रो. हरिदत्त शर्मा द्वारा ‘लसल्लतिका’ गीति-संग्रह से संकलित है।

हिन्दी-अनुवादः – हे भाई ! पक्षियों की सामूहिक ध्वनि से गूंजते हुए वनप्रदेश की ओर चलो। (जो) नगर के शोरगुल से भयभीत लोगों के लिए सुख का सन्देश धारण किया हुआ है अर्थात् नगर के शोरगुल से थके हुए लोगों के लिए यहाँ सुख मिलता हैं। चकाचौंध करने वाला यह जगत् (कहीं) जीवन के सुख का अपहरण न कर ले। अतः शुद्ध पर्यावरण ही एकमात्र शरण (आश्रय) है।

7. प्रस्तरतले लतातरुगुल्मा नो भवन्तु पिष्टाः।
पाषाणी सभ्यता निसर्गे स्यान्न समाविष्टा ॥
मानवाय जीवनं कामये नो जीवन्मरणम्। शुचि…..॥7॥

अन्वयः – लतातरुगुल्मा प्रस्तरतले पिष्टाः न भवन्तु। पाषाणीसभ्यता निसर्गे समाविष्टा न स्यात्। मानवाय जीवनं कामये नो जीवन् मरणम्। शुचिपर्यावरण शरणम्।

शब्दार्थाः – लतातरुगल्माः = लताः च तरवः च गल्माः च (बेल, वक्ष और झाडियाँ), प्रस्तरतले = शिलातले (पत्थरों के नीचे), पिष्टाः = प्रविष्टाः, दमिताः (दबी हुई), निसर्गे = प्रकृतौ (प्रकृति में), पाषाणी सभ्यता = पाषाणकालीन सभ्यता (पाषाण काल की सभ्यता), समाविष्टाः = समावेशः, पेषणम् (समावेश, पिसना-कुचलना), न स्यात् = न भवेत् (नहीं होना चाहिए), जीवनम् = जीवितं (जीवन की), कामये = कामनां करोमि (कामना करता हूँ)। नो = अस्माकम् (हमारी), जीवन मरणम् = जीवत् मरणम् च (जीते-मरते), मानवाय = मनुष्याय (मानव के लिए), शुचि = शुद्धम् (शुद्ध), पर्यावरणम् = वातावरणम् (पर्यावरण ही), शरणम् = आश्रयः (आश्रय है)।

सन्दर्भ-प्रसङ्गश्च – यह गीतांश हमारी शेमुषी’ पाठ्य-पुस्तक के ‘शुचिपर्यावरणम्’ पाठ से लिया गया है। यह पाठ प्रो. हरिदत्त शर्मा द्वारा रचित ‘लसल्लतिका’ गीति-संग्रह से संकलित है। इस पाठ में कवि भारतीय संस्कृति संरक्षण के आधार पर कहता है।

हिन्दी-अनुवादः – लता, वृक्ष और झाड़ियाँ पत्थरों के तल पर नष्ट नहीं होनी चाहिए। अर्थात् पाषाणकालीन सभ्यता प्रकृति में समाविष्ट नहीं होनी चाहिए। मैं मानव के जीवन की कामना करता हूँ, जीवन के नष्ट होने की नहीं। अतः शुद्ध पर्यावरण की शरण में जाना चाहिए।

Jharkhand Board JAC Class 10 Science Solutions Chapter 6 Life Processes Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 6 Life Processes

Jharkhand Board Class 10 Science Life Processes Textbook Questions and Answers

Question 1.
The kidneys in human beings are a part of the system of ……………….
A. nutrition
B. respiration
C. excretion
D. transportation
Answer:
excretion

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: transport of water

Question 3.
The autotrophic mode of nutrition requires ……………….
A. carbon dioxide and water
B. chlorophyll
C. sunlight
D. all of the given
Answer:
all of the given

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:
mitochondria

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
Large fat globules break into small fine droplets by the effect of bile salts of bile juice. This is called emulsification of fats.

Pancreatic lipase acts on emulsified fats to break it and finally intestinal lipase digests fats into fatty acids and glycerol. This process take place in small intestine.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
Saliva contains salivary amylase (ptyalin) which digests starch into sugar.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by-products?
Answer:
The necessary conditions for autotrophic nutrition are:

• Presence of chlorophyll
• Absorption light energy
• Splitting of water molecules
• Reduction of carbon dioxide to carbohydrates.

Oxygen is the by-product.

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
1. O2 is used in this process.	1. O2 is not used in this process.
2. At the end of this process CO2 and H2O are produced.	2. At the end of this process in medium of plant origin Ethanol and CO2 are produced and in medium of animal origin only lactic acid is produced and no CO2.
3. In aerobic respiration complete oxidation of glucose molecules occurs, in which one mole of glucose on oxidation releases much greater energy.	3. In anaerobic respiration glucose molecules are incompletely oxidized, so one mole of glucose releases less energy along with the organic by products.
4. There are two phases in aerobic respiration, the first phase occurs in the cytoplasm and does not utilize O2. The second phase occurs in the mitochondria and utilizes O2.	4. There is only one phase in anaerobic respiration. It occurs in the cytoplasm. It occurs entirely in the absence of O2.

Question 9.
How are the alveoli designed to maximise the exchange of gases?
Answer:
The alveoli are located at the terminal ends of bronchioles. They are balloon-like structures provides large surface area for exchange of gases with an extensive network of blood vessels.

Question 10.
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
A deficiency of haemoglobin in our bodies leads to a disease called anaemia. Due to this, cells of our body do not get sufficient oxygen for cellular respiration, which may lead to release less energy. Weakness, fatigue, tiredness, etc. conditions may arise.

Question 11.
Describe double circulation of blood in human beings. Why is it necessary?
Answer:
Blood passes through the heart twice during each cycle in human beings. This is called double circulation.

[Deoxygenated blood from different organs is drained and finally through vena cava it is poured in right atrium. Prom here this blood is transported to lungs via right ventricle. In lungs, blood become oxygenated and is again transported to left atrium. From here it transports in left ventricle and then by aorta to different body parts.]

It is necessary because it allows a highly efficient supply of oxygen to the body cells, which fulfill the high energy need of body.

Question 12.
What are the differences between the transport of materials in xylem and phloem?
Answer:

In xylem	In phloem
1. Water and minerals are transported.	1. Food especially carbohydrate, sucrose is translocated.
2. Transpiration pull becomes the major driving force for transport in xylem.	2. Osmotic pressure is responsible for translocation in phloem.
3. Generally ATP is not used for transport of material in xylem.	3. Phloem tissue uses ATP for translocation of food materials.
4. Vessels and tracheids are involved in transport.	4. Sieve tube and companion cells are involved in translocation.

Question 13.
Compare the functioning of alveoli in the lungs to their structure and functioning.
Answer:

Alveoli	Nephrons
1. It is a structural and functional unit of lung.	1. It is a structural and functional unit of kidney.
2. Alveoli are balloon-like structures at the terminal region of bronchioles.	2. Nephrons are long-coiled tube-like structures having Bowman’s capsule at the tip.
3. The alveoli provide a surface for exchange of gases.	3. Filtration of blood for removing of nitrogenous wastes take place in nephron units.
4. The wall of the alveoli contain an extensive network of blood vessels.	4. A cluster of blood capillaries associated with Bowman’s capsule called glomerulus and tubular part of nephron is surrounded by network of blood capillaries.

Jharkhand Board Class 10 Science Life Processes InText Questions and Answers

Question 1.
Why is diffusion insufficient to meet the oxygen requirement of multicellular organisms like humans?
Answer:
In multicellular organisms like human, all the cells are not in direct contact with the s surrounding environment. The body structure is more complex and the body size is also large. Therefore, simple diffusion will not be sufficient to s send oxygen to every cell. It has been estimated that a period of 3 years would be needed to carry a molecule of O₂ from our lungs to reach our toes through diffusion.

So, diffusion is insufficient to meet the oxygen requirement of multicellular organisms like humans.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
Movement, growth, breathing, cell-structure, etc. are the criteria we use to decide whether something is alive.

Question 3.
What are outside raw materials used for by an organisms?
Answer:

Name of outside raw materials	Used for
1. CO2, H2O	Photosynthesis by plants
2. Carbon based food source, O2	Respiration by aerobic organisms

Question 4.
What processes would you consider essential for maintaining life?
Answer:
The processes essential for maintaining life are nutrition, respiration, transport, excretion, etc.

Question 5.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Autotrophic Nutrition	Heterotrophic Nutrition
1. It occurs in green plants and some bacteria.	1. It occurs in animals and fungi.
2. In such mode of nutrition. food is synthesised from inorganic components, i.e., CO2 and H2O.	2. In such mode of nutrition food is consumed from other organisms.
3. Photosynthesis is important process for autotrophic nutrition.	3. Food digestion is important for such nutrition.

Question 6.
Where do plants get each of the raw materials required for photosynthesis?
Answer:
Raw materials required for photosynthesis :

• CO₂ : Plants get it from atmosphere.
• H₂O : Plants root absorbed it from soil.
• Energy : Plants get it directly from sun.

Question 7.
What is the role of the acid in our stomach?
OR
What are the functions of the acid in our stomach?
Answer:
Role or Functions of the acid:

• Acid destroys the bacteria and other microorganisms that enter the stomach, along with the food.
• It converts the inactive enzyme pepsinogen into active enzyme pepsin.
• It provides acidic medium required for the action of pepsin. Pepsin can digest proteins present in food only in acidic medium.
• Insoluble mineral salts get dissolved in acid.

Question 8.
What is the function of digestive enzymes?
Answer:
Digestive enzymes hydrolyse / digest complex component (carbohydrates, lipids, proteins) of food into simple, soluble and absorbable form of nutrients.

Question 9.
How is the small intestine designed to absorb digested food?
Answer:
Small intestine is long tubular structure. The inner lining of the small intestine has numerous finger like projection called villi, which increase the surface area for absorption.

Question 10.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
The amount of dissolved oxygen is fairly low in aquatic environment as compared to the amount of oxygen in the air. So, terrestrial organisms fulfil their oxygen demand with low breathing rate as compared to aquatic organisms.

Question 11.
What are the different ways in which glucose is oxidised to provide energy in various organisms?
Answer:
There are three different ways in which glucose is oxidised to provide energy in various organisms.

Question 12.
How is oxygen and carbon dioxide transported in human beings?
Answer:
In human beings, the respiratory pigment haemoglobin has high affinity for oxygen, so it is mostly transported by haemoglobin. Carbon dioxide is more soluble in water than oxygen and hence it is mostly transported in dissolved form in our blood.

Question 13.
How are the lungs designed in human beings to maximise the area for exchange of gases?
Answer:
The respiratory passage in the lungs, divides into smaller and smaller tubes which finally terminate in balloon – like structures alveoli. The alveoli present in lungs provide the maximum area for exchange of gases in human beings.

Question 14.
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	Functions
1. Blood Plasma Red blood corpuscles White blood corpuscles Platelets	Acts as fluid transport medium of various material. Transport of food, CO2, salts and nitrogenous wastes. Transport of O2. Fight with invading pathogens. Help in clotting mechanism during injury.
2. Heart	It acts as a blood pumping organ.
3. Blood vessels Arteries Veins Capillaries	Carry blood away from the heart. Carry blood from different organs and bring it back to the heart. Exchange of material between the blood and surrounding cells.
4. Lymph	Carries digested and absorbed fat from intestine and drain excess fluid from intercellular spaces back into the blood.

Question 15.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
It is necessary to separate oxygenated s and deoxygenated blood in mammals and birds because it allows a highly efficient supply of oxygen to the body and this is useful in their high energy needs for to maintain constant body temperature.

Question 16.
What are the components of the i transport system in highly organised plants?
Answer:
Xylem tissue (vessels and tracheids) and phloem tissue (sieve tube and companion cell) are the components of the transport system in highly organised plants.

Question 17.
How are water and minerals transported in plants?
Answer:
Water conducting channels : Xylem consists of vessels and tracheids, which form continuous water conducting channel.

Absorption of water by the roots : The root cells actively take up ions from soil. 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.

Column of water : To eliminate the concentration difference between the soil and the root, the steady movement of water into root creates column of water.

Conduction of water by root pressure : Due to absorption of water by root cells, a pressure is generated to push water in xylem element.

This pressure is insufficient to move water over the heights of plants. So, plants use another strategy to move water in xylem upwards to the highest points of the plant body.

Conduction of water by transpiration pull : The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

The water which is lost through the stomata is replaced by water from the xylem vessels in the leaf. Evaporation of water molecules from the cells of a leaf creates a sunction which pulls water from the xylem cells of roots.

During the day when the stomata are open, the transpiration pull becomes the major driving force in the movement of water in the xylem.

At night effect of root pressure is necessary for the upward flow of water.

Thus, transpiration helps in absorption and upward movement of water and minerals dissolved in it from roots to the leaves.

Question 18.
How is food transported in plants?
Answer:
The transport of soluble products of photosynthesis is called translocation.

• Translocation 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 especially delivered to roots, fruits, seeds and to growing organs where it is stored.
• 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.
• The translocation in phloem is achieved by utilising energy.
• Sucrose (Sugar / Carbohydrate) is transferred into phloem tissue using energy from ATP. This increases the osmotic pressure of the tissue causing water to move into it.
• This pressure moves the material in the phloem to tissues having less pressure. The phloem thus moves material according to the plant’s need.

Example: In the spring, sugar stored in root or stem tissue is transported to the buds which need energy to grow.

Question 19.
Describe the structure and functioning of nephrons.
Answer:
Nephron is a basic filtration unit in the kidneys.

• Each kidney has large numbers of nephrons packed closely together.
• Nephron is a long-coiled tubular structure which begins with a cup-shaped end. called Bowman’s capsule and it ends in collecting tubule.
• A cluster of very thin-walled blood capillaries seen in the Bowman’s capsule is called glomerulus.
  

The purpose of making urine is to filter out waste products from blood.

• Nitrogenous wastes such as urea, uric acid, etc. are removed from blood in the kidneys.
• Urine is produced by filtration units i.e., nephrons.
• Cup-shaped Bowman’s capsule collects the filtrate.
• Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water are selectively reabsorbed as the urine flows along the coiled tube.
• The amount of water reabsorbed depends on the amount of water present in the body and amount of dissolved waste which is to be excreted.
• Thus, urine is formed in both kidneys. [In a normal healthy adult, the initial filtrate in the kidneys is about 180 L daily. However, the s volume of excreted urine is only a litre or two per day. The remaining filtrate is reabsorbed ; in the kidney tubules.]

Question 20.
What are the methods used by plants to get rid of excretory products?
Answer:
Unlike animals, the plants do not possess any special organs or system for excretion. However, the plants excrete their wastes in different ways:

• O₂ produced during photosynthesis by the green plants is set free directly in the atmosphere.
• Plants remove surplus water by the process of transpiration through the stomata.
• Sometimes plants store certain wastes in the cells of their leaves which are ultimately shed off.
• Certain plants store wastes in the cellular vacuoles of their cells.
• Other waste products such as resin and gum are stored especially in old xylem.
• Plants excrete some waste substances into the soil around them.

Question 21.
How is the amount of urine produced regulated?
Answer:
The amount of urine formed depends on how much excess water there is in the body, and on how much of dissolved waste there is to be excreted. More water and dissolved wastes in the body will produce more urine. On the other hand, less water and less dissolved wastes will produced less urine.

Activity 6.1 [T. B. Pg. 96]

To demonstrate that chlorophyll is essential for photosynthesis.

Materials : Potted plant (money plant or croton), beaker, water bath, alcohol, iodine.

Procedure :

• Take a potted plant of money plant or croton with variegated leaves.
• Keep the potted plant in complete darkness for three days.
• After three days expose the plant to bright sunlight for about 6 hours.
• Pluck one such leaf of this plant which is green in certain parts and white in the other remaining parts.
• Mark the green areas in it and trace them on a papersheet.
• Boil this plucked leaf in a beaker full of alcohol and kept in boiling water bath for some time, till it becomes colourless.
• Wash and clean this colourless leaf with water and dip it, for a few minutes, in dil. iodine solution.
• Observe the colour of the leaf and compare this with tracing of the leaf done in the beginning.
  

Questions:

Question 1.
What happens when the plant is kept in s a dark for 2-3 days?
Answer:
All the starch gets used up when the plant is kept in a dark for 2-3 days.

Question 2.
What happens to the colour of the leaf when keep in boiling alcohol?
Answer:
When the leaf is kept in boiling alcohol, it gets decolourised.

Question 3.
What is the colour of the alcohol solution when leaf is taken out from it?
Answer:
The colour of the alcohol solution is green when leaf is taken out from it.

Question 4.
State the use of iodine solution.
Answer:
Iodine solution is used to know the presence of starch.

Question 5.
What can you conclude about the presence of starch in various areas of the leaf?
Answer:
We conclude that the presence of starch is detected with iodine solution in only such areas where chlorophyll present.

Question 6.
What will you infer from this activity?
Answer:
This activity explains that chlorophyll is essential for photosynthesis and surplus glucose stored in form of starch as internal reserve energy.

Activity 6.2 [T. B. Pg. 97]

To demonstrate that carbon dioxide (CO₂) necessary for photosynthesis

Materials : Potted plants, Bell-jar, watch glass, KOH (potassium hydroxide), alcohol, iodine solution.

Procedure:

• Take two potted plants. Place them in dark for three days. This will de-starch them.
• After three days place each of the pots on a smooth glass plate.
• Label one pot as A and the other as B.
• Keep a small watch glass or petri dish containing pellets of potassium hydroxide (KOH) on the glass plate near the pot A.
• Cover both the potted plants under two separate glass bell-jars.
• Apply a thick coat of vaseline on the bottom of the bell-jar close to the glass plate in order to make both the jars air-tight.
• Expose both these plants to sunlight for about 2 to 3 hours.
• Thereafter pluck one leaf each from the plants A and B and examine these leaves separately for the presence of starch therein.
  

Questions :

Question 1.
Which gas is absorbed by KOH from the air in the bell-jar?
Answer:
Carbon dioxide is absorbed by KOH from the air in the bell-jar.

Question 2.
Which potted plant (A or B) shows the presence of starch in its leaf?
Answer:
B potted plant shows the presence of starch in its leaf.

Question 3.
Why is starch not formed in the leaf of the plant kept under the bell-jar with KOH kept along with it?
Answer:
Starch is not formed in the leaf of the plant kept under the bell-jar with KOH kept along with it because KOH absorbs COa from air. So, CO₂ is not available for the photosynthesis and thus starch is not formed.

Question 4.
Write your inference based on your observation and study.
Answer:
The potted plant B showed normal photosynthesis in presence of CO₂. This shows that CO₂ is necessary for photosynthesis.

Question 5.
In order to prevent the increase of which gas in the atmosphere, the conservation of plant organisms (trees) are important?
Answer:
To prevent the increase of CO₂ gas in the atmosphere, the conservation of plant organisms (trees) are important.

Question 6.
Do both the leaves show the presence of same amount of starch?
Answer:
No.

Activity 6.3 [T. B. Pg. 99]

To check the effect of saliva on starch. Materials : Two test tubes, starch solution, iodine.

Procedure:

• Take 1 mL starch solution in two test tubes (A and B).
• Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20 – 30 minutes.
• Now add a few drops of dilute iodine solution to the test tubes.
  

Questions:

Question 1.
In which test tube do you observe a colour change?
Answer:
In test tube B colour change is observed.

Question 2.
What does this indicate about the presence or absence of starch in the two test tubes?
Answer:
Test tube B solution showed colour change means there is presence of starch. While in test tube A colour is not changed that means starch is absent.

Question 3.
What is the difference in colour of solution, A and B after adding iodine solution?
Answer:
Colour of solution A is yellow and colour of solution B is blue or black after adding iodine solution.

Question 4.
The colour of solution in tube A does not change to blue or black. Why?
Answer:
The colour of solution in test tube A does not change blue or black because starch is hydrolysed by the effect of enzymes present in saliva.

Question 5.
Which component of food is digested (partially) when the food is chewed in the mouth?
Answer:
Starch (Carbohydrate) is partially digested when the food is chewed in the mouth.

Question 6.
Why does the bread or chapati tastes sweet when chewed for a longer time?
Answer:
When bread or chapati chewed for a longer time, starch is converted to simple sugar by salivary amylase. The sugar formed produces sweet taste in mouth.

Question 7.
State which of the following edible substances in our food are the source of starch: Potato, Lettuce leaf. Wheat, Maize, Sweet pea. Groundnut, Ghee.
Answer
Potato, Wheat, Maize, etc. are the sources of starch in our food.

Activity 6.4 [T. B. Pg. 101]

To demonstrate that CO₂ is exhaled by us during breathing.

Materials : Two test tubes, Rubber tube, Pichkari, Lime water.

Procedure :

• Take two clean glass test tubes. Label one of the tubes as (a) and the other as (b).
• In each test tube add about 10 mL of freshly prepared lime water (Ca(OH)₂ solution).
• Use a syringe or pichkari to pass air through some fresh lime water taken in test tube (a).
• Blow air through tube in a lime water taken in test tube (b).
• Note how long it takes for this lime water to turn milk in each test tube.
  

Questions :

Question 1.
What change is observed in test tube (a) and test tube (b)? Why?
Answer:
Lime water turns milky in test tube (a) and test tube (b). Because of presence of CO₂ the colour of lime water is changed.

Question 2.
In which tube does the change occur more rapidly? Why?
Answer:
In test tube (b), the change occurs more rapidly because the air that we expell contain more CO₂.

Question 3.
What does this activity tell us about the amount of carbon dioxide in the air that we breathe out?
Answer:
This activity tells us that the amount of carbon dioxide is more in the air that we breathe out.

Question 4.
What you conclude from this activity?
Answer:
Carbon dioxide is produced in the process of respiration.

Activity 6.5 [T. B. Pg. 101]

To demonstrate that CO₂ is produced during fermentation.

Materials : Test tube. Bent glass tube, Fruit juice, Yeast powder. Lime water. One holed cork.

Procedure :

• Take some fruit juice or sugar solution and add some yeast to this mixture.
• Take this mixture in a test tube fitted with one holed cork.
• 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.
  

Questions :

Question 1.
What change is observed in the colour of Ca(OH)₂ solution?
Answer:
The colour of Ca(OH)₂ solution turns milky.

Question 2.
How much time does it take for the change of colour?
Answer:
It takes long time for the change of colour of lime water.

Question 3.
Which product of fermentation is responsible for bringing about the change in colour of lime water?
Answer:
CO₂ is a product of fermentation is responsible for bringing about the change in colour of lime water.

Question 4.
What does this activity tell us about the products of fermentation?
Answer:
This activity tells us about the products of fermentation which are CO₂ and ethanol.

Activity 6.6 [T. B. Pg. 103]

To compare the breathing rate of fish and human.

Materials : An aquarium

Procedure:

• Observe moving fish in an aquarium.
• Count the number of times the fish opens and closes its mouth in a minute.
• Count the number of times you breathe in and out in a minute.
• Compare the breathing count of fish with yours.

Questions :

Question 1.
What is an operculum?
Answer:
An operculum is folded covering of gill slits in some fishes.

Question 2.
By which organ do fish respire?
Answer:
Fish respires with gills.

Question 3.
Are the timing of the opening and closing of the mouth and gill-slits in fish coordinated in some manner?
Answer:
Yes, in fish when mouth opens, gill-slits close and vice versa.

Question 4.
How do fish respire?
Answer:
Fishes take in water through their mouths and force it over the gills where the dissolved oxygen is taken up by blood capillaries by diffusion and CO₂ is released in water. Such water is discarded through gill slits.

Question 5.
What is an average breathing rate per minute in experimental fish?
Answer:
An average breathing rate 66-78 per minute in experimental fish.

Question 6.
What is an average breathing rate per minute in human beings?
Answer:
An average breathing rate 12-16 per minute in human beings.

Question 7.
Why in aquatic animal the breathing rate is much faster than that seen in terrestrial animals?
Answer:
Since the amount of dissolved oxygen is fairly low in water as compared to the amount of oxygen in the air, the rate of breathing in aquatic animal like fish is much faster than that seen in terrestrial animals.

Activity 6.7 [T. B. Pg. 105]

To know the haemoglobin content in human beings and in animals like buffalo, cow, etc.

• Visit a health centre in your locality and find out the normal range of haemoglobin content in human beings.
• Visit a veterinary in your locality. Find out the normal range of haemoglobin content in an animal like buffalo, cow, etc.
• Compare the difference seen in male and female human beings and animals.

Questions :

Question 1.
What is the normal range of haemoglobin content in human beings?
Answer:
The normal range of haemoglobin content 12-18 g/ decilitre.

Question 2.
Is it the same for children and adult?
Answer:
No, in children (3 month to 12 years) haemoglobin content 11.0 ± 1.5 g / decilitre.

Question 3.
Is there any difference in the haemoglobin levels for men and women?
Answer:
Yes, Men ⇒ 13-18 g / decilitre
Women ⇒ 12-16 g / dl

Question 4.
Is the haemoglobin content different in calves, male and female animals?
Answer:
Yes.

Question 5.
What is the normal range of haemoglobin content in cow and buffalo?
Answer:
Cow-10 to 15 g /decilitre
Buffalo -12.5 to 14.5 g / decilitre

Activity 6.8 [T. B. Pg. 109]

To demonstrate the physiological process of transpiration in plant.

Materials : A pot with growing plant, a pot with similar size with same amount of soil. a stick, plastic sheet.

Procedure:

• Take two small pots of approximately the same size and having the same amount of soil.
• One pot labelled as (a) with a plant.
• Place a stick of the same height in other pot labelled as (b).
• Cover the soil in both pots with a plastic sheet.
• Cover both pots, with plastic sheets and place them in bright sunlight for half an hour.

Questions :

Question 1.
Why the soil in both pots is covered with a plastic sheet?
Answer:
The soil in both pots is covered with a plastic sheet to prevent evaporation and loss of moisture.

Question 2.
What do you observe after half an hour?
Answer:
After half an hour, small water droplets are observed on the inner surface of a plastic sheet of pot (a).

Question 3.
What do you conclude from your observation?
Answer:
From our observation, we concluded that there is a water loss from aerial part of a plant in the form of vapour. This process is called transpiration.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 10 Circles Ex 10.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 10 Circles Exercise 10.1

Question 1.
How many tangents can a circle have?
Solution :
A circle can have infinite number of tangents.

Question 2.
Fill in the blanks:
(i) A tangent to a circle intersects it in ________ point (s).
(ii) A line intersecting a circle in two points is called a _______.
(iii) A circle can have _________ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called ________.
Solution :
(i) one,
(ii) sccant,
(iii) two,
(iv) point of contact.

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is:

(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm.
Solution :
PQ² = OQ² – OP² (Using Pythagoras theorem)
= (12)² – (5)²
= 144 – 25
= 119.
PQ = \(\sqrt{119}\) cm.

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution :
AB is the given line. CD is the secant and PQ is the tangent to the circle at point R.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory:

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	No. of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:

Weight (in kg)	Frequency	Cumulative frequency
36 – 38	0	0
38 – 40	3	3
40 – 42	2	5
42 – 44	4	9
44 – 46	5	14 = cf
46 – 48	14 = f	28
48 – 50	4	32
50 – 52	3	35
	n = 35

\(\frac{\mathrm{n}}{2}=\frac{35}{2}=17.5\)
Plot the points (38, 0) (40, 3) (42, 5) (44, 9) (46, 14) (48, 28) (50, 32) (52, 35)
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 46 + \(\left[\frac{17.5-14}{14}\right]\) × 2
= 46 + \(\frac{3.5 \times 2}{14}\)
= 46 + 0.5= 46.5.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village:

Change the distribution to a more than type distribution, and draw its ogive.
Solution:

Production yield (in kg/hec)	Number of farms	c.f.
More than 50	2	100
More than 55	8	98
More than 60	12	90
More than 65	24	78
More than 70	38	54
More than 75	16	16

∴ Co-ordinate points are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of the cube = a³ = 64 cm³
Side of the cube \(\sqrt[3]{64}\) = a = 4 cm.

Surface area of the cuboid = 2(lb + bh + lh)
= 2[(8 × 4) + (4 × 4) + (8 × 4)]
= 2[32 + 16 + 32]
= 2 × 80
= 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

π = \(\frac{22}{7}\), radius of the hemisphere = 7 cm,
height of the hemisphere = 7 cm,
height of the cylinder, h = 13 – 7 = 6 cm.
Inner area of the vessel = Inner area of the hemisphere vessel + Inner area of the cylinder
= 2πr² + 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 7 + 2 × \(\frac{22}{7}\) × 7 × 6
= 2 × 22 × 7+2 × 22 × 6
= 2 × 22(7 + 6)
= 44 × 13 = 572 cm².

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:

π = \(\frac{22}{7}\), r = 3.5, l = ?
Total surface area of the toy = C.S.A. of the cone + C.S.A. of the hemisphere
= πrl + 2πr²
lv = r² + h²
= (3.5)² + (12)²
= 12.25 + 144
= 156.25
l = \(\sqrt{156.25}\)
= 12.5 cm.

h = Ax – Ox
= 15.5 – 3.5
= 12 cm.

Surface area of the toy = πrl + 2πr²
= \(\frac{22}{7}\) × 3.5 × 12.5 + 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= \(\frac{22}{7}\) × 3.5[12.5 + 2 × 3.5]
= 22 × 0.5[12.5 + 7]
= 11[12.5 + 7]
= 11 × 19.5 = 214.5 cm².

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:

Greatest diameter = 7 cm.
Surface area of the block = T.S.A. of the cube – Base area of the hemisphere + C.S.A. of the hemisphere
= 6 × 72 – πr² + 2πr²
= 6 × 49 + πr²
= 6 × 49 + \(\frac{22}{7}\) × 3.5 × 3.5
= 294 + 11 × 3.5
= 294 + 38.5
= 332.5 cm².

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:

Surface area of the remaining solid = Surface area of the cube + Surface area of the hemisphere
= 6l² + 2πr²
= 6l² + 2π\(\left(\frac{l}{2}\right)^2\)
= 6l² + 2π\(\frac{l^2}{4}=\frac{l^2}{4}\)(24 + 2π) sq. units.

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution:

Height of the cylindrical portion = 14 – 2.5 – 2.5 = 9m = h
r = 2.5 m.
Surface area of the capsule = Surface area of the cylindrical portion + Areas of the hemispherical regions
= 2πrh + 2πr² + 2πr²
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 2.5 + 2.5)
= 2 × \(\frac{22}{7}\) × 2.5 (9 + 5)
=2× \(\frac{22}{7}\) × 2.5 × 14
= 44 × 5
= 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:
r = \(\frac{4}{2}\) = 2m, h = 2.1, l = 2.8

Area of the canvas used = C.S.A. of the cylindrical portion + C.S.A. of the conical region
= 2πrh + πrl
= πr(2h + l)
= \(\frac{22}{7}\) × 2(2 × 2.1 + 2.8)
= \(\frac{22}{7}\) × 2(4.2 + 2.8)
= \(\frac{22}{7}\) × 2 × 7
= 44 m².
Cost of the canvas = Area × Rate
= 44 × 500
= Rs. 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:
Height of conical part = Height of cylindrical part h = 2.4 cm.
Diameter of cylindrical part = 1.4 cm, so, the radius of cylindrical part r = 0.7 cm

Slant height of cylindrical part l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{(0.7)^2+(2.4)^2}\)
= \(\sqrt{0.49+5.76}\)
= \(\sqrt{6.25}\)
= 2.5
The total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of base of cylinder
= 2πrh + πrl + πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 2.4 + \(\frac{22}{7}\) × 0.7 × 2.5 + \(\frac{22}{7}\) × 0.7 × 0.7
= 4.4 × 2.4 + 2.2 × 2.5 + 2.2 × 0.7
= 10.56 + 5.50 + 1.56
= 17.60 cm².

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution:

Total surface area of the article = C.S.A. of the cylinder + Surface area of the hemisphere at the top + Surface area of the hemisphere at the bottom
= 2πrh + 2πr² + 2πr²
= 2πr(h + r + r)
= 2 × \(\frac{22}{7}\) × 3.5 (10 + 3.5 + 3.5)
= 2 × 22 × 0.5 × 17
= 22 × l × 17
= 374 cm².

Jharkhand Board JAC Class 10 Science Important Questions Chapter 6 Life Processes Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 6 Life Processes

Additional Questions and Answers

Question 1.
Distinguish between :
(1) Gastric juice and Bile
Answer:

Gastric juice	Bile
1. It is a mixture of secretions from gastric glands located in the inner wall of the stomach.	1. It is secreted from the liver cells.
2. It is not stored anywhere in stomach.	2. It is stored in the sac-called gall bladder.
3. It is secreted from the gastric glands in stomach and acts in the stomach itself.	3. It is secreted from the liver cells and acts in the duodenum (small intestine).
4. It is an acidic digestive juice.	4. It is an alkaline digestive juice.
5. It contains dil. HCl, enzyme pepsin and mucus.	5. It does not contain any enzyme.

(2) Herbivores animals and Carnivores animals
Answer:

Herbivores animals	Carnivores animals
1. These animals take in only plant material as food.	1. These animals take in only animal flesh and bones as well as blood as their food.
2. Their small intestine is relatively much longer than in carnivores.	2. Their small intestine is relatively much shorter than in herbivores.
3. They are consumers of the first order.	3. They are consumers of the second and third order.
4. The digestion of cellulose of the plant cells is quite complex and takes time.	4. The digestion of flesh of the animals is quite easy and rapid.
5. Example: rabbit, cow, buffalo, goat, etc.	5. Example: tiger, lion, leopard, wolf, etc.

(3) Respiration in plants and Respiration in animals
Answer:

Respiration in plants	Respiration in animals
1. In plants, exchange of gases is carried out individually by different organs (roots, stem and leaves in the process of respiration.)	1. In animals, exchange of gases is carried out by some definite parts or organs of the body, meant for respiration.
2. The flow of respiratory gases from one part of the body to the other, is a slow process.	2. The flow of respiratory gases from one part of the body to the other, is a rapid process.
3. The respiration in plants is slow.	3. The respiration in animals is quite rapid.
4. In anaerobic respiration in plants the end product is ethanol and CO2.	4. In anaerobic respiration in animals the end product is lactic acid.

(4) Xylem tissue and Phloem tissue
Answer:

Xylem tissue	Phloem tissue
1. It transports water and mineral Ions absorbed by the roots to different parts of the plant.	1. It transports organic products of photosynthesis from the leaves to different parts of the plant.
2. The principal conducting elements of xylem are tracheids and tracheae (vessels).	2. The principal conducting elements of phloem are sieve cells and sieve tubes with companion cells.
3. The conduction occurs only in upward direction.	3. The conduction occurs in both upward and downward directions.
4. It conducts only water and mineral ions.	4. Along with the sugars it conducts amino acids, plant hormones and several other substances.
5. For conduction of water In the xylem, the suction force due to transpiration is the principal force.	5. For conduction of organic food substances the energy required is obtained from ATP.

(5) Conduction of water in plants and Translocation of food in plants
Answer:

Conduction of water in plants	Translocation of food in plants
1. It occurs through the xylem tissue.	1. It occurs through the phloem tissue.
2. It occurs from the roots to the stem, leaves and flowers.	2. It occurs from the leaves to different parts of the plant.
3. It occurs only from below upwards.	3. It occurs from above downwards as well from below upwards.
4. A continuous water column is formed in the plant from root upwards which is pulled up due to suction force.	4. No food column is formed but the difference of pressure causes translocation which requires energy from ATP.

(6) Atria and Ventricles
Answer:

Atria	Ventricles
1. The upper two chambers of the heart are atria.	1. The lower two chambers of the heart are ventricles.
2. They are relatively thin-walled.	2. They are quite thick-walled.
3. Atria receive blood from different parts of the body and is poured in the ventricles.	3. Ventricles receive blood from the auricles and force the blood towards different parts of the body.
4. In atrium, the blood pressure is relatively low.	4. In ventricles, the blood pressure is quite high.

(7) Artery and Vein
Answer:

Artery	Vein
1. The blood vessel that carries blood from the heart to different organs is called an artery.	1. The blood vessel that carries blood from any organ towards the heart is called a vein.
2. In artery, the blood flows under higher pressure.	2. In vein, the blood flows under somewhat low pressure.
3. The wall of the artery is relatively thick and elastic.	3. The wall of the vein is relatively thin and less elastic.
4. The artery divides into several arterioles and numerous fine blood capillaries in the organs and tissues.	4. In the organs and tissues, the veins are formed by the union of numerous blood capillaries and several venules.
5. Arteries carry oxygenated blood (exception -Pulmonary artery).	5. Veins carry deoxygenated blood (exception -Pulmonary vein).

(8) Blood and Lymph
Answer:

Blood	Lymph
1. It is red coloured fluid connective tissue.	1. It is a colourless liquid connective tissue.
2. It contains liquid blood plasma and freely floating blood corpuscles.	2. It contains a certain amount of blood plasma, proteins and some blood cells (except red blood corpuscles).
3. It flows in the heart, arteries, veins and blood capillaries.	3. It flows in the intercellular spaces, larger lymph capillaries and in lymph ducts.
4. It is an independent liquid connective tissue.	4. It arises by the diffusion from the thin walls of the blood capillaries and after circulation in the body, it is poured back in the blood.

(9) Breathing and Respiration
Answer:

Breathing	Respiration
1. It is a physical / mechanical process.	1. It is a physiological process.
2. It occurs through the respiratory organs aided by accessory respiratory organs.	2. It occurs in each and every living cell of the body.
3. The mechanism of breathing is not necessarily found in all the living organisms.	3. The process of respiration occurs invariably in each and every living cell of all the living organisms.
4. It includes the physical processes of inhalation or inspiration expelled (taking in of atmospheric air) and exhalation or expiration (throwing out air contaihing CO2, into the atmosphere).	4. It includes the physiological (biochemical) processes of glycolysis and Krebs cycle and also oxidative phosphorylation.
5. There are no subtypes of breathing.	5. Aerobic and anaerobic respiration are the two different types of respiration.
6. Energy is utilized in this process.	6. Energy is released in this process.

Question 2.
Give scientific reasons for the following statements:
(1) Proper transportation (conducting) system is necessary in higher plants.
Answer:
The green leaves of plants obtain CO₂ from the atmosphere and synthesize carbohydrates. The plants, through their roots, absorb water and other raw mineral elements essential for the constitution of the body, from the soil.

In higher plants the distance between the roots and the leaves being more, the water, mineral elements and the products of photosynthesis cannot be sent to all the different parts of the plant body, merely by diffusion from cell to cell. Therefore, in order to distribute all these substances rapidly and timely, a proper transportation (conducting) system is necessary in higher plants.

(2) In very tall plants, the suction force created due to transpiration is the main conducting force for water and mineral ions through the xylem.
Answer:
The xylem tissue in all the organs of a plant remains connected to each other and forms a continuous path for the flow of water, etc. Thus, a continuous water column is formed therein.

Mere root pressure, created in small herbs, is not sufficient to push water and minerals to the great height of very tall plants. The plants adopt another way to reach the target of fulfilling the water requirement. Evaporation of water molecules in the form of vapour occurs through stomata.

Due to that a suction force arises in the cells of leaves. This suction force comes into being from the cells of the leaves and is gradually experienced in the xylem of roots. As a result, the water column in the xylem rises up. Hence in very tall plants, the suction force created due to transpiration is the main conducting force for water and mineral ions through the xylem.

(3) Translocation in the phloem takes place in both upward and downward directions.
Answer:
The phloem transports amino acids, various plant hormones and other organic substances in addition to the products of photosynthesis.

Carbohydrates are synthesized in the leaves due to photosynthesis. These carbohydrates are transported to the roots and stem through phloem. The plant hormones synthesized in shoot apex flow downwards through the phloem and the plant hormones synthesized in the root apex and the food reserve stored in roots are transported upwards through the phloem. Thus, the translocation in the phloem takes place in both, upward and downward directions.

(4) The right side chambers of the heart have deoxygenated blood and left side chambers have oxygenated blood in them.
Answer:
The four-chambered heart, in man, is formed of two atria and two ventricles. All the four chambers of the heart are separated from each other by septa.

Deoxygenated blood from different organs of the body (except lungs) is brought through s superior and inferior vena cava and poured in the right atrium and then into the right ventricle, Similarly oxygenated blood from the two lungs is brought through pulmonary veins and poured in the left atrium and then into the left ventricle.

The four-chambered heart prevents the mixing of oxygenated blood with deoxygenated blood. Hence, the right side chambers of the heart have deoxygenated blood and the left side chambers have oxygenated blood in them.

(5) Lymph separates from the blood and remixes with the blood.
Answer:
The lymph oozes out through the pores in thin walls of the blood capillaries, as a fluid from the blood flowing through the capillaries. It flows very slowly in the intercellular spaces between the tissue cells. The intercellular spaces have no walls of their own and are called lymph capillaries.

These lymph capillaries meet and join with each other to form larger ones which finally open in a vein to pour its contents. Thus, lymph, as a colourless watery fluid collects in a large lymph vessels that finally open in particular veins S in the body to pour its contents back in blood.

(6) The wall of the artery is thick and elastic while that of vein is relatively thin.
Answer:
The arteries carry blood from the heart towards different organs. When the ventricles contract, the blood is pushed in the arteries under high pressure. In order to withstand this pressure, the arteries must have thick and elastic walls. The veins receive blood from different organs and carry it to the heart. The blood in the veins flows at relatively low pressure. Hence, the wall of the veins is relatively thin and less elastic.

(7) The organisms possessing chlorophyll are autotrophs.
Answer:
The organisms, possessing chlorophyll, can trap and utilize the solar energy to synthesize their own food using CO₂ and water. This process of trapping the solar energy for synthesis of ones own food is called photosynthesis and the mode of nutrition of such organisms is called autotrophic. In photosynthesis, the food synthesized is the simplest hexose sugar-glucose, which is utilized for obtaining energy. The surplus glucose is stored as reserve food in the form of starch. Hence, the organisms possessing chlorophyll are autotrophs.

(8) The stomata in leaves keep on opening and closing.
Answer:
On one or both the surfaces of the leaves of flowering plants, there are numerous stomata as minute pores. Each of these pores is guarded by a pair of guard cells. The opening and closing of the stomata is controlled by these guard cells, which contain chloroplasts.

When water enters the guard cells, the latter swell and cause the opening of stomata and when the guard cells lose water, the guard cells contract and cause the closing of stomata. Thus, the stomata in leaves keep on opening and closing due to entry and exit of water in the guard cells.

(9) The parasitic mode of nutrition is harmful for the host organism.
Answer:
In parasitic nutrition, one organism depends fully for obtaining its nutritional needs, directly on other living organism. The latter is called a host from whom the parasite directly obtains food. The parasitic organism keeps close contact with the host and sucks or absorbs nutrients from its body. The host goes on becoming weaker physically and physiological. The health of host thus is affected. Thus, the parasitic mode of nutrition is harmful for the host organism.

(10) HCl (Hydrochloric acid) is an important constituent of gastric juice.
Answer:
For the chemical digestion of food in stomach, the stomach secretes gastric juice from its gastric glands. HCl is one of the constituents of gastric juice.

HCl destroys the bacteria and other micro¬organisms entering along with the ingested food and thereby prevent the decay of food in stomach. HCl provides acidic medium for the action of gastric enzyme. HCl converts inactive enzyme pepsinogen into an active enzyme pepsin. Pepsin acts in acidic medium on proteins and starts their digestion and convert them into preoteoses and peptones. Thus, HCl is an important constituent of gastric juice.

(11) The length of small intestine of herbivorous is relatively much longer than that of carnivorous.
Answer:
The length of small intestine is different in different animals and that depends upon the nature of food taken by the animal. The carnivorous eat flesh. The digestion of flesh as food is quite easy and rapid and there is very small amount of roughage.

Hence these animals have short small intestine. The herbivorous eat grass and other vegetation. The cellulose of the plant cells is a complex substance and its complete digestion needs more space and time and hence longer small intestine and longer large intestine. Hence, the length of small intestine of herbivorous is relatively much longer than that of carnivorous.

(12) Bile is an important digestive juice though it does not contain any digestive enzymes.
Answer:
Bile is a greenish yellow alkaline digestive juice secreted from the liver cells. Bile contains bile salts, certain bile pigments but does not contain any digestive enzymes.

The bile salts turn the acidic food from stomach, alkaline and thereby provide alkaline medium for further reactions in intestine. Pancreatic enzymes and intestinal enzymes need alkaline medium. Bile salts emulsify the large fat globules into a very large number of very minute fat droplets and thereby greatly increase the exposed surface area of fat for the rapid action of lipases. Hence bile is an important digestive juice though it does not contain any digestive enzymes.

(13) Respiration is important to keep the organism in living state.
Answer:
The living cells of the body need energy for performing various vital functions. The energy is obtained by the biological oxidation of organic nutrients in the cell.

The process of breakdown of food sources for cellular needs either using oxygen or without oxygen is called respiration. The energy, so released is for continuation of various functions and thereby ‘ maintaining the living state of the organism. Thus, respiration is essential for life.

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

Questions :

1. Identify label x and state any two process that occur through it.
2. Identify label y and state the name of process and equation that occurs in it.
3. Identify label z and which situation you think for the given diagram?
4. What you think about transportation during day from label x in given diagram?

Answer:

1. x – stomatal pore, exchange of gases and transpiration occur through it.
2. y – chloroplast, photosynthesis process occurs in it.
   
3. z – guard cells, the guard cells swell when water flows into them, causing the stomatal pore to open.
4. During the day when stomata are open, the transpiration pull becomes the major driving force in the movement of water in xylem.

Questions :

1. Identify x and state the name of enzyme secreted in it and a medium required for its action.
2. Which juice is secreted from y? Where does it show its action and state the name of process which occurs by it?
3. State the name of specific finger-likc projections located in z and its functions.
4. Which other finger-like projection do you know and where can you observe it?

Answer:

1. x – stomach, name of enzyme is pepsin and acidic medium is required for its action.
   Bile juice secreted from y. It shows its action in small intestine and emulsification (breakdown of fat globules) process occurred by it.
2. Specific finger-like projections in the walls of z are villi. It increases surface area for absorption of food.
3. We know other finger-like projections are pseudopodia. We can observe it on the cell surface of amoeba.

Questions :

1. Which type of blood circulation you see in this diagram, what it means?
2. Which type of blood flows through x? Which one is exceptional for this?
3. Which type of blood flows through y? Why?
4. How our body gets highly efficient supply of oxygen?

Answer:

1. Double circulation. It means blood goes through the heart twice during each cycle.
2. Oxygenated blood flows through x. Pulmonary artery is exceptional because it transports deoxygenated blood.
3. Oxygenated blood flows through y. Because it carries blood from lungs to heart and in lungs blood becomes oxygenated.
4. The separation of the right side and the left side of the heart is useful to keep oxygenated and deoxygenated blood from mixing.

Questions:

1. Which structure is shown in the above diagram? Which nitrogenous wastes is removed from blood by this structure?
2. Identify x and mention its shape and function.
3. Identify y and z.
4. Compare the blood flowing through u and v blood vessles.

Answer:

1. Structure of nephron shown in the diagram. Nitrogenous wastes such as urea and uric acid are removed from blood by it.
2. x – Bowman’s capsule. It is cup-shaped and collects the filtrate.
3. y – collecting duct, z – glomerulus.
4. u – It transports oxygenated blood containing more nitrogenous wastes, v – It transports deoxygenated blood with lesser nitrogenous wastes due to filtration.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) Which inorganic substances are used as raw materials by autotrophic organisms?
Answer:
Water and CO₂ are the inorganic substances used as raw material for the synthesis s of organic food by autotrophic organisms.

(2) What is the mode of nutrition in fungi?
Answer:
The fungi show heterotrophic nutrition, in which breakdown the food material outside the body and then absorb it.

(3) Name one organism each, having saprophytic, parasitic and holozoic modes of nutrition.
Answer:

Mode of nutrition	Name of organism
Saprophytic	Most of fungi
Parasitic	Tapeworm, Ascaris
Holozoic	Amoeba, Human

(4 ) In addition to carbon dioxide and water, state two other conditions necessary for the process of photosynthesis.
Answer:
Presence of chlorophyll in the cells and <: the presence of sunlight are also necessary, in addition to CO₂ and water, for photosynthesis.

(5) Why there is a controversy about whether viruses are truly alive or not?
Answer:
There is a controversy about whether viruses are truly alive or not, because viruses do not show any molecular movement in them unless they infect specific host cell.

(6) Why are molecular movements needed for life?
Answer:
Molecular movements are needed for life because all the structures of living cells are made up of molecules and they must move molecules around all the time.

(7) On what the survival of heterotrophs depend? Give example of heterotrophic organisms?
Answer:
The survival of heterotrophs depends directly or indirectly on autotrophs.
Example of heterotrophic organisms : Animals and fungi.

(8) Which form of carbohydrate is stored in green plants and in human beings?
Answer:
Carbohydrates in the form of starch is stored in green plants while in human beings it is stored as glycogen.

(9) How desert plants perform process of photosynthesis?
Answer:
Desert plants take up carbon dioxide at night and prepare an intermediate which is acted upon by the energy absorbed by the chlorophyll during the day.

(10) Where does the exchange of gases occur in plants other than stomatal pores?
Answer:
The exchange of gases occurs across the surface of stems, roots and leaves other than in stomatal pores of plants.

(11) What is the function of guard cells?
Answer:
The opening and closing of the stomatal pore is a function of the guard cells.

(12) What is the essentiality of nitrogen element?
Answer:
Nitrogen is an essential element in the synthesis of protein and other compounds.

(13) Give the name of organisms that use parasitic nutritive strategy.
Answer:
The name of organisms that use parasitic nutritive strategy are cuscuta, ticks, lice, leech, tapeworm, etc.

(14) How a food vacuole is formed in amoeba?
Answer:
Amoeba takes in food using temporary finger-like extensions called pseudopodia of the cell surface which fuse over the food particle forming a food vacuole.

(15) What is peristaltic movement?
Answer:
The rhythmic movement shown by the contractions of the muscles of lining of alimentary canal which push the food only in one direction is called peristaltic movement.

(16) What causes acidity in adults?
Answer:
Acidity is caused due to excess secretion of hydrochloric acid in stomach.

(17) State the name of enzyme involved in digestion of protein and its location.
Answer:

Name of enzyme involved in digestion of protein	Location
(1) Pepsin	Gastric juice
(2) Trypsin	Pancreatic juice

(18) In which type of respiration more energy is released?
Answer:
In aerobic respiration more energy is released.

(19) Which part of root is involved in the exchange of respiratory gases?
Answer:
The root hairs, formed from the epidermal cells of the root, are involved in the exchange of respiratory gases.

(20) Name the respiratory organ of fish.
Answer:
The fish possesses pharyngeal gills as respiratory organs.

(21) In which pathway of breakdown of glucose CO₂ is not produced?
Answer:
Anaerobic respiration in our muscle cells

(22) Why the air passage does not collapse in our body?
Answer:
Rings of cartilage are present in the trachea so that the air passage does not collapse in our body.

(23) What are the characteristics of respiratory surface?
Answer:
Respiratory surface is very fine, delicate, moist and contains an extensive network of blood vessels and it remains in contact with atmosphere.

(24) Why the lungs always contain a residual volume of air?
Answer:
Thle lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for carbon dioxide to be released.

(25) Give the name, location and function of respiratory pigment in human beings.
Answer:
In human beings, respiratory pigment

(26) What is called single circulation?
Answer:
Blood goes only once through the heart in the fish body during the once circulation.

(27) What is the function of blood capillaries?
Answer:
Exchange of material between blood and the surrounding cells takes place across the thin wall of blood capillaries.

(28) What is the significance of lymph?
Answer:
Lymph carries digested and absorbed fat from small intestine and drains excess fluid from intercellular space back into the blood.

(29) Who forms conducting tubes in higher plants? What are transported through it?
Answer:
Xylem and phloem forms conducting tubes in higher plants. Xylem transports water and minerals, phloem transports products of photosynthesis.

(30) How sunction is created in xylem? What is it significance?
Answer:
Evaporation of water molecules from the cells of a leaf creates a sunction in xylem. It pulls water from the xylem cells.

(31) State any two points of importance of transpiration in plants?
Answer:
Importance of transpiration in plants:

• It helps in the absorption and upward movement of water and minerals dissolved in it from roots to the leaves.
• It helps in temperature regulation.

(32) State the name of forces important for the movement of water in the xylem during day and at night respectively?
Answer:
Transpiration pull during day and root pressure at night are important forces for the movement of water in the xylem.

(33) Which substances are transported through phloem?
Answer:
Sucrose, amino acids and other substances are transported through phloem.

(34) Which component of phloem shows translocation of food? In which direction does s it take place?
Answer:
The translocation of food takes place in the sieve tubes with the help of adjacent companion cells of phloem and in both upward and downward directions.

(35) Explain translocation of sugar in the spring season in plants.
Answer:
In the spring season, sugar stored in root or stem tissue is translocated to the buds which need energy to grow.

(36) Which substances are selectively reabsorbed from initial filtrate in the tubular part of nephron?
Answer:
Glucose, amino acids, salts and a major amount of water are selectively reabsorbed from initial filtrate in the tubular part of nephron.

(37) Until when urine is stored in the urinary bladder?
Answer:
Urine is stored in the urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra.

(38) Why we can usually control the urge to urinate?
Answer:
We can usually control the urge to urinate because the bladder can store urine and it is under voluntary nervous control.

(39) State name and location of any three structures which are richly supplied with blood vessels.
Answer:

Structure	Location
(1) Villi	Wall of intestine
(2) Alveoli	Terminale of bronchioles in lungs
(3) Nephron	In the kidneys

Question 2.
Define : OR Explain the terms :
(1) Nutrition
Answer:
A process of transfer of a source of energy from outside the body of the organism to the inside is called nutrition.

(2) Photosynthesis
Answer:
A process of synthesis of simple form of carbohydrate, i.e. glucose with the use of solar energy, water and carbon dioxide in presence of chlorophyll is called photosynthesis.

(3) Autotrophs
Answer:
Those organisms which utilise simple inorganic sources in the form of carbon dioxide and water and synthesise complex food are called autotrophs.

(4) Heterotrophs
Answer:
Those organism which utilise complex food material prepared by other organisms are called heterotrophs.

(5) Digestion
Answer:
A process by which complex food components are transformed into simple, soluble and absorbable form with the help of enzymes is called digestion.

(6) Respiration
Answer:
A process of breakdown of food source such as glucose, in presence or in absence of oxygen inside the living cell to provide energy for cellular need is called respiration.

(7) Breathing
Answer:
A process of inhalation and exhalation is called breathing.

(8) Transpiration
Answer:
A process of loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

(9) Excretion
Answer:
The biological process involved in removal of nitrogenous metabolic wastes from the body is called excretion.

Question 3.
Fill in the blanks :

1. The substance, used in cellular respiration, is ………………..
2. The effect of ……………….. in transport of water is more important at night.
3. Carbohydrate is synthesized by the reduction of COa in the process of ………………..
4. The secretions from the liver and pancreas are poured in the ………………..
5. Human beings show ……………….. mode of nutrition.
6. The enzyme ……………….. digests starch and converts it into maltose.
7. Pepsin is an enzyme that can act only in ……………….. medium.
8. The digestion of food particle in Amoeba occurs in ………………..
9. The ……………….. in the wall of small intestine greatly increase the surface area for absorption.
10. The conversion of glucose into ……………….. during the first phase of the respiratory process occurs in the cytoplasm.
11. The conduction of the photosynthetic products is called ………………..
12. The ……………….. increases when the sucrose is transported through the phloem tissue.
13. The suction force created due to ……………….. is the main force for the conduction of water in the xylem.
14. The artery emerging from the left ventricle is called ………………..
15. There is always ……………….. blood in the right auricle.
16. As compared to blood the lymph contains ……………….. proteins.
17. The exchange of materials between the blood and tissue cells of the body takes place through ………………..
18. The terminal end of the excretory unit opens in the ………………..
19. The wall of the arteries are thick and ………………..
20. ……………….. During photosynthesis, ……………….. is evolved as by-product.

Answer:

1. glucose
2. root pressure
3. photosynthesis
4. small intestine
5. heterotrophic
6. amylase
7. acidic
8. food vacuole
9. vIlli
10. pyruvate
11. translocauon
12. osmotic pressure
13. transpiraüon
14. aorta
15. deoxygenated
16. less
17. capillaries
18. collectIng duct
19. elastic
20. oxygen

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

1. Euglena is an autotrophic animal.
2. In human body the carbohydrates are stored in the form of glycogen.
3. In photosynthesis the carbon dioxide is oxidized to form carbohydrates.
4. The control and regulation of the opening and closing of stomata is done by the chloroplasts.
5. Liver and pancreas produce digestive juices which help in digestion in small intestine.
6. The liver secretes acidic bile.
7. The cellulose in the cells of grass, can be digested by herbivores animals.
8. The inner wall of the stomach possesses tubular glands which secrete gastric juice.
9. Cuscuta is a plant, harmful for the host plant.
10. The rate of breathing in terrestrial animals is much faster than that seen in aquatic animals.
11. Cilia help in ingestion of food in paramoecium.
12. The enzymes pepsin and trypsin digest carbohydrates and fats respectively.
13. An artificial kidney is a device to remove nitrogenous wastes from the blood through dialysis.
14. The bronchus ends in the alveolus.
15. The diaphragm bends (moves) downwards at the time of expiration.
16. The blood flows from the heart to other organs under pressure.
17. In unicellular animals, the excretory substances are removed by diffusion in the surrounding water.
18. The blood vessels absorb fat through the villi of the ileum.
19. In certain plants, the useless waste substances are stored in cellular vacuoles.
20. The wall of the blood capillaries is bilayered and thick.
21. The body temperature is maintained by using energy in animals of classes Mammalia and Aves.
22. The pulmonary arteries cany oxygenated blood.
23. In plants, the conduction of water is in both upward and downward directions.
24. The phloem tissue transports carbohydrate, amino acids and plant hormones.
25. Blood is a red coloured, non-living liquid connective tissue.

Answer:

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

Question 5.
Match the following:
(1)

Column I	Column II
1. Algae	p. Saprophytic nutrition
2. Cuscuta	q. Holozoic nutrition
3. Fungi	r. Autotrophic nutrition
4. Amoeba	s. Parasitic nutrition

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

(2)

Column I	Column II
1. Salivary glands	p. Beginning of protein digestion
2. Liver	q. Enzyme trypsin
3. Pancreas	r. Alkaline bile
4. Stomach	s. Secretion of amylase

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

(3)

Column I	Column II
1. Amoeba	p. Omnivores
2. Paramoecium	q. Gills
3. Human being	r. Pseudopodia
4. Fish	s. Cilia

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

(4)

Column I	Column II
1. Villi	p. Exchange of gases
2. Cartilagenous ring	q. Absorption
3. Alveolus	r. Helps in breathing
4. Diaphragm	s. Trachea

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

(5)

Column I	Column II
1. Phloem	p. Urine formation
2. Excretory unit	q. Upward and downward conduction
3. Pulmonary vein	r. Deoxygenated blood
4. Renal vein	s. Oxygenated blood

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

(6)

Column I	Column II
1. Human heart	p. Cup-shaped
2. Human kidney	q. Four-chambered
3. Nephron	r. Bean-shaped
4. Bowman’s capsule	s. Long-coiled tubule

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

(7)

Column I	Column II
1. Villi	p. Right auricle
2. Bowman’s capsule	q. Less protein
3. Lymph	r. Glomerulus
4. Vena cava	s. Small intestine

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

(8)

Column I	Column II
1. Chloroplast	p. Stomatal pore
2. Mitochondria	q. Digestion
3. Guard cells	r. Reduction of CO<sub>2</sub>
4. Food vacuole	s. Breakdown of pyruvate using O<sub>2</sub>

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

(9)

Column I	Column II
1. Photosynthesis	p. Temperature regulation
2. Respiration	q. Energy stored
3. Transpiration	r. Sucrose
4. Translocation	s. Energy released

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

Question 6.
Chart – diagram based questions:

Identify x in diagram and state which system of plant does it indicate.
Answer:
x-phloem, xylem vascular bundle, it indicates transportation system of plant.

2.

Why starch test is negative in any leaf of plant? What you conclude?
Answer:
Starch test is negative in any leaf of plant s because CO₂ is not available as the plant is in bell-jar in which KOH absorbed CO₂.
CO₂ is a raw material essential for s photosynthesis.

State the lable x and y in given diagram and also mention which life process do they indicate.
Answer:
x – pseudopodia, y – food vacuole
Life process : nutrition in amoeba

4. Fill the blanks in given table with reference to digestion process in human:

Answer:

1. Amylase
2. Protein
3. Trypsin
4. Fatty acid and glycerol

What change occurs in the solution in test tube? What is responsible for such change?
Answer:
The lime water in test tube turns milky.
We breath out CO₂ which is responsible for lime water to turn milky.

6. Fill the blanks in given chart:

Answer:

1. Xylem
2. Transpiration pull
3. Root pressure

7.

Observe the diagram and indicate which parts shows function of storage of urine and filtration of blood?
Answer:
y – kidney – filtration of blood
z – urinary bladder – storage of urine

Question 7.
Select the correct alternative from those given below each question:
1. Which of the following organism breaks down food material outside of body?
A. Mushroom
B. Cuscuta
C. Leech
D. Lice
Answer:
A. Mushroom

2. Where does the digestion of protein start in human being?
A. Mouth
B. Stomach
C. Small intestine
D. Colon
Answer:
B. Stomach

3. Which of the following type has longest small intestine?
A. Tiger
B. Human
C. Cow
D. Rat
Answer:
C. Cow

4. Which one of the following organisms can live without oxygen or air?
A. Amoeba
B. Leech
C. Green plant
D. Yeast
Answer:
D. Yeast

5. Which is the exact site for gaseous exchange s in human beings?
A. Brochus
B. Alveoli
C. Villi
D. Skin
Answer:
B. Alveoli

6. What is the product from glucose in the first phase of respiration?
A. Ethanol
B. Lactic acid
C. Pyruvic acid
D. CO₂
Answer:
C. Pyruvic acid

7. What is the ultimate purpose of digestion?
A. Transportation
B. Absorption
C. Respiration
D. Assimilation
Answer:
B. Absorption

8. Which cells surround the pore of the stomata?
A. Epidermal cells
B. Companion cells
C. Sieve cells
D. Guard cells
Answer:
D. Guard cells

9. Which organ stores bile?
A. Liver
B. Pancreas
C. Small intestine
D. Gall bladder
Answer:
D. Gall bladder

10. The enzyme acting in acidic medium is ………………..
A. amylase
B. pepsin
C. trypsin
D. both B and C
Answer:
B. pepsin

11. In plants, photosynthetic products are transported through ………………..
A. vessel and sieve tube
B. tracheid and vessel
C. sieve tube and companion cell
D. sieve tube and tracheid
Answer:
C. sieve tube and companion cell

12. In which part of the body blood is oxygenated?
A. Heart
B. Liver
C. Kidney
D. Lungs
Answer:
D. Lungs

13. At which part the actual process of filtration of blood takes place in kidney?
A. Bowman’s capsule
B. Collecting duct
C. Tubular part of nephron
D. Capillaries of nephron
Answer:
A. Bowman’s capsule

14. From which part of the human heart only oxygenated blood always flows?
A. Both atria
B. Both ventricles
C. Left atrium and left ventricle
D. Right atrium and right ventricle
Answer:
C. Left atrium and left ventricle

15. As compared to blood, the lymph contains ………………..
A. fewer RBCs
B. less amount of water
C. less amount of metabolic waste
D. less amount of proteins
Answer:
D. less amount of proteins

16. Which of the following brings oxygenated blood into left atrium in heart?
A. Pulmonary vein
B. Pulmonary artery
C. Vena cava
D. Aorta
Answer:
A. Pulmonary vein

17. What happens in the process of photo-synthesis?
A. Transformation of solar energy into functional energy
B. Transformation of solar energy into chemical energy
C. Transformation of chemical energy into functional energy
D. Transformation of functional energy into chemical energy
Answer:
B. Transformation of solar energy into chemical energy

18. Which is the main force for the conduction of water through the xylem in plants?
A. Absorption of water through roots
B. Absorption of ions through roots
C. Sufficient availability of water from the soil
D. Suction due to transpiration
Answer:
D. Suction due to transpiration

19. Which of the following alternative shows the correct path of oxygenated blood flow in human beings?
A. Lungs → Pulmonary veins → Left auricle → Left ventricle → Various organs of the body
B. Lungs Pulmonary artery → Left auricle → Left ventricle → Various organs of the body
C. Lungs → Pulmonary artery → Right auricle → Right ventricle → Various organs of the body
D. Various organs of the body Right auricle → Right ventricle → Pulmonary artery → Lungs
Answer:
A. Lungs → Pulmonary veins → Left auricle → Left ventricle → Various organs of the body

20. Select the correct pair :
A. Stomata – transpiration
B. Translocation – glucose
C. Villi – egestion of feaces
D. Trachea – cartilaginous rings
Answer:
A. Stomata – transpiration
D. Trachea – cartilaginous rings

21. Which process is important for osmoregulation?
A. Nutrition
B. Circulation
C. Breathing
D. Excretion
Answer:
D. Excretion

22. Which one is the false statement for arteries?
A. Arteries carry blood from the heart towards other organs.
B. In arteries the blood flows under high pressure.
C. All arteries carry oxygenated blood.
D. The arterial walls are thick and elastic.
Answer:
C. All arteries carry oxygenated blood.

23. Resin and gum are the substances of plants.
A. excretory
B. nutritive
C. constitutional
D. reserve
Answer:
A. excretory

24. Which is the circulatory pathway of blood in the human heart?
A. Right auricle → Left auricle → Lungs → Right ventricle → Left ventricle → Different organs
B. Right auricle → Right ventricle → Lungs → Left auricle → Left ventricle → Different organs
C. Right auricle → Right ventricle Different organs → Left auricles Left ventricle → Lungs
D. Right auricle → Left auricle → Different organs → Right ventricle → Left ventricle → Lungs
Answer:
B. Right auricle → Right ventricle → Lungs → Left auricle → Left ventricle → Different organs

25. Which substance does not get reabsorbed in nephron during urine formation?
A. Glucose
B. Urea
C. Amino acid
D. Uric acid
Answer:
B. Urea, Uric acid

26. In which of following process. ATP is used?
A. Translocation of food
B. To maintain body temperature in human
C. Respiration
D. Simple diffusion in human
Answer:
Translocation of food, To maintain body temperature in human

27. Statement A: The rate of breathing is much faster in fishes.
Reason R : The blood goes only once through the heart in the fish during one cycle of passage through the body.
Which is the correct option for Statement A and Reason R?
A. Both A and R correct and R is explanation of A.
B. Both A and R correct but R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
E. Both A and R incorrect.
Answer:
A. Both A and R correct and R is explanation of A.

28. Statement A : ATP is the energy currency for cellular processes.
Reason R: Complete oxidation of pyruvate in presence of oxygen occurs in mitochondria.
Which is the correct option for Statement A and Reason R?
A. Both A and R correct and R is explanation of A.
B. Both A and R correct but R is not explanation of A.
C. A is correct, R is incorrect.
D. A is incorrect, R is correct.
E. Both A and R incorrect.
Answer:
B. Both A and R correct but R is not explanation of A.

29. Which animals tolerate some mixing of oxygenated and deoxygenated blood streams?
A. Fishes
B. Birds
C. Mammals
D. Amphibians
Answer:
D. Amphibians

30. The diagram is labelled, which part secretes digestive juice?

A. 1, 3, 5, 7
B. 2, 4. 6, 8
C. 2, 3, 5, 8
D. 1, 4, 6, 8
Answer:
A. 1, 3, 5, 7

Question 8.
Answer as directed : (Miscellaneous)
(1) Which pigment has a very high affinity for oxygen? Where is it present in human body?
Answer:
Haemoglobin in red blood cells

(2) State the equation of photosynthesis.
Answer:

(3) Name of instrument used to measure blood pressure.
Answer:
Sphygmomanometer

(4) About which organism there is a controversy regarding whether it is a living being or non-living entity?
Answer:
Viruses

(5) Give the full form of ATP.
Answer:
Adenosine TriPhosphate

(6) Identify me : I am a cup-shaped structure with glomerulus and I carry out filtration.
Answer:
Bowman’s capsule

(7) Find mismatched pair :
1. One cell thick – blood capillary
2. Ring of cartilage – trachea
3. Phloem tissue – transport of sucrose
4. Platelet cells – transport of respiratory gases
Answer:
4. Platelet cells – transport of respiratory gases

(8) State the normal blood pressure of human.
Answer:
Systolic pressure 120 mm Hg and diastolic pressure 80 mm Hg

(9) Identify me : A component of gastric juice protect the inner lining of the stomach from the action of hydrochloric acid under normal condition.
Answer:
Mucus

(10) Sketch the respiratory pathway which does not produce CO₂.
Answer:

(11) Which of the following organisms show parasitic nutritive strategy?
Lion, lice, bread mould, cuscuta, leech, yeast, tick
Answer:
lice, cuscuta, leech, tick

(12) Find mismatched pair:
1. Paramoecium – Fermentation
2. Peristaltic movement – All along the gut
3. Emulsifying action – Bile
4. Trachea – Windpipe
Answer:
1. Paramoecium – Fermentation

(13) Which treatment do you suggest to a patient whose both kidneys have stopped functioning?
Answer:
Hemodialysis

(14) Transpiration helps to create osmotic pressure for translocation of sucrose. State whether s this sentence true or false.
Answer:
False

Value Based Questions With Answers

Question 1.
Your younger brother complains about pain in teeth. You often notice that he frequently eats chocolates and pastries. Even he likes to eat sweets.
Questions:

1. What do you think about the pain in teeth?
2. Which advise will you give to your brother?
3. What will happen if this problem is untreated?

Answer:

1. Sugar content is very high in chocolates, pastries and sweets. Bacteria act on sugars and produce acidic substances. Acids soften? the enamel, i.e.. cause dental caries or tooth decay. Masses of bacterial cells together with food particles stick to the teeth to form dental plaque. This is responsible for pain in teeth.
2. Brushing the teeth after eating, will remove plaque.
3. If this problem is untreated, microorganisms may invade the gums causing inflammation and infection

Question 2.
Your uncle often complains about indigestion of after having oily food. He consults a doctor and? is diagnosed with stone in gall bladder. Doctor advised him to remove gall bladder surgically.
Questions:

1. What is the function of gall bladder?
2. Which process initiates the digestion of oils?
3. After surgery, which type of food should be given to uncle?

Answer:

1. Gall bladder stores bile juice.
2. Emulsification (i.e.. bile salts breakdown large 5 oil globules into fine small droplets) process initiate the digestion of oils.
3. After surgery, food with low fat content, i.e., less oil, ghee, butter, etc. is advisable.

Question 3.
Your neighbour is a chain-smoker. He often suffers from cough and lung infection. His relatives often tell him to leave smoking.

Questions:

1. How inhaled air is filtered in the upper part of respiratory tract?
2. Which is the effect of smoking on the upper part of respiratory tract?
3. Why do you call smoking as injurious to health?

Answer:

1. The upper part of respiratory tract is provided with small hair-like structures called cilia, which help to remove germs, dust and other harmful particles from inhaled air.
2. Smoking destroys hair like cilia due to which germs, dust, smoke and other harmful chemicals enter lungs and causes harm.
3. Smoking reduces the breathing efficiency of lungs, may cause various infection and even lung cancer. So, called it is injurious to health. Cigarette contains nicotine which can cause cancer to the respiratory organs.

Question 4.
Your subject teacher arranges a visit to a hospital, where you can observe haemodialysis.

Questions :

1. Which conditions may lead to kidney failure?
2. What is the use of artificial kidney?
3. Explain how artificial kidneys work?

Answer:

1. Several factors such as infections, injury or restricted blood flow to kidney reduce the activity of kidneys. This leads to accumulation of toxic nitrogenous substances, gradually may lead to kidney failure.
2. An artificial kidney is a device to remove s nitrogenous waste products from the blood through dialysis.
3. Artificial kidneys contain a number of tubes with a semi-permeable lining, suspended in a tank filled with dialysing fluid. This fluid has same osmotic pressure as blood but nitrogenous wastes are absent.

As shown in diagram, patient’s blood is passed through tubes. During this, nitrogenous waste products from blood pass into dialysing fluid by diffusion. The purified blood is pumped back into the patient’s vein.

Practical Skill Based Questions With Answers

Question 1.
Observe experiment arranged in your school laboratory as shown in figure.

Questions :

1. State the colour of the stem and veins of the leaves.
2. Does the volume of solution in the beaker get reduced? Why?
3. In the T.S. of stem, which part is observed to be reddish in colour? Why?
4. State your inference.

Answer:

1. Reddish
2. Yes, because plant absorbed solution from beaker through its roots.
3. Xylem tissue becomes reddish in colour in T.S. of stem because water moves up through xylem.
4. Water and minerals absorbed by root move in upward direction through xylem vessels.

Question 2.
Take a potted plant.

• Insert one of its branches in a large, thin and transparent plastic/polythene bag and tie the bag at its open end with the branch.
• Add adequate amount of water to the clay in the pot, and keep the pot exposed to sunlight for a few hours.
• Observe the inner side of the bag after a few hours.
  

Questions :

1. Why are small droplets of water seen on the inner surface of the plastic/polythene bag?
2. State the role played by sunlight in the formation of water droplets.
3. State the passage of flow of water droplets through the potted plant.

Answer:

1. Loss of water vapour from leaves condense and show water droplets.
2. Sunlight is responsible for transpiration. Water is lost in the form of vapour, which creates suction and transpiration pull in upward direction.
3. Soil root xylem → stem-xylem → leaf xylem → stomata → water vapour

Question 3.
Place two fingers of your right hand on the left wrist and feel the pulse beats.

• Count the number of pulse beats felt by your right hand fingers in exactly one minute.
• Repeat the counting of pulse beats twice or thrice for accuracy. Find out the mean of all readings.
• Now run fast for a short distance for about one or two minutes, or climb the steps of a staircase quite rapidly twice or thrice.
• And thereafter, again measure your pulse s beats.

Questions:

1. State the normal pulse rate.
2. State the relation of pulse rate with the rate of heart beats.
3. State the number of pulse beats after a little running or climbing the staircase.
4. What is the change found in the pulse rate after running, as compared to the normal pulse rate? Why?

Answer:

1. 60- 100
2. The pulse rate is similar to heart beats.
3. 120 to 130 times in a minute.
4. Pulse rate increases after riming because body requires more oxygen and to fullfil it, heart beats increase.

Memory Map:

Jharkhand Board JAC Class 10 Science Solutions Chapter 5 Periodic Classification of Elements Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 5 Periodic Classification of Elements

Jharkhand Board Class 10 Science Periodic Classification of Elements Textbook Questions and Answers

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table :
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides become more acidic.
Answer:
The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCl₂, which is a solid with a high melting point. X would most likely be in the same group of the periodic table as…
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
Mg

Question 3.
Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second? shell as in its first shell?
Answer:
(a) Neon (2, 8)
(b) Magnesium (2, 8, 2)
(c) Silicon (2, 8, 4)
(d) Boron (2, 3)
(e) Carbon (2, 4)

Question 4.
(a) What property do all elements in the same column of the periodic table as boron have in common?
(b) What property do all elements in the same column of the periodic table as fluorine have in common?
Answer:
(a) In the modern periodic table, Boron is an element of group 13. Its valency is 3. Thus, all other elements of this group have valency 3.

(b) In the modern periodic table, fluorine is an element of group 17. All the elements of this group have 7 electrons in their valence shells. Therefore, the valency of all elements of this group is 1.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)
N (7) F (9) P (15) Ar (18)
Answer:
(a) The atomic number of this element is 17(2 + 8 + 7).
(b) F (9) [ ∵ Electronic configuration of F is 2, 7.]

Question 6.
The position of three elements A, B and C in the periodic table are shown below:

Group	Group 17
–	–
–	A
–	–
B	C

(a) State whether A is a metal or non-metal.
(b) State whether C is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
(a) Element A is an element of group 17. There are 7 electrons in their valence shell and thus by gaining one more electron it acquire a complete octet. Thus, an element A is a non-metal.

(b) On going down in a group, the atomic s size increases. Therefore, the force of attraction of the nucleus on the incoming electron decreases. As a result, reactivity decreases down the group. Since element C has larger atomic size than A, the element C is less reactive than the element A. In reference of forming positive ion, element C is more reactive than the element A.

(c) Elements B and C belong to the same period. On moving left to right in a period, atomic size (volume) decreases. Thus, the atomic size of C is smaller than B.

(d) Since element A has 7 electrons in the valence shell, it has a tendency to gain one electron to complete its octet. Thus, element A forms an anion.
A + \(\overline{\mathrm{e}}\) → A^–

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
The electronic configuration of nitrogen and phosphorus are as follows:

Element	Atomic number	Electronic configuration
		K	L	M
Nitrogen (N)	7	2	5	–
Phosphorus (P)	15	2	8	5

Nitrogen is more electronegative than phosphorus because electronegative character decreases on moving down a group.

Question 8.
How does the electronic configuration of an atom relate to its position in the modern periodic table?
Answer:
In the periodic table, position of an element depends on its electronic configuration. The position of an element can be determined by knowing the number of valence electron in its electronic configuration.
For example,

Possesses one electron in its valence shell. Hence, it belongs to group 1.
→ The number of shells in the electronic configuration of an element determines its position in a period.
For example,

has three shells (K, L and M). So, it belongs to 3rd period of the periodic table.

Question 9.
In the modern periodic table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?
Answer:

Element	Atomic number	Electronic configuration
		k	L	M	N	O
Calcium	20	2	8	8	2	–
Magnesium	12	2	8	2	–	–
Potassium	19	2	8	8	1	–
Scandium	21	2	8	8	3	–
Strontium	38	2	8	18	8	–

Elements with atomic number 12 and 38 have 2 electrons in their last shell like calcium. So, they will resemble Ca in their chemical properties.

Question 10.
Compare and contrast the arrangement of elements in Mendeleev’s periodic table and the modern periodic table.
Answer:

Mendeleev’s periodic table	Modern periodic table
1. Mendeleev’s periodic table consists of seven periods and eight groups.	1. Modern periodic table consists of seven periods and eighteen groups.
2. Transition elements are not separated in the Mendeleev’s periodic table.	2. Transition elements are placed in a separate groups in the modern periodic table.
3. In Mendeleev’s periodic table, elements are arranged in increasing order of their atomic masses.	3. In the modern periodic table, elements are arranged in increasing order of their atomic numbers.
4. Period number and group number of an element cannot be predicted.	4. Period number and group number of an element can be determined easily.
5. Mendeleev’s periodic table has descripancies and limitations.	5. Modern periodic table is almost errorless.
6. Periodicity in the properties of elements cannot be explained.	6. Periodicity in the properties of elements can be explained.

Jharkhand Board Class 10 Science Periodic Classification of Elements InText Questions and Answers

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ octaves? Compare and find out.
Answer:
Dobereiner’s triads also exist in the columns of Newlands’ octaves.

• Lithium, sodium and potassium forms a Dobereiner’s triad.
• Lithium, the first element of this triad is considered as the first element as Newlands’ octave, then the eighth element from it is sodium. These elements possesses similar properties according to the law of triads and the law of octaves.
• Similarly, if sodium is considered as the first element then the eighth element from it is potassium.  Moreover, sodium and potassium possess similar properties according to both the laws.
• Apart from these, some other elements beryllium (Be), magnesium (Mg) and calcium (Ca) obeys law of triads and the law of octaves.
• Thus, Dobereiner’s triads also exist in the columns of Newlands’ octaves.

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:
The limitations of Dobereiner’s classification are as follows:

• All the elements known at that time could not be arranged as Dobereiner’s triad. Hence this method of classification of elements into triads was not found to be successful.
• Three elements nitrogen (N), phosphorus (P) and arsenic (As) were then known elements, but these elements could not be classified as Dobereiner’s triad.

Question 3.
What were the limitations of Newlands’ law of octaves?
Answer:

• Newlands’ law of octaves was applicable only to lighter elements having atomic masses up to 40 u.
  The law of octaves was applicable only s upto calcium, because after calcium every eighth element did not possess properties similar to that of the first element.
• Newlands assumed that only 56 elements s existed in nature and no new elements would be discovered in the future. But, later on, several new elements were discovered whose properties did not fit into the law of octaves.
• In order to fit elements into his table, Newlands adjusted two elements in the same slot, but also put some unlike elements under the same column.

For example, cobalt (Co) and nickel S (Ni) are placed in the same slot and these are placed in the same column as fluorine, chlorine and bromine which have very different properties S than these elements. Iron, which resembles cobalt and nickel in properties, has been placed far away from these elements.

Question 4.
Use Mendeleev’s periodic table to predict the formulae for the oxides of the following elements:
K, C, Al, Si, Ba
Answer:

Element	Group number (Valency)	Molecular formula of oxide
K	1	K<sub>2</sub>O
C	4	CO2
Al	3	Al2O3
Si	4	SiO2
Ba	2	BaO

Question 5.
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his periodic table? (any two)
Answer:
Besides gallium, Mendeleev had left gaps for germanium and scandium in his periodic table.

Question 6.
What were the criteria used by Mendeleev in creating his periodic table?
Answer:
The following criteria was used by Mendeleev in creating his periodic table :

• The properties of elements are the periodic function of their atomic masses.
• Elements with similar properties are arranged in the same group.
• The formula of oxides and hydrides formed by an element.

Question 7.
Why do you think the noble gases are placed in a separate group?
Answer:
Noble gases like helium (He), neon (Ne) and argon (Ar) are chemically very inert and are present in extremely low concentrations in our atmosphere. Hence they are placed in a separate s group.

Question 8.
How could the modern periodic table remove various anomalies of Mendeleev’s periodic table?
Answer:
The modern periodic table removed three main anomalies of Mendeleev’s periodic table as discussed below:
(1) Position of isotopes : All the isotopes of an element have the same atomic number. Therefore, they are placed at one place in the same group of the periodic table.

(2) Anomalous position of some pairs of elements: In the Mendeleev’s periodic table, elements with similar properties are placed in the same group. For example, cobalt (atomic mass 58.9 u) placed first and nickel (atomic mass 58.7 u) placed later while in the modern periodic table elements are arranged in increasing order of their atomic numbers, therefore, cobalt with atomic number 27 placed first and nickel with atomic number 28 placed later.

(3) Uncertainty in discovery of new elements: Since atomic masses do not increase in a regular manner in going from one element to the next, therefore, in Mendeleev’s periodic table, it was not possible to predict as to how many new elements could be discovered between two known elements.

Since, modern periodic table is formed on the basis of atomic numbers of elements, discovery of new element become easy.

Question 9.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?
Answer:
In the modern periodic table, elements having same number of electrons in the valence shell show similar chemical properties.
Magnesium has two electrons in the valence shell, hence all the elements such as beryllium (Be), calcium (Ca) and strontium (Sr) having two electrons in the valence shell show similar chemical properties.

Elements of group 2
Element	Atomic number	Electronic configuration
		K	L	M	N	O
Beryllium (Be)	4	2	2
Magnesium (Mg)	12	2	8	2
Calcium (Ca)	20	2	8	8	2
Strontium (Sr)	38	2	8	18	8	2

Question 10.
Name :
(a) three elements that have a single electron in their outermost shells.
(b) two elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Lithium (Li), sodium (Na) and potassium (K)
(b) Magnesium (Mg) and calcium (Ca)
(c) Neon (Ne), argon (Ar) and krypton (Kr)

Question 11.
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the? atoms of these elements?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?
Answer:
(a) Lithium, sodium and potassium are alkali metals which react with water to form metal hydroxides with the release of hydrogen gas.
2M + 2H₂O → 2MOH + H₂
Where, M = Li, Na and K
All these metals have one electron in their respective valence shells.

(b) Helium and neon are noble gases. Both the elements have their outermost shells completely filled. Helium has only K shell which is complete with 2 electrons while neon has two shells, K and L. Both these shells are complete, i.e., K s shell has 2 electrons and L shell has 8 electrons.

Question 12.
In the modern periodic table, which are the metals among the first ten elements?
Answer:
The first ten elements of the modern periodic table are as follows :
₁H, ₂He, ₃Li, ₄Be, ₅B, ₆C, ₇N, ₈O, ₉F and ₁₀Ne.
Among these elements Li and Be are metals.

Question 13.
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristics?
Ga Ge As Se Be
Answer:
Among the given elements, Be and Ga will show maximum metallic characteristics.
The arrangement of given elements in different groups and periods is as follows :

Activity 5.1 [T. B. Pg. 84]

Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a : correct position in Mendeleev’s periodic table.
To which group and period should hydrogen be assigned ?

Discussion:

• Hydrogen is an element having lowest atomic number (Z = 1) and lowest atomic mass (1.008 u).
• The electronic configuration of hydrogen resembles with alkali metals.
• Like alkali metals, hydrogen combines with halogens, oxygen and sulphur to form compounds having similar molecular formulae. Hence, hydrogen can be placed along with alkali metals of group (IA).
• Hydrogen exists as a diatomic molecule like halogens and it combines with alkali metals to form ionic compounds and with non-metals to form covalent compounds.
• Thus, hydrogen should be placed along with halogens in group (VII).

Conclusion :
The position of hydrogen in the periodic table is controversial, however, it would be more appropriate to place it in group I and period I.

Activity 5.2 [T. B. Pg. 85]

• Consider the isotopes of chlorine, Cl-35 and Cl-37.
• There are two known isotopes of chlorine : Cl-35 and Cl-37.
• The atomic number of these two isotopes is 17, hence they have similar electronic configuration and chemical properties, but their atomic masses are different.

Questions :

Question 1.
Would you place them in different slots because their atomic masses are different?
Answer:
According to Mendeleev’s periodic table, “The properties of elements are the periodic function of their atomic masses.”

• According to Mendeleev, if these two isotopes are arranged in the order of increasing atomic masses, then their position should be before K(39.1u), but there is no vacant position available for Cl-37 in between Cl-35 and K(39.1u). Therefore they cannot be placed at the different position.

Question 2.
Would you place them in the same position because their chemical properties are the same?
Answer:
As they have similar chemical properties, they should be placed at the same position in group 17.

Activity 5.3 [T. B. Pg. 85]

Questions:

Question 1.
How were the positions of cobalt and nickel resolved in the modern periodic table?
Answer:

• In modern periodic table, elements are arranged in increasing order of their atomic numbers.
• The atomic number of cobalt and nickel are 27 and 28 respectively. Hence, on the basis of increasing order of their atomic numbers, cobalt is placed s in group 9 and nickel is placed in group 10.

Question 2.
How were the positions of isotopes of various elements decided in the modern periodic table?
Answer:
In the modern periodic table, positions of isotopes of different elements are not fixed separately. Since the various isotopes of an element have the same atomic number, they are assigned the same position in the modern periodic table.

Question 3.
Is it possible to have an element with atomic number 1.5 placed between hydrogen and helium?
Answer:

• The atomic number of an element is always definite and whole number.
• In the modern periodic table, elements are arranged in increasing order of their atomic numbers.
• The atomic number cannot be represented in fraction number. Thus, an element with atomic number 1.5 cannot be placed between hydrogen S and helium.

Question 4.
Where do you think should hydrogen be placed in the modern periodic table?
Answer:
Hydrogen should be placed in period and group 1 in the modern periodic table, $ since it has atomic number one.

Activity 5.4 [T. B. Pg. 87]

Questions:

Question 1.
Look at the group 1 of the modern periodic table and name the elements present in it.
Answer:
The names of the elements present in group 1 are: hydrogen (H), lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs) and francium (Fr).

Question 2.
Write down the electronic configuration of the first three elements of group 1.
Answer:
The electronic configuration of the first three elements are as follows :

Element	H	Li	Na
Shell	K	K, L	K, L, M
Electronic configuration	1	2, 1	2, 8, 1

Question 3.
What similarity do you find in their electronic configurations?
Answer:
These elements possesses same number of electrons in their valence shell.

Question 4.
How many valence electrons are present in these three elements?
Answer:
These three elements have one electron in their respective valence shell.

Activity 5.5 [T. B. Pg. 87]

Questions:

Question 1.
If you look at the modern periodic table, you will find that the elements Li, Be, B, C, N, O, F and Ne are present in the second period. Write down their electronic configurations.
Answer:

Element	3Li	4Be	5B	6C	7NN	8O	9F	10Ne
Shell	K, L	K, L	K, L	K, L	K, L	K, L	K, L	K, L
Electronic configuration	2, 1	2, 2	2, 3	2, 4	2, 5	2, 6	2, 7	2, 8

Question 2.
Do these elements also contain the same number of valence electrons?
Answer:
These elements do not contain the same number of valence electrons.

Question 3.
Do they contain the same number of shells?
Answer:
These elements contain the same number of shells (2, K and L).

Activity 5.6 [T. B. Pg. 88]

Questions :

Question 1.
How do you calculate the valency of an element from its electronic configuration?
Answer:
The valency of an element is determined by the number of valence electrons present in the outermost shell of the atom.
For elements of group 1, 2, 13 and 14, the valency is equal to the number of valence electrons and for elements of group 15, 16, 17 and 18, the valency is equal to 8 minus number of valence electrons.

Question 2.
What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16?
Answer:
Electronic configuration of Mg with atomic number 12 is K L M 2 8 2
∴ Valency of Mg = 2
Electronic configuration of S with atomic number 16 is K L M 2 8 6
∴ Valency of S = 8 – 6 = 2

Question 3.
Find the valency of the first twenty elements.
Answer:

Element	Atomic number	Group number	Electronic configuration				Number of valence electrons	Valency of the element
			K	L	M	N
H	1	1	1				1	1
He	2	18	2				2	2-2 = 0
Li	3	1	2	1			1	1
Be	4	2	2	2			2	2
B	5	13	2	3			3	3
C	6	14	2	4			4	4
N	7	15	2	5			5	8-5 = 3
O	8	16	2	6			6	8-6 = 2
F	9	17	2	7			7	8-7=1
Ne	10	18	2	8			8	8-8 = 0
Na	11	1	2	8	1		1	1
Mg	12	2	2	8	2		2	2
Al	13	13	2	8	3		3	3
Si	14	14	2	8	4		4	4
P	15	15	2	8	5		5	8-5 = 3
S	16	16	2	8	6		6	8-6 = 2
Cl	17	17	2	8	7		7	8-7=1
Ar	18	18	2	8	8		8	8-8 = 0
K	19	1	2	8	8	1	1	1
Ca	20	2	2	8	8	2	2	2

Question 4.
How does the valency vary in a period on going from left to right?
Answer:
In a period, the valency first increases and then decreases. In a period, the valency first increases from 1 to 4 and then decreases from 4 to 0.

Question 5.
How does the valency vary in going down a group?
Answer:
On moving down the group, the valency does not change. It remains constant.

Activity 5.7 [T. B. Pg. 88]

Questions:

Question 1.
Atomic radii of the elements of the second period are given below:

Period II elements:	B	Be	O	N	Li	C
Atomic radius (pm):	88	111	66	74	152	77

Arrange them in decreasing order of their atomic radii.
Answer:
Decreasing order of the atomic radii:
Period II elements : Li > Be > B > C > N > O
Atomic radius (pm) : 152  111 88 77 74 66

Question 2.
Are the elements now arranged in the pattern of a period in the periodic table?
Answer:
Now, the above elements are arranged in the pattern of a period in the periodic table.

Question 3.
Which elements have the largest and the smallest atoms?
Answer:
Lithium (Li) has the largest and oxygen (O) has the smallest atoms.

Question 4.
How does the atomic radius change as you go from left to right in a period?
Answer:
The atomic radii decreases as we move from left to right in a period.

Activity 5.8 [T. B. Pg. 89]

Questions:

Question 1.
Study the variation in the atomic radii of ? first group elements given below and arrange them in an increasing order.

Group I elements :	Na	Li	Rb	Cs	K
Atomic radius (pm):	186	152	244	262	231

Answer:
Increasing order of the atomic radii:
Group I elements : Li < Na < K < Rb < Cs
Atomic radius (pm): 152 186 231 244 262

Question 2.
Name the elements which have the smallest and the largest atoms.
Answer:
Lithium (Li) has the smallest atoms while Cesium (Cs) has the largest atoms.

Question 3.
How does the atomic size vary as you go down a group ?
Answer:
As we go down a group, the atomic size increases gradually.

Activity 5.9 [T. B. Pg. 89 ]

Questions:

Question 1.
Examine elements of the third period and classify them as metals and non-metals.
Answer:
The elements of the third period :

Question 2.
On which side of the periodic table do you find the metals?
Answer:
Metals are present on the left side of the periodic table.

Question 3.
On which side of the periodic table do you find the non-metals?
Answer:
Non-metals are present on the right side of the periodic table.

Activity 5.10 [T. B. Pg. 89]

Questions:

Question 1.
How do you think the tendency to lose electrons changes in a group?
Answer:
As we move from top to bottom in a group, the tendency to lose electrons increases.

Question 2.
How will this tendency change in a period?
Answer:
As we move from left to right in a period, the tendency to lose electrons decreases.

Activity 5.11 [T.B.Pg. 90]

Questions:

Question 1.
How would the tendency to gain electrons change as you go from left to right across a period?
Answer:
As we go from left to right across a period, the tendency to gain electrons increases.

Question 2.
How would the tendency to gain electrons change as you go down a group?
Answer:
As we go down a group, the tendency to gain electrons decreases.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 15 Probability Ex 15.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1.
Complete the following statements:
i) Probability of an event E+ Probability of the event ‘not E’ =
ii) The probability of an event that cannot happen is …………….. Such an event is called ……………
iii) The probability of an event that is certain to happen is …………. Such an event is called …………….
iv) The sum of the probabilities of all the elementary events of an experiment is ……………
v) The probability of an event is greater than or equal to ………………. and less than or equal to ………………
Solution:
i) 1,
ii) 0, impossible event,
iii) 1, sure or certain,
iv) 1,
v) 0, 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
iii) A trial made to answer a true-false question. The answer is right or wrong.
iv) A baby is born. It is a boy or a girl.
Solution:
(iii) and (iv) have equally likely outcomes. Only two possibilities are there in each of these cases.

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
When a coin is tossed, head or tail are equally likely possible events. So the result of an individual coin toss is unpredictable.

Question 4.
Which of the following cannot be the probability of an event:
A) \(\frac{2}{3}\)
B) -1.5
C) 15%
D) 0.7?
Solution:
(B) Because, probability of an event cannot be negative.

Question 5.
If P(E)= 0.05, what is the probability of ‘not E’?
Solution:
P(E) = 0.05.
[P(\(\overline{\mathrm{E}}\)) = Probability of not an event]
We know that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(E) = 1 – 0.05 = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Solution:
i) P(orange flavoured candy) = 0. Impossible event.
ii) P(Lemon flavoured candy) = 1. Sure event.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E = Event of 2 students not having the same birthday
∴ P(E) = 0.992
∴ P(E) + P(\(\overline{\mathrm{E}}\)) = 1
∴ 0.992 + P(\(\overline{\mathrm{E}}\)) = 1
⇒ P(\(\overline{\mathrm{E}}\)) = 1 – 0.992
= 0.008
So, the probability of two students having the same birthday is 0.008.

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?
Solution:
Total number of balls, n(S) = 3 + 5 = 8.
Let E = Event of drawing 1 red ball
∴ n(E) = 3
(i) Probability of drawing a red ball = \(\frac{n(E)}{n(S)}=\frac{3}{8}\)
(ii) Probability of not drawing a red ball = 1 – P(Drawing a red ball)
= \(1-\frac{3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total possible outcomes = 5 + 8 + 4 = 17.
P(R) = \(\frac{5}{17}\), P(W) = \(\frac{8}{17}\)
P(Not green) = P(R + W) = \(\frac{5}{17}+\frac{8}{17}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50 p coins, fifty Re. 1 coins, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a Rs. 5 coin?
Solution:
Total possible outcomes: 100 + 50 + 20 + 10 = 180.
P(50 paise coin) = \(\frac{100}{180}=\frac{5}{9}\)
P(Not Rs. 5 coin) = \(\frac{100}{180}+\frac{50}{180}+\frac{20}{180}\)
= \(\frac{170}{180}=\frac{17}{18}\).

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Total number of fish in an aquarium = 5 male fish + 8 female fish = 13 fish
∴ Probability of taking out a male fish = \(\frac{\text { Number of male fish }}{\text { Total number of fish }}=\frac{5}{13}\)

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at
i) 8?
ii) an odd number?
iii) a number greater than 2?
iv) a number less than 9?

Solution:
Total possible outcomes = 8.
i) P(8) = \(\frac{1}{8}\)
ii) P(odd number) = \(\frac{4}{8}=\frac{1}{2}\)
iii) P(no. > 2) = \(\frac{6}{8}=\frac{3}{4}\)
iv) P(no. < 9) = \(\frac{8}{8}\) = 1

Question 13.
A die is thrown once. Find the probability of getting
(i) a prime number, (ii) a number lying between 2 and 6, (iii) an odd number.
Solution:
Total possible outcomes 1, 2, 3, 4, 5, 6 = 6
P(Prime number) (2, 3, 5) = \(\frac{3}{6}=\frac{1}{2}\)
P(Number between 2 and 6) = \(\frac{3}{6}=\frac{1}{2}\)
P(Odd number) = \(\frac{3}{6}=\frac{1}{2}\)

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
i) a king of red colour
ii) a face card
iii) a red face card
iv) the jack of hearts
v) a spade
vi) the queen of diamonds.
Solution:
Total number of cards in one deck, n(S) = 52.
i) Let E₁ = Event of getting a king of red colour
∴ n(E₁) = 2
(∵ In a deck of cards, 26 cards are red and 26 cards are black. There are four kings in a deck in which two are red and two are black)
Probability of getting a king of red colour = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{52}=\frac{1}{26}\)

ii) Let E₂ = Event of getting a face card
∴ n(E₂) = 12
(∵ In a deck of cards, there are 12 face cards – 4 king, 4 jack, 4 queen)
Probability of getting a face card = \(\frac{n\left(E_2\right)}{n(S)}=\frac{12}{52}=\frac{3}{13}\)

iii) Let E₃ = Event of getting a red face card
∴ n(E₃) = 6
(∵ In a deck of cards, there are 12 face cards – 6 red, 6 black)
Probability of getting a red face card = \(\frac{n\left(E_3\right)}{n(S)}=\frac{6}{52}=\frac{3}{26}\)

iv) Let E₄ = Event of getting a jack of hearts
∴ n(E₄) = 1
(∵ There are four jacks in a deck- 1 heart, 1 club, 1 spade, 1 diamond)
Probability of getting a jack of hearts = \(\frac{n\left(E_4\right)}{n(S)}=\frac{1}{52}\)

v) Let E₅ = Event of getting a spade
∴ n(E₅) = 13
(∵ In a deck of cards, there are 13 spades, 13 clubs, 13 hearts, 13 diamonds)
Probability of getting a spade = \(\frac{n\left(E_5\right)}{n(S)}=\frac{13}{52}=\frac{1}{4}\)

vi) Let E₆ = Event of getting a queen of diamond
∴ n(E₆) = 1
(∵ In 13 diamond cards, there is only one queen)
Probability of getting a queen of diamond = \(\frac{n\left(E_6\right)}{n(S)}=\frac{1}{52}\).

Question 15.
Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their face downwards. One card is then picked up at random.
i) What is the probability that the card is the queen?
ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
i) Total possible outcomes = 5
P(Queen card) = \(\frac{1}{5}\)
ii) If the queen card is put aside, total possible outcomes = 4.
iii) P(ace) = \(\frac{1}{4}\)
iv) P(queen) = \(\frac{0}{2}\) = 0.

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Total possible outcomes = 132 + 12 = 144
No. of good pens = 132.
P(good pen) = \(\frac{132}{144}=\frac{11}{12}\)

Question 17.
i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
i) Total possible outcomes = 20
P(Defective bulbs) = \(\frac{4}{20}=\frac{1}{5}\)

ii) Total possible outcomes = 20 – 1 = 19
No. of defective bulbs = 4
No. of good bulbs = 15
P(Not defective bulb) = \(\frac{15}{19}\)

Question 18.
A box contains 90 dises which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.
Solution:
S = {1, 2, 3, 4, 5, … 90}
∴ Total possible outcomes n(S) = 90
i) Number of 2-digit numbers = 90 – 9 = 81
P(a 2-digit number) = \(\frac{81}{90}=\frac{9}{10}\)

ii) Event A = {A perfect square number}
A = (1, 4, 9, 16, 25, 36, 49, 64, 81) = n(A) = 9
Probability of the event P(A) = \(\frac{n(A)}{n(S)}=\frac{9}{90}\)

iii) A number divisible by 5, i.e., multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90 = 18.
P(a no. divisible by 5) = \(\frac{18}{90}=\frac{1}{5}\).

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Total possible outcomes = 6
No. of A’s = 2
No. of D’s = 1
P(A) = \(\frac{2}{6}=\frac{1}{3}\)
P(D) = \(\frac{1}{6}\)

Question 20.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
i) she will buy it?
ii) she will not buy it?
Solution:
Total number of possible outcomes = 144
No. of good pens = 144 – 20 = 124.
P(of buying) = \(\frac{124}{144}=\frac{31}{36}\)
P(of not buying) = \(\frac{20}{144}=\frac{5}{36}\)

Question 21.
(i) Two dice, one blue and one grey, are thrown at the same time. Write down all possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is (i) 8, (ii) 13, (iii) less than or equal to 12? Complete the following table:

Solution:


1) Sum of 2 dice = 2 (1 + 1)
P(Sum 2) = \(\frac{1}{36}\)

2) Sum of 2 dice = 3 (1 + 2) (2 + 1)
P(Sum 3) = \(\frac{2}{36}\)

3) Sum 4 (1, 3) (2, 2) (3, 1)
P(Sum 4) = \(\frac{3}{36}\)

4) Sum 5 (1, 4) (2, 3) (3, 2) (4, 1)
P(Sum 5) = \(\frac{4}{36}\)

5) Sum 6 (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)
P(Sum 6) = \(\frac{5}{36}\)

6) Sum 7 (1,6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)
P(Sum 7) = \(\frac{6}{36}\)

7) Sum 8 (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)
P(Sum 8) = \(\frac{5}{36}\)

8) Sum 9 (3, 6) (4, 5) (5, 4) (6, 3)
P(Sum 9) = \(\frac{4}{36}\)

9) Sum 10 (4, 6) (5, 5) (6, 4)
P(Sum 10) = \(\frac{3}{36}\)

10) Sum 11 (5, 6) (6,5)
P(Sum 11) = \(\frac{2}{36}\)

11) Sum 12 (6,6)
P(Sum 12) = \(\frac{1}{36}\)

ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.
Solution:
Total possible outcomes of throwing the two dice, S =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6,3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36

a) Let E₁ = Sum of two dice is 3 = {(1, 2), (2, 1)}
n(E₁) = 2
∴ P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}\)

b) Let E₂ = Sum of two dice is 4 = {(1, 3), (2, 2), (3, 1)}
n(E₂) = 3
∴ P(E₂) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{3}{36}\)

c) Let E₃ = Sum of two dice is 5 = {(1, 4), (2, 3), (3,2), (4, 1)}
n(E₃) = 4
∴ P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{4}{36}\)

d) Let E₄ = Sum of two dice is 6 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
n(E₄) = 5
∴ P(E₄) = \(\frac{5}{36}\)

e) Let E₅ = Sum of two dice is 7 = {(1, 6), (2, 5), (3, 4), (4,3), (5, 2), (6, 1)}
n(E₅) = 6
∴ P(E₅) = \(\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}\)

f) Let E₆ = Sum of two dice is 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6,2)}
n(E₆) = 5
∴ P(E₆) = \(\frac{n\left(E_6\right)}{n(S)}=\frac{5}{36}\)

g) Let E₇ = Sum of two dice is 9 = {(3, 6), (4, 5), (5, 4), (6,3)}
n(E₇) = 4
∴ P(E₇) = \(\frac{n\left(E_7\right)}{n(S)}=\frac{4}{36}\)

h) Let E₈ = Sum of two dice is 10 = {(4, 6), (5, 5), (6, 4)}
n(E₈) = 3
∴ P(E₈) = \(\frac{n\left(E_8\right)}{n(S)}=\frac{3}{36}\)

i) Let E₉ = Sum of two dice is 11 = {(6,5), (5, 6)}
n(E₉) = 2
∴ P(E₉) = \(\frac{n\left(E_9\right)}{n(S)}=\frac{2}{36}\)

j) Let E₁₀ = Sum of two dice is 12 = {(6, 6)}
n(E₁₀) = 1
∴ P(E₁₀) = \(\frac{n\left(E_{10}\right)}{n(S)}=\frac{1}{36}\)
No. The eleven events are not equally likely.

Question 22.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Total possible outcomes (H + T)³
= H³ + 3H²T + 3HT² + T³
HHH HHT HTH THH HTT THT TTH TTT = 8.
Possible losses HHT HTH THH HTT THT TTH = 6
P(of losses) = \(\frac{6}{8}=\frac{3}{4}\)

Question 23.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
i) Total number of cases, n(S) = 6² = 36
Let \(\overline{\mathrm{E}}\) = Event that 5 will come up either time
= {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
⇒ n(\(\overline{\mathrm{E}}\)) = 11
and E = Event that 5 will not come up either time
n(E) = 36 – 11 = 25
∴ Probability that 5 will not come up either time = \(1-\frac{11}{36}=\frac{36-11}{36}\)
= \(\frac{25}{36}\)
ii) Probability that 5 will come up at least once = 12 – 1 = \(\frac{1}{2}\).

Question 24.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\).
Solution:
i) Incorrect: We can classify the outcomes like this but they are not then ‘equally likely’. The reason is that ‘one of each’ can result in two ways – from head on first coin and tail on the second coin or from tail on the first coin and head on the second coin. This makes it twice as likely as 2 heads or 2 tails.

ii) Correct. The two outcomes considered in the question are equally likely. Both have the same probability. i.e., \(\frac{1}{2}\).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the media, mean and mode of the data and compare them.

Monthly consumption (in units)	No. of consumers
65 – 85	4
85 – 105	5
105 – 125	13
125 – 145	20
145 – 165	14
165 – 185	8
185 – 205	4

Solution:


Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Class interval	Frequency
0 – 10	5
10 – 20	x
20 – 30	20
30 – 40	15
40 – 50	y
50 – 60	5
Total	60

Solution:

Class interval	Frequency	Cumulative frequency
0 – 10	5	5
10 – 20	x	5 + x
20 – 30	20	25 + x
30 – 40	15	40 + x
40 – 50	y	40 + x + y
50 – 60	5	45 + x + y
	60

n = 60, 45 + x + y = 60
x + y = 60 – 45
x + y = 15
The median is 28.5. It lies in the class interval 20 – 30.
∴ l = 20, f = 20, cf = 5 + x, h = 10

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	No. of policyholders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution:

Class interval	No. of policyholders	c.f.
Below 20	2	2
20 – 25	4	6
25 – 30	18	24
30 – 35	21	45
35 – 40	33	78
40 – 45	11	89
45 – 50	3	92
50 – 55	6	98
55 – 60	2	100
	n = 100	\(\frac{n}{2}\) = 50

l = 35, \(\frac{n}{2}\) = 50, cf = 45, f = 33, h = 5

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

Length (in mm)	Number of leaves
118 – 126	3
127 – 135	5
136 – 144	9
145 – 153	12
154 – 162	5
163 – 171	4
172 – 180	2

Find the median length of the leaves.
(Hint: The data need to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5).
Solution:
The data have to be converted to continuous classes for finding the median since the formula. assumes continuous classes.

Class interval	No. of leaves	Cumulative frequency (cf)
117.5 – 126.5	3	3
126.5 – 135.5	5	8
135.5 – 144.5	9	17
144.5 – 153.5	12	29
153.5 – 162.5	5	34
162.5 – 171.5	4	38
171.5 – 180.5	2	40

n = 40, \(\frac{n}{2}\) = 20
The median lies in the class interval 144.5 – 153.5.
l = 144.5, \(\frac{\mathrm{n}}{2}=\frac{40}{2}\) = 20, cf = 17, f = 12, h = 9.
Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h
= 144.5 + \(\left[\frac{20-17}{12} \times 9\right]\)
= 144.5 + \(\left[\frac{27}{12}\right]\)
= 144.5 + 2.25
= 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Lifetime (in hours)	Number of lamps
1500 – 2000	14
2000 – 2500	56
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

Find the median lifetime of a lamp.
Solution:

Lifetime in hours (CI)	No. of lamps (l)	Cumulative frequency
1500 – 2000	14	14
2000 – 2500	56	70
2500 – 3000	60	130
3000 – 3500	86	216
3500 – 4000	74	290
4000 – 4500	62	352
4500 – 5000	48	400

The median lies in the class interval 3000 – 3500.
\(\frac{\mathrm{n}}{2}=\frac{400}{2}=200\)
l = 3000, \(\frac{n}{2}\) = 200, cf = 130, f = 86, h = 500.
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\left[\frac{70}{86}\right]\) × 500
= 3000 + \(\frac{35000}{86}\)
= 3000 + 406.976
Median life of a lamp is 3406.98 hours.

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

No. of letters	No. of surnames
1 – 4	6
4 – 7	30
7 – 10	40
10 – 13	16
13 – 16	4
16 – 19	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.
Solution:

Hence the modal size of the surnames is 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Solution:

Weight in kg.	No. of students	Cumulative frequency (c.f.)
40 – 45	2	2
45 – 50	3	5
50 – 55	8	13
55 – 60	6	19
60 – 65	6	25
65 – 70	3	28
70 – 75	2	30

\(\frac{n}{2}\) = 15
The median lies in the class 55 – 60.
l = 55, \(\frac{n}{2}\) = 15, c.f. = 13, f = 6, h = 5.
Median = l + \(\frac{1}{2}\) × h
= 55 + \(\left[\frac{15-13}{6}\right]\) × 5
= 55 + \(\frac{2}{6}\) × 5
= 55 + \(\frac{5}{3}\)
= 55 + 1.666 = 56.666
∴ Median = 56.67 kg.
Hence, the median weight of the students is 56.67 kg.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Exercise 9.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Solution :

sin \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
sin 30° = \(\frac{1}{2}\)
\(\frac{1}{2}\) = \(\frac{AB}{20}\)
AB = \(\frac{20}{2}\) = 10 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution :

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:

(i) Figure (a) shows the slide for children below the age of 5 years.
Let BC = 1.5 m be the height of the slide. Slide AC is inclined at CAB = 30° to the ground.
In right angled ΔABC, sin 30° = \(\frac{BC}{AC}\)
\(\frac{1}{2}=\frac{15}{AC}\)
⇒ AC = 3 m.

(ii) Figure (b) shows the slide for elder children. Let RQ = 3 m be the height of the slide. Slide PR is inclined at ∠RPQ = 60° to the ground.
In right angled ΔPQR, sin 60° = \(\frac{RQ}{PR}\) ⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{3}{PR}\)
PR = \(\frac{3 \times 2}{\sqrt{3}}\) = 2\(\sqrt{3}\) m.

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution :

tan C = \(\frac{AB}{CB}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
tan 30° = \(\frac{h}{30}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{30}\)
h\(\sqrt{3}\) = 30
h = \(\frac{30}{\sqrt{3}}=\frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{30 \sqrt{3}}{3}\)
h = 10\(\sqrt{3}\)mts.
Hence, height of the tower is 10\(\sqrt{3}\) mts.

Question 5.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution :

Length of the string is 40\(\sqrt{3}\) mts.

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution :

Let the boy be standing at point B initially. He walks towards the building and reaches point D. From the figure, the distance walked by the boy towards the building is BD.
AC = AG – CG
AC = 30 – 1.5 = 28.5
Now, in ΔABC, we have
tan 30 = \(\frac{AC}{BC}\)
BC = \(\frac{AC}{tan 30}\)
BC = 28.5\(\sqrt{3}\)
Again, in ΔADC, we have
tan 60 = \(\frac{AC}{DC}\)
DC = \(\frac{AC}{tan 60}\)
DC = \(\frac{28.5}{\sqrt{3}}\)
DC = \(\frac{28.5 \sqrt{3}}{3}\) = 9.5\(\sqrt{3}\)
BD = BC – DC
BD = 28.5\(\sqrt{3}\) – 9.5\(\sqrt{3}\)
BD = 19\(\sqrt{3}\)
The distance walked by the boy towards the building is 19\(\sqrt{3}\) m.

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Solution :

tan 60° = \(\sqrt{3}\)
tan 45° = 1
In ΔADC, A\(\hat{\mathrm{D}}\)C = 60
tan 60° = \(\frac{AC}{AD}\)
\(\sqrt{3}\) = \(\frac{x+20}{DA}\)

In ΔBDA, A\(\hat{\mathrm{D}}\)B = 45°
tan 45° = \(\frac{AB}{AD}\) = 1
AB = AD = 20 mts.
DA\(\sqrt{3}\) = x + 20
\(\sqrt{3}\)DA = x + 20
\(\sqrt{3}\)(20) = x + 20
x = 20\(\sqrt{3}\) – 20
= 20(\(\sqrt{3}\) – 1)
Height of the tower B = 20(\(\sqrt{3}\) – 1) mts.

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution :

tan 60° = \(\sqrt{3}\)
In ΔADC, tan 60° = \(\frac{AC}{CD}\)
\(\sqrt{3}\) = \(\frac{AC}{CD}\)
AC = \(\sqrt{3}\)CD

In ΔBDC, tan B\(\hat{\mathrm{D}}\)C = tan 45° = \(\frac{BC}{CD}\)
1 = \(\frac{BC}{CD}\)
CD = BC.

A = AC – CB
= AC – CD (∵ CB = CD)
= \(\sqrt{3}\)CD – CD
1.6 = CD(\(\sqrt{3}\) – 1)

Height of the pedestal = 0.8(\(\sqrt{3}\) + 1) mts.

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution :

Let the height of the building CD be h.
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan \(\hat{\mathrm{C}}\) = \(\frac{AB}{AC}\)
tan 60° = \(\frac{50}{AC}\)
\(\sqrt{3}\) = \(\frac{50}{AC}\)
AC\(\sqrt{3}\) = 50°
AC = \(\frac{50}{\sqrt{3}}\)mts.

Height of the building = 16\(\frac{2}{3}\) mts.

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Solution :

Let AE = x.
EB = 80 – x
tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)

Question 11.
A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Solution :

tan 60° = \(\sqrt{3}\)
tan 30° = \(\frac{1}{\sqrt{3}}\)
In ΔABC, tan 60° = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
\(\frac{h}{x}\) = \(\sqrt{3}\)
∴ h = \(\sqrt{3}\)x
In ΔADB, tan 30° = \(\frac{AB}{DB}\)

Height = x\(\sqrt{3}\) = 10\(\sqrt{3}\) mts. Width of the canal = 10 mts.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution :

In ΔDBC, tan 60° = \(\frac{DC}{BC}\)
\(\sqrt{3}\) = \(\frac{DC}{BC}\)
= \(\frac{DC}{AE}\) = \(\frac{DC}{7}\) (BC = AE).

In ΔABE, tan 45° = \(\frac{AB}{AE}\)
1 = \(\frac{7}{AE}\)
∴ AE = 7 mts.
AB = BC = 7 mts.
DC = 7\(\sqrt{3}\)
∴ DE = DC + CE
= 7\(\sqrt{3}\) + 7 = 7(\(\sqrt{3}\) + 1) mts.

Question 13.
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Solution :

Hence, the distance between the two ships is 75(\(\sqrt{3}\) – 1) m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Solution :

Let the initial position of the balloon be A and final position be B.
Height of the balloon above the girl’s height = 88.2 m – 1.2 m = 87m
Distance travelled by the balloon = DE = CE – CD

Distance travelled by the balloon, DE = CE – CD
= (87\(\sqrt{3}\) – 29\(\sqrt{3}\)) m
= 29\(\sqrt{3}\) (3 – 1) m
= 58\(\sqrt{3}\) m.

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution :

In ΔBCA, tan 30° = \(\frac{h}{CA}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac{h}{CA}\)
CA = h\(\sqrt{3}\) mts.

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution :

Let AB be the tower. ∠ABC = x
∴ ∠ADB = 90° – x
In ΔABC tan x = \(\frac{AB}{BC}\)
tan x = \(\frac{AB}{4}\) ……………(i)
In ΔADB tan (90° – x) = \(\frac{AB}{9}\)
cot x = \(\frac{AB}{9}\) ……………(ii)
(i) × (ii)
tan x × cot x = \(\frac{AB}{4}\) × \(\frac{AB}{9}\)
tan x × \(\frac{1}{tan x}\) = \(\frac{\mathrm{AB}^2}{36}\)
1 = \(\frac{\mathrm{AB}^2}{36}\)
AB² = 36
AB = ± \(\sqrt{36}\)
AB = ± 6
∴ Height of the tower AB = 6 m.
Note: C and D can be taken on the same side of AB.