Plus One

HSSLive.Guru has great pleasure in presenting HSSLive Plus One Malayalam Textbook Questions and Answers, Plus One Malayalam Chapters Summary, Malayalam Chapter Wise Notes, SCERT Malayalam Guide Pdf, Malayalam Study Materials, Plus One Malayalam Handbook, Malayalam Textbook Kerala Syllabus PDF free download according to the latest syllabus prescribed by SCERT under the activity-oriented grading pattern for class 11.

Kerala Plus One Malayalam Textbook Questions and Answers, Notes, Chapters Summary HSSLive

Unit 1 Kinav

• Chapter 1 Sandarsanam
• Chapter 2 Ormayude Njaramb
• Chapter 3 Verukal Nashtapedunnavar
• Chapter 4 Matsyam

Unit 2 Kalca

• Chapter 1 Kayalarikathu
• Chapter 2 Cinemayum Samoohavum Kalavupoya Cyclum
• Chapter 3 Kazhinjupoya Kalaghattavum
• Chapter 4 Kaippad Kelkkunnundo?

Unit 3 Ullariv

• Chapter 1 Kavyakalaye Kurichu Chila Nireekshanangal
• Chapter 2 Oonjalil
• Chapter 3 Anargha Nimisham
• Chapter 4 Lathiyum Vediyundayum

Unit 4 Uravu

• Chapter 1 Peeli Kannukal
• Chapter 2 Anukampa
• Chapter 3 Muhyadheen Mala
• Chapter 4 Vasanavikrithi
• Chapter 5 Samkramanam
• Chapter 6 Shasthrakriya

We hope that HSSLive Plus One Malayalam Study Material will help you improve your quality of learning as per the new grading system and to get an A⁺ grade in Malayalam. Suggestions for further improvement of this material are always appreciated.

Plus One Malayalam Previous Year Question Papers and Answers

HSSLive Plus One

Kerala State Board New Syllabus Plus One Hindi Textbook Answers Unit 3 Chapter 9 आनंद की फूलझडियाँ Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Hindi Textbook Answers Unit 3 Chapter 9 आनंद की फूलझडियाँ

Anand Ki Fuljhadiya Plus One Hindi Chapter 9  प्रश्न 1.
पात्र और घटनाओं का सही मिलान करें।

उत्तर:

आनंद की फुलझड़ियाँ Plus One Hindi Chapter 9 प्रश्न 2.
संक्षेपण करें: “एक बूढ़ा आदमी, जिसके बाल सफ़ेद हो गए थे इसके फल मेरे नाती-पोते खाएँगे।”
उत्तर:

सत्कृत्य:
बूढ़े आदमी से नवजवान ने समझा कि पेड़ पौधे लगाने से प्रकृति सुन्दर हो जाती है और भविष्य में उपयोग में आता है।

Hindi Plus आनंद की फुलझड़ियाँ Chapter 9 प्रश्न 3.
संभ्रांत महिला रेलगाड़ी से कुछ चीजें बाहर फेंकती जा रही है तब सहयात्री और संभ्रांत महिला के बीच का संभावित वार्तालाप तैयार करें।
उत्तर:
संहयात्री : यह आप क्या फेंक रही हैं?
महिला : मैं…….?
सहयात्री : हाँ…. हाँ…….
महिला : तुम देखते नहीं?
सहयात्री : इसलिए तो पूछता हूँ।
महिला : ये तो बीज हैं।
सहयात्री : बीज?
महिला : हाँ… हाँ… फल-फूलों के बीज हैं।
सहयात्री : इनको खिड़की से क्यों फैंकती हैं?
महिला : इनमें कुछ जड़ पकड़ लेंगे।
सहयात्री : तो फिर?
महिला : तब फायदा होगा।
सहयात्री : फायदा? किस प्रकार?
महिला : फूलेंगे, फलेंगे।
सहयात्री : तब?
महिला : मनुष्य के लिए उपयोगी होंगे।
सहयात्री : अरे बापरे! आप तो महान कार्य कर रही हैं।
महिला : यह लो….आप भी फेंकिए।
सहयात्री : हाँ…… हाँ….. दीजिए।

Apoorv Anubhav Class 11 Summary In Hindi Chapter 9 प्रश्न 4.
मान लें, रेलगाड़ी में सफर करनेवाली वृद्ध संभ्रांत महिला की नज़र डिब्बे में चिपके हुए विज्ञापन पर पड़ती है जो रक्तदान के महत्व को रेखांकित करता है। संकेतों के सहारे वह विज्ञापन तैयार करें।

• समभाव
• सहिष्णुता
• मानव-प्रेम
• जीवनदान

उत्तर:

स्वास्थ्य मंत्रालय का विज्ञापन
‘रक्तदान महादान है।’

भाईयो,…… बहनो,…..
रोगावस्था में पीड़ित भाई-बहनों से समभाव रखिए । सहिष्णुता और अनुकंपा रखकर जान बचाने के लिए रक्तदान करके सहायता दीजिए। रक्तदान जीवनदान ही है!! सरकारी रक्तदान केन्द्रों में जाकर खुशी से रक्तदान कीजिए!! आपका रक्त कटेगा नहीं बढ़ेगा!! दूसरों की जान बचेगी।

Plus One 9 Hindi आनंद की फुलझड़ियाँ प्रश्न 5.
‘आज भी वह रक़म अमेरिका में ज़रूरतमंदों के हाथों में घूम रही है’ – मान लें, वह रक़म अपने वर्तमान अनुभवों का आत्मकथा के रूप में ज़िक्र करती है। वह आत्मकथांश लिखें।
उत्तर:

मैं हूँ जरूरतमंदों के सामने

मैं रकम हूँ। मैं साधारण रकम नहीं। मैं एक अपूर्व रकम हूँ। मैं जन्म से अमेरिकी हूँ। अमेरिका के प्रसिडेंट बेंजमिन फ्रैंकलीन के हाथों से मेरा जन्म हुआ। मुझे प्रसिडेंटजी ने एक गरीब विद्यार्थी की सहायता में दिया था। मैंने उसे भाग्यवान् बनाया। विद्यार्थी ने मुझे वापस करने के लिए प्रसिडेंट के पास गया। लेकिन, बेंजमिनजी ने उससे बतायाः “आप इसे अपने ही पास रखिए और जब आपके पास कोई ऐसा ही ज़रूरतमंद आये तो उसे यह दे दीजिए” । उस व्यक्ति ने ऐसा ही किया। आज भी मैं अमेरिका के ज़रूरतमंदों के हाथों में घूम रही हूँ। मैं कितना सौभाग्यवान हूँ! ज़रूरत पड़े, मुझे बुलाईए। मैं ज़रूर आऊँगा।

मेरे इतने जीवनकाल से मैंने समझा कि समाज की पूँजी धनवानों के हाथों में है। पूँजी का समुचित विवरण से समाज का संतुलन होता है। जरूरतमंदों के हाथों में पूँजी का सौगुना मूल्य होगा। मैं रकम, पूँजी समान है। मैं जरूरतमंदों को जीवनदान करता हूँ।

Plus One Hindi आनंद की फूलझडियाँ Important Questions and Answers

प्रश्न 1.
ज़मीन कौन खोद रहा था?
उत्तर:
एक बूढ़ा आदमी।

प्रश्न 2.
बूढ़े आदमी क्या बो रहे हैं?
उत्तर:
आम की गुठलियाँ।

प्रश्न 3.
नौजवान के प्रश्न पर बूढ़े का उत्तर क्या था?
उत्तर:
भविष्य में दूसरों की भलाई के लिए आम की गुठलियाँ बो रहा हूँ।

प्रश्न 4.
पूर्वजों के मनोवृत्ति का फल क्या है?
उत्तर:
वर्तमान के लोगों के लिए भलाई होती है।

प्रश्न 5.
लेखक ने सुंदर स्वभाव की परिभाषा कैसे दी है?
उत्तर:
अशा दूसरों को सुख और आनंद पधुंचानेवाले सात्विक आनंद का स्वभाव।

प्रश्न 6.
हमारे पूर्वजों की इसी मनोवृत्ती का फल है, जो हम जगह-जगह अमराई देखते हैं। कौन-सी मनोवृत्ति?
उत्तर:
अपने स्वार्थी जीवन को त्यागकर दूसरों को सुख और आनंद पहुँचाने के सुंदर स्वभाव की मनोवृत्ति ।

प्रश्न 7.
बूढ़ा आदमी का ‘आम की गुठलियाँ बोना’ घटना का मुख्य आशय क्या है?
उत्तर:
हमें दूसरों को सुख और आनंद पहूँचानेवाले सात्विक आनंद के स्वभाव अपनाना चाहिए।

प्रश्न 8.
मान लें, रेलगाड़ी में सफर करनेवाली वृद्ध संभ्रांत महिला की नज़र डिब्बे में चिपके हुए विज्ञापन पर पड़ती है जो रक्तदान के महत्व को रेखांकित करता है। संकेतों के सहारे वह विज्ञापन तैयार करें।

• समभाव
• सहिष्णुता
• मानव-प्रेम
• जीवनदान

उत्तर:

स्वास्थ्य मंत्रालय का विज्ञापन
‘रक्तदान महादान है।’

भाईयो,…… बहनो,…..
रोगावस्था में पीड़ित भाई-बहनों से समभाव रखिए। सहिष्णुता और अनुकंपा रखकर जान बचाने के लिए | रक्तदान करके सहायता दीजिए। रक्तदान जीवनदान ही है!! सरकारी रक्तदान केन्द्रों में जाकर खुशी से रक्तदान कीजिए!! आपका रक्त कटेगा नहीं बढ़ेगा!! दूसरों की जान बचेगी।

प्रश्न 9.
संभ्रान्त महिला किस उम्मीद से फल और फूलों के बज़ फेंक रही है?
उत्तर:
उनमें से कुछ भी अगर जड़ पकड़ लेगें तो लोगों का इससे कुछ फायदा होगा।

प्रश्न 10.
निम्नलिखित गद्यांश पढ़ें और प्रश्नों का उत्तर लिखें।

एक वृद्ध संभ्रांत महिला रेलगाड़ी से सफ़र कर रही थी। वे खिड़की के पास बैठकर, बीच-बीच में अपनी मुट्ठी से कुछ चीज़ बाहर फेंकती जा रही थीं। एक सहयात्री ने, जो यह देख रहा था, पूछा, “यह आप क्या कर रही हैं?” उस महिला ने जवाब दिया, “ये सुंदर फलों और फूलों के बीज़ हैं। मैं इन्हें इस उम्मीद से फेंक रही हूँ कि इनमें से कुछ भी अगर जड़ पकड़ लेंगे तो लोगों का इससे कुछ फायदा होगा । पता नहीं इस रास्ते से फिर गुजरूँ या न गुजरूँ, इसलिए क्यों न मैं इस संधि का उपयोग क लूँ?”

i. रेलगाड़ी से कौन यात्रा कर रहीं थी?
उत्तर:
एक वृद्ध संभ्रात महिला।

ii. वे क्या कर रही थीं?
उत्तर:
अपनी मुट्ठी से सुंदर फलों और फूलों के बीज़ बाहर फेंकती जा रही थीं।

iii. सहयात्री ने क्या पूछा?
उत्तर:
सहयात्री ने पूछा : यह आप क्या कर रही हैं?

iv. गद्यांश का संक्षेपण करें।
उत्तर:
एक वृद्ध संभ्रात महिला रेलगाड़ी से सफर करते समय सुंदर फलों और फूलों के बीज़ बाहर फेंकती जा रही थीं। उनका उद्धेश्य था कि उनमें से कुछ भी अगर जड़ पकड़ लेंगे तो, लोगों का उससे कुछ फायदा होगा।

v. संक्षेपण केलिए उचित शीर्षक दें।
उत्तर:
जन-सेवा।

प्रश्न 11.
हमें किसके परे रहना चाहिए?
उत्तर:
घर भौतिकवाद और भोग-विलास की हाय-हाय से परे ।

प्रश्न 12.
किस प्रकार के लोगों को देखकर मानव जाति के भविष्य पर श्रद्धा और विश्वास कर सकते हैं?
उत्तर:
जमाने के अंधकार में भी आनंद की फुलझड़ियों से प्रकाश फैलाते रहनेवालों को देखकर ।

प्रश्न 13.
बेंचमिन फ्रांक्लिन ने विद्यार्थी की मदद कैसे की?
उत्तर:
उन्होंने विद्यार्थी को बीस डॉलर देकर मदद की।

प्रश्न 14.
कुछ दिनों के बाद विद्यार्थी डॉलर लौटाने आए तो फ्रांक्लिन ने क्या कहा?
उत्तर:
उन्होंने कहा : “मुझे याद तो नहीं है कि मैंने यह रक़म आपको कब दी। लेकिन खैर, आप इसे अपने ही पास रखिए और जब आपके पास कोई ऐसा ही जरूरतमंद आए तो उसे यह दे दीजिए।”

प्रश्न 15.
बीस डॉलर किनके हाथों में घूम रही है?
उत्तर:
जरूरतमंदों के हाथों में।

प्रश्न 16.
ज़रूरतमंद कौन-कौन हो सकता है?
उत्तर:
सहायता मिलने के लिए व्याकुल रहनेवाले सभी लोग ज़रूरतमंद होते हैं।

प्रश्न 17.
हमारा जीवन मुसीबतों से भरा पड़ा है । कैसे?
उत्तर:
अब लड़ाई, गरीबी, महंगाई और गुलामी से।

प्रश्न 18.
दुनिया रहने लायक कैसे बनी?
उत्तर:
निस्वार्थ और आदर्श-प्रिय लोगों की उपस्थिति से।

प्रश्न 19.
पैसा वापस देने आया छात्र औह बेंजमिन फ्रैंकलीन के बीच का वार्तालाप तैयार करें?
उत्तर:
छात्र : नमस्कार जी!
फ्रैंकलीन : नमस्कार।
छात्र : आप मुझे जानते हैं?
फ्रैंकलीन : याद नहीं, लगता है कि देखा है।
छात्र : मैंने आप से कुछ डॉलर माँग लिया था।
फ्रैंकलीन : कब?
छात्र : कुछ साल पहले मेरी पढ़ाई केलिए। अब मेरी पढ़ाई खतम हुई। वह डॉलर वापस देने आया हूँ।
फ्रैंकलीन : यह डॉलर मुझे नहीं, किसी ज़रूरतमंद को दो।
छात्र : आप का मन कितना अच्छा है!
फ्रैंकलीन : तुम से कोई ज़रूरतमंद व्यक्ति माँगे है तो उसे यह डॉलर दो। जाओ।
छात्र : ठीक है, धन्यवाद ।

प्रश्न 20.
किसके लिए छीना-झपटी होती थी?
उत्तर:
टिकट लेने के लिए।

प्रश्न 21.
टिकट बाबू किन बातों को सुना-अनसुना करके अपना काम करते रहे?
उत्तर:
टिकट लेने के लिए छीना-झपटी करनेवालों के परिहासों और धमकियों को।

प्रश्न 22.
टिकट बाबू की परेशानी का कारण क्या था?
उत्तर:
टिकट लेने के लिए बड़ी भीड़ थी। वहाँ पर छीना-झपटी होती थी। छोटी-सी खिड़की से टिकट के लिए घुसानेवाले अनेक हाथों को टिकट देने केलिए टिकटबाबू अकेला था।

प्रश्न 23.
टिकट बाबू पर इन शब्दों का अजीब असर पड़ा । क्यों?
उत्तर:
लेखक ने टिकट बाबू की परेशानी समझकर बड़ी सहानुभूति से व्यवहार करके मुसाफिरों को शांत किया।

प्रश्न 24.
टिकट बाबू को नई ताकत कैसे मिली?
उत्तर:
लेखक की सहानुभूति के शब्दों से टिकट बाबू को नयी … ताकत मिली।

प्रश्न 25.
टिकट बाबू का हृदय कब मोम-सा हो गया?
उत्तर:
टिकट बाबू के प्रति लेखक सहानुभूति दर्शाने पर।

प्रश्न 26.
टिकट बाबू से संबंधित घटना का ज़िक्र करते हुए लेखक अपने मित्र को पत्र लिखता है। वह पत्र तैयार करें।
उत्तर:

स्थान,
तारीख,

प्रिय मित्र रामू,

तुम कैसे हो? ठीक हो न? यहाँ पर मैं ठीक ही हूँ।
रामू, कुछ दिन पहले अपने काम से मुंबई जाने के लिए मैं टिकट लेने गया। वहाँ पर बड़ी छीना-झपटी होती थी। टिकट बाबू बड़ी परेशानी में था। लोग हल्ला करते थे, टिकट बाबू का परिहास करते थे। कुछ लोग उनको धमकी भी करता था। लेकिन टिकट बाबू इन बातों को सुना- अनसुना करके आपना काम करते रहे। मैं पास ही खड़ा था। मैंने बड़ी सहानुभूति से टिकट बाबू के बारे में यात्रियों से बात की। मेरी सहानुभूति देखकर टिकटबाबू को बड़ा आश्वास मिल गया। उन्होंने जल्दी मेरा टिकट दे दिया और एक नए उत्साह से अन्य यात्रियों को वे टिकट देने लगे।

प्रिय मित्र, दुसरों की कठिनाइयों को समझकर हमें व्यवहार करना चाहिए। हमारे कर्म और वचन से दूसरों को आश्वास मिला चाहिए।
यहाँ पर तुम कब आवोगो? तुम्हारी प्रतीक्षा में मित्र,

(हस्ताक्षर)
शेवड़े।

पताः
नाम

प्रश्न 27.
क्लर्क का तमाम दिन कैसे बीतता है?
उत्तर:
बैंक के रूखे आँकड़ों से माथापच्ची करते -करते क्लर्क का तमाम दिन बीतता है।

प्रश्न 28.
क्लर्क के किस गुण का सम्मान किया गया?
उत्तर:
अच्छी हस्तलिपि का।

प्रश्न 29.
क्लर्क का चेहरा प्रसन्नता से क्यों खिल उठा?
उत्तर:
लेखक के थोड़े-से शब्दों द्वारा क्लर्क के जीवन में किंचितमात्र सुख पहुँचने पर।

प्रश्न 30.
लेखक को क्यों प्रसन्नता महसूस हुई?
उत्तर:
प्रय लेखक को थोड़े-से शब्दों द्वारा बैंक के क्लर्क के जीवन में थोड़ा सुख पहूँचते देखकर लेखक को प्रसन्नता महसूस हुई।

प्रश्न 31.
बैंक का क्लर्क अपनी हस्तलिपि की तारीफ सुनने पर बहुत खुश हुआ। घर आने पर पत्नी उसकी खुशी का कारण जानना चाहती है-दोनो के बीच का संभावित वार्तालप लिखें।
उत्तर:
पत्नी : आज आप बहुत खुश है ….
क्लर्क : हाँ-हाँ….
पत्नी : कारण क्या है?
क्लर्क : एक कारण है।
पत्नी : मुझे भी बताओ….
क्लर्क : तुम जानना चाहती हो?
पत्नी : क्यों नहीं?
क्लर्क : तुम अनंत गोपाल शेवड़े को…
पत्नी : ओहो… सुप्रसिद्ध लेखक?
क्लर्क : जानती हो उन्हें?
पत्नी : सुनी तो है।
क्लर्क : आज उन्होंने मेरे बैंक में…
पत्नी : बैंक में?
क्लर्क : आये थे।
पत्नी : तो ?
क्लर्क : उन्होंने मेरी हस्तलिपि की …..
पत्नी : प्रशंसा की?
क्लर्क : हाँ…हाँ…
पत्नी : बड़ी बात है।
क्लर्क : हाँ…हाँ…

प्रश्न 32.
बैंक क्लर्क ने अपनी डायरी में क्या लिखा होगा? वह डायरी तैयार करें।
उत्तर:
तारीख
आनंद नगर :
आज का दिन बहुत अच्छा है। आज मुझे मिली प्रशंसा एक पंखुडी के समान मुझे अब भी सहलाती है। वह आदमी कितना अच्छा है! आज पहली बार बैंक के बोरिंग समय में कुछ राहत मिली। उस आदमी ने खाता खोलने के लिए आया था। मैं उनका नाम और पता लिखते समय उन्होंने मेरी हस्तलिपि पर ध्यान दिया और कहा ‘ब्यूटिफुल’! मेरा मन खुशी से भरा। सब लोगों ने मेरी हस्तलिपि देखी थी, पर किसीने भी मुझ से कुछ नहीं कहा था । अब मैं गर्व का अनुभव करने लगा। बड़ी खुशी से मैंने उनसे बातें कीं। पत्नी से भी यह बात कही। यह दिन में कभी नहीं भूलूंगा।

प्रश्न 33.
आनंद की फुलझड़ियाँ निबन्ध में लेखकने रेल विभाग के एक टिकट बाबू से करुणा प्रकट की थी। उस दिन के टिकट बाबू की डायरी तैयार करें।
उत्तर:

तारीख

आनंद शहर :
आज का दिन अच्छा दिन था। एक आदमी ने मुझे आज ठीक समझा है। यह बात मेरे मन में खुशी भरती है। यात्रियों की भीड़, उनकी परेशानियाँ, धमकियाँ आदि – आदि ने मुझे बहुत परेशानियाँ देती थीं। टिकट कौंटर में हर दिन अकेला रह गया हूँ। मेरी कठिनाइयों पर किसी ने ध्यान नहीं किया था। पर आज एक सज्जन ने मुझपर ध्यान देकर मेरी मदद की। टिकट लेने केलिए खड़े लोगों से मेरी परेशानियों के बारे में बताने की कृपा उन्होंने की। यह एक अजीब बात थी। उनकी सहानुभूति देखते वक्त मेरा हृदय मोम जैसा बन गया। वे शब्द मेरे मन में सांत्वना देने लायक थे। उस घटना के बाद मैं शांत भाव से टिकट देने में समर्थ हुआ।

वे कौन होंगे? जाते वक्त उन्होंने कहा कि फिर मिलें। ज़रूर उनसे मिलना चाहिए। ऐसे सज्जनों से परिचय पा लेना कितनी अच्छी बात है! आज का दिन मैं कभी नहीं भूलूंगा।

प्रश्न 34.
निम्नलिखित अर्थों के समानार्थी मुहावरों को लेख से छाँटकर लिखें।
(नष्ट होना, विपत्ति के दिनों के बाद सुख का दिन आना, सहन करना, हिम्मत करना, किसी के अच्छे काम की न्यायदृष्टि से प्रशंसा करना)
उत्तर:
रसातल में जाना = नष्ट होना
दिन फिरना = विपत्ति के दिनों के बाद सुख का दिन आना
ताना कसना = सहन करना
दिल कडा करना = हिम्मत करना
दाद देना = किसी के अच्छे काम की न्यायदृष्टि से प्रशंसा करना

प्रश्न 35.
सूचनाः यह गद्यांश पढ़िए और नीचे दिए प्रश्नों के उत्तर लिखिए:

नाविक भोलाराम रेलगाड़ी के वातानुकूलित डिब्बे में बैठ रहे थे। उनके पास काफ़ी रुपये थे। एक लड़की भी उस डिब्बे में आ बैठी। उसने नाविक से बातें शुरू की। उसने अपनी गरीबी का जिक्र किया तो नाविक ने पूछा कि इतनी गरीबी में भी वातानुकूलित डिब्बे में तुम क्यों यात्रा कर रही हो? तब लड़की ने कहा कि उसकी शादी तय हो चुकी है। और अपने ससुरालवालों को प्रभावित. करने केलिए वह इस डिब्बे में यात्रा कर रही है। फिर उस लड़की ने नाविक से कुछ रुपया माँग लिया तो उसने देने से इनकार कर दिया। तब लड़की ने उसको धमकी देकर कहा कि मुझे बीस हज़ार रुपये दें दो, नहीं तो मैं तुम्हारे ऊपर झूठे इल्ज़ाम लगाउँगी। गाड़ी रुकी तो लड़की ने पुलिस से कहा कि नाविक ने अपने हाथ से मेरा मुँह बंद कर लिया और दूसरे हाथ से मुझे खींचकर मेरी इज्जत लूटने की कोशिश की है। लेकिन जाँच करने पर पुलिस को मालूम हुआ कि नाविक के दोनों हाथ कटे हुए हैं। झूठे इल्ज़ाम लगाने के अपराध में लड़की पकड़ी गयी।
(इल्ज़ाम – आरोप, इज्जत लूटना – अपमानित करना)

i) इस गद्यांश से कौन सा सन्देश मिलता है?
उत्तर:
हमें कभी भी कपट न होना चाहिए।

ii) इस गद्यांश का संक्षेपण करें और उचित शीर्षण लिखे।
उत्तर:
कपटता :
एक लड़की ने भोलाराम नामक एक बिना हाथवाला नाविक के साथ रेलगाड़ी में यात्रा करते समय धोखा देने की कोशिश की। पुलिस आकर लड़की को पकड़ा।

प्रश्न 36.
हिंदी भारत की राजभाषा एवं राष्ट्रभाषा है- ‘सीखें हिंदी, सिखाएँ हिंदी’ – इस विषय पर निबंध लिखिए।
उत्तर:

सीखें हिंदी, सिखाएँ हिंदी

भारत में अनेक भाषाएँ हैं। भाषाओं को उपभाषाएँ और प्रादेशिक भाषाएँ भी हैं। लेकिन भारत के अधिकांश लोगों से बोलनेवाली भाषा हिंदी है। इसलिए हिंदी को भारत की संपर्क भाषा के रूप में माना जाता है। हिंदी भारत की राजभाषा एवं राष्ट्रभाषा है। हिंदी एक सरल भाषा है। हमें यह जानने से खुशी होगी कि संसार में सबसे ज्यादा बोलनेवाली तीसरी भाषा हिंदी है। आज हिंदी अन्तर्देशीय भाषा के रूप में प्रचलित होती जाती है।

भारत विविधता का देश है। लेकिन हिंदी एकता की कड़ी है। भारत की संस्कृति हिंदी से जुड़ी रहती है। हिंदी समृद्ध साहित्य से भी संपन्न है।

हिंदी का प्रचार करना प्रत्येक भारतीय का दायित्व है। हिंदी के प्रचार से भारत में एकता बढ़ेगी। इससे यह .मतलब नहीं है कि अन्य प्रादेशिक भाषाएँ महत्वपूर्ण नहीं है और वे तिरस्कृत हो जायें। प्रादेशिक भाषाओं का भी संरक्षण होना चाहिए। हिंदी के प्रचार से भारत की अखंडता सदा सुरक्षित रखें।

आनंद की फूलझडियाँ Previous Years Questions & Answers

प्रश्न 1.
निम्नलिखित सहायक बिंदु के आधार पर वार्तालाप तैयार कीजिए।
संभ्रान्त महिला रेलगाड़ी से कुछ चीजें बाहर फेंकती जा रही थी। तब सहयात्री और संभ्रान्त महिला के बीच का संभावित वार्तालाप तैयार कीजिए। सहायक बिंदुः

• रेलगाड़ी से चीजें बाहर फेंकना
• सहयात्री द्वारा पूछा जाना
• निःस्वार्थ सेवा
• दुनिया में रहने लायक

उत्तर:
सहयात्री : यह आप क्या फेंक रही हैं?
महिला : मैं……?
सहयात्री : हाँ…. हाँ…….
महिला : तुम देखते नहीं?
सहयात्री : इसलिए तो पूछता हूँ।
महिला : ये तो बीज हैं।
सहयात्री : बीज?
महिला : हाँ… हाँ… फल-फूलों के बीज हैं।
सहयात्री : इनको खिड़की से क्यों फैंकती हैं?
महिला : इनमें कुछ जड़ पकड़ लेंगे।
सहयात्री : तो फिर?
महिला : तब फायदा होगा।
सहयात्री : फायदा? किस प्रकार?
महिला : फूलेंगे, फलेंगे।
सहयात्री : तब?
महिला : मनुष्य के लिए उपयोगी होंगे।
सहयात्री : अरे बापरे! आप तो महान कार्य कर रही हैं।
महिला : यह लो….आप भी फेंकिए।
सहयात्री : हाँ…… हाँ….. दीजिए।

प्रश्न 2.
मान लीजिए, आनंद की फूलझड़ियाँ इस निबंध का लेखक आत्मकथा लिखता है। आत्मकथा में टिकट बाबू के प्रसंग का उल्लेख है। निम्नलिखित सहायक बिंदु के आधार पर वह आत्मकथांश तैयार कीजिए।
सहायक बिंदुः

• लेखक का मुंबई जाना।
• टिकट काऊंटर के पास भीड़ लगना ।
• टिकट बाबू का परेशान होना ।
• लेखक द्वारा टिकट बाबू के प्रति सहानुभूति प्रकट करना।

उत्तर:
आत्मकथा  :
कुछ साल पहले की बात है। मुझे जल्द ही मुंबई पहूँचना था। मैं टिकट लेने केलिए टिकट काउंटर पहूँचा । लड़ाई के कारण गाड़ियों की संख्या कम थी। इसलिए काउंटर के पास बहुत भीड़ लगी हुई थी। टिकट बाबू परेशानी से टिकट बनाते थे। लेकिन भीड़ ज्यादा होने के कारण उसे ठीक तरह से सभी लोगों को टिकट बन नहीं पा रहे थे। लोगों की ओर से कई प्रकार के बुरे टिप्पणियाँ उन पर हो रहे थे। इसका असर उपनर और बुरी तरह से हो रहे थे। बाबू कितने ही ईमानदारी से काम करें लोग उनपर शक से बात करते थे। यह सब देखकर मुझे बहुत दुख हुआ। वह पूरी ताकत से काम कर रहे थे, फिर भी लोग उनपर गालियाँ दे रहे थे। मैं वहाँ के लोगों को समझाया कि बाबू बहुत कोशिश कर रहे हैं और उसे कुछ समय दीजिए। मैं बाबु से ऐसे बातें किया कि उसे कुछ आश्वास मिला। मेरा सहानुभूति का असर उनपर हुआ। वह फिर पूरी कोशिश की और मुझे धन्यवाद भी अदा की। मुझे पूरा यकीन था कि अच्छे वाक्यों का अच्छा असर हो जायेगा।

आनंद की फूलझडियाँ Summary in Malayalam

आनंद की फूलझडियाँ शब्दार्थ

Students can Download Chapter 3 Liberalisation, Privatisation and Globalisation – An Appraisal Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Economics Notes Chapter 3 Liberalisation, Privatisation and Globalisation – An Appraisal

Background of the economic reforms
India introduced economic reforms in 1991. It was due to several reasons. Important among them are:

• Policies such as MRTP and FEMA prevented large scale domestic and foreign investments.
• Reserving certain sectors exclusively for the public sector prevented private investment less attractive for such sectors.
• Gulf war and subsequent events created a severe foreign exchange crisis in our country.
• Import bill of petroleum products increased alarmingly leading to BoP deficit.
• Political instability.

Plus One Economics Chapter 3 Notes Liberalisation
Liberalization implies liberating trade from unwanted government controls and restrictions. Indian economy prior to the nineties was following a restrictive policy and excessive government interferences in all economic activities. This interference created the license-permit-raj as indicated earlier. This has led to extensive corruption, red-tapism, undue delay, and inefficiency. Most of the policies such as the licensing system, FERA, MRTP hindered economic growth, and industrialisation. The aim of the liberalization policy was very comprehensive, promoting economic growth by reducing factors hindering it and makes the economy very competitive at international standards.

Liberalisation policies included reforms in the following sectors.

• Industrial sector reforms
• Financial sector reforms
• Tax reforms
• Foreign exchange reforms

Plus One Economics Chapter 3 Notes Pdf Privatisation
Privatisation refers to any process that reduces the participation of the state/public sector in the economic activities of a country. In other words, the conversion of ownership or management of a government-owned enterprise into a private enterprise is known as privatization or denationalization. India started privatization as part of the Structural Adjustment Programme (SAP). The process of privatisation can take place either by the withdrawal of government ownership and management of public sector companies or by the outright sale of public sector companies (disinvestment).

Economics Notes Class 11 Kerala Syllabus Aims of disinvestment:

1. Better performance of public sector units (PSUs) through better management techniques
2. Enforcing financial discipline and improving financial performance
3. Enhancing the ability of companies to raise financial resources from the market
4. Raising revenue of the government from sale of equity
5. A strong impetus to the flow of FOI (Foreign Direct Investment)

Plus One Economics Notes Chapter 3 Globalisation
Globalisation is a complex phenomenon. The term globalisation indicates the opening up of domestic economy for the world market, or integration of an economy with global economy. It involves creation of network and activities transcending economic, social and geographical boundaries. It attempts to establish links in such a way that the happening in India can be influenced by events happening miles away. Integration of economies is possible through interlinking domestic market with world market through foreign trade. Therefore, it is treated as a very complex phenomenon.

Plus One Economics Notes Pdf Outsourcing
Outsourcing is an important feature of globalisation. It is practice where a company hires regular service from external sources mostly from other countries which previously provided internally or within the country.

Plus One Economics Malayalam Notes World Trade Organisation (WTO)
WTO was founded in 1995 replacing GATT. GATT was established in 1948. Following are the aims of WTO.

• Provides equal opportunities to all participating nations in international trade.
• To ensure optimum utilization of world resources and protect the environment.
• Remove of tariffs (tax) and non-tariffs (quota). This leads to the removal of restrictions on trade thereby facilitating free-entry and free exit of goods
• To encourage multi-lateral trade (more than two nations) rather than bilateral trade (two countries).
• Extension of a trade by including trade in services like banking, insurance communication.
• To include Trade-Related Intellectual Property Rights (TRIPs), commonly known as Patent Rights and Trade-Related Investment Measures (TRIMs) within the span of international trade.

Students can Download Chapter 3 Motion in a Straight Line Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 3 Motion in a Straight Line

Plus One Physics Chapter 3 Notes Pdf Summary
Motion In A Straight Line
In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity, acceleration and relative velocity. We also develop a set of simple equations called Kinematic equations.

• Motion: Motion is change in position of an object with time.
• Rectilinear motion: The motion along a straight line is called rectilinear motion.
• Point object: If the distance travelled by the body is very large compared with its size, the size of the body may be neglected. The body under such a condition may be taken as a point object. The point object can be represented by a point.

Example:

• The length of bus may be neglected compared with the length of the road it is running.
• The size of planet is ignored compared with the size of the orbit in which it is moving.

Position, Path Length And Displacement
1. Reference point, Frame of reference:
In order to specify position of object, we take reference point and a set of axes. Consider a rectangular coordinate system consisting of three mutually perpendicular axes, labelled x, y, and z axes. The point of intersection of these three axes is called origin (O) and serves as the reference point.

The coordinates (x, y, z) of an object describe the position of the object. To measure time, we place a clock in this coordinate system. This coordinate system along with a clock is called a frame of reference.

Straight-line motion in coordinate system

To describe the motion along a straight line we can choose x-axis. The position of a carat different time are given in the figure 3.1. The position to the right of 0 is taken as positive and to the left of 0 as negative. The position coordinates of point P and Q are +360m +240m. The position coordinate of R is-120m.

2. Path Length (Distance):
The total length of the path travelled by an object is called path length.
Explanation:
Consider a car moving along straight line. The positions of car at different time are given in the x-axis. (See figure 3.1)
Case-1:
The car moves from 0 to P. In this case the distance moved by car is OP = +360.
Case-2:
The car moves from 0 to P and then moves back from P to Q.
In this case, the distance travelled is OP + PQ = +360 + (+120) = +480m.

3. Displacement:
The distance between initial point and final point is called displacement.

OR

The change of position of the particle in a particular direction is called displacement.
Explanation:
Consider a car moving along a straight line. The positions of car at different time is given in the x-axis.
See figure (3.1)
Let us take two cases
Case-1:
The car moves from 0 to P, in this case displacement = (360 – 0) = 360
Case-2:
The car moves from 0 to P and moves back from P to Q.
In this case,
Displacement = 240m
Let x₁ and x₂ be the positions of an object at time t₁ and t₂. Then displacement in time Dt = (t₂ – t₁) can be written as Dx = x₂ – x₁
If x₁ < x₂, Dx is positive and if x₂ < x₁, Dx is negative.
Note: The magnitude of displacement may or may not be equal to the path length traversed by an object.

4. Position Time Graph:
Motion of an object can be represented by a position-time graph.
Position time graph for a stationary object:
For a stationary object, the position does not change with time. Hence the position time graph will be a straight line parallel to time axis.

Position time graph in a uniform motion:
Uniform motion:
A body is said to be uniform motion, if it undergoes equal displacements in equal intervals of time. In uniform motion velocity is constant The figure below shows the positiontime graph of such a motion.

Plus One Physics Chapter 3 Notes Question 1.
The position-time of a car is given below. Analyze the graph and explain the motion of car.

Answer:
The car starts from rest a time t=0s from the origin 0 and picks up speed till t=10s. After 10 sec, the car moves with uniform speed till t=18 sec. Then the brakes are applied and the car stops at t = 20s and x = 296m.

Class 11 Physics Chapter 3 Notes Question 2.
Draw the position-time for an object

1. moving with positive velocity
2. moving with negative velocity.

Answer:
1.

2.

Average Velocity And Average Speed
1. Average Velocity:
The average velocity of a particle is the ratio of the total displacement to the time interval.

Explanation:
To explain average velocity, consider a position time graph of a body given below.

Let x₁ be the position of body at a time t₁ and x₂ be the position at t₂.
The average velocity during the time interval Dt = (t₂ – t₁)

where Dx = x₂ – x₁, and Dt = t₂ – t₁,
\(\overline{\mathbf{v}}\) is the average velocity.

Motion In A Straight Line Class 11 Notes Pdf Question 3.
Find the slope of position time graph given below of uniform motion and explain the result.

Answer:

Slope of displacement time graph gives average velocity.

Class 11th Physics Chapter 3 Notes Question 4.
Displacement time graph of a car is given below.

1. Find the average velocity during the time interval 5 to 7 sec.
2. Find the average velocity by taking slope in the interval 5 to 7 sec.

Answer:
1.

2. Slope, tan q

In this case, slope and average velocity are equal in the same interval.

2. Average Speed:
Average speed of a particle is the ratio of the total distance to total time taken.

Physics Chapter 3 Class 11 Notes Question 5.
A car is moving along a straight line. Say OP in figure. It moves from 0 to P in 18s and returns from P to Q in 6s. What are the average velocity and average speed of the car in going?

1. From 0 to P? and
2. from 0 to P and back to Q. (See Figure 3.1)

Answer:
1. Average velocity

Average speed

In this case the average speed is equal to the magnitude of the average velocity.

2. In this case
Average velocity

Average speed

In this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction. So that the distance is greater than displacement.
Note: In general, the velocity is always less than or equal to speed.

Instantaneous Velocity And Speed
Nonuniform Motion:
A body is said to be nonuniform motion, if it undergoes unequal displacements in equal intervals of time.

OR

A body moving with varying velocity is called nonuniform motion.

1. Instantaneous Velocity:
Chapter 3 Physics Class 11 Notes Question 6.
Why the concept of instantaneous velocity is introduced?
Answer:
In nonuniform motion the average velocity tells us how fast the object has been moving over a given interval. But it does not tell us how it moves at different instants during that interval. For this we define instantaneous velocity. The velocity at an instant is called instantaneous velocity.
Explanation:
Position-time of a body moving along a straight line is given below.

Let us find average velocity in the interval 2 sec (3s to 5s), centered at t = 4 sec. In this case, the slope of line P₁P₂ give the value of average velocity, ie. Slope of P₁P₂,

Decrease the value of Dt from 2.to 1 sec. (ie. 3.5 to 4.5 sec). Then line P₁P₂ becomes Q₁Q₂. Then the slope of gives average velocity overthe interval 3.5 sec to 4.5sec.
ie. slope of Q₁Q₂

In the limit Dt ® 0, gives the instantaneous velocity at t = 4sec and its value is nearly 3.84m/s.

Motion In A Straight Line Class 11 Notes Question 7.
When average velocity of a body becomes instantaneous velocity?
Answer:
In the limit, Dt goes to zero, the average velocity becomes instantaneous velocity.

But lim \(\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{d x}{d t}\)
\Instantaneous velocity,

Here \(\frac{d x}{d t}\) is the differential coefficient of x with respect to time. It is the rate of change of position with respect to time at an instant.

Class 11 Physics Notes Chapter 3 Question 8.
The table given below gives the value of \(\frac{\Delta x}{\Delta t}\) for Dt equal to 2s, 1s, 0.55, 0.1s and 0.01s centered at t = 4 sec. (See figure given above). What conclusions can be made from this table?
Answer:
The value of average velocity \(\left(\frac{\Delta x}{\Delta t}\right)\) becomes instantaneous velocity (3.8 m/s), in the limit of Dt goes to zero, (ie Dt is infinitesimally small).

Class 11 Physics Chapter 3 Notes Pdf Download Question 9.
The position of an object moving along x-axis is given by x = a + bt² where a = 8.5m, b = 2.5 m/s² and t is measured in seconds

1. What is the velocity at t = 0s and t = 2s.
2. What is the average velocity between t = 2s and t = 4s?

Answer:
1.

when t = 0
we get v = 2 × 2.5 × 0
v = 0
when t = 2sec
v = 2 × 2.5 × 2 v = 10m/s.

2. The average velocity

Note: If a body is moving with constant velocity, the average velocity is the same as instantaneous velocity at all instants.

2. Instantaneous Speed:
The speed at an instant is called instantaneous speed.
Note:

• The average speed over a finite interval of time is greater or equal to the magnitude of the average velocity.
• Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant.

Motion In Straight Line Notes Pdf Acceleration
1. Average Acceleration:
Average acceleration of a particle is ratio of the change in velocity to the time interval.

Explanation
Consider a body moving along a straight line. Let v₁ and v₂ be the instantaneous velocities at time t₁ and t₂ respectively.


where Dv = change in velocity, Dt = Time interval

2. Instantaneous Acceleration:
Acceleration at any instant is called instantaneous acceleration.
Explanation
In the limit Dt ® 0, (Dt goes to zero) the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Instantaneous acceleration is the rate of change of velocity with respect to time.

3. Uniform Acceleration:
A body is said to be in uniform acceleration, if velocity changes equally in equal intervals of time.

Motion In A Straight Line Class 11 Pdf Question 10.
The velocities of two bodies A and B are given in the tables. From this table, find which body is moving with uniform acceleration. Explain.
Body A

Body B

Answer:
The body A is moving with uniform acceleration be-cause the velocity of body increases at the rate of 2 m/s².
The body B is moving with constant velocity. Hence this motion is called uniform motion.

4. Velocity-Time Graph For Uniformly Accelerated Motion:

An example for velocity-time of a uniformly accelerated motion is given in the above figure.
Let v_t1 and v_t2 be the velocities at instants t₁ and t₂respectively.
The slope of graph in the interval (t₂ – t₁) can be written as,

∴ tan q = acceleration
Thus the slope of the velocity-time gives the acceleration of the particle.

Motion In A Straight Line Notes Pdf Question 11.
Velocity-time of a body is given below. From this graph draw corresponding acceleration time graph.

Answer:
The slope of velocity-time graph increases in the interval (0 – 10) sec which means that acceleration of the body increases in this interval.

Velocity is constant in the interval (10 – 18) sec. Hence ’ the slope is zero which means that acceleration is zero in this range.

The slope in the interval (18 – 20) sec is constant and negative. Hence acceleration in this is a negative value. The acceleration – time graph for the above motion is given below.

Motion In Straight Line Class 11 Notes Question 12.
The position-time graph of a car is given below.

1. Draw corresponding velocity-time graph. Explain the reason for your answer.
2. From velocity-time graph draw acceleration-time graph and identify the regions of

• positive acceleration
• Negative acceleration
• zero acceleration.

Answer:
1. In the time interval (0 – t₁) sec, the slope of x – t graph increases which means that velocity is increasing in this time interval.

In the time interval (t₁ – t₂) sec, slope is constant. Hence velocity remains constant in this time interval.

In the time interval (t₂ – t₃) sec, the slope is decreasing and finally becomes zero. Which means that velocity decreases to zero.

2. Slope is constant throughout the interval (0 – t₁) sec which means that acceleration constant.

In the interval (t₁ – t₂) sec, slope is zero. Which means that acceleration is zero in this region.

Slope is constant (but negative) in the interval (t₂ – t₃)sec. Hence acceleration is constant and negative in this time interval.

Motion In A Straight Line Class 11 Notes Pdf Download Question 13.
Find the region of

1. positive acceleration
2. zero acceleration
3. negative acceleration from the above x-t graph

Answer:

1. Region OA – Positive acceleration
2. Region AB – zero acceleration
3. Region BC – Negative acceleration

Straight Line Class 11 Notes Question 14.
Match the following.

Answer:
1) – d, 2) – c, 3) – b, 4) – a.

5. Area Under Velocity-Time Graph:
Area under velocity-time graph represents the displacement over a given time interval.
Explanation
Consider a body moving with constant velocity v. Its velocity-time graph is given below.

The area of the rectangle has height v and bast t. Therefore,

Note: The acceleration and velocity of a body cannot change values abruptly at an instant. Changes are always continuous.

Kinematic Equations For Uniformly Accelerated Motion
For uniformly accelerated motion, we can derive some simple equations.

1. Velocity-time relation
2. Position-time relation
3. Position-velocity relation

These equations are called kinematic equations for uniformly accelerated motion.
1. Velocity-Time Relation:

Consider a body moving along a straight line with uniform acceleration ‘a’. Let ‘u’ be initial velocity and ‘v ‘ be the final velocity at time t.
We know acceleration a = \(\frac{\text { Change in velocity }}{\text { Time interval }}\)
a = \(\frac{v-u}{t}\)
at = v – u

2. Position-Time Relation:

Consider a body moving along a straight line with uniform acceleration a. Let ‘u’ be initial velocity and ‘v’ be the final velocity. ‘S’ is the displacement travelled by the body during the time interval ‘t‘.
Displacement of the body during the time interval t,
S = average velocity × time
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at ____(2)
Substitute eq.(2) in eq.(1), we get

3. Position-Velocity Relation:
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at
\(\frac{v-u}{a}\) = t _____(2)
Substitute eq.(2) in eq.(1)

2as = v² – u²
v² – u² = 2as

Free-fall:
An object released (near the surface of earth) is accelerated towards the earth. If air resistance is neglected, the object is said to be in free fall. The acceleration due to gravity near the surface of earth is 9.8 m/s².
Note: Free-fall is a case of motion with uniform acceleration.

Question 15.
A body is allowed to fall freely. Draw the following graph.

1. Acceleration-time
2. Velocity-time
3. Position-time

Answer:
1.

2.

3.

Stopping distance of vehicles:
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.

Question 16.
Derive an expression for stopping distance of a vehicle in terms of initial velocity (u) and retardation (a).
Answer:
Let the distance travelled by the vehicle before it stops be ‘s’.
Then we can find ‘s’ using the formula
v² = u² + 2as
0 = u² + -2as

3.7 Relative Velocity
Suppose the distance between two bodies changes with time in magnitude, or in direction or in both. Then each body is said to have a velocity relative to the other.

For example, consider two cars A and B moving in the same direction with equal velocities. To a person in A, the car B would appear to be rest.

Hence the velocity of B relative to A is zero.
ie. V_BA = 0
Similarly, the velocity of A with respect to B is zero.
or V_AB = 0
Let A be moving with a velocity V_A and B be moving with a greater velocity V_B in the same direction. Then the person in A feels that the car B is moving away from him with a velocity V_BA. The velocity of B relative to A

For an observer in B, the car A is going back with a velocity. The velocity of A relative to B
V_AB = -(V_B – V_B).

Question 17.
The position-time graph of two bodies A and B (at different situations) are given in the following graphs. Find the relative velocities of the following graph.

Answer:
a) The slope of Aand B are equal. Hence velocity of A and B are equal. So velocity of A with respect to B, V_AB = 0

b) The body A and B meet at t = 3sec

Velocity of A w.r. to B, V_AB = V_A – V_B
= 20-10 = 10 m/s Velocity of B w.r. to A, V_BA = V_B – V_A
= 10 – 20 = -10 m/s

c) The body A and B meet at t = 1 sec.
The velocity of body in the interval t = 1 sec,

Velocity of A w. r. to B,
V_AB = V_A – V_B
= 20 – 10 = 30 m/s
Similarly velocity of B w.r. to A,
V_BA = V_B – V_A
= 10 – +20 = -30 m/s
The magnitude of V_BA or V_AB (=30 m/s) is greater than the magnitude of velocity A or that of B.

Students can Download Chapter 1 Sets Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 1 Sets

Plus One Maths Sets Three Mark Questions and Answers

Plus One Maths Chapter Wise Previous Questions And Answers Question 1.
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical A, 50 to chemical B and 30 to both chemical A and B, Find the number of individuals exposed to

1. Chemical A but not chemical B. (1)
2. Chemical B but not.chemical C. (1)
3. Chemical A or chemical B. (1)

Answer:
1. Given; n(U) = 200; n(A) = 120;
n(B) = 50; n(A∩B) = 30
n (Chemical A but not chemical B)
= n(A ∩ B’) = n(A) – n(A ∩ B) = 120 – 30 = 90

2. n (Chemical B but not chemical A)
= n(A’ ∩ B) = n(B) – n(A ∩ B) = 50 – 30 = 20

3. n (Chemical A or chemical B)
= n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 120 + 50 – 30 = 140.

Plus One Maths Sets Previous Questions And Answers Question 2.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer:
Let A – Apple juice; O – Orange juice be the sets.
Given; n(U) = 400; n(A) = 100;
n(O) = 150; n(A ∩ O) = 75
n (neither apple juice nor orange juice)
= n(A’ ∩ O’) = n((A ∪ O)’)
= n(U) – n(A ∪ O)
= 400 – [n(A) + n(O) – n(A ∩ O)]
= 400 – [100 + 150 – 75] = 400 – 175 = 225.

Plus One Maths Sets Questions And Answers Pdf Download Question 3.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speaks at least one of these two languages?
Answer:
Let F – French; S – Spanish be the sets.
Given; n(F) = 50; n(S) = 20;w(F ∩ S) = 10
n (speaks at least one of these two languages)
= n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
= 50 + 20 – 10 = 60.

Plus One Maths Textbook Questions And Answers Question 4.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis, how many like tennis only and not cricket? How many like tennis?
Answer:
Let C – Cricket; T – Tennis be the sets.
Given;
n(C’ ∪ T) = 65; n(C) = 40; n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) – 10 ⇒ n(T) = 35
n (tennis only and not cricket)
= n(T ∪ C’) = n(T) – n(T ∩ C) = 35 – 10 = 25.

Plus One Maths Chapter Wise Questions And Answers Pdf Question 5.
Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find

1. n(B) (1)
2. n(B – A) (1)
3. n(A – B) (1)

Answer:

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
   ⇒ 42 = 20 + n(B) – 4 ⇒ n(B) = 26
2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22
3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

Plus One Maths Chapter Wise Questions And Answers Question 6.
A = {x: x is a natural number less than 8}

1. Write in roster form. (1)
2. Write a subset of A containing all even numbers in A. (1)
3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]? (1)

Answer:

1. A = {1, 2, 3, 4, 5, 6, 7}
2. {2, 4, 6} or {2, 4, 6, 7}
3. 10. (since other are powers of 2.)

Plus One Maths Sets Four Mark Questions and Answers

Plus One Maths First Chapter Questions And Answers Question 1.
Observe the Venn diagram.

1. Write in roster form. (1)
2. Verify that (A – B) ∪ (A ∩ B) = A (2)
3. Find (A ∩ B)’ (1)

Answer:

1. A = {1, 3, 4, 8} ; B = {2, 3, 5}
2. A – B = {1, 4, 8}; A ∩ B = {3}
   ⇒ (A – B) ∪ (A ∩ B) = {1, 3, 4, 8}
   Hence; (A – B) ∪ (A ∩ B) = A
3. (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 8, 9}

Plus One Maths Sets Practice Problems Questions and Answers

Plus One Maths Chapter Wise Questions Question 1.
Write the following sets in roster form.

1. A = {x: x xis an integer and -3 < x < 7}
2. B = {x: x ∈ N; x ≤ 6}
3. C = {x : x is a vowel in English alphabet}
4. D = {x : x is a two-digit natural number such that the sum of its digits is 8}
5. E = {x: x ∈ Z; \(-\frac{1}{2}<x<\frac{9}{2}\)}

Answer:

1. A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
2. B = {1, 2, 3, 4, 5, 6}
3. C = {a, e, i, o, u}
4. D = {17, 71, 26, 62, 35, 53}
5. E = {1, 2, 3, 4}

Plus One Maths Text Book Questions And Answers Question 2.
Write the following sets in Set builder form.

1. A = {3, 6, 9, 12}
2. B = {2, 4, 8, 16, 32}
3. C = \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}\)
4. D = {5, 25, 125, 625}
5. E = {2, 4, 6……..} (1 score each)

Answer:

1. A = {x: x = 3n, n ∈ N, n< 4 }
2. B = {x: x = 2ⁿ; n ∈ N; n < 5 }
3. C = {x: x = \(\frac{n}{n+1}\); n ∈ N, n ≤ 6}
4. D = {x: x = 5ⁿ; n ∈ N, n ≤ 4 }
5. E = {x: x is an even number}

Plus One Maths Chapter Wise Questions And Answers Pdf Hsslive Question 3.
Write the following in interval form.

1. {x: x∈ R, -4 < x ≤ 6}
2. {x: x∈ R, 0 ≤ x < 7 }
3. {x: x∈ R, 3 ≤ x ≤ 4 } (1 score each)

Answer:

1. (-4, 6]
2. [0, 7)
3. [3, 4]

Plus One Maths Sets Questions And Answers Pdf Question 4.
Write the following in set builder form.

1. [0, 10]
2. [-2, 7)
3. (3, 4)

Answer:

1. {x: x∈ R, 0 ≤ x ≤ 10 }
2. { x: x ∈ R, -2 ≤ x < 7 }
3. {x: x ∈ R, 3 < x < 4 }

Plus One Maths Questions And Answers Question 5.
Find Set A, B and Universal set U (1 score each)

Answer:
A = {e, f, d}; B = {a, b, c, d} and U = {a, b, c, d, e, f, g, h}

Hsslive Maths Textbook Answers Plus One Question 6.
Write all subset of the following

1. {1, 2}
2. {a, b, c}
3. Φ (1 score each)

Answer:

1. Φ, {1}, {2}, {1, 2}
2. Φ, {a}, {b}, {c} ,{a, b}, {a, c}, {b, c}, {a, b, c}
3. Φ

Plus One Mathematics Questions And Answers Question 7.
Let A = {1, 2, {3, 4}, s, d, θ} , Which of the following statements are true/false and why?

1. 3 ∈ A
2. {1, {3, 4}} ∈ A
3. {1, 2, 3} ⊂ A
4. Φ ∈ A
5. 1 ⊂ A (1 score each)

Answer:

1. False
2. True
3. False
4. False
5. False

Important Questions For Class 11 Maths Sets Question 8.
If A = {1, 2, 4, 6, 7, 8}; B = {2, 5, 7, 9, 10} and C = {4 , 5, 9, 10}. Find

1. A ∪ B
2. B ∪ C
3. A ∪ C
4. A ∩ B
5. B ∩ C
6. A∪ B ∪ C
7. A ∩ B ∩ C
8. (A ∩ B) ∪ (C ∩ A) (1 score each)

Answer:

1. A ∪ B = {1, 2, 4, 5, 6, 7, 8, 9, 10}
2. B ∪ C = {2, 4, 5, 7, 9, 10}
3. A ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
4. A ∩ B = {2, 7}
5. B ∩ C = {5, 9, 10}
6. A ∪ B ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
7. A ∩ B ∩ C = Φ
8. (A ∩ B) ∪ (C ∩ A) = {2, 7} u {4} = {2, 4, 7}

Plus One Maths Important Questions And Answers Question 9.
If A = {2, 4, 6, 7, 8, 12}; B = {2, 7, 9, 10} and C = {5, 9, 10, 12}. Find

1. A – B
2. B – C
3. A – C
4. B – A
5. C – A
6. (A ∪ B) – C
7. A – {B ∩ C)
8. (A ∩ B) – (C ∩ A) (1 score each)

Answer:

1. A – B = {4, 6, 8, 12}
2. B – C = {2, 7}
3. A – C = {2, 4, 6, 7, 8}
4. B – A = {9, 10}
5. C – A = {5, 9, 10}
6. (A ∪ B) – C = {2, 4, 6, 7, 8, 9, 10, 12} – {5, 9, 10, 12} = {2, 4, 6, 7, 8}.
7. A – (B ∩ C) = {2, 4, 6, 7, 8, 12} – {9, 10} = {2, 4, 6, 7, 8, 12}
8. (A ∩ B) – (C ∩ A) = {2, 7} – {12} = {2, 7}

Plus One Maths Previous Questions And Answers Question 10.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; A = {2, 7, 9, 10} B = {5, 9, 10, 12} and C = {1, 4, 5, 7, 11}. Find

1. A’
2. B’
3. A’ – C
4. (B – A)’
5. B’ ∩ C’
6. (A ∪ B)’
7. A’ ∩ B’ (1 score each)

Answer:

1. A’ = { 1, 3, 4, 5, 6, 8, 11, 12}
2. B’ = { 1, 2, 3, 4, 6, 7, 8, 11}
3. A’ – C = {1, 3, 4, 5, 6, 8, 11, 12} – {2, 3, 6, 8, 9, 10, 12} = {1, 4, 5, 11}
4. (B – A)’ = {5, 12}’ = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11}
5. B’ ∩ C’ = {1, 2, 3, 4, 6, 7, 8, 11} – {2, 3, 6, 8, 9, 10, 12} = {2, 3, 6, 8}
6. (A ∪ B)’ = {2, 5, 7, 9, 10, 12}’ = {1, 3, 4, 6, 8, 11}
7. A’ ∩ B’ = {1, 3, 4, 5, 6, 8, 11, 12} ∩ {1, 2, 3, 4, 6, 7, 8, 11} = {1, 3, 4, 6, 8, 11}

Plus One Maths Text Book Answers Question 11.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have?
Answer:
Given; n(X ∪ Y) = 50; n(X) = 28; n(Y) = 32
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 50 = 28 + 32 – n(X ∩ Y)
⇒ n(X ∩ Y) = 60 – 50 = 10

Students can Download Chapter 4 Motion in a Plane Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 4 Motion in a Plane

Summary
Motion In A Plane Class 11 Notes Pdf Introduction
In this chapter, we will study, about vector, its ’ addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.

Plus One Physics Chapter 4 Notes Scalars And Vectors

a. Scalars:
A quantity which has only magnitude and no direction is called a scalar quantity.
Eg: length; volume, mass, time, work etc.

b. Vectors:
(i) The need for vectors:
In one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.

(ii) Vector:
A physical quantity which has both magnitude and direction is called a vector quantity.
Eg: Displacement, Velocity, Acceleration, Force, momentum.

1. Position and Displacement Vectors:
Position vector:
Consider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.

A vector representing the position of an object P with respect to an origin O is called position vector \(\overrightarrow{\mathrm{OP}}\) of the object. This position vector may be represented
by an arrow with tail at O and head at P.

The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\(\overrightarrow{\mathrm{OP}}\)|.

Displacement vector:

Consider the motion of an object in a plane. Let P be the position of a moving object at a time t and p¹ that at a later time t¹. \(\overrightarrow{\mathrm{OP}}\) and \(\overrightarrow{\mathrm{OP}^{1}}\) are the position vectors at time t and t¹ respectively. So the vector \(\overrightarrow{\mathrm{PP}^{1}}\) is called displacement vector corresponding to the motion in the time interval (t – t¹).

2. Equality of vectors:
Two vectors are said to be equal if they have the same magnitude and direction.

The above figure shows two vectors \(\vec{A}\) and \(\vec{B}\) having the same magnitude and direction.
∴ \(\vec{A}\) = \(\vec{B}\).

Motion In A Plane Class 11 Notes Hsslive Question 1.
Observe the following figures (a) and (b) and find which pair does represents equal vectors?

Answer:
Figure a represent that A and B are equal vectors. Two vectors A¹ and B¹ are unequal, because they were in different directions.

Multiplication Of Vectors By Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number I gives a vector whose magnitude is changed by the factor λ.

The direction λ\(\vec{A}\) is the same as that of \(\vec{A}\).
Examples:

A vector \(\vec{A}\) and the resultant vector after multiplying \(\vec{A}\) by a positive number 2.

A vector A and resultant vector after multiplying it by a negative number-1 and -1.5.

Addition And Subtraction Of Vectors – Graphical Method
Vectors representing physical quantities of the same dimensions can be added or subtracted. The sum of two or more vectors is known as their resultant.

1. When two vectors are acting in the same direction:

2. When two vectors act in opposite direction:
In this case, the angle between the vectors is 180°.

The resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.

3. When two vectors are inclined to each other:
The sum of two vectors inclined at an angle q can be obtained either by

• the law of triangle of vectors
• the parallelogram law of vectors

(i) Triangle method:
This law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

(ii) Parallelogram law of vector addition:
This law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

To find \(\vec{A}\) + \(\vec{B}\), we bring theirtails to a common origin Q as shown below.


The diagonal of parallelogram OQSP, gives the resultantof (\(\vec{R}\) = \(\vec{A}\) + \(\vec{B}\)) of two vectors \(\vec{A}\) and \(\vec{B}\).
Note: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.

4. Substraction of vectors:

To substract \(\vec{B}\) from \(\vec{A}\), reverse the direction of \(\vec{B}\).

Then add –\(\vec{B}\) with \(\vec{A}\) using parallelogram law or tri¬angle law.

The resultant of \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{R}\).
Null vector or zero vector:
A vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \(\vec{O}\). Since the magnitude is zero, we don’t have to specify its direction.
Properties of null vector:

Class 11 Physics Motion In A Plane Notes Pdf Question 2.
Explain a zero vector using an example.
Answer:
Suppose that an object which is at P at time t, moves to p¹ and then comes back to P. In this case displacement is a null vector.

Resolution Of Vectors Unit Vectors
A vector divided by its magnitude is called unit vector along the direction of that vector. A unit vector in the direction of \(\vec{A}\) is written as \(\hat{A}\).

Orthogonal unit vectors:

In the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) respectively and are known as orthogonal unit vectors.
For unit vectors

Resolution of vector into rectangular components:
The components of a vector in two mutually perpendicular directions are called its rectangular components.
Explanation

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in x-y plane as shown in figure. To resolve \(\overrightarrow{\mathrm{A}}\), draw lines from the head of \(\overrightarrow{\mathrm{A}}\) perpendicularto the coordinate axes as shown below.

The quantities A_x and A_y are called x and y components of the vector \(\overrightarrow{\mathrm{A}}\). Hence the vector \(\overrightarrow{\mathrm{A}}\) can be written in terms of rectangular components as

Magnitude of \(\overrightarrow{\mathrm{A}}\):

From the figure, the magnitude of \(\overrightarrow{\mathrm{A}}\) can be written as,

Vectors Physics Class 11 Notes Pdf Question 3.
A vector \(\overrightarrow{\mathrm{A}}\) in xyz plane is given below. A_x, A_y and A_z are the perpendicular components in x,y and z directions respectively.

1. Write \(\overrightarrow{\mathrm{A}}\) in terms of rectangular components.
2. Write the magnitude of \(\overrightarrow{\mathrm{A}}\).

Answer:

The magnitude of vector \(\overrightarrow{\mathrm{A}}\) is

Vector Addition – Analytical Method
The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use analytical method to add vectors.
Explanation

The vectors obey commutative and associative laws. Hence

Projectile Motion Class 11 Notes Question 4.
Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{A}}\) in terms of their magnitudes and angle between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle q. Using the parallelogram method of
vectors, the resultant vector \(\overrightarrow{\mathrm{R}}\) can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosq and SN = Bsinq
Substituting these values in eq.(1), we get
OS² = (OP + Bcosq)² + (Bsinq)²
But OS = R and OP = A
R² = (A + Bcosq)² + B²sin²q
= A² + 2ABcosq + B²cos²q + B²sin²q
R² = A² + 2 ABcosq + B²

The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \overrightarrow{\mathrm{A}}. From the right angled triangle OSN,

But SN = Bsinq PN = Bcosq

Motion In A Plane

1. Position vector and displacement vector Position vector:

Consider a small body located at P with reference to the origin O. The position vector of the point ‘P’

Displacement vector


where Dx = x¹ – x¹, Dy = y¹ – y
Velocity:
If Dt is the time taken to reach from P to P¹
The average velocity, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\Delta r}}{\Delta \mathrm{t}}\) ____(3)
Substitute eq.(2) in eq.(3), we get

The direction of average velocity is the same as that of \(\overrightarrow{\Delta r}\).
The instantaneous velocity can be written as

Acceleration:
If the velocity of an object changes from \(\overrightarrow{\mathrm{v}} \text { to } \overrightarrow{\mathrm{v}^{1}}\) in time Dt, then its average acceleration is given by

Instantaneous acceleration:
The acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Motion In A Plane With Constant Acceleration
Consider an object moving in xy plane with constant acceleration ‘a’. Let \(\vec{u}\) be the initial velocity at t=0 and \(\vec{v}\) be the final velocity at time t.
Then by definition acceleration

In terms of components
v_x = u_x + a_xt
v_y = u_y + a_yt
Displacement in a plane
If \(\overrightarrow{\mathrm{r}_{0}}\) and \(\vec{r}\) be position vectors of particle at t = 0 and time t respectively, then
displacement = \(\vec{r}-\vec{r}_{0}\) _______(1)
For uniformly accelerated motion, displacement,

In terms of components
x = x₀ + u_xt + 1/2 a_xt²
y = y₀ + u_yt + 1/2 a_yt²
The eq.(2) shows that, the above motion in xy plane can be treated as two separate one dimensional motions along two perpendicular directions.

Relative Velocity In Two Dimensions
Consider two bodies A and B moving along a plane with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\). Then velocity of A relative to that of B is,

Similarly velocity of B relative to that of A

Projectile Motion
Projectile:
A body is projected into air and is allowed to move under the influence of gravity is called projectile.

Consider a body which is projected into air with a velocity u at an angle q. The initial velocity ‘u’ can be divided into two components ucosq along horizontal direction and using along vertical direction.

1. Time of flight:
The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinq, we get
0 = usinqt – 1/2gt²
1/2 gt² = usinqt

2. Vertical height:
Vertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v² = u² + 2as
When we substitute v=0, a = -g, s = H and u = usinq, we get
0 = (usinq)² + 2 – g × H
2gH = u²sin²q

3. Horizontal Range:
If we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

The eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q₀ = 45°.
The maximum horizontal range

Equation for path of projectile

Motion In A Plane Class 11 Pdf Question 5.
What is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
Answer:
S = ut+ 1/2at²
y = usinqt – 1/2gt²
But we know horizontal displacement, x = ucosq × t

In this equation g, q and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

Uniform Circular Motion
The motion of an object along the circumference of a circle is called circular motion.
Uniform circular motion:
When an object follows a circular path at a constant speed, the motion is called uniform circular motion.
Period:
The time taken by the object to complete one full revolution is called the period.
Frequency:
The number of revolutions completed per second is called the frequency u of the circular motion.
If the period of a circular motion isT, its frequency

Angular Displacement (Dq):
The angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
Angular velocity:
The rate of change of angular displacement is called the angular velocity.

If T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
Therefore in one second it sweeps out an angle \(\frac{2 \pi}{T}\).
∴ Angular velocity of the object

Expression for velocity and acceleration in uniform circular motion:

The direction of velocity is in the direction of tangent at that point. The change in velocity vectors \((\overrightarrow{\Delta v})\) is obtained by triangle law of vector as shown in figure (b).

a. Speed and angular speed in uniform circular motion:
Let the Dq be the angle constructed by the body during the time interval ∆t. The angular velocity can be written as

If the distance travelled by the object during the time Dt is Dr (ie. PP¹ = Ds) then speed

But Ds = RDq
where R = \(|\vec{r}|=|\overrightarrow{r^{\prime}}|\)
Substituting Dr = RDq in eq.(1)
we get

b. Acceleration in uniform circular motion:

The direction of this acceleration should be in the direction of \(\overrightarrow{\Delta V}\). The fig(b) shows that \(\overrightarrow{\Delta V}\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration.

The force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as
F = ma_c

But ω = \(\frac{V}{R}\). Hence we get K

Students can Download Chapter 1 Introduction to Accounting Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 1 Introduction to Accounting

Plus One Accountancy Introduction to Accounting One Mark Questions and Answers

Plus One Business Studies Chapter 1 Notes Pdf Question 1.
Who among the following are not a user of accounting information?
(a) Management
(b) Investors
(c) Advertisers
(d) Lenders
Answer:
(c) Advertisers

Plus One Business Studies Chapter 1 Notes Question 2.
Spot the odd one out and state reason
(a) Loose Tools
(b) Copy Write
(c) Patent
(d) Goodwill
Answer:
(a) Loose Tools, is a Fixed assed, all others are intangible assets.

Plus One Business Studies Chapter 1 Question 3.
A Person who owes money to the business is a
(a) Debtor
(b) Investor
(c) Creditor
(d) Borrower
Answer:
(a) Debtor

Plus One Business Studies Notes Question 4.
Book-keeping is concerned with
(a) Analysis of transaction
(b) Recording of transaction
(c) Classification of transaction
Answer:
(b) Recording of transaction

Plus One Business Studies Notes In English Question 5.
Amount spent for purchasing fixed asset is a
(a) Revenue Expenditure
(b) Capital Expenditure
(c) Deferred Revenue Expenditure
Answer:
(b) Capital Expenditure.

Plus One Business Studies Chapter 1 In Malayalam Question 6.
Which quantitative characteristics of accounting in-formation is reflected when accounting information is clearly presented?
(a) Understandability
(b) Relevance
(c) Comparability
(d) Reliability
Answer:
(a) Understandability.

Plus One Business Studies Notes Pdf Question 7.
Which of the following is an example of a business transaction?
(a) Appointed Mr.Ram as the Manager of the business with a salary of Rs. 15,000.
(b) Obtain a loan of Rs. 1,00,000 to the business from Bank of India.
(c) Sent a quotation to Matha Traders worth Rs. 20,000.
Answer:
(b) Obtained a loan of Rs. 1,00,000 from Bank of India. Loan is taken meant for business. So it is a business transaction.

Hsslive Plus One Business Studies Notes Question 8.
Find the odd one out and state the reasons,
(a) share capital
(b) Debentures
(c) Sundry creditors
(d) Long-term loans.
Answer:
(c) Sundry creditors, all others are long term liabilities.

Business Studies Plus One Notes Question 9.
Value of goods remaining unsold at the end of an accounting period is termed as …………
Answer:
Closing Stocks

Plus One Business Studies Chapter Wise Notes Question 10.
Arun, a sole trader, draw Rs. 500 from the business for paying tuition fees to his child. This amount is termed as ……….
Answer:
Drawings.

Nature And Purpose Of Business Notes Pdf Question 11.
Assets minus liabilities are called ……….
Answer:
Capital.

Nature And Purpose Of Business Class 11 Notes Pdf Download Question 12.
………….. assets are those assets, which do not have any real value.
Answer:
Fictitious Assets.

Hss Live Business Studies Plus One Notes Question 13.
The amount earned by a business concern through sale of its products or providing services to customers is called ……
Answer:
Revenues.

Plus One Business Studies Notes In English Pdf Question 14.
The assets bought for long-term use in the business are termed as …………….. assets
Answer:
Fixed.

Hsslive Business Studies Plus One Notes Question 15.
Analysis of recorded data to bring entries of similar nature to one plane is called …………..
Answer:
Classifying.

Hsslive Plus One Business Studies Malayalam Notes Question 16.
A Person who is entitled to get money from the business is termed as …………
Answer:
Creditor

Question 17.
Information in financial reports is based on …………….. transaction.
Answer:
Economic.

Question 18.
All claims against the business are called ………………..
Answer:
Equity

Question 19.
The transaction is one wherein payment or receipt of money is postponed for a future date.
Answer:
Credit transaction.

Question 20.
Mr. Ismail, who is the owner of a Provision shop, took 50 kg. of rice worth Rs. 600 for his house-hold use. He should record this as
Answer:
Drawings.

Question 21.
Ravi, a trader purchased 100 notebooks from ‘Shyni stores’ on credit. How is Shyni stores related to Ravi?
Answer:
Shyni stores is the creditor of Ravi.

Question 22.
Who was the inventor of double-entry bookkeeping?
Answer:
Luca Pacioli

Question 23.
Expand AICPA
Answer:
The American Institute of Certified Public Accounts.

Question 24.
Identify the events not used to the accounting treatment.
(a) Commenced business with cash
(b) Bought Machinery for cash
(c) Cash Purchase of goods.
(d) The firm appointed an efficient Manger.
Answer:
(d) The firm appointed an efficient Manger.

Plus One Accountancy Introduction to Accounting Two Mark Questions and Answers

Question 1.
Define Accounting.
Answer:
According to American Institute of certified Public Accountants, “Accounting is the art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are in part, at least, of a financial character and interpreting the results thereof.”

Question 2.
‘Accounting is the language of the business’. Why?
Answer:
Accounting is the language of the business:
The performance of business in terms of profit or losses is conveyed to users of accounting information in a systematic manner. The financial position of the business concerned is revealed through accounting information.

Question 3.
“Raju sold goods to Rahim on credit”. What relation exists between them? What are the accounting terms involved in it?
Answer:

• Rahim – Debtor
• Raju – Creditor

Question 4.
You are the accountant of a firm. What are the functions to be performed by you?
Answer:
Accounting provides information regarding the financial status of a business and results of its operations. The following are the important functions of accountant of a firm.

1. Recording transactions by referring source documents.
2. Preparing journal, subdivision of journal.
3. Preparing ledger accounts.
4. Summarising.
5. Making statements of interpretation 0 Reporting to Management.

Question5.
Users of accounting are classified as under:

1. Internal users – Management, Investor, Creditor, Bank, Employees, Stock exchange.
2. External Users – Customers, Government, Researchers, Lenders.

Do you agree with this classification, if not correct it?
Answer:
No,
1. Internal Users:

• Management
• Employees

2. External Users:

• Investors
• Creditors
• Bank
• Stock Exchange
• Government
• Customers
• Lenders
• Researchers

Question 6.
All business transactions are events. But all events are not business transactions. Comment.
Answer:
Events can be anything, some events can be expressed in monetary terms while others are not. Only those events which can be expressed in money, terms are business transactions. Transaction is an event or economic activity of a businessman in his business having exchange of money or money’s worth. While events is part of the business transaction.

Question 7.
Classify the following expenses into capital expenditure and revenue expenditure.

• Machinery purchased
• Rent paid
• Interest paid
• Purchased building
• The amount for repair of the building

Answer:
1. Capital Expenditure:

• Machinery Purchase
• Purchased building

2. Revenue Expenditure:

• Rent paid
• Interest
• paid Amount for repair of building

Question 8.
How will you define Revenue and Expenses?
Answer:
1. Revenue:
These are the amounts earned by a business concern through sale of its products or providing services to customers. The common items of revenues are sales, commission received, rent received, interest received, etc.

2. Expenses:
The amount spent in the process of earning revenue is termed as expenses. Examples are Wages, Salaries, rent, Interest paid, electricity charges, etc.

Question 9.
What is a capital expenditure? Give some examples?
Answer:
Capital expenditure represents the amount spent for the acquisition of assets, the benefit from which is derived over a period that extends beyond the accounting year. It is long term in nature.
Examples: Furniture purchased, Land Purchased, Building purchased, etc.

Question 10.
Explain the meaning of Gain and Profit.
Answer:
1. Profit: The excess of revenues of a period over its related expenses during an accounting year is profit. Profit increases the investment of the owners.

2. Gain: A profit that arises from events or transactions which are incidental to business such as sale of fixed assets, winning a court case, receipt of interest and dividend, etc. Gain is irregular in nature. Gains are part of capital receipt. Gains are also known as “non-operating income.”

Question 11.
Match the following.

Answer:

Question 12.
Name the branches of accounting.
Answer:

1. Financial Accounting
2. Cost Accounting
3. Management Accounting

Plus One Accountancy Introduction to Accounting Three Mark Questions and Answers

Question 1.
Classify the following assets into suitable head Goodwill, Building, Land, Patent, Cash, Oilwell, Copy-write, Debtors, Stock, mines, Bill receivable, Preliminary expenses.
Answer:

Question 2.
Define assets, Liabilities, and capital.
Answer:
1. Assets:
Assets are properties and things of value owned by the business which can be expressed in monetary terms. Examples of Machinery, Buildings, Stock, Debtors, Furniture, etc.

2. Liabilities:
Liabilities are the obligations that an enterprise owes. These represent the amount payable by the business in the future. They represent the claim against the asset of business. Examples Loans, Creditors, Bills payable, etc.

3. Capital:
Capital is the investment made by the owners for use in the business. It is owner’s claim on the total assets of the business and is also called “owners equity”.

Question 3.
Distinguish between:

1. Goods and Assets
2. Expense and Loss

Answer:
1. Goods and Assets:

• Goods refer to things in which the trader deals. But assets refers to things with which the trader deals.
• Goods are meant for resale, while assets are kept in the business permanently with the help of which the business is carried on.

2. Expense and Loss:

• Costs incurred by a business in the process of earning revenue are known as expense.
• The excess of expenses of a period over its related revenues is termed as loss. It decreases in owner’s equity.

Plus One Accountancy Introduction to Accounting Four Mark Questions and Answers

Question 1.
Accounting has certain objectives to business Explain.
Answer:
The following are the important objectives of accounting.

1. Keeping of records of business transactions.
2. Ascertainment of Profit or Loss.
3. Ascertainment of financial position of business enterprises.
4. Providing meaningful information to different groups of people having interest in the business.

1. Keeping of records of business transaction:
The main purpose of accounting is to identify business transactions of financial nature and enter into appropriate books of accounts. The accounting records should be made properly and systematically, so that requisite information may be obtained at a glance.

2. Ascertainment of Profit or Loss:
The result of business (Profit or Loss) is available from the statement prepared for ascertaining it, called the Profit and Loss Account.

3. Ascertainment of financial position:
At the end of an‘ accounting year, a position statement known as the ‘Balance Sheet’ is prepared. The value of assets and liabilities are depicted in the Balance Sheet. The Balance sheet gives a true and fair view of the state of affairs of the concern.

4. Providing meaningful information to different groups of people having interest in the business:
Accounting records provide meaningful information to different groups of people having interest in the business.

Question 2.
Accounting information must possess certain qualitative characteristics. What are they?
Answer:
The following are the qualitative characteristic of accounting information.

1. Reliability: Accounting information will be reliable if it is free from error and faithfully represents what it seeks to represent.
2. Relevance: Information should be relevant and must be available in.time.
3. Understandability: Accounting information that is relevant must be capable of being understood by all its users.
4. Timeliness: Information must be available timely. If not, it loses its ability to influence decision.
5. Comparability: Accounting information should facilitate inter-firm comparison as well as interfirm comparison.
   Maximum Cputtishers

Question3.
Accounting and Book-keeping are viewed as distinct functions. Mention any four differences between Accounting and Book Keeping.
Answer:

Question 4.
“Accounting gives number of advantages to the business”. What are the important advantages of accounting?
Answer:
The following are the advantages of accounting
1. Provide Quantitative information:
Accounting helps in gathering quantitative information on profits earned by the business or loss sustained by them.

2. Helps in ascertaining financial position of the business that is, total assets owned and total liabilities owed.

3. Helps in making a systematic record of transactions, which can be used for future reference and appropriate retrieval.

4. Acts as an information system:
It provides adequate information to the interested users in a processed form.

5. Beneficial to different interested users of accounting information.

Question 5.
Accounting has certain ‘Limitations’. Explain.
The following are the limitations of accounting.
Answer:
1. It records only transactions which can be recorded in monetary terms:
Qualitative aspects like managerial skill, Services of experts, etc. are not recorded.

2. Accounting is a post mortem survey:
It records events as they have taken place. For example, expenses are recorded as incurred, assets are recorded at their cost of purchase. There is no scope for ascertaining what the appropriate expenditure or cost of acquisition should have been.

3. Effect of price level changes are not considered:
Transactions are always recorded in the books at cost price and not at market price.

4. Inexactness:
Accounting transactions are not exact. Different firms have their own different methods, so the results of the business will change in the practice.

Plus One Accountancy Introduction to Accounting Six Mark Questions and Answers

Question 1.
What are the different types of assets? Explain briefly.
Assets are things of value owned. They may be subdivided into the following.
Answer:
1. Fixed Assets:
Fixed Assets are assets held on long term basis, such as land, buildings, machinery, plant, furniture, etc. These assets are used for the normal operations of the business.

2. Current Assets:
These are assets held on a short-term basis such as debtors, bills receivables, stock, cash in hand, cash at bank, etc. It is also known as “Floating asset.”

3. Fictitious Assets:
These are those assets, which do not have a physical form. They do not have any real value. Actually, they are not the real assets but they are called assets on legal and technical grounds. Examples are preliminary expenses, discount on issue of shares or debentures, etc.

4. Tangible Assets:
Assets having physical existence which can be seen, touched are known as tangible assets. These assets are land, building, plant, equipment, etc.

5. Intangible Assets:
These assets have no physical existence which cannot be touched, seen or felt. Examples are Goodwill, trademark, patent, copyright.

6. Wasting Assets:
Assets, whose value goes on declining with the passage of time, are known as wasting assets. Mines, oilwells, quarries are its examples.

7. Liquid Assets:
Liquid assets are those assets, which can be converted into cash at short notice. The examples of liquid assets are cash in hand, cash at bank, debtors, bills receivable, etc. Liquid assets = Current Assets – (Stock + Prepaid Expenses)

Question 2.
“Accounting provides information to various users.” Discuss accounting as an information system.
Answer:
Accounting plays a significant role in society by providing information to management at all levels (internal users) and to those having a direct financial interest in the enterprise (external users), such as present and potential investors, creditors. Accounting information is also important to those having an indirect financial interest, such as regulatory agencies, tax authorities, customers, labour unions, stock exchange, and others.

Internal users, mainly management, need timely information on cost of sales, profitability, etc. for planning, controlling and decision making. External users who have limited authority, ability and resources to obtain the necessary information have to rely on financial statements. The external users are interested in the following.

1. Investors and Potential investors:
Information on the risks and returns on investments.

2. Suppliers and Creditors:
Information on whether amounts owed will be repaid when due and on the continued existence of the business.

3. Customers:
Information on the continued existence of the business and thus the profitability of a continued supply of products, parts, and after-sales services.

4. Employees:
They are interested in getting their salary, welfare measures, bonus, working conditions, etc. which are all related to financial performance of the business.

5. Lenders:
Information on the creditworthiness of the business and its ability to repay loans and pay interest.

6. Government and other regulators:
Information on the allocation of resources and the compliance to regulators.

Students can Download Chapter 1 Nature and Purpose of Business Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 1 Nature and Purpose of Business

1st Standard Malayalam Textbooks Chapter 1 Contents

• Economic and non-economic activities
• Business and its characteristics
• Business, Profession and Employment
• Classification of business activities
• Industry and commerce
• Objectives of business
• Business risk
• Factors to be considered before starting a business.

Activities which human beings undertake are known as human activities. We can divide these activities into two categories.

1. Economic activities
2. Non-economic activities

1. Economic Activities:
The human activities that are undertaken with an objective to earn money or livelihood is known as economic activities. eg: A worker working in a factory, a doctor operating in his hospital, a manager working in the office, a teacher teaching in a school etc.

2. Non-economic activities:
Activities are undertaken to derive psychological satisfaction are known as non economic activities. eg: Mother preparing food for her children, Praying, listening to radio or watching television, playing football with friends, etc.

Types of economic activities:

• Business
• Employment
• Profession

1st Standard Malayalam Chapter 1 Business:
Business may be defined as an economic activity involving the production or purchase and sale of goods and services with the main object of earning profit by satisfying human needs in the society.
Characteristics of business

1. Business is an economic activity with the object of earning profit.
2. Business includes all the activities concerned with the production or procurement of goods and services.
3. There should be sale or exchange of goods and services for the satisfaction of human needs.
4. Business involves dealings in goods or services on a regular basis. Normally, one single transaction of sale or purchase is not treated as business.
5. One of the main objectives of business is to earn maximum profit.
6. Business involves risk and uncertainty of income. Risk means the possibility of loss due to change in consumer taste and fashion, strike, lockout competition, fire, theft etc.

Malayalam Text Book Std 1 Pdf Download Employment:
Employment refers to that type of economic activity in which people engage in some work for others regularly and get salary or wages in return of their services.
Characteristics of Employment

1. There must exist employer-employee relationship.
2. There must be a service contract between the employer and employee.
3. Employees get salary or wages for their services
4. Regularity in service.

First Standard Malayalam Textbooks Chapter 1 Profession:
Profession refers to an occupation which requires specialized knowledge, skip and training. Its objective is to provide service to the society.
Characteristics of Profession

1. A profession requires specialized knowledge, training and skill
2. The membership of a professional body is a must
3. Professionals have a code of conduct
4. They charge fee in return of their service.

Comparison of Business, Profession and Employment:

Classification of Business Activities: Business activities may be classified into two categories

• Industry
• Commerce.

Chart showing business activities

1 Standard Malayalam Textbooks Chapter 1 Industry:
Industry refers to economic activities, which are connected with conversion of resources into useful goods. Industries may be divided into 3 categories.
They are
1. Primary industries:
Primary industries are connected with the extraction and production of natural resources and reproduction and development of living organisms, plants, etc. Such industries are further divided into two.
(i) Extractive industries:
These industries extract products from natural resources. eg: mining, farrqing, hunting, fishing etc.

(ii) Genetic industries:
These industries are engaged in activities like rearing and breeding of animals, birds and plants. eg: diary faming, paultry farming, floriculture, pisciculture etc.

2. Secondary industries:
Secondary industries deal with materials extracted at the primary stage. Such goods may be used for consumption or for further production. Secondary industries are classified into two.
They are:
(i) Manufacturing industries:
Manufacturing industries engage in converting raw materials into finished goods. eg: Conversion of rubber into cotton, timber into furniture rubber into tyres etc. Manufacturing industries may be further divided into four categories. They are,

• Analytical industry which analyses and separates different elements from the same materials. eg: Oil refinery
• Synthetical industry which combines various ingredients into a new product. eg: cement
• Processing industry which involves successive stages for manufacturing finished products. eg: Sugar and paper industry.
• Assembling industry which assembles different component parts to make a new product. eg: television, car, computer, etc.

(ii) Construction industries:
These industries are involved in the construction of buildings, dams, bridges, roads etc.

3. Tertiary industries:
These are concerned with providing support services to primary and secondary industries. eg: Transport, banking, insurance, warehousing, communication, advertising etc.

1st Standard English Book Chapter 1 Commerce:
Commerce is defined as all activities involving the removal of hindrances in the process of exchange of goods. It includes all those activities, which are necessary for the free flow of goods and services from the producer to the consumer. Commerce includes trade and auxiliaries to trade.
Commerce = Trade + auxiliaries to trade
Functions of commerce:

Malayalam 1st Standard Chapter 1 Trade:
Trade refers to sale, transfer or exchange of goods. Trade may be classified into two broad categories
They are:

1. Internal trade
2. External trade

1. Internal, domestic ot dome trade:
is concerned with the buying and selling of goods and services within the geographical boundaries of a country. This may further be divided into two. They are:-
(a) Wholesale trade:
Under wholesale trade, the trader purchases goods in large quantities from the producers, and sells them in smaller quantities to the retailers.

(b) Retail trade:
Under the retail trade, the trader buys in comparatively smaller quantities from the wholesalers or producers and sells them to ultimate consumers.

2. External or Foreign trade:
Foreign trade consists of exchange of goods and services between two or more countries. Foreign trade may be divided in to three.

• Import trade: If goods are purchased from a foreign country, it is called import trade.
• Export trade: When goods are sold to a foreign country, it is known as export trade.
• Entrepot trade: When goods are imported for export to other countries, it is known as entrepot trade.

Auxiliaries to Trade (Aids to trade):
Activities which assist trade are called aids to trade or auxiliaries to trade.

1. Transport & Communication:
Transport facilitates the movement of raw material to the place of production and the finished products from factories to the place of consumption. Communication helps the producers, traders and consumers to exchange information with one another.

2. Banking & Finance:
Banking helps business activities to overcome the problem of finance. Commercial banks lend money in the form of overdraft, cash credit, loans and advances etc… and they also provide many services required for the business activity.

3. Insurance:
The goods may be destroyed while in production process or in transit due to accidents, or in storage due to fire or theft, etc. Insurance provides protection in all such cases.

4. Warehousing:
The goods should be stored carefully from the time they are produced till the time they are sold. This function is performed by warehouses.

5. Advertising:
Advertising helps in providing information about available goods and services and create in them a strong desire to buy the product.

Multiple Objectives of Business:
The main objectives of a business are:
1. Market standing:
A business firm can succeed only when it has a good market standing. Market standing refers to the position of an enterprise in relation to its competitors.

2. Innovation:
Innovation means developing new product or services orfinding new ideas and new methods of production and distribution. Innovation accelerates the growth of an enterprise.

3. Productivity:
Productivity is ascertained by comparing the value of output with the value of input. Every enterprise must aim at greater productivity through the best use of available resources.

4. Physical and financial resources:
The business must aim at maximum utilization of available physical and financial resources, i.e. men, material, money and machine in the best possible manner.

5. Earning Profit:
Earning maximum profit is the primary objective of every business. Profit is required for survival and growth of a business.

6. Manager performance and development:
Efficient managers are needed to conduct and co-ordinate business activities. So it is the objective of an enterprise to implement various programs for motivating the managers.

7. Worker performance and attitude:
Every enterprise must aim at improving its workers performance by providing fair salary, incentives, good working conditions, medical and housing facilities.

8. Social responsibility:
It refers to the obligation of business firm to contribute resources for solving social problems and work in a socially desirable manner.

Malayalam Text Book Std 1 Chapter 1 Business Risks:
The term ‘business risks’ refers to the possibility of inadequate profits or even losses due to uncertainties or unexpected events. Business enterprises may face two types of risk, i.e. speculative risk and pure risk.

Speculative risks involve both the possibility of gain as well as the possibility of loss. It arise due to change in demand, change in price etc. Pure risks involve only the possibility of loss or no loss. The chance of fire, theft or strike is examples of pure risks.
Nature of Business Risks:

1. Business risks arise due to uncertainties.
2. Risk can be minimized, but cannot be eliminated. It is an essential part of business.
3. Degree of risk depends mainly upon the nature and size of business.
4. Profit is the reward for risk taking.

Causes of Business Risks: Business risks arise due to a variety of causes.
They are:

1. Natural Causes: it includes natural calamities like flood, earthquakes, lightning, heavy rains, famine, etc.
2. Human Causes: Human causes include dishonesty, carelessness or negligence of employees, strikes, riots, management inefficiency, etc.
3. Economic causes: These include change in demand, change in price, competition, technological changes etc.
4. Political Causes: Change in Govt, policies, taxation, licensing policy etc.

Starting a Business – Basic Factors:
Factors to be considered for starting a business:
1. Selection of line of business:
The first thing to be decided by any entrepreneur of a new business is the nature and type of business to be undertaken.

2. Size of the firm:
If the market conditions are favorable, the entrepreneur can start the business at a large scale. If the market conditions are uncertain and risks are high, a small size business would be better choice.

3. Choice of form of ownership:
The selection of a suitable form of business enterprise i.e. Sole proprietorship, Partnership or a Joint stock company is an important management decision. It depends on factors like nature of business, capital requirements, liability of owners, legal formalities, continuity of business etc.

4. Location of business enterprise:
Availability of raw materials and labour, power supply and services like banking, transportation, communication, warehousing, etc., are important factors while making a choice of location.

5. Financing:
Proper financial planning must be done to determine (a) the requirement of capital, (b) source from which capital will be raised and (c) the best ways of utilizing the capital in the firm.

6. Physical facilities:
Availability of physical facilities including machines and equipment, building and supportive services is a very important factor to be considered at the start of the business.

7. Plant layout:
Layout means the physical arrangement of machines and equipment needed to manufacture a product.

8. Competent worked force:
Every enterprise needs competent and committed employees to perform various activities so that physical and financial resources are converted into desired outputs.

9. Tax planning:
The promoter must consider in advance the tax liability under various tax laws and its impact on business decision.

Students can Download Chapter 1 Biological Classification Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 1 Biological Classification

Two kingdom classification:

• It was proposed by Linnaeus, include two Kingdoms-Plantae & Animalia.
• In two kingdom classification following things are not considered

Cell structure, nature of wall, mode of nutrition, habitat methods of reproduction and evolutionary relationship .

Five kingdom classification:

• It was first proposed by R H.Whittaker (1969).
• It includes Monera, Protista, Fungi, Plantae and Animalia
• Blue green algae are placed in kingdom Monera.
• Chlamydomonas, Chlorella with Paramoecium and Amoeba are placed in kingdom protista
• Chlorophyll less and non cellulosic (chitin)type plants are placed in kingdom Fungi
• All photosynthetic plants are placed in-kingdom plantae
• All animals with mode of nutrition(ingestion) placed in kingdom animalia.
• The characteristics of classification are

Kite Victers Plus One Botany Notes Pdf Kingdom monera:
Types of bacteria:
Based on shape, bacteria are of 4 types

1. Spherical – Coccus
2. Rod – shaped – Bacillus
3. Comma – shaped – Vibrium
4. Spiral – Spirillum.

Based on nutrition, bacteria are of 3 types:

1. Photosynthetic autotrophic: Bacteria can synthesise their own food by using chlorophyll in the presence of light.
2. Chemosynthetic autotrophic: Bacteria can synthesise their own food from inorganic substrates.
3. Heterotrophs: They are depend on other organisms for food.

Plus One Botany Notes Archaebacteria:

• These bacteria can live in extreme conditions.
• Its cell wall structure is different from other bacteria

Types of archaebacteria:

1. Halophiles: They are found in salty areas
2. Thermoacidophiles: They are found in hot springs.
3. Methanogens: They are found in marshy areas and guts of ruminant animals eg-cows, buffaloes etc.

Importance in industry:
They are responsible for the production of methane (biogas) from the dung.

Eubacteria Cyanobacteria (blue-green algae):

• They have chlorophyll a similar to green plants called as photosynthetic autotrophs.
• Some of them found in polluted water bodies.

Significance of cyanobacteria:
They can fix atmospheric nitrogen in specialised cells called heterocysts and increases fertility of soil eg: Nostoc and Anabaena

Chemosynthetic autotrophs:
They oxidise various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for their ATP production.

Significance chemosynthetic autotrophs:
They play a great role in the recycling of nutrients like nitrogen, phosphorous, iron and sulphur.

Heterotrophic bacteria:

• They depend upon others for getting energy.
• Most of them are decomposers.
• They are helpful in making curd from milk, production of antibiotics, fixing nitrogen in legume roots, etc.

Disease caused by bacteria:
Some are pathogens causing disease to plants and animals, eg: Cholera, typhoid, tetanus, and citrus canker in plants.

Cell division in bacteria:
Reproduction:

• Bacteria reproduce mainly by fission.
• Some bacteria produce spores during unfavourable conditions.
• In sexual reproduction, transfer of DNA from one bacterium to the other takes place

Mycoplasmas:
Salient features:

• They are the smallest living cells can survive without oxygen
• They are pathogenic in animals and plants
• They lack a cell wall.

Plus One Botany Chapter Wise Notes Kingdom Protista:
Salient features:

• They are single-celled eukaryotes.
• Their cell body contains a well defined nucleus and other membrane-bound organelles.
• They are mainly aquatic.

Types of protist:
The kingdom include chrysophytes, Dianoflagellates, Euglenoids, Slime moulds and Protozoans.

Plus One Botany Chapters Chrysophytes:

• They are diatoms(chief ‘producers’ in the oceans) and golden algae (desmids).
• Most of them are photosynthetic.

Salient features:

• In diatoms the cell walls form two thin overlapping shells,which fit together as in a soap box.
• Their cell wall contain silica

Economic value:
Their cell wall deposited in ocean floor over billions of years in large amount called as ‘Diatomaceous earth’. It is used in polishing, filtration of oils and syrups.

Plus One Botany Notes Hsslive Dinoflagellates:
Salient features:

• They are marine and photosynthetic
• Most of them have two flagella one lies longitudinally and the other transversely.

Harmful effects:

• Red dianoflagellates -Gonyaulax undergoes rapid multiplication and sea appear red (red tides).
• Toxins released by them kill other marine animals such fishes.

Plus One Botany Victers Notes Euglenoids:
Salient features:

• They are fresh water organisms found in stagnant water.
• Pigments of Euglenoids are identical to those of higher plants.
• They have a protein rich layer called pellicle which makes their body flexible.
• They have two flagella, a short and a long one.

Mode of nutrition:
They are photosynthetic in the presence of sunlight (autotrophic) and predating on other smaller organisms in the absence of sunlight (heterotrophs). Hence nutrition is mixotrophic. Example: Euglena.

Plus One Botany Notes Victers Slime Moulds:
Salient features:

1. They are saprophytic protists.
2. They form an aggregation during favourable conditions called plasmodium which may grow and spread over several feet.
3. The plasmodium differentiates and forms fruiting bodies bearing spores at their tips during unfavourable conditions.
4. They are resistant and survive for many years under adverse conditions.

Plus One Botany Notes Pdf Protozoans:
They are heterotrophs and live as predators or parasites.
Type of protozoans:
1. Amoeboid protozoans:

• They live in freshwater or sea water
• They capture their prey by pseudopodia (false feet) as in Amoeba.
• Some are parasites as in Entamoeba

2. Flagellated protozoans:

• They possesss flagella.
• The parasitic forms cause diseases such as sleeping sickness. Example: Trypanosoma.

3. Ciliated protozoans:

• They are aquatic and have a cavity (gullet) that collect food from outside.
• They move with the help of cilia. Example: Paramoecium.

4 Sporozoans:
They are infectious due to the spore-like stage in their life cycle. eg: Plasmodium (malarial parasite) which causes malaria.

Hsslive Plus One Botany Notes Kingdom Fungi:
They are heteterotrophs, mainly 2 types

1. Saprophytes: They can absorb soluble organic matter from dead substrates
2. Parasites: They are depend on living plants and animals

Symbiotic associations:

1. Lichens-Fungi forms an association with algae
2. Mycorrhiza-Fungi forms association with roots of higher plants.

Disease caused bv fungi:
They cause diseases in plants and animals. eg: wheat rust disease by Puccinia.

Salient features:

• Yeast is unicellular fungus but others are multicellular.
• Mycelium: lt is body of fungi contains many hyphae
• Hyphae It is the long, slender thread-like structures
• Some hyphae with multinucleated cytoplasm are called coenocytic hyphae.
• Others have septae or cross walls in their hyphae. eg: Penicillium.
• The cell walls of fungi are composed of chitin and polysaccharides

Types of Reproduction:

1. Vegetative method: It takesplace by fragmentation, fission and budding.
2. Asexual method: It takesplace by spores called conidia orsporangiospores or zoospores.
3. Sexual reproduction: It takesplace by oospores, ascospores and basidiospores.
4. The spores are produced in special structures called fruiting bodies.

Steps of sexual cycle:

In this, two haploid hyphae come together and fuse results in diploid cells (2n).

Dikarvotic stage in funai:

1. In ascomycetes and basidiomycetes, after plasmogamy dikaryotic stage (n + n ) occurs for some time. Later these nuclei fuse and the cells become diploid.
2. The fungi form fruiting bodies in which reduction division occurs and forms haploid spores.

Hss Live Plus One Botany Notes Phvcomycetes:
Salient features:

• They are found in aquatic habitats and on decaying wood.
• Their mycelium is aseptate and coenocytic.

Types of reproduction:
1 Asexual reproduction:
It takes place by zoospores (motile) or by aplanospores (non-motile). These spores are produced in sporangium.

2 Sexual reproduction:
It is the fusion of gametes have similar structure (isogamous) or dissimilar structure (anisogamous or oogamous) and after fusion zygospore is formed. Examples are Mucor, Rhizopus (the bread mould) and Albugo (the parasitic fungi on mustard).

Hss Live Plus One Biology Notes Ascomycetes:
Salient features:

• They are commonly known as sac-fungi, eg-unicellular- yeast (Sacharomyces) or multicellular Penicillium.
• Some are coprophilous (growing on dung).
• Mycelium is branched and septate.
• The asexual spores are conidia produced on conidiophores.
• Sexual spores are called ascospores which are produced in sac like ascus.
• Spores are arranged in fruiting bodies called ascocarps.
• Examples are Aspergillus, Claviceps and Neurospora.

Economic value:

• Neurospora is used in genetic studies.
• Edible members are morels and buffles.

Plus One Botany Notes Chapter 1 Basidiomycetes:
Salient features:

• Their mycelium is branched and septate.
• Some members grow as parasites and disease causing organisms e.g. rusts and smuts
• Common basidiomycetes are mushrooms, bracket fungi or puffballs

Reproduction:

1. The asexual spores and sex organs are not found but vegetative reproduction by fragmentation.
2. In sexual reproduction, plasmogamy occur by fusion of two vegetative cells of different strains. It results dikaryotic mycelia which gives rise to basidium. Later, karyogamy and meiosis takeplace in the basidium and producing four basidiospores.
3. The basidia are arranged in fruiting bodies called basidiocarps.
4. Examples -Agaricus (mushroom) Ustilago (smut) and Puccinia (rust fungus).

Hsslive Botany Plus One Notes Deuteromvcetes:
Salient features:

• They are called as imperfect fungi because perfect stage or sexual reproduction is absent.
• Their asexual reproduction takes place with the help of conidia. ,
• The mycelium is septate and branched.

Economic value:

• Some members are decomposers play an important role in the mineral cycling.
• Examples are Alternaria, Colletotrichum and Trichoderma.

Hsslive Botany Notes Plus One kingdom Plantae

1. Majority members are eukaryotic, chlorophyll-containing organisms.
2. Few members are partially heterotrophic- insectivorous plants or parasite. eg: Bladder wort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.

Different types of plant group:
The kingdom Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.

Life cycle:
It has two distinct phases – the diploid sporophytic and the haploid gametophytic – that alternate with each other.

Plus One Botany Biological Classification Notes Kingdom Animalia:
Salient features:

• It includes heterotrophic eukaryotic organisms.
• They are multicellular and their cells lack cell wall.
• Their mode of nutrition is holozoic (ingestion of food).
• Most of them are capable of locomotion.

Plus One Botany Kite Victers Notes Pdf Viruses, Viroids And Lichens:
R.H. Whittaker not placed acellular organisms such as viruses, viroids and lichens in five kingdom classification.

VIRUSES:
Historical aspects and Discovery

1. Name virus that means venom or poisonous fluid was given by Pasteur D.J. Ivanowsky.
2. Extract of the infected plants of tobacco could cause infection in healthy plants and called the fluid as Contagium vivum fluidum (infectious living fluid).It was identified by M.W. Beijerinek (1898)
3. Viruses could be crystallised and crystals consist of proteins outside. It was identified by W.M. Stanley. (1935)

Salient features:

• Viruses are non living particle outside the living cell.
• It has an inert crystalline structure .
• They have living state inside the host and multiply by using host cell machinery-Ribosome. So they are called as obligate parasites.
• They are smaller than bacteria because they passed through bacteria-proof filters.

Structure of viruses:
Viruses contain proteins coat outside, either RNA or DNA inside, (i.e either single or double stranded RNA or double stranded DNA).

Symptoms and disease caused by viruses:
Disease:
Mumps, smallpox, herpes, influenza and AIDS
Symptoms in plants:

Viroids:
Strucure:
It has free RNA without protein coat.

Discovery:
T .O .Diener found that this infectious agent was smaller than viruses.

Disease:
It causes potato spindle tuber disease

Lichens:
They are symbiotic associations between algae and fungi.

Types of component:

1. The algal component is called phycobiont: (autotrophic) prepare food for fungi
2. The fungal component is called mycobiont: (heterotrophic) provide shelter and absorb mineral nutrients and water for algae.

Significance:
Lichens are very good pollution indicators i.e they do not grow in polluted areas.

Ncert Supplementary Syllabus
Six kingdom classification:
It was proposed by Carl Woese. It includes kingdoms like Archaebacteria, Eubacteria Protista, Mycota, Plantae and Animalia.

Students can Download Chapter 2 Structure of Atom Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 2 Structure of Atom

Plus One Chemistry Structure of Atom One Mark Questions and Answers

Plus One Chemistry Structure Of Atom Questions Question 1.
Which of the following is not true for cathode rays?
a) They possess kinetic energy
b) They are electromagnetic waves
c) They produce heat
d) They produce mechanical pressure
Answer:
b) They are electromagnetic waves

Previous Hse Questions From The Chapter Structure Of Atom Question 2.
The mass of the electron = ________ kg
Answer:
9.11 × 10^-31 kg

Plus One Chemistry Chapter 2 Questions And Answers Question 3.
Bohr’s orbits are called stationary states because
a) Electrons in them are stationary
b) Their orbits have fixed radii
c) The electrons in them have fixed energy
d) The protons remain in the nuclei and are stationary
Answer:
c) The electrons in them have fixed energy

Structure Of Atom Class 11 Questions And Answers Pdf Question 4.
The metal which gives photoelectrons most easily is
a) Lithium
b) Sodium
c) Calcium
d) Cesium
Answer:
d) Cesium

Questions On Structure Of Atom Class 11 Question 5.
The orbitals having same energy are called ________ orbitals.
Answer:
degenerate

Questions On Quantum Numbers Class 11 Question 6.
Match the following:

1. Spherically symmetrical – d
2. Dumb-bell – s
3. Doubly dumb-bell – p

Answer:

1. Spherically symmetrical – s
2. Dumb-bell – p
3. Doubly dumb-bell – d

Class 11 Chemistry Chapter 2 Important Questions With Answers Question 7.
Which of the following set of quantum numbers is correct for an electron in 4/ orbital
a) n = 4 l = 4 ml = -4 ms = +14
b) m = 4 l = 3 ml = +4 ms = +14
c) n = 4 l = 3 ml = -3 ms =-14
d) n = 4 l = 3 ml = +4 ms = -14
Answer:
c) n = 4 l = 3 ml = -3 ms = -14

Structure Of Atom Class 11 Questions And Answers Question 8.
The limiting line of Balmer Series has the frequency of ________ .
Answer:
8.23 × 10¹⁴

Class 11 Chemistry Chapter 2 Important Questions Question 9.
The number of orbitals and the maximum number of electrons that can be accommodated in a principal quantum level are
Answer:
n² & 2n²

Structure Of Atom Important Questions Question 10.
If the uncertainty in position and momentum of a particle like electrons are equal the uncertainty in velocity is ________
Answer:
\(\triangle V=\frac { 1 }{ 2m } \sqrt { \frac { h }{ \pi } } \)

Important Questions Of Atomic Structure Class 11 Question 11.
The total energy of an electron in a Bohr orbit is given by ________
Answer:
\(\frac{-Z e^{2}}{8 \pi \varepsilon_{0} r}\)

Plus One Chemistry Structure of Atom Two Mark Questions and Answers

Question 1.
Match the following:

Answer:

Question 2
Of the following which is/are correct? Give justification.
a) n = 2 l = 1 m = 0 s = +½
b) n = 3 l = 3 m = 2 s =-½
c) n = 4 l = 3 m = 1 s = +½
d) n = 3 l = 2 m = 3 s = +½
Answer:
a) n = 2 f = 1 m = 0 s = +½
c) n = 4 f = 3 m=1 s= -½
Option b) is wrong because when n = 3, ^ =0, 1, 2
Option d) is wrong because when l = 2, m = -2, -1, 0, 1, 2

Question 3.
Match the following:
1. Anode rays – Nucleus
2. Cathode rays – Plum-pudding model
3. J.J Thomson – Proton
4. Thea-particle scattering experiment – Electron
Answer:
1. Anode rays – Proton
2. Cathode rays – Electron
3. J.J Thomson – Plum-pudding model
4. Thea-particle scattering experiment – Nucleus

Question 4.
Calculate the uncertainty in the determination of velocity of a ball of mass 200 g, if the uncertainty in the determination of position is 1 A.
[h=6.626 × 10^-34 J s]
Answer:

Question 5.
Which among the following sets of quantum numbers is/are not possible?
a) n = 3, l = 2, m = 0, s = +½
b) n = 2, l= 1, m = 0, s = +½
c) n = 1, l= 0, m = 0, s = -½
d) n = 4, l = 2, m = 2, s = -½
Answer:
All sets are possible.

Question 6.
1. How many sub-shells are associated with n = 4?
2. How many electrons will be present in the sub-shells having m_s value of –\(\frac{1}{2}\) for n = 4?
Answer:
1. For n = 4, l can have values 0, 1, 2, 3. Thus, there are four sub-shells in n = 4 energy level.
These four sub-shells are 4s, 4p, 4d and 4f.

2. For n = 4, the number of orbitals = (4)² = 16.
Each orbital can have one electron with m_s = –\(\frac{1}{2}\).
Thus, there are 16 electrons in sub-shells having n = 4 and m_s = –\(\frac{1}{2}\)

Question 7.
i) Name the principle which restricts the pairing of electrons in degenerate orbitals. .
ii) How many electrons can be accomodated in the sub-shell having n = 4 and l = 2?
Answer:
i) Hund’s rule of maximum multiplicity,
ii) 10 electrons (4d sub-shell).

Question 8.
If the electron is to be located within 5 x 10^-5A°, what will be the uncertainty in its velocity?
(Mass of the electron = 9.1 x 10^-31 kg).
Answer:
According to Heisenberg’s uncertainty principle,

Question 9.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴per second, calculate the power of this laser.
Answer:

Plus One Chemistry Structure of Atom Three Mark Questions and Answers

Question 1.
Fill in the blanks suitably by studying the relationship of the given pairs:

1. Lyman : Ultraviolet:: Balmer: ……………..
2. s-subshell:spherical:: p-subshell: …………….
3. Rydberg’s formula: \(\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\) :: de Broglie relation:

Answer:

1. Balmer: Visible
2. psubshell: dumb-bell
3. de Broglie relation: \(\lambda=\frac{h}{m v}\)

Question 2.
Fill in the blanks:

Answer:

Question 3.
Filling of electrons in the orbitals on a ground state atom is governed by three rules.
a) Which are the three rules?
b) State any one of them.
Answer:
a) 1) Aufbau principle
2) Pauli’s exclusion principle
3) Hund’s rule of maximum multiplicity

b) Hund’s rule of maximum multiplicity – electron pairing in orbitals of same energy will not take place until each degenerate orbital of a given subshell is singly occupied.

Question 4.
During a class room discussion, one of your friends argued that, “we can’t determine both position and velocity of an electron”.

1. Is it true?
2. Which principle is behind your answer?
3. State it.

Answer:

1. Yes. It is true.
2. Heisenberg’s uncertainty principle.
3. Heisenberg’s uncertainty principle states that it is not possible to determine simultaneously both position and momentum of a microscopic moving particle such as electron with absolute accuracy.

Question 5.
The arguments of two students is as given:
Student 1 : “We need four quantum numbers to represent an electron in a multi-electron atom.”
Student 2 : “We need only first three quantum numbers to represent an electron in a multi-electron atom.”
a) Which are the four quantum numbers?
b) Who is correct? Why?
c) Write the possible four quantum numbers of the valence electron of Na atom.
Answer:
a) The four quantum numbers are:
i) Principal quantum number (n)
ii) Azimuthal quantum number (l)
iii) Magnetic quantum number (m_l)
iv) Spin quantum number (m_s)

b) The argument of student 1 is correct. If there are two electrons in the same subshell the first three quantum numbers become the same. Hence we need fourth quantum number to identify the electron.

c) n = 3, l = 0, m = 0, s = +½

Question 6.
a) Identify the experiment associated with the following figure:

b) Explain the experiment done by Rutherford and give its observations.
c) Write the concludions of the experiment.
Answer:
a) It is the figure of α -particle scattering experiment (gold foil experiment).

b) In this experiment, a stream of high energy α – particles from a radioactive source was directed at a thin foil of gold metal which had a circular fluorescent zinc sulphide screen around it. Whenever α-particles struck the screen, a tiny flash of light was produced at that point. The α-particles striking the gold foil were analysed. It was observed that:

1. Most of the α – particles passed through gold foil undeflected.
2. A small fraction of the α – particles are deflected by small angles.
3. A very few α – particles(1 in 20,000) bounced back i.e., deflected by nearly 180°.

c) Rutherford drew the following conclusions regard¬ing the structure of atom from this experiment:

1. Most of the space in the atom is empty as most of the α – particles passed through the foil undeflected.
2. The positive charge of the atom is concentrated in a very small volume (called nucleus) that repelled and deflected the positively charged α – particles.
3. Volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.

Question 7.
The 4s subshell has more energy than 3p subshell.
a) Is it true? Justify your answer.
b) StateAufbau principle.
Answer:
a) Yes. For both 4s and 3p subshells the (n+l) value is 4. But 4s, with high value of ‘n’ has higher en¬ergy.
b) Aufbau principle – In the ground state of the atoms, the orbitals are filled in order of their increasing energies.

Question 8.
Two students were analysing the electronic configurations of the first 30 elements of the Periodic Table as part of an assignment. They found that two elements showed difference from other twenty eight elements.

1. Which are the two elements?
2. Write their electronic configurations.
3. Why they show this anomalous behaviour?

Answer:
1. ₂₄Cr and ₂₉Cu.
2. ₂₄Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ and ₂₉Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
3. This is due to the fact that exactly half filled and completely filled orbitals (i.e., d⁵, d¹⁰) have extra stability due to symmetrical distribution of electrons and maximum exchange energy.

Question 9.
Three box diagrams of 2p³ configuration are given below:
i)

1. Which one is correct?
2. Name the principle behind your answer.
3. State the principle.

Answer:

1. The correct one is (ii).
2. Hund’s rule of maximum multiplicity.
3. Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.

Question 10.
J.J. Thomson proposed his atom model in 1898.

1. Explain Thomson’s model of atom.
2. Why Thomson’s atom model is called plum pudding model or watermelon model?
3. What is the limitation of Thomson’s atom model?

Answer:
1. J.J. Thomson proposed that an atom possess a spherical shape in which the positive charge is uniformly distributed. The electrons are embedded into it in such a manner as to give the most stable electrostatic arrangement. The mass of the atom is assumed to be uniformly distributed over the atom. This model explained the overall neutrality of the atom.

2. Thomson’s model of atom can be visualised as a pudding or watermelon of positive charge with electrons embedded into it like the plums or seeds.

3. Thomson’s model was not consistent with the results of later experiments. It failed to explain the observations of Rutherford’s α-particle scattering experiment.

Question 11.
A student argued that the 3d orbitals will be filled only after the 4s orbital is completely filled in accordance with aufbau principle. Then another student opposed by saying that it is not true for certain elements like Cr and Cu.
a) Whose argument is correct?
b) Write electronic configurations of ₂₄Crand ₂₉Cu. Justify your answer.
Answer:
a) Both arguments are correct.
b) Cr and Cu have anomalous electronic configurations. This is because half filled and completely filled sub-shells have extra stability due to the symmetrical distribution of electrons maximum exchange energy.
₂₄Cr = 1 s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ OR [Ar]3d⁵ 4s¹ This is due to the extra stability of half filled 3d⁵ sub-shell.

₂₉Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ OR [Ar]3d¹⁰ 4s¹ This is due to the extra stability of completely filled 3d¹⁰ sub-shell.

Question 12.
a) What are the atomic numbers of elements whose
outermost electronic configurations are given by
i) 3s¹
ii) 3p⁵?
b) Which of the following are isoelectronic species?
Na⁺, K⁺, Mg²⁺,Ca²⁺, S^2-,Ar
c) What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 ms^-1? m
Answer:
a) i) 3s¹-Atomic number is 11 (Na)
ii) 3p⁵ -Atomic number is 17 (Cl)

b) Na⁺, Mg²⁺ (both of them have same no. of electrons, i.e., 10 each)

Question 13.
Quantum numbers are a set of four numbers used to designate electron in an atom.

1. How many electrons in an atom can have the following quantum numbers, n = 1, l = 0 ?
2. Give the quantum numbers of the valence electron of an atom with atomic number 13.
3. Draw the shape of orbital having n = 1 and l = 0.

Answer:
1. 2 electrons (i.e., 1s orbital)
2. The element with atomic number 13 is aluminium. ₁₃Al ⇒ [Ne] 3s²3p¹⁺
The valence electron is in the 3p orbital. Hence, the quantum numbers for the valence electron are n = 3, l= 1, m = -1, 0 +1
3. It is the 1s orbital

Question 14.
a) i) What is meant by line spectra or atomic spectra?
ii) Name the series of lines in the hydrogen spectrum belonging to the visible region.
iii) What is the wave length of light emitted when the electron in a hydrogen atom undergoes transmission from n = 4 to n = 2? (RH= 109677 cm-1)
b) State the principles/rules for filling of orbitals in atoms.
Answer:
a) i) Line spectra or atomic spectra are the spectra obtained from excited atoms due to emission of radiation. The emitted radiation is identified by the appearance of bright lines.
ii) Balmer series
iii) n₁ = 2, n₂ = 4

b)

1. Aufbau principle: In the ground state of the atoms, the orbitals are filled in order of their increasing energies.
2. Pauli’s exclusion principle: No two electrons in an atom can have the same set of four quantum numbers.
3. Hund’s rule of maximum multiplicity: Pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each i.e., it is singly occupied.

Question 15.
Line emission spectra are often called finger print of atoms.
a) Justify the above statement.
b) Yellow light emitted from a sodium lamp has a wave length (X) of 580 nm. Calculate the frequency and wave number of this yellow light.
Answer:
a) Each element has a unique line emission spectrum. The characteristic lines in atomic spectra can be used in chemical analysis to identify unknown atoms in the same way as finger prints are used to identify people.

Question 16.
1. How many orbitals are possible in a p-subshell. Which are they?
2. What is the shape of p-orbital?
3. Sketch the boundary surface diagrams of 2p orbitals.
Answer:
1. Three.
These are p_x, p_y and p_z orbitals.
2. Dumb-bell shaped.
3.

Question 17.
Electronic configuration of an element written by a student is given below:

a) Which rule is violated here?
b) Give the correct configuration.
c) What is the uncertainty in position of an electron if the uncertainty in its velocity is 1.159×107 m/s?
Answer:
a) Hund’s rule of maximum multiplicity.

Plus One Chemistry Structure of Atom Four Mark Questions and Answers

Question 1.
Bohr’s model of hydrogen atom is a modification of
Rutherford’s model.
a) Write any two merits of Bohr’s model.
b) Write any two demerits of Bohr’s model.
Answer:
a)

1. Bohr’s model could explain the stability of an atom.
2. Bohr’s model could explain the atomic spectrum of hydrogen.

b)

1. Failed to explain the finer details of hydrogen atom spectrum observed by using sophisticated spectroscopic techniques.
2. It could not explain the ability of atoms to form molecules by chemical bonds.

Question 2.
Consider the statement, “The two electrons of He atom have the same set of quantum numbers.”

1. Do you agree?
2. Name the principle applied here.
3. State the principle.
4. Write the all quantum numbers of outer electrons of the atom.

Answer:

1. No.
2. Pauli’s exclusion principle.
3. No two electrons in an atom can have same set of four quantum numbers.
4. Forthe1st electron: n = 2, l = 0, m = 0, s = +½.
   For the 2nd electron: n = 2, l = 0, m = 0, s = -½

Question 3.
Complete the following table with respect to the va¬lence electron of each element.

Answer:

Question 4
Analyse the following figure showing transitions of electrons in the hydrogen atom.

a) Name the series (a), (b), (c), (d) and (e).
b) Mention the region of the spectrum in which each series belongs to.
c) Explain how they are obtained.
d) Calculate the wave length of the first line of series (b).
[R_H = 109677 cm^-1]
Answer:
(a) = Lyman series,
(b) = Balmer series,
(c) = Paschen series,
(d) = Brackett series,
(e) = Pfund series

b)
Lyman series – UV region
Balmer series – Visible region
Paschen series – Infrared region
Brackett series – Infrared region
Pfund series – Infrared region

c) In hydrogen atom there is one electron which is present in first orbit in ground state. When energy is supplied this electron may be excited to some higher energy level. Since in a sample of hydrogen there are large number of atoms, the electrons in different atoms absorb different amounts of energies and are excited to different higher energy levels. Now, from excited states, the electron may return to ground state in one or more jumps. These different downward jumps are associated with different amounts of energies and hence result in the emission of radiations of different wavelengths which appear as different lines in the hydrogen spectrum.
Series – Obtained when electron jumps from any of the higher energy levels to
Lyman series – 1^st energy level
Balmer series – 2^nd energy level
Paschen series – 3^rd energy level
Brackett series – 4^th energy level
Pfund series – 5^th energy level

d) n₁ = 2, n₂ = 3, R_H = 109677 cm^-1

Question 5.
Quantum numbers are the address of an electron in an atom. Justify the statement by explaining different quantum numbers.
Answer:
Quantum numbers are certain numbers which are used to identify an electron in an atom.
The following four quantum numbers are used for this purpose.
1. Principal quantum number (n):
It determines the size and to large extent the energy of the orbital. It also identifies the shell. The value of ‘n’ ranges from 1 to a. With increase in the value of ‘n’, the number of allowed orbitals increases and are given by ‘n2’. All the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by letters K (n = 1), L (n = 2), M (n = 3), N (n = 4) etc. The size and energy of the orbital will increase with increase of ‘n’.

2. Azimuthal/Orbital angular momentum/Subsidiary quantum number (l): It defines the three dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n-1). It gives an idea regarding the subshell in which the electrons are present.

3. Magnetic quantum number (m_l):
It gives information about the spatial orientation of the orbital with respect to standard set of coordinate axis. For any sub-shell, (2l+1) values of m_l are possible ranging from -l to +l including zero. The permitted values of m_l gives the number of orbitals in that sub-shell.

4. Spin quantum number (m_s) :
It refers to the orientation of the spin of the electron. An orbital can have two electrons. If an electron is spinning in the clockwise direction, it is given a spin quantum number value of+1/2 and if an electron is spinning in the anti-clockwise direction it is given spin quantum number value of -1/2. These are called two spin states of the electron and are represented by two arrows ↑ (spin up) and ↓. (spin down). Thus, the two electrons in an orbital should have opposite spins.

Question 6.
Dual nature of matter was proposed by Louis de Broglie.
a) Calculate the de Broglie wavelength associated with an electron with velocity equal to that of light.
b) State Heisenberg’s uncertainty principle and give its mathematical expression.
Answer:

b) It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.
Mathematically, it can be given as in the equation

∆x ⇒ uncertainty in position of the particle
∆px ⇒ uncertainty in momentum of the particle
∆vx ⇒ uncertainty in velocity of the particle
m ⇒ mass of the particle and
h ⇒ the Planck’s constant

Question 7.
Rutherford’s atom model had strong similarity to a small scale solar system. (4)
a) What are the important features of Rutherford’s. nuclear model of atom?
b) What are the drawbacks of Rutherford’s model of atom?
Answer:
a) i) The positive charge and most of the mass of the atom is densely concentrated in extremely small region of the atom called nucleus.
ii) The nucleus is surrounded by electrons that ’ move around the nucleus with a very high speed in circular paths called orbits.
iii) Electrons and the nucleus are held together by electrostatic forces of attraction.

b) i) It failed to explain the stability of atom,
ii) It says nothing about the electronic structure of atoms.

Question 8.
a) Calculate the momentum of a particle which has de Broglie wave length of 250 pm. (4)
b) The distribution of electron into orbitals of an atom is called its electronic configuration.
i) Give the valence shell electronic configuration of chromium atom.
ii) Which among the following configurations is more.stable, d4 ord5? Justify your answer.
Answer:
a) According to the de Broglie matter wave equation

b) i) ₂₄Cr → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
ii) d⁵ is more stable.
The d⁵ configuration is half filled. It has symmetrical distribution of electrons and maximum exchange energy. Hence, it has extra stability compared to d⁴ configuration.

Question 9.
a) Name and state the principle, which restricts the maximum number of electrons in an orbital to be two.
b) Using s, p, d, f notation represent the sub-shell with the following quantum numbers.
i) n = 1, l = 0
ii) n = 4, l=3
c) The uncertainty in the position and velocity of a particle are 10 cm and 5.27 × 10³m/s respectively. Calculate the mass of the particle (h=6.626 × 10^-34 J s).
Answer:
a) Pauli’s exclusion principle
No two electrons in an atom can have the same set of four quantum numbers.
b) i) 1s ii) 4f
c) According to Heisenberg’s uncertainty principle,

Question 10.
The photoelectric effect was first observed by H.Hertz.
a) What is photoelectric effect?
b) What are the observations of photoelectric effect experiment?
Answer:
a) It is the phenomenon of ejection of electrons when certain soft metals like potassium, rubidium, cae-sium, etc. are exposed to a beam of light.
b) i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface of metal.
ii) The number of electrons ejected is proportional to the intensity or brightness of light.
iii) For each metal, there is a characteristic minimum frequency (υ₀), known as threshold frequency below which photoelectric effect is not observed.

Question 11.
A mathematical representation is given below:
\(\Delta x \times \Delta p_{x} \geq \frac{h}{4 \pi}\)
a) Which principle is illustrated by this equation?
b) If the position of the electron is measured within an accuracy of ±0.002 nm, calculate the uncertainty in the momentum of the electron.
c) Using s, p, d notation represent the sub-shell with the following quantum numbers:
i) n = 3 l = 2
ii) n = 5 l = 1
Answer:
a) Heisenberg’s uncertainty principle.

Plus One Chemistry Structure of Atom NCERT Questions and Answers

Question 1.

1. Calculate the number of electrons which will together weigh one gram.
2. Calculate the mass and charge of one mole of electrons.

Answer:
1. Mass of one electron = 9.11 × 10^-31 kg
∴ Number of electrons in one gram = \(\frac{10^{-3} \mathrm{kg}}{9.11 \times 10^{-31} \mathrm{kg}}=1.098 \times 10^{27}\)

2. Mass of one electron = 9.11 × 10^-31 kg
∴ Mass of 1 mole of electrons = 9.11 × 10^-31 kg × 6.022 × 10²³ = 5.486 x 10^-7 kg
Charge on one electron = 1.602 × 10^-19 C
∴ Charge on one mole of electrons
= (1.602 × 10^-19C) × (6.022 × 10²³)
= 9.65 × 10⁴ C.

Question 2.
Write the complete symbol forthe atom with the given atomic number (Z) and atomic mass (A):
i) Z = 17, A = 35
ii) Z = 92, A = 233
iii) Z = 4, A = 9
Answer:

Question 3.
Electro.magnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer:
Ionisation energy of sodium = Energy of one photon of radiation of wavelength.
Energy of photon = hυ

Question 4.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J energy?
Answer:
Suppose N photons of the light with wavelength 4000 pm can provide 1 J of energy.
Energy of N photons = Nhυ

Question 5.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷m S^-1. (2)
Answer:

Students can Download Chapter 1 The Living World Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Plus One The Living World One Mark Questions and Answers

Plus One Zoology Exercise Answers Chapter 1 Question 1.
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics
(a) Will decrease
(b) Will increase
(c) Remain same
(d) May increase or decrease
Answer:
(a) Will decrease

Plus One Zoology Chapter Wise Questions And Answers Pdf Question 2.
Which of the following ‘suffixes’ used for units of classification in plants indicates a taxonomic category of ‘family’.
(a) -Ales
(b) -Onae
(c) -Aceae
(d) none of the above
Answer:
(c) -Aceae

Zoology Chapter Wise Questions And Answers Question 3.
The term ‘systematics’ refers to:
(a) Identification and classification of plants and animals
(b) Nomenclature and identification of plants and animals
(c) Diversity of kinds of organisms and their relationship
(d) Different kinds of organisms and their classification
answer:
(c) Diversity of kinds of organisms and their relationship

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Question 4.
Genus represents
(a) An individual plant or animal
(b) A collection of plants or animals
(c) Group of closely related species of plants or animals
(d) None of these
Answer:
(c) Group of closely related species of plants or animals

Plus One Zoology Previous Question Papers Chapter Wise  Question 5.
The taxonomic unit ‘Phylum’ in the classification of animals is equivalent to which hierarchical level in classfication of plants.
(a) Class
(b) Order
(c) Division
(d) Family
Answer:
(c) Division

Plus One Zoology Chapter Wise Questions And Answers Question 6.
Correct and rewrite the following if there is any mistake.
Sativa Oriza, Tigris Panthera
Answer:
Oriza sativa, Panthera tigris

Plus One Zoology Chapter Wise Previous Questions And Answers Question 7.
Kingdom – Carnivora
Phylum – Mammalia
Class – Felidsffe
Order – Chordata
Family – Tigris
Genus – Animalia
Sps – Panthera

Correct the mistakes of the flow chart to get the hierarchical arrangement of tiger in the ascending order.
Answer:
Kingdom – Animalia
Phylum – Chordata
Class – Mammalia
Order – Carnivora
Family – Felidae
Genus – Panthera
Sps – Tigris

Hsslive Plus One Zoology Chapter Wise Questions And Answers Question 8.
Match the following:

1. A Unit of classification – kingdom
2. Lowest taxonomic category – Genus
3. Panthera – Species
4. Highest Taxonomic category – Taxon

Answer:

1. A Unit of classification – Taxon
2. Lowest taxonomic category – Species
3. Panthera – Genus
4. Highest Taxonomic category – kingdom

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Question 9.
Expand the term ICBN and ICZN.
Answer:
1. ICBN: International Code for Botanical Nomenclature.
2. ICZN: International Code for Zoological Nomenclature.

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Question 10.
Given below is the scientific name of Frog. Identify the correctly written name.
(a) Rana Tigrina
(b) Rana tigrina
Answer:
(b) Rana tigrina

Plus One Zoology Animal Kingdom Questions Question 11.
Rearrange the order of classification.
Genus, Family, Phylum, Species, Class, Kingdom, Order
Answer:
Species, Genus, Family, Order, Class, Phylum, Kingdom

The Living World Class 11 Questions And Answers Question 12.
Arrange the given terms in their taxonomic hierarchy.
Primata, Homosapien, Chordata, Mammalia, Hominidae.
Answer:
Homosapiens, Hominidae, Primata, Mammalia, Chordata.

Plus One The Living World Two Mark Questions and Answers

The Living World Class 11 Important Questions With Answers Question 1.
You are provided with a stuffed rabbit and a dried leaf. As a student of biology which taxonomic aid will you choose to store them.
Answer:
1. Stuffed rabbit – Museum
2. Dried leaf – Herbarium.

Question 2.
Find the odd one out and give reason.
Herbarium, botanical garden, museum, aquarium
Answer:
Aquarium – Aquarium is not a taxonomical aid.
All other are taxonomical aids.

Question 3.
Once you visited a museum. There are different kinds of animals and plants are preserved. How these plants and animals all preserved in a museum.
Answer:
Plants and animals are preserved in the containers or jars in preservative solutions. These may also be preserved as dry specimen. Insects are preserved in insect box.

Larger animals like birds and mammals are usually stuffed and preserved. Museums often have collections of skeleton of animals too.

Question 4.
Zoological parks are different from museums. Give reason?
Answer:
Zoological parks are the places where wild animals are kept in protected environments, that the conditions similar to their natural habitats. Museums have collection of preserved plants and animal specimens for study and reference.

Question 5.
Raju collected a skull of an animal and a living rare animal during a study tour. Select the suitable location for each from the list given in the brackets. (Botanical garden, Zoological park, Herbarium, Museum)
Answer:
Skull of an animal – museum
Living rare animal – Zoological park

Question 6.
Distinguish between taxonomic category and taxonomic hierarchy.
Answer:
Classification involves hierarchy of steps in which each step represents a rank or category. Since category is a part of overall taxonomic arrangement, it is called taxonomic category. All the categories together constitute taxonomic hierarchy.

Question 7.
Write the scientific name of following animals.

1. Lion
2. Frog
3. Housefly
4. Tiger

Answer:

1. Pantheraleo
2. Ranatigrina
3. Musca domestica
4. Panthera tigris

Question 8.
Fill in the blank spaces in the table given below.

Answer:

(a) Live
(b) Museum

Question 9.
Panthera jeo is the scientific name of Lion. List the rules you follow to write this scientific name.
Answer:

1. Biological names are generally in Latin and written in italics.
2. When handwritten, the words in a biological name are separately underlined.
3. The first word in a biological name represents the genus and second word represents the species name.
4. The first name (Genus) starts with capital letter and the second name (species) starts with small letter.

Question 10.
List the advantages of

1. Taxonomical key
2. Herbarium

Answer:
1. axonomical key is a taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities. Key are generally analytical in nature.

2. Herbarium is a store house of collected plant specimens that are dried, pressed and preserved on sheets. These specimens, along with their descriptions on herbarium sheets, became a store house or repository for future use. Herbarium serves as quick reference system in taxonomical studies.

Question 11.
Give the terms.

1. The actual account of habital and distribution of plants of a given area.
2. Providing information for identification of names of species found in an area.
3. Contain information on any one taxon.
4. Identification of plants and animals based on the similarities and dissimilarities.

Answer:

1. Flora
2. Manuals
3. Monographs
4. Taxonomical key

Question 12.
Define a taxon. Give some examples of taxon at different hierarchical levels.
Answer:
Each category in the taxonomical hierarchy is considered as a taxonomic unit and is known as a taxon. The taxon used in the classification of animals are kingdom, phylum, class, order, family, genus and species.

Plus One The Living World NCERT Questions and Answers

Question 1.
What do we learn from identification of individuals and populations?
Answer:
In a diverse country like India can learn following things from identification of individuals and population:

1. Native place
2. MotherTongue
3. Costumes
4. Cuisine
5. Religion
6. Caste
7. Socio-economic Background

Question 2.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
As clear from the table in previous answer, (a) and (c) are showing the correct order.

Question 3.
Why are the classification system changing every now and then? ,
Answer:
In any branch of science nothing is written in concrete. Theories keep on changing as more relevant and correct theories are being discovered. In case of living beings certain species become extinct and some new species is being formed in every era.

This process of addition and deletion of species necessitates the continuous change of the classification system.

Question 4.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
A taxon is a particular level of hierarchy in the system of classification of living beings. The following figure gives taxa at different hierarchical levels:

Plus One The Living World Multiple Choice Questions and Answers

Question 1.
Binomial nomenclature is described in the book
(a) Genera Plantarum
(b) Historia Plantarum
(c) Systema Naturae
(d) Flora Japonica
Answer:
(c) Systema Naturae

Question 2.
Which of the following is not the main criteria for five kingdom system of classification?
(a) Cell structure and thallus organization
(b) Mode of nutrition and reproduction
(c) Phylogenetic relationship.
(d) Gram staining
Answer:
(d) Gram staining

Question 3.
Two plants can be conclusively said to belong to the same species if they
(a) Can reproduce freely with each other and form seeds
(b) have more than 90 percent similar genes
(c) look similar and possess identical secondary metabolites
(d) have same number of chromosomes
Answer:
(a) Can reproduce freely with each other and form seeds

Question 4.
ICZN stands for
(a) International code of Botanical Nomenclature
(b) International code of Zoological Nomenclature
(c) international code of Viral Nomenclature
(d) International code of Zoo Nomenclature
Answer:
(b) International code of Zoological Nomenclature

Question 5.
In Mangifera indica L. Generic epithet is
(a) Indica
(b) Mangifera
(c) Linnaeus
(d) None of these
Answer:
(b) Mangifera

Question 6.
ICBN stands for
(a) Indian Congress of Biological Names
(b) International Code of Botanical Nomenclature
(c) International Congress of Biological Names
(d) Indian Code of Botanical Nomenclature
Answer:
(b) International Code of Botanical Nomenclature

Question 7.
Taxon is the
(a) taxonomic group of any rank
(b) procedure to assign a scientific name
(c) process of classification
(d) process by which anything is grouped into convenient categories based on characters
Answer:
(a) taxonomic group of any rank

Question 8.
Which of the following is famous for stating that “Population increases much faster than its food supply”?
(a) Fredrick Losch
(b) R Vircow
(c) T R Malthus
(d) Karl Von Baer
Answer:
(c) T R Malthus

Question 9.
Reproduction is the characteristic feature of living organisms. Which of the following can not reproduce?
(a) Amoeba and Paramecium
(b) Fungi and filamentous algae
(c) Humans and Ayes
(d) Mules and worker bees
Answer:
(d) Mules and worker bees

Question 10.
Nicotiana sylvestris flowers only during long days and N.tobacurh flowers only during short days. If raised in the laboratory under different photoperiods, they can be induced to flower at the same time and can be cross fertilized to produce self-fertile offspring. What is the best reason for considering N. sylvestris and N. tobocum to be separate species?
(a) They are physiologically distinct
(b) They are morphologically distinct
(c) They cannot interbreed in nature
(d) They are reproductively distinct
Answer:
(c) They cannot interbreed in nature

Question 11.
Which of the following book is not associated with Carolus Linnaeus, the father of Taxonomy and Nomenclature?
(a) Systema Naturae
(b) Genera Plantarum
(c) Species Plantarurn
(d) Historia Generalis Plantarum
Answer:
(d) Historia Generalis Plantarum

Question 12.
Philosophie Zoologique, a book written by Jean Baptiste de Lamarck is based on
(a) Survival of the fittest
(b) Natural Selection
(c) Inheritance of acquired characters
(d) Biogenetic law
Answer:
(c) Inheritance of acquired characters

Question 13.
The book “Philosophic Zoologique” was written by
(a) Lamarck
(b) Mendel
(c) Haeckel
(d) Hugo deVries
Answer:
(a) Lamarck

Question 14.
Binomial nomenclature was introduced by
(a) Linnaeus
(b) Darwin
(c) Aristotle
(d) deCandoile
Answer:
(a) Linnaeus

Students can Download Chapter 2 Animal Kingdom Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 2 Animal Kingdom

Plus One Animal Kingdom One Mark Questions and Answers

Plus One Zoology Animal Kingdom Questions Chapter 2 Question 1.
Fill in the blanks.

Answer:
i) B – Nereis
ii) B – Ctenoplanna
C – Ctenophora
iii) A – radula

Plus One Zoology Chapter Wise Questions And Answers Pdf Chapter 2 Question 2.
Write any two members of the Phylum Aschelminthes which are found parasitic on Human beings.
Answer:

1. Ascaris (Roundworm)
2. Wuchereria (Filaria worm)

Plus One Zoology Previous Question Papers Chapter Wise Chapter 2 Question 3.
In some animal groups, the body is found divided into compartments with at least some organs/ organ repeated. This characteristic feature is named
(a) Segmentation
(b) Metamerism
(c) Metagenesis
(d) Metamorphosis
Answer:
(b) Metamerism

Presence Of Column And Metamerism Are The Most Important Characters In Chapter 2 Question 4.
Given below are types of cells present in some animals. Each one is specialized to perform a single specific function except
(a) Choanocytes
(b) Interstitial cells
(c) Gastrodermal cells
(d) Nematocytes
Answer:
(b) Interstitial cells

Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 5.
Which one of the following sets of animals share a four chambered heart?
(a) Amphibian, Reptiles, Birds
(b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles
(d) Lizards, Mammals, Birds
Answer:
(b) Crocodiles, Birds, Mammals

Hsslive Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 6.
Which of the following pairs of animals has non glandular skin.
(a) Snake and Frog
(b) Chameleon and Turtle
(c) Frog and Pigeon
(d) Crocodile and Tiger.
Answer:
(c) Frog and Pigeon

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Chapter 2  Question 7.
Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin
(b) Alimentary canal with some modification
(c) Viviparity
(d) Warm blooded nature
Answer:
(d) Warm blooded nature

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Chapter 2 Question 8.
Note the relationship between the first two words and find a suitable word for the fourth place,

1. Coelenterata: radial symmetry, platyhelminthes, _______
2. Lizard: Poikilothermous, crow, _________

Answer:

1. bilaterally symmetrical
2. Homoiothermous

Plus One Zoology Chapter Wise Previous Questions And Answers Chapter 2 Question 9.

1. Annelida: Parapodia :: __________ : Comb plates
2. _________: Water vascular system :: Coelenterata : cnidoblast

Answer:

1. Ctenophora
2. Echinodermata

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Chapter 2 Question 10.
Malpighian tubule is the excretory organ of which phylum?
(a) Phylum porifera
(b) Phylum arthropoda
(c) Phylum Coelenterata
(d) Phylum mollusca
Answer:
(b) Phylum Arthropoda

Pick Out The Acoelomate Organisms From The Following Chapter 2 Question 11.
A chordate animal having flame cells as the excretory organ.
Answer:
Amphioxus

Animal Kingdom Class 11 Important Questions And Answers Chapter 2 Question 12.
From the pictures given below, find out the poikilothermic animals.

Answer:
B and D are poikilothermic animals.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
Answer:
(c) Annelida

Question 14.
Shark has to swim continuously, otherwise, it will sink down. Give reason.
Answer:
Due to absence of air bladder.

Plus One Animal Kingdom Two Mark Questions and Answers

Question 1.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthesip
Answer:

• Presence of milk producing mammary gland.
• Presence of hair on skin.
• External ear or pinnae is present.
• Different types of teeth are present in the jaw.

Question 2.
Complete the blanks.

Answer:

Question 3.
Triploblastic animals are more complex than diploblastic animals. Do you agree with this statement? Justify.
Answer:
Triploblastic animals have more cell layers so they have the possibility of greater degree of cellular specialisation.

Question 4.
All vertebrates are chordates but all chordates are not vertebrates. Justify.
Answer:
Notochord is present in all vertebrates but vertebral column is present only in vertebrates and not in all chordates.

Question 5.
Copy and complete the table.

Answer:

Question 6.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthes
Answer:
Porifera → Cnidaria → Ctenophora→ Platyhelminthes → Aschelminthes → Annelida → Arthropoda → Mollusca → Echinodermata

Question 7.
Categorise and classify the following organisms and arrange them in a table with separate columns and provide appropriate headings.
Exocoetus, Physalia, Ascaris, Apis, Locusta, Corvus, Pita, Hydra, Sepia, Ancylostoma.
Answer:

Question 8.

1. Identify the phylum which exhibit metagenesis or alternation of generation.
2. What is meant by alternation of generation?

Answer:

1. Cnidaria
2. Alternation of sexual and asexual forms of organism: ie., Polyp asexually produce medusa, medusa sexually product polyp.

Question 9.
Name the following.

1. Phylum in which flatworms are included
2. Excretory organs of Annelids.
3. Largest phylum.
4. An oviparous mammal.

Answer:

1. Platyhelminthes
2. Nephridia
3. Arthropoda
4. Platypus

Question 10.
Apis, Prawn, Locust, Spider
Following animals have different habit and habitat. But they have many common characters.

1. Mention the common characters.
2. Identify their phylum.

Answer:

1. Joint footed animals
   • Metameric segmentation
   • Chitinous exoskeleton
2. Arthropoda

Question 11.
A list of animals are given below. Arrange them according to increase in complexity of organization. Scorpion, Earthworm, Liver fluke, Pigeon, Seaanemon, Sycon, Elephant, Anabas.
Answer:
Sycon, Seaanemon, Liver fluke, Earthworm, Scor¬pion, Anabas, Pegeon, Elephant.

Question 12.
Nithin is Studying in Std. XI. He collected some specimens during the field trip conducted by the Science Club of his School. Help Nithin to Classify the Animal in respective Phylum.
Prawn, Slug worm, Butterfly, Pila, Grass Hopper, Crab
Answer:

Question 13.
During a field trip Raju has collected some organisms with the following characters. Help him to identify the phyla of those organisms.

1. Metamerically segmented body.
2. Body covered with calcareous shell.
3. Dorso-ventrally flattened leaf like body.
4. Body divided into head, thorax and abdomen.

Answer:

1. Annelida
2. Mollusca
3. Platyhelminthes
4. Arthropoda

Question 14.
Categorise the following fishes into Osteichthyes and Chondrichthyes?

1. Exocoetus
2. Trygon

Answer:

1. Exocoetus – Oesteichthyes
2. Trygon – Chondrichthyes

Question 15.
Arrange the following terms in two columns correctly. Malpighian tubules, radula, metamerism, Bioluminescence, choanocytes, nematocytes, Phylum-coelenterate, phylum-Arthropoda, phylum- ctnophora, phylum-Mollusca, Phylum-Porifera.
Answer:

Question 16.
Match the following

Answer:

Question 17.
Observe the given organisms

1. Place these animals in proper phylum.
2. Segmentation in the body is first observed in which of the above phylum?

Answer:

1.  i) Arthropoda,
   ii) Porifera,
   ii) Annelida
2. Annelida

Question 18.
Observe the table given below and fill the blank columns A, B, C and from the animals given in brackets. (Ascaris, Starfish, Fasciola, Earthworm)

Answer:

Question 19.
1. Identify the animal given below.

2. Write one well marked property of the above animal.
Answer:

1. Pleurobrachia
2. Bioluminescence is well marked property of pleurobrachia

Question 20.
Representatives of some vertebrate classes are introducing themselves. Write down the name of the class in which they belong.

1. My gills are covered by operculum. I have bony endoskeleton.
2. I give birth to young ones. My body is covered by hair.
3. My skin is glandular? Have trilocular heart.
4. I live only in marine water. My endoskeleton is made up of cartilage.

Answer:

1. Osteichthyes
2. Mammalia
3. Amphibia
4. Chondrichthyes

Question 21.
During classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Justify.
Answer:
NO. Sponges are asymmetrical and body is formed of loose aggregate of cells. Cells are not organised to from tissues and organs. Cnidarians are radially symmetrical and tissue grade of organisation. So cnidarians are more complex than sponges.

Question 22.
Due to the absence of air bladder, fishes belonging to the class Chondrichthyes have to swim constantly. How important is the presence of air bladder in these fishes?
Answer:
Due to the absence of air bladder in chondrichthyes, they have to swim constantly to avoid sinking. If air bladder is present which regulates buoyancy.

Question 23.
Match the column A, B &C in the table given below:

Answer:

Question 24.
Identify the phylum whose larvae are bilaterally symmetrical, but adults are radially symmetrical.

1. Annelida
2. Arthropoda
3. Mollusca
4. Echinodermata

Mention two salient features of the phylum.
Answer:
4. Echinodermata
Salient features Presence of Echinodermata:

• Water vascular system
• Spiny bodies, Endoskeleton of Calcareous ossicles

Question 25.
Write the name of phylum.

1. Diploblastic, tissue grade of organisation, radially symmetrical, polymorphic animals.
2. Soft bodied, Unsegmented, Bilaterally symmetrical animals with open type circulation.
3. Triploblastic, bilaterally symmetrical, true coelomic animals with metamerism, closed circulation.
4. Triploblastic, chitinous exoskeleton and open circulation.

Answer:

1. Cnidaria
2. Mollusca
3. Annelida
4. Arthropoda

Question 26.
Observe the table given below and fill the blank columns a, b, c, and d from the animals given in brackets.

(Hydra, Shark, Spongilla, Obelia)
Answer:
(a) Hydra/Obelia
(b) Shark
(c) Spongilla
(d) Hydra/Obelia

Question 27.
Copy and complete the table.

Answer:

Question 28.
You are provided with two fishes Catla (Bony fish) and Shark (Cartilagenous fish). Prepare a table showing difference in:
(a) Position of mouth
(b) Air bladder
(c) Scales
(d) Fertilization
Answer:

Question 29.
Prepare a list of some animals that are found parasitic on human beings.
Answer:
Tapeworm (Taenia), Ascaris (Roundworm), Wuchereria (Filaria worm), Ancylostoma(Hookworm)

Question 30.

1. Identify the 2 forms of Cnidarians.
2. Mention any 2 difference between them.

Answer:

1. A – polyp
   B – medusa
2. Polyp: Asexual, sessile, mouth upwards Medusa: Sexual, Free swimming, Mouth downwards

Plus One Animal Kingdom Three Mark Questions and Answers

Question 1.
Classify the given organisms and arrange them in the order of their phylum.
Limulus, Corvus, Spongilla, Ascaris, Physalia, Nereis, Catla, Sepia, Echinus, Taenia, Pleurobrachia, Tiger, Viper, Toad.
Answer:

Question 2.
Match the following.

Answer:

Question 3.
Observe the diagram

1. Identify the phylum of this hypothetical organism.
2. List out the features that helps in identifying it.
3. Write about the fate of notochord in Urochordata, Cephalochordata and Chordata.

Answer:

1. Chordata
2. Notochord, Dorsal nerve cord, Pharyngeal gill slits, Post anal tail
3. In Urochordatarfiotochond is present only in larval tail. In Cephalochondata, notochord extends from head to tail region is persistent throughout their life. In vertebrata, the notochord is replaced by a cartilagenous or bony vertebral column in the adults.

Question 4.
Select the following items into their appropriate phylum.
Radula, Parapodia, Comb plate, Nephridia, Choanocytes, Flame cells.
Answer:

• Radula – Mollusca
• Parapodia – Annelida
• Comb plate – Ctenophora
• Nephridia – Annelida
• Choanocytes – Porifera
• Flame cells – Platyhelminthes

Question 5.

(a) Fill and complete the chart given below.

(b) Write any two fundamental characters of the phylum chordata.
(c) Classify Tetrapoda into classes:
Answer:
Male accessory ducts store and transport the sperms from testis to the outside through urethra. Male accessory glands secrete seminal Plasma, Which is rich in fructose, citrate, prostaglandins and certain enzymes. The secretion of cowper’s glands lubricate the penis.

Question 6.
Write down the functions of the following (any two) structures and assign their phyla.

1. Radula
2. Flame cells
3. Parapodia

Answer:

1. Radula: File like rasping organ for feeding.
   Phylum: Mollusca
2. Flame cells: Osmoregulation and excretion
   Phylum: Platyhelminthes
3. Parapodia: help in swimming
   Phylum: Annelida

Question 7.
1. Identify the organism A and B.

2. Which features of these organism enable you to identify them?
Answer:

1. A-Tapeworm
   B – Earthworm
2. Features of these organism:
   • Tapeworm: Scolex is present, suckers and hooks are present. No true segmentation.
   • Earthworm: True segmentation, Absence of Scolex, hook and suckers, Clitellum is present.

Question 8.

1. Which of the following animals exhibit metagenesis? (Ascaris Obelia Earthworm Crab)
2. To which phylum does it belong?
3. Write any two features of the phylum.

Answer:

1. Obelia
2. Cnidaria
3. Two features of the phylum:
   • Presence of Cnidoblasts
   • Cnidarians exhibit two basic forms called polyp and medusa.

Question 9.
Pick out the appropriate one from the term given within bracket and put against the corresponding phylum.

1. Porifera
2. Coelenterata
3. Platyhelminthes
4. Annelida
5. Arthropoda
6. Mollusca
7. Echinodermata
8. Chordata

(Hirudin, Flame cell, Choanocytes, Cnidoblast, Jointed legs, Radula, Notochord and Dermal Ossicles)
Answer:

1. Porifera – Choanocytes
2. Coelenterata – Cnidoblast
3. Platyhelminthes – Flame cells
4. Annelida – Hirudin
5. Arthropoda – Jointed legs
6. Mollusca – Radula
7. Echinodermata -Dermal ossicles
8. Chordata – Notochord

Question 10.

1. Identify the organism (A) and (B) and which class do they belong?
2. On which basis do you classify these animals?

(Hint: Write any 2 Identifying characters)
Answer:

1. Organism in (A) and (B)
   • A – Bony fish – Osteichthyes
   • B – Cartilagenous fish – Chondrichthyes

2.

Question 11.
From a fish market, you got a fish, on a close watching you friend says it is a cartilaginous fish.

1. Which characters helped him to identify it as a cartilaginous fish, (any four characters.)
2. Name the class it belongs.

Answer:

1. Characters:
   • a – Gillslits are separate and without operculum
   • b – Placoidscale
   • c – Mouth is located ventrally
   • d – heterocercal caudal fin
2. Chondrichthyes

Question 12.
Presence or absence of a cavity between the body wall and the gut wall is very important in classification.

1. Identify the different types of body cavities.
2. Give examples to each

Answer:

1. Different types of Body Cavities:
   • a – Coelomate
   • b – Pseudocoelomate
   • c – Acoelomate
2. Examples:
   • Coelomate – Chordates
   • Pseudocoelomate – Aschelminthes
   • Acoelomate – Platyhelminthes

Question 13.
From the following general characters find out corresponding/Class with an Example

1. Exclusively marine, triploblastic, spines on the skin, radially symmetrical in the adult and bilaterally symmetrical in the larval stage.
2. Marine, they migrate towards freshwater for spawning, then their larvae return to ocean after metamorphosis.
3. Triploblastic, bilaterally symmetrical, coelomate and metamerically segmented animals.

Answer:

1. Echinodermata
   • eg: Starfish
2. Class – Cyclostomata
   • eg: Petromyzon
3. Annelida
   • eg: Earthworm

Question 14.
Arrange the following points in a two-column table and give suitable heading for each column.

1. Notochord present
2. Post anal tail absent
3. Pharynx Perforated by gill slits
4. Notochord absent
5. Posts anal tail present
6. Gill slits are absent

Answer:

Question 15.
Observe the figure and answer the questions.

1. Identify the structure.
2. Name the phylum which possess this structure.
3. How this structure help the organism?

Answer:

1. Cnidoblastornematocyst
2. Cnidaria
3. Capture of Prey
   • Defense organ.

Question 16.
Note the relationship between first two words and suggest suitable words for the 4th place.

1. Planaria: Flartfe cells: Earthworm: _________
2. Jaw present: Gnathostomata:: Jaw absent: ___________

Answer:

1. Nephridia
2. Agnatha

Question 17.
Give reasons for the following.

1. Respiratory and circulatory system are absent in parasitic platyhelminthes and Aschelminthes.
2. Arthropods are the most successful invertebrate.
3. Body of endoparasites are covered with cuticle.

Answer:

1. Parasitic Plalyhelminthes and Aschelminthes lives in anaerobic condition. So respiratory and circulatory systems are absent in these parasitic forms.
2. Arthropods are most successful, because of the presence of unique chitinous cuticle.
3. In these parasites, the cuticle helps in escaping from the action of digestive enzymes.

Question 18.
Arthropodes are organisms with chitinous exoskeleton. Suppose exoskeleton is absent in arthropods. List the difficulties arthropodes has to face.
Answer:

1. Their body will be dried up due to evaporation
2. They couldn’t escape from predators.
3. They couldn’t live in all environments.

Question 19.
Suppose during your field visit for collection from a rocky seashore you have got some live specimens such as sea urchin, sea cucumber, sea anemone. Is it possible to keep them on an aquarium in your school. Give reasons for your answer.
Answer:
No. It is not possible.
Marine animals cannot live on freshwater because it leads to endosmosis and death occurs.

Question 20.

1. Why nematocysts are more concentrated on the oral end and tentacles of cnidarians?
2. What are the difficulties that coelenterate have to face if nematocysts were absent in body.

Answer:

1. Nematocysts are concerned with defence and offence. Tentacles are usually used for defence, offence and food collection. Hence nematocysts are more concentrated in the oral end.
2. The major difficulties cnidarians has to face in the absence of nematocysts are for food collection and escaping from enemies.

Question 21.

1. Which of the following show the body cavity of earthworm?
2. Identify the names of germlayers ‘a’ and ’b’.

Answer:

1. Figure C is the body cavity of earthworm. Because it is a true coelom
2. The names of germlayers
   • a-ectoderm
   • b-endQderm

Question 22.
Identify the characters listed below and put (✓) mark on appropriate places

Answer:

Question 23.

1. Identify the organisms.
2. Which feature of these organisms enable you to identify them?

Answer:

1. The organisms:
   • A-Tape worm
   • B – Ascaris
   • C – Earthworm
2. Features:
   • Tapeworm – Body is dorso-ventrally flattened and segmented.
   • Ascaris – Body is cylindrical and vermiform
   • Earthworm – Body is divided into similar segments and clitellum is present.

Question 24.
During a classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Jusftfy.
Answer:
1. Sponges: Cellular grade of organisation and cell aggregate body plan.

2. Cnidarians: Tissue level of organisation and blind sac body plan. So Cnidarians are more complex than sponges.

Question 25.
While comparing the digestive system of a roundworm and flatworm, a boy noted some differences. List out the differences.
Answer:
1. Flatworms: Digestive system incomplete, has only a single opening, ingestion and egestion occurs through the same opening.

2. Round worm: Complete digestive system, it has both mouth and anus. A muscular pharynx is also present in their digestive system.

Question 26.

1. Identify the invertebrate whose mouth is on ventral or lower side and anus is on dorsal or upper side.
2. Mention the phylum.
3. Comment on its General characters.

Answer:

1. Starfish
2. Echinodermata
3. Its General characters are:
   • Water vascular system
   • Tube feet
   • Coelomate
   • Spiny body

Question 27.
Match the following.

Answer:

• Arthropoda – Bombyx – Joint footed animals
• Annelida – Earthworm – Little rings
• Echinodermata – Antedon – Spiny bodies animals
• Mollusca – Pila – Soft bodies animals

Question 28.
Animals showing metameric segmentation are included under annelida. Body of tapeworm has numerous segments, but the animal is not included under annelida. How will you account for it?
Answer:
Tapeworm show false segmentation. In true segmentation number of segments is fixed and age of the segments are all same.

Question 29.
Complete the given flow chart showing the flow of water through canal system in sponges.

Answer:

Question 30.
Mention a single word for the following.

1. Sexes are not separate.
2. Body wall with three layers.
3. File like rasping organs in Mollusca.
4. Segmentation of animal body.
5. The property of a living organism to emit light.
6. Alternation of generation in Cnidarians.

Answer:

1. Hermaphrodite
2. Triploblastic
3. Radula
4. Metamerism
5. Bioluminescence
6. Metagenesis

Question 31.
The birds are well adapted for flying.

1. Write the general characters of Aves.
2. Give some of the flight adaptations seen in birds.

Answer:

1. General characters of Aves:
   • Presence of feathers
   • Presence of beak
   • Forelimbs are modified into wings.
   • Warm blooded
   • Respiration by lungs with air sacs.
   • Oviparous
2. Flight adaptations:
   • Forelimbs are modified into wings
   • Long bones are hollow with air cavities (Pneumatic)
   • Air sacs connected to lungs supplement respiration
   • Constant body temperature

Plus One Animal Kingdom NCERT Questions and Answers

Question 1.
What is the difference between direct and indirect development?
Answer:
1. Direct Development:
In direct development the young animal resembles an adult. There is no intermediate stage.

2. Indirect Development:
In indirect development there is intermediate stage, like larval stage. For example, frog before being developed into adult passthrough a tadpole stage. This is the case of indirect development.

Question 2.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
In parasitic platyhelminthes hooks and suckers are present. Suckers help the parasite, in sucking the blood from the host.

Question 3.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods are the first phylum to have well developed systems to carry out different activities. There is distinct system for respiration, locomotion and reproduction. Their survival capacity is great because of elaborate system. This has helped them survive in diverse conditions. They can live in water, on land and in air.

This can be one of the reasons why arthropods are the largest group among the animal kingdom. Another reason is their early development compared to animals of higher phylum.

Question 4.
Water vascular system is the characteristic of which group of the following:
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
(c) Echinodermata

Question 5.
“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Answer:
All chordates have notochord present in some stage of life.
The difference between vertebrates and nonvertebrates is as follows:
In vertebrates the notochord is present in the embryonic stage. This is replaced by a vertebral column during the adult stages.

Question 6.
How important is the presence of air bladder in Pisces?
Answer:
Presence of air-bladder in Pisces helps in buoyancy. This means that members of pisces don’t have to keep on swimming to remain floating.

Question 7.
What are the modifications that are observed in birds that help them fly?
Answer:
Following modification in birds help them fly:

1. Pneumatic or hollow bones make for a light weight skeleton.
2. Fore limbs are modified into wings to assist in flight.
3. Excertion of urine and faeces is through single opening facilitating weight reduction.
4. Aerodynamic body helps in flying.

Question 8.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
Usually number of eggs produced by oviparous mothers is greater than number of young ones produced by viviparous mothers. The main reason for this is the need of resources required for development of the embryo.
In oviparous the major part of development of the embryo takes place outside the uterus. This makes lesser burden on the mother.

On the other hand in viviparous animals the development takes place inside the uterus so lesser number of young ones can be successfully incubated. Moreover, once eggs are outside they are at risk of getting eaten by some predator because of their immobility, so need of more eggs is there to ensure continuity of progeny.

Plus One Animal Kingdom Multiple Choice Questions and Answers

Question 1.
Calcareous skeleton is found in
(a) echinoderms
(b) some sponges
(c) mollusca
(d) all the above
Answer:
(d) all the above

Question 2.
Which cannot be the character of cnidaria
(a) musculoepithelial cells
(b) gastrovascular cavity
(c) nerve cells and process
(d) organ grade organization
Answer:
(d) organ grade organization

Question 3.
A non-matching set in the following is
(a) sepia – cuttle fish
(b) octopus – devilfish
(c) limulus – king crab
(d) ancylostoma – pinworm
Answer:
(d) ancylostoma – pinworm

Question 4.
A character common to Echinoderms and chordates
(a) marine
(b) benthonic
(c) deuterostome
(d) none of the above
Answer:
(c) deuterostome

Question 5.
Largest animal in the world that feed on smallest plankton is the
(a) dolphin
(b) killer whale
(c) blue whale
(d) sea cow
Answer:
(c) blue whale

Question 6.
Which is common to amphibian, reptelia and fishes
(a) nucleated RBC
(b) dermal scales
(c) poikelothermic condition
(d) both a and d
Answer:
(d) both a and d

Question 7.
Which of the following has pseudocoelomate tube within a tube body plan
(a) hydra
(b) planaria
(c) ascaris
(d) pheretima
Answer:
(c) ascaris

Question 8.
Ink gland associated with alimentary canal is found in
(a) sepia
(b) earthworm
(c) starfish
(d) cockroach
Answer:
(a) sepia

Question 9.
Which is common to all tetrapods
(a) epidermal scales
(b) red coloured blood
(c) 12 pairs of cranial nerve
(d) ureotelism
Answer:
(b) red coloured blood

Question 10.
Which one of the following is not a characteristic of phylum annelida?
(a) closed circulatory system
(b) segmentation
(c) pseudocoelom
(d) ventral nerve cord
Answer:
(c) pseudocoelom

Question 11.
Respiratory pigment of mollusc is
(a) haemocyanin
(b) haemoglobin
(c) haemoerythrin
(d) both a and b
Answer:
(a) haemocyanin

Question 12.
Select the character that can be attributed to chondrithytes
(a) persistened notochord
(b) placoid scales
(c) poikelothermic body
(d) all the above
Answer:
(d) all the above

Question 13.
The central cavity of sponge is called
(a) spongocoel
(b) coelocentron
(c) canal system
(d) spongilla
Answer:
(c) canal system

Question 14.
Flame cell are excretory organ of
(a) hydra
(b) cockroach
(c) planaria
(d) frog
Answer:
(c) planaria

Question 15.
Pneumatic skeleton is a feature of
(a) amphibians
(b) reptiles
(c) fishes
(d) birds
Answer:
(d) birds

Question 16.
The number of gills present in osteichthyes is
(a) 2 pairs
(b) 6-15 pairs
(c) 5 pairs
(d) 4 pairs
Answer:
(d) 4 pairs

Students can Download Chapter 6 Work, Energy and Power Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 6 Work, Energy and Power

Plus One Physics Work, Energy and Power One Mark Questions and Answers

Plus One Physics Work Energy Power Questions Question 1.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Plus One Physics Chapter Wise Questions And Answers Question 2.
What is the work done by the tension in the string of simple pendulum?
Answer:
Zero

Plus One Physics Work Energy Power Question 3.
When is the exchange of energy is maximum during an elastic collision?
Answer:
When mass of two colliding bodies are same, there will be maximum exchange of energy.

Work Energy And Power Questions And Answers Question 4.
In atom, an electron is revolving around the nucleus. What is the work done?
Answer:
Work done is zero because work done by centripetal force is zero.

Plus One Physics Important Questions And Answers Pdf Question 5.
What is the type of collision when macroscopic particles collide?
Answer:
Perfectly inelastic collision.

Hsslive Plus One Physics Chapter Wise Questions And Answers Question 6.
Name the parameter which is a measure of degree of elasticity of a body.
Answer:
Coefficient of restitution.

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Question 7.
What is the source of kinetic energy for falling rain drops?
Answer:
Gravitational potential energy.

Plus One Physics Work, Energy and Power Two Mark Questions and Answers

Work Energy And Power Class 11 Important Questions With Answers Question 1.
The law of conversation of energy states that energy can neither be created nor be destroyed but can only change from one form into another. A bus and a car, moving with the same kinetic energy are brought to rest by applying an equal retardation force by the breaking systems. Which one will come to rest at a shorter distance? Give the reason behind your answer.
Answer:
Change in K.E. = Force × Displacement
1/2 mv² = F × S
ie. KE α s
KE_car α S_car _____(1)
KE_bus α S_bus ______(2)

ie. S_car = S_bus
Both will travel equal distance.

Work Energy And Power Questions And Answers Pdf Question 2.
A body constrained to move alomg the Z-axis of a co-ordinate system is subjected to a constant force \(\bar{F}=(\hat{i}+2 \hat{\jmath}+3 \hat{k}) N\)

1. What is the magnitude of force along z direction.
2. What is the work done by this force in moving the body over a distance of 4m along z-axis.

Answer:

1. 3N
2. Work done = Force × Displacement = 3 × 4 = 12J.

Physics Questions On Work Energy And Power Question 3.
Match the following

Answer:
Collision of two balls – inelastic – TE and momentum Collision of two molecules – elastic – KE, TE, and momentum.

Plus One Physics Work, Energy and Power Three Mark Questions and Answers

Questions From Work Energy And Power Question 1.
A car of mass 1000kg moving with a speed 18mk/h on a horizontal road collides with a horizontally mounted spring of spring constant 6.25 × 10³N/m

1. What do you mean by Spring constant.
2. What is the maximum compression of the spring?

Answer:
1. Spring constant is the force required to stretch the spring by a unit distance.

2. \(\frac{1}{2}\)mv² = \(\frac{1}{2}\)kx², 18km/h = 5m/s.

x = 2m.

Question 2.
A man tries to lift a mass 200kg with a force 100N

1. Is he doing work? Explain.
2. If yes, find the amount of work done If No, find the force required to lift it.
3. If it is lifted to 2m in 10 seconds, find his power.

Answer:

1. No work is done, as there is no displacement, 100N force is insufficient to raise 200kg.
2. Force required to lift 200 kg = 200 × 9.8 = 1960N
3. Power = \(\frac{m g h}{t}=\frac{200 \times 9.8 \times 2}{10}\) = 392W.

Question 3.
Two cricket balls are colliding each other.

1. Name the collision
2. Say whether law of conservation of Kinetic Energy hold good in this case. Why?
3. State and prove the other conservation law applicable here.

Answer:

1. Inelastic collision
2. No, Total KE before collision is not equal to total KE after collision.
3. Proof and statement of law of conservation of momentum.

Plus One Physics Work, Energy and Power Four Mark Questions and Answers

Question 1.
Two cars A and B travelling with speeds 20m/s and 10m/s respectively applies breaks,.so that A comes to rest in 15 second and B in 7.5s

1. From the graph determine which of the two cars travelled further after brakes were applied and by how much distance it travelled?
2. Draw the velocity time graph of A and B in the same graph.
3. In the above process ,the wear and tear of which the car gets affected more ?

Answer:
1. The area of velocity time graph gives displacement distance travelled by the car A, S_A = 1/2 × 20 × 15 = 150 m
distance travelled by the car B, S_B = 1/2 × 10 × 7.5 = 37.5.

2.

3. Wear and tear gets affected more for the car A.

Question 2.
A sphere of mass m is moving with a velocity u and makes a head on collision with another identical mass which is at rest. It is observed that the stationary mass starts moving with a lesser velocity than u, after the collision.

1. Which physical quantity is conserved here?
2. Define coefficient of restitution.
3. Determine the ratio of the velocities of the two spheres after elastic collision if ‘e’ is the coefficient of restitution.

Answer:
1. conservation of linear momentum.

2. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

3. Coefficient of restitution,
e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
in this cos u₁ = u, u₂ = 0, v₁ = 0, v₂ = u
∴ e = \(\frac{u-0}{u-0}\)
e = 1.

Question 3.
From the table given below

1. Draw the force displacement curve
2. Analyse the graph & find the type of force involved
3. Estimate the workdone

Answer:
1.

2. Workdone by a variable force.

3. Workdone = Area of the graph
= \(\frac{1}{2}\)bh
= \(\frac{1}{2}\)5 × 10 = 25J.

Question 4.
Raju increased the speed of moving mass ‘50 kg’ from 2 m/s to 4m/s.

1. How much force will be required, if velocity change takes place with in 0.2 sec?
2. How much work is done by Raju?

Answer:
1. F = mass × acceleration
= 50 × \(\frac{(4-2)}{0.2}\)
= 500N.

2. w = \(\frac{1}{2}\)mv² – \(\frac{1}{2}\)mu².
=\(\frac{1}{2}\)50 (4² – 2²)
= 300 J.

Plus One Physics Work, Energy and Power Five Mark Questions and Answers

Question 1.
A car and a truck have the same kinetic energies at a certain instant while they are moving along two parallel roads. (Assume that the truck is heavier than the car)

1. Which one will have greater momentum?
2. Write the relationship between kinetic energy and linear momentum.
3. If the mass of truck is 100 times greater than that of the car, find the ratio between their velocities.

Answer:
1. Kinetic energy, of car, K.E_c = \(\frac{P_{c}^{2}}{2 m_{c}}\)
Kinetic energy of truck,

Since m_c < m_t
Hence P_t > P_c
∴ momentum of truck is greater than car.

2. KE = P²/2m.

3. KE_c = KE_t
1/2m_c V_c² = 1/2 x m_tV_t²
But m_t = 100m_c
\(V_{t}^{2}=\frac{V_{c}^{2}}{100}\)
Velocity of truck, Vt = \(\frac{V_{c}}{10}\)
ratio of velocity, 10V_t = V_c
10:1.

Question 2.
Raju dropped a rubber ball of mass m from a height h to the ground. He observed that the ball rebounds vertically and along the same line to a height h₁, which is less than h.

1. Is it an elastic or inelastic collision?
2. Find the velocity with which it strikes the ground?
3. If it is replaced by a solid aluminium ball, then what happens to the height of rebound?
4. If the rubber ball is allowed to fall on a spring placed on the ground then what change will Raju notice in the height of rebound?

Answer:
1. Inelastic collision.

2. The velocity with which ball strikes the ground,
v² = u² + 2as
v² = 0 + 2g × h
v = \(\sqrt{2 g h}\).

3. Height of rebound decreases.

4. Height of rebound depends on the state of potential energy stored in the spring. If ball falls on a compressed spring, the height of rebound increases due to potential energy given by the spring to bail.

Question 3.
A man tries to lift a mass 200kg with a force 100N.

1. Is he doing work? Explain.
2. If it is lifted to 2m in 10s, find the power.
3. Show that total mechanical energy is conserved fora freely falling body.

Answer:
1. No. Force required to lift the body is 2000N (w = mg = 200 × 10). But the applied force is 100N. Hence there is no displacement due to this applied force.

2. Power P = \(\frac{w}{t}=\frac{m g h}{t}=\frac{200 \times 10 \times 2}{10}\) = 400 watt.

3.

Consider a body of mass ‘m’ at a height h from the ground.
Total energy at the point A
Potential energy at A,
PE = mgh
Kinetic energy, KE = \(\frac{1}{2}\) mv² = 0
(since the body at rest, v = 0).
∴ Total mechanical energy = PE + KE
= mgh + 0 = mgh
Total energy at the point B
The body travels a distance x when it reaches B. The velocity at B, can be found using the formula.
v² = u² + 2as
v² = 0 + 2 gx

∴ KE at B, = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) m2gx
= mgx
P.E. at B, = mg(h – x)
Total mechanical energy = PE + KE
= mg(h – x) + mgx
= mgh
Total energy at C
Velocity at C can be found using the formula
v² = u² + 2as
v² = 0 + 2 gh
∴ KE at C, = \(\frac{1}{2}\)mv²
= \(\frac{1}{2}\)m2gh
= mgh
P.E. at C = 0
Total energy = PE + KE = 0 + mgh = mgh.

Question 4.
An elevator of total mass 1800kg is moving up with a constant speed of 2m/s. A frictional force of 400N acts on this motion.

1. The direction of frictional force is______
   • Opposite to direction of motion
   • In the direction of motion
2. What is the work done by gravitational force.
3. What is the total work done by the elevator?

Answer:
1. Opposite to direction of motion

2. w = F × V
w = mg × 2
= 1800 × 10 × 2
w = 36000J

3. P = P_total × V
= (mg + F_fricti0n) × V
= (1800 × 10 + 4000) × 2
= (18000 + 4000) × 2
P = 44000w.

Question 5.
A stone of mass ‘m’ is to be thrown to a height h

1. What is the acceleration of the stone?
2. With what minimum velocity should it be thrown.
3. At what height does the KE and PE become equal?
4. Find the velocity at that height

Answer:
1. ^–g or ^–9.8m/s.

2. v = 0, a = -g, S = h
Substitute this values in
V² = u² + 2as we get
0 = u² – 2gh
u = \(\sqrt{2 g h}\)

3. at \(\frac{h}{2}\)., KE and PE are equal.

4. V² = U² + 2aS
= U² – 2g \(\frac{h}{2}\) (u² = 2gh) = U² – \(\frac{U^{2}}{2}\),
V² = \(\frac{U^{2}}{2}\),
V = \(\frac{U}{\sqrt{2}}\).

Question 6.
A toy gun, with a spring compresser 3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 7.
Find the odd one out and find the relation connecting the remaining quantities. Joule, Calorie, Kilowatt, electron volt.
Answer:
kilowatt, unit of power
1 calorie = 4.2 joule
1 electron volt = 1.6 × 10^-19J.

Question 8.
Atoy gun, with aspring compresser3cm is used to project a stone of mass 50gm to a height of 10m.

1. What is the potential energy of spring.
2. How much it should be compressed to throw the stone to a height 5m.
3. Find out the physical constant associated with the spring.

Answer:
1. PE of the spring = PE of the mass at the height h
= mgh = 0.050 × 9.8 × 10
= 5 × 9.8 = 4.9J

2. \(\frac{1}{2}\)kx₁² = mgh₁
\(\frac{1}{2}\)kx₂² = mgh₂

3. physical constant associated with the spring constant
\(\frac{1}{2}\)kx² = mgh

Question 9.

A graph paper is fitted on a board as shown in figure. Near to the graph paper a spring is placed. A pencil is attached to the end of the spring as shown in figure. The pencil is free to move on the graph paper. A stone of mass 50 gm is placed 1m above the spring. [Spring constant k = 98N/m]

1. The energy possessed by the stone due to its height is called_______
2. If this stone falls on the spring, find the length of mark that produced on the graph paper due to pencil [The change in P E of stone due to compression may be negleted]
3. What will happen to the length of mark, if spring having smaller spring constant is used? Justify.

Answer:
1. Potential energy.

2. \(\frac{1}{2}\)kx² = mgh
\(\frac{1}{2}\) × 98 × x² = 50 × 10^-3 × 9.8 × 1
x² = 100 × 10⁴
x = 10cm.

3. The length of mark will be decreased. Compression of spring depends on spring constant.

Plus One Physics Work, Energy and Power NCERT Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
2. work done by gravitational force in the above case.
3. work done by friction on a body sliding down an inclined plane.
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.
5. work done by the resistive force of air on vibrating pendulum in bringing it to rest.

Answer:

1. +ve
2. -ve
3. -ve
4. +ve
5. -ve

Question 2.
The potential energy function fora particle executing linear simple harmonic motion is given by
V(x) = \(\frac{k x^{2}}{2}\), where k is the force constant of the oscillator, For k = 0.5N nm^-1, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2m.
Answer:
We know that maximum potential energy = total energy
∴ (\(\frac{1}{2}\)kx²) max = 1 joule or \(\frac{1}{2}\) × 0.5 × (x²)_max = 1
or (x²)_max = 4 or (x)_max = ± 2m.

Question 3.
Choose the correct alternative:

1. When a conservative force does positive work on a body, the potential energy of the body increases/ decreases/remains unaltered.
2. Work done by a body against friction always results in a loss of its kinetic/potential energy.
3. The rate of change of total momentum of a many particle system is proportional to the external force/sum of the internal forces on the system.
4. In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/ total energy of the system of two bodies.

Answer:

1. decreases
2. kinetic energy
3. external force
4. total linear momentum and also total energy (if the system of two bodies is isolated).

Question 4.
State if each of the following statements is true or false.

1. In an elastic collision of two bodies, the momentum and energy of each body is conserved.
2. Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False
2. False
3. False
4. False (true usually but not always).

Question 5.
A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) untill at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10ms^-1?
Answer:
r = 2 × 10^-3m,
volume = \(\frac{4}{3} \times \frac{22}{7}\) (2 × 10^-3)³ m³
p = 1000kgm^-3,
h = 250m
W= \(\frac{4}{3} \times \frac{22}{7}\) × 8 × 10^-9 × 1000 × 9.8 × 250J = 0.082J
Data reamains unchanged in the next half.

Question 6.
A bullet of mass 0.012kg and horizontal speed 70ms-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
If V be the velocity of the block after collision, the using law of conservation of momentum, we get
0.012 × 70 + 0 = (0.012 + 0.4)V
or V = \(\frac{0.012 \times 70}{0.412}\) ms^-1 = 2.04ms^-1
If h be the height through which block rises, then
(M + m) gh = \(\frac{1}{2}\) (M + m)V²
or h = \(\frac{v^{2}}{2 g}\) or
h = \(\frac{2.04 \times 2.04}{2 \times 9.8}\)m = 0.212m = 21.2 cm
Amount of heat produced in the block = loss of K.E.
= \(\frac{1}{2}\) × 0.012 × 70 × 70 – \(\frac{1}{2}\) × 0.412 × 2.04 × 2.04
= 29.4J – 0.857J = 28.543J.

Students can Download Chapter 1 Fundamentals of Computer Questions and Answers, Plus One Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Application Chapter Wise Questions Chapter 1 Fundamentals of Computer

Plus One Computer Application Fundamentals of Computer 1 Mark Questions and Answers

Plus One Computer Application Textbook Questions And Answers Question 1.
________ is a collection of unorganized fact.
Answer:
Data

Plus One Computer Application Chapter Wise Questions And Answers Question 2.
Data can be organized into useful ______.
Answer:
Information

Plus One Computer Application Textbook Solutions Question 3.
________ is used to help people to make decision.
Answer:
Information

Plus One Computer Application Chapter 1 Questions And Answers Question 4.
Processing is a series of actions or operations that convert inputs into _______.
Answer:
Output

Plus One Computer Application Textbook Answers Question 5.
The act of applying information in a particular con-text or situation is called ________.
Answer:
Knowledge

Plus One Computer Application Chapter Wise Questions And Answers Pdf Question 6.
What do you mean by data processing?
Answer:
Data processing is defined as a series of actions or operations that converts data into useful information.

Plus One Commerce Computer Application Textbook Answers Question 7.
Odd man out and justify your answer.
(a) Adeline
(b) 12
(3) 17
(d) Adeline aged 17 years is in class 12.
Answer:
d) This is information. The others are data.

Computer Application Exam Questions And Answers Question 8.
Raw facts and figures are known as ________.
Answer:
data

Plus One Computer Application Textbook Question 9.
Processed data is known as ______.
Answer:
Information

Plus One Computer Application Chapter Wise Notes Question 10.
Which of the following helps us to take decisions ?
(a) data
(b) information
(c) Knowledge
(d) intelligence
Answer:
(b) information

All In One Computer Application Class 10 Pdf Question 11.
Manipulation of data to get information is known as _________.
Answer:
Data processing

Class 7 Computer Chapter 1 Number System Question 12.
Arrange the following in proper order .
Process, Output, Storage, Distribution, Data Capture, Input.
Answer:
a) Data Capture
b) Input
c) Storage
d) Process
e) Output
f) Distribution

Question 13.
Pick the odd one out and give reason
a) Calculation
b) Storage
c) Comparison
d) Categorization
Answer:
b) Storage
It is one of the data processing stage the others are various operations in the stage Process.

Question 14.
Information may act as data. State true or False.
Answer:
False

Question 15.
Complete the Series.
a) 1012,1112,10012 ______,_______.
b) 10112,11102,100012 , ______,______.
Answer:
a) 1011, 1101
b) 10101,10111

Question 16.
What are the two basic types of data which are stored and processed by computers?
Answer:
Characters and number

Question 17.
The number of numerals or symbols used in a number system is its _____.
Answer:
Base

Question 18.
The base of decimal number system is _____.
Answer:
Base

Question 19.
MSD is _______.
Answer:
Most significant digit

Question 20.
LSD is ________.
Answer:
Least significant digit

Question 21.
Consider the number 627. Its MSD is _____.
Answer:
6

Question 22.
Consider the number 23.87. Its LSD is ______.
Answer:
7

Question 23.
The base of Binary number system is ________.
Answer:
2

Question 24.
What are the symbols used in Binary number system?
Answer:
0 and 1

Question 25.
Complete the following series.
(101)2, (111)2, (1001)2, ……..
Answer:
1011, 1101

Question 26.
State True or False.
In Binary, the unit bit changes either from 0 to 1 or 1 to 0 with each count.
Answer:
True

Question 27.
The base of octal number system is
Answer:
8

Question 28.
Consider the octal number given below and fill in the blanks.
0, 1,2, 3, 4, 5,6, 7, _
Answer:
10

Question 29.
The base of Hexadecimal number system is
Answer:
16

Question 30.
State True or False.
In Positional number system, each position has a weightage.
Answer:
True

Question 31.
In addition to digits what are the letters used in Hexa decimal number system.
Answer:
A(10), B(11), C(12), D(13), E(14), F(15)

Question 32.
Convert (1110.01011)2 to decimal.
Answer:
1110.01011 = 1 x 23 + 1 x 22 + 1 x 21 + 0 x 20 + 0 x 2- 1 + 1 x 2 – 2 + 0 x 2 – 3 + 1 x 2 – 4 + 1 x 2 – 5
= 8 + 4 + 2 + 0 + 0 + 0.25 + 0 + 0.0625 + 0.03125
= (14.34375)₁₀

Question 33.
1 KB is ______ bytes.
(a) 2⁵
b) 2¹⁰
c) 2¹⁵
d) 2²⁰
Answer:
d) 2²⁰

Question 34.
The base of hexadecimal number system is _______ Hexadecimal number system .
Answer:
16

Question 35.
A computer has no _______.
(a) Memory
(b) l/o device
(c) CPU
(d) IQ
Answer:
IQ

Question 36.
Real numbers can be represented in memory by using ______.
Answer:
Exponent and Mantissa

Question 37.
Consider the number 0.53421 x d) 10^-8 Write down the mantissa and exponent.
Answer:
Mantissa : 0.53421
Exponent:- 8

Question 38.
Characters can be represented in memory by using ______.
Answer:
ASCII Code

Question 39.
ASCII Code of ‘A’ is
Answer:
(100 0001)₂= 65 .

Question 40
ASCII Code of ‘a’ is
Answer:
(110 0001)₂= 97

Question 41.
Find MSD in the decimal number 7854.25
Answer:
Because it has the most weight

Question 42.
Which is the MSB of representation of-80 in SMR?
Answer:
It is 1 because In SMR if the number is negative then the MSB is 1.

Question 43.
Write 28.756 in Mantissa exponent form.
Answer:
28756 = .28756 x 100
= .28756 x10₂
= .28756 E + 2

Question 44.
ASCII stands for ______.
Answer:
American Standard Code for Information Interchange

Question 45.
List any two image file formats.
Answer:
BMP, GIF

Question 46.
Name the character representation coding scheme developed in India and approved by the Bureau of Indian Standards (BIS).
Answer:
lSCII(lndian Standard Code for Information Interchange)

Question 47.
Fill the series. Series
(151)₈, (153)₈, (155)₈ _____,_____.
Answer:
(157)₈, (161)₈

Question 48.
Meaningful and processed form of data is known as _______ Process
Answer:
Information

Question 49.
Choose the correct number system from the following to which the number 121 (one hundred and twenty one) belongs.
a) Octal and Decimal
b) Binary only
c) Binary, Octal, Decimal and Hexadecimal
d) Decimal only
Answer:
d) Decimal Only OR a) Octal and Decimal

Question 50.
Which one of the following is considered as brain of the computer?
a) Central Processing Unit
b) Control Unit
c) Arithmetic Logic Unit
d) Monitor
Answer:
Central Processing Unit

Question 51.
Which one of the following CPU resister helds address of next instruction to be executed by the processor?
a) Accumulator
b) Instruction Register (IR)
c) Memory address Register
d) Program Counter (PC)
Answer:
d) Program Counter (PC)

Question 52.
Processed data is known as ______.
a) facts
b) figures
c) information
d) raw material
Answer:
c) information

Plus One Computer Application Fundamentals of Computer 2 Marks Questions and Answers

Question 1.
Why do we store information?
Answer:
Normally large volume of data has to be given to the computer for processing so the data entry may be taken more days, hence we have to store the data. After processing these stored data, we will get Information as a result that must be stored in the computer for future references.

Question 2.
Which is the final stage in data processing?
Answer:
Distribution of information is the final stage in data processing

Question 3.
What is source document.
Answer:
Acquiring the required data from all the sources for the data processing and by using this data design a document, that contains all relevant data in proper order and format. This document is called source document.

Question 4.
Convert (106)₁₀ = ( )₂?
Answer:

Question 5.
Convert (106)₁₀ = ( )₈
Answer:

Question 6.
(106)₁₀ = ( )₁₆?
Answer:

Question 7.
Convert (55.625)₁₀ = ( )₂
Answer:

Question 8.
Convert (55.140625)₁₀ = ( )₈
Answer:

Question 9.
(55.515625)₁₀ = ( )₁₆
Answer:

Question 10.
Convert (101.101)₂ = ( )₁₀?
Answer:

Question 11.
Convert (71.24)₈ = ( )₁₀?
Answer:

Question 12.
Convert (AB.88)₁₆ = ( )₁₀
Answer:

Question 13.
Convert (1011)₂ = ( )₈?
Answer:

Question 14.
Convert (110100)₂ = ( )₁₆?
Answer:

Question 15.
(72)₈ = ( )₂?
Answer:

Question 16.
Convert (AO)₁₆ = ( )₂ ?
Answer:

Question 17.
Convert (67)₈ = ( )₁₆ ?
Answer:

Question 18.
Convert (A1)₁₆ = ( )₈ ?
Answer:

Question 19.
Write short notes about Unicode Unicode
Answer:
it is like ASCII Code. By using ASCII, we can represent limited, number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
Eg:- Malayalam, Hindi, Sanskrit …….

Question 20.
What is the use of the ASCII Code ?
Answer:
ASCII means American Standard Code for Information Interchange. It is a 7 bit code. Each and every character on the, keyboard is represented in memory by using ASCII Code.
Eg:- A’s ASCII Code is 65 (1000001).
a’s ASCII Code is 97 (1100001)

Question 21.
Define the term’bit’?
Answer:
A bit stands for Binary digit. That means either 0 or 1.

Question 22.
Convert the decimal number 31 to binary
Answer:

Question 23.
Find decimal equivalent of (10001 )₂
Answer:

Question 24.
If (X)₈ =(101011 )₂ then find X.
Answer:
Divide the binary number into groups of 3 bits and write down the corresponding octal equivalent.

Question 25.
Fill the blanks
(____)₂ = (AB)₁₆
Write down the 4 bit equivalent of each digit
Answer:

Question 26.
Represent-60 in 1’s complement form – 60 am 1’s complement form
Answer:
Change all 1 to 0 and all 0 to 1 to get the 1’s complement.
– 60 is in 1’s complement is 11000011

Question 27.
Define Unicode.
Answer:
The limitations to store more characters is solved by the introduction of Unicode.
It uses 16 bits so 2¹⁶ =65536 characters(i.e,world’s all written language characters) can store by using this.

Question 28.
Find the smallest number in the list.
a) (1101)₂
b) (A)₁₆
c) (13)₈
d) (15)₁₀
Answer:
Convert all the numbers into a common base i.e. to decimal

Question 29.
Represent – 83 in 1’s complement form.
Answer:
Divide the number 82 by 2 successively and write down the remainders from bottom to top + 83=01010011
To take 1 ‘s complement of a binary number change all 1 s to 0 and all 0’s to 1.
Hence – 83 is 10101100

Question 30.
a) Write the two’s complement form of the decimal number-119.
b) State the benefit of using two’s complement representation as compared to one’s complement form.
Answer:
Binary equivalent of 119 in 8 bit is (0111 0111)2.
To find the 2’s complement of -119. First find the 1’s complement and Odd 1 to it.

b) If a computer uses 8 bit word length, 1 ’s complement method can represent numbers from – 127 to + 127. That means only 127 + 128 = 255 numbers can represent. But 2’s complement method can represent numbers from – 128 to + 127. That means we can represent a total of 256 numbers. We can represent one number more in 2’s complement representation.

Question 31.
Write a short note on Unicode.
Answer:
It is like ASCII Code. By using ASCII, we- can represent limited number of characters. But using Unicode we can represent all of the characters used in the written languages of the world.
Eg:- Malayalam, Hindi, Sanskrit,….

Question 32.
“Central Processing Unit (CPU) is the brain of the computer”. What is the role of Control Unit (CU) in the CPU?
Answer:
All the activities of a computer is controlled by the control unit. That means the function of key board, mouse, monitor, memory etc. are controlled by the control unit.

Question 33.
Storage of data, capturing of data, processing of data, input of data, and output of data are the different stages in data processing. Write these stages in correct order?
Answer:
1) Capturing of data
2) Input of data
3) Storage of data
4) Processing of data
5) Output of data

Question 34.
There is a memory inside the CPU. What is its name? Write down its purpose In the computer.
Answer:
It stores data, intermediate results, Address, instructions etc for CPU to process temperarily.

Question 35.
Convert the hexadecimal (A2D)₁₆ into its octal equivalent.
Answer:
Step 1: First convert the number into binary for this do the following.

Plus One Computer Application Fundamentals of Computer 3 Marks Questions and Answers

Question 1.
Briefly explain data, information and processing with real life example.
Answer:
Consider the process of making coffee.
Here data is the ingredients – water, sugar,milk and coffee powder.
Information is the final product- i.e, Coffee Processing is the series of steps to convert the in-gradients into final product, Coffee. That is mix the water,sugar and milk and boil it. Finally pour the coffee powder.

Question 2.
Differentiate manual data processing and electronic data processing?
Answer:
In manual data processing human beings are the processors. Our eyes and ears are input devices. We get data either from a printed paper, that can be read using our eyes or heard with ears. Our brain is the processor and it can process the data, and reach in a conclusion known as result. Our mouth and hands are output devices.
In electronic data processing the data is processing with the help of a computer. In a super market, key board and hand held scanners are used to in. put data, the CPU process the data, monitor and printers (Bill) are output devices.

Question 3.
Complete the series.
(a) 324₈,327₈,332₈ ,______,______.
(b) 567₈,576₈,605₈ ,______,_______.
Answer:

Question 4.
Fill up the missing digits
(a) (4___)₈=(___110)₂
(b) ( __7___ )₈ = (100 ___ 110)₂
Consider the following:

Answer:
a) 4 ……..100
and 110 ……… 6
So (46)₈ = (100 110)₂
b) 100 ……… 4
7 ………. 111
110 …….. 6
So (476)₈ = (100 111 110)₂

Question 5.
Fill up the missing numbers.
(a) (A ___)₁₆ = ( ___ 1001)₂
(b) ( __ B ___ )₁₆ = (1000 ___ 1111)₂
consider the following:

Answer:

Question 6.
Complete the Series.
(a) 6ADD , 6ADF , 6AE1 ___,____.
(b) 14A9,14AF , 14B5 , ___,____.
Answer:
a) Consider the Seguence.-
6ADD, 6ADF, 6AE1,
Here the ’numbers’ are
0,1,2, 3, 4, 5,6, 7, 8, 9,A,,B, C, D, E, F, 10, 11, ………
The difference between 6ADD & 6ADF is 2

So Add 2 to 6AE1 we will ge 6AE3 Then add 2 to 6AE3 we will get 6AE5 Therefore the missing terms 6AE3, 6AE5
b) Consider the sequence.

So the missing terms are 14BB and 14C1

Question 7.
Find the octal numbers corresponding to the following numbers using shorthand method.
Short hand method
(a) (ADD)₁₆
(b)(DEAD)₁₆
Answer:
a) Step 1 : Write down the 4 bit binary equivalent of each digit.

Step 2 : Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

b) Step 1 : Write down the 4 bit binary equivalent of each digit.

Step 2 :Divide this number into groups of 3 bits starting from the right and write down the octal equivalent.

Question 8.
The numbers in column A have an equivalent number in another number system of column B.
Find the exact match.

Answer:

Question 9.
ASCII Is used to represent characters in memory. Is it sufficient to represent all characters used in the written languages of the world ? Propose a solution. Justify.
Answer:
No It is not sufficient to represent all characters used in the written languages of the world because , it is a 7 bit code so it can represent 2⁷ = 128 possible codes. To represent all the characters Unicode is used because it uses 4 bytes, so it can represent 232 possible codes.

Question 10.
If (126)x = (56)y , then find x and y.
Answer:

Question 11.
If (102)x = (42)y then (154)x = (___) y.
Answer:

Question 12.
a) Name various number systems commonly used in computers.
b) Include each of the following numbers into all possible number systems
Answer:
a) The number system are binary, octal, decimal and hexa decimal.
b) 123 Octal, decimal and hexa decimal
569 Decimal, hexa decimal
1101 Binary, Octal, Decimal, Hexa decimal

Question 13.
Fill up the missing digit. (Score 1)
(41)₈ = ( )₁₆
Answer:
Step 1 : Divide the number into one each and write doWn the 3 bits equivalent.
Step 2: Then divide the number into group of 4 bits starting from the right then write its equivalent hexa decimal.’

So the answer is 21.

Question 14.
Fill up the missing digit. (Score 2)
If (220)_a = (90)_b then (451 )_a = ( )₁₀
Answer:
It contains 2 & 9, so a and b 2, b 8. The values of a can be 8 or 19. The values of b can be 10 or 16, L.H.S > R.H.S. a The possible values of a and b are given below

Question 15.
Fill up the missing digit. . (Score.3)
If (121)_a = (441)_b then (121)_b = ()₁₀
Answer:

Question 16.
Fill up the missing digit. (Score 3)
If (128)_a = (450)_b then (16)_a = ()₁₀
Answer:

Question 17.
Fill up the missing digit.
(3A.6D)₁₆ = ( )₈
Answer:
Step I: Write down the 4 bits equivalent of each digit.

Step II : Divide this number into groups of 3 bit starting from the right side of the left side of the decimal point and starting from the left side of the right side of the decimal point.
So 00/111/010.011/011/010
Step III: Write the octal equivalent of each group: So we will get. (72.332)₈
(3A.6D)₁₆ = (72.332)₈

Question 18.
What are the various ways to represent integers in computer?
Answer:
There are three ways to represent integers in computer. They are as follows:
1) Sign Magnitude Representation (SMR)
2) 1’s Complement Representation
3) 2’s Complement Representation .
1) SMR : Normally a number has: two parts sign and magnitude, eg:- Consider a number +5. Here + is the sign and 5 is the magnitude. In SMR the most significant Bit (MSB) is used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is – ve.

Here MSB is used for sign then the remaining 7 bits are used to represent magnitude. So we can represent 27 = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to + 127 and 0 to – 127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.
2) 1’s Complement Representation : To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but + ve number has no 1’s complement,
eg:- To find the 1 ‘s complement of 21 +21 = 00010101
To get the 1 ‘s complement change all 0 to 1 and all 1 to 0.
– 21 = 11101010
1’s complement of 21 is 11101010
3) 2’s Complement Representation : To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg:- To find the 2’s complement of 21
+21 = 00010101
First lake the 1’s complement for this change all 1 to 0 and all 0 to 1

2’s complement of 21 is 1110 1011

Question 19.
Match the following.

Answer:
1-b
2-f
3-a
4-c
5-d
6-e

Question 20.
Pick invalid numbers from the following.
i) (10101 )₈
ii) (123)₄
iii) (768)₈
iv) (ABC)₁₆
Answer:
i) (10101 )₈ – Valid
ii) (123)₄ – Valid
iii) (768)₈ – Invalid. Octal number system does not contain the symbol 8
iv) (ABC)₁₆ – Valid

Question 21.
Find the largest number in the list
i) (1001 )₂
ii) (A)₁₆
iii) (10)₈
iv) (11)₁₀
Answer:
Convert all numbers into decimal

Question 22.
If (11011)₂ = (A)₈ = (B)₁₆ = (c)₁₀.
Find the value of A, B and C.
Answer:
i) To find the value of A, First divide the binary number (11011)₂ into groups of 3 bits (starting from the right) Then write down the corresponding octal number of each group for this insert a zero (0) in the left side of the binary number.

ii) To find the value of B, First divide the binary number (11011)₂ into groups of 4 bits (starting from the right). Then write down the corresponding.
Hexa decimal equivalent of each group. For this insert 3 zeroes in the left side of the binary number.

iii) To find the value of C find the decimal equivalent so as to do the following.

Question 23.
(i), Binary representation of +38 is 00100110. Which of the following is the 2’s compliment representation of -38?
a) 11011001
b) 00100111
c) 11011010
d) 11011011
ii) If the Octal representation of decimal number X is (64)8. Find the hexa decimal equivalent of X.
Answer:
i) (c) 11011010
ii) Step 1 : First convert the octal number (64)₈ into binary for this write down the 3 bit binary equivalent

Step II:- The obtained binary number is divided into groups of 4 bits, starting from the right. For this insert 2 zeroes in the left side. After that write down the corresponding Hexa decimal number .

Question 24.
Convert the decimal number 29 into binary. Using sign and magnitude form arid 1 ’s complement form represent +29 and -29 in memory in 8-bit word length?
Answer:

A computer with 8 bit word length, sign and magnitude representation of +29 is (00011101)₂ and -29 is (10011101)₂.
1 ’s complement form of (-29) is (11100010)₂.
Note : To find the 1 ’s complement of (+29), change all ones to zeroes and all zeroes to ones.

Question 25.
Data processing refers to the activities performed on dafa to generate information. List the stages of data processing.
Answer:
Data processing phases (6)
a) Capturing data – In this step acquire or collect data from the user to input into the computer.
b) Input- It is the next step. In this step appropriate data is extracted and feed into the computer.
c) Storage – The data entered into the computer must be stored before starting the processing.
d) Processing /Manipulating data – It is a laborious work. It consists of various steps like computations, classification, comparison, summarization etc. that converts input into output.
e) Output of information – In this stage we will get the results as information after processing the data.
f) Distribution of information – in this phase the information(result) will be given to the concerned persons / computers.

Question 26.
a) Convert (1010.11)₂ to decimal.
b) Find the missing terms in the following series.
Answer:

Plus One Computer Application Fundamentals of Computer 5 Marks Questions and Answers

Question 1.
Explain the components of Data processing
Answer:
Data processing consists of the techniques of sorting, relating, interpreting and computing items of data in order to convert meaningful information. The components of data processing are given below.
a) Capturing data – In this step acquire or collect data from the user to input into the computer.
b) Input – It is the next step. In this step appropriate data is extracted and feed into the computer.
c) Storage – The data entered into the computer must be stored before starting the processing.
d) Processing / Manipulating data – It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
e) Output of information – In this stage we will get the results as information after processing the data.
f) Distribution of information – In this phase the information(result) will be given to the concerned persons / computers.

Question 2.
Define computer. What are the characteristics?
Answer:
A computer is an electronic device used to perform operations at very high speed and accuracy. Following are the characteristics of the computer.
1) Speed : It can perform operations at a high speed.
2) Accuracy : It produces result at a high degree . of accuracy.
3) Diligence: Unlike human beings, a computer is
free from monotony, tiredness, lack of concentration etc. We know that it is an electronic machine. Hence it can work four hours without making any errors.
4) Versatility: it is capable of performing many tasks. It is useful in many fields.
5) Power of Remembering: A computer consists of huge amount of memory. So it can store and recall any amount of information. Unlike human beings it can store huge amount of data and can be retrieved when needed.

Disadvantages of computer

(1)No. IQ : It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t take its own decisions.
(2) No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers)

Students can Download Chapter 2 Forms of Business Organisation Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 2 Forms of Business Organisation

Plus One Business Studies Chapter 2 Notes Contets

• Sole Proprietorship – Meaning – Features Advantages & Disadvantages
• Joint Hindu Family Business (H.U.F) – Meaning – Features – Advantages & Disadvantages
• Partnership – Meaning – Features – Advantages & Disadvantages – Types of Partners – Types of Partnership – Partnership Deed – Registration
• Co operative Society – Meaning – Features Advantages & Disadvantages – Types of Co-operative Societies
• Joint Stock Company- Meaning – Features – Advantages & Disadvantages – Types of Companies-Choice of form of Business organisation

Various forms of business organisations are:

(a) Sole proprietorship,
(b) Joint Hindu family business,
(c) Partnership,
(d) Cooperative societies, and
(e) Joint stock company.

Plus One Business Studies Chapter 2 Sole proprietorship:
Sole proprietorship refers to a form of business organization which is owned, managed and controlled by an individual who is the recipient of all profits and bearer of all risks. It is the most common form of business organization.
Features:

1. The sole trader is the single owner and manager of the business.
2. The formation of a sole proprietorship is very easy. There are no legal formalities to form and close a sole proprietorship.
3. The liability of a sole trader is unlimited, i.e. in case of loss, his personal properties can be used to pay the business liabilities.
4. The entire profit of the sole trading business goes to the sole proprietor. If there is any loss it is also to be borne by the sole proprietor alone.
5. The sole trader has full control over the affairs of the business. So he can take quick decisions.
6. A sole trading concern has no legal existence separate from its owner.
7. The death, insolvency etc. of a sole trader causes discontinuity of business.

Merits:
1. Easy formation:
The formation of a sole proprietorship is very easy. There are no legal formalities to form and close a sole proprietorship.

2. Quick Decision:

The sole trader has full control over the affairs of the business. So he can take quick decisions and prompt actions in all business matters.

3. Motivation:
The entire profit of the sole trading business goes to the sole proprietor. It motivates him to work hard.

4. Secrecy:
A sole trader can keep all the information related to business operations and he is not bound to publish firm’s accounts.

5. Close Personal Relation:

The sole proprietor can maintain good personal contact with the customers and employees and thus, business runs smoothly.

Advantage And Disadvantage Of Cooperative Society Limitations

1. Limited capital: A sole trader can start business only on a small scale because of limited capital.
2. Lack of Continuity: Death, insolvency or illness of a proprietor affects the business and can lead to its closure.
3. Limited managerial ability: A sole proprietor may not be an expert in every aspect of management.
4. Unlimited liability: The liability of a sole trader is unlimited, i.e. in case of loss, his personal properties can be used to pay off the business liabilities.
5. Suitability: Sole proprietorship is suitable in the following cases.
   • Where the market is limited, localized and customers demand personalized services. Eg. tailoring, beauty parlour etc.
   • Where goods are unstandardized like artistic jewellery.
   • Where lower capital, limited risk & limited managerial skills are required as in case of retail store.

Joint Hindu Family Business (HUF):
It refers to a form of organisation where in the business is owned and carried on by the members of a joint Hindu family. It is also known as Hindu Undivided Family Business (H.U.F). It is governed by Hindu succession Act, 1956. It is found only in India.

The business is controlled by the head of the family who is the eldest member and is called karta. All members have equal ownership right over the property of an ancestor and they are known as co-parceners.

Features
1. Formation:
For a Joint Hindu family business there should be at least two members in the family and some ancestral property to be inherited by them.

2. Membership:

Membership by virtue of birth in the family.

3. Liability:
The Karta has unlimited liability. Every other coparcener has a limited liability up to his share in the HUF property.

4. Control:
The control of the family business lies with the karta. He takes all the decisions and is authorised to manage the business.

5. Continuity:

The business is not affected by the death of the Karta as in such cases the next senior male member becomes the Karta.

6. Minor Members:
The basis of membership in the business is birth in the family. Hence, minors can also be members of the business.

Merits
1. Effective control:
The karta has absolute decision making power. This avoids conflicts among members

2. Continuity of business:
The death of the karta will not affect the business as the next eldest member will then take up the position

3. Limited liability of members:
The liability of all the co-parceners except the karta is limited to their share in the business.

4. Increased loyalty:
Members are likely to work with dedication, loyalty and care, because the work involves the family name.

Limitation
1. Limited capital:
The capital of HUF is limited since the ancestral property only can be used for the business. This reduces the scope for business growth.

2. Unlimited liability:
The liability of Karta is unlimited. His personal property can be used to repay business debts.

3. Dominance of karta:
There is a possibility of differences of opinion among the members of the Joint Family. It may affect the stability of the business.

4. Limited managerial skills:
The karta may not be an expert in all areas of management. It may affect the profitability of the business.

Advantages And Disadvantages Of Cooperative Society Partnership:
The Indian Partnership Act, 1932 defines partnership as “the relation between persons who have agreed to share the profit of the business carried on by all or any one of them acting for all.”
Features

1. Formation: For the formation of a partnership, agreement between partners is essential.
2. Liability: The partners of a firm have unlimited liability. The partners are jointly and individually liable for payment of debts.
3. Risk bearing: The profit or loss shall be shared among the partners equally or in agreed ratio.
4. Decision making and control: The activities of a partnership firm are managed through the joint efforts of all the partners.
5. Lack of Continuity: The retirement, death, insolvency, insanity etc of any partner brings the firm to an end.
6. Membership: There must be at least two persons to form a partnership. The maximum number of persons is ten in banking business and twenty in non banking business.
7. Mutual agency: In partnership, every partner is both an agent and a principal.

Merits of Partnership:

1. Easy formation and closure:
A partnership firm can be formed and closed easily without any legal formalities.

2. Balanced decision making:
In partnership, decisions are taken by all partners. So they can take better decisions regarding their business.

3. Division of labour:
Division of labour is possible in partnership firm. Duties can be assigned to different partners according to their ability.

4. Large funds:
In a partnership, the capital is contributed by a number of partners. So they can start business on a large scale.

5. Sharing of risk:
The risks involved in running a partnership firm are shared by all the partners. This reduces the anxiety, burden and stress on individual partners..

6. Secrecy:
A partnership firm is not legally required to publish its accounts and submit its reports. Hence it can maintain confidentiality of information relating to its operations.

Limitations of Partnership:
1. Unlimited liability:
The partners of a firm have unlimited liability. The partners are jointly and individually liable for payment of debts.

2. Limited resources:
There is a restriction on the number of partners. Hence capital contributed by them is also limited.

3. Possibility of conflicts:
Lack of mutual understanding and co-operation among partners may affect the smooth working of the partnership business.

4. Lack of continuity:
The retirement, death, insolvency, insanity etc of any partner brings the firm to an end.

5. Lack of public confidence:
A partnership firm is not legally required to publish its financial reports. As a result, the confidence of the public in partnership firms is generally low.

Business Studies Class 12 Notes Chapter 2 Types of Partners:
1. Active partner:
A partner who contribute capital and takes active part in the business is called an active partner.

2. Sleeping (Dormant partner):
Partners who do not take part in the day to day activities of the business are called sleeping partners. He contributes capital, share profits and losses and has unlimited liability.

3. Secret partner:
A secret partner is one whose association with the firm is unknown to the general public. He contributes to the capital, takes part in the management, shares its profits and losses, and has unlimited liability.

4. Nominal partner (Quasi Partner):
A nominal partner neither contributes capital nor takes any active part in the management of the business. He simply lend his name to the firm. But, he is liable to third parties for all the debts of the firm.

5. Partner by estoppel:
If a partner by his talk or action leads others to believe that he is a partner in a firm, then he is known as partner by estoppel. However, he is liable to third parties.

6. Partner by holding out:
if a partner declares that a particular person is a partner of their firm, and such a person does not disclaim it, then he/she is known as ‘Partner by Holding out’. Such partners are not entitled to profits but are liable to third parties.

7. Minor Partner:
A minor can be admitted to the benefits of a partnership firm with the mutual consent of all other partners. In such cases, his liability will be limited to the extent of the capital contributed by him.

He will not be eligible to take an active part in the management of the firm. But, a minor can share only the profits and cannot be asked to bear the losses. However, he can inspect the accounts of the firm.

Class 11 Business Studies Chapter 2 Notes Types of Partnerships:
On the basis of duration, there are two types of partnerships:

1. Partnership at will
2. Particular partnership

1. Partnership at will:
This type of partnership exists at the will of the partners. It can continue as long as the partners want and is terminated when any partner gives a notice of withdrawal from partnership to the firm.

2. Particular partnership:
Partnership formed for the accomplishment of a particular project or for a specified time period is called particular partnership.

On the basis of liability, the two types of partnerships are:

1. General partnership
2. Limited partnership

1. General Partnership:
In general partnership, the liability of partners is joint and unlimited. Registration of the firm is optional. The existence of the firm is affected by the death, lunacy, insolvency or retirement of the partners.

2. Limited Partnership:
In limited partnership, the liability of at least one partner is unlimited whereas the rest may have limited liability. Registration of such partnership is compulsory. Such a partnership does not get terminated with the death, lunacy or insolvency of the limited partners. This form of partnership is permitted in India after the introduction of Small Enterprise Policy in 1991.

Class 11th Business Studies Chapter 2 Notes Partnership Deed:
The written agreement which specifies the terms and conditions that govern the partnership is called the partnership deed;
Contents

1. Name of firm
2. Nature of business and location of business
3. Duration of business
4. Investment made by each partner
5. Profit sharing ratio
6. Rights, duties and powers of the partners
7. Salaries and withdrawals of the partners
8. Terms governing admission, retirement and expulsion of a partner
9. Interest on capital and interest on drawings
10. Procedure for dissolution of the firm
11. Preparation of accounts and their auditing
12. Method of solving disputes

Registration of partnership:
According to Indian Partnership Act 1932, registration of a partnership is not compulsory, it is optional. However, they can register with the Registrar of firms of the state in which the firm is situated.
Procedure for Registration:

1. 1. Submission of application in the prescribed form to the Registrar of firms. The application should contain the following particulars:
   • Name of the firm
   • Location of the firm
   • Names of other places where the firm carries on business
   • The date when each partner joined the firm
   • Names and addresses of the partners
   • Duration of partnership. This application should be signed by all the partners.
2. Deposit of required fees with the Registrar of Firms.
3. The Registrar after approval will make an entry in the register of firms and will subsequently issue a certificate of registration. The consequences of non-registration of a firm are as follows:
   • A partner of an unregistered firm cannot file suit against the firm or other partner.
   • The firm cannot file a suit against third party.
   • The firm cannot file a case against its partner.

Business Studies Class 11 Notes Chapter 2 Co-operative Society:
The cooperative society is a voluntary association of persons, who join together with the motive of welfare of the members. The basis of co-operation is self help through mutual help, the motto is “each for all and all for each”.

The cooperative society is compulsorily required to be registered under the Cooperative Societies Act 1912. At least ten persons are required to form a society. The capital of a society is raised from its members through issue of shares.

Features:
The important features of a co-operative society are:

1. Voluntary membership: The membership of a cooperative society is voluntary. Membership is open to all, irrespective of their religion, caste, and gender.
2. Legal status: Registration of a cooperative society is compulsory.
3. Limited liability: The liability of the members of a cooperative society is limited to the extent of the amount contributed by them as capital.
4. Control: Management and control lies with the managing committee elected by the members.
5. Service motive: ‘Self help through mutual help’ or ‘each for all and’ all for each’ is the foundation of co-operative society.

Merits:

1. Equality in voting status: The principle of ‘one man one vote’governs the cooperative society.
2. Limited liability: The liability of members of a cooperative society is limited to the extent of their capital contribution.
3. Stable existence: Death, insolvency or insanity of the members do not affect continuity of a cooperative society.
4. Economy in operations: Co-operative society aims to eliminate middlemen. This helps in reducing cost.
5. Support from government: A co-operative society gets support from the government in the form of low taxes, subsidies and low interest rates on loans.
6. Easy formation: The cooperative society can be started with a minimum often members. Its registration procedure is simple involving a few legal formalities

Limitations:

1. Limited resources:
Resources of a cooperative society consists of limited capital contributions of the members.

2. Inefficiency in management:
Cooperative societies are unable to attract and employ expert managers because of their inability to pay them high salaries.

3. Lack of secrecy:
As a result of open discussions in the meetings of members it is difficult to maintain secrecy about the operations of a cooperative society.’

4. Government control:
cooperative societies have to comply with several rules and regulations related to auditing of accounts, submission of accounts, etc. It affects its freedom of operations.

5. Differences of opinion:
The different viewpoints of members in a co-operative society may lead to difficulties in decision making.

Class 11 Bst Chapter 2 Notes Types of co-operative society:
1. Consumer’s cooperative societies:
The consumer cooperative societies are formed to protect the interests of consumers. The society aims at eliminating middlemen to achieve economy in operations. It purchases goods in bulk directly from the wholesalers and sells goods to the members at the lowest price.

2. Producer’s cooperative societies:
These societies are set up to protect the interest of small producers. It supplies raw materials, equipment and other inputs to the members and also buys their output for sale.

3. Marketing cooperative societies:
Such societies are established to help small producers in selling their Products. It collects the output of individual members and sell them at the best possible price. Profits are distributed to members.

4. Farmer’s cooperative societies:
These societies . are established to protect the interests of farmers by providing better inputs at a reasonable cost. Such societies provide better quality seeds, fertilizers, machinery and other modern techniques for use in the cultivation of crops.

5. Credit cooperative societies:
Credit cooperative societies are established for providing easy credit on reasonable terms to the members. Such societies provide loans to members at low rates of interest.

Joint Stock company:
A company may be defined as a voluntary association of persons having a separate legal entity, with perpetual succession and a common seal. It is an artificial person created by law. The companies in India are governed by the Indian . Companies Act, 1956.

The capital of the company is divided into smaller parts called ‘shares’ which can be transferred freely, (except in a private company). The shareholders are the owners of the company. The company is managed by Board of Directors, elected by shareholders.
Features:
1. Incorporated association:
A company is an incorporated association, i.e. Registration of a company is compulsory under the Indian Companies Act, 1956.

2. Separate legal entity:
A company is an artificial person created by law. Company has a separate legal entity apart from its members. It can enter into contracts, own property, sue and be sued, borrow and lend money etc.

3. Formation:
The formation of a company is a time consuming, expensive and complicated process.

4. Perpetual succession:
A company has a continuous existence. Its existence not affected by death, insolvency or insanity of shareholders. Members may come and go, but the company continues to exist.

5. Control:
The management and control of the affairs of the company is in the hands of Board of directors who are elected the representatives of the shareholders.

6. Liability:
The liability of the shareholders is limited to the extent of the face value of shares held by them.

7. Common seal:
The Company being an artificial person acts through its Board of Directors. All documents issued by the company must be authenticated by the company seal.

8. Transferability of shares:
Shares of a joint stock company are freely transferable except in case of a private company.

Merits:

1. Limited liability:
The liability of the shareholders is limited to the extent of the face value of shares held by them. This reduces the degree of risk borne by an investor.

2. Transferability of shares:
Shares of a public company are freely transferable . It provides liquidity to the investor.z

3. Perpetual existence:
A company has a continuous existence. Its existence not affected by death, insolvency or insanity of shareholders.

4. Scope for expansion:
A company has large financial resources. So it can start business on a large scale.

5. Professional management:
A company can afford to pay higher salaries to specialists and professionals. This leads to greater efficiency in the company’s operations.

6. Public confidence:
A company must publish its audited annual accounts. So it enjoys public confidence.

Limitations:

1. Difficulty in formation:
The formation of a company is very difficult. It requires greater time, effort and extensive knowledge of legal requirements.

2. Lack of secrecy:
It is very difficult to maintain secrecy in case of public company, as company is required to publish its annual accounts and reports.

3. Impersonal work:
It is difficult for the owners and top management to maintain personal contact with the employees, customers and creditors.

4. Numerous regulations:
The functioning of a company is subject to many legal provisions and compulsions. This reduces the freedom of operations of a company.

5 Delay in decision making:
A company takes important decisions by holding company meetings. It requires a lot of time.

6. Oligarchic management:
Theoretically, a company is democratically managed but actually it is managed by few people, i.e board of directors. The Board of Directors enjoy considerable freedom in exercising their power which they sometimes ignore the interest of the shareholders.

7. Conflict in interests:
There may be conflict of interest amongst various stakeholders of a company. It affects the smooth functioning of the company.

8. Lack of motivation:
The company is managed by board of directors. They have little interest to protect the interest of the company.

Types of Companies:
A company can be either a private or a public company.
Private Company:
A private company means a company which:

1. restricts the right of members to transfer its shares
2. has a minimum of 2 and a maximum of 50 members
3. does not invite public to subscribe to its share capital
4. must have a minimum paid up capital of Rs.1 lakh

It is necessary for a private company to use the word private limited after its name.

Privileges of a private company:

1. A private company can be formed by only two members.
2. There is no need to issue a prospectus
3. Allotment of shares can be done without receiving the minimum subscription.
4. A private company can start business as soon . as it receives the certificate of incorporation.
5. A private company needs to have only two directors.
6. A private company is not required to keep an index of members.
7. There is no restriction on the amount of loans to directors in a private company.

Public Company:
A public company means a company which is not a private company. As perthe Indian Companies Act, a public company is one which:

1. has a minimum paid-up capital of Rs. 5 lakhs
2. has a minimum of 7 members and no limit on maximum members
3. can transfer its shares
4. can invite the public to subscribe to its shares.

A private company which is a subsidiary of a public company is also treated as a public company. A public company’must use the word limited after its name

Difference between a public company and private company:

A Comparative assessment of different forms of business organisation:

Choice of business organisation:
The important factors determining the choice of organization are:
1. Cost and Ease of formation:
From the point of view of cost, sole proprietorship is the preferred form as it involves least expenditure and the legal requirements are minimum. Company form of organisation, is more complex and involves greater costs.

2. Liability:
In case of sole proprietorship and partnership firms, the liability of the owners/ partners is unlimited. In cooperative societies and companies, the liability is limited. Hence, from the point of view of investors, the company form of organisation is more suitable as the risk involved is limited.

3. Continuity:
The continuity of sole proprietorship and partnership firms is affected by death, insolvency or insanity of the owners. However, such factors do not affect the continuity of cooperative societies and companies. In case the business needs a permanent structure, company form is more suitable.

4. Management ability:
If the organisation’s operations are complex in nature and require professionalized management, company form of organisation is a better alternative.

5. Capital:
If the scale of operations is large, company form may be suitable whereas for medium and small sized business one can opt for partnership or sole proprietorship.

6. Degree of control:
If direct control over business and decision making power is required, proprietorship may be preferred. But if the owners do not mind sharing control and decision making, partnership or company form of organisation can be adopted.

7. Nature of business:
If direct personal contact is needed with the customers, Sole proprietorship may be more suitable. Otherwise, the company form of organisation may be adopted.