Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal

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Adisthana Padavali Malayalam Standard 9 Guide Unit 2 Chapter 2 Velichathinte Viralukal

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Velichathinte Viralukal Hassan Character Sketch In Malayalam 9th

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Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal 19
Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal 20

Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal 21
Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal 22
Adisthana Padavali Malayalam Standard 9 Solutions Unit 2 Chapter 2 Velichathinte Viralukal 23

Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने

You can Download आई एम कलाम के बहाने Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने (फ़िल्मी लेख)

आई एम कलाम के बहाने Text Book Activities & Answers

आई एम कलाम के बहाने विश्लेषणात्मक प्रश्न

I Am Kalam Ke Bahane Question Answer Kerala Syllabus 10th प्रश्ना 1.
‘हमारा सौदा था खेल घंटी में खाने की अदला-बदली का।’ इस तरह की अदला-बदली से हम क्या समझ सकते हैं?
I Am Kalam Ke Bahane Question Answer Kerala Syllabus 10th
उत्तर:
बच्चों के मन में हमेशा प्यार रहता है। उनके बीच ऊँच-नीच या गरीब-धनी का कोई भेद-भाव नहीं होता। वे एक दूसरे को कुछ भी देने को तैयार होते हैं। चाहे वह खाना हो या और कुछ। निरीह बच्चों के इस अदला-बदली से उनकी मानवीयता का बोधमिल जाता है।

I Am Kalam Ke Bahane Kerala Syllabus 10th प्रश्ना 2.
‘रविवार की छुट्टी का दिन उनके लिए बफ़्ते का सबसे बुरा दिन हुआ करता।’ ऐसा क्यों कहा गया है?
I Am Kalam Ke Bahane Kerala Syllabus 10th
उत्तर:
मोरपाल स्कूल जाकर खूब पढ़ना चाहता है। बडा होकर कुछ बनने की आकांक्षाएँ उसके मन में होंगी। इसलिए स्कूली दिन उनके लिए उत्साह के दिन और रविवार छुट्टी का दिन सबसे बुरा दिन लग जाता है।

Iam Kalam Ke Bahane Kerala Syllabus 10th प्रश्ना 3.
‘बाकी निन्यानवे कहानियों को कभी भूलना नहीं चाहिए जो हमारे बचपनों में है। – लेखक ने ऐसा क्यों कहा है?
Iam Kalam Ke Bahane Kerala Syllabus 10th
उत्तर:
कलाम की कहानी बस एक फिल्म की कहानी है। फिल्म में वह अपनी मंज़िन पाता भी है। . लेकिन लेखक के जीवन में हुई अन्य कहानियों का अंत दर्दनाक है। वे बच्चे अपनी मंज़िल पाये बिना बहुत कष्ट उठाकर जी रहे हैं।

Sslc Hindi Unit 2 Chapter 1 Kerala Syllabus प्रश्ना 4.
‘लेकिन छोटू सिर्फ छोटू होकर नहीं जीना चाहता’ – इससे आपने क्या समझा?
Sslc Hindi Unit 2 Chapter 1 Kerala Syllabus
उत्तर:
छोटू स्कूल जाकर खूब पढ़ना चाहता है। राष्ट्रपति अब्दुलकलाम के शब्दों से प्रभावित होकर कलाम बनना चाहता है। इसलिए उसने स्वयं ही अपना नाम कलाम रखा है। अपना भविष्य उज्ज्वल बनाना चाहता है।

I Am Kalam Ke Bahane Notes Kerala Syllabus 10th प्रश्ना 5.
‘लेकिन कलाम फिर कलाम है’ – लेखक के इस प्रस्ताव पर चर्चा करें।
I Am Kalam Ke Bahane Notes Kerala Syllabus 10th
उत्तर:
कलाम एक ईमानदार लड़का है। चोरी का आरोप लगाते समय भी वह सबकुछ सहता है। कुँवर रणविजय को सज़ा मिलने के डर से उनसे हुई दोस्ती के बारे में भी कुछ नहीं कहना। सच्ची दोस्ती यही है। दोस्ती के लिए कुरबान करने वाला ही असली दोस्त निकलता है।

आई एम कलाम के बहाने Text Book Activities and Answers

I Am Kalam Ke Bahane Malayalam Meaning 10th प्रश्ना 1.
लिखें, प्रत्येक पात्र क्या करता था?
I Am Kalam Ke Bahane Malayalam Meaning 10th
आई एम कलाम के बहाने Kerala Syllabus 10th
उत्तर:
Iam Kalam Ke Bahane Question Answer Kerala Syllabus

आई एम कलाम के बहाने Kerala Syllabus 10th प्रश्ना 2.
संबंध पहचानें, लेख से उचित शब्द चुनकर रिक्त स्थान की पूर्ति करें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 28
I Am Kalam Ke Bahane In Malayalam Kerala Syllabus
उत्तर:
Iam Kalam Ke Bahane Notes Kerala Syllabus 10th

Iam Kalam Ke Bahane Question Answer Kerala Syllabus प्रश्ना 3.
संबंध पहचानकर सही मिलान करें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 29
आई एम कलाम के बहाने Meaning In Malayalam 10th
उत्तर:
Sslc Hindi I Am Kalam Notes Kerala Syllabus 10th

I Am Kalam Ke Bahane In Malayalam Kerala Syllabus प्रश्ना 4.
यह प्रसंग पढ़ें, मिहिर और मोरपाल के जीवन अनुभवों के आधार पर टिप्पणी लिखें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 13
‘जिस स्कल में बिताए समय को मैं अपने बचपन का सबसे खराब समय समझ करता था, शायद वही मोरपाल के लिए उसके जीवन का सबसे अच्छा समय होता था।’
उत्तर:
समाज के दो पहलू
मिहिर एवं मोरपाल दोस्त थे। दोस्ती इतनी घनी थी कि खाने की चीज़ भी आपस में बाँट लेते थे। मोरपाल गरीब घराने का लड़का है। उसके माँ-बाप खेत-मजूरी करते थे। घर में उसे कमर तोड़ मेहनत है। फिर भी वह रोज़ स्कूल आना पसंद करता है। पंद्रह किलोमीटर दूर किसी गाँव से रोज़ साइकिल चलाता स्कूल आता था। रोज़ छाछ का डिब्बा लेकर आता था। स्कूल आकर लेखक (मिहिर) का राजमा-चावल बडी चाव से खाता था। क्योंकि उसने पहले कभी यह देखा ही नहीं। स्कूल के प्रति उसका लगाव कुछ खास है इतना कि बिना नागा रोज़ स्कूल आता था। रविवार की छुट्टी का दिन उसके लिए हफ़्ते का सबसे बुरा दिन था। किसी शादी में भी स्कूल के नीली-खाकी यूनीफॉर्म पहनते दिखाई पड़ता था। बेचारे के पास जो कमीज़पैंट का एक ही जोड़ा था, वह नीली-खाकी यूनीफॉर्म था। मोरपाल का स्कूल आठवीं के बाद छूट जाता है। वह आज भी अपने पिता की तरह वहीं खेती-मजूरी करता है।

मिहिर पांडेय धनी परिवार का है। घर मे सारी सुविधाएँ हैं। वह स्कूल जाने से रोया करता था। रोज़ नए बहाने बनाता था। स्कूल से इतना नफ़रत करता था कि अगर स्कूल के रास्ते में पानी भर जाने के कारण स्कूल की छुट्टी हो जाए तो घर में वह नाचता रहता था। उसे स्कूल की नीली-खाकी यूनीफॉर्म से घृणा थी। उसे पहनना हमेशा टाल देता था। स्कूल में बिताए समय को वह अपने बचपन का सबसे खराब समझा जाता था। अपने राजमा-चावल के बदले मोरपाल का छाछ का डिब्बा अपना लेता था।

समाज के दो विभिन्न पहलुओं को मिहिर और मोरपाल के माध्यम से हमारे सामने पेश किया है। मोरपाल के लिए घर में कोई सुविधा नहीं है, फिर भी वह पढ़ना चाहता है। जबकि मिहिर के लिए सुविधा होते हुए भी स्कूल जाना पसंद नहीं करता। स्कूल एवं यनीफॉर्म से गहरा प्यार रखते हुए भी मोरपाल को आठवीं से स्कूल छोडना पडता है और वह पिताजी की तरह खेतीमजूरी करने को विवश होता है। मगर विशेष रुचि न रखते हुए भी मिहिर ऊँचे पद पर पहूँचता है। यहाँ सामाजिक असमता का तस्वीर खींच लिया है।

Iam Kalam Ke Bahane Notes Kerala Syllabus 10th प्रश्ना 5.
पढ़ें, डायरी लिखें।
वह तय करता है कि वह अपनी चिट्ठी सीधे अपने हमनाम डॉ कलाम को दिल्ली जाकर खुद देगा। और वह अकेला ही निकल पड़ता है। रास्ते में मुश्किलें हैं। लेकिन कथा के अंत में कलाम को अपनी मंजिल मिलती है।
फ़िल्म के अंत में छोटू उर्फ कलाम का सपना साकार होता है। अपनी सफलता की बात वह अपनी डायरी में लिखता है। संभावित डायरी लिखें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 14
उत्तर:
20 मई 2019
आज कैसा दिन रहा ! सालों का मेरा सपना आज पूरा हुआ। मेरी खुशी का ठिकाना नहीं। स्कूल में मेरी भर्ती हुई। दोस्त रणविजय के साथ मैं भी स्कूल जा रहा हूँ। सारे लोगों ने मुझे चोर बुलाया था। उस दिन मैं कितना रोया था। मुझे लूसी मैडम से मिलने था। इसलिए सीधे दिल्ली गया था। रास्ते में कितनी मुश्किलें सहना पड़ा था। न खाना, न रहन-सहन। संयोग से हुई मुलाकात की वजह से दोस्त रणविजय एवं रिश्तेदारों को मिला। इसलिए आज स्कूल जाने का सौभाग्य प्राप्त हुआ है। आगे मन में यही वादा है खूब पढूँगा और मंज़िल तक पहुँच जाऊँगा। सीधे डॉ. कलाम से मिलूँगा और खूब बातें करेगा। गाँववालों की मदद करूँगा। काश मैं कलाम बन जाएँ.

आई एम कलाम के बहाने Meaning In Malayalam 10th प्रश्ना 6.
‘हम’ शब्द सर्वनाम के रूप में जाना जाता है। मगर ‘हम’ पूर्व प्रत्यय के रूप में विभिन्न प्रसंगों में विशेष प्रकार का अर्थ पैदा करता है। नीचे दिए आरेख का विश्लेषण करें, ‘हम’ से जुड़े अन्य पदों को ढूँढ़ निकालें और रिक्त स्थानों की पूर्ति करें।
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 15
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 16
उत्तर:
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 17

Sslc Hindi I Am Kalam Notes Kerala Syllabus 10th प्रश्ना 7.
पढ़ें,
1. वह अपनी चिट्ठी कलाम को दिल्ली जाकर खुद देगा।
2. लूसी मैडम वादा करती हैं कि वे उसे अपने साथ दिल्ली लेकर जाएँगी। रेखांकित शब्दों पर ध्यान दें।
चर्चा करें,
1. रेखांकित शब्द क्रिया के किस समय के होने की सूचना देते हैं?
2. इन वाक्यों में क्रिया की अन्विति किन शब्दों से है
उत्तर:
1. रेखांकित शब्द क्रिया के सामान्य भाविकाल की सूचना देते हैं।
2. इन वाक्यों में क्रिया की अन्विति कर्ता के साथ है। (उदा : देगा – वह, जाएँगी – वे)

Hss Live Guru 10th Hindi Kerala Syllabus 10th प्रश्ना 8.
पढ़ें,
1. मैं स्कूल जाने में रोया करता।
2. पानी भर जाने से छुट्टी हो जाया करती।
क्रिया रूपों की विशेषता पर
चर्चा करें।,
रेखांकित क्रिया रूपों से क्रिया के होने से संबंधित कौन-सा आशय मिलता है?
उत्तर:
1. क्रिया रूप नित्यता बोधक क्रिया रूप हैं।
2. नित्य करनेवाले काम को इन क्रियारूपों से घोतित करते हैं। हमेशा मुख्य क्रिया के भूतकाल रूप के साथ ‘जाना’ क्रिया का प्रयोग करते हैं।

I Am Kalam Ke Bahane Summary Kerala Syllabus 10th प्रश्ना 9.
इस प्रकार के अन्य वाक्य चुनकर लिखें ।
उत्तर:

  • रोज़ नए बहाने बनाया करता ।
  • पानी भर जाने से छुट्टी हो जाया करती ।
  • मैं घर पर नाचा करता।
  • मैं स्कूल की नीली-खाकी यूनीफॉर्म से हमेशा चिढ़ा करता ।
  • उसे पहनना हमेशा टाला करता।
  • रविवार की छुट्टी का दिन उनके लिए हफ़ते का सबसे बुरा दिन हुआ करता।

आई एम कलाम के बहाने Additional Questions and Answers

Sslc Hindi Iam Kalam Ke Bahane Kerala Syllabus  प्रश्ना 1.
‘हमारा सौदा था खेल घंटी में खाने की अदला-बदली का’ । खेलघंटी के समय मिहिर और मोरपाल के बीच का संभावित वार्तालाप लिखें।
उत्तर:
मिहिर : घंटी बज गई, चलो हम कुछ खाएँ।
मोरपाल : हाँ ठीक है।
मिहिर : अरे मोरपाल, तुम्हारे पास क्या है?
मोरपाल : बस छाछ का डिब्बा।
मिहिर : हाय! वह मुझे दे दो, यह तुम लो।
मोरपाल : यह क्या है भाई?
मिहिर : अरे राजमा – चावल।
मोरपाल : राजमा – चावल ! मुझे बहुत पसंद है।
मिहिर : पहले कभी देखा नहीं क्या?
मोरपाल : नहीं तो, मैं पहली बार देख रहा हूँ।
मिहिर : पेट भर खाओ।
मोरपाल : कल भी यही लाना। तुम्हारे लिए छाछ पक्का।
मिहिर : ठीक है।

I Am Kalam Ke Bahane In Malayalam Meaning 10th प्रश्ना 2.
स्कुल में भाषण-प्रतियोगिता चलाई गई। उसमें भाग लेते हुए कुँवर रणविजय को प्रथम स्थान प्राप्त हुआ। मान लें अगले दिन के समाचार-पत्र में प्रस्तुत घटना का रपट आ रहा है। वह रपट तैयार करें।
उत्तर:
भाषण प्रतियोगिता-कुँवर को प्रथम स्थान
जैसलमेर : कल जैसलमेर के कोसमोस हायर सेंकन्टरी स्कूल में एक भाषण प्रतियोगिता चलाई गई। ढाणी के राणा सा का बेटा कुँवर रणविजय को प्रथम स्थान प्राप्त हुआ। पुरस्कार प्राप्ति के बाद उसने कहा कि उसके दोस्त कलाम ने यह भाषण तैयार किया था। इसलिए यह पुरस्कार उसके लिए है। कुँवर की हिंदी उतनी अच्छी नहीं थी। इसलिए दोस्त ने तैयार किया था। उनके बीच की दोस्ती की अनूठी निशानी भी है यह पुरस्कार प्राप्ति। पुरस्कार वितरण स्कूल के प्रधानाध्यापक ने किया। ढाणी में कुँवर के विजय पर खुशी मनाई गई।

आई एम कलाम के बहाने  Summary in Malayalam and Translation

Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 18
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 19
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 20
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 21
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 22
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 23
Kerala Syllabus 10th Standard Hindi Solutions Unit 2 Chapter 1 आई एम कलाम के बहाने 24

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Adisthana Padavali Malayalam Standard 10 Solutions Unit 1 Chapter 2 Oro Viliyum Kathu

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Adisthana Padavali Malayalam Standard 10 Solutions Unit 2 Chapter 1 Kochu Chakkarachi

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Kerala Padavali Malayalam Standard 10 Solutions Unit 1 Chapter 1 Lakshmana Santhwanam

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Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away

Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away English Medium

Select the correct answer.

Sslc Biology Chapter 4 Question Paper Question 1.
Hepatitis get transmitted through
a. contact with contaminated water and soil.
b. animals
c. blood components
d. mosquitoes
Answer:
c. blood components

Question 2.
The pathogen of wilt disease of brinjal.
a. Virus
b. Bacteria
c. Fungus
c. Aphids
Answer:
b. Bacteria

Question 3.
Analyse the statements.
i. Filariasis is spread by culex mosquitoes.
ii. Diphtheria is a bacterial disease.
a. (i) correct (ii) incorrect
b. (i) incorrect (ii) correct
c. (i) and (ii) incorrect
d. (i) and (ii) correct
Answer:
d. (i) and (ii) correct

Question 4.
BCG vaccine is used against
a. AIDS
b. Tuberculosis
c. Dengue fever
d. Malaria
Answer:
b. Tuberculosis

Question 5.
Fill up the blanks by observing the relationship between the first pair.
a. AIDS : Through body fluids
Ringworm : …………..
Answer:
Through contact

b. Nipah : Virus
Tuberculosis : …………….
Answer:

Bacteria
c. Block of blood flow in brain : Stroke Deficiency of insulin or its malfunctioning : ……………………
Answer:
Diabetes

Sslc Biology Chapter 4 Model Question Paper Question 6.
Find the odd one out and note down the common features of others.
a. Diabetes, Fatty liver, Heart attack Tuberculosis
Answer:
Tuberculosis
Others are lifestyle diseases.
b. AIDS, Diphtheria, Nipah, Hepatitis
Answer:
Diphtheria
Others are viral diseases.

Question 7.
The symptom of a genetic disease is given below.
Excess blood is lost even through minor wounds
a. Identify the disease..
b. Write the reason for this disease.
c. Write remedy for this disease.
Answer:
a. Haemophilia

b. Blood clots with the help of some proteins present in blood plasma. The synthesis of protein fails when the genes that control protein synthesis become defective.

c. A complete cure is not possible at present. Temporary relief is brought in by injecting the deficient protein identified through clinical diagnosis.

Question 8.
Observe the graph and answer the questions given below.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 1
a. Name the common name for these diseases.
b. Write the cause of each disease.
Answer:
a. Life style diseases.
b.

  • Fatty liver: Deposition of excess fat in the liver.
  • Stroke : Rupture of blood vessels in brain, block of blood flow.
  • Diabetes : Deficiency of insulin or its malfunctioning.
  • Heart attack : Block of blood flow due to the deposition of fat in coronary arteries, which carry blood to heart.

Question 9.
Observe the picture and answer the questions given below.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 2
a. Identify the organism.
b. Which is the disease caused by this organism?
c. What are the ways by which this organism spreads?
Answer:
a. HIV
b. AIDS
c. Through sexual contact with HIV infected person, from HIV infected mother to the foetus, by sharing needle and syringe contaminated with HIV components, through the reception of blood and organs contaminated with HIV.

Question 10.
The symptoms of some diseases are given. Analyse and answer the questions given below.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 3
a. Identify the diseases.
b. Name the pathogens.
c. How are they spread?
Answer:
a. A. Tuberculosis, B. Malaria, C. Ringworm
b.

  • Tuberculosis: Mycobacterium tuberculosis
  • Malaria: Protozoa
  • Ringworm: Fungus

c.

  • Tuberculosis : through air
  • Malaria: through female anopheles mosquito
  • Ringworm : through contact.

Question 11.
Find the indicators related to virus only.
a. No cell organelles as seen in normal cell.
b. There are pathogenic and beneficial ones.
c. Multiply through binary fission.
d. The toxins produced by these damage living cells.
Answer:
a. No cell organelles as seen in normal cell.

Question 12.
What is cancer? How do normal cells get transformed to cancer cells?
Answer:
Cancer is caused by the uncontrolled division of cells and their spread to other tissues. Environmental factors, smoking, radiations, virus, hereditary factors and alternations in genetic material lead to the transformation of normal cells into cancer cells.

Keeping Diseases Away Previous Year Questions Question 13.
Complete the illustration.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 4
Answer:
i. Bacteria
ii. Bunchy top
iii. Fungus
iv. Quickwilt
v. Paddy

Question 14.
Observe the illustration.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 5
a. Identify the disease indicated in the illustration.
b. Write the reason for this disease.
Answer:
a. Sickle cell anaemia.
b. The defects of genes may also cause deformities in the sequencing of amino acids which are the building blocks of haemoglobin. As a result of this, the structure of haemoglobin changes.

Question 15.
Observe the graph , illustrating the data of the diseases affected by the plants ¡n Ravi farm. Analyse the graph and answer the given questions.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 6
a. Identify the plants which are affected.
b. Which is the most affected crop?
c. Note down the causative organism of each disease.
Answer:
a.

  • Blight – Paddy –
  • Bunchy top – banana
  • Quick wilt – pepper
  • Bud rot – coconut

b. Coconut

c.

  • Blight – Bacteria
  • Bunchy top – Virus
  • Quickwilt, Bud rot – Fungus

Question 16.
Select the statement related to Athlete’s foot from the statements given below.
a. Fungal disease.
b. Appearance of reddish scaly rashes that cause itching is the major symptom.
c. Bacterial disease.
d. Manifests as round red blisters on the skin.
e. Pathogens enter through the toes when they come in contact with contaminated water and soil.
Answer:
a. Fungal disease.
b. Appearance of reddish scaly rashes that cause itching is the major symptom.
e. Pathogens enter through the toes when they come in contact with contaminated water and soil.

Sslc Biology Chapter 4 Pdf Question 17.
The symptom of a disease is given below.
The pathogen multiplies using the genetic mechanism of lymphocytes and the number of lymphocytes decreases considerably and reduces the immunity of the body.
a. Identify the disease.
b. Identify the pathogen.
c. Explain how this disease does not spread.
Answer:
a. AIDS
b. HIV
c.

  • By touch, shaking hands, coughing, sneezing, etc.
  • Through insects like mosquitoes, houseflies, etc.
  • By staying together and sharing food.
  • By using same toilets.
  • By taking bath in the same pond.

Question 18.
Observe carefully the graph, illustrating the data of a survey conducted in certain areas. Analyse the graph and answer the given questions.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 7
a. Which type of mosquito dominates in town B?
b. Identify the disease which is likely to be spread in town A. Name the pathogen which causes the disease.
c. Write an important symptom of this disease.
Answer:
a. Anopheles mosquito
b. Filariasis, filarial worms
c. Swelling in the lymph ducts in the leg

Question 19.
How does smoking affect the following body parts?
a. Brain
b. Lungs
Answer:
a. Brain – Stroke, addiction to nicotine
b. Lungs – Lung cancer, emphysema, bronchitis

Keeping Diseases Away Questions And Answers Question 20.
Choose the diseases transmitted through mosquitoes from the following.
Tuberculosis, Diabetes, Dengue fever, Chicken pox, Chikungunya, Botulism, Malaria, Cholera
Answer:
Dengue fever, Chikungunya, Malaria

Question 21.
Rearrange columns B and C suiting the pictures in column A.
Kerala SSLC Biology Chapter 4 Questions Keeping Diseases Away 8
Answer:
i. b,R
ii. c,P
iii.a,Q

Kerala SSLC Biology Chapter Wise Questions and Answers

Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis

Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis English Medium

Select the correct answer.

Sslc Biology Chapter 3 Question Paper Question 1.
Location of adrenal gland
a. Below thalamus
b. In the scrotum
c. Above the kidneys
d. Just below the sternum
Answer:
c. Above the kidneys

Question 2.
The normal level of glucose in blood
a. 80 – 130 mg/100 ml
b. 70 – 110 mg/100 ml
c. 9-11 mg/100 ml
d. 60 – 100 mg/100 ml
Answer:
b. 70-110 mg/100 ml

Question 3.
The ‘Youth hormone ’ is
a. Thymosin
b. Epinephrine
c. Melatonin
d. Oxytocin
Answer:
a. Thymosin

Question 4.
The hormone in gaseous form
a. Auxin
b. Ethylene
c. Gibberellins
d. Cytokinin
Answer:
b. Ethylene

Sslc Biology Chapter 3 Questions And Answers Pdf Question 5.
Fill up the blanks by observing the relationship between the first pair.
a. Calcium : Calcitonin
Glucose : …………………
b. Breaks up stored food : Gibberellins. Promoting the growth of terminal buds : ………………….
c. Norepinephrine : Medulla Aldosterone : ………………..
Answer:
a. Insulin
b. Auxin
c. Cortex

Question 6.
Find out the odd one and note down the common features of others.
a. Cortisol, Aldosterone, Epinephrine,Sex hormones.
b. TSH, ACTH, GTH, ADH
c. Cretinism, Gigantism, Dwarfism, Acromegaly
Answer:
a. Epinephrine
Others are produced by the cortex of adrenal glands.
b. ADH
Others are tropic hormones.
c. Cretinism
Others are related to the production of somatotropin.

Question 7.
The quantity of urine excreted by a person in different seasons is given below. Analyze it and answer the following questions.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 1
a. Which is the coldest season?
b. Which hormone is responsible for the variation in quantity of urine?
c. How does this hormone regulate the water level of the body?
Answer:
a. Season 3
b. Vasopressin (ADH)
c. If the production of vasopressin decreases, the reabsorption of water in kidneys decreases. As a result. the quantity of water level in urine increases.

Question 8.
Given below is an incomplete table showing reproduction related both’ changes, controlling hormones, their glands and functions. Fill the missing gaps.

HormoneGlandFunction
ProgesteroneaMaintenance of embryo
bOvaryGrowth of sex organs
TestosteroneCd
Prolactinef
ghFacilitating child

Answer:
a. Ovary
b. Oestrogen
c. Testis
d. Sperm production, controls secondary sexual characters in males such as change in voice, growth of hair, etc.
e. Pituitary gland
f. Production of milk
g. Oxytocin
h. Hypothalamus

Question 9.
All hormones do not act upon all cells. Why?
Answer:
Hormones reach every cell in the body as they are transported by blood. But each hormone acts only upon those cells which have specific receptors. The cells which are acted upon by hormones are their target cells

Question 10.
What are the functions of thyroxine? How do the abnormalities in the production of thyroxine affect the body?
Answer:

  • Functions of thyroxine – Raises the rate of metabolism, accelerates the growth and development of the brain in the foetal stage and infancy, increases energy production, regulates growth in children.
  • The deficiency of thyroxine during the foetal stage or infancy – cretinism.
  • Lack of thyroxine in adults – myxoedema.
  • The condition in which all life activities controlled by thyroxine are accelerated due to the excessive production of thyroxine – hyperthyroidism.

Sslc Biology Chapter 3 Questions And Answers Pdf Question 11.
Select the answers for the statements from the box given below.
Auxin, Gibberellins, Ethylene, Cytokinin,
Abscisic acid
a. Helps in the ripening of fruits.
b. Helps to sustain the plant in adverse conditions.
c. Promotes cell growth and differentiation.
d. Breaks up stored food.
e. Cell growth, cell elongation, promoting the growth of terminal buds
Answer:
a. Ethylene
b. Abscisic acid
c. Cytokinin
d. Gibberellins
e. Auxin

Question 12.
Identify the disease. Write the reason for this.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 2
Answer:
Goitre.
Iodine is essential for the production of thyroxine. The production of thyroxine is obstructed in the absence of iodine. In an attempt to produce more thyroxine, the thyroid gland enlarges.

Question 13.
Find out the peculiarities of column A and rearrange columns B and C.

ABC
OxytocinParathyroidPrevents the rise of calcium
ProlactinThyroidFacilitates lactation
ParathormoneHypothalamusProduction of milk
PituitaryPrevents the lowering of calcium

Answer:

  • Oxytocin – Hypothalamus – Facilitates lactation
  • Prolactin – Pituitary – Production of milk
  • Parathormone – Parathyroid – Prevents the lowering of calcium

Question 14.
Hints related to a patient are given below.
Analyze and answer the questions given below.

  • The blood glucose level is more than normal level.
  • Always tired
  • The production of insulin is normal in his body.

a. Identify the disease.
b. Write the reason for this disease.
c. How is the disease related to tiredness?
Answer:
a. Diabetes
b. Malfunctioning of insulin.
c. Insulin enhances cellular uptake of glucose molecules. If the insulin is malfunctioned, it cannot perform this function. The glucose level in cell decreases leading to low energy production. So he feels tired.

Question 15.
Observe the illustration.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 3
Complate A,B,C,D,E
Answer:
A. Cortex
B. Norepinephrine
C. Aldosterone
D. Sex hormones
E. Cortisol

Sslc Biology Chapter 3 Pdf Question 16.
“Isn’t this merely a storage and distribution centre? Is it sensible to call this a gland?” This is a doubt raised by Hari, pointing to a particular portion of a gland in the picture of the brain.
a. Which part of which gland may be the basis of Hari’s doubt?
b. Why does Hari have such a doubt? Is his doubt genuine? Why?
Answer:
a. Pituitary – Posterior lobe
b. Sensible. The hormone produced by hypothalamus is stored and released from here. The posterior lobe of pituitary itself does not produce hormone.

Question 17.
Use of artificial plant hormones contributed a lot to the progress of the agricultural sector.
Justify this statement.
(Hints: Artificial gibberllins, artificial abscisic acid, Ethylene)
Answer:
Artificial gibberllins – It is used for increasing fruit size in grapes and apple and also for preventing ripening of fruits to assist marketing.

Artificial abscisic acid – It is used for harvesting fruits at the same time.
Ethylene- Ethylene is used for the flowering of pineapple plants at a time and for the ripening of tomato, lemon, orange, etc. Ethyphon, a chemical which is available in liquid form, gets transformed into ethylene, when used in rubber trees and it increases the production of latex.

Question 18.
Arrange columns B. C according to the data given in column A.

ABC
Diabetes mellitusExcessive production of somatotropin.Low metabolic rate, hypertension. sleeplessness
AcromegalyLack of thyroxineIncreased appetite and thirst and frequent urination.
CretinismDecreased production of insulin.High metabolic rate, rise in body temperature.
Production of somatotropin decreasesGrowth of the bones on face, jaws and fingers.

Answer:

ABC
Diabetes mellittusDecreased production of insulin.Increased appetite and thirst and frequent urination.
AcromegalyExcessive production of somatotropin.Growth of the bones on face, jaws and fingers.
CretinismLack of thyroxineLow metabolic rate, hypertension, sleeplessness.

Question 19.
Observe the graph and answer the questions given below.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 4
a. Who has the normal level of calcium in blood?
b. Name the hormones which help to regulate the blood calcium level in other persons. Explain the action of these hormones.
Answer:
a. Mini
b.
Raju- The level of calcium in blood is more.
The calcitonin secreted by thyroid gland helps in maintaining the level of calcium in blood by depositing the excess calcium in bones and by preventing the mixing of calcium with blood from the bones.

Jose – The level of calcium in blood is less. The parathormone secreted by parathyroid gland helps in maintaining the level of calcium in blood by the reabsorption of calcium in the blood from kidneys and also prevents the deposition of calcium in bones.

Sslc Biology Chapter Wise Questions And Answers 2023 Question 20.
Some hormones are given below. Make them into four pairs. Give reasons for pairing.
Oxytocin, Thyroxine, Insulin, Epinephrine, Vasopressin. Glucagon, Cortisol, Calcitonin
Answer:

  • Oxytocin, Vasopressin – Secreted by hypothalamus
  • Thyroxine, Calcitonin -Secreted by thyroid gland
  • Insulin, Glucagon – Secreted by pancreas.
  • Epinephrine, Cortisol – Secreted by adrenal gland

Question 21.
Manu rubbed strongly across the path of the row of moving ants. All the ants moved in different directions immediately.
a. Which is the substance responsible for this?
b. Write another use of this substance.
Answer:
a. Pheromones
b. Pheromones help in attracting mates, informing the availability of food, determining the path of travel, informing about dangers, etc.

Question 22.
Analyze the illustration and answer the questions.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 5
a. What is indicated by A (Write the lobe also)?
b. ‘A’ produces the hormones of same type. Do you agree with this statement ? Why?
Answer:
a. The anterior lobe of pituitary.
b. No. It produces tropic hormones (TSR. ACTh,GTH, STH) and prolactin.

Sslc Biology Chapter 3 Questions And Answers Question 23.
Arrange columns B and C according to the data given in column A.

ABC
VasopressinAdrenalOn either side of trachea, just below the larynx
CortisolPituitaryJust below the sternum
ThymosinHypothalamusAbove kidneys
ThymusBelow thalamus

Answer:

ABC
VasopressinHypothalamusBelow thalamus
CortisolAdrenalAbove kidneys
ThymosinThymusJust below the sternum

Question 24.
Arrange the statements suitably ¡n the table given below
a. Low metabolic rate.
b. High metabolic rate
c. Emotional imbalance
d. Increase in body weight
e. Oedema
f. Excessive sweating

HypothyroidismHyperthyroidism
.

.

.

.

.

.

Answer:

HypothyroidismHyperthyroidism
Low metabolic rateHigh metabolic rate
Increase in body weightEmotional imbalance
OedemaExcessive sweating

Question 25.
Observe the illustration and answer the question given below
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 6
a. Identify the hormones indicated by i, ii, iii, iv
b. Write the functions of prolactin and oxytocin.
c. Name the condilion caused by the decreased production of somatotropin?
Answer:
a. i. Vasopressin
ii. Oxytocin
iii. Tropic hormones
iv. Somatotropin
b.

  • Prolactin – Production of milk
  • Oxytocin – Facilitates child birth by stimulating the contraction of smooth muscles in the uterine wall, facilitates lactation,

c. Dwarfism

Sslc Biology Chapter Wise Questions And Answers Pdf Question 26.
Some statements related to endocrine system are given below.
A. Hormones are the secretions of endocrine glands.
B. Hormones are transported through lymph.
C. Hormones are transported through blood.
D. All the hormones produced by the endocrine glands are proteins.
a. Choose the correct statement.
b. Imagine that a particular hormone is not entering a particular cell. What may be the reason?
Answer:
a. A.Hormones are the secretions of endocrine glands.
C.Hormones are transported through blood.

b. Receptors of that hormones are not in the cell.

Question 27.
Case sheets of two patients are given below. Analyse them and answer the questions.

Case -1Case -2
Age – 4 yrsAge – 42 yrs
Mental retardationHigh metabolic rate
Stunted growthIncreased heart beat
Emotional imbalance

a. Which are the diseases whose symptoms are indicated above?
b. Write the reasons for the diseases.
Answer:
a. Case – 1 Cretinism
Case – 2 Hyperthyroidism

b. Case – 1
Deficiency of thyroxine during foetal stage and infancy.
Case – 2
Excessive production of thyroxine.

Question 28.
Observe the diagram of the endocrine gland given below and answer the questions.
Kerala SSLC Biology Chapter 3 Questions Chemical Messages for Homeostasis 7
a. Name the parts indicated as A and B.
b. Name the hormones synthesized by A. Explain their action.
Answer:
a. A-Medulla B – Cortex
b. Epinephrine, Norepinephrine
Epinephrine – Helps to tide over emergency situations.
Norepinephrine – acts along with epinephrine.

Question 29.
Analyse the statements given below and write the reason.
a. Oxytocin is injected in pregnant women during child birth (delivery).
b. Feels sleepy during night, wakes up when day breaks.
Answer:
a. Facilitates child birth by stimulating the contraction of smooth muscles in the uterine wall.
b. When the level of melatonin increases at night, we feel sleepy. We wake up when the level of melatonin decreases during the day.

Sslc Biology Chapter Wise Questions And Answers Question 30.
A farmer named Balan cultivated oranges in his orchard. Now the trees are full of oranges. The price of oranges is Rs. 80/kg.
A. This farmer wants to harvest all fruits together.
B. Ripen them together.
a. Suggest two artificial plant hormones to satisfy the A, B needs of the farmer.
b. Uncontrolled use of plant hormones must
be controlled. Evaluate this statement.
Answer:
a. A – Abscisic acid.
B – Ethylene
b.

  • Control uncontrolled use.
  • Environmental issues.
  • Health issues.

Question 31.
Indicators related to the endocrine glands are given below. Analyse them and answer the questions.

  • Situated just below the sternum.
  • Active during infancy.

But constricts at puberty.
a. Name this endocrine gland.
b. Which is the hormone synthesised by this gland?
c. Write the function of this hormone.
Answer:
a. Thymus gland
b. Thymosin
c. Controls the activities and maturation of lymphocytes which help to impart immunity.

Kerala SSLC Biology Chapter Wise Questions and Answers

Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge

Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge and Responses English Medium

Select the correct answer.

Sslc Biology Chapter 2 Question 1.
The transparent anterior part of sclera
a. Conjunctiva
b. Cornea
c. Iris
d. Pupil
Answer:
b. Cornea

Question 2.
The liquid which helps in maintaining the shape of eye
a. Aqueous humour
b. Cerebrospinal fluid
c. Tears
d. Vitreous humour
Answer:
d. Vitreous humour

Question 3.
The visual pigment in rod cells
a. Rhodopsin
b. Retinal
c. Photopsin
d. Iodopsin
Answer:
a. Rhodopsin

Question 4.
The part that connects middle ear and pharynx
a. Oval window
b. Eustachian tube
c. Round window
d. Tympanum
Answer:
b. Eustachian tube

Question 5.
Fill up the blanks by observing the relationship between the first pair.
a. Planaria : Eyespot :: House fly : ……………..
b. Cochlea : Hearing :: Vestibule : ……………
c. The outer layer made up of connective tissues : Sclera :: The inner layer which has photoreceptors : ………………….
Answer:
a. Ommatidia
b. Body balancing
c. Retina

Question 6.
Find out the odd one and note down the common features of others.
a. Sclera, Choroid, Optic nerve, Retina ‘
b. Pinna, Malleus, Incus, Eustachian tube
c. Vestible, Semicircular canals, Iris, Vestibular nerve
Answer:
a. Optic nerve
Others are layers of eye.
b. Pinna
Others are parts of middle ear.
c. Iris
Others are parts related to body balancing.

Windows of Knowledge Class 10 Questions And Answers  Question 7.
The ciliary muscles of a person do not contract.
a. How does this affect his vision?
b. Does this affect the action of photoreceptors of retina? Why?
Answer:
a. He cannot see near objects.
b. It does not affect the action of photoreceptors of retina. It is because when light ray reaches the retina, the photoreceptors are stimulated.

Question 8.
The diagram given below shows the structure of ear.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 1
a. Redraw the picture.
b. Label the parts A, B, C, D.
c. Write down the name and function of E and F.
Answer:
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 2
b. A. Tympanum, B. Auditory nerve, C. Ear ossicles, D. Eustachian tube
c.

  • E-Semicircular canals. It helps in the maintenance of balancing of body.
  • F-Cochlea. Helps in hearing.

Question 9.
Observe the figure and answer the questions given below.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 3
a. Identify the part labelled as A, B, C, D.
b. Write the peculiarities of part mentioned as A, C.
c. Write the functions of part indicated as B, D.
Answer:
a. A-Cornea, B- Sclera, C-Pupil, D-Optic nerve
b.

  • A-Slightly projected transparent anterior part of the sclera.
  • C – The aperture seen at the centre of the iris. The size increases and decreases depending on the intensity of light.

c.

  • Sclera – Gives firmness to the eye
  • Optic nerve – Transmits impulses from photoreceptors to the visual centre in the brain.

Biology Chapter 2 Class 10 Kerala Syllabus Question 10.
Arrange the statements suitably in the table given below.
a. Ligaments stretch.
b. Focal length decreases.
c. Ciliary muscles relax.
d. Ligaments relax.
e. Ciliary muscles contract.
f. Curvature of lens decreases.
g. Focal length increases
h. Curvature of lens increases.

While viewing near objectsWhile viewing distant objects
.

.

.

.

Answer:

While viewing near objectsWhile viewing distant objects
Focal length decreasesFocal length increases
Ligaments relaxLigaments stretch
Ciliary muscles contractCiliary muscles relax
Curvature of lens increasesCurvature of lens decreases

Question 11.
The flowchart related to the process of hearing is given below complete it.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 4
Answer:
a. Tympanum
b. Ear ossicles
c. Cochlea
d. Auditory nerve
e. Cerebrum

Windows Of Knowledge Class 10 Question Paper Question 12.
Make suitable word pairs by using the words given in the box.
Jacobson’s organ, Housefly, Eye spot, Snake, Shark, Planaria, Ommatidia, Lateral line
Answer:

  • Snake – Jacobson’s organ
  • Housefly – Ommatidia
  • Shark – Lateral line
  • Planafia – Eye spot

Question 13.
The stages related to maintain the equilibrium of the body are given below. Arrange them in correct order.
a. Generates impulses
b. Maintains the equilibrium of the body
c. Body movements create movement in the fluid inside the vestibule and semicircular canals.
d. The impulses are transmitted by the vestibular nerve.
e. Cerebellum enables muscular movements.
f. Creates movements of the sensory hair cells.
Answer:
c. Body movements create movement in the fluid inside the vestibule and semicircular canals.
f. Creates movements of the sensory hair cells.
a. Generates impulses
d. The impulses are transmitted by the vestibular nerve.
e. Cerebellum enables muscular movements.
b. Maintains the equilibrium of the body

Sslc Biology Chapter 2 Questions And Answers Pdf Question 14.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 5
a. Construct two questions related to this figure and see whether you can give any explanation.
b. Give suitable explanation of the questions.
Answer:
a.

  • Two images from an object are formed in two eyes. But we see only one image of the object. Why?
  • Draw a flow chart of light rays from the object to the centre of vision in brain.

b

  • Two images of the same object formed in the two eyes reach the visual area of the brain. The brain combines the two images together to give a single vision.
  • Light from the object → Eyes → Sensory nerve → Brain → Centre of vision.

Question 15.
“Receptors are modified neurons”. Justify the statement with examples of receptors in different sense organ.
Answer:
Receptors are modified neurons. They are of different types. Rods and cones of eye, sound receptors in ear, taste receptors on the tongue, olfactory receptors in nose and receptors in skin are examples.

Sslc Biology Chapter Wise Questions And Answers Pdf Question 16.
Glaucoma is a serious eye defect. Analyze the statement.
Answer:
If the reabsorption of aqueous humour does not occur, it causes an increase in the pressure inside the eyes. This causes damage to the retina and the photoreceptor cells and ultimately leads to blindness.

Sslc Biology Chapter 2 Questions And Answers Pdf Question 17.
Arrange columns B and C according ro ¡he daa given in column A.

ABC
PupilSeen around lensPlenty of photoreceptors are present
Ciliary muscleThe part of retina where the optic nerve beginsSize of aperture regulated depending on the intensity of light
Blind spotPart of choroid seen behind the corneaThe contraction and relaxation of these alter curvature of lens
The aperture seen at the centre of the iris.Photoreceptors are absent

Answer:

ABC
PupilThe aperture seen at the centre of the iris.Size of aperture regulated depending on the intensity of light
Ciliary muscleSeen around lensThe contraction and relaxation of these alter curvature of lens ,
Blind spotThe part of retina where the optic nerve beginsPhotoreceptors are absent

Biology Class 10 Kerala Syllabus Chapter 2 Question 18.
Observe the figure and answer the questions given below.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 6
a. Identify the situation indicated by the figure.
b. Identify the parts indicated as A, B.
c. How do A and B act in the situation?
Answer:
a. While viewing nearby objects
b. A – Ciliary muscles, B – lens
c. The contraction of ciliary muscles and the relaxation of ligaments help to increase the curvature of lens.

Sslc Biology Chapter Wise Questions And Answers Pdf Question 19.
Prepare a flowchart related to the sense of vision by selecting suitable words given in the box.

Cornea, Auditory nerve, Blind spot, Aqueous humour. Retina, Eustachian tube, Light, Impulse, Pupil, Choroid, Vitreous humour, Cerebrum, Cochlea, Optic nerve, Sense of vision, Lens
Answer:
Light → Cornea → Aqueous humour → Pupil → Lens → Vitreous humour → Retina → Impulse → Optic nerve → Cerebrum → Sense of vision

Question 20.
Observe the picture and answer the questions – given below.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 7
a. Identify the picture
b. Identify the parts indicated as A, B, C, D.
c. What are the functions of C, D?
Answer:
a. Internal ear.
b. A- Semicircular canals B – Cochlea C – Autidory nerve D – Vestibular nerve
c.

  • Auditory nerve – Transmits impulses related to hearing to cerebrum.
  • Vestibular nerve – Transmits impulses related to body balancing to cerebellum.

Sslc Biology Chapter 2 Questions And Answers Pdf English Medium Question 21.
Prepare a flowchart related to the experience of taste.
Answer:
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 8

Question 22.
Find out the peculiarities of column A and rearrange columns B and C.

A

Organs

B

Secretions

C

Functions

EyeEndolymphExcretion of waste and regulation of temperature
NoseCerebrospinal fluidProtection, nutrition, removal of urea
EarAqueous fluidHelps in functioning of olfactory receptors
SkinMucusNutrients and oxygen to the tissues
SweatTransmission of auditory impulses

Answer:

A

Organs

B

Secretions

C

Functions

EyeAqueous fluidNutrients and oxygen to the tissues
NoseMucusHelps in functioning of olfactory receptors
EarEndolymphTransmission of auditory impulses
SkinSweatExcretion of waste and regulation of temperature

Question 23.
Binocular vision gives us three dimensional vision. Then why do we close one eye when we aim at an object?
Answer:
Here distance is not important. Location of the object and eye must come in a straight line.

Question 24.
Ear canal → Cochlea → Tympanum → Auditory nerve
a. Correct the mistake, if any, in the above flowchart.
b. Modify the flow chart by adding the terms ‘oval window’ and ‘ear ossicles’
Answer:
a. Ear canal → Tympanum → Cochlea → Auditory nerve
b. Ear canal → Tympanum → Ear ossicles → Oval window → Cochlea →Auditory nerve

Question 25.
The flow chart related to the experience of smell is given below. Complete the flowchart.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 9
Answer:
a.

  • A – Aromatic particles dissolve in the mucus inside the nostrils, stimulate olfactory receptors.
  • B – Generate impulses
  • C – Impulses reach cerebrum through olfactory nerve

Sslc Biology Chapter 2 Questions And Answers Pdf English Medium Question 26.
The symptoms of two below. eye defects are given
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 10
Identify the defects. Write how these defects can be rectified.
Answer:
A – Glaucoma. Laser surgery
B – Cataract. Replace the lens with an artificial one.

Question 27.
Observe the picture and answer the questions given below.
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 11
a. Identify the photoreceptor.
b. Write the pigment and components of this pigment.
Answer:
a. Rod cell
b. Rhodopsin – opsin, retinal

Question 28.
The features of a fluid are given below.

  • Seen between retina and lens
  • Jelly like substance

a. Identify the fluid.
b. Write its function.
Answer:
a. Vitreous humour
b. Helps in maintaining the shape of the eye.

Question 29.
Given below are the two diseases of eyes.
i. Colour blindness
ii. Night blindness
a. Which of these is caused by malnutrition?
b. Deficiency of which nutrient causes this disease?
c. Write the main symptom of this disease.
Answer:
a. Night blindness
b. Vitamin A
c. Cannot see objects clearly in dim light.

Question 30.
Statements related to sense organs are given below:
Choose the correct ones.
a. Taste buds are the chemoreceptors seen in the papillae.
b. Receptors are uniformly distributed all over the skin.
c. Impulses from the olfactory receptors reach the cerebrum through the olfactory nerve.
d. We experience taste when impulses from the taste buds reach the cerebellum.
Answer:
a. Taste buds are the chemoreceptors seen in the papillae.
c. Impulses from the olfactory’ receptors reach the cerebrum through the olfactory nerve.

Question 31.
Light rays which reflect from the object are focussed on the retina and an image is formed,
a. Write the peculiarities of this image.
b. How do the. images formed in the two eyes combine? What is its advantage?
Answer:
a. Small, inverted, real.
b.

  • Three dimensional image of the object is formed due to the activity of the brain.
  • Binocular vision becomes possible.

Question 32.
Vision is enabled when the impulse from the retina reaches the cerebrum through the optic nerve.
a. Draw a flow chart showing the pathway of light from cornea to retina.
b. There is no vision at the point where the optic nerve starts. Why?
Answer:
a. Light → Cornea → Aqueous humour → Pupil → Lens Vitreous humour → Retina → Optic nerve
b. Rod cells and cone cells are absent in the part from where the optic nerve begins / Photoreceptors are absent.

Question 33.
Rhodopsin Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 12 Retinal + Opsin
a. How is this reaction related to vision?
b. How does the deficiency of vitamin A cause poor vision in dim light?
Answer:
a.

  • When light rays fall, rhodopsin dissociates into retinal and opsin.
  • Generates impulses.
  • When impulses reach the cerebrum, vision is experienced.

b

  • Amount of retinal decreases.
  • Causes a decrease in the level of rhodopsin.
  • Resynthesis of rhodopsin gets blocked.
  • Cannot see objects clearly in dim light / night blindness.

Question 34.
Observe the figure given below and answer the questions: .
Kerala SSLC Biology Chapter 2 Questions Windows of Knowledge 13
a. Which is the receptor seen in the figure?
b. In which sense organ is this receptor seen?
c. What is the function of this receptor?
Answer:
a. Olfactory receptor
b. Nose
c. Gets stimulated by aromatic particles and generates impulses.

Kerala SSLC Biology Chapter Wise Questions and Answers

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Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

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

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Plus Two Chemistry Solutions One Mark Questions and Answers

Question 1.
Which of the following is a liquid in solid type solution?
(a) glucose dissolved in water
(b) Camphor in N2
(c) amalgam of Hg with Na
(d) Cu dissolved in Au
Answer:
(c) amalgam of Hg with Na

Question 2.
The concentration term used when the solute is present in trace quantities is _______
Answer:
ppm (parts per million)

Question 3.
A binary solution of ethanol and n-heptane is an example of
(a) ideal solution
(b) Non-ideal solution with +ve deviation
(c) Non-ideal solution with -ve deviation
(d) unpredictable behaviour
Answer:
(b) Non-ideal solution with +ve deviation

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 4.
A solution which has higher osmotic pressure as compared to other solution is known as _____
Answer:
Hypertonic

Question 5.
Which of the following solutions will have the highest boiling point at 1 atm pressure?
(a) 0.1M FeCl3
(b) 0.1MBaCl2
(c) 0.1MNaCl
(d) 0.1Murea
Answer:
(a) 0.1M FeCl3

Question 6.
1 kilogram of sea water sample contains 6 mg of dissolved O2. The concentration of O2 in the sample in ppm is
Answer:
6.0

Question 7.
The amount of solute (molar mass 60 g mol-1) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer:
60 g

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Pure Aqua provides users with an osmotic pressure calculator to gain better insight on your osmotic pressure requirements.

Question 8.
The correct equation for the degree of association (a) of an associating solute ‘n‘ molecules of which undergoes association in solution is
Answer:
a = \(\frac{n(i-1)}{1-n}\)

Question 9.
A solution is prepared by dissolving 10 g NaOH in 1250 ml of a solvent of density 0.8 ml/g. The molarity of the solution is _______
Answer:
0.25

Question 10.
If the elevation in boiling point of a solution of non volatile, non electrolyte in a solvent (Kb = xk. kg mol-1) is 7 K, then the depression in freezing point would be kf = ZK kg mol-1
Answer:
\(\frac{Y Z}{x}\)

Plus Two Chemistry Solutions Two Mark Questions and Answers

Question 1.
Match the terms of list A with those in list B.

AB
Raoult’s Law.Colligative property.
Henry’s Law.Ideal solution.
Elevation of boiling point.Solution of gases in liquids.
Benzene + Toluene.Vapour pressure of solutions.

Answer:

AB
Raoult’s Law.Vapour pressure of solutions.
Henry’s Law.Solution of gases in liquids.
Elevation of boiling point.Colligative property.
Benzene + Toluene.Ideal solution.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 2.
At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm of Hg respectively. Two moles of A and 3 Moles of B are mixed to form an ideal solution. What is the vapour pressure for the solution?
Answer:
A = 120mm of Hg
B= 180 mm of Hg
χA = 2/5
χB =3/5
PA = P°A × χA
= 120 × 2/5 = 48 mm of Hg
PB = P°B × χB
= 180 × 3/5 = 108 mm of Hg
Ps = PA + PB
= 108 + 48 = 156 mm of Hg

Question 3.
Find the volume of H2O that should be added to 300 mL of 0.5 M NaOH so as to prepare a solution of 0.2 M.
Answer:
M1V1 = M2V2
300 × 0.5 = V2 × 0.2
V2 \(\frac{300 \times 5}{2}\) = 750 mL
H2O to be added = 750 mL – 300 mL = 450mL

Question 4.
Calculate the osmotic pressure of 5% solution of urea at 27°C?
Answer:
Mass of urea, WB = 5 g
R = 0.0821 L atm K-1 mol-1
T = 273 + 27°C = 300 K
Molecular mass of urea, MB = 60 g mol-1
V = 100 mL = 0.1L
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions two marks q4 img 1
= 20.53 atm

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 5.
Osmotic pressure of 1M solution of NaCl is approximately double than that of 1M sugar solution. Why?
Answer:
Osmotic pressure is a colligative property and it depends on the number of solute particles present in the solution. In solution, each NaCl unit undergoes dissociation to form two particles (NaCl → Na+ + Cl) and hence osmotic pressure of 1M NaCl is twice that of 1M sugar solution. Sugar molecules does not undergo association or dissociation in solution.

Question 6.
What do you mean by ideal solution?
Answer:
Ideal solution is a solution which obeys Raoult’s law over the entire range of concentration and temperature, i.e., for an ideal solution having two volatile components A and B.
PA = P°A χA, PB = P°B χB,
Ps = PA + PB =P°A χA + P°B χB

Question 7.
Many countries use desalination plants to meet their potable water requirements. Comment on the phenomenon behind it.
Answer:
This is based on reverse osmosis. When a pressure more than osmotic pressure is applied, pure water is squezed out of the sea water through a semi-permeable
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions two marks q7 img 2

Question 8.
Movement of solvent molecules through a semipermeable membrane from pure solvent to the solution side is called osmosis.
What are the following

  1. Isotonic solution
  2. Hypertonic solution?

Answer:
1. Isotonic solution:
If the osmotic pressure of the two solutions are equal, they are called isotonic solutions.

2. Hypertonic solution:
A solution having higher osmotic pressure than another solution is called hypertonic solution.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 9.
What is meant by azeotrope?
Answer:
Liquid mixtures which distil without change in composition are called azeotropic mixtures or azeotropes.

Plus Two Chemistry Solutions Three Mark Questions and Answers

Question 1.
A student was asked to define molality. Then he answered that it is the number of gram moles of the solute dissolved per litre of the solution.

  1. Is it correct?
  2. Can you help the student to define molality?
  3. Calculate the molality of a solution containing 20 g of NaOH in 250 g of H2O.

Answer:

  1. It is not correct.
  2. It is the number of moles of the solute present in 1000 g of the solvent. Molality can be determined by using the formula.
    Molality, m = \(\frac{\text { Mass of the solute in gram } \times 1000}{\text { Molar mass of the solute } \times \text { Mass of the solvent in gram }}\)
  3. Molality, m = \(\frac{20 \times 1000}{40 \times 250}\) = 2 m

Question 2.
The graph of non-ideal solution showing -ve deviation as drawn by a student is given below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 3

  1. Is this diagram correct?
  2. Substantiate your argument with a suitable example.

Answer:

  1. No.
  2. Because this graph shows the non-ideal solution showing +ve deviation and not -ve deviation. Consider a solution obtained by mixing chloroform and acetone. Here chloroform molecule is able to form hydrogen bond with acetone molecule as shown below:

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 4
As a result of this, vapour pressure of the solution decreases. Due to this, boiling point increases. The volume of the solution is less than the expected value. The mixing process is exothermic. So the actual graph is as given below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q2 img 5

Question 3.
Some words are missed in the following paragraph. Add suitable words in the blanks:
If osmotic pressure of 2 solutions are equal they are called ______(a)_____ solution. The solution which is having ______(b)______ osmotic pressure is called hypertonic solution and the solution which is having lower osmotic pressure is called ______(c)_____ solution.
Answer:

  1. Isotonic
  2. Higher
  3. Hypotonic

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 4.
The solubility of gases depends upon some factors.

  1. Can you suggest the factors?
  2. Which is the law behind it?
  3. What are the limitations of this law?

Answer:

  1. Nature of the gas, Nature of the solvent, Pressure, Temperature
  2. Henry’s law. It states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure applied on it.
  3. Henry’s law is valid only under the following conditions:
    • The pressure of the gas is not too high.
    • The temperature is not too low.
    • The gas is not highly soluble.

Question 5.
A solution of 12.5 g of an organic solute in 170g of H2O a boiling point elevation of 0.63 K. Calculate the molecular mass of the solute (K2=0.52 K/m).
Answer:
Kb for water = 0.52 K Kg mol-1
WA = 170g
ΔTb = 0.63 K
WB = 12.5 g
MB = ?
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q5 img 6

Question 6.
Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H20. (Kf for H2O = 1.86 K/m).
Answer:
Mass of glucose, WB = 3.6 g
Molecular mass of glucose = 180 g mol-1
Mass of solvent, WA = 50 g
Kf for H2O = 1.86 K/m = 1.86 K kg-1 mol-1
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q6 img 7
i.e., ΔTf = T°f – Tf= 0.744 K
∴ Freezing point of the solution, Tf = T°f – ΔT
= 273 K – 0.744 K = 272.3 K

Question 7.
Raw mangoes shrivel when pickled in brine solution.

  1. Name the process behind this.
  2. Define that process.

Answer:

  1. Osmosis
  2. When a solution is separated from its solvent by a semipermeable membrane, the solvent flows into the solution through the semipermeable membrane. This process is called osmosis.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 8.
200 cm3 of an aqueous of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q8 img 8

Question 9.
What type of deviation from Rauolt’s law is exhibited by a mixture of phenol and aniline? Explain with the help of graph.
Answer:
Negative deviation.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q9 img 9
In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules. Thus, escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

Question 10.
Wilted flowers revive when placed in fresh water.

  1. Identify and define the phenomenon.
  2. Mixing of acetone and CHCl3 results in a reduction in volume. What type of deviation from Raoult’s law is observed here? Give reason.
  3. Benzoic acid dissolved in benzene shows double of its molecular mass. Explain.

Answer:
1. Osmosis. It is the phenomenon of flow of solvent from pure solvent into a solution or from a solution of lower concentration into a solution of higher concentration through a semi-permeable membrane.

2. Negative deviation. Here chloroform molecule is able to form hydrogen bond with acetone molecule.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q10 img 10
Thus escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

3. Benzoic acid undergoes association in solution.
2C6H5COOH \(\rightleftharpoons \) (C6H5COOH)2 Thus, the number of particles as well as colligative properties decreases. So molecular mass increases.

Question 11.
A solution is obtained by mixing 300 g of 25 % solution and 400 g of 40 % solution by mass. Calculate the mass percentage of the resulting solution.
Answer:
Mass of solute in 300 g of 25 % solution
\(\frac{300 \times 25}{100}\) = 75 g
Mass of solute in 400 g of 40 % solution
\(\frac{400 \times 40}{100}\) = 160 g
Total mass of solute = (75 + 160) g = 235 g
Total mass of solution = (300 + 400) g = 700 g
Mass % of solute in resulting solution = \(\frac{235 \times 100}{700}\) = 33.57%
Mass % of solvent (water) in resulting solution
= 100 – 33.57 = 66.43%

Question 12.
A graph showing vapour pressure against mole fraction of an ideal solution with volatile components A and B are shown below:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q12 img 11

  1. Which law is supported by the graph?
  2. State the law.
  3. Resketch the graph if the attraction between A and B is greater than that between A-A and B-B.

Answer:

  1. Raoult’s law.
  2. Raoult’s law states that vapour pressure of a volatile component in a solution is the product of vapour pressure of that component in the pure form and mole fraction of that component in the solution.
  3. If the A-B attraction is greater than A-A and B-B attractions the liquid mixture behaves as a non-ideal solution with negative deviation.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q12 img 12

Question 13.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q13 img 13

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 14.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Answer:
Dissolution of gases is exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent \(\rightleftharpoons \) Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution.

Question 15.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= \(\frac{1000}{18}\) = 55.5
∴ χH2O = \(\frac{55.5}{55.5+1}\)
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 16.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer:
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
\(\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}\)
= 5.27 × 10-3 atm

Question 17.
Solution of sucrose is prepared by dissolving 34.2 g of it in 1000 g of water. Find out the freezing point of the solution, if Kf of water is 1.86 K/kg/mol? (Molecular mass of sucrose is 342 g/mol).
Answer:
ΔTf = kf × m
=1.86 K kg mol-1 × \(\frac{34.2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{342 \mathrm{g} \mathrm{mol}^{-1} \times 1000 \mathrm{g}}\) = 0.186K
Tf = OK – 0.186 K
Freezing point of the solution = – 0.186 K

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 18.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
In the first case, π1 = C1RT
i.e., 4.98 bar = \(\frac{36 \mathrm{g} \times \mathrm{R} \times 300 \mathrm{K}}{180 \mathrm{g} \mathrm{mol}^{-1}}\) ——– (1)
In second case, π2 = C2RT
i.e., 1.52 bar = C2R × 300 K ———- (2)
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q18 img 14

Question 19.
A solution containing 12.5 g of non-electrolytic substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molecular mass of the substance? (Kb for water = 0.52 K kg mol-1)
Answer:
Molecular mass of the solute, MB = \(\frac{1000 \mathrm{K}_{\mathrm{b}} \mathrm{W}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \Delta \mathrm{T}_{\mathrm{b}}}\)
\(\frac{1000 \mathrm{g} \mathrm{kg}^{-1} \times 0.52 \mathrm{K} \mathrm{kg}^{-1} \mathrm{mol}^{-1} \times 12.5 \mathrm{g}}{175 \mathrm{g} \times 0.70 \mathrm{K}}\)
= 53.06 mol-1

Question 20.

  1. “For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. Name the law stated above.
  2. Study the graph. What phenomenon it denotes? Based on your observation predict the reason for the greater volatility of a mixture of carbon disulphide and acetone?

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q20 img 15

Answer:

  1. Raoult’slaw
  2. Positive deviation from Raoult’s law. In a solution formed by adding CS2 to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules. Thus, the escaping tendency of the particles increases and the solution shows +ve deviation.

Question 21.

  1. To get hard boiled eggs, common salt is added to water during boiling. Give reason.
  2. Which colligative property is more suitable for the determination of molecular mass of polymers? Give the expression to determine molecular mass by this method.

Answer:

  1. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg and thus get hardened.
  2. Osmotic pressure method.

Question 22.
18 g of glucose is dissolved in 1kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1)
Answer:
Number of moles of glucose = 18/180 = 0.1 mol.
Mass of solvent = 1 kg.
Morality of glucose , m = \(\frac{n_{8}}{W_{A}}\)
= \(\frac{0.1 \mathrm{mol}}{1 \mathrm{kg}}\) = 0.1 mol/Kg
Elevation of boiling point ΔTb = Kb × m
= 0.52 K kg/mol × 0.1 mol/kg = 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 K + 0.052 K = 373.202 K

Plus Two Chemistry Solutions Four Mark Questions and Answers

Question 1.
Colligative properties are exhibited by dilute solutions.

  1. What do you mean by colligative properties?
  2. Which are the four colligative properties?

Answer:

  1. Colligative properties are those properties of dilute solutions of non-volatile solutes whose value depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.
  2. The four colligative properties
    • Relative lowering of vapour pressure
    • Elevation of boiling point
    • Depression of freezing point
    • Osmotic pressure

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 2.
PA = P°A χA
PB = P°B χB
ΔmixV = O

  1. Arun said to Subin that it is the condition for a type of solutions.
  2. Identify the type of solutions.
  3. What are the differences between ideal and non-ideal solutions?

Answer:
1. Ideal solutions.

2. Vapour pressure of a volatile component in the solution is the product of vapour pressure of pure component and mole fraction of that component in the solution.

3.

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q2 img 16

Question 3.

  1. What are the characteristics of a non-ideal solution?
  2. Explain +ve and -ve deviations from Raoult’s law with suitable examples.

Answer:
1. The characteristics of a non-ideal solution

  • Does not obey Raoult’s law over the entire range or concentration. Fora non-ideal solution having two volatile components A and B,
    PA ≠ P°A χA
    PB ≠ P°B χB
  • Volume of mixing not equal to zero, DmixV10
  • Enthalpy of mixing not equal to zero, DmixH10

2.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q3 img 17

Question 4.
Osmosis and osmotic pressure are two important terms related to solutions.

  1. Explain these terms.
  2. How will he determine the molar mass of a substance by this method?

Answer:
1. The phenomenon of the spontaneous flow of a solvent from a solution of lower concentration to higher concentration, separated by a semipermeable membrane is called osmosis.

The excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semipermeable membrane is called osmotic pressure.

2. Osmotic pressure (p) is proportional to the molar concentration/molarity (C) of the solution at a given temperature (T K).
π = CRT, where R is the gas constant.
π = \(\frac{n_{\mathrm{B}}}{V}\) RT, where nB is the number of moles of the solute and V is the volume of the solution in litres.
π V = nBRT
π V = \(\frac{w_{B}}{M_{B}}\)RT, where WB is the mass of the solute and MB is the molar mass of the solute.
Or MB = \(\frac{\mathrm{W}_{\mathrm{B}} \mathrm{RT}}{\pi \mathrm{V}}\)
Osmotic pressure measurement is widely used to determine molar mass of proteins, polymers and other macro molecules.

Question 5.
The value of molecular mass determined by colligative property measurement is sometimes abnormal.

  1. Explain these abnormalities in the case of benzoic acid in benzene and KCl in water.
  2. What is van’t Hoff factor?

Answer:
1. This is caused by dissociation in the case of KCl and association in the case of acetic acid. KCl in aqueous solution undergoes dissociation as KCl → K+ + Cl
Molecules of ethanoic acid (acetic acid) dimerises in benzene due to hydrogen bonding. As a result of dimerisation the actual number of solute particles in solution is decreased. As colligative property decreases molecular mass increases.
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions four marks q5 img 18

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 6.
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. This is the commonly used law for expressing the solubility of gas in liquid.

  1. Name the law. Write its mathematical expression.
  2. What are the factors affecting the solubility of a gas in a liquid? Explain.

Answer:
1. Roult’s law.
For an ideal solution containing two volatile components A and B,
PA = P°A χA,
PB = P°B χB and
P[Total] = PA + PB = P°A χA + P°B χB

2. The factors affecting the solubility of a gas in a liquid:

  • Nature of the gas and the liquid – Each gas has a characteristic solubility in a particular liquid at a particular temperature and pressure.
  • Temperature – solubility of a gas in a liquid is an exothermic process. Hence according to Le Chatelier’s principle solubility of gases in liquids decreases with rise in temperature.
  • Pressure – According to Henry’s law, solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Question 7.
Concentration of solution may be expressed in different ways.

  1. Mention any one of the concentration terms.
  2. What are colligative properties?
  3. Show that elevation in boiling point is a colligative property.

Answer:
1. Molarity – It is the number of moles of the solute present in one litre of the solution.

2. Colligative properties are those properties which depends only on the number of solute particles.

3. ΔTb = Kbm
= \(\mathrm{k}_{\mathrm{b}} \frac{\mathrm{n}_{\mathrm{B}} \times 1000}{\mathrm{W}_{\mathrm{A}}}\)
i.e., ΔTb α nB i.e., elevation of boiling point depends on number of moles of solute. Hence, it is a colligative property.

Plus Two Chemistry Solutions NCERT Questions and Answers

Question 1.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?
Answer:
Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions three marks q13 img 13

Question 2.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Answer:
Dissolution of gases is an exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent \(\rightleftharpoons \) Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Question 3.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer:
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= \(\frac{1000}{18}\) = 55.5
∴ χH2O = \(\frac{55.5}{55.5+1}\)
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 4.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Answer:
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
\(\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}\)
= 5.27 × 10-3 atm

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

You can Download The Last Leaf (Story) Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 2 Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

A The Trio (Story) Textual Questions and Answers

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5 Question 1.
Observe figure try to write down the activities shown in them.
Answer:

  1. A man pushes a trolly.
  2. Batting of a cricket ball.
  3. Pushing a wall.

Kerala Syllabus 9th Standard Physics Notes Chapter 5 Question 2.
Write down more activities familiar to you
Answer:

  1. A man carrying a load
  2. Throwing a ball
  3. Lifting the bag on to the shoulder
  4. Pushing a car into motion

Hsslive Guru 9th Physics Kerala Syllabus Chapter 5 Question 3.
You have understood that a force is to be applied on a body to do an activity. Find out the source of applied force for every activity and note them down in the table.
Kerala Syllabus 9th Standard Physics Notes Chapter 5
Hsslive Guru 9th Physics Kerala Syllabus Chapter 5
Answer:

ActivitySource of applied force
Falling of a mangoThe earth
A trolley being pushedThe person pushing
Batting of a cricket ballBatsman
Pushing a wallThe person pushing
Lifting a loadThe person lifting

Objects undergo displacement only when the force is applied on it them.

Displacement takes place in the direction of force appliedNo displacement for the body in the direction of force applied
1. A cricket ball when hit by a bat1. A wall is pushed
2. A trolley is being pulled.2. A car is pushed by sitting inside the car
3. Climbing a ladder with a load on head.
4. Falling of a mango from mango tree.

Work :

Work is said to be done only when a body under¬goes displacement in the direction of the applied force.
Kerala Syllabus 9th Physics Notes Chapter 5

work energy and power Question 4. Observe figure and write down the situation where work is said to be done
Answer:
The person loaded on head and goes upward.

Kerala Syllabus 9th Physics Notes Chapter 5 Question 5.
A boy pushed an object of mass 30Kg horizontally across the floor through 50m. Another boy pushed an object of mass 50kg across the same floor through 50m. Both of them gave the same speed for moving the objects.
a) Who applied greater force here?
b) In which case was the work greater?
c) Write down a factor influencing work.
Answer:
a) Second boy applied greater force
b) More work is done in the second situation
c) The Factor influencing work is force.

Learn Mass Transfer MCQ questions & answers are available for a Chemical Engineering students.

Hss Live Guru Physics 9th Kerala Syllabus Chapter 5 Question 6.
A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.
a) Who pushed a greater distance here?
b) What about the force applied
c) Who did the greater work?
d) Which is the factor influencing work here?
Answer:
a) Second boy
b) Force is same
c) Second boy done greater work
d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)
Equation for calculating work done Work W = Fs
Unit of work is Joule (J), Kilojoule (KJ)
1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Get the free “Nuclear Equation Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Physics Notes For Class 9 Kerala Syllabus Chapter 5 Question 7.
When a force of 10N is applied continuously on a body it undergoes a displacement of 2m find out the magnitude of the work done?
Answer:
F = 10N, s = 2m
Work W= Fs = 10 × 2 = 20 Nm
Look at the Figure
Hss Live Guru Physics 9th Kerala Syllabus Chapter 5

Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5 Question 8.
A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer:
Weight of the body applies downwards and the table applies an equal force upwards.

Hss Live 9th Physics Kerala Syllabus Chapter 5 Question 9
In which directions do these forces act?
Answer:
Forces acting on both directions, upward direction and downward direction.

9th Physics Notes Kerala Syllabus Chapter 5 Question 10.
A book of mass 100g is raised to the top of a table of height 1 m. Find the magnitude of the work done by the force applied against the gravitational force(g = 10m/s2)
Answer:
m = 100g = 0.1kg
g = 10m/s2
h = 1m
W = mgh
= 0.1 × 10 × 1 = 1J
1 J is the amount of work done to raise a body of mass 100 g through a height of 1m.

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Question 11.
If a force of 50N is applied on a body and it under¬goes a displacement of 2m in the direction of the force, calculate the amount of work done.
Answer:
F = 50N
s = 2m
W = Fs = 50 x 2 = 100J

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5 Question 12.
a) If a force of 200N is applied on a table of mass 50kg, it undergoes a displacement of 0.5m in the direction of force. Calculate the amount of work done.
b) If the same table is raised by 3 m, what would be the work done against the gravitational force?
Answer:
a) F= 200N, s = 0.5m
W = Fs = 200 x 0.5 = 100J
b) m = 50 kg, g = 10m/s2
h = 3 m
W = mgh = 50 x 10 x 3 = 1500J
Observe Figure
Physics Notes For Class 9 Kerala Syllabus Chapter 5
Let a body mass m be pulled by a force F. If the body has a displacement s in the direction of the force, then the work done by the force F, Wf = Fs Here the displacement produced is in the direction of the force itself.

Kerala Syllabus Class 9 Physics Solutions Chapter 5 Question 13.
Write whether this work is negative or positive.
Answer:
Work is positive

9th Class Physics Notes Kerala Syllabus Chapter 5 Question 14.
The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer:
If the displacement is in opposite direction work done by frictional force is negative.

Class 9 Physics Kerala Syllabus Chapter 5 Question 15.
In which direction is the force of gravity on the body?
Answer:
Direction of the gravitational force will be in the downward direction

9th Standard Physics Notes Kerala Syllabus Chapter 5 Question 16.
Is there a displacement for the body in the direction of the gravitational force?
Answer:
No displacement occurs in the direction of the gravitational force.
When a body on a floor is pulled and if it is displaced in the direction of the applied force, the work done by the applied force will be positive and the work done by the frictional force exerted by the floor will be negative.

Energy

Question 17.
What is the work to be done to raise a body of mass m kg through h meter?
Physics chapter 5 work and energy Answer:
w = mgh
Energy is the capacity to do work. Unit is Joule (J)

work energy book Question 18. In daily life, we use different forms of energy for various activities. List the forms of energy familiar to you
Answer:

  1. Mechanical energy
  2. Heat energy
  3. Electrical energy
  4. Chemical energy
  5. Light energy
  6. Nuclear energy

There are two type of Mechanical energy Kinetic energy and potential energy

Kinetic Energy

Pulling the toy car backwards a little and allow it to hit the plastic ball
Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5

Question 19.
What happened to the ball when the moving car hit it?
Answer:
Ball moves forward

10 to the 9th power Question 20.
How did the car get the energy to move the ball forward?
Answer:
The energy is obtained from the motion of toy car
Conclusion:

  • Moving objects possess energy
  • The energy possessed by a body by virtue of its motion is the Kinetic energy

Experiment:
Let’s do another activity. Allow a powder tin to slide down a polished, inclined plane as shown in the figure and let it hit a toy car. Try to measure the displacement of the toy car. Repeat the experiment by increasing the height of the inclined plane and filling the tin with sand.
Hss Live 9th Physics Kerala Syllabus Chapter 5
Observation:

  • The displacement of the toy car is greater When it is hit by the powder filled with sand
  • Also the displacement is greater when it is dropped from a greater height

Conclusion: Kinetic energy depends on mass (m) and velocity (v)
Derivation of the equation for calculating K.E
Work W = Fs
As per II law of motion
F = ma
∴ W = mas
According to third equation of motion
v2 = u2 + 2as = 0 + 2as (u = 0) = 2as
∴ as = \(\frac{v^{2}}{2}\)
W = mas , if we put \(\frac{v^{2}}{2}\) instead of as
9th Physics Notes Kerala Syllabus Chapter 5
Work is equal to the magnitude of kinetic energy
i.e, Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2,
M – mass, V – Velocity
When a body of mass m moves with a velocity v, its kinetic energy will be K = \(\frac { 1 }{ 2 }\) mv2

Question 21.
A man having a mass of 70kg is riding a scooter of mass 80kg. What is the total kinetic energy if the velocity of the scooter is 10m/s?
Answer:
m = 70kg + 80kg = 150kg
v = 10m/s
Kerala Syllabus 9th Standard Physics Solutions Chapter 5

Question 22.
A car of mass 1500 kg is moving at a velocity of 20m/s. What is its kinetic energy?
Answer:
m = 1500 kg
v = 20m/s
K = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 1500 × 202
= 300000J = 300kJ

Question 23.
A boy of mass 50kg is riding a bicycle with a speed 2m/s. The bicycle has a mass of 10kg. Calculate the total kinetic energy?
Answer:
m = 50kg + 10kg = 60kg
v = 2m/s
K = \(\frac { 1 }{ 2 }\) mv2
1/2 × 60 × 22 = 120J

Potential Energy

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5
When we lift a body in perpendicular direction, the work is said to be done against gravitational force. According to figure, maximum work is done when it reaches at a height fb from the group, That means as the height increases work done also increases.

  • The energy received by the body is equal to the work done on it.
  • The body attains more energy when the height from the ground level increase

The energy possessed by a body due to its position is the potential energy.
ie, Potential energy
U = mgh
m – mass, g – acceleration due to gravity
h = height from the ground

Question 24.
Identify more situation in which potential energy is acquired by virtue of position
Answer:

  • Coconut in a coconut tree
  • Water stores in huge reservoirs
  • Objects placed above the buildings,

Inference:
Height increases, potential energy increases

Question 25.
Write down situations in which potential energy varies.
Answer:

  • Falling of a coconut from a coconut tree
  • Pumping water to a tank at a height
  • Climbing a ladder

Question 26.
Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground?
Answer:
Mass m = 1 kg, Acceleration due to gravity
g = 10m/s2
h = 6 m,
U = mgh = 1 × 10 × 6 = 60J

Question 27.
A bird of mass 0.5 kg is flying at the same speed at the same height of 5m. In this state, if its Kinetic energy and potential energy are equal.
a) What is the potential energy of the bird?
b) What is the velocity of the bird?
Answer:
a) m = 0.5kg
g = 10m/s2
h = 5m
u = mgh
= 0.5 × 10 × 5 = 25J
b) Kinetic energy K = 25J (∴U = K)
Kerala Syllabus Class 9 Physics Solutions Chapter 5

Question 28.
Write other examples for getting potential energy due to strain
Answer:
9th Class Physics Notes Kerala Syllabus Chapter 5
Class 9 Physics Kerala Syllabus Chapter 5

  • A Stretched bow
  • An elongated rubber band
  • Compressed spring
  • Compressed spring in a toy car

Conclusion: The factor which gives potential energy are position and strain

Law Of Conservation Of Energy

9th Standard Physics Notes Kerala Syllabus Chapter 5
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 16

Question 29.
What form of energy does the flower pot have when it is on the sunshade of a building?
Answer:
Potential energy

Question 30.
While the flower pot is falling down, what forms of energy does it possess?
Answer:
potential energy and kinetic energy

Question 31.
Answer:
Does its potential energy increase/decrease when the pot falls down.
Answer:
Potential energy decreases

Question 31.
Will the kinetic energy increase/decrease at that time?
Answer:
Kinetic energy increases

Question 33.
What energy transformation takes place just before the flower pot reaches the ground?
Answer:
Potential energy is converted completely into kinetic energy

Question 34.
Let the mass of the flower pot 15kg and the height of the sunshade 4m.
a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2).
b) When it is on the sunshade, what is its kinetic energy?
c) If so, what is its total energy?
Answer:
U = mgh
= 15 × 10 × 4
= 600J
b) Kinetic energy will be zero
c) Total energy = PE + KE
= 600J + 0 = 600J

Question 35.
While falling, when the flower pot is at a height of 2m from the ground, what will be its kinetic energy?
Answer:
K = \(\frac { 1 }{ 2 }\) mv2
u = 0, g = 10m/s2 ,
s = 4 – 2 = 2m
v2 = u22 + 2as
=0 + 2 × 10 × 2 = 40
K = 1/2 × 15 × 40
= 300J

Question 36.
What is the potential energy when it is at a height of 2m from the ground? What is the total energy now?
Answer:
Potential energy at 2m height
U = mgh = 15 × 10 × 2 = 300J
∴ Total energy = 300 + 300 = 600J

Question 37.
What is the kinetic energy of the flower pot just before it touches the ground?
Answer:
K = \(\frac { 1 }{ 2 }\) mv2
v2 = u2 + 2as
= 0 + 2 x 10 x 4 = 80
Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 15 × 80 = 600J

Question 38.
The potential energy
U = mgh
= 15 x 10 x 10 x = 0.
What will be the total energy?
Answer:
Total energy = 600 + 0 = 600J

Question 39.
To sum up the amount of energy at each situation:
a) When on the sunshade
b) When at a height of 2m from the ground
c) Just before hitting the ground
Answer:
a) When on the sunshade = 600J
b) At a height of 2m from the ground = 600J
c) Just before hitting the ground = 600J
Energy can neither be created nor be destroyed. It can be transformed from one from to another
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 17
Sun is the major source of energy. We utilize solar energy in different ways. It is renewable source of energy Plants prepare food by using sunlight. From these fossil fuels, firewood are formed. Windmills works by utilizing kinetic energy of wind energy from infra¬red rays. Tidal energy is obtained due to gravitation. When we consider any energy sources, they were originated from solar energy.

Power

Question 40.
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s2).
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 18
Answer:
b) 150000J
c) 150000J

Question 41.
Is the amount of work done by the pump to fill water in the three tanks equal?
Answer:
Amount of work done is equal

Question 42.
Find the amount of work done per second by each pump.
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 19
Answer:

PumpWork doneTime (s) (J)Work done per second (J/S)
A1500001001500
B150000200750
C150000400375

Amount of work done per second is referred as power of the pump.
Work done per unit time or rate of doing work is power
Power = \(\frac{\text { work }}{\text { time }}, P=\frac{w}{t}\)
Units of power
watt (W), kilowatt (KW), Horse power (H.P)
1 kW= 1000W
1 HP = 746W

Question 43.
If a man of mass 70kg climbs up a mountain of height 30m in 5 minutes, what is his power?
Answer:
m = 70kg, g = 10m/s2
h = 30m
Work W = mgh = 70 × 10 × 30 = 21000J
Time t = 5mt = 5 × 60 = 300s
Power p = \(\frac { w }{ t }\) = \(\frac { 21000 }{ 300 }\) = 70W

Question 44.
If a man of mass 50kg takes 60s to climb up 20 steps, each 15cm high, calculate his power.
Answer:
m = 50kg, g = 10m/s2
h = 15cm × 20 = 300cm = 8m
t = 60s
work W = mgh = 50 × 10 × 3 = 1500J
Power p = \(\frac { w }{ t }\) = \(\frac { 1500 }{ 60 }\) = 25W

Let Us Assess

Question 1.
A boy is trying to push the concrete pillar of the building using a force of 300N. Calculate the amount of work done by the boy.
Answer:
F= 300N, s = 0
w = Fs = 300 × 0 = 0

Question 2.
From what you have learnt of potential energy and kinetic energy write down the from of energy possessed by the bodies given below.
a) water in a dam
b) Stretched rubber band
c) Mango falling from a tree
Answer:
a) Potential energy
b) Potential energy
c) Potential energy lesser. Kinetic energy greater

Question 3.
Calculate the kinetic energy of an athlete of mass 60kg running with a velocity 10 m/s
m = 60kg
v= 10m/s
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 20

Question 4.
A stone of mass 2kg is thrown upwards from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.
Answer:
u = 3m/s v = 0
a = g = -10m/s2 (Velocity of a body moving upwards is decreasing so acceleration is negative)
v2 = u2 + 2as
0 = 32 + 2 × -10 × s
0 = 9 – 20s
20s = 9, s = \(\frac { 9 }{ 20 }\)
U = mgh = 20 × 10 × \(\frac { 9 }{ 20 }\) = 90J

Question 5.
The heart of a healthy person beats 72 times per minute and each beat uses up about 1J of energy. Calculate the power of the heart.
Answer:
W= 1 x 72 = 72J
t = 1mt = 60s
P = \(\frac { W }{ t }\) = \(\frac { 72 }{ 60 }\) = 1.2w

Question 6.
Which among the following is a vector quantity?
a) work
b) momentum
c) power
d) energy
Answer:
momentum

Question 7.
If the velocity of an object is doubled, its kinetic energy becomes
a) 2 times
b ) 1/2
c) 4 times
e) 1/4
Answer:
4 times

Question 8.
An object of mass 1 kg is falling from a height 10 m. What be the work done while falling?
a) 10 J
b) 1J
c) 100 J
d)1000 J
Answer:
w = mgh = 1 × 10 × 10 = 100J

Question 9.
Which one among the following is correct?
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 21
Answer:
W = p × t

Question 10.
A roller weighing 1 tonne is being dragged along a road. What is the work done against gravity? Why?
Answer:
Zero, No displacement in the direction of force.

Question 11.
Is it possible for an object to possess energy without momentum? Give one example for such a situation.
Answer:
Yes, possible. A coconut on a coconut tree has potential energy, but it has no momentum.

Question 12.
Say whether the following are positive work or negative work.
1) The work done by a person drawing water from a well using a rope without a pulley.
2) Work done by gravitational force in this situation
3) Work done by the frictional force while and object is sliding down along an inclined plane.
4) Work done by the force is moving along a plane surface.
Answer:

  1. Positive work
  2. Negative work
  3. Negative work
  4. Positive work

Question 13.
How much joule is 1 kWh?
Answer:
1 kWh = 1000 × 60 × 60 = 3600000J

Question 14.
Find out the work done against the gravitational force in the situations given below.
1. A child is standing still with a bundle of books of mass 5 kg
2. With the same bundle of books, she travels 1m along a plane surface with a speed 5 m/s
3. The bundle of books is lifted onto the top of a cupboard having 1m height (g=10m/s2).
Answer:
1. 0
2. 0
3. W = mg = 5 × 10 × 1 = 50 J

Question 15.
A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m/s. Calculate its kinetic energy and potential energy after 1 s.
(Hint: v + at, s = ut + 1/2at2)
Answer:
v = u + at
= 14 + 10 × 1
= 24 m/s
KE = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 0.4 × 24
= 115.2 J
S = ut + 1/2 at2
= 14 × 1 + 1/2 × 10 × 12
= 14 + 5 = 19 m
E = mgh = 0.4 × 10 × 19
= 76 J

Question 16.
An object of mass 1000 kg is travelling with a velocity 72 km/h. Calculate the work done to bring it rest.
Answer:
m = 1000 kg
v = 72 km/h = 20 m/s
Work = difference of kinetic energy
= \(\frac { 1 }{ 2 }\) mv2
= \(\frac { 1 }{ 2 }\) × 1000 × 20 × 20 = 400000J

Question 17.
Estimate the work done on a object of mass 80 kg to change its velocity from 5 m/s to 10 m/s
Answer:
m = 80 kg
Work = difference of kinetic energy
Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power 22

Work, Energy and Power More Questions

Question 1.
Fill in the blanks
a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ……….
b) The two factors related to potential energy to body are………….
c) 1HP = ……… Watt
Answer:
1 Joule
b) Position, Strain
c) 746W

Question 2.
Classify the following into work done and work is not done
1. A mango falling from a mango tree
2. Pushing a table while sitting on it
3. Pushing a wall
4. Kicking a football
5. A trolley is moving forward
6. Standing with a load above the head
Answer:
Work done:
1. A mango falling from a mango tree
2. Kicking a football
3. A trolling is moving forward
Work is not done:
1. Pushing a table while sitting on it
2. Pushing a wall
3. Standing with a load on the head

Question 3.
When Lekshmi is applied a force of 50N on an object it undergoes a displacement of 2m. When Vinitha applied the same force on itundergoesadisplacementof3m if so
a) Which person done more work?
b) Give reason
c) Calculate the work done by each person?
Answer:
a) Vinitha
b) Displacement is greater
c) Work done by Lekshmi
W = Fs = 50N × 2m = 100J
Work done by Vinitha
W = Fs = 50N × 3m = 150J

Question 4.
Calculate work done when a boy of mass 40kg climb¬ing a staircase of height 50cm, g = 10m/s2
Answer:
W= Fs
F = mg = 40 × 10 = 400N
S = h = 0.5m
W= 400 × 0.5 = 200J

Question 5.
a) What is mean by kinetic energy?
b) How mass and velocity affect kinetic energy?
c) A body of mass 20kg is moving with a velocity of 5m/s. Calculate the kinetic energy?
Answer:
a) The energy possessed by a body by virtue of its motion is called kinetic energy
b) Mass increases, kinetic energy increases mass . is doubled kinetic energy also doubled. Velocity increases, kinetic energy increases, Velocity is doubled kinetic energy becomes four times.
c) Kinetic energy KE = \(\frac { 1 }{ 2 }\) mv2
= 1/2 × 20 × 52 = 1/2 × 20 × 25 = 250J

Question 6.
a) Write the equation for finding potential energy?
b) indicate the representation of each letter
Answer:
a) U = mgh
b) M – mass,
g – acceleration
h – height from the ground

Question 7.
State law of conservation of energy?
Answer:
Energy can neither be created nor be destroyed. It can be transformed from one form to another.

Question 8.
A store of mass 5kg was raised from the ground to the second floor of height 7m and from there to the third floor of height 3m from the second floor. Calcu¬late the potential energy of the store with respect to the ground floor and the second floor?
Answer:
Mass m = 5kg
Height from ground to third floor
h = 7m + 3m
g = 10m/s2
Potential energy with respect to the ground
U = mgh = 5kg × 10m/s2 × 10 = 500J
Height from second floor to third floor h = 3m
Potential energy with respect to the second Floor
U = mgh = 5kg × 10m/s × 3 = 150J

Question 9.
a) What is power?
b) The time takes to move an object at a distance of 5m is 10s is the force applied is 30N, what is the power?
Answer:
a) Rate of doing work is power
b) Power P = E
W = Fs = 30 × 5 = 150J
t = 10sec.
Power P = \(\frac { 150 }{ 10s }\) = 15W

Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion

Students can Download Chapter 5 Law of Motion 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 5 Law of Motion

Plus One Physics Law of Motion One Mark Questions and Answers

Plus One Physics Laws Of Motion Questions Chapter 5 Question 1.
Which one of the following is not a force?
(a) Impulse
(b) Tension
(c) Thrust
(d) Weight
Answer:
(a) Impulse
Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force.
Impulse = Force × Time duration.

Plus One Physics Laws Of Motion Questions And Answers Chapter 5 Question 2.
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is an example for
(a) Inertia of motion
(b) Second law of motion
(c) Third law of motion
(d) Inertia of rest
Answer:
(a) Inertia of motion
A passenger getting down from a moving bus, falls in the direction of the motion of the bus. This is because his feet come to rest on touching the ground and the remaining body continues to move due to inertia of motion.

Plus One Physics Important Questions And Answers Pdf Chapter 5  Question 3.
Which one of the following is not a contact force?
(a) Viscous force
(b) Magnetic force
(c) Friction
(d) Buoyant force
Answer:
(b) Magnetic force

Plus One Physics Chapter Wise Questions And Answers Chapter 5 Question 4.
A jet engine works on the principle of
(a) Conservation of linear momentum
(b) Conservation of mass
(c) Conservation of energy
(d) Conservation of angular momentum
Answer:
(a) Conservation of linear momentum
A jet engine works on the principle of linear momentum.

State And Prove Impulse Momentum Theorem Chapter 5 Question 5.
Newton’s second and third laws of motion lead to the conservation of
(a) linear momentum
(b) angular momentum
(c) potential energy
(d) kinetic energy
Answer:
(a) linear momentum
Newton’s second and third laws lead to the conservation of linear momentum.

Hsslive Plus One Physics Chapter Wise Questions And Answers Chapter 5 Question 6.
A large force is acting on a body for a short time. The impulse imparted is equal to the change in
(a) acceleration
(b) momentum
(c) energy
(d) velocity
Answer:
(b) momentum
If a large force F acts for a short time dt, the impulse imparted is
I = F.dt, = \(\frac{d p}{d t}\).dt
I = dp = change in momentum.

Laws Of Motion Class 11 Questions With Solutions Pdf Chapter 5 Question 7.
When a shell explodes, the fragments fly apart though no external force is acting on it. Does this violate Newton’s first law of motion?
Answer:
No. The explosion takes place due to the internal force. The internal force does not change the position of centre of mass.

Prove Impulse Momentum Theorem Chapter 5 Question 8.
In taking a catch, a cricket player moves his hands backward on holding the ball. Why?
Answer:
We know F = \(\frac{\Delta P}{\Delta t}\)
When ∆t increases, the force acting on hand decreases.

350 degrees f to c … T C T F 32 x 59 is the method for converting degrees Fahrenheit to degrees Celsius.

Plus One Physics Important Questions And Answers Chapter 5 Question 9.
Name the factor on which inertia depends.
Answer:
Mass

Laws Of Motion Class 11 Test Paper Chapter 5 Question 10.
Why does a swimmer push the water backwards?
Answer:
A swimmer pushes the water backward in order to be pushed forward (Newton’s third law).

Laws Of Motion Previous Year Questions Chapter 5 Question 11.
Rocket works on the principle of conservation of_______.
Answer:
Momentum

Motion Questions And Answers Pdf Chapter 5 Question 12.
A man experience a backward jerk, while firing bullet from gun. Which law is applicable here? Answer:
Conservation of momentum.

Plus One Physics Laws Of Motion Notes Chapter 5 Question 13.
If you jerk a piece of paper under a book quick enough, the book will not move. Why?
Answer:
This is due to inertia of rest.

Class 11 Physics Chapter 5 Important Questions Chapter 5 Question 14.
Why it is difficult to walk on a slipper road?
Answer:
We will not get required reaction from slippery road.

Laws Of Motion Class 11 Important Questions Chapter 5 Question 15.
A stone, when thrown on a glass window, smashes the window pan to pieces. But a bullet fired from the gun passes through it making a hole why?
Answer:
This is due to inertia of rest of glass window.

Question 16.
Why an athlete runs some distance before taking a jump?
Answer:
An athletic runs some distance before taking a jump to gain some initial momentum. It helps the athlete to jump more.

Question 17.
Why a horse can not pull a cart and run in empty space?
Answer:
The horse-cart system moves forward due to reaction of ground on the feet of horse. In free space, there is no reaction. So it can not pull cart.

Question 18.
Why parachute descends slowly?
Answer:
Parachute has large surface area. This increases fluid friction and slows down the motion of parachute.

Question 19.
Sand is thrown on tracks with snow. Why?
Answer:
The presence of snow on tracks reduces friction and driving is not safe. If sand is thrown, friction will be increased and driving becomes safe.

Question 20.
It is difficult to move a cycle along a road with its brakes on. Explain.
Answer:
When the cycle is moved with its brakes on, wheels can only skid. There will be sliding friction. The sliding friction is more compared to rolling friction. Hence it is difficult to move a cycle with its brakes on.

Plus One Physics Law of Motion Two Mark Questions and Answers

Question 1.
Two masses are in the ratio 1:5

  1. What is inertia.
  2. What is the ratio of inertia of above case?

Answer:

  1. The inability of a body to change it’s state of rest or uniform motion is called inertia.
  2. Mass is a measure of inertia. Hence ratio of inertia is 1:5.

Question 2.
More force is required to push a body than pull to get same speed on a ground with some friction. Why?
Answer:
When we push, the action on the surface and normal reaction on the body increases. (Friction is directly proportional to normal reaction).

As a result more force is required to push the body. When we pull, normal reaction decreases. Hence friction decreases. Hence less force is required to pull the body.

Question 3.
A lift in a multistoried building is moving from ground floor to third floor. What will happen to weight of a person sitting in side of the lift.

  1. A When starts to move up from ground floor.
  2. When the lift moves with constant speed.

Answer:

  1. A weight increases weight w = mg + ma
  2. weight is constant ie. w = mg

Question 4.
Why it is advisable to hold a gun tight to one’s shoulder when it is being fired?
Answer:
The recoiling gun can hurt the shoulder. If gun is held tightly against the shoulder, the body and gun act a system. This will reduce recoil velocity as it is inversly proportional to mass of system.

Question 5.
Why shockers are used in vehicles?
Answer:
When there is a jerk or jump, the time for which force acts (∆t) increases. As the product of force and time for which force acts (F∆t) remains constant, increase in At will reduce the force. This provide smooth motion.

Plus One Physics Law of Motion Three Mark Questions and Answers

Question 1.
Give the magnitude and direction of net force on

  1. a drop of rain falling down with a constant velocity.
  2. a stone of mass 0.1 kg just after it dropped from the window of a tram accelerating at 1 ms-2.

Answer:
1. Net force is zero

2. When stone is dropped, gravitational force will act on the stone.
Gravitational force F = mg
= 0.1 × 10
= 1 N downward.

Question 2.
An external force is always required to break the inertia of a body which is either in the state of rest or state of uniform motion.

  1. Which law governs this statement?
  2. Can all forces produce acceleration? Why?
  3. A boy holding a spring balance in his hand suspend a mass 2kg from it. If the balance slips from his hand and falls down, find the reading of the balance while it is in the air.

Answer:

  1. Newtons first law of motion.
  2. No. If resultant force acting on the body is zero, the body will move with constant velocity or remain at rest.
  3. Zero

Question 3.
A man weighs 70 kg. He stands on a weighing scale in a lift which is moving.

  1. upward with a uniform speed of 10 m/s.
  2. downward with an uniform acceleration of 5 m/s2.
  3. upward with an uniform accelerate of 5 m/s2. (Take g = 10m/s2). Find weight in each case.

Answer:
1. Weight W = mg
= 70 × 10 = 700 N.

2. W = mg – ma
= 70 × 10 – 70 × 5
= 700 – 350
= 350 N

3. W = mg + ma
= 70 × 10 + 70 × 5
= 700 + 350
= 1050N.

Question 4.
A body of mass ‘m’ is placed on a rough inclined plane having coefficient of friction µs. The inclination of plane is given as ‘θ’.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 1

  1. Which component of weight brings the body towards the bottom along the plane.
  2. Find how much force is required to pull the body along the plane.

Answer:
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 2

  1. mg Sinθ brings the body downwards
  2. When the body moves upwards the frictional force (Fs) acts downwards
    Total pulling Force = mg Sinθ + Frictional force (Fs) (u, mgCosθ).

Question 5.
Four person sitting in the back seat of a car at rest, is pushing on the front seat.

  1. Does the car move. Why?
  2. State the law which help you to answer above question.
  3. Long jumpers take a long run before the jump. Why?

Answer:

  1. No. Action and reaction cancel each other.
  2. Newtons third law of motion.
  3. To get large inertia of motion.

Question 6.
A Cricket player lowers his hands while catching a Cricket ball to avoid injury.

  1. What do you mean by impulsive force?
  2. Prove impulse – momentum theorem.

Answer:
1. The forces which acton bodies for short time are called impulsive forces.
Example:

  • In hitting a ball with a bat
  • In firing a gun

2. F = \(\frac{d p}{d t}\)
F∆t = dp
impulse = change in momentum.

Plus One Physics Law of Motion Four Mark Questions and Answers

Question 1.
A bead sliding on a wire A moves to C through B as shown in the figure. The bead at A has a speed of200cms
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 3

  1. what is speed at B?
  2. To what height will it rise before it returns?
  3. Why the ball moves up even after reaching the bottom most point B?

Answer:
1. mgh = 1/2 mv2
m × 10 × 0.8 = 1/2 mv2
V2 = 2 × 10 × 0.8
V = \(\sqrt{2 \times 10 \times 0.8}\)
V = 4 m/s.

2. 80 cm (if friction is neglected).

3. when the ball reaches at B, the potential energy is converted into kinetic energy. Due to this kinetic energy the ball raises to the point c.

Question 2.
Figure shows a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 4

  1. Obtain formula for the acceleration of the system and tension in the cord.
  2. If m1 and m2 interchanges its position, will it affect the tension of the string?
  3. What is the acceleration of the system if m1 = 5 kg and m2 = 2kg?

Answer:
1. When the body m2 moves in down ward direction.
m2g – T = m1 a
T = m2g – m1a.

2. New tension can be found from the relation
m1g – T = m2a
T = m1g – m2 a.

3. Acceleration of system, a
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 5

Question 3.
The collision of two ice hockey players are shown in figure.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 6
Analyse the data given in the figure and answer the following questions.

  1. Which conservation law is applicable in this case.
  2. In which direction and at what speed do they travel after they stick together.
    [Hint – towards right can be taken us +ve direc¬tion and vice versa]
  3. If we assume the friction of playing ground is zero, predict the nature of motion and the point at which they come to rest.

Answer:
1. Conservation of linear momentum.

2. Total momentum before collision = Total momentum after collision.
110 × 4 + 90 × -6 = (110 + 90)v
v = 0.5 m/s
-ve direction, (in the direction of man mass 90 kg).

3. Uniform motion They will not stop.

Question 4.
A circular track of radius 300m is kept with outside of track raised to make 5 degree with the horizontal.

  1. Name the process in which outside of the road is raised little above the inner.
  2. Obtain an expression for the optimum speed to avoid skidding (considering to friction)

Answer:
1. Banking of roqd

2.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 7
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 8
Consider a vehicle along a curved road with angle of banking θ. Then the normal reaction on the ground will be inclined at an angle θ with the vertical.
The vertical component can be divided into N Cosθ (vertical component) and N sinθ (horizontal component). The frictional force can be divided into two components. Fcosθ (horizontal component) and F sinθ (vertical component).
From the figure
N cos θ = F sinθ + mg
N cosθ – F sinθ = mg ______(1)
The component Nsin0 and Fsinθ provide centripetal force. Hence
N sinθ + F cos θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}\) ______(2)
eq (1) by eq (2)
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 9
Dividing both numerator and denominator of L.H.S by N cosθ. We get
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 10
This is the maximum speed at which vehicle can move over a banked curved road.
Optimum speed:
Optimum speed is the speed at which a vehicle can move over a curved banked road without using unnecessary friction. Putting µ = 0 in the above equation we get
v0 = \(\sqrt{\mathrm{Rg} \tan \theta}\).

Question 5.
A circular track of radius 400m is kept with outer side of track raised to make 5° with the horizontal (coefficient of friction 0.2)
(a) Name such track?
(b) What is optimum speed to avoid wear and tear of type?
(c) What is the maximum permissible speed to avoid skidding?
Answer:
(a) Banking.

Plus One Physics Law of Motion Five Mark Questions and Answers

Question 1.
A horse pulls a cart with constant force so that the cart moves with a constant speed.

  1. Does it violate Newtons second law of motion?
  2. If not, how will you account for the non acceleration of the cart?
  3. Will the speed of the cart increase, decrease or remain the same if the horse applied more force?
  4. A body of mass 5kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of the acceleration of the body.

Answer:
1. No.

2. The force applied by the car is balanced by the frictional force. Hence the cart moves with constant velocity.

3. If the horse is applied more force, the speed of the cart increases.

4. The resultant force,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 11
F = 10N
We know, F = ma
10 = 5 × a
acceleration, a = \(\frac{5}{10}\) = 2 m / sec2
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 12
The angle of resultant force,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 13
θ = tan-1 6/8
θ = 36°521
The angle of acceleration θ = 36°521.

Question 2.

  1. Friction is the force which opposes the relative motion between two surfaces in contact with each other. What is a limiting static friction? State the laws related to this.
  2. Show that the coefficient of friction is equal to the tan of the angle between the resultant and normal reactions.
  3. For a body of mass 5kg on a plane at a limiting static friction of 30 degrees. What is the force of friction?

Answer:
1. The maximum value of static friction is called limiting static friction.

  • The magnitude of the limiting friction is independent of the area of contact between the surfaces.
  • The limiting static friction is directly proportional to the normal reaction R.

ie f α R
fs = µsR.

2. Angle of friction is the angle whose tangent gives the coefficient of friction.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 14
Consider a body placed on a surface. Let N be the normal reaction and limit is the limiting friction. Let ‘θ’ be the angle between Resultant vector and normal reaction. From the triangle OBC,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 15
∴ tanθ = µ.

3. Tangent of the angle cient of friction.
µs = tanθ
µs = tan 30
µs = \(1 / \sqrt{3}\)
Friction F = µsmg
= \(1 / \sqrt{3}\) × 5 × 10
F = \(\frac{50}{\sqrt{3}}\)N.

Question 3.
The rate of change of linear momentum of a body is directly proportional to the external force applied on it, and takes place always in the direction of force applied.

  1. Name this law.
  2. Using this law obtain the expression for force.
  3. The motion of a particle of mass m is described by y = ut + \(\frac{1}{2}\) gt2. Find the force acting on it.

Answer:
1. Newton’s Second Law.

2. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 16
Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 17

3.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 18
Hence force F = mg.

Question 4.
Recoil of gun is based on the principle of conservation of momentum.

  1. State the principle of conservation of momentum.
  2. Explain the reoil velocity of gun.
  3. A bullet of mas 100g is fired from a rife of mass 200 kg with a spped of 50 m/s. Calculate the recoil velocity of the rife.

Answer:
1. According to law of conservation of linear momentum, if the external force acting on a body is zero, total linear momentum remains constant. According to Newton’s second law.
F = \(\frac{d p}{d t}\)
If F = 0, \(\frac{d p}{d t}\) = 0 i.e; P is constant.

2. Let M, m be the mass of gun and bullet respectively. Let V and ν be the velocities of gun and bullet after firing.
According to consevation of momentum
Total momentum before firing = Total momentum after firing
∴ O = MV + m ν
-MV = mν
The above equation shows that when bullet moves in forward direction, the gun moves in back direction. This motion of gun is called recoil of gun.

3. M = 200kg, m = 100g = 0.1kg
ν = 50 m/s, V = ?
MV = mν
200 × V = 0.1 × 50
V = \(\frac{0.1 \times 50}{200}\)m/s.

Question 5.
While firing a bullet, the gun must be held tight to the shoulder.

  1. Which conservation law helps you to explain this
  2. “In the firing process, the speed of the gun is very low compared to the speed of the bullet.” Substantiate the above statement using mathematical expressions.
  3. A shell of 20kg moving at 50m/s bursts in to two parts of masses 15kg and 5kg. If the larger part continues to move in the same direction at 70 m/s. What is the velocity and direction of motion of the other piece.

Answer:
1. Conservation of momentum.

2. Total momentum is conserved
∴ mu + MV = 0
V = \(\frac{-m u}{M}\) M is very large. Hence v is small

3. MV = m1 u1 + m2 u2
20 × 50 = 5u1 + 15 × 70
5u1 = 50
u1 = 10m/s.

Question 6.
While firing a bullet, the gun must be held tight to the shoulder.

  1. This is a consequence of______
  2. Show that recoil velocity is opposite to the muzzle velocity of the bullet.
  3. A gun of mass 5 kg fire a bullet of mass 5g, vertically upwards to a height of 100m. Calculate the recoil velocity of gun.

Answer:
1. Conservation of linear momentum.

2. Let M be the mass of gun and m be the mass of bullet. When gun fires, the gun and bullet acquire velocities V and v respectively.
According to conservation of momentum.
Total momentum before firing = Total momentum afterfiring
m × o + M × o = mu + MV
O = mv + MV
ie. – MV = mv
V = \(\frac{-m v}{M}\)

3. M = 5kg, m = 5 × 10-3 kg, h = 100m
v2 = u2 + 2as
0 = u2 + 2 × 10 × 100
Velocity of bullet,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 19

Question 7.
A standing passenger falls backwards when the bus starts suddenly.

  1. Explain why this happens?
  2. Which Newtons law gives the above concept. State the law.
  3. Obtain an expression for force using Newtons law.

Answer:
1. Due to inertia of rest, the body continues in the state of rest.

2. Newtons first law:
Everybody continues in its state of rest or of uniform motion along a straight line unless it is compelled by an external unbalanced force to change that state:

3. Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal ∆t. Due to this force the momentum is changed from \(\vec{p}\) to p + ∆p. Then according to Newtons second law, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 20
Where K is a constant pf proportionality. When we take the limit ∆t → 0, we can write
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 21

Question 8.
According to Newton’s law of motion rate of change of momentum is directly proportional to applied force.
a. Impulse has the unit similarto that of

  1. Momentum
  2. force
  3. time
  4. Energy

b. A man falling from certain height receives more injuries when he falls on a marble floor than when he falls on a heap of sand. Explain. Why?
c.
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 22
Force – time graph for a body starting from rest is shown in the figure. What is the velocity of the body at the end of 12 second? (Mass of the body is 5 kg)
Answer:
a. 1. Momentum.

b. When a man falls on a marble floor, the momentum is reduced to zero in lesser time. Due to this, the rate of change of momentum is large. So greater force acts on a man falls on marble floor.

c. The area of force – time graph gives change in momentum.
ie. change in momentum,
mv = 1/2 × (12 – 4) × (20 -10)
mv = 40

Plus One Physics Law of Motion NCERT Questions and Answers

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed.
(b) a cork of mass 10g floating on water
(c) a kite skillfully held stationary in the sky
(d) a car moving with a constant velocity of 30km h-1 on a rough road
(e) a high – speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
Applying Newton’s first law of motion, we find that no net force acts in any of the situations, (a) to (d). Again, no force in situation (e). This is because electron is far away from all material agencies producing electromagnetic and gravitational forces.

Question 2.
A constant retarding force of 50 N is applied to a body of mass 20kg moving initially with a speed of 15ms-1. How long does the body take to stop?
Answer:
Acceleration, a = –\(\frac{50 \mathrm{N}}{20 \mathrm{kg}}\) = -2.5ms-2
[Negative sign indicates retardation]
u = 15ms-1, v = 0, t = ?
v = u + at
0 = 15 – 2.5t or 2.5t = 15 or
t = \(\frac{15}{2.5}\)s = 6.0s.

Question 3.
A constant force acting on a body of mass 3.0kg changes its speed from 2.0ms-1 to 3.5 ms-1 in 25s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
m = 3kg; u = 2ms-1; v = 3.5 ms-1;
t = 25s ; F = ?
v = u + at
3.5 = 2 + 25a or a = 0.06 ms-2
F = ma = 3kg × 0.06 ms-2 = 0.18N.
The direction of force is along the direction of motion.

Question 4.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the string is cut when the bob is

  1. at one of its extreme positions.
  2. at its mean position.

Answer:

  1. At the extreme position, the speed of the bob is zero. If the string is cut, it will fall vertically down wards.
  2. At the mean position, the bob has a horizontal velocity. If the string is cut, it will fall along a parabolic path.

Question 5.
A man of mass 70kg stands on a weighing scale in a lift which is moving

  1. upwards with a uniform speed of 10ms-1
  2. downwards with a uniform acceleration of 5ms-2
  3. upwards with a uniform acceleration of 5ms-2 What would be the readings on the scale in each case?
  4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:

  1. a = 0, R = mg = 70 × 10 N = 700N
  2. mg – R = , ma ; R – mg – ma = (g – a)
    = 70(10 – 5) N = 350N
  3. R – mg = ma or R = m(g + a)
    = 70(10 + 5)N = 1056 N
  4. In the event of free fall, it is a condition of weight lessness.

Question 6.
A nucleus is at rest in the laboratory frame of reference. Show that if it dist integrates into two smaller nuclei, the products must move in opposite directions.
Answer:
Applying principle of conservation of momentum,
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 23
The negative sign indicates that the products move in opposite directions.

Question 7.
A shell of mass 0.020 kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms-1, what is the recoil speed of the gun?
Answer:
m = 0.02kg, M = 100kg, v = 80ms-1, V = ?
Plus One Physics Chapter Wise Questions and Answers Chapter 5 Law of Motion - 24
= -0.016ms-1 = -1.6cm s-1
Negative sign indicates that gun moves in a direction opposite to the direction of motion of the bullet.

Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images

You can Download Moving Images Questions and Answers, Kerala SSLC 10th IT Theory Questions and Answers Chapter 9 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus SSLC IT Theory Questions and Answers Chapter 9 Moving Images

Moving Images Questions and Answers

Section -1

Question 1.
The animation software in IT @ School GNU/Linux is
a) Synfig studio
b) Tupi
c) Gimp
d) Paint
Answer:
a) Synfig studio

HSSLive.Guru

Question 2.
Which type of animation software is synfig studio.
a) Proprietary
b) not able to take copy
c) pay and use
d) free software
Answer:
d) free software

Question 3.
To prepare a film, How many times images appear one after the other continuously in front of our eyes in one second?
a) 48
b) 12
c) 24
d) 36
Answer:
c) 24

Question 4.
Say the important stage of an animation film?
a) character designing
b) background
c) characters
d) action
Answer:
a) character designing

Question 5.
Construction of a storyboard is a preparation of
a) Drawing a picture
b) finding time zones
c) producing an animation
d) character designing
Answer:
c) Producing an animation

Question 6.
Synfig Studio is a free-animation software
a) Drawing pictures
b) Studio
c) Three dimensional
d) Two dimensional
Answer:
d) Two dimensional

Question 7.
Synfig studio software is designed by
a) Stall men
b) Robert B Quattlebaum
c) Leslie Lamport
d) Donald Knuth
Answer:
b) Robert B Quattlebaum

Question 8.
What versions of animation software can run in GNU/Linux and Microsoft Windows?
a) Grass
b) Notepad
c) Synfig studio
d) Pencil
Answer:
c) Synfig studio

Question 9.
Which tool is used to fill colors to the objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 1
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 2

Question 10.
Which tool is used to draw rectangular objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 3
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 4

Question 11.
Which tool is used to move the objects in synfig studio?
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 5
Answer:
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 6

Question 12.
Expand FPS
a) Frames per system
b) Frames per second
c) File properties settings
d) Frames per hour
Answer:
b) Frames per second

Question 13.
Animation is done by fast and continuous movements of images in a two-dimensional canvas. The images are known as
a) characters
b) FPS
c) Frames
d) Scene
Answer:
c) Frames

Question 14
Which menu is used to studio
a) File
b) view
c) window canvas
d)canvas
Answer:
d)canvas

Question 15.
To change the FPS in Synfig studio, by clicking
a) canvas → properties → time
b) view → pause
c) window → toolbox
d) canvas properties → image
Answer:
a) canvas → properties → time

Question 16.
We can import Images into synfig studio and use them
a) odf
b) vector
c) Pdf
d) Bitmap
Answer:
d) Bitmap

Question 17.
FPS = 24, Time =5, for an animation. Find the total number of frames in that animation
a) 24,
b) 48
c) 120
d) 920
Answer:
c) 120

Question 18.
The frames that represent important positions are known as
a) Tweening
b) current time
c) key
d)keyframe
Answer:
d) keyframe

Question 19.
What happens when you press
Kerala SSLC IT Theory Questions and Answers Chapter 9 Moving Images 7
play button
a) Animation works
b) Animation stops
c) no change
d) to save
Answer:
a) Animation works

Question 20.
The software fills up the frames in between two keyframes is called.
a) Interpolation
b) keyframe
c) Tweening
d) Edit
Answer:
c) Tweening

Question 21.
Rey frame utility is set in one of the panels in synfig audio. Give the names of that panel
a) Time track panel
b) layers panel
c) parameters panel
d) panel
Answer:
c) parameters panel

Question 22.
Name the extension of project file in synfig studio?
a). svg
b) .pdf
c) .sifz
d) .ods
Answer:
c) .sifz

Question 23.
In which panel to give the number of frames in current time?
a) Time track panel
b) layers panel
c) meters panel
d) panel
Answer:
a) Time track panel

HSSLive.Guru

Question 24.
What is the current time to animation from first frame
a) 0 f
b) 60 f
c) 120f
d) 121 f
Answer:
a) 0 f

Question 25.
What is the name for the frame with current time is ‘o’ f
a) first keyframe
b) last keyframe
c) middle keyframe
d) keyframe
Answer:
a) first keyframe

Question 26.
Give the order of activity to export a project file on synfig studio
a) File → Render
b) File → Save
c) File → Export
d) File → Save as
Answer:
a) File → Render

Question 25.
Which is not a member related to synfig studio?
a) parameters panel
b) time track panel
c) panel
d) layers panel
Answer:
c) panel

Question 28.
The utility to animate the flap its wings of a bird is
a) joining wings
b) adding time loop layer
c) copy of the first wing
d) adding the layer
Answer:
b) adding time loop layer

Question 29.
Write the activity to include an image to synfig studio
a) file → open
b) file → new
c) file → import
d) file → save
Answer:
c) file → import

Question 29.
in which menu import facility is available”?
a) Edit
b) file
c) canvas
d) windows
Answer:
b) file

Question 30.
Select ofie not group
a) . dv
b) .flv
c) . mpeg
d). svg
Answer:
d) .svg

Question 32.
What is the maximum value of current time?
a) 20 f
b) 60 f
c) 120 f
d) 1 f ,
Answer:
c) 120 f

Question 33.
What is the least value of current time?
a) 60 f
b) 120 f
c) 40 f
d) Of
Answer:
d)0f ‘

Question 34.
Keyframe are constructed in different type or same type
a) same type
b) different type
c) not constructing a keyframe
d) Equal type
Answer:
b) different type

Question 35.
Select any two free animation software’s
a) Anim studio
b) pencil
c) synfig studio
d) adobe flash
Answer:
b) pencil
and
c) synfig studio

Question 36.
An important stage in preparing animation is character designing. What is the meaning of character designing?
a) characters
b) Draw the characters
c) bringing character to life with humanity & personality
d) Giving life to the story
Answer:
b) Draw the characters
and
c) bringing character to life with humanity & personality

Question 37.
What are the uses of multi-colored buttons on the handle of a image?
a) To switch on the Animate editing mode
b) To adjust size
c) To take copy
d) To rotate if needed.
Answer:
b) To adjust size
and
d) To rotate if needed

Question 38.
Give the two activities done before starting animation from first keyframe
a) Current time is 60 f
b) Current time is 0 f
c) Before editing the motion, animate the edit mode button is active
d) press the play button to see the animation
Answer:
b) Current time is 0 f
and
c) Before editing the motion, animate the edit mode button is active

Question 39.
Sky and one star are drawn in synfig studio canvas. In the layers panel we can see a rectangle 001, and star 001. What are the tools use dot draw them
a) Circle tool
b) Fill tool
c) Star tool
d) Rectangle tool
Answer:
c) Star tool
and
d) Rectangle tool

Question 40.
Animation became much easier in film industry, with the arrival of new technologies. What are they?
a) computers
b) Radio
c) Animation software’s
d) office software.
Answer:
a) computers
and
c) Animation software’s

HSSLive.Guru

Question 41.
Create animations is one of the stages in animation film. There are two other stages before creating an animation. What are they?
a) To save animation project
b) Character designing
c) Playing an animation
d) Preparation of storyboards
Answer:
b) Character designing
and
d) Preparation of storyboards

Question 42.
Utilities of synfig studio windows are given below. Select one of each set and make a list.

Set-I

a) Database panel
b) Work area
c) Layers Panel
d) Task area
Answer:
c) Layers Panel

Set -II

a) Audio track
b) Video track
c) Time track panel
d) Fill
Answer:
c) Time track panel

Set-III

a) Parameters panel
b) Table
c) Storyboard
d) Path
Answer:
a) Parameters panel

Set -IV

a) Database window
b) tooIX
c) Editor window
d) Window
Answer:
b) tooIX

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Question 43
Select one video formats from each set

Set I

a) PNG
b) gif
c) xcf
d) sifz
Answer:
b) gif

Set II

a) dv
b) ods
c) odf
d) odp
Answer:
a) dv

Set III

a) jpg
b) flv
c) jpg
d) py
Answer:
b) flv

Set -IV

a) Html
b) ph
c) htm
d) mpeg
Answer:
d) mpeg

Question 44.
There are a lots of job opportunities related to animation in Government/public sectors of India and abroad. A list of job opportunities are given below. Select one from each set.

Set-I

a) Cinema production
b) Still photograph
c) Newspaper
d) Shops
Answer:
a) Cinema production

Set-II

a) Schools
b) advertising agencies
c) building construction
b) advertising agencies
Answer:
d) Library

Set-III

a) Radio
b) FM radio
c) TV
d) Tape recorder
c) TV

Set-IV

a) mobile
b) computer
c) calculator
d) computer games
Answer:
d) computer games

Question 45.
Images are included in synfig studio through import menu what are the uses of handles on the image? Select one form each set.

Set-1

a) To join the parts of an image
b) To arrange its position
c) To animation
d) Frame
Answer:
b) To arrange its position

Set – II

a) To arrange its size
b) To reduce its size
c) To increase its size
d) Can’t change its size
Answer:
a) To arrange its size

Set-II

a) To rotate
b) Can’t rotate
c) To remove the image
d) To include and image
Answer:
a) To rotate

Set-IV

a) To add a background image
b) Creating a scene
c) To adjust the views of the images
d) To change its background-color
Answer:
c) To adjust the views of the images

HSSLive.Guru

Question 46.
The layers are seen in the layers panel of a synfig studio some activities related to the layers are given in sets. Select correct activities from each set.

Set-1

a) Can’t change its order
b) can change its order
c) Arrange the order before including the layer
d) Layers are not visible
Answer;
b) can change its order

Set -II

a) Can’t group together
b) Objects
c) Can group together
d) Can’t change its color
Answer:
c) Can group together

Set-III

a) To take a copy of a layer
b) Layer is fixed
c) Can edit animation
d) Can’t delete the layer
Answer:
a) To take a copy of a layer

Set-IV

a) Removing the layer may effect he other
b) Can delete a layer
c) Can edit animation
d) Can’t delete the layers
Answer:
b) Can delete a layer

Plus One Physics Model Question Paper 4

Kerala Plus One Physics Model Question Paper 4

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

  • There is a ‘cool off time’ of 15 minutes in addition to the writing time.
  • Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
  • Read the instructions carefully.
  • Read questions carefully before you answering.
  • Calculations, figures, and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Plus One Physics Previous Year Question Papers and Answers 2018 1

Answer any four questions from question numbers 1 to 5. Each carries one score.
Plus One Physics Model Question Papers Paper 4 1

Engineering Physics MCQ with answers in PDF format.

Question 1.
Name the weakest force among the fundamental forces.
Plus One Physics Model Question Papers Paper 4 2

Question 2.
The work done during an isochoric process is …………….
Plus One Physics Model Question Papers Paper 4 3

The harmonic sequence formula is a sort of average calculator that is estimated by dividing the number of utilities.

Question 3.
Highway police detect over speeding vehicles by using ……………….
a. Magnus effect
b. Pascals law
c. Doppler effect
d. Bernoulli’s theorem
Plus One Physics Model Question Papers Paper 4 4

Question 4.
Two forces 3N and 4N are acting perpendicular to each other. The magnitude of the resultant force is
Plus One Physics Model Question Papers Paper 4 5

Question 5.
Say true/false: “Trade winds are produced due to conduction.”
Plus One Physics Model Question Papers Paper 4 6

Answer any five questions from question numbers 6 to 11. Each carries two scores.
Plus One Physics Model Question Papers Paper 4 7Plus One Physics Model Question Papers Paper 4 8

Question 6.
The displacement (S) of a body in time ‘t’ is given by S = at2 + bt. Find the dimensions of a and b.
Plus One Physics Model Question Papers Paper 4 9

Question 7.
Give the magnitude and direction of the net force on a stone of mass 0.1 kg.
a. Just after it is dropped from the window of a train accelerating 1 ms2.
b. Lying on the floor of a train which is accelerating with 1 ms-2, the stone being at rest relative to the train.
Plus One Physics Model Question Papers Paper 4 10

Question 8.
A body is rolling on a horizontal surface. Derive an equation for its kinetic energy,
Plus One Physics Model Question Papers Paper 4 11

Question 9.
The stress-strain graphs for two materials A and B are shown below (the graphs are drawn using the same scale) Which one is more elastic? Why?
Plus One Physics Model Question Papers Paper 4 12
Plus One Physics Model Question Papers Paper 4 13

Question 10.
“A heavy and a light body have the same kinetic energy.” Which one has greater momentum? Why?
Plus One Physics Model Question Papers Paper 4 14

Question 11.
The following figures refer to the steady flow of a nonviscous liquid. Which of the two figures is correct? Why?
Plus One Physics Model Question Papers Paper 4 15

Answer any five questions from question num 1.5 m numbers 12 to 17. Each carries three scores.
Plus One Physics Model Question Papers Paper 4 16

Question 12.
The side of a cube is measured as 3.405 cm.
a. How many significant figures are there in the measurement?
b. If the percentage error in the measurement of the side of the cube is 3%, find; the percentage error in its volume.
Plus One Physics Model Question Papers Paper 4 17

Question 13.
According to the conservation of energy “energy can neither be created nor be destroyed”
a. Prove law of conservation of mechanical energy in the case of a freely falling body.
b. The bob of a pendulum of length 1.5 m is released from the position A shown in the figure. What is the speed with which the bob arrives at the lowermost point B, given that 5% of its initial energy is dissipated against air resistance?
Plus One Physics Model Question Papers Paper 4 18

Question 14.
Acceleration due to gravity on earth changes with depth and height.
a. What is the weight of a body placed at the center of the earth? Why?
b. Find the height at which the acceleration due to gravity is 1/4th that at the surface of the earth.
Plus One Physics Model Question Papers Paper 4 19
Plus One Physics Model Question Papers Paper 4 20

Question 15.
A metal sphere of density ‘p’ and radius ‘a is falling through an infinite column of liquid of density ‘o’ and coefficient of viscosity Ty
a. Name any two forces acting on the; sphere.
b. With the help of Stokes theorem, derive an equation for the terminal velocity of I the sphere.
Plus One Physics Model Question Papers Paper 4 21

Question 16.
Conduction is the mode of transfer of heat in solids. of Write the unit of thermal conductivity.
b. “Burns produced by steam is severe than that produced by boiling water”Why?
Plus One Physics Model Question Papers Paper 4 22

Question 17.
A gas has ‘f’ degrees of freedom.
a. Calculate its Cp, Cv, and γ.
b. Define the mean free path.
Plus One Physics Model Question Papers Paper 4 23

Answer any five questions from question numbers 18 to 22. Each carries two scores.
Plus One Physics Model Question Papers Paper 4 24

Question 18.
A satellite moves in a circular orbit of radius ‘r’ with an orbital velocity.
a. Derive an equation for the orbital velocity of a satellite.
b. The time taken by Saturn to complete one orbit around the Sun is 29.5 times the earth year. If the distance of the earth from the Sun is 1.5 × 108km, then what will be the distance of the Saturn from the Sun?
Plus One Physics Model Question Papers Paper 4 25

Question 19.
In the simple harmonic motion, force is directly proportional to the displacement from the mean position.
a. Give an example of a harmonic oscillator.
b. Derive equations for the kinetic and potential energies of a harmonic oscillator.
c. Show graphically the variation of kinetic energy’ and potential energy of a harmonic oscillator.
Plus One Physics Model Question Papers Paper 4 26

Question 20.
A stretched string can be used as a musical instrument.
a. What is the fundamental frequency of a stretched string?
b. With neat diagrams, derive equations for the second and third harmonics of a stretched string.
Plus One Physics Model Question Papers Paper 4 27
Plus One Physics Model Question Papers Paper 4 28

Question 21.
A body having an initial velocity ‘v0’ has an acceleration ‘a’.
a. Using the velocity-time graph, derive an equation for displacement of the above body.
b. Draw the velocity Time graph and speed Time graph of a body thrown vertically in the air.
Plus One Physics Model Question Papers Paper 4 29

Question 22.
A javelin is thrown with an initial velocity ‘ V0‘ at an angle ” with the horizontal.
a. What are the horizontal and vertical velocities of the body
i. At the point of projection
ii. At maximum height
b. Find the angle of projection at which the maximum height attained by the javelin is equal to the horizontal range.
Plus One Physics Model Question Papers Paper 4 30

Answer any three questions from question numbers 23 to 26. Each carries five scores.
Plus One Physics Model Question Papers Paper 4 31

Question 23.
a. What is meant by ‘banking of roads’?
b. With a neat diagram, derive an equation for the maximum velocity of a car on a banked road.
c. What is the optimum speed of the car along the banked road?
Plus One Physics Model Question Papers Paper 4 32

Question 24.
The moment of inertia of a thin rod of mass M and length 1 about an axis perpendicular to the rod at its midpoint is \(\frac { { Ml }^{ 2 } }{ 12 }\).
a. What is the radius of gyration in the above case?
b. A student has to find the moment of inertia of the above rod about an axis (AB) perpendicular to the rod and passing through one end of the rod. Name and state the law used for this case.
c. Using the theorem, find the moment of inertia of the rod about AB.
Plus One Physics Model Question Papers Paper 4 33

Question 25.
Small drops of water assume spherical shape due to surface tension.
a. Define surface tension.
b. Derive an equation for the excess pressure inside a liquid drop of radius ‘R’ having surface tension σ.
c. Why do farmers plow the fields before summer?
Plus One Physics Model Question Papers Paper 4 34

Question 26,
Carnot engine is considered as an ideal heat engine.
a. Draw the PV graph of Carnot’s cycle.
b. Derive an equation to find the work done during an adiabatic process.
c. Calculate the efficiency of a heat engine working between ice point and steam point.
Plus One Physics Model Question Papers Paper 4 35

Answers

Answer 1.
Doppler effect

Answer 2.
Zero

Answer 3.
Doppler effect

Answer 4.
7N

Answer 5.
False

Answer 6.
[S] = [L]
[at2] = [L]
a = [LT-2]
[bt] = [L]
[b] = [LT-1]

Answer 7
a. Only force is gravitational. F = mg = 0.1 × 9.8 = 9.8 N downward j
b. Gravitational force is cancelled by normal I reaction.
∴ F2 = ma = 0.1 × 1 = 0.1 N, direction of motion of train.

Answer 8.
Plus One Physics Model Question Papers Paper 4 36

Answer 9.
In the two graphs, the slope of a graph of material A is greater than the slope of a graph of material B. So material A is more elastic than B. For material A the break-even point (D) is higher.

Answer 10.
Plus One Physics Model Question Papers Paper 4 37
Momentum is greater for a heavy body.

Answer 11.
Figure b is correct. According to an equation of continuity, the speed of liquid is larger at a smaller area. From Bernoulli’s theorem due to larger speed, the pressure will be lower at a smaller area and therefore the height of liquid column will also be at lesser height, while in Fig(a) height of liquid column at the narrow area is higher.

Answer 12.
Plus One Physics Model Question Papers Paper 4 38

Answer 13.
a. Law of conservation of energy. Energy can neither be created nor be destroyed, but it can be transformed from one form into another. Consider a body of mass’s’ placed at
Plus One Physics Model Question Papers Paper 4 39
b. Changing in PE after dissipation.
Plus One Physics Model Question Papers Paper 4 40

Answer 14.
Plus One Physics Model Question Papers Paper 4 41
Plus One Physics Model Question Papers Paper 4 42

Answer 15.
a. i. Weight, F, = mg acting downward
ii. Viscous force, F2 acting upward,
b. By strokes, formula F = 6πrηV Viscous force = Apparent weight of sphere in the solid
Plus One Physics Model Question Papers Paper 4 43

Answer 16.
a. W m-1K-1
b. Boiling water contains only a specific amount of heat energy required for it to boil. However, as steam is formed from boiling water, it contains the heat energy of boiling water, along with the latent heat of vaporization.i.e., 1kg of steam at 100°C contains 22.6 × 105 J more heat than 1 kg of water at 100°C. Hence, as steam has more heat energy, it can cause more severe burns than boiling water.

Answer 17.
Plus One Physics Model Question Papers Paper 4 44
b. Mean free path is an average distance between two successive collisions.

Answer 18.
a. It is the velocity required to put the satellite into its orbit around the earth.
Plus One Physics Model Question Papers Paper 4 45
The gravitational force on the satellite
Plus One Physics Model Question Papers Paper 4 46
The centripetal force required by the satellite to stay in this orbit is
Plus One Physics Model Question Papers Paper 4 47
in this orbit is In equilibrium the centripetal force is given by the gravitational force
Plus One Physics Model Question Papers Paper 4 48

Answer 19.
a. Oscillation of simple pendulum Oscillation of loaded spring
b. Let m be the mass of the particle executing SHM. Let v be the velocity at any instant,
Plus One Physics Model Question Papers Paper 4 49
Potential energy is the work required to take a particle against the restoring., force. Let a particle be displaced through a distance x from the mean position. Then restoring force, F = – kx, where k is the force constant. Now if we displace the particle further through a distance dx, Small work done, dw = – Fdx = kx dx Total work done from 0 to x
Plus One Physics Model Question Papers Paper 4 50

Answer 20.
a. Fundamental mode (or) First harmonic: If the string is plucked in the middle and released, then it vibrates in one segment with nodes at its ends and an antinode in the middle.
Plus One Physics Model Question Papers Paper 4 51
This is the lowest frequency with which string vibrates.
b. Second harmonic If the string is pressed in the middle and plucked at one-fourth of its length, then the string vibrates in two segments.
Plus One Physics Model Question Papers Paper 4 52
Plus One Physics Model Question Papers Paper 4 53
Third harmonic If the striping is pressed at one-third of its length from one end and plucked at one-sixth its length, it will vibrate in three segments.
Plus One Physics Model Question Papers Paper 4 54
Thus a collection of all possible mode is called harmonic series and n is called harmonic number.

Answer 21.
Plus One Physics Model Question Papers Paper 4 55
The area under the velocity-time graph gives the displacement of the body. Displacement, x = area OABD x = area of triangle ABC+ area of rectangle OACD.
Plus One Physics Model Question Papers Paper 4 56

Answer 22.
a.
i. Horizontal Vx = V0 cosθ Vertical Vy = V0sinθ
ii. Horizontal V’x = VO cosθ
Plus One Physics Model Question Papers Paper 4 57

Answer 23.
a. To avoid skidding and damage to tires of vehicles, the outer part of a road is slightly raised than the inner part. This is known as banking of roads.
Plus One Physics Model Question Papers Paper 4 58
The forces on the car are:
1. The weight of the car vertically downwards.
2. Normal reaction Racing normal to the road.
3. Frictional force acting parallel to the road.
Since there is no vertical acceleration,
R cosθ = mg + F sinθ
or R cosθ – F sinθ = mg …(1)
Now for maximum speed, F = μ, R
The centripetal force is provided by horizontal components of Rand Fas shown in the figure.
Plus One Physics Model Question Papers Paper 4 59
Plus One Physics Model Question Papers Paper 4 60

Answer 24.
a. The radius of gyration (k). It is the defined as the distance from an axis of rotation at which, if the whole mass of the body was concentrated, then its moment of inertia about that point would be the same as the moment of inertia of actual distribution of mass. l = Mk2
The radius of gyration (k) of a body is the square root of a ratio of the moment of inertia and a total mass of the body.
ie., a radius of gyration, k= \(k=\sqrt { \frac { l }{ M } }\)

b. Theorem of parallel axes: This theorem is good for any shape. The moment of inertia of the body about any axis is equal to the sum of a moment of inertia of a.parallel axis passing through the center of mass and product of its mass of the body and square of the distance between the two parallel axes.
Plus One Physics Model Question Papers Paper 4 61
where I am the moment
c. Using parallel axes theorem, the moment of inertia about AB,
Plus One Physics Model Question Papers Paper 4 62

Answer 25.
a. Surface tension (a) is the property due to which the free surface of a liquid at rest behaves like an elastic stretched membrane tending to contract so as to occupy a minimum surface area.
Plus One Physics Model Question Papers Paper 4 63
Thus it is measured as the force acting per unit length of an imaginary line drawn on the liquid surface, the direction of force being perpendicular to this line and tangential to the liquid surface.

b. Consider a liquid drop of radius R and surface tension o. Let P be the excess pressure inside the drop. The work done by the force due to excess pressure is
Plus One Physics Model Question Papers Paper 4 64
c. On plowing, the gap between sand particles act as a capillary tube, so that groundwater reaches the surface easily due to capillary rise.

Answer 26.
Plus One Physics Model Question Papers Paper 4 65
b. Work was done in the adiabatic process: We have a small amount of work done when volume changes through at pressure P.
Plus One Physics Model Question Papers Paper 4 66
(volume changes from v1 to v2 diabolically)
Plus One Physics Model Question Papers Paper 4 67
Plus One Physics Model Question Papers Paper 4 68
Plus One Physics Model Question Papers Paper 4 69

Plus One Physics Previous Year Question Papers and Answers