AP State Important Questions – Page 45

AP 9th Class Telugu Important Questions Chapter 9 భూమి పుత్రుడు

March 30, 2023July 29, 2021 by Srinivas

These AP 9th Class Telugu Important Questions 9th Lesson భూమి పుత్రుడు will help students prepare well for the exams.

AP State Syllabus 9th Class Telugu 9th Lesson Important Questions and Answers భూమి పుత్రుడు

9th Class Telugu 9th Lesson భూమి పుత్రుడు Important Questions and Answers

I. అవగాహన – ప్రతిస్పందన

క్రింది అపరిచిత పద్యాలను చదివి, ప్రశ్నలకు జవాబులు రాయండి.

తన కోపమె తన శత్రువు
తన శాంతమె తనకు రక్ష, దయ చుట్టంబౌ
తన సంతోషమె స్వర్గము
తన దుఃఖమె నరకమండ్రు తథ్యము సుమతీ.
ప్రశ్నలు – జవాబులు:
1. శత్రువు ఎవరు?
జవాబు:
కోపం

2. ఏది రక్ష?
జవాబు:
శాంతం

3. దయ ఎలాంటిది?
జవాబు:
చుట్టము

4. స్వర్గ నరకాలు అంటే ఏవి?
జవాబు:
సంతోషం, దుఃఖం

2. లావు గల వాని కంటెను
భావింపగ నీతిపరుడు బలవంతుండౌ
గ్రావంబంత గజంటును
మావటివాడెక్కినట్లు మహిలో సుమతీ.
ప్రశ్నలు – జవాబులు:
1. బలవంతుడు ఎవరు?
జవాబు:
నీతిపరుడు

2. ఏనుగు నడిపేవాడు?
జవాబు:
మావటివాడు

3. సుమతీ శతక కర్త?
జవాబు:
బద్దెన

4. ‘గ్రావం’ అర్థం?
జవాబు:
కొండరాయి

3. ఈ కింది సమీక్షనుచదివి ఇచ్చిన ప్రశ్నలకు జవాబులు రాయండి. (S.A. II – 2017-18)

మిద్దెతోటల పెంపకం ఇలా

మిద్దెతోటల పెంపకం సాగులో సేంద్రియ పద్ధతుల్ని ప్రోత్సహిస్తున్న రైతు నేస్తం ఫౌండేషన్ ప్రచురించిన ఈ పుస్తకాన్ని మిద్దెతోట సాగుచేస్తున్న తుమ్మేటి రఘోత్తమరెడ్డి తమ అనుభవాన్ని రంగరించి రాశారు. దీనిలో మిద్దెతోటల పెంపకం గురించి సూచనలిచ్చారు. అటువంటి రైతులకు మంచిసూచనలిచ్చారు. మిద్దెతోట పుస్తకం వెల రూ. 349/-
ప్రశ్నలు:
1. ‘మిద్దెతోట’ అనేది ఏమిటి ?
2. ‘మిద్దెతోట’ను ఎవరు ప్రచురించారు?
3. ‘మిద్దెతోట’ ఖరీదెంత?
4. పై సమీక్ష వలన ఎవరికి ప్రయోజనం?
జవాబులు:
1. భవనం పైన గల ఖాళీస్థలంలో ఏర్పాటు చేసుకున్న కుండీల మొదలైన వాటిలో చేసే మినీ వ్యవసాయం.
2. రైతు నేస్తం ఫౌండేషన్
3. రూ. 349/
4. మిద్దెతోట రైతులకు.

II. స్వీయరచన

క్రింది ప్రశ్నలకు నాలుగైదు వాక్యాలలో సమాధానాలు రాయండి.

ప్రశ్న 1.
తన కష్టంతో లోకానికి భుక్తిని పంచే భూమి పుత్రుని గూర్చి విశదపరచిన కవిని గూర్చి రాయండి. (S.A. II – 2017-18)
(లేదా)
అన్నదాతయైన భూమి పుత్రుడు’ ఔన్నత్యాన్ని అభివర్ణించిన కవిని పరిచయం చేయండి. (S.A. II – 2015-16)
జవాబు:
కవి : శ్రీ దువ్వూరి రామిరెడ్డి
కాలం : 9. 11. 1895 నుండి 11.9.1947
జన్మస్థలం : నెల్లూరు
రచనలు : నలజారమ్మ, వనకుమారి, కృషీవలుడు, జలదాంగన, యువక స్వప్నం, కడపటి వీడ్కోలు, పానశాల, నక్షత్రశాల, నైవేద్యం, భగ్న హృదయం, పరిశిష్టం, ప్రథమ కవిత్వం.
బిరుదు : కవికోకిల
శైలి : సరళ సుందరంగా ఉంటుంది. విశ్వశాంతి, దేశభక్తి, మానవతావాదం, అభ్యుదయం వీరి రచనల్లో కనిపిస్తాయి.

ప్రశ్న 2.
రైతుతో ఎవరెవరు సాటిరారని కవి అన్నారు?
జవాబు:
రైతును తమ్ముడా ! అని సంబోధిస్తూ, లోకంలో కొందరు చిత్రంగా ఉంటారు. వీరిలో కొందరు చిన్నతాడు కట్టిన చిన్న చెంబుతో నేల నూతిలో నీళ్ళు తోడేవారు (ఉపయోగం లేని పని), కొందరు తలకు, మోకాలకీ ముడి పెట్టేవారు (సందర్భ శుద్దిలేని పని), ఇంకొందరు చిటికెలతో పందిళ్ళు అల్లేవారు (కబుర్లే పని), అంటే వీళ్ళంతా కేవలం మాటల చమత్కారంతో అరచేతిలో స్వర్గం చూపించేవారు. కానీ చేతులతో సమాజ సేవ చేస్తున్న నీకు వీరెవ్వరూ సాటిరారని కవి అన్నారు.

ప్రశ్న 3.
రైతుకు ఏవి కొరత?
జవాబు:
సమాజం సుఖసంతోషాలతో ఉండటానికి రైతే కారణం. కానీ అతని కష్ట ఫలితాన్ని ఇతరులు అనుభవించి సుఖపడతున్నారు. రైతు క్షేమాన్ని, శ్రేయస్సును కోరేవారు ఎవరూ లేరు. కనీసం కన్నెత్తి అయిన చూడరు. ఆప్యాయంగా పలకరించరు. చివరకు తిండికీ, బట్టకు ఎప్పుడూ కొరతే.

ప్రశ్న 4.
“అట్టి కృతఘలన్………… పద్యం ద్వారా రైతు ఎలాంటి వాడని అర్థమైంది?
జవాబు:
చేసిన మేలు మరచేవారిని రైతు అసలు పట్టించుకోడని ఈ పద్యం ద్వారా అర్థమైంది. మరియు పొలం పనులలో అతని శరీరం ఎముకలగూడుగా మారినా, వానలు ముంచెత్తినా, కరవు పీడించినా వాటిని లెక్కచేయడని తెలిసింది. ఇంకా కాయకష్టాన్నే నమ్మి, స్వార్జితమైన పట్టెడన్నమే తిని రైతు నిజంగా ‘భూమి పుత్రుడె’ అని గ్రహించాను.

ప్రశ్న 5.
‘భూమి పుత్రుడు’ ప్రక్రియను గూర్చి రాయండి.
జవాబు:
‘భూమి పుత్రుడు’ పాఠ్యభాగం ‘కావ్యం’ ప్రక్రియకు చెందినది. కవి యొక్క కర్మము – కావ్యము. దీనిలో వర్ణనయే ప్రధానాంశముగా కల్గి, మనసుకు హత్తుకునేలా రచన సాగుతుంది.

ఈ క్రింది ప్రశ్నలకు పది లేక పన్నెండు వాక్యాలలో సమాధానాలు రాయండి.

ప్రశ్న 1.
రైతును ఆదర్శంగా తీసుకొని ప్రజలు జీవించడం అవసరం ఎంతైనా ఉంది. దీనిని నీవు సమర్థిస్తావా ? వివరించండి.
జవాబు:
‘రైతే దేశానికి వెన్నెముక’, ‘పల్లెలే దేశానికి పట్టుకొమ్మలు’ – అన్న మాటలు అందరూ అనే మాటలు, వినే మాటలు. రైతు, పల్లెలోని గొప్పదనాన్ని మాటల్లో చెప్పడం తప్ప ఎవరూ వారికి సాయం చేతల్లో చూపించరు. పల్లె సౌందర్యాన్ని ఆస్వాదిస్తామేగాని, అక్కడి ప్రజల బాగోగులు చూడము. పల్లె ప్రజల్లో ఇచ్చి పుచ్చుకొనే తత్వం ఉంటుంది. ఒకరికొకరు పనులలో సాయం అంది పుచ్చుకుంటారు. రైతును ఆదర్శంగా తీసుకోవడం అంటే భేషజం లేని జీవితం గడపటమే. ఉన్నా లేకపోయినా ఒకేలా ఉండడం రైతు జీవితం. నలుగురి క్షేమం కోరేవాడు. ఈ లోకంలో రైతు తప్ప ఇంకెవరుంటారు. మనం రైతులాగా నిస్వార్థంగా, తృప్తిగా జీవించగలిగితే మనమున్న చోటే స్వర్గం అవుతుంది.

రైతు తాను పండించిన పంటను గిట్టుబాటు ధర రాకపోయినా తృప్తిపడి, మరుసటి సంవత్సరం పంట ఇంకా బాగా మొదలుకొని, చిరవకు పంట చేతికి వచ్చే దాకా పండించాలని తాపత్రయపడతాడు. పంట వేయడానికి ముందు పొలం దున్నటం రైతు గుండె ఎంతగా అల్లాడుతుందో ఎప్పుడైనా మనం ఆలోచిస్తామా. పంట పదును మీదున్నప్పుడు వానో, వరదో వస్తుందనే ఊహే ప్రాణాన్ని విలవిలలాడిస్తుంది. అయినా వీటన్నింటిని భరించి, తోటివాళ్ళమైన మనందరి ఆకలి తీర్చే రైతు మనందరికి భగవంతుడు ఇచ్చిన సోదరుడు.

మనం గుర్తించినా, గుర్తించకపోయినా తన సంసారాన్ని ఒక ప్రక్క వ్యవసాయాన్ని ఒక ప్రక్క నడుపుతూ , సమాజాన్ని నడిపిస్తున్నాడు. నిస్వార్థం అతని మనసు, సంతృప్తి అతని ఆలోచన, అందరూ బాగుండాలి అనేది అతని ఆకాంక్ష. మనం గమనిస్తే ఏదైనా సమస్య వచ్చినపుడు పెద్దల సమక్షంలో చర్చకు వస్తే అప్పుడు మధ్యమ మార్గంగా తీర్పు చెప్పడానికి “రైతు పద్ధతిలో మాట్లాడుకుందాం” అంటారు. దీనిని బట్టే మనం అర్థం చేసుకోవచ్చు రైతు ఎంత గొప్ప వ్యక్తో.

ప్రశ్న 2.
రైతు దేశానికి వెన్నెముక అంటారు కదా! అంతటి ప్రాధాన్యత వహించిన భూమి పుత్రుడుని గురించి దువ్వూరి రామిరెడ్డి గారెలా ఆవిష్కరించారో మీ స్వంత మాటల్లో రాయండి. (S.A. II – 2018-19)
జవాబు:
రైతు దేశానికి వెన్నెముక. నలుగురి క్షేమం కోరేవాడు. ఈ లోకంలో రైతు తప్ప ఇంకెవరుంటారు. మనం రైతులాగా నిస్వార్థంగా, తృప్తిగా జీవించగలిగితే మనమున్నచోటే స్వర్గం అవుతుంది. రైతు తాను పండించిన పంటకు గిట్టుబాటు ధర రాకపోయినా తృప్తిపడి, మరుసటి సంవత్సరం పంట ఇంకా బాగా పండించాలని తాపత్రయ పడతాడు. పంట వేయడానికి ముందు పొలం దున్నడం మొదలుకొని ధాన్యం ఇంటికి తెచ్చేవరకు రైతు గుండె ఎంతగా అల్లాడుతుందో ఆలోచిస్తేనే గుండె జారిపోతుంది.

ఉన్నా లేకపోయినా ఒకేలా ఉండడం రైతు జీవితం. పంట పదును మీదున్నప్పుడు వానో, వరదో వచ్చినప్పుడు అతని మానసిక స్థితి స్థిరంగా ఉంటుంది. అతని ధ్యాస పంటను రక్షించడమే, లేకపోతే నలుగురికి అన్నం లేకుండా చేసినవాణ్ణి అవుతానని బాధ్యత పడతాడు. సృష్టి స్థిల కారులలో విష్ణువు స్థితికర్త. అంటే మనల్ని పోషించేవాడని అర్థం. ప్రస్తుత కాలంలో మనకు రైతే స్థితికర్త,

మనం గమనిస్తే ఏదైనా సమస్య వచ్చినప్పుడు పెద్దల సమక్షంలో చర్చకు వస్తే అప్పుడు మధ్యమ మార్గంగా తీర్పు చెప్పడానికి ‘రైతు పద్ధతిలో మాట్లాడుకుందాం’ అంటారు. దీనిని బట్టి మనం అర్థం చేసుకోవచ్చు. రైతు ఎంత గొప్ప వ్యక్తో. అందుకే దువ్వూరి రామిరెడ్డిగారు “చేతులతో సమాజసేవ చేస్తున్న నీకు వేరెవ్వరూ సాటిరారని” అన్నారు.

III. భాషాంశాలు (పదజాలం, వ్యాకరణం)

1. పర్యాయపదాలు:

ఈసు : అసూయ, ఈర్య
కన్ను : అక్షి, నేత్రం, నయనం
మనుజుడు : మానవుడు, నరుడు, మనుష్యుడు
కృషి : వ్యవసాయం, సేద్యం, కరిసనం
నుతి : పొగడ్త, ప్రశంస
క్షామం : కరవు, అనావృష్టి
తాత : తండ్రి తండ్రి, పితామహ
క్ష్మా : ధారణి, నేల, భూమి

2. వ్యుత్పత్త్యర్థాలు :

కావ్యం : కవి యొక్క కర్మము (గ్రంథం)
అతిథి : తిథి, వార, నక్షత్రము నియమాలు లేక ఇంటికి భోజనానికి వచ్చేవాడు
కృతఘ్నుడు : చేసిన మేలు మఱచువాడు
క్ష్మా : భారమును వహించుటయందు క్షమ (ఓర్పు) కలది (భూమి)
సత్యం : సత్పురుషులయందు పుట్టునది (నిజం)
పుత్రుడు : పున్నామ నరకం నుండి రక్షించువాడు (కుమారుడు)

3. నానార్థాలు :

ఆత్మ : మనస్సు, పరమాత్మ, బుద్ధి, దేహం
రసము : చారు, పాదరసం, శృంగారాది రుచి, కోరిక
కాలము : సమయం, నలుపు, చావు

4. ప్రకృతి – వికృతులు :

భూమి – బూమి
మృత్తిక – మట్టి
కాంక్ష – కచ్చు
కష్టము – కసుటు
భోగం – బోగం (సుఖం)
విద్య – విద్ధియ, విద్దె
పుత్రుడు – బొట్టె, బొట్టియ, పట్టి
గౌరవం – గారవం
బ్రధ్న – పొద్దు
శ్రీ – సిరి
విశ్వాసం – విసువాసం
స్పర్థ – పంతం

5. సంధులు :

హిత + అర్థ = హితార్థ – సవర్ణదీర్ఘ సంధి
దైనిక + ఆవశ్యకం = దైనికావశ్యకం – సవర్ణదీర్ఘ సంధి
కష్ట + ఆర్జితం = కష్టార్జితం – సవర్ణదీర్ఘ సంధి
రస + ఆస్వాద = రసాస్వాద – సవర్ణదీర్ఘ సంధి
దుర్భర + అవస్థ = దుర్భరావస్థ – సవర్ణదీర్ఘ సంధి
కన్నెత్తియున్ + చూతురే = కన్నెత్తియుంజూతురే – సరళాదేశ సంధి
తోపు + తోపు = తోదోపు – ప్రాతాది సంధి
పస్తు + ఉన్న = పస్తున్న – ఉత్వసంధి
ప్రొద్దు + పొడిచిన = ప్రొద్దువొడిచిన – గసడదవాదేశ సంధి
ప్రొద్దు + క్రుంకు = ప్రొద్దుగ్రుంకు – గసడదవాదేశ సంధి
జీవ + కట్టి = జీవగట్టు – ఉత్వసంధి
కన్ను + ఎత్తి = కన్నెత్తి – ఉత్వసంధి
శ్రమ + ఆర్జితం = శ్రమార్జితం – సవర్ణదీర్ఘ సంధి

6. సమాసాలు:

భూమిపుత్రుడు = భూమి యొక్క పుత్రుడు – షష్ఠీ తత్పురుష సమాసం
ధారుణీపతి = ధరణికి పతి – షష్ఠీ తత్పురుష సమాసం
పవిత్రమూర్తి = పవిత్రమైన మూర్తి – విశేషణ పూర్వపద కర్మధారయ సమాసం
శూరమణి = శూరుల అందు శ్రేష్ఠుడు – సప్తమీ తత్పురుష సమాసం
జీవన స్పర్థ = జీవనమునందు స్పర్థ – సప్తమీ తత్పురుష సమాసం
జీవన సంగ్రామం = జీవనమనే సంగ్రామం రూపక సమాసం
హాలిక వర్య – రైతులలో శ్రేష్ఠ – షష్ఠీ తత్పురుష సమాసం

7. అలంకారాలు:

జీవన సంగ్రామం – రూపకాలంకారం. ఉపమాన ఉపమేయాలకు అభేదం చెప్పుట.
జీవనం – ఉపమేయం
సంగ్రామం – ఉపమానం
ఈ రెండింటికి అభేదం చెప్పబడినది. కనుక ఇది రూపకాలంకారం.

9th Class Telugu 9th Lesson భూమి పుత్రుడు 1 Mark Bits

1. ఆధునిక కాలంలో కృషి చేయడానికి ఎవరూ కృషి చేయడం లేదు – గీత గీసిన పదాలకు నానార్థపదాలు గుర్తించండి. (S.A. II – 2017-18)
ఎ) కష్టం – కారణం
బి) వ్యవసాయం – సాయం
సి) వ్యవసాయం – వ్యవహారం
డి) వ్యవసాయం – ప్రయత్నం
జవాబు:
డి) వ్యవసాయం – ప్రయత్నం

2. లక్ష్మి అనుకున్న కర్జము నెరవేరింది. (ప్రకృతి పదం గుర్తించండి) (S.A. II – 2017-18)
ఎ) కారణం
బి) కార్యం
సి) కయ్యం
డి) కాలం
జవాబు:
బి) కార్యం

3. ‘మనిచిరి నీ పితామహులమాంద్య సుశీలురు సర్వవృత్తిపా’. (ఏ పద్యపాదమో గుర్తించండి.) (S.A. III – 2016-17)
ఎ) మత్తేభము
బి) శార్దూలము
సి) ఉత్పలమాల
డి) చంపకమాల
జవాబు:
డి) చంపకమాల

4. అఖిల వాణిజ్యములు సిరికాట పట్లు. (ఆధునిక వచనాన్ని గుర్తించండి) (S.A. II. 2017-18)
ఎ) అఖిలమైన వాణిజ్యంబులు సిరికాట పట్లు
బి) అఖిలంబైన వాణిజ్యమ్ములు సిరికినాట పట్లు
సి) అఖిల వాణిజ్యాలు సిరికాట పట్లు
డి) అఖిల వాణిజ్యముల్ సిరికి నాటపట్టులు
జవాబు:
సి) అఖిల వాణిజ్యాలు సిరికాట పట్లు

5. “చిన్నప్పటి నుండీ నాకు బోటనీ అభిమాన విషయం” అన్నాడు రచయిత. (పరోక్ష కథనంలోకి గుర్తించండి) (S.A. II – 2017-18)
ఎ) చిన్నప్పటి నుండీ తనకు బోటనీ అభిమాన విషయమని అన్నాడు రచయిత.
బి) రచయితకు బోటనీ అభిమాన విషయమన్నాడు.
సి) రచయిత బోటనీ నాకు అభిమాన విషయమన్నాడు.
డి) బోటనీ అభిమాన విషయమని రచయిత అన్నాడు.
జవాబు:
ఎ) చిన్నప్పటి నుండీ తనకు బోటనీ అభిమాన విషయమని అన్నాడు రచయిత.

భాషాంశాలు (పదజాలం, వ్యాకరణం)

1. ఆర్థాలు :

6. అన్ని వృత్తులలో పావనమైనది వ్యవసాయం – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) ధర్మం
B) పవిత్ర
C) మలినం
D) న్యాయం
జవాబు:
B) పవిత్ర

7. శ్రమ పడకుండా ఫలములు తమంతట తాముగా రావు – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) దారులు
B) పండ్లు
C) దేవతలు
D) ఫలితాలు
జవాబు:
D) ఫలితాలు

8. బావులకు ఉగ్గాలు ఏర్పాటు చేసేవారు – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) బకెట్లు
B) బిందెలు
C) చేదలు
D) గంగాళాలు
జవాబు:
C) చేదలు

9. రాజు చేతిలోని ధర్మదండం కన్నా నీ చేతి హలం గొప్పది – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) నాగలి
B) కొడవలి
C) గొడ్డలి
D) కర్ర
జవాబు:
A) నాగలి

10. ఇరుగుపొరుగు వారి సంపదకై ఈసు పొందవు – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) ప్రేమ
B) అభిమానం
C) కోపం
D) ఈర్ష్య
జవాబు:
D) ఈర్ష్య

11. నీ హృదయ కళిక ఎంతో పవిత్రమైనది – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) పువ్వు
B) దీపం
C) మొగ్గ
D) బంగారం
జవాబు:
C) మొగ్గ

12. కృషి సకల పరిశ్రమలకు మూలము – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) పశువు
B) వ్యవసాయము
C) పక్షి
D) కష్టం
జవాబు:
B) వ్యవసాయము

13. సంపదయే సుఖాలను పొందడానికి జీవగఱ్ఱ – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) జీవనౌషధం
B) జీలకట్ట
C) కారణం
D) ఆధారం
జవాబు:
A) జీవనౌషధం

14. నీకు మాత్రం తిండికి, బట్టకు ఎప్పుడూ కఱవె – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) నిండు
B) సమం
C) క్షామమె
D) ఎక్కువ
జవాబు:
C) క్షామమె

15. పండ్లనిచ్చిన వృక్షమును గూర్చి ఆలోచించరు – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) కొమ్మ
B ) మొక్క
C) మొగ్గ
D ) చెట్టు
జవాబు:
D ) చెట్టు

16. వ్యవసాయాన్ని చేయడంలో నీ శరీరం అస్థిపంజరంగా మారింది – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) ఎముక
B) బోను
C) ఎముకల గూడు
D) పుర్రె
జవాబు:
C) ఎముకల గూడు

17. నీకు కొదవ ఏముంది? – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) లోపం
B) స్థాయి
C) స్థానం
D) హీనం
జవాబు:
A) లోపం

18. బ్రతకడంకోసం స్పర్థ సహజమైన కాలం ఇది – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) పందెం
B) పోటీ
C) తగాదా
D) యుద్ధం
జవాబు:
B) పోటీ

19. జీవితం అనే సంగ్రామంలో విజయం పొందాలి – గీత గీసిన పదానికి అర్థం గుర్తించండి.
A) పందెం
B) పోటీ
C) యుద్ధం
D) తిట్టు
జవాబు:
C) యుద్ధం

2. పర్యాయపదాలు :

20. ‘వారి సంపదకై యీసు గూరబోవవు’ – గీత గీసిన పదానికి సమానార్థక పదాన్ని గుర్తించండి.
A) ఆశ
B) ఈర్ష్య
C) వాంఛ
D) ప్రేమ
జవాబు:
B) ఈర్ష్య

21. ఈ ఏడాది నీరు లేక క్షామం వచ్చింది – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) క్షారం, కాయం
B) కామం, కారం
C) కరవు, అరువు
D) అనావృష్టి, కరవు
జవాబు:
D) అనావృష్టి, కరవు

22. అసూయ మనిషిని రాక్షసుణ్ణి చేస్తుంది – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) అనసూయ, ఈసు
B) ఈర్ష్య, ఈసు
C) ఈర్ష్య, ద్వేషం
D) కోపం, క్రోధం
జవాబు:
B) ఈర్ష్య, ఈసు

23. శ్రుతిమించి నుతి కూడదు – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) బావి, నూయి
B) చెరువు, బావి
C) ప్రశంస, పొగడ్త
D) ధర్మం, దానం
జవాబు:
C) ప్రశంస, పొగడ్త

24. కన్నులున్న వారిని సైతం గుడ్డివారిని చేస్తున్నది అంధకారం – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) అక్షి, కుక్షి
B) నేత్రం, నయనం
C) ఆత్రం, నయనం
D) నేత్రం, నయం
జవాబు:
B) నేత్రం, నయనం

25. మా తాత అంటే మాకెంతో ఇష్టం – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) తండ్రి తండ్రి, పితామహుడు
B) తల్లి తండ్రి, పితామహి
C) బ్రహ్మ, తండ్రి
D) విధాత, తాత
జవాబు:
A) తండ్రి తండ్రి, పితామహుడు

26. రాయలు గొప్ప క్ష్మా పాలకుడు – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) భూమి, రాజు
B) నేల, రేడు
C) ధరణీ, మంత్రి
D) వసుధ, పృథ్వి
జవాబు:
D) వసుధ, పృథ్వి

27. మనదేశం వ్యవసాయం ప్రధాన వృతిగా గల దేశం – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) సాగు, బాగు
B) సేద్యం, కృషి
C) కరిసనం, కూలీ
D) సేద్యం, మద్యం
జవాబు:
B) సేద్యం, కృషి

28. నీ హలము కన్నను కవి కలము గొప్పదగునె? – గీత గీసిన పదానికి పర్యాయపదాలు గుర్తించండి.
A) అరక, కత్తి
B) పార, ఖడ్గము
C) నాగలి, సీరము
D) గునపము, నాగలి
జవాబు:
C) నాగలి, సీరము

29. ‘నేల నూతులకుగ్గాలు నిలుపువారు’ – గీత గీసిన పదానికి పర్యాయపదాలు ఏవి?
A) బావులు, కూపములు
B) గోతులు, పాతరలు
C) తాళ్ళు, నూతులు
D) చేలు, పొలములు
జవాబు:
A) బావులు, కూపములు

30. ‘కావున కృషీవలా నీవె కారణమవు’ – గీత గీసిన పదానికి పర్యాయపదాలు ఏవి?
A) రైతు, కార్మికుడు
B) కర్షకుడు, సైరికుడు
C) రైతు, పనివాడు
D) శ్రామికుడు, కార్మికుడు
జవాబు:
B) కర్షకుడు, సైరికుడు

31. వృక్షములు మానవుల పాలిటి ప్రత్యక్ష దైవాలు – గీత గీసిన పదానికి పర్యాయపదాలేవి?
A) చెట్టు, గుట్టు
B) పైరు, పచ్చ
C) తరువు, చెట్టు
D) తీగ, పాదు
జవాబు:
C) తరువు, చెట్టు

32. ‘జీవన సంగ్రామం అనే పోరాటంలో శ్రామికుడికే విజయం ‘ – గీత గీసిన పదాలకు పర్యాయపదం గుర్తించండి.
A) పరిశ్రమ
B) కృషి
C) రణము
D) ప్రయత్నం
జవాబు:
C) రణము

3. వ్యుత్పత్యర్థాలు :

33. ‘కావ్యం’ వ్యుత్పత్తి గుర్తించండి.
A) కవికర్త
B) కవి కర్మము
C) కవి క్రియ
D) కవి హేతువు
జవాబు:
B) కవి కర్మము

34. తిథి, వార, నక్షత్ర, నియమం లేక భోజనానికి వచ్చేవాడు – వ్యుత్పత్త్యర్థం గుర్తించండి.
A) చుట్టం
B) మిత్రుడు
C) అతిథి
D) హరిదాసు
జవాబు:
C) అతిథి

35. చేసిన మేలు మఱచువాడు నరకానికి పోతాడు – గీత గీసిన వానికి వ్యుత్పత్త్యర్థం గుర్తించండి.
A) ధన్యుడు
B) ధర్మాత్ముడు
C) పుణ్యశీలి
D) కృతఘ్నుడు
జవాబు:
D) కృతఘ్నుడు

36. సత్పురుషులయందు పుట్టు మాటలు శిరోధార్యాలు – గీత గీసిన వానికి వ్యుత్పత్త్యర్థం గుర్తించండి.
A) సత్యం
B) ప్రాణం
C) జీవితం
D) గుండె
జవాబు:
A) సత్యం

37. ‘పున్నామ నరకం నుండి కాపాడువాడు’ – దీని వ్యుత్పత్త్యర్థం గుర్తించండి.
A) అల్లుడు
B) తమ్ముడు
C) పుత్రుడు
D) మిత్రుడు
జవాబు:
C) పుత్రుడు

38. ‘భారమును వహించుట యందు క్షమ కలది’ – వ్యుత్పత్యర్థం గుర్తించండి.
A) క్షా
B) క్యా
C) క్ష్వా
D) క్ష్మా
జవాబు:
D) క్ష్మా

39. ‘కృషీవలుడు’ పదానికి వ్యుత్పత్తిని గుర్తించండి.
A) కృషి చేసేవాడు
B) భూమిని దున్ని బ్రతికేవాడు
C) పొలంపని చేసేవాడు
D) కార్మికుడు
జవాబు:
B) భూమిని దున్ని బ్రతికేవాడు

4. నానార్థాలు :

40. మానవుడు కాల మాన పరిస్థితులకు అనుగుణంగా ఉండాలి – గీత గీసిన పదానికి నానార్థాలు గుర్తించండి.
A) సమయం, నలుపు
B) చావు, మరణం
C) నలుపు, తెలుపు
D) సమయం, సాయం
జవాబు:
A) సమయం, నలుపు

41. ఆత్మ, పరమాత్మ వేరని ద్వైత సిద్ధాంతం చెబుతుంది – గీత గీసిన పదానికి నానార్థాలు గుర్తించండి.
A) మనస్సు, మనసు
B) బుద్ధి, పరమాత్మ
C) దేహం, శరీరం
D) బుద్ధి, బుద్ధుడు
జవాబు:
A) మనస్సు, మనసు

42. రసములు తొమ్మిది – గీత గీసిన పదానికి నానార్థాలు గుర్తించండి.
A) చారు, సాంబారు
B) పాదరసం, హసరసం
C) రుచి, కోరిక
D) శృంగారాది, హాస్యం
జవాబు:
C) రుచి, కోరిక

43. నేడు ధరకు విపరీతంగా ధర పెరిగింది – గీత గీసిన పదాలకు నానార్థాలు రాయండి.
A) ఖరీదు, ప్రియము
B) నేల, నెల
C) ధరణి, వెల
D) రేటు, గోటు
జవాబు:
C) ధరణి, వెల

44. సరియైన వర్షం లేక పంటలు పండలేదు – గీత గీసిన పదం నానార్థాలు ఏవి?
A) వాన, సంవత్సరం
B) వర్షం, హర్షం
C) వాన, నాన
D) ఏడు, పంట
జవాబు:
A) వాన, సంవత్సరం

5. ప్రకృతి – వికృతులు :

45. పూల కాంక్ష చెట్టు తల్లి పాదాల చెంత రాలిపోవాలని – గీత గీసిన పదానికి వికృతిని గుర్తించండి.
A) కచ్చు
B) కోరిక
C) ఇచ్చ
D) వాంఛ
జవాబు:
A) కచ్చు

46. కష్టము చేసినవాడు ఫలితం తప్పక పొందుతాడు – గీత గీసిన పదానికి వికృతి పదం గుర్తించండి.
A) కసము
B) కసుట
C) కసట
D) కసటము
జవాబు:
B) కసుట

47. ఆత్మవిశ్వాసం ఎప్పుడు విడిచిపెట్టకూడదు – గీత గీసిన పదానికి వికృతి పదం గుర్తించండి.
A) నమ్మకం
B) విశవాసం,
C) విసువాసం
D) విసాసం
జవాబు:
C) విసువాసం

48. స్పర్థా వర్తతే విద్యా – గీత గీసిన పదానికి వికృతి పదం గుర్తించండి.
A) పోటీ
B) పందెం
C) యుద్ధం
D) పంతం
జవాబు:
D) పంతం

49. మట్టి పిసుక్కొనే వారిని హీనంగా చూడకు – గీత గీసిన పదానికి ప్రకృతి పదం గుర్తించండి.
A) మర్యం
B) మృత్తిక
C) నేల
D) భూమి
జవాబు:
B) మృత్తిక

50. పుత్రుడు లేనివారికి మోక్షపదం రాదా? – గీత గీసిన పదానికి వికృతి పదం గుర్తించండి.
A) పుతుడు
B) సుతుడు
C) బొట్టె
D) కొడుకు
జవాబు:
C) బొట్టె

51. వారి సంపదకై ఈసు గూరబోవు – గీత గీసిన పదానికి ప్రకృతి పదం ఏది?
A) ఈస
B) ఈర్ష్య
C) అసూయ
D) ద్వేషం
జవాబు:
B) ఈర్ష్య

52. అఖిల వాణిజ్యములు సిరి కాటపట్టు – గీత గీసిన పదానికి ప్రకృతిని గుర్తించండి.
A) సిరీ
B) హరీ
C) శ్రీ
D) హరి
జవాబు:
C) శ్రీ

53. ఎంత నిర్మలమోయి నీ హృదయ కళిక – గీత గీసిన పదానికి వికృతిని గుర్తించండి.
A) ఎద
B) డెందము
C) చిత్తము
D) గుండె
జవాబు:
A) ఎద

6. సంధులు :

54. ‘కషార్జితం’ – పదాన్ని విడదీయుము.
A) కష్ట + ఆర్జితం
B) కష్ట + అర్జితం
C) కష్టా + ఆర్జితం
D) కష్టా + అర్జితం
జవాబు:
A) కష్ట + ఆర్జితం

55. ‘తో దోపు’ పదాన్ని విడదీయుము.
A) తో + తోపు
B) తోపు + తోపు
C) తో + దోపు
D) తోపు + దోపు
జవాబు:
B) తోపు + తోపు

56. ‘కన్ను + ఎత్తి’ – సంధి పేరేమిటి?
A) ఇత్వసంధి
B) అత్వసంధి
C) ఉత్యసంధి
D) గుణసంధి
జవాబు:
C) ఉత్యసంధి

57. కింది వానిలో గసడదవాదేశ సంధికి ఉదాహరణను గుర్తించండి.
A) వస్తున్న
B) దుర్భరావస్థ
C) హితార్థ
D) ప్రొద్దు గ్రుంకు
జవాబు:
D) ప్రొద్దు గ్రుంకు

58. ‘ద్రుత ప్రకృతికము మీది పరుషములకు సరళములగు’ – ఈ సూత్రానికి సంబంధించిన ఉదాహరణను కింది వానిలో గుర్తించండి.
A) తోదోపు
B) కన్నెత్తియుం జూతురే
C) జీవగడ్డ
D) ప్రొద్దువొడిచిన
జవాబు:
B) కన్నెత్తియుం జూతురే

59. ‘దుర్భరావస్థ’ అనే పదాన్ని విడదీయండి.
A) దుర్భ + రావస్థ
B) దుర్భరా + వస్థ
C) దుర్భరము + అవస్థ
D) దుర్భర + అవస్థ
జవాబు:
D) దుర్భర + అవస్థ

60. ‘భోగోపలబ్ది’ – ఈ పదంలో గల సంధి ఏది?
A) ఉత్వ సంధి
B) గుణ సంధి
C) సవర్ణదీర్ఘ సంధి
D) వృద్ధి సంధి
జవాబు:
B) గుణ సంధి

61. ‘ఉత్కటము + దుర్బరావస్థ’ – సంధి జరిగిన పిమ్మట ఏర్పడిన పదం ఏది?
A) ఉత్కటపు దుర్భరావస్థ
B) ఉత్కటంపు అవస్థ
C) ఉత్కట దుర్భరావస్థ
D) ఉత్కటావస్థ
జవాబు:
A) ఉత్కటపు దుర్భరావస్థ

62. “సిరి కాటపట్టు’ – విడదీసి, సంధిని గుర్తించండి.
A) సిరి + కాటపట్టు (ఇత్వ సంధి)
B) సిరిక + ఆటపట్టు (సవర్ణదీర్ఘ సంధి)
C) సిరికిన్ + ఆటపట్టు (ఇత్వ సంధి)
D) సిరికాట + పట్టు (అత్వ సంధి)
జవాబు:
C) సిరికిన్ + ఆటపట్టు (ఇత్వ సంధి)

7. సమాసాలు :

63. భూమి పుత్రుడు’ లోని విగ్రహవాక్య విభక్తిని గుర్తించండి.
A) చేత
B) వలస
C) యొక్క
D) అందు
జవాబు:
C) యొక్క

64. “జీవన సంగ్రామం’ సమాసం పేరేమిటి?
A) రూపకం
B) షష్టి
C) ద్వంద్వం
D) బహువ్రీహి
జవాబు:
A) రూపకం

65. ‘హృదయకళిక’ లోని విభక్తిని గుర్తించండి.
A) మైన
B) అనెడి
C) లో
D) అందు
జవాబు:
B) అనెడి

66. ‘శూరులందు శ్రేషుడు’ – సమాసం పేరేమిటి?
A) షష్టీ
B) తృతీయా
C) బహువ్రీహీ
D) సప్తమీ
జవాబు:
D) సప్తమీ

67. విశేషణ పూర్వపద కర్మధారయ సమాసానికి ఉదాహరణను గుర్తించండి.
A) జీవన స్పర్థ
B ) పవిత్రమూర్తి
C) ధరణీపతి
D) హాలిక వర్య
జవాబు:
B ) పవిత్రమూర్తి

68. ‘జీవన సంగ్రామము’ పదానికి విగ్రహవాక్యం గుర్తించండి.
A) జీవనము చేత సంగ్రామం
B) జీవనం కొఱకు సంగ్రామం
C) జీవనము అనే సంగ్రామం
D) జీవనము, సంగ్రామము
జవాబు:
C) జీవనము అనే సంగ్రామం

69. ‘హృదయ కళిక‘ వికసించినది – గీత గీసిన పదం ఏ సమాసం?
A) రూపక సమాసం
B) ద్విగు సమాసం
C) ద్వంద్వము
D) ఉపమాన ఉత్తరపద కర్మధారయం
జవాబు:
A) రూపక సమాసం

70. ‘చిటికెలతో పందిళ్ళు’ – సమాస పదంగా కూర్చండి.
A) చిటికెల పందిళ్ళు
B) చిటికె పందిళ్ళు
C) పందిరి చిటికెలు
D) చిటిక పందిళ్ళు
జవాబు:
A) చిటికెల పందిళ్ళు

8. గణాలు:

71. ‘హితార’ గురులఘువులు గుర్తించండి.
A) III
B) IIU
C) IUI
D) UII
జవాబు:
C) IUI

72. ‘గౌరవం’ అనేది ఏ గణం?
A) మ గణం
B) ర గణం
C) న గణం
D) భ గణం
జవాబు:
B) ర గణం

73. ‘శ్రమలు’ గురులఘువులు గుర్తించండి.
A) III
B) UII
C) IUI
D) IIU
జవాబు:
A) III

74. ‘న, జ, భ, జ, జ, జి, ర’ గణాలు ఏ వృత్తానికి చెందినవి?
A) ఉత్పలమాల
B) మత్తేభం
C) శార్దూలం
D) చంపకమాల
జవాబు:
D) చంపకమాల

75. ‘1 సూర్యగణం, 2 ఇంద్రగణాలు, 2 సూర్యగణాలు’ – ఇవి ఏ పద్యానికి చెందిన గణాలు (S.A. II – 2017-18)
A) ఆటవెలది
B) తేటగీతి
C) కందం
D) సీసం
జవాబు:
B) తేటగీతి

76. మత్తేభ వృత్తంలోని యతి స్థానం
A) 11
B) 10
C) 14
D) 13
జవాబు:
C) 14

77. ‘భ,ర,న,భ,భ,ర,వ’ గణాలు ఏ వృత్తానికి చెందినవి?
A) తేటగీతి
B) ఆటవెలది
C) కందము
D) ఉత్పలమాల
జవాబు:
D) ఉత్పలమాల

78. తేటగీతి పద్యపాదంలో ఉండే గణాలు ఏవో గుర్తించండి.
A) 3 సూర్య, 2 ఇంద్ర గణాలు
B) 1 సూర్య, 2 ఇంద్ర, 2 సూర్య గణాలు
C) 5 సూర్య గణాలు
D) భరనభభరవ
జవాబు:
B) 1 సూర్య, 2 ఇంద్ర, 2 సూర్య గణాలు

79. ‘సంగ్రామం’ అనేది ఏ గణం?
A) భ గణం
B) ర గణం
C) త గణం
D) మ గణం
జవాబు:
D) మ గణం

9. అలంకారాలు :

80. ‘జీవన సంగ్రామం’ రూపకాలంకారానికి చెందిన ఉదాహరణ – దీనిలో ఉపమానం గుర్తించండి.
A) జీవనం
B) సంగ్రామం
C) రెండూ
D) ఏదీకాదు
జవాబు:
B) సంగ్రామం

81. ‘హృదయ కళిక’ దీనిలోని అలంకారం గుర్తించండి.
A) ఉపమా
B) అతిశయోక్తి
C) రూపకం
D) శ్లేష
జవాబు:
C) రూపకం

82. ‘జింకలు బిత్తరి చూపులు చూస్తూ చెవులు నిగిడ్చి చెంగు చెంగున దూకుతున్నాయి’ – ఈ వాక్యంలో గల అలంకారమును గుర్తించండి.
A) ఛేకానుప్రాస
B) స్వభావోక్తి
C) అతిశయోక్తి
D) శ్లేష
జవాబు:
B) స్వభావోక్తి

83. వాక్యాలకు బింబ ప్రతిబింబత్వం ఉన్నట్లయితే అది ఏ అలంకారం?
A) స్వభావోక్తి
B) దృష్టాంతం
C) ఉపమా
D) అర్థాంతరన్యాస
జవాబు:
B) దృష్టాంతం

10. ఆధునిక వచనాన్ని గుర్తించడం:

84. ‘శ్రమలు లేకయె ఫలములు దుముకబోవు’ – దీనికి ఆధునిక వాక్యం ఏది?
A) శ్రమ పడకుండా ఫలితాలు దుముకవు.
B) శ్రమలు లేకుండా ఫలాలు రావు
C) శ్రమ లేనిదే ఫలితాలు అవే రావు
D) శ్రమే లేకపోతే ఫలాలు ఎక్కడివి
జవాబు:
A) శ్రమ పడకుండా ఫలితాలు దుముకవు.

85. ‘సిరియె భోగోపలబ్ధికి జీవగట్టి’ – ఆధునిక వాక్యం గుర్తించండి.
A) సిరి సుఖాలను పొందడానికి మందు
B) సిరి భోగోపలబ్దికి జీవగట్టు
C) సంపదయే సుఖాలన్నిచ్చే మందు
D) సిరియె సుభాలనిచ్చే జీవనౌషధం
జవాబు:
B) సిరి భోగోపలబ్దికి జీవగట్టు

11. కర్తరి, కర్మణి వాక్యాలను గుర్తించడం :

86. రైతు పంట పండించాడు – కర్మణి వాక్యము గుర్తించండి.
A) రైతు పంట పండించబడింది
B) రైతు చేత పంట పండించాడు
C) రైతుచే పంట పండించబడింది
D) రైతు పంటచేత పండించాడు
జవాబు:
C) రైతుచే పంట పండించబడింది

87. ‘నీవు చెప్పిన విషయం పరిశీలించబడుతుంది’ – ఈ కర్మణి వాక్యానికి కర్తరి వాక్యం గుర్తించండి.
A) నీవు చెప్పిన విషయాన్ని పరిశీలిస్తారు.
B) నీవు చెప్పిన విషయం పరిశీలిస్తాము.
C) నీవు చెప్పినది పరిశీలించరు.
D) నీ చేత చెప్పిన విషయం పరిశీలిస్తారు.
జవాబు:
A) నీవు చెప్పిన విషయాన్ని పరిశీలిస్తారు.

12. ప్రత్యక్ష, పరోక్ష కథనాలను గుర్తించడం:

88. “నీకు సుఖం ఉందా” అని రైతును కవి అడిగాడు – పరోక్ష కథనం గుర్తించండి.
A) సుఖం ఉందాని రైతును కవి అడిగాడు
B) సుఖంగా ఉన్నావాయని రైతును కవి అడిగాడు
C) సుఖం ఉందాయని కవితో రైతు అడిగాడు
D) రైతుతో సుఖం ఉందాని అన్నాడు కవి.
జవాబు:
A) సుఖం ఉందాని రైతును కవి అడిగాడు

89. వాని చేతిలోని నాగలి గొప్పదని దువ్వూరి అన్నారు – ప్రత్యక్ష కథనం గుర్తించండి.
A) నా చేతిలోని నాగలి గొప్పది అని దువ్వూరి అన్నారు.
B) నీ చేతిలోని నాగలి గొప్పది” అని దువ్వూరి అన్నారు.
C) అతని చేతిలోని నాగలి గొప్పది అని దువ్వూరి అన్నారు.
D) నీ చేతిలోని నాగలి గొప్పదే కదా అని దువ్వూరి అన్నారు.
జవాబు:
B) “నీ చేతిలోని నాగలి గొప్పది” అని దువ్వూరి అన్నారు.

90. “చిన్నప్పటి నుండి నాకు బోటనీ అభిమాన విషయం” అన్నాడు రచయిత-దీనిని పరోక్ష వాక్యాన్ని గుర్తించండి.
A) చిన్నప్పటి నుండి నీకు బోటనీ అభిమాన విషయ మని రచయిత అన్నాడు.
B) చిన్నప్పటి నుండి తనకు బోటనీ అభిమాన విషయ మని రచయిత అన్నాడు.
C) చిన్నప్పటి నుండి ఆమెకు బోటనీ అభిమాన విషయ మని రచయిత అన్నాడు.
D) చిన్నప్పటి నుండి ఆయనకు బోటనీ అభిమాన విషయమని రచయిత అన్నాడు.
జవాబు:
B) చిన్నప్పటి నుండి తనకు బోటనీ అభిమాన విషయ మని రచయిత అన్నాడు.

13. వ్యతిరేకార్థక వాక్యాన్ని గుర్తించడం :

91. రైతుకు తిండి లేదు – వ్యతిరేక వాక్యం గుర్తించండి. చెందినదో గుర్తించండి?
A) తిండి ఉంది
B) రైతుకు తిండి ఉంది
C) రైతుకు తిండి పెట్టు
D) రైతుకు ఆకలి లేదు
జవాబు:
B) రైతుకు తిండి ఉంది

92. పళ్ళు తినేవారు చెట్టును చూడరు – వ్యతిరేక వాక్యం గుర్తించండి.
A) పళ్ళు లేనివారు చెట్టును చూస్తారు
B) చూస్తారు
C) పళ్ళు తినేవారు చెట్టును చూస్తారు
D) చూడరు
జవాబు:
C) పళ్ళు తినేవారు చెట్టును చూస్తారు

93. ‘ఒకే ఒక్క ఆవు తిరిగి రాలేదు’ – వ్యతిరేకార్థక వాక్యాన్ని గుర్తించండి.
A) ఒకే ఒక్క ఆవు తిరిగి వచ్చింది.
B) ఆవులన్నీ తిరిగి వచ్చాయి.
C) ఒకే ఒక్క ఆవు తిరిగి రాదు.
D) ఒక్క ఆవు మాత్రం వచ్చింది.
జవాబు:
A) ఒకే ఒక్క ఆవు తిరిగి వచ్చింది.

14. వాక్యరకాలను గుర్తించడం:

94. రైతు మనస్సు స్వచ్ఛమైంది. రైతు మనస్సు అసూయలేనిది – సంయుక్త వాక్యం గుర్తించండి.
A) రైతు మనస్సు స్వచ్చమైంది, అసూయలేనిది.
B) రైతు మనస్సు స్వచ్ఛమైంది, అనసూయలేనిది
C) స్వచ్చమైంది మనస్సు, అసూయలేనిది రైతు
D) స్వచ్ఛమైంది, అసూయ ఉంది రైతు మనస్సు
జవాబు:
A) రైతు మనస్సు స్వచ్చమైంది, అసూయలేనిది.

95. బుద్దుడు వటవృక్ష చ్ఛాయకు వచ్చాడు. అష్టాంగ ధర్మ ప్రవచనం ప్రారంభమైంది – ఈ వాక్యాలలో సంయుక్త వాక్యాన్ని గుర్తించండి.
A) బుద్ధదేవుడు వటవృక్ష చ్ఛా యకు వచ్చి అష్టాంగ ధర్మ ప్రవచనం ప్రారంభం చేశాడు.
B) బుద్ధదేవుడు వటవృక్ష చ్ఛాయకు రాగానే అష్టాంగ ధర్మ ప్రవచనం ప్రారంభమైంది.
C) బుద్ధదేవుడు వటవృక్ష చ్ఛాయకు వచ్చాడు వెంటనే అష్టాంగ ధర్మ ప్రవచనం ప్రారంభమైంది.
D) బుద్ధదేవుడు వచ్చిన వెంటనే అష్టాంగ ధర్మ ప్రవచనం ప్రారంభం అయ్యింది.
జవాబు:
C) బుద్ధదేవుడు వటవృక్ష చ్ఛాయకు వచ్చాడు వెంటనే అష్టాంగ ధర్మ ప్రవచనం ప్రారంభమైంది.

15. ప్రక్రియలను గుర్తించడం :

96. ‘శ్రమ చేయకుండా ఫలితాలు రావు’ – ఇది ఏ ప్రక్రియకు –
A) చేదర్థకం
B) ప్రశ్నార్థకం
C) శత్రర్థకం
D) క్యార్ధకం
జవాబు:
A) చేదర్థకం

97. ‘రైతు ఉదయం నుండి సాయంత్రం వరకు కష్టపడతాడు” -ఏ ప్రక్రియ?
A) ఆశ్చర్యార్థకం
B) సామర్థ్యార్థకం
C) నిషేధార్థకం
D) హేత్వర్ణకం
జవాబు:
B) సామర్థ్యార్థకం

98. పండ్లు ఇచ్చిన చెట్టు గూర్చి ఎప్పుడైనా ఆలోచిస్తారా? -ఏ ప్రక్రియ?
A) సందేహార్థక
B) విధ్యర్థకం
C) ప్రశ్నార్ధకం
D) అనుమత్యర్థకం
జవాబు:
C) ప్రశ్నార్ధకం

99. నీ గొప్పతనాన్ని నీవు తెలుసుకో – ఏ ప్రక్రియ?
A) ప్రార్ధనార్థకం
B) ఆశీర్వాద్యర్థకం
C) సామర్థ్యార్థకం
D) ప్రేరణార్థకం
జవాబు:
D) ప్రేరణార్థకం

100. ఈ కింది వాక్యంలోని అసమాపక క్రియ ఏదో గుర్తించండి. ‘కాపలా కాస్తూ హాయిగా తిని కూర్చో’.
A) చేదర్థకం
B) శత్రర్థకం
C) తద్ధర్మార్థకం
D) ప్రశ్నార్థకం
జవాబు:
B) శత్రర్థకం

AP 9th Class Physical Science Important Questions Chapter 11 Sound

March 30, 2023July 29, 2021 by Srinivas

These AP 9th Class Physical Science Important Questions 11th Lesson Sound will help students prepare well for the exams.

AP State Syllabus 9th Class Physical Science 11th Lesson Important Questions and Answers Sound

9th Class Physical Science 11th Lesson Sound 1 Mark Important Questions and Answers

Question 1.
Mention two important devices used in SONAR system?
Answer:
Sonar system consists of a transmitter and a detector.

Question 2.
A sound of wavelength 0.6 cm is travelling in air with a velocity of 300 m/s. Is this sound audible?
Answer:
We know that v = nλ

This sound is not in the audible range. Hence this is not audible.

Question 3.
How do we get the sensation of sound?
Answer:
Sound travels in air due to the to and fro motion of the air particles, which act upon the ear and produce the sensation of sound.

Question 4.
On which factor does the pitch of a tuning fork depend?
Answer:
The pitch of the tuning fork depends on the length of the prongs.

Question 5.
How does the sound travel?
Answer:
Sound travels^in the form of waves.

Question 6.
What are longitudinal waves?
Answer:
If the particles of the medium vibrate along the direction of wave, the wave is called a longitudinal wave.

Question 7.
Which wave involves change in the density of the medium?
Answer:
Longitudinal wave involves change in the density of the medium.

Question 8.
What do you call the regions formed in longitudinal wave propagation?
Answer:
The regions formed during the longitudinal wave propagation due to change in density are compressions and rarefactions.

Question 9.
When do we call a wave as transverse wave?
Answer:
If the particles of the medium vibrate perpendicular to the direction of wave, then the wave is called a transverse wave.

Question 10.
What is an echo?
Answer:
A reflection of sound, arriving at the listener in more than 0.1 sec after direct sound is called an echo.

Question 11.
What is reverberation?
Answer:
A reflection of sound, arriving at the listener, in less than 0.1 sec after direct sound is called reverberation.

Question 12.
What is the audible range of an ordinary person?
Answer:
20 Hz to 20 K.Hz.

Question 13.
Give examples to the animals which produce infrasonics.
Answer:
Animals such as elephants and whales produce infrasonics.

Question 14.
Expand SONAR.
Answer:
SONAR stands for Sonographic Navigation and Ranging.

Question 15.
Define a compression and a rarefaction.
Answer:

Compressions are the regions where density, as well as pressure of particles, is high.
Rarefactions are the regions where the density, as well as pressure of particles, is low.

Question 16.
Define time period of the sound wave.
Answer:
The time taken to complete one oscillation of the density of the medium is called the time period of the sound wave, denoted by T, and unit in S.I. system is second.

Question 17.
Define speed of a sound wave.
Answer:
The distance by which a point on the wave, such as a compression or rarefaction, travels in unit time is called speed of sound wave.

Question 18.
In which of the three media air, water, or iron sound travel the faster at a particular temperature?
Answer:
Sound travels faster in iron because speed of sound is maximum in solids then liquids and least in gases.

Question 19.
What is audiable range of the average human ear?
Answer:
The audiable range of the average human ear is 20 Hz to 20 KHz.

Question 20.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen why?
Answer:
Speed of sound is less (i.e., 342 m/s) is less than speed of light (i.e., 3 x 10⁸ m/s). So, thunder is heard a few seconds after the flash is seen.

Question 21.
The frequency of source of sound is 100 Hz. How many times does it vibrates in 1 minute?
Answer:
Frequency (υ) = 100 Hz = 100 vibrations/ second.
Number of vibrations in minute = 100 × 60 = 6000 vibrations/ minute.

Question 22.
Which fundamental particles were discovered by
i) Thomson
ii) Goldstein
iii) Chadwick.
Answer:
i) Thomson – Electron
ii) Goldstein – Proton
iii) Chadwick – Neutron

Question 23.
What is a nucleus?
Answer:
The small positively charged central part of an atom is called nucleus.

Question 24.
What is an orbit?
Orbit is the path of the electron around the nucleus.

Question 25.
What will the addition of a neutron to the nucleus of an atom do?
Answer:
It will increase the atomic mass of the atom.

Question 26.
Which constituent particles of the atom determine the following
a) size of the atom
b) change on the nucleus?
Answer:
a) Electrons.
b) Protons.

Question 27.
Magnesium atoms has 12 electrons. Which energy shell is incomplete?
Answer:
M – shell is incomplete.

Question 28.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
K – 2 ; L – 8.
2 + 8 = 10 is the total number of electrons if K and L – shells are full.

Question 29.
An atom of an element has 7 electrons in its L shell
a) What is its atomic number?
b) State its valency.
Answer:
a) Atomic number = 9 [∴ K : 2 ; L : 7]
b) Valency is ‘I’.

Question 30.
The atomic number and mass number of an element are 11 and 23 respectively. Find the number of neutrons in its nucleus.
Answer:
Number of neutrons (N) = Mass number (A) – Atomic number (Z) = 23 – 11 = 12.

Question 31.
The mass number of chlorine atom is 35 and its atomic number is 17. How will this chlorine atom be represented?
Answer:
\({ }_{17}^{35} \mathrm{Cl}\)

Question 32.
What is an anion?
Answer:
When an atom gains one or more electrons, it becomes negatively charged and is known as anion.

Question 33.
Give one example each of diatomie and triatomic molecules.
Answer:

Diatomic molecule – Oxygen (O₂)
Triatomic molecule – Ozone (O₃)

Question 34.
If Z = 3, what would be the valency of the element?
Answer:

If Z = 3, the distribution of electrons is (2,1).
Thus the valency of the Element is 1.

Question 35.
Fluorine atom has 9 electrons and 9 protons. How many energy shells it has?
Answer:
Two energy shells. [∴ K = 2 ; L = V]

Question 36.
Which subatomic particle is not present in an ordinary hydrogen atom?
Answer:
Neutron.

Question 37.
What is the maximum number of electrons which can be accommodated in the
a) innermost shell of an atom?
b) outermost shell of an atom?
Answer:
a) 2
b) 8

Question 38.
What is the usual symbol for a) an electron b) a proton and c) a neutron?
Answer:
a) electron = e^–
b) proton = p⁺
c) neutron = n°

Question 39.
Name the shell of an atom which can accommodate a niaximum of a) 8 electrons b) 18 electrons.
Answer:
a) L-shell

Question 40.
Name the negatively charged particle present in the atones of all the elements.
Answer:
electron (e^–).

Question 41.
Which part of an atom was discovered by Ruthorford’s alpha particle scattering experiment?
Answer:
Nucleus.

Question 42.
State the relative mass and charge of a proton.
Answer:

Relative mass = 14
Relative charge = + 1

Question 43.
Name the radioactive isotope which is used in the treatment of cancer.
Answer:
Cobalt – 60.

Question 44.
Which radioactive isotope is used to determine the activity of thyroid gland?
Answer:
Iodine – 131 radioisotope

Question 45.
What name is given to the pair of atoms such as \({ }_{7}^{14} \mathrm{N}\) and \({ }_{7}^{15} \mathrm{N}\)?
Answer:
Isotopes.

Question 46.
Which noble gas has less than 8 electrons in the valence shell of its atom?
Answer:
Helium.

Question 47.
What is the general name of the elements having 8 electrons in the valence of shell of their atoms?
Answer:
Noble gases.

9th Class Physical Science 11th Lesson Sound 2 Marks Important Questions and Answers

Question 1.
How can you say that the sound waves are longitudinal?
Answer:

When sound wave passes through air, the layers in the medium are alternately pushed and pulled.
Thus the particles of the medium move to and fro along the direction of propagation.
Therefore sound waves in air are longitudinal.

Question 2.
What are the characteristics of a sound wave?
Answer:
The characteristics of a sound wave, which play an important role in describing the nature of a wave are

Wavelength (λ)
Amplitude (A)
Frequency (υ)
Wave speed (v)

Question 3.
Deduce a relationship between time period and frequency of a sound wave.
Answer:

Let the time taken for o oscillations = 1 sec.
The time taken for one oscillation = \(\frac{1}{υ}\) sec.
But the time taken for one oscillation is called the time period (T) and the number of oscillations per second is called the frequency (υ).
Hence frequency and time period are related as T = \(\frac{1}{υ}\) (or) υ = \(\frac{1}{T}\)

Question 4.
What is the speed of sound wave in air, water and iron at 20 °C?
Answer:
The speed of sound in air at 20°C is 343.2 m/s.
The speed of sound in water at 20°C is 1484 m/s.
The speed of sound in iron at 20°C is 5120 m/s.

Question 5.
Define echo and reverberation.
Answer:
Echo :
A reflected sound arriving at the position of listener in more than 0.1 s after the direct sound is called an echo’.

Reverberation :
A reflection of sound, arriving at the listener in less than 0.1 s after the direct sound is called reverberation’.

Question 6.
Explain the working of a megaphone and a horn.
(Or)
How multiple reflections of sound are used in working of mega phone and a horn?
Answer:

In megaphone and a horn, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.

Question 7.
Describe the working of a stethoscope.
Answer:

Stethoscope is a medical instrument used for listening to sounds produced with in the body, chiefly in heart or lungs.
In stethoscopes the sound of the patient’s heartbeat reaches the doctor’s ears by multiple reflection and amplifying the sound.

Question 8.
How the concert hails and cinema halls are designed to use multiple reflections of sound?
Answer:

Generally the ceilings of concert halls, conference halls and cinema halls are designed such that the sound after reflection reaches all corners of the hall as shown in the figure.

In some halls, a curved ceiling is arranged in such a way that the sound after reflecting from the ceiling spreads evenly across the hall.

Question 9.
What are the medical applications of ultrasound?
Answer:
1) Imaging of organs :
Ultrasonics are useful in Electro Cardio Graphy (ECG) to form an image of heart.

Ultrasonography is used in the formation of images of organs such as liver, gall bladder, uterus, etc. to identify the abnormalities and tumors. Ultrasonography is also used to monitor the growth of a foetus inside the mother’s womb.

2) Surgical use of ultrasound :
Ultrasounds are used in cataract removal, breaking up of stones in kidneys, etc. without surgery.

Question 10.
Show the frequencies of various musical notes in a table.
Answer:
In musical terms, the pitch of the note determines the position of the note on the musical scale which is denoted as :

The tuning fork set is prepared based on the above frequencies.

Question 11.
Draw a diagram to show wavelength and amplitude of a wave.
Answer:

Question 12.
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer:

Question 13.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer:
Speed of sound (v) = 342 m/s, Time taken to heard echo, t = 3 s.
Distance travelled by sound = v × t = 342 × 3 = 1026 m.

Question 14.
What are the range of frequencies associated with (a) infrasonic sound, b) ultrasonic sound?
Answer:
a) For infrasonic sound frequency is less than 20 Hz.
b) For ultrasonic sound frequency is greater than 20 KHz.

Question 15.
Which characteristic of sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
The pitch of voice differs for person to person which inturn depends on frequency. So, the frequency is the characteristic of sound which helps to identify my friends voice.

Question 16.
A person has hearing range from 20 Hz to 20 KHz. What are the typical wavelength of sound waves in air corresponds to these two frequencies? Take speed of sound in air as 344 m/s.
Answer:

Question 17.
A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm. What is the frequency of the wave? Will it be audiable?
Answer:

It is more than 20000 Hz (20 KHz). So the sound is not audiable.

Question 18.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound if the distance of an object from the submarine is 3625 m.
Answer:
Distance of an object from the submarine (d) = 3625 m.
The time taken to receive signal (t) = 5s.

Question 19.
Fill in the following blanks in respect of an atom of an element.

Answer:
Mass number = 23
Atomic number =11
Number of electrons =11
Valency = 1

Question 20.
Write the electronic configurations of the following elements.
a) Carbon
b) Neon
c) Chlorine
d) Calcium
Answer:
a) 2, 4
b) 2, 8
c) 2, 8, 7
d) 2, 8, 8, 2

Question 21.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
1. Mass number = No. of protons + No. of neutrons
⇒ 4 = 2 + No. of neutrons
⇒ No. of neutrons = 4 – 2 = 2
2. Thus, the helium atom has 2 neutrons.

Question 22.
Fill in the following blanks.

Answer:
1. Protons : 10
2. Neutrons : 12
3. Electrons : 10
4. Symbol : Ne

Question 23.
Write the three isotopes of carbon and hydrogen.
Answer:

Question 24.
Three different atoms of oxygen are represented as \({ }_{8}^{16} O,{ }_{8}^{17} O, \text { and }{ }_{8}^{18} O\).
a) What do the lower figures and upper figures represent ?
b) Give the nuclear composition of \({ }_{8}^{18} \mathrm{O}\).
Answer:
a) Lower figures → Atomic number
Upper figures → Mass number

b) Composition of \({ }_{8}^{18} \mathrm{O}\) → Protons : 8
Neutrons : 18 – 8 = 10

Question 25.
An atom of an element has 6 electrons in L shell.
a) What is the atomic number of the element?
b) State its valency.
c) Identify the element and write its name. [ K = 2 ; L = 6 Given]
Answer:
a) Atomic number = 2 + 6 = 8
b) Valency = 2
c) Oxygen atom.

Question 26.
State any two similar properties of isotopes.
Answer:

Isotopes of an element have same atomic number.
Isotopes of an element have similar number of electrons.

Question 27.
If chlorine atom is available in the form of Say, two isotopes \({ }_{17}^{35} \mathrm{Cl}\) (75%) and \({ }_{17}^{35} \mathrm{Cl}\) (25%), calculate the average atomic mass of chlorine atom.
Answer:

Question 28.
Write the name of any two radioactive isotopes.
Answer:
1. Uranium – 235
2. Cobalt – 60

9th Class Physical Science 11th Lesson Sound 4 Marks Important Questions and Answers

Question 1.
“Sound travels in the form of waves”. Justify your argument.
Answer:

Sound is a form of energy which travels through the air and reaches our ears to give the sensation of sound.
There may be two possible ways by which transfer of energy from the source of sound to our ears take place.
a) Source of sound produces disturbances in air and they strike our ears.
b) Some particles are shot off from the source of sound and they reach our ears.
If the second explanation is correct, the vibrating body would gradually lose its weight as particles are continuously shot off from it.
This is impossible because it would lead to vanishing of the object.
Hence the first explanation that the sound travels through disturbances in the form of waves is correct.

Question 2.
Mention the industrial applications of ultrasonic waves.
Answer:
1) Drilling holes and making cuts of desired shapes :
a) Holes can also be drilled using ultrasonic vibrations.
b) Ultrasonic cutting and drilling are very effective for fragile materials like glass, etc.

2) Ultrasonic cleaning :
a) Ultrasonics help in cleaning the parts located in hard-to-reach places.
b) The high frequency ultrasonic vibrations knocks off all dirt and grease particles from the objects.

3) Ultrasonic detection of defects in metals :
The defects in the metallic structures, which are not visible from the outside, can be detected by ultrasonic waves.

Question 3.
Draw the graphical representation of
a) Lower pitch, higher pitch
b) Louder sound, soft sound
Answer:

Question 4.
Draw the graphical representation of sound wave produced by a tuning fork, a violin and a piano playing the same note with equal loudness.
Answer:

Question 5.
Composition of the nuclei of two atomic species X and Y are given as under.

a) Give the mass number of X and Y.
b) What is the relation between the two species ?
Answer:
a) Mass number of X = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14

b) Since X and Y both have atomic numbers as 6 but mass numbers are different.
∴ These are isotopes to each other.

Question 6.
How will you find the valency of Magnesium, Sulphur, and Chlorine?
Answer:
Electronic configuration

Valency of Magnesium = 2
Valency of Sulphur = 8 – 6 = 2
Valency of Chlorine = 8 – 7 = 1

Question 7.
For the symbol H, D and T tabulate three sub-atomic particles found in each of them.
Answer:

Question 8.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:

Question 9.
Observe the following table and answer the following questions.

	X	Y
Electrons	8	8
Protons	8	8
Neutrons	8	9

a) What is the mass number of X?
b) What is the mass number of Y?
c) What is the relation between X and Y?
d) Which element do they represent?
Answer:
a) Mass number of X = 16 (∵ number protons + number neutrons)
b) Mass number of Y = 17
c) Isotopes
d) Oxygen (¹⁶₈O)

Question 10.
Draw the diagrams of the three isotopes of hydrogen and label the parts.
Answer:

Question 11.
Explain why, Sodium ion, Na⁺ has completely filled K and L shells.
Answer:

A sodium ion, Na⁺ has 10 electrons in it.
Now, the maximum capacity of K shell is 2 electrons and that of L shell is 8 electrons.
Taken together, the maximum capacity of K and L shells is 2 + 8 = 10 electrons.
A sodium ion Na⁺ has completely filled K and L shells because its 10 electrons can completely fill up K and L shells.

AP 8th Class Social Important Questions Chapter 10 Landlords and Tenants under the British and the Nizam

March 29, 2023July 29, 2021 by Mahesh

These AP 8th Class Social Studies Important Questions 10th Lesson Landlords and Tenants under the British and the Nizam will help students prepare well for the exams.

AP State Syllabus 8th Class Social 10th Lesson Important Questions and Answers Landlords and Tenants under the British and the Nizam

Question 1.
Did the Zamindar own all the lands of the villages in the Mughal times?
Answer:
No. They owned some of the village lands. They had the power to collect revenue from the villagers. Rest of the land was owned by peasants and others.

Question 2.
What did the Zamindars do for the Mughal government and what did they get in return?
Answer:
The Zamindars collected the revenue from the peasantry on behalf of Mughal emperors. In return they received a share of revenue collected. Sometimes they also had right to collect some small local taxes.

Question 3.
How would the changes have affected the position of the farmers who tilled the land for several generations?
Answer:

This settlement inadvertently converted all the peasants into the tenants.
Since rent collected was much higher than the revenue, the cultivators could not pay and sometimes deserted the lands. These were the affects.

Question 4.
What is the difference between revenue and rent?
Answer:
Revenue: Tax on land which is paid to the Zamindars or government.
Rent: Amount paid to the owners of the land for using and cultivating it.

Question 5.
In what ways did the government invest in agriculture during the early British rule? Do you think it could have been done by farmers themselves?
Answer:

The Britishers made necessary advances to the ryots to purchase seeds, implements, bullocks and to repair old wells or dig new ones,
They constructed anicuts & canals. It could not have been done by farmers themselves.

Question 6.
How was a ‘dora’ different from an ordinary landlord?
Answer:
The large landlords were called Doras. They had a large retinue of servants and soldiers. Ordinary landlords were under the control of these ‘doras’. Landlords should follow the orders of ‘Doras’. This is the difference between them.

Question 7.
How do you think it was possible for the zamindars to keep increasing their incomes without increasing investment in the land?
Answer:
The prices of grains in the market rose and cultivation slowly expanded. So there was an increase in the income of Zamindars without increasing investment in the land.

Question 8.
Who do you think would have benefited from the Ryotwari Settlement – the farmers, the landlords or the British? Give reasons.
Answer:
The landlords would have benefited from the Ryotwari settlement.
Reasons:

There was an increase in the number of landlords in the Ryotwari area.
They showed much interest in giving their lands on rents rather than cultivation.

Question 9.
Why do you think the ryots did not invest in improving agriculture or extending agriculture?
Answer:
The prices rose faster than the land revenue and the ryots found it more profitable to employ tenants to cultivate their lands and receive rent from them. So they did not invest in improving agriculture or extending agriculture.

Question 10.
Who ultimately profited from production for export market and why?
Answer:
Moneylenders profited ultimately from production for export market. They lent more and more money to farmers on high-interest rates and profited.

Question 11.
Discuss the changes in the lives of traditional crafts persons and village artists.
Answer:
Traditional crafts persons are adding technology to their tradition. As the prices are high, there is no demand for their products. Village artists had enjoyed the encouragement of the Kings and Zamindars. Now they lack it. So their lives are in worse condition.

Question 12.
How are we paying the taxes nowadays?
Answer:
At present revenue is paid in the form of money.

Question 13.
Have you heard of any similar rapid rise or decline in prices of any agricultural produce in our own times? Find out about its impact on the farmers.
Answer:
In recent past, there was an increase in the prices of rice and red gram. Later they were reduced by the government. Peasants grew some crops thinking of high profits. If the demand fell, they would get losses that year.

Question 14.
Why do you think the Zamindars kept the soldiers and small forts?
Answer:
Zamindars had some villages under their control. They were the mediators between the rulers and the ruled. They had much income also. So they lived in small forts.
To protect their people from thefts and attacks, to collect revenues they needed soldiers. So they maintained them.

Question 15.
Why did the high revenue rates prevent zamindars and farmers from improving agriculture?
Answer:
As they paid heavy rates of rents and revenue from their income, they were not able to improve agriculture. They used the remaining amounts for their household purposes.

Question 16.
During the freedom movement the Zamindars normally supported the British. Can you explain why?
Answer:
The policies of Zamindars created a gap between them and their people. They earned more assets in the British reign. So they normally supported the British during the freedom movement.

Question 17.
Who introduced the Permanent Settlement of Tax and when?
Answer:
Cornwallis introduced this in 1793.

Question 18.
Why did the war in America lead to increase in prices of cotton in India?
Answer:
In 1861, there was Civil war in America and British factories turned to India for supply of cotton. As a result cotton prices soared high.

Question 19.
Why were famines caused under British rule? Do you think it was because of failure of rains or floods?
Answer:
The rains and floods caused one or two famines during the British rule. Most of the famines occurred due to their dictatorial rule. They exported food grains in the famines. They had no interference when the merchants created artificial scarcity by hoarding food grains. They didn’t work for the welfare of the people in India. So the famines were caused.

Question 20.
Write the names of any two peasant movements.
Answer:
The Deccan riots of 1860s, Rampa fituries, Moppila agitations etc.

Question 21.
Match the following.
1) 1865/66 ( ) A) Dhawaleshwaram anicut
2) 1860 ( ) B) Anicut over the Krishna river at Vijayawada
3) 1793 ( ) C) Ganjam famine
4) f1849 ( ) D) Permanent settlement
5) 1854 ( ) E) The Deccan riots
Answer:
1) C
2) E
3) D
4) A
5) B

Question 22.
Imagine that you live in the ‘Ceded Districts’, find put about the Palegars who fought with the British.
Answer:
Many wars were broken out between Palegars and the British from 1798 to 1805. First war was fought between Veerapandya and the British in 1799. Second main war was fought between a group of Palegars and the British. After a long and expensive campaign the British finally defeated the revolting Palegars of whom many were beheaded and hanged while others were deported to the Andaman Islands. Of the Palegars who submitted to the British some of them granted Zamindari status, which had only tax collection rights and disarmed them completely.

AP 10th Class Biology Important Questions Chapter 8 Heredity

March 29, 2023July 29, 2021 by Mahesh

These AP 10th Class Biology Important Questions and Answers 8th Lesson Heredity will help students prepare well for the exams.

AP State Syllabus 10th Class Biology 8th Lesson Important Questions and Answers Heredity

10th Class Biology 8th Lesson Heredity 1 Mark Important Questions and Answers

Question 1.
How did you get the characters from your parents and grandparents?
Answer:
By Genes

Question 2.
Why man is called a moving museum of vestigial organs?
Answer:

The organs which are not useful in animals are called ‘vestigial organs’. There are nearly 180 vestigial organs in human beings,
Hence, human being is said to be a moving museum of vestigial organs.

Question 3.
Observe the given Flow-chart and answer the question.
Who decides the sex of the baby – mother or father? How?
Answer:
Father decides the sex of the baby.
Mother has XX chromosomes. Father has ‘XY’ chromosomes. Y chromosome is determining factor. So father is responsible.

Question 4.
What examples you will give to prove that Lamarckism is not correct?
Answer:
August Weismann, tested the theory proposed by Lamarck by experiments on rats. He removed tails of parental rats. He observed for twenty two generations but still off springs are normal with tails.

Question 5.
Which chromosomes determine the sex in human beings?
Answer:
Allosomes or Sex chromosomes. They are xx (girls) and xy (boys).

Question 6.
Why do we call appendix as a vestigial organ?
Answer:

Vestigial organ is the organ of our body which is smaller and simpler than those in related species they have lost their original function.
Appendix is highly developed in ruminants which helps in the digestion of cellulose. But, in human beings the cellulose is eliminated as undigested food. Hence in human beings appendix has no role in cellulose digestion. So, we call it as a vestigial organ.

Question 7.
What are variations?
Answer:
Differences in characters within very closely related groups of organisms are referred to as variations.

Question 8.
Who is known as father of genetics?
Answer:
Gregor Johann Mendel is known as father of genetics.

Question 9.
Why Mendel has chosen garden pea plant as material for his experiments?
Answer:
Pea plant has following advantages.

Well defined characters
Bisexual flowers
Predominently self fertilization
Early hybridization
Annual plant.

Question 10.
What are the vitamins present in pea plant?
Answer:
The vitamins present in pea plant are ‘A, C, E, K and B’.

Question 11.
What is F₃ generation?
Answer:
F₃ generation represents the offsprings produced from the individuals of F₂ generation.

Question 12.
What is a factor?
Answer:
The determining agent responsible for each trait is called a factor.

Question 13.
WTiat is law of dominance?
Answer:
According to Mendel, among a pair of alleles for a character, only one expresses itself in the first generation as one of the allele is dominant over the other. This is known as law of dominance.

Question 14.
What is phenotype ratio?
Answer:
The characters which can be seen is known as phenotype and their ratio is called phenotype ratio.

Question 15.
What is the phenotype ratio in F₁ generation of monohybrid cross?
Answer:
The phenotype ratio in F₁ generation of monohybrid cross is 3 : 1.

Question 16.
What is genotype ratio?
Answer:
The genetic makeup of an individual with reference to a specific character under consideration is called genotype and their ratio is called genotype ratio.

Question 17.
What is the genotype ratio in F₂ generation of monohybrid cross?
Answer:
The genotype ratio in F₂ generation of monohybrid cross is 1 : 2 : 1.

Question 18.
What are genes?
Answer:
Genes are the factors which are responsible for characters or traits of an organism. These are the units of heredity that are transferred from a parent to offspring. These are small segments of DNA on a chromosome.

Question 19.
What is an allele?
Answer:
The pair of genes which are responsible for character is called allele.

Question 20.
What are homozygous alleles?
Answer:
If an organism has two copies of the same allele for example TT or tt it is homozygous for that trait.

Question 21.
What are heterozygous allele?
Answer:
If an organism has one copy of two different alleles for example Tt, then it is heterozygous.

Question 22.
What is law of independent assortment?
Answer:
In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assorted independently of the other pair. This is known as “Law of independent assortment”.

Question 23.
Wliat is the law of segregation?
Answer:
The law of segregation states that every individual possesses a pair of alleles for any particular trait and that each parent passes a randomly selected copy of only one of these to its offspring.

Question 24.
What are heritable traits?
Answer:
Traits that may be passed on from one generation to the next are called as heritable traits.

Question 25.
What is heredity?
Answer:
The process of acquiring characters or traits from parents is called heredity.

Question 26.
What is inheritance?
Answer:
The process in which traits are passed from one generation to another generation is called inheritance.

Question 27.
What is genetic drift?
Answer:
Change in the frequency of genes in small populations is called genetic drift.

Question 28.
Who was the first person to propose the theory of evolution?
Answer:
Jean Baptist Lamarck was the first person to propose the theory of evolution.

Question 29.
What are acquired characters?
Answer:
The characters developed during the lifetime of an organism are called acquired characters.

Question 30.
What is inheritance of acquired characters?
Answer:
Lamarck proposed that the acquired characters are passed to its offsprings i.e., to next generation. This is known as inheritance of acquired characters.

Question 31.
Who wrote the book “principles of geology” of evolution?
Answer:
The book “principles of geology” of evolution was written by Charles Lyell.

Question 32.
What is micro evolution?
Answer:
Small changes within the species is known as micro evolution.

Question 33.
What is macro evolution or speciation?
Answer:
The process of evolution through which new species are going to be formed is known as speciation or macro evolution.

Question 34.
The double helix structure of DNA was discovered by?
Answer:
James Watson and Francis Crick discovered the double helix structure of DNA.

Question 35.
What are autosomes?
Answer:
Chromosomes whose number and morophology do not differ between males and females of a species are called autosomes.

Question 36.
What are allosomes?
Answer:
The chromosomes that determine sex of the organism are called allosomes.

Question 37.
What is meant by survival of the fittest?
Answer:
According to Darwin’s theory of natural selection, nature only selects or decides which organism should survive or perish in nature. This is the meaning of survival of the fittest.

Question 38.
Write the expanded form of DNA.
Answer:
The expanded form of DNA is Deoxyribo Nucleic Acid.

Question 39.
What is the basis of evolution?
Answer:
Selection of variants by environmental factors forms the basis of evolution.

Question 40.
What are analogous organs?
Answer:
The organs which have similar appearance and functions but have different structure and origin. Ex : Wings of a butterfly, bat and a bird.

Question 41.
What is embryology?
Answer:
Embryology is the study of the development of an organism from egg to adult stage.

Question 42.
What is Palaeontology?
Answer:
The study of fossil is called Palaeontology.

Question 43.
How palaeontologists determine the age of fossil?
Answer:
Palaeontologists determine the age of fossil by using carbon dating method.

Question 44.
Where do scientists collected fossil of dinosaurs?
Answer:
Scientists collected fossils of dinosaurs from Yamanapalli in Adilabad district in Telangana State.

Question 45.
What are connecting links?
A. The organisms which bear the characters of two different groups are called connecting links.

Question 46.
Which organism is recognised as the connecting link between aves and reptiles?
Answer:
Archeopteryx is recognised as the connecting link between aves and reptiles.

Question 47.
What is human evolution?
Answer:
Human evolution is the evolutionary process leading upto the appearance of modern human beings.

Question 48.
How do variations occur?
Answer:
Variations develop during reproduction in organisms. Sexual reproduction and errors in DNA copying lead to variations in offspring in a population.

Question 49.
What is divergent evolution?
Answer:
The evolutionary process through which homologous organs develop is called divergent evolution.

Question 50.
What is convergent evolution?
Answer:
The evolutionary process through which analogous organs develop is called convergent evolution.

Question 51.
What are vestigial organs?
Answer:
Organs which are not useful in animal are called vestigial organs.

Question 52.
Who proposed the theory of inheritance of acquired characters?
Answer:
Jean Baptist Lamarck proposed the theory of inheritance of acquired characters.

Question 53.
Who proved that the bodily changes which may occur due to environment won’t be passed to its offsprings?
Answer:
Augustus Weisemann proved that the bodily changes which may occur due to environment won’t be passed to its offsprings.

Question 54.
How the study of fossil is considered significant?
Answer:
The study of fossil is considered significant because

Fossils provide direct evidence of past life and
Fossils provide convincing proof of organic evolution.

Question 55.
What are the nitrogen bases present in DNA?
Answer:
The nitrogen bases present in DNA are adenine, guanine, thymine and cytosine.

Question 56.
How do embryological studies provide evidence for evolution?
Answer:
The similarities in embryonic development reinforce the idea of evolution from common ancestors. The sequence of embryonic development in different vertebrates shows similarities.

Question 57.
Why Mendel selected garden pea plant for his experiments? Give a reason.
Answer:
Mendel selected garden pea plant for his experiments as these can be self pollinated.

Question 58.
What is the evolutionary significance of the fossil archaeopteryx?
Answer:
Archaeopteryx serves as a connecting link between birds and reptiles. It is the fossil evidence to show that birds have evolved from reptiles.

Question 59.
How does the creation of variation in a species ensure survival?
Answer:
The creation of variations in a species enable them to adapt according to the changes and the new needs thus they will enable the survival of the species.

Question 60.
Define evolution.
Answer:
Evolution is the sequence of gradual changes which takes place in the primitive organisms over millions of years in which new species are produced.

10th Class Biology 8th Lesson Heredity 2 Marks Important Questions and Answers

Question 1.
Write Phenotypic and Genotypic ratio of table given at side.
(OR)
Write the Phenotypic and Genotypic ratio when heterogygous (Yy) pea plant is hybridised with the same kind of plant.
(OR)
Write the phenotypic and genotypic ratios of Mendel’s cross-pollination experiments in pea plants with heterozygous yellow seeds (Yy) with that of the same type, i.e., Yy.
Answer:
Phenotypic ratio -3:1
Genotypic ratio -1:2:1

Question 2.
Define and explain Variations with examples.
Answer:
Variations: Differences in characters within very closely related groups of organisms are referred to as variations.
(OR)
Differences among living beings are called variations.
Ex:

Earlobes in some humans are free and in others attached.
Colour of eyes (cornea) in some people are blue and in others black.
Colour of skin is black or white.

Question 3.
“Human being is said to be a moving museum of vestigial organs”. How can you support this statement?
Answer:

During the course of evolution, some organs remain in organisms. For example, appendix in the digestive system.
In human beings it has no role to play in the process of digestion.
But in herbivores like rabbit appendix play important role.
Such type of organs which are not useful in animal are called vestigial organs.
There are nearly 18 vestigial organs in human beings.
For example pinna, hair on skin, mammary glands in man, etc.
That’s why human being is said to be a moving museum of vestigial organs.

Question 4.
Fill the given table and write the genotypic ratio basing on table.

Answer:

The genotypic ratio is 1: 2: 1

Question 5.
Who decides the sex of the baby, mother or father? Explain with a flow chart.
(OR)
Draw a flow-chart showing the sex determination in human beings.
Answer:
Father decides the sex of the baby.

Question 6.
Define the terms phenotype and genotype.
Answer:
Phenotype: The observable properties of an organism that are produced by the interaction of the genotype and the environment. These characters can be seen.
Genotype: The genotype is the genetic make-up an individual usually with reference to a specific characteristic consideration.

Question 7.
What questions you will ask a palaeontologist about fossils?
Answer:

What are fossils?
How do they preserve?
What can be the actual remains?
How do they form?
What do we call the study of fossil?
Can you tell some examples of fossils?
How do the palaeontologists determine the age of fossils?
What are dinosaurs and ketosaurs? In which years they belong?
Where did they collect the fossil of dinosaurs? What is the length of this fossil?
Where did they preserve the fossil of dinosaurs?

Question 8.
How does the embryological evidences support that Evolution has taken place?
Answer:
Evidences:

Remarkable similarities in the Embryos of different animals from fish to man.
Tadpole of frog resembles the fish more than the frog.
Life history of every individual exhibits the structural features of their ancestors.
The resemblance is so close at an early stage, it is difficulty to distinguish one embryo from other.

Question 9.
Observe the checker board and answer the following questions.
i) Write phenotypic ratio of monohybrid cross.
ii) How many heterozygous plants are present in the checker board?
Answer:
i) Phenotypic ratio 3 : 1
ii) Two heterozygous plants – (Yy, yY)

Question 10.
What happens if there is no evolution?
Answer:

Evolution is a continuous and comprehensive process.
If it does not take place, there is no formation of new species.
Variations do not take place and hence desirable traits are not developed.
All the Earth would be with the primitive species without any changes.

Question 11.
If you cross a plant with pure yellow seeds (YY) with a plant with pure green seeds (yy), what would be the colour of the seeds in F₂ generation? Show in a checker board.
Answer:
All pea plants are yellow in F₁ generation on self pollination in F₁ generation.
In F₂ generation, we can observe that 75% are yellow seed producing pea plants and 25% are green ones.
Hence the phenotypic ratio is 3 : 1 the genotypic ratio is 1 : 2 : 1

Question 12.
If you meet a historian to clarify your doubt on ‘Man has first born in African conti¬nent’, what type of questions will you ask him / her?
Answer:

What is human evolution?
When were early man like forms appeared on land?
When did the fossil of the homosapiens appeared on earth?
Where did the early man lived?
Where can we trace the earliest members of the human race homosapiens?
When do some of our ancestors left Africa?
How the residents of Africa migrated to other places?
Why did the residents of Africa migrated to other places?
Are all humans evolved from single ancestor?

Question 13.
What is the difference between Phenotype and Genotype?
Answer:

Phenotype	Genotype
1. The characters which can be seen is known as phenotype.	1. The genetic make up of an individual is known as genotype.
2. We can’t determine the internal factors by phenotype.	2. Genotype itself is the indication of internal factors.
3. It tells about only the dominating characters which express externally.	3. It tells about both dominant and recessive characters present within.
4. The phenotype ratio in monohybrid cross is 3 : 1.	4. The genotype ratio in mono-hybrid cross is 1 :2 : 1.

Question 14.
What are the differences between homozygous and heterozygous ?
Answer:

Homozygous	Heterozygous
1. It refers to having two identical alleles for a single trait.	1. It refers to having two different alleles for a single trait.
2. For example, for long pea plant homozygous combination is ‘TT’	2. For example, for long pea plant heterozygous combination is ‘Tt’
3. On self pollination homozygous plants produce homozygous plants only.	3. On self pollination heterozygous plants produce homozygous dominant, heterozygous dominant and homozygous recessive plants in 1: 2 :1 ratio respectively.

Question 15.
How do traits get expressed according to Mendel?
Answer:

Mendel hypothesised that each character or trait is expressed due to a pair of factors or alleles.
Now these alleles are known as genes.
Gene is made up of a segment of DNA which provide information of protein.
Protein is needed for biochemical process.
If the proteins work efficiently, the traits get expressed in better way.
Thus genes control the traits or characters.

Question 16.
What is speciation? How it occurs?
Answer:

Origin of new species from existing one is called speciation.
It may occur due to a) mutations and b) natural selection.

Question 17.
What is sex chromosome? Name the two types of sex chromosomes. Mention the chromosomes present in male and female?
Answer:

The chromosome which determine the sex of a person is called sex chromosome.
The two types of sex chromosomes are X chromosome and Y chromosome.
Sex chromosomes in male is XY.
Sex chromosomes in female is XX.

Question 18.
Write a short notes on the law of “inheritance of acquired characters”.
Answer:

Law of inheritance of acquired characters was proposed by Jean Baptist Lamarck.
He thought that the characters acquired by an organism in its life time are passed to its offsprings.
He thought that at same point of time in the history, the size of giraffe was equal to that of deer.
Due to shortage of food material on the ground and lower branches of trees giraffes started stretching their necks.
Because of continuous usage of neck, after several generations giraffes obtained longer necks.
Such characters that are developed during the lifetime of an organism are called acquired characters.

Question 19.
Write a short notes on the theory of “Natural selection”.
Answer:

The theory of natural selection was proposed by Charles Darwin.
This theory states that nature only selects or decides which organism should survive or perish in nature.
The organisms with useful traits will survive.
The organisms having harmful traits are going to be perished or eliminated from its environment.

Question 20.
Write briefly about “Survival of the fittest”.
Answer:

Variations which are useful are retained, while those which are harmful are removed.
In a population where there is a struggle, the “fittest” will be survived.
Nature favours only useful variations.
Each species tends to produce large number of offspring.
They compete with each other for food, space, mating and other species.
In this struggle for existence, only the fittest can survive. This is called survival of the fittest.
Over long period of time this leads to formation of new species.

Question 21.
How are new species evolved?
Answer:

Sexual reproduction and errors in DNA copying leads to variations in offspring in a population.
Organisms contain variations that help to adapt to its environment going to be survived more efficiently.
But in the same population, the organisms which contain the trait which may not help to adapt in its environment may be perished or eliminated slowly.
These small changes within the population due to variations is called micro evolution.
When organisms of the same species with variations are separated by some cause for long years, lot of variations may take place in these species.
These accumulated variations make them unable to mate and produce new offsprings.
Thus new species form and this is known as speciation or macro evolution.

Question 22.
Write a brief note on homologous organs.
Answer:

Homologous organs are the organs which perform different functions but have similar structure and origin.
For example forelimb of a whale (swimmer), bat (flyer), horse (runner), mole (digger) and man (grasping).
If we carefully observe the anatomy of all these animals, they have a common pattern in the arrangement of bones.
Even though their external form and functions are different, they are similar internally.
Thus it indicates that all the vertebrates have evolved from a common ancestor.

Question 23.
What are fossils? Write a short note on their formation.
Answer:

Fossils are evidence of ancient life forms or ancient habitats which have been preserved by natural processes.
Fossil evidence is typically preserved within sediments deposited beneath water and land.
They can be actual remains of once lived such as bones or seeds or even traces of past event such as dinosaurs foot print or ripple marks on a pre-historic shore.
Usually when organisms die, their bodies will be decomposed and lost.
But sometimes the body or at least some parts may be in an environment that does not let it decompose completely.
For example if a dead insect get caught in mud, it will not decompose quickly and the mud will eventually harden and retain the impression of the body parts of insect.
All such preserved traces of living organisms are called fossils.

Question 24.
How would you appreciate Jean Baptist Lamarck for his contribution to the biology?
Answer:

Jean Baptist Lamarck was the first person to propose the theory of evolution.
He proposed that the acquired characters are passed to its offspring i.e., to next generation.
This is known as inheritance of acquired characters.
For example elongation of neck and forelimbs in giraffe.
Even though this theory was disproved, his contribution to biology was appreciable because it changed the belief of the people of olden days that the organisms on the earth have not undergone any change.

Question 25.
How did August Weisemann disprove the theory of “Inheritance of acquired characters” proposed by Lamarck? (OR)
What example will you give to prove that Lamarckism is not correct?
Answer:

August Weisemann, tested the theory of “Inheritance of acquired characters” proposed by Lamarck by an experiment on rats.
He removed tails of parental rats.
He observed its offsprings which have normal tails.
He has done it again for twenty two generations but still offsprings are normal with tails.
He proved that the bodily changes are not inherited. So they won’t be passed to its offspring.
Thus he disproved the theory of “Inheritance of acquired characters”.

Question 26.
Some organisms or species adapt better and survive in a community of organisms. Why do you think this may happen?
Answer:

This may happen due to the variations posessed by the organisms which are suitable to that habitat.
The variations that help the organism to collect food to escape from their enemies, increase the chance of survival for the organism than the other organisms.
In general, variations come during sexual reproduction or mutation.
If the variations are useful, that organisms can adapt better and survive.
These organisms can be selected by the nature.

Question 27.
What do you understand about pure breeds?
Answer:

Pure breed is that expresses the selected character over several generations.
A pure breed will have both the factors of the same type.
It means all the pure breeds are homozygous.
All the gamates produced by them will have same type of factor.
Pure breed on self pollination will give pure breed again.

Question 28.
What do you understand about F₁ generation?
Answer:

F₁ generation or first filial is the offspring of first generation parents.
Cross pollination of pure breeds will give F₁ generation.
All the individuals produced in F₁ generation are heterozygous.
Only the dominant characters are expressed in this generation.

Question 29.
What are the differences between F₁ generation and F₂ generation of mono hybrid cross?
Answer:

F₁ generation

F₂ generation

1) F₁ generation or first filial is the offspring of first generation parents.

2) Cross pollination of pure breeds will give F₁ generation.

3) All the individuals produced in F₁ generation are heterozygous.

4) Only the dominant characters are expressed in this generation.

5) All the individuals produced in F₁ generation are same, both phenotypically or genotypically.

1) F₂ generation is the offspring of second generation parents.

2) Self or cross-pollination of F₁ generation will give F₂ generation.

3) Individuals produced in F₂ generation may be homozygous or heterozygous.

4) Homozygous recessive plants express recessive characters.

5) In F₂ generation individuals, the phenotype ratio is 3:1 and the genotype ratio is 1 : 2 : 1.

Question 30.
What are the differences between monohybrid cross and dihybrid cross ?
Answer:

Monohybrid cross

Dihybrid cross

1) In monohybrid cross only one pair of contrasting characters are taken into consideration.

2) In monohybrid cross, the phenotype ratio of F₂ generation individuals is 3:1.

3) In monohybrid cross, the genotype ratio of F₂ generation individuals is 1:2:1.

1) In dihybrid cross two pairs of contrasting characters are taken into consideration.

2) In dihybrid cross, the phenotype ratio of F₂ generation individual is 1:2: 1.

3) In dihybrid cross, the genotype ratio of F₂ generation individuals is 9 : 3 : 3 : 1.

Question 31.
Write a short note on fossils of dinosaurs, ketosaurs collected in Telangana state.
Answer:

A rare and magnificient fossil of the dinosaurs, ketosaurs were collected in Yamanapalli in Adilabad district of Telangana district.
They belong to the lower jurassic age going back to about 160 million years.
This fossil has 14 meters length and 5 meters height.
This fossil is preserved in BM Birla Science Centre. Hyderabad.

Question 32.
Write a short notes on vestigial organs.
Answer:

During the course of evolution some organs remain in organisms, even though they don’t have any work to do.
For example appendix in the digestive system of human beings has no role to play in the process of digestion.
But in herbivores like rabbit appendix plays important role.
Such type of organs which are not useful in animal are called “vestigial organs”.
There are nearly 180 vestigial organs in human beings such as pinna, hair on skin, mammary glands in human, etc.
That’s why human being is said to be a moving museum of vestigial organs.

10th Class Biology 8th Lesson Heredity 4 Marks Important Questions and Answers

Question 1.
Fossils are the precious evidences preserved by the nature to help us knowing about ancient life forms. Write the information you collected about fossils.
Answer:

Fossils are the evidences of ancient life forms or ancient habitates which have been preserved by natural processes.
Fossils provide information about what lived in the past.
Palentologists determine the age of fossils by using carbon-dating method.
They convey us about genetic condition, heredity characters through inactive chromosomes which are present in them.
They give a detailed information about their diet, life styles, shape of body, etc.
Fossils provide the information about how species have changed across long periods of the earth history.

Question 2.
a) If a sperm with ‘X’ chromosome fertilizes with an ovum with ‘X’ chromosome, what will be the gender of the baby?
b) Who determines the sex/gender of the baby, mother or father?
c) Is it correct to blame the mother for giving birth to a baby girl?
d) Are all our characters resembles our parents?
Answer:
a) Female
b) Father
c) Not correct
d) No, some variations occur.

Question 3.
Observe the diagram and answer the following questions. Vamsi and Priya are newly married couple. They want to give birth to a male child.

a) Draw a probable diagram showing transfer of chromosomes from parents to give birth to male child.
Answer:
b) Who determines the sex of the baby? How can you say ?
Answer:
Father (Vamsi) determines the sex of the baby. Because the chromosome ‘Y’ that determines male sex is present in males.

Question 4.

i) What does the given flow chart indicate?
Answer:
Sex determination in human beings

ii) What will happen if the sperm containing ‘X’ chromosomes fertilises the ovum?
Answer:
Baby will be a girl

iii) Who decides the sex of the baby – Mother or Father?
Answer:
Father

iv) How many pairs of chromosomes are present in off-spring?
Answer:
23 pairs

Question 5.
Write a brief note on Homologous and Analogous organs.
Answer:
Homologous organs: Organs which are structurally similar but functionally different are known as “Homologous organs”.
Forelimbs of a whale – swimming
Wings of a bat – flying
Forelegs of cheetah – running

Analogous: Organs which are structurally different but functionally similar are known as “Analogous organs”.
Eg : Wings of a bird – flying
Wings of a bat – flying

Question 6.
Observe the flow – chart and answer the following.
i) What does the flow – chart represent?
Answer:
The flow – chart represents a monohybrid hybridisation between a pure breed Tall (T) and a pure breed dwarf (t) plants resulting first filial generation. On self pollinating with F₁ generation the new breed have any combinations of T, t came in F₂ generation.

ii) What is the phenotype characters in F₁ generation?
Answer:
In the phenotypic characters in F₁ generation all are dominant that is (T) Tall,

iii) What is the Genotype, Phenotype ratio of F₂ generation?
Answer:
Genotype ratio in F₂ generation is 1 : 2 : 1
Phenotype ratio in F₂ generation is 3 : 1

iv) What laws of inheritance did you understand by this flow – chart?
Answer:
understand that i) the law of dominance ii) law of segregation proposed by Mendel.

Question 7.
Write the Darwin’s theory of evolution in a nutshell.
Answer:

Any group of population of an organism has variations and all members of group are not identical.
Variations maybe passed from parent to offspring through heredity.
The natural selection over abundance of offspring leads to a constant struggle for their survival in any population.
Individuals with variations that help them to survive and reproduce tend to live longer and have more offsprings than organisms with less useful features.
The offsprings of survivors inherit the useful variations, and the same process happens with every new generation until the variation becomes a common feature.
As the environment changes, the organism within the environment adapt and changes to the new living conditions.
Over a long period of time, each species of organism can accumulate so many changes that it becomes a new species, similar to but distinctly different from the original species. All species on the earth arise in this way.
Evolution is a slow and continuous process that involves several thousands of generations.

Question 8.
What are Mendel’s laws of inheritance? What are the reasons to choose pea plant for his experiment?
Answer:
Mendel’s Laws of inheritance:

Law of Dominance : Among a pair of closely related ‘alleles’ or factors, only one expresses itself. In the first generation as one of the allele is dominant over the other. This is called as Mendel’s Law of dominance.
Law of Segregation : The law of segregation states that every individual possesses a pair of alleles for any particular trait that each parent posses a randomly selected copy only one of these to its off-spring.
Law of Independent assortment : In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assort independently of the other pairs. This is known as ‘Law of Independent assortment’.

Mendel has chosen garden pea as material for his experiment because:

It has well developed characters.
It is a bisexual flower.
Predominently self pollinating.
Suitable for cross pollination.
It is an annual plant.

Question 9.
What is Phenotype and Genotype? Explain them with the help of Mendel’s Monohybrid cross.
Answer:
Phenotype: Expression of visible character of an individual is called phenotype.
Genotype: Genetic constitution of an individual for any character is called Genotype or Probable nature of factors is known as genotype.
Cross Pollinating a pure breed of yellow coloured pea seeds (YY) and green coloured pea seeds (yy) give F₁ generation. All pea seeds were yellow in F₁ generation. So, yellow colour is phenotype. ‘Yy’ is genotype of all pea seeds in F₁ generation.

F₂ Generation : Self pollination of F₁ pea plants (Yy)

Phenotype Ratio is 3:1 Genotype Ratio is 1:2:1.

Question 10.
Explain in brief any two evidences of Evolution.
Answer:
Some of the evidences of evolution are

Homologous and analogous organs
Evidences from embryology and
Evidences from fossils.

I. Homologous and analogous organs :

Organs which have common fundamental anatomical plan and similar embryonic origin, whatever varied functions they may perform are regarded as homologous organs.
For example forelimb of a whale, wing of bat, leg of leopord, claw of mole and hand of man.
They indicate that all the vertebrates are evolved from common ancestor.
Organs which are structurally different but functionally similar are known as “Analogous organs”.
For example wings of bats and wings of birds.
The designs of the two wings, their structure and components are different but they look similar because they have a common use for flying but their origin is not common.

II. Evidences from embryology:

There are remarkable similarities in the embryos of different animals from fish to man.
The resemblance is so close that at an early stage even an experienced embryologist would find difficulty to distinguish one embryo from the other.
This strengthens the view of the existence of a common ancestor from which all these have evolved.

III. Evidences from fossils:

Fossils are evidences of ancient life forms or ancient habitats which have been preserved by natural processes.
Palaeontologists determine the age of fossils by using carbon dating method.
These fossils provide evidences of presence of extinct animals like dinosaurs and how the evolution occurred on the earth, etc.

Question 11.
Observe the given flow-chart and answer the following questions:

i) Name the chromosomes that determine the sex of an individual.
Answer:
Y Chromosome of father.

ii) Show given information in the form of Checker Board.
Answer:

iii) In this situation, which principle of Mendel is applicable?
Answer:
Law of dominance

iv) “Mother determines the sex of the baby”. Is this statement correct or not? Why?
Answer:
The statement is not correct. Because the sperm of father that carries Y chromosome fertilize with ovum of mother that contains X chromosome the resultant will be XY – Male baby.

Question 12.
Keep in mind Mendel’s experiments and write what you know about the following concepts?
a) Pure breed b) Phenotype c) Genotype d) Alleles
Answer:
a) Pure breed: These are the plants that expresses a selected character over several generations. Such plants according to Mendel were pure breed for that character.

b) Phenotype: The characters which can be seen is known as phenotype. We cannot determine the internal factors by phenotype. It tells about only the dominating char-acters which express externally. The phenotype ratio in monohybrid cross is 3 : 1.

c) Genotype: The genetic make up of an individual is known as genotype. Genotype itself is the indication of internal factors. It tells about both dominant and recessive characters present within. The genotype ratio in monohybrid cross is 1 : 2: 1.

d) Alleles: Alleles are corresponding pairs of genes located at specific positions in chromosomes. Together they determine the genotype of their host organism. Every individual possesses a pair of alleles for any particular trait and that each parent passes a rondomly selected copy of only one of these to an offspring. The offspring then receives its own pair of alleles for that trait one each from both parents.

Question 13.
Competition among organisms, variations, natural selection survival of the fittest.
Which theory explains all these aspects? Describe them in a orderly manner.
Answer:

Competition: Every living organism in this world reproduces itself. The rate of multiplication and existence of organisms are more or less common for all living things. But the food supply and space remain unchanged, in other words they are limited. Under these conditions, there is a competition among the organisms to fulfil their needs of food and space. This is known as struggle for existence. It may be interspecific, intraspecific and the environment factors.
Variations: Every organism has its own specific characteristic and relationship with its environment. The variability caused by variations may be large or small. Large variations are known as macro variation and small variations called micro variations. Variations may be favourable or harmful. Those with useful variations survive while the others perish.
Natural selection: The organism with favourable variations are best adapted to the environment in which they live. They have a better chance of survival and perpetuation of race. This principle is called natural selections.
Survival of the fittest: Variations which are useful are retained, while those are harmful are lost. In a population where there is a struggle, the fittest will be survived and the less adjusted will be perished. This is known as the elimination of the unfit.

Question 14.
What is genetic drift? Explain how it provides diversity in the population.
Answer:

Changes in the frequency of genes in small populations, due to accidents is known as “Genetic drift”.’
Let us consider a colour variation occurred in red colour beetles which are living on green coloured leaves of bushes.
It results in “blue” colour beetles instead of ‘red’ colour beetles and passed its colour to its progeny.
Initially in the population, there are few blue beetles, but most are red.
Imagine at this point, an elephant comes by and stamps on the bushes where the beetles live.
This kills most of the beetles but by chance a few beetles survived are mostly blue.
Again the beetle population slowly increases, but in the beetle population most of them are in blue colour.
Thus genetic drift provides diversity in the population.

Question 15.
Have the apparent groups of human beings (races) evolved differently?
Answer:

No, there is no biological basis to the notion of human races. All humans are a single species.
Regardless of where we have lived for the past few thousand years, we all came from Africa.
The earliest members of human species, Homosapiens, can be traced there.
Our genetic footprints can be traced back to our African roots.
A couple of hundred thousand years ago, some of our ancestors left Africa while others stayed on.
While the residents spread across Africa, the migrants slowly spread across the planet.
They did not go in a single line. They went forwards and backwards, with groups, sometimes separating from each other, even moving in and out of Africa.
Like all other species on planet, they had come into being as an accident of evolution, and were trying to live their lives the best they could.

Question 16.
How would you appreciate Gregor Johann Mendel’s contribution to the genetics?
(OR)
Why Gregor Johann Mendel is considered as the father of genetics?
Answer:

Gregor Johann Mendel worked on the problem of how variations were passed from one generation to the other.
As he was a monk, he did his experiments with interest in the garden of the monastery.
He worked for over seven years after which he presented conclusions from his experimental data in a form of a detailed research paper.
Mendel made many careful observations of pea plants and asked himself questions about what he observed and then planned and designed experiments to find the answers.
He had worked on nearly 10,000 pea plants of 34 different varieties choosing 7 distinguishing forms of characters.
His experimental outcomes gave the idea how the variations were passed on from one generation to another.
He was an exemplary person for his observation, planning, patience and experimental skills.
His efforts to know the secrets of nature was really appreciable.

Question 17.
What are the hypothesis assumptions and outcomes of Mendel’s experiments with pea plants?
Answer:
Regarding his experiments with pea plants, Mendel hypothesised that

Characters were carried as traits and an organism always carried a pair of factors for a character.
The distinguishing traits of the same character were present in the population of an organism.
The traits shown by the pea plants must be in the seeds that produced them.
The seeds must have obtained by the traits from the parent plants.

His assumptions made to explain his observations are:
Assumption 1: Every pea plant has two ‘factors’ which are responsible for producing a particular property or trait.
Assumption 2: During reproduction one ‘factor’ from each parent is taken to form a new pair in the progeny.
Assumption 3: One of these will always dominate the other if mixed together.

Laws made from his experiments:

Law of Dominance: Among a pair of alleles for a character, only one expresses itself in the first generation as one of the allele is dominant over the other.
Law of Segregation: Every individual possesses a pair of alleles for any particular trait and that each parent passes a randomly selected copy of only one of these to its offspring.
Law of Independent Assortment: In the inheritance of more than one pair of characters the factors for each pair of characters assorts independently of the other pairs.

AP 9th Class Social Important Questions Chapter 15 Industrialisation and Social Change

March 30, 2023July 29, 2021 by AP Board

These AP 9th Class Social Studies Important Questions 15th Lesson Industrialisation and Social Change will help students prepare well for the exams.

AP State Syllabus 9th Class Social 15th Lesson Important Questions and Answers Industrialisation and Social Change

9th Class Social 15th Lesson Industrialisation and Social Change 1 Mark Important Questions and Answers

Question 1.
In what way did Canal Transportation aid Industrial Revolution? (SA-I : 2019-20)
Answer:
Canals were the answer to moving heavy objects to large distances

9th Class Social 15th Lesson Industrialisation and Social Change 2 Marks Important Questions and Answers

Question 2.
Why did modem industrialisatfon first occured in Britain only? (SA-III : 2016-17)
(OR)
Why do you think Britain was the first country to Industrialize?
Answer:

Britain was blessed with modest climate to suit cotton industry.
There was no scarcity of power.
There was availability of sufficient raw material.
Coal and Iron were available in abundance side by side.
It was convenient to transport the raw material by water.
There was no dearth of capital and Bank of England was also helpful in raising capital.
Workers were available in large numbers.
It was ready to sea trade with a large number of ports.

Question 2.
20th century inventions were hugely influenced by major developments in technology and resources, enabling the inventions of key items and devices which changed the way we live today.
Mention any two inventions one each in transportation and medicine and mention briefly how they changed the way we live today. (SA-II : 2018-19)
Answer:
Transportation :
Aeroplanes, Motor vehicles, Railways, etc.

Medicine:
Anti-biotics, vaccines, chloroforms, Anti-virus, etc. Now-a-days world become a big village by a fastest means of transport. Even long distance also reached in a short time. Due to medicines, the life expectancy of human being is increased.

9th Class Social 15th Lesson Industrialisation and Social Change 4 Marks Important Questions and Answers

Question 1.
The road-builder John Metacalf, who personally surveyed surfaces for roads and planned them, was blind. The canal builder James Brindley was almost illiterate, with such poor spelling that he could never spell the word ‘navigation’, but he had tremendous powers of memory, imagination and concentration.
Read the paragraph given above and comment on that. (SA-III : 2016-17)
Answer:
According to the above para, if we want to invent a new thing we need more determination, interest, curiosity, even luck, than the application of scientific knowledge. Along with the above, memory, imagination, and concentration also need.

Question 2.
“Good transportation facilities leads to rapid industrialization” – Justify the statement. (SA-I : 2018-19)
Answer:

Transport system really helps the industry.
Without transportation, the products made by the industry cannot be sold in distant places.
The industry has a particular transport system to send their products to places and sell them for profit.
Without proper transport system, there is risk of damage of products.
The invention of steam engine reduced the cost of transportation.
Development of aircraft was useful for fastest means of transport.
Good transport system leads to rapid industrialization.

Question 3.
Mobile phone – Internet – Computers – Airways
Choose any TWO of the inventions listed above Explain how they changed people’s lives. (SA-I : 2019-20)
Answer:
Mobile Phone:

Mobile phones have completely changed the way people interact.
Today, the mobile phone has become part and parcel of many people’s lives.
You can call, send text messages, read emails, play games as well as read and edit documents.
Through mobile phone, we can send money quickly to others who are in need.
It create new market places.
It help people find jobs.
It help the farmers to get weather information.

Internet:

The Internet creates new ways for us to communicate, work and share information.
The main advantage of the internet is it ability to connect billions of computers and devices to each other.
By it, we get more information.
Enjoy virtual field trips.
It creates assignment convenience and flexibility.

Computer:

It act as highly reliable scientific equipment.
They receive and store data and carries out logical and mathematical calculations too to provide fast and accurate results.
It can store more data.

Airways :

Through air ways we can reach our destination quickly.
It transports people, goods, the military equipment around the world.
It is the fastest means of transport.
During the time of floods and calamities we can use more airways only to reach the place.
It is supported to go to remote areas.
It creates convenience to passengers.

Question 4.
Explain how the condition of the workers steadily declined in the early twentieth century Europe. (SA-I : 2019-20)
Answer:
Conditions of the worker.
1. The aboudance of labour in the market affected.

2. Seasonality of work :
It meant prolonged periods without work.

3. Low Real wages :
Wages are very low.

4. Poverty and Unemployment:
At the mid of 19th century, about 10 percent of Urban population was extremely poor.

The Unemployment rate was very high.

9th Class Social 15th Lesson Industrialisation and Social Change Important Questions and Answers

Question 1.
What is called ‘First Industrial Revolution?
Answer:
The transformation of industry and the economy in Britain between the 1780s and the 1850s is called the “First Industrial Revolution.”

Question 2.
Who used the term ‘Industrial Revolution’?
Answer:
The term ‘Industrial Revolution’ was used by European scholars – Georges Michelet in France and Friedrich Engels in Germany. It was used for the first time in English by the philosopher and economist Arnold Toynbee (1852-83), to describe the changes that occurred in British industrial development between 1760 and 1820.

Question 3.
What were the two main features of cotton industry?
Answer:
From the 1780s, the cotton industry symbolised British industrialisation in many ways. This industry had two features which were also seen in other industries. Raw cotton had to be entirely imported and a large part of the finished cloth was exported to other countries. Britain had its control over the sources of raw cotton as well as the markets. Where they sold cloth. This helped to increase its control over the colonies.

Question 4.
What was the new power source in England?
Answer:
Steam

Question 5.
Which country experienced modern industrialisation for the first time?
Answer:
Britain experienced modern industrialisation for the first time.

Question 6.
Name any bank in England.
Answer:
The Bank of England.

Question 7.
Who were used in coal mines to reach deep coal faces?
Answer:
Children were used in coal mines to reach deep coal faces.

Question 8.
Which countries lacked colonial resources?
Answer:
Germany and France lacked colonial resources.

Question 9.
What is Industrial revolution? Who coined this name?
Answer:

The economy and industries in Britain changed completely between 1780s and 1850s.
This phase is known as ‘First Industrial Revolution’.
The term ‘Industrial Revolution’ was used by European Scholars George Michelet in France and Fredrich Engels in Germany.
It was used for the first time in English by the Philosopher and economist Arnold Tonybee.

Question 10.
What do you know about the picture?

Answer:
This was the Cast Iron Bridge. It was near Coalbrookdale. It was designed by the third Darby.

Question 11.
What was it? How did it look like?

Answer:
This was James Watt’s steam engine. It was just looking like our present day tractor.

Question 12.
Compare the effects of the coming of the railways in different countries in the world.
Answer:

The first steam locomotive, Stephenson’s Rocket, appeared in 1814. Railways emerged as a new means of transportation that was available throughout the year, both cheap and fast, to carry passengers and goods.
They combined two inventions, the iron track which replaced the wooden track in the 1760s, and haulage along it by steam engine.
The invention of the railways took the entire process of industrialisation to a second stage.
The first railway line connected the cities of Stockton and Darlington in 1825, a distance of 9 miles that was completed in two hours at speeds of up to 24 kph (15 mph).
The next railway line connected Liverpool and Manchester in 1830. Within 20 years, speeds of 30 to 50 miles an hour were usual.

Britain :
The railways changed British Society in numerous ways. It is nonetheless clear that the railways had a sizeable impact in many spheres of economic activity and daily life of the people.

Canada :
In Canada, the national government strongly supported railway construction for political goals.

India :
India provides an example of the British empire pouring its money into a very well built system designed for military reasons and with the hope that it would stimulate industry. The system was over built and much too elaborate and expensive for the small amount of freight traffic it carried.

Belgium :
Belgium took the lead in Industrial revolution on the continent starting in the 1820s. It provided an ideal model for showing the value of the railways for speeding the industrial revolution.

France:
In France, railways became a national medium for the modernisation of backward regions. Although starting the whole system at once was politically expedient, it delayed completion and forced even more reliance and temporary experts brought in from Britain.

Germany:
The take-off stage of economic development came with the railroad revolution in the 1840s. As a means of national defence, it facilitates the concentration, distributional direction of the army. It is a means to the improvement of the culture of the nation.

Russians:
It was Americans who brought the technology of railway construction to Russia in 1842. Russia was in need of big transportation system and geographically suited to railroads.

Thus the railways in different countries helped the industrial revolution in transportation.

Question 13.
Read the following passages and answer the given questions.

The Krupp Family

The Krupp family established what was going to become the world’s largest arms factory. This first factory specialised in field gun manufacture and, by 1887, it supplied arms to forty six different countries.

During World War I the Krupp factories made guns for the German artillery.

The Krupps supported Hitler in the German general election of 1933. As Nazi Germany occupied neighbouring countries, Alfred Krupp seized new land to make more factories. Many of these factories used slave-labour from the Nazi concentration camps.

1) What was the name of the family?
Answer:
The ’Krupp’ family.

2) What was the first factory?
Answer:
Gun manufacturing factory.

3) Which World War was mentioned here?
Answer:
First World War.

4) Whom did the Krupps support?
Answer:
The Krupps supported Hitler in the German general election of 1933.

5) Who were the workers in the factories?
Answer:
The slave – labour from the Nazi concentration camps.

Question 14.
Read the following passages and answer the given questions.
Steam power was first used in mining industries. As the demand for coal and metals expanded, efforts to obtain them from ever-deeper mines intensified. Flooding in mines was a serious problem and steam engines were used to drain the mines. But the technology was still very imperfect to be used on a large scale.

James Watt’s (1736-1819) invention converted the steam engine from being a mere pump into a ‘prime mover’ capable of providing energy to power machines in factories. Backed by the wealthy manufacturer Matthew Boulton (1728-1809), Watt created the Soho Foundry in Birmingham in 1775. From this foundry James Watt’s steam engines were produced in steadily growing numbers. By the end of the eighteenth century, Watt’s steam engine was beginning to replace hydraulic power. In 1840, British steam engines were generating more than 70 per cent of all European power.
1) Which power was first used in mining industries?
Answer:
Steam power

2) What was a serious problem?
Answer:
Flooding in mines was a serious problem.

3) Who was the wealthy manufacturer?
Answer:
Matthew Boulton.

4) In which foundry were the Watt’s engines produced?
Answer:
In Soho Foundry in Birmingham.

5) Appreciate the British steam engines.
Answer:
In 1840, British steam engines were generating more than 70% of all European power.

Question 15.
Read the following passages and answer the given questions.
Transportation
The need to transport raw materials and manufactured products led to the improvement of roads and the digging of canals in England. Me Adam devised the method of making ‘pakka’ or ‘macadamamised’ roads.

Canals were initially built to transport coal to cities. This was because the bulk and v/eight of coal made its transport by road much slower and more expensive than by barges on canals. The demand for coal, as industrial energy and for heating and lighting homes in cities, grew constantly. The making of the first English canal, the Worsley Canal (1761) by James Brindley (1716-72), had no other purpose than to carry coal from the coal deposits at Worsley (near Manchester) to that city. After the canal was completed the price of coal fell by half. Thousands of kilometres of canals were built by 1830 and were used to transport commodities cheaply. They were mostly built by landowners to enhance the value of their properties.
1) Which led to the improvement of roads and the digging of canals in England?
Answer:
The need to transport raw materials and manufactured products led to the improvement of roads and the digging of canals in England.

2) Which were initially built?
Answer:
Canals were initially built to transport.

3) Which grew constantly?
Answer:
The demand for coal, as industrial energy and for heating and lighting homes in cities, grew constantly.

4) What is the name of first English canal?
Answer:
Worsley Canal.

5) What was the interest of land owners in building the first canal?
Answer:
They were interested in enhancing the values of their properties.

6) Who built this canal?
Answer:
James Brindley.

Question 16.
Observe the following picture and answer the given questions.

Woman in gilt-button factory, Birmingham. In the 1850s, two thirds of the workforce in the button trade was women and children. Men received 25 shillings a week, women 7 shillings and children one shilling each, for the same hours of work.
1) Who was she?
Answer:
She was a woman in gilt-button factory.

2) Where was the factory?
Answer:
It was in Birmingham.

3) Which year did the picture belonged to?
Answer:
It belonged to the 1850s.

4) What were the earnings of the workers?
Answer:
Men – 25 shillings a week
Women – 07 shillings a week
Children – 01 shilling a week

5) Is there any difference in the working time?
A. No, it was equal to all.

Question 17.
Read the following passages and answer the questions given below.

The Workers

A survey in 1842 revealed that the average lifespan of workers was lower than that of any other social group in cities: it was 15 years in Birmingham, 17 in Manchester, 21 in Derby. More people died, and died at a younger age, in the new industrial cities, than in the villages they had come from. Half the children failed to survive beyond the age of five. The increase in the population of cities was because of immigrants, rather than by an increase in the number of children born to families who already lived there.

Deaths were primarily caused by epidemics of disease that sprang from the pollution of water, like cholera and typhoid, or of the air, like tuberculosis. More than 31,000 people died from an outbreak of cholera in 1832. Until late in the nineteenth century, municipal authorities were negligent in attending to these dangerous conditions of life. The medical knowledge to understand and cure these diseases was unknown.
1) When was the survey conducted?
Answer:
In 1842.

2) What was the average lifespan?
Answer:
It was 15 years in Birhingham, 17 in Manchester, and 21 in Derby.

3) What was the cause of the increase in population?
Answer:
The increase in the population of cities was because of immigrants, rather than by an increase in the number of children born to families who already lived there.

4) What are the diseases mentioned here?
Answer:
Cholera, Typhoid, and Tuberculosis.

5) Why were the municipal authorities negligent in attending to these diseases?
Answer:
The medical knowledge to understand and cure these diseases was unknown.

Question 18.
Read the lesson and prepare any four questions relating to it.
Answer:

How can we say that England was fortunate?
What was the serious problem in mines?
Whom was the first locomotive connected to?
Why did the Krupp family establish the world’s largest arms factory?

Question 19.
What were the relative advantages of canal and railway transportation?
Answer:

Both were used to transport commodities cheaply.
Both enhanced the values of the properties.
The invention and usage took the entire process of industrialisation to a second stage.
Both reduced the time taken to travel between two places.

Question 20.
Why was Britain the first country to experience modern industrialisation?
Answer:
Reasons:

Political stability
Modest climate
Availability of continuous water power
Transportation facilities
Abundant wealth
Availability of labour
Invention of machinery.

Question 21.
Write the favourable conditions which helped the inventions.
Answer:
It is interesting to read about the individuals who brought about these changes during industralisation. Few of them were trained scientists. Education in basic sciences like physics or chemistry was extremely limited until the late nineteenth century, even after the technological inventions described above. Since these inventions did not require a full knowledge of the laws of physics or chemistry on which they were based, advances in science could be and were made by brilliant, intuitive thinkers and persistent experimenters.

They were helped by the fact that England had certain features which other European countries did not. Dozens of scientific journals and published papers of scientific societies appeared in England between 1760 and 1800. There was a widespread thirst for knowledge even in the smaller towns. This was met by the activities of the Society of Arts (founded in 1754), by travelling lecturers, or in ‘coffee houses’ that multiplied through the eighteenth century.

Question 22.
Appreciate the efforts of German to become an Industrialised state.
Answer:

Germany (erstwhile Prussia) exploited its rich coal fields (Silesia and Rhineland – the Rhur) and iron deposits (Bohemia) to flourish in steel industry.
With the invention of electric dynamo, they lead electric industry in the world.
Unification in 1871 made access to iron from Lorraine fields of iron from France.
Government encouraged industry by providing road/rail transports and markets also.
It also focused on arnament industry.
Large banks provided the capital needed.
By the beginning of 20^th century, Germany had developed a powerful industrial base.

AP 10th Class Social Important Questions Chapter 9 Rampur: A Village Economy

March 29, 2023July 29, 2021 by Mahesh

These AP 10th Class Social Studies Important Questions 9th Lesson Rampur: A Village Economy will help students prepare well for the exams.

AP State Syllabus 10th Class Social 9th Lesson Important Questions and Answers Rampur: A Village Economy

10th Class Social 9th Lesson Rampur: A Village Economy 1 Mark Important Questions and Answers

Question 1.
What is the main factor of production in the agricultural sector?
Answer:
The main factor of production in the agriculture sector is land.

Question 2.
Why were usually men paid more wages than women for the same work?
Answer:

Because ours is a male-dominated society.
Women are paid less wages because of gender bias.

Question 3.
Write any two aspects of Rampur village’s economy which you like.
Answer:

In Rampur no land is left idle.
The non-agricultural sector also developed.
Due to the development of the transport sector, other sectors of the economy also developed.

Question 4.
What is multiple cropping?
Answer:
To grow more than one crop on a piece of land during the same year is called multiple cropping.

Question 5.
What are the main non-farm production activities taking place in your region?
Answer:

Transportation
Dairy farming
Brick making
Carpentry
Basket making
Poultry
Shopkeeping
Fishing, etc.

Question 6.
Who are called as labour force?
Answer:
Labour force: People, in the age group of 15 – 59 are labour force, (or)
People who are employed are called labour force.

Question 7.
How do the small farmers procure the capital needed for farming?
Answer:
The small farmers procure the capital needed for farming by borrowing money from

Large farmers
Moneylenders
The traders who supply various inputs for cultivation.

Question 8.
Create a slogan on the importance of agriculture.
Answer:
Without agriculture – there is no future.
No farmer-No food.

Question 9.
Which is the main production activity in villages across India?
Answer:
Farming is the main production activity in villages across India.

Question 10.
Name any two non-farming activities in Rampur village.
Answer:
Small scale manufacturing, dairy farming, transport services, etc.

Question 11.
Who owns the majority of land in Rampur village?
Answer:
80 upper-caste families own the majority of land in Rampur village.

Question 12.
What educational facilities are available in Rampur village?
Answer:
Rampur has two primary schools and one high school.

Question 13.
What health facilities are available in Rampur village?
Answer:
Rampur has a primary health centre run by the government and one private dispensary where the sick are treated.

Question 14.
What is the main aim of production?
Answer:
The main aim of production is to produce goods and services required by the people.

Question 15.
What is a marketable surplus?
Answer:
The difference between the quantity of output that a farmer produces during a year and the quantity that he keeps with himself for his own and family’s consumption is called a marketable surplus.

Question 16.
Who are small farmers?
Answer:
Farmers who own less than 2 hectares of land are known as small farmers.

Question 17.
Who are the medium farmers?
Answer:
Farmers who own more than 2 hectares and less than 10 hectares of land are called medium farmers.

Question 18.
Who are large farmers?
Answer:
The farmers who own more than 10 hectares of land are known as large farmers.

Question 19.
What is the basic constraint in raising farm production?
Answer:
Land area under cultivation is the basic constraint in raising farm production because it is fixed and scarce.

Question 20.
What is the minimum wage rate for a farm labourer?
Answer:
The minimum wages for a farm labourer set by the government is Rs. 60/- per day.

Question 21.
Which is the most abundant factor of production?
Answer:
Labour is the most abundant factor of production.

Question 22.
What do medium and large farmers do with their earnings from surplus farm produce?
Answer:
A part of the earning from surplus farm produce is saved and kept for buying capital for the next season. Another part may be utilized for lending to small farmers who require a loan.

Question 23.
How did the spread of electricity help the people of Rampur?
Answer:
a) Electricity helped farmers in running tubewells in the fields.
b) It is also used for carrying out various small scale business in the village.

Question 24.
What are the sources of irrigation?
Answer:

Canals
Tubewells
Tanks and
Rains.

Question 25.
What is fixed capital?
Answer:
Tools, machines and buildings are called ‘fixed’ capital because these can be used in production for several years.

Question 26.
What is working capital?
Answer:
Raw materials and money form part of working capital. The capital which is used for production is known as working capital.

Question 27.
Classify the capital investment on various items in production.
Answer:
There are two types of capital investments in production.

Physical or fixed capital
Working capital

Question 28.
Name the items/factors that are needed for production.
Answer:
There are four factors of production. They are:

land 2
labour
capital and
organisation.

10th Class Social 9th Lesson Rampur: A Village Economy 2 Marks Important Questions and Answers

Question 1.
What are the aims and objectives that are directed by the landless rural workers or labourers?
Answer:
The aims and objectives that are directed by the landless rural workers or labourers:

Some more days of work.
Still better wages.
Better education to their children in the local govt, schools.
No social discrimination.
Chances of leadership activities, etc.

Question 2.
Study the information given below and write your observations.

Answer:

87% of the farmers are small farmers are holding only 48% of the cultivated land.
13% are the medium and large farmers.
There is unequal land distribution in India.

Question 3.
Convert the information given below into a pie-chart (rough diagram). Write your observation.

Type of farmers	Percentage of cultivated area
Small farmers	48%
Medium and large farmers	52%

Answer:
Cultivated Area

Observation: While a major part of the land (52%) is under control of medium and large farmers, minor part of land (48%) is under the control of small farmers.

Question 4.
Write a letter to the Tahasildar (MRO), explaining the problems faced by the farmers because of famine.
Answer:

Kalyandurg,
Date : xx xx xxxx.

To,
The Tahsildar,
Mandal Revenue office,
Anantapur

Sub: Problems faced by farmers due to famine – request for write off old loans.

Respected Sir,

I am from Kalyandurg I would like bring to your notice the following one and for favourable help. The rains are very less in the before year. So, crops left much loss. The farmers get into debt. They can’t repay the loans. They need money for purchasing seeds, fertilizers, pesticides and to cultivate the land. It will be highly difficult to get a loan again from bank.

Hence, I request you place it in front of the government, write – off the old loans and give them the new loans. I hope you provide immediate help to the farmers without loss to them to grow the crop this year.

Thanking you sir,

Yours truly,
x x x x x

Question 5.
Mention the factors of production.
(OR)
Explain any one factor of production.
Answer:
Land: An area of ground, especially when used for a particular purpose such as farming or building; A necessary factor for production.
Labour: Workers especially people who do practical work with their hands.
Working capital: The requirement of raw material and money which are used up in the production cycle.
Fixed capital: Physical Capital = Tools, machine and buildings, which are not used up or consumed immediately in the production process.
Knowledge: It is essential to use all the above in a meaningful way to produce some goods or services.

Question 6.
Today, why is there a need for expansion of non-farm activities in rural areas?
Answer:
Nowadays farming is not providing regular employment to rural people and they are not getting sufficient income also. That’s why the expansion of non – farm activities in rural areas is needed. Nowadays people with some amount of capital can set up non-farm activities.

It requires very little land. Banks provide loans to self-employment purpose. People can get market facilities to sell their goods. Neighbouring towns and cities provide more demand to village goods.
Ex: Milk, jaggery, broomsticks etc.

Question 7.
How do the farmers get their capital for agriculture?
Answer:
For capital, farmers face many troubles. They get their capital by borrowing from large farmers and village money lenders. They take loans from traders who supply various inputs for cultivation. They get from banks also but very less and rare.

Question 8.
What can be done so that more non-farm production activities can be started in the villages?
Answer:

At present, non-farm production activities engage less number of people in villages.
Many things can be done so that more non-farm production activities can be started in the villages.
It is important to see that loan available to people, who are interested in setting up non-farm production activities, at low rate of interest.
It is good to have markets where these goods and services produced can be sold.
To connect more and more villages to towns and cities through good roads.
As these require little land and some capital, entrepreneurs should be encouraged.

Question 9.
Describe the work of a farmer with 1 hectare of land.
Answer:

The farmer with 1 hectare of land is considered as a small farmer.
He usually keeps large share of production for household needs.
He does not have much saving for capital.
When he thinks of inputs expenditure, he may borrow money from large farmer or moneylender.
In case of borrowing from large farmer, he not only pays high rate of interest but he is also forced to work as a labourer in the large farmer’s fields.
In case of borrowing from money lender, he has to pay high rates of interests.

Question 10.
Modern farming methods require more inputs which are manufactured in industry. Do you agree?
Answer:
Yes, modern farming methods like the use of fertilizers, pesticides, high yielding variety of seeds, power-driven tubewells of irrigation, latest tools and implements like tractors, harvesters, threshers, etc. require more inputs which are manufactured in industry.

Question 11.
Is it important to increase the area under irrigation? Why?
Answer:

It is very important to increase the area under irrigation because it is not wise to depend on monsoon rains which were uncertain and erratic.
To increase production we have to bring more area under irrigation.
It helps us to improve the farming pattern or adopt modern farming methods in those areas.

Question 12.
Give the characteristics of Rural industries.
Answer:

Rural industries are small scale manufacturing units.
They involve very simple production methods.
The output is very small.
The work is usually carried out by the members of the family.
The work is carried out at home and not in workshops.
The profits earned are also less.

Question 13.
Explain ‘Multiple Cropping’ in your own words.
Answer:

Growing more than one crop on a piece of land during the year is known as multiple cropping.
It is the most common and traditional practice to increase production on a given piece of land.
All farmers in Rampur grow at least two main crops.

Question 14.
Which changes have taken place in the way of farming practised in India?
Answer:

Traditional seeds have been replaced by HYVS.
Natural manures were replaced by chemical fertilizers.
Use of pesticides.
Use of farm machinery.
Use of tubewells for irrigation, instead of Persian wheels.

Question 15.
“Excessive use of chemical Fertiliser and pesticides declines the Fertility of land”.
Give your opinion on this.
Answer:

Experience shows fertility of the land is declining due to overuse, excessive use of chemical fertilisers and pesticides.
The situation of water is equally alarming.
Underground water tables have fallen rapidly across the country.
The use of natural resources also not always been judicious while increasing production.

Question 16.
“Labour in an essential item for production”. Read the statement and interpretate.
Answer:

Labour means people who do the work either highly trained and educated or who can do manual work.
Each worker is providing necessary labour for production.
Skilled or semi-skilled or unskilled labour – all are providing labour for production.
Thus labour is an essential item in production.

Question 17.
The expenditure on tools, machines and buildings is called as physical capital. Ex¬plain why.
Answer:

The expenditure on tools, machines and building is called physical capital.
They are not used up or consumed immediately in the process of production.
They help to produce goods over many years.
They require some repair and maintenance so that they remain useful and can be used year after year.
That is why they are called fixed or physical capital.

Question 18.
Ground-level waters are rigorously declining nationwide. Suggest some alternatives.
Answer:

Underground water levels are deteriorating at a faster rate throughout India.
To conserve water schemes like a watershed, soaking pit, check dams, afforestation, bund construction, etc., should be taken up.
Digging borewells should be allowed only for drinking water but not irrigation.
Farmers should look for alternate crops which use lesser water.

Question 19.
Name some Dairy activities in your area.
Answer:

Many families in my region are engaged in milk production.
Various types of grass, maize, bajra, etc., is used to feed the cattle.
Two tradesmen established milk collection and storing points.
Family labour and especially women take care of the rearing of animals.

Question 20.
Explain the business tactics of a small businessman in your area.
Answer:

Once Mr. Venkateswara Rao organised a cloth business in my region.
At the beginning, he used to wander every household to sell cloths and collects money weekly.
By conducting his business reliably and supplying quality items he became popular.
Later he set up his own business firm and till date it is running well.

10th Class Social 9th Lesson Rampur: A Village Economy 4 Marks Important Questions and Answers

Question 1.
Based on the pie chart and the particulars given below, answer the questions.

Which kind of houses are more in Rampur Village?
Answer:
Huts and thatched houses are more in Rampur village.
In which category does 60% of Rampur village belong?
Answer:
60% of Rampur village belong to the poor class.
State approximately, the number of middle-class people in Rampur village.
Answer:
25% people of Rampur belong to the middle class.
Which category of people live in the terraced houses built with cement and bricks?
Answer:
Rich people live in the terraced houses built with cement and bricks.

Question 2.
Read the table given below and answer the following questions.

Workers	Ploughing	Sowing	Weeding	Transplanting	Harvesting	Winnowing	Threshing	Picking Cotton
Male	214	197	215	–	164	168	152	–
Female	–	152	130	143	126	124	118	136

What are the works done only by male workers?
Answer:
Ploughing.
What are the works done only by female workers?
Answer:
Transplanting, Picking cotton.
In which works, the wages are different for male and female?
Answer:
Sowing, Weeding, Harvesting, Winnowing, Threshing.
What are the reasons for paying less wages to female workers than male workers?
Answer:
The concept existing in the society is that, the women do less work than men.

Question 3.
Read the given paragraph and interpret.

Labour being the most abundant factor of production, it would be ideal if the new ways of farming used much more labour. Unfortunately, such a thing has not happened. The use of labour on farms is limited. The labour, looking for opportunities is thus migrating to neighbouring villages, towns and cities. Some labour has entered the non – farm sector in the village.

Answer:

Labour means people who do the work either highly trained and educated or who can do manual work.
Each worker is providing necessary labour for production.
Skilled or semi-skilled or unskilled labour – all are providing labour for production.
Thus labour is essential item in production.

Conclusion: If government provides loans to landless labour for agriculture purpose, motor pumpsets, engines, etc. more labour we will find in the agricultural sector.

Question 4.
Mention the factors of production and explain any two of them.
Answer:

The aim of production is to produce the goods and services that we want.
There are four requirements for the production of goods and services.

They are:

Land
Labour
Capital – (a) Physical or fixed capital (b) Working capital
Knowledge and enterprise

1. Land: The first requirement is land, and other natural resources such as water, forests, minerals.
It is a free gift of nature and it is neither created nor destroyed.
2. Labour: It is a factor which helps in production. There are three types of labour – skilled, semi-skilled and unskilled. Some production activities require highly educated workers to per¬form the necessary task. Other activities require manual work. Each worker provides the labour necessary for production.

Question 5.
Observe the following table and analyse it.
Data refers to the land cultivated by farmers).

Type of farmers	Size of plots	% of farmers	% of cultivated area
Small farmers	Less than 2 hectares	87%	48%
Medium and Large farmers	More than 2 hectares	13%	52%

Answer:

The given data refers to the land cultivated by farmers.
As per the data, 87% of the total farmers, i.e. small farmers are cultivating only 48% of the total land.
The remaining 13% of the farmers are medium and large farmers. They are cultivating the remaining 52% of the land.
This indicates that there is no equal distribution of land among farmers.
I opine that the distribution of cultivated land is unequal in India.

Question 6.
Read the following paragraph and write your opinion.

The land is the most crucial factor necessary for farm production. Land area under cultivation, however^practically fixed. In Rampur, since 1921 there has been no expansion in land area under cultivation. By then, the nearby forests had been cleared and some of the wastelands in the village Were converted to cultivable land. There is no further scope to increase agriculture production by longing new land under cultivation.

Answer:

The given paragraph says that land is essential for agriculture.
The cultivable land is not expanding.
In some areas a few forest lands are cleared and the waste land is being converted to agriculture land. There is no other way of expansion of land.
My opinion is that the issue discussed in this paragraph is correct. The basic factor of production is land.
The population is increasing day by day.
The food grain production is not increasing or expanding in proportion to the population. It is the time for searching for new ways of increasing the production of food grains.
Green revolution is also to be encouraged.
Rampur village is a model for us.
In conclusion I would like to say that there is much land without irrigation.
Due to some other reasons like power cut, lack of capital, non-availability of ground water the farmers are not able to involve in farm activities.
The government should concentrate on these issues and try to solve them so as to increase the cultivable land and increase the food grains.

Question 7.
Read the following paragraph and write your comments.

Most small farmers have to borrow money to arrange for the working capital. They borrow from large farmers or the village moneylenders or the traders who supply various inputs for cultivation. The rate of interest on such loans is very high. They are put to a great distress to repay the loan.

Answer:

According to the given paragraph, small farmers in villages take loans from the moneylenders.
They don’t have sufficient capital for farming activities.
As they are charged a high rate of interest, they are put to distress.
This is the real situation of small farmers.
I do agree with this. My comments on this paragraph are as follows.
Generally, large farmers have sufficient capital for next crop.
Small farmers depend on others for loans.
The government should concentrate on providing a loan facility to the small farmers without collateral.
Farmers’ cooperative societies are to be encouraged.
Minimum facilities and needs are to be identified and fulfilled.

Question 8.
Read the following paragraph and interpret.

There is often a loss inform activities, especially when crops are damaged because of floods, pests, etc. The other risk is sudden fall in price of agricultural produce. In such situations, farm¬ers find it difficult to recover the working capital they had spent.

Answer:

According to the paragraph given it is clearly understood that when there is a natural hazard like floods or pests the farmer’s life will be in risk.
In those conditions, they hardly survive.
Many of the farmers commit suicide because of this situation.
Farmers normally desire to grow more food grains.
They hope the yield must be more. They spend much on agriculture.
They depend on moneylenders for loans. Sometimes they take loans from banks also. They have to pay all these loans when the yield comes. Due to the pests, floods or drought, they don’t have money to pay the loans or for their survival.
Many a time, they commit suicides.
When they face this type of situation the government should give them a helping hand.
They should be saved from losing their lives.
New schemes should be launched for the sake of farmers who lose because of these natural hazards.

Question 9.
From your field visit find out at what rate of interest do farmers borrow when they need money for inputs. Compare with interest charged by the bank.
Answer:

Farmers need money for inputs like seeds, fertilizers, pesticides and repair of tools, etc.
Small farmers cannot mobilise that money.
They borrow from medium and large farmers or moneylenders.
They charge higher rate of interest, usually 36% per annum.
Whereas banks provide crop loan to farmers.
But banks charge only 8% per annum.
Thus there is great variation in the rate of interest.

Question 10.
Groundwater levels are deplected dangerously day by day. If this is continued, what would be the consequences?
Answer:

The groundwater levels are depleted dangerously day by day.
Excessive use of groundwater for irrigation and industrial purposes has resulted in a decrease of groundwater levels.
It affects the stock of water that would be available for future generations.
We will face the problem of scarcity of water.
Water will not be available for domestic purposes also.
Hence we should not over-extract water from tube wells.
Water should be allowed to sink into the ground.
Under any cost we should conserve groundwater for future generations.

Question 11.
Locate the following in the given map of India.

Capital of Maharashtra
Answer:
Mumbai
The birth place of river Godavari.
Answer:
Triyambak
The birth place of river Krishna.
Answer:
Mahabaleswar
Santal tribal people are in this state.
Answer:
Odisha
Uttar Pradesh
Kerala
Punjab
Jammu & Kashmir
Assam
Tripura

AP 10th Class Maths Important Questions Chapter 10 Mensuration

March 29, 2023July 29, 2021 by Srinivas

These AP 10th Class Maths Chapter Wise Important Questions Chapter 10 Mensuration will help students prepare well for the exams.

AP State Syllabus 10th Class Maths 10th Lesson Important Questions and Answers Mensuration

Question 1.
Find the volume of a sphere of radius 21cm.(Take π = 22/7)
Solution:
Volume of the sphere = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 21 × 21 × 21 = 38,808(cm)³

Question 2.
Find the total surface area of a hemisphere, whose radius is 7 cm.
Solution:
Radius of Hemisphere, r = 7 cm
T.S.A of Hemisphere = 3πr²
= 3 × \(\frac{22}{7}\) × 7 × 7 = 462 (cm)²

Question 3.
Find the volume of right circular cone with radius 3 cm. and height 14 cm.
Solution:
Volume of right circular cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 14 = 132 cm³

Question 4.
Find the curved surface area of cylinder, whose radius is 7 cm. and height is 10 cm.
Solution:
Radius of a cylinder = r = 7 cm
Height of a cylinder = h – 10 cm
CSA of a cylinder = A = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 10
= 44 × 10 = 440 sq.cm

Question 5.
The surface area of a football is 616 cm², then find the radius of that ball. ( π = 22/7).
Solution:
Surface area of a football (sphere) = 4πr².
4πr² = 616
πr² = 154
r² = 154 × \(\frac{7}{22}\)
r² = 7 × 7 = 49
∴ r = 7 cm

Question 6.
The volume of a cylinder is 448π cm³ and height is 7 cm. What is its radius?
Solution:
Volume of the cylinder = πr²h = 448π
Here h = 7cm ,r = r
then πr² × 7 = 448π
r² = \(\frac{448}{7}\) = 64
∴ Radious (r) = 8 cm

Question 7.
Find the surface area of a sphere of radius 14 cm. (Take π = \(\frac{22}{7}\))
Solution:
Radius of given sphere = (r) = 14 cm
Formula for surface area of sphere = 4πr²
∴ Surface area of given sphere = 4 × \(\frac{22}{7}\) × 14 × 14 = 88 × 28 = 2464 cm²

Question 8.
Draw a rough diagram of a solid, showing the combination of a cone and cylinder, whose base radii are same.
Answer:

AB is radius of cylinder as well as cone.

Question 9.
Find the area of the paper sheet required to make a conical joker cap with radius 7 cm and slant height 14 cm.
Solution:
Given, radius = 7 cm
Slant height = 14 cm
Area of paper sheet = πrl
= \(\frac{22}{7}\) × 7 × 14
= 22 × 14 = 308 cm²

Question 10.
If the diameter of a sphere is equal to the side of a cube, find the ratio of their volumes.
Solution:
Let diameter of the sphere = d = side of the cube
∴ Radius of the sphere (r) = \(\frac{\mathrm{d}}{2}\)
volume of the sphere = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\)π(\(\frac{\mathrm{d}}{2}\))³ = \(\frac{4}{3}π\)\(\frac{\mathrm{d}^{3}}{8}\) = \(\frac{\pi}{6}\) d³
And, volume of the cube = d³
∴ The ratio of volume of sphere to volume of cube = \(\frac{\pi}{6}\) d³ : d³ = \(\frac{\pi}{6}\) : 1

Question 11.
Find the volume and total surface area of a hemisphere whose radius is 35 cm?
Solution:
Radius of the hemisphere (r) = 35 cm.
Volume of the hemisphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3} \times \frac{22}{7}\) × 35 × 35 × 35

Total surface area = 3πr²
= 3 × \(\frac{22}{7}\) × 35 × 35 = 11550 cm²

Question 12.
A solid iron rod has a cylindrical shape. Its height is 11 cm and base diameter is 7 cm. Then find the total volume of 50 rods.
Solution:
The height of the cylinderical
Rod (h) = 11 cm,
Raidus (r) = 7/2 cm
Volume = πr²h = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 11
= 423.5 (cm)³
Volume of 50 rods = 50 × 423.5 = 21175 (cm)³

Question 13.
Two cubes each of volume 125 cm³ are joined end to end together. Find the total surface area of the resulting cuboid.
Solution:
Volume of each cube = 125 cm³
∴ a³ = 125 cm³ ⇒ a = 5 cm

By joining these two cubes we get a cuboid, whose
length (l) = 10 cm
breadth (b) = 5 cm
height (h) = 5 cm
then the total surface area
= 2 (lb + bh + lh)
= 2(10(5) + 5(5) + 5(10))
= 2(50 + 25 + 50)
= 2(125) = 250 cm²

Question 14.
The base area of a cone is 616 sq.cm and its height is 48 cm. Find its total surface area.
Solution:
The base area of a cone = πr² = 616 cm²
Given height of cone = 48 cm = h

∴ Total surface area
= C.S.A + base area
= 2200 + 616 = 2816 cm²

Question 15.
Find the volume of largest right circular cone that can be cut out from a cube, whose edge is 7 cm.
Solution:
Volume of the cone (V) = \(\frac{1}{3}\) πr²h
[r = \(\frac{s}{2}=\frac{h}{2}\)]
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 7
∴ V = 89.83 cm³

Question 16.
If half of the vertical angle of a cone of height 3 cm is 60°. Find its volume.
Solution:

Let ‘B’ be the centre of base of the cone and BC is radius.
AB is vertical height (i.e.,)
AB = 3 cm
Given that ∠BAC = 60°
ΔABC is a right angled triangle, then
tan 60° = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ √3 = \(\frac{\mathrm{BC}}{3}\) ⇒ BC = 3√3 cm
Now we have h = 3 cm, r = 3√3 cm
Volume of the cone = \(\frac{1}{3}\) πr²h
\(\frac{1}{3}\) π(3√3)² × 3 = 27π cm³
(or)
= \(\frac{594}{7}\) = 84.86 cm³

Question 17.
Find the volume and total surface area of a hemisphere of diameter 7 cm. Take π = \(\frac{22}{7}\)
Solution:
Diameter of hemisphere = 7 cm
∴ Radius of hemisphere = (r) = \(\frac{d}{2}=\frac{7}{2}\) cm
Then volume of hemisphere = V = \(\frac{2}{3}\)πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{22 \times 49}{12}\) = 89.83
Volume = 89.83 cm³
Total surface area of hemisphere = 3πr²
= 3 × \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77 \times 3}{2}\) = 115.5 cm²

Question 18.
The length, breadth and height of a cuboid are (log 125 + log 8),
(log 1000 – log 10) and log 10 respectively. Find the total surface area of the cuboid.
Solution:
Length of cuboid (l) = (log 125 + log 8)
Breadth of cuboid (b)
= (log 1000 – log 10)
Height of cuboid (h) = log 10
∴ (l) = log (125 × 8)
= log 1000
= log 10³ = 3 log 10 = 3

(b) = log 1000 – log 10
= log \(\frac{1000}{10}\)
= log 100 = log 10² = 2
(h) = log 10 = 1
Then total surface area of cuboid
= 2(lb + bh + lh)
= 2(3 × 2 + 2 × 1 + 1 × 3)
= 2 (6 + 2 + 3) = 2(11)
= 22 square units.

Question 19.
The radius of a sphere is 3.5 cm. Find it’s surface area.
Solution:
Given, r = 3.5 cm
Surface area of sphere = 4πr²
= 4 × \(\frac{22}{7}\) × 3.5 × 3.5
= \(\frac{88 \times 12.25}{7}\) = 154 cm²

Question 20.
Express the volume of a cone in terms of volume of right circular cylinder of the same base and height and explain how you arrived at it.
Solution:
We know that,
Volume of cone = \(\frac{1}{3}\) πr²h
Volume of cylinder = πr²h
Hence,
volume of cone : volume of cylinder
= \(\frac{1}{3}\) πr²h : πr²h
Hence, volume of cone = \(\frac{1}{3}\) × volume of cylinder.

Question 21.
The radius of a conical tent is 5m and its height is 12m. Calculate the length of the canvas used in making the tent if width of canvas is 2cm.
Solution:
Radius of the conical sent (r) = 5 m.
Height of the tent (h) = 12 m.
∴ Slant height of the cone
(l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{5^{2}+12^{2}}\)
= \(\sqrt{25+144}\) = \(\sqrt{169}\) 13 m.
Now, surface area of the tent = πrl
= \(\frac{22}{7}\) × 5 × 13 = \(\frac{1430}{7}\) m²
Area of the canvas used = \(\frac{1430}{7}\) m²
It is given that the width of the canvas = 2m.
Length of the canvas used = \(\frac{\text { Area }}{\text { Width }}\)
= \(\frac{1430}{7} \times \frac{1}{2}\)
= 102.14 m

Question 22.
How many spherical balls can be made out of a solid cube of lead, whose edge measures 66 cm. and each ball being 3 cm in radius ?
Solution:
Side of the cube (s) = 66 cm
Radius of the spherical ball = r = 3 cm
Let the number of spherical balls that can be made = n
n × volume of a spherical ball = Volume of the cube
n × \(\frac{4}{3}\)πr³ = s³
n × \(\frac{4}{3} \times \frac{22}{7}\) × 3 × 3 × 3 = (66)³
n = 66 × 66 × 66 × \(\frac{3}{4} \times \frac{7}{22} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\)
= 2541
Number of spherical balls can be made = 2541

Question 23.
An oil drum is in the shape of cylinder, whose diameter is 2 m and height is 7 m. The painter charges ? ₹ 5 per m² to paint the drum. Find the total charges to be paid to the painter for
10 drums.
Solution:
It is given that diameter of the (oil drum) cylinder = 2 m.
Radius of cylinder = \(\frac{\mathrm{d}}{2}=\frac{2}{2}\) = 1 m.
Total surface area of a cylindrical drum = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 1(1 + 7)
= 2 × \(\frac{22}{7}\) × 8 = \(\frac{352}{7}\)m² = 50.28 m².
So, the total surface area of a drum = 50.28 m²
Painting charge per 1 m² = ₹ 5.
Cost of painting of 10 drums
= 50.28 × 5 × 10
= ₹ 2514

Question 24.
A hemisphere is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the Cube. Determine the total surface area of the remaining block.
Solution:
Let diameter of the hemisphere be ‘l’ = 21 cm
Radius of the hemisphere = \(\frac{l}{2}=\frac{21}{2}\) cm
Also, length of the edge of the cube = l = 21 cm
∴ Total surface area of the remaining solid .

Question 25.
Two metallic spheres of radii 6 cm and 8 cm are melted along with another sphere and made into a big sphere of radius 12 cm. Find the radius of the third sphere.
Solution:
Radii of two metallic spheres
= r₁ = 6 cm
r = 8 cm
Volume of the metallic spheres

⇒ 728 + x³ – 12³
⇒ x³ = 12³ – 728 = 1728 – 728
= 1000 cm
⇒ x = 10 cm.

Question 26.
Two cubes each of volume 216 cm³ are joined end to end together. Find the total surface area of the resulting cuboid.
Solution:
Given, volume of the cube V = a³ = 216 cm³
∴ a³ = 6 × 6 × 6 = 6³
Hence a = 6 cm

When two cubes are added, the length of cuboid = 2a = 2 × 6 = 12 cm,
breadth = a = 6 cm,
height = a = 6 cm is formed.
∴ T.S.A. of the cuboid = 2(lb + bh +lh)
= 2(12×6 + 6×6 + 12×6)
= 2(72 + 36 + 72)
= 2 × 180 = 360 cm²
∴ The surface area of resulting cuboid is 360 cm².

Question 27.
What is area of required cloth to make 10 conical hats having 7 cm ground radius and 24 cm height ?
Solution:

Ground radius of cone (r) = 7 cm
Height of cone (h) = 24 cm
Then its slant height
l = \(\sqrt{r^{2}+h^{2}}\)
= 7² + 24²
= \(\sqrt{49+576}\)
l = \(\sqrt{625}\) = 25 cm
Then C.S.A. of the cone = Area of cloth required
= πrl = \(\frac{22}{7}\) × 7 × 25 = 550 cm²
Then the cloth required for 10 hats = 550 × 10 = 5500 cm²

Question 28.
A circle having 21 cm radius is cut into 3 equal parts to make 3 equal circular cones. Then what will be the radius of such cone ?
Solution:
Radius of the circle = 21 cm
Then the circumference of circle
= 2πr = 2 × \(\frac{22}{7}\) × 21 = 132 cm
\(\frac{1}{3}\) of this circumference of circle = Circumference of the ground of the
cone
= \(\frac{1}{3}\) × 132 = 44 cm
Now let the ground radius of new cone = r
Then its ground circumference = 2πr = 44

So ground radius of cone = 7 cm.

Question 29.
Define “Regular cone”. Deduce formula for slant height of a regular cone.
Solution:

A solid having a circular base and vertex above the centre of circular base is called a regular cone.
A → Vertex
‘O’ centre of ground
OB is the radius of cone (r) and
0A is the height of cone (h)
AB is slant height (l)
Now ΔAOB is a right angled triangle.
Where ∠AOB = 90 and AB is hypotenuse.
Then from Pythagorus theorem
AB² = OA² + OB² = r² + h²
∴ AB = \(\sqrt{r^{2}+h^{2}}\)
∴ Slant hight (l) = \(\sqrt{r^{2}+h^{2}}\)

Question 30.
Draw a cone and label them.
Solution:

In the above cone
OB is the radius of the ground.
AO is the height of cone.
AB is slant height of cone.

Question 31.
Which kind of cones are formed by rotating on their axis of following triangles?
a) Equilateral
b) Right angled
c) Scalene
Solution:
a) Equilateral triangle : Having the altitude of this equilateral as rotat¬ing axis, on rotation we get a regular cone, (right cone)

b) Right angled triangle : Having a leg (a side other than hypotenuse) as rotating axis, on rotation of this right angled triangle we get a right circular cone.

c) Scalene triangle : Having a particular altitude from a vertex as rotating axis, on rotation we get a
cone

Question 32.
A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes. (AS₄) [Hint: Diameter of the sphere is equal to the heights of the cylinder and the cone.] A sphere, a cylinder and a cone have the same radius and same height.
Solution:
Sphere Cylinder Cone
Radius = r r r
Height = r 2r = h 2r = h

Now C.S.A of spehre = 4πr² …………….(1)
C.S.A of cylinder = 2πrh
= 2πr.2r-2r = 4πr² ……………….. (2)
Now for cone, slant height = l
= \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{r^{2}+4 r^{2}}\) ( ∵ h = 2r)
= \(\sqrt{5 r^{2}}\) = √5(r)
Then C.S.A of cone = πrl = πr. √5r
= √5 πr² ……………. (3)
∴ Ratio of C.S.A of sphere, cylinder and cone = 4πr² : 4πr² : √5 πr²
= 4 : 4 : √5

AP 8th Class Biology Important Questions Chapter 4 Reproduction in Animals

March 29, 2023July 29, 2021 by Mahesh

These AP 8th Class Biology Important Questions 4th Lesson Reproduction in Animals will help students prepare well for the exams.

AP State Syllabus 8th Class Biology 4th Lesson Important Questions and Answers Reproduction in Animals

8th Class Biology 4th Lesson Reproduction in Animals 1 Mark Important Questions and Answers

Question 1.
What is meant by external fertilization?
Answer:
The process of fertilization that occurs outside of an organism is called External Fertilization. E.g.: Frog, Fish, Star fish etc.

Question 2.
What is meant by Internal Fertilization?
Answer:
The process of fertilization that takes place inside the body of females is called Internal Fertilization. E.g.: Animals, Human beings.

Question 3.
Which are called Viviparous animals?
Answer:
Animals which give birth to their offsprings are called Viviparous animals. E.g.: Animals, human beings.

Question 4.
What are oviparous animals?
Answer:
Animals which lay eggs are called oviparous animals.
E.g. Hen, Duck, Pigeon etc.

Question 5.
Which is formed by the fusion of male and female gametes?
Answer:
Zygote is formed by the fusion of male and female gametes.

Question 6.
Which gamete is that has a tail and travel to fuse with female gamete?
Answer:
Male gamete or sperm or spermatozoa. The structure of sperm has a head, a middle piece and a tail.

8th Class Biology 4th Lesson Reproduction in Animals 2 Marks Important Questions and Answers

Question 1.
Complete this flow chart.

Answer:
1 – Testes
2 – Sperms
3 – Ovary
4 – Ovum
5 – Embryo
6 – Foetus

Question 2.
What is meant by cloning if Who invented this?
Answer:

Cloning is the reproduction of an exact copy of the cell, any other living part or complete organism from somatic cells.
Cloning of an animal was successfully performed for the first time by lan Wilmut in Scotland.

8th Class Biology 4th Lesson Reproduction in Animals 4 Marks Important Questions and Answers

Question 1.
Complete the cross word puzzle using the hints given below.
Across:
1. The process of the fusion of the gametes.
6. The type of fertilization in hen.
7. The term used for bulges observed on the sides of the body of Hydra.
8. Eggs are produced here.
Down:
2. Sperms are produced in these male reproductive organs.
3. Another term for the fertilized egg.
4. These animals lay eggs.
5. A type of fission

Answer:

Question 2.
Write a note on “test tube babies”.
Answer:

The babies who born through IVF (In Vitro Fertilization) technique are called “Test tube babies”.
In some women, oviducts are blocked due to some reasons. In these cases fertilization will not takes place. Doctors collect freshly released egg and sperms and keep them for few hours.
In case, fertilization occurs, the zygote is allowed to develop for about a week and then it is placed in the mother’s uterus.
Babies born through this technique are called ‘Test tube babies’ which does not mean babies growing in test tubes.

8th Class Biology 4th Lesson Reproduction in Animals Important Questions and Answers

Question 1.
What are Oviparous and Viviparous animals ? Write their characters and give some examples.
Answer:

The animals giving birth to their off springs are called Viviparous animals.
The animals which lay eggs are called Oviparous animals.
Viviparous animals have epidermal hair on their skin and external ears. Examples are cow, elephant, dog, etc.
Oviparous animals do not have epidermal hair or external ears. Examples are fish, frog, crow, peacock, snake, lizard etc.

Question 2.
Kshitija found a fish like larva and put it in an aquarium supposing it as a fish. What would she find after some days?
Answer:

Kshitija had brought tadpole larva, which looks like a fish. It was the larva of the frog.
After somedays, this tadpole larva undergoes metamorphosis and develops into a frog.
Hence, Kshitija would observe a frog in place of tadpole larva after some days in the aquarium.

Question 3.
You want to clarify some doubts about test tube babies what questions do you ask the doctor?
Answer:

Where do test tube babies born?
What differences may we find between normally born baby and a test tube baby?
What problems will test tube babies face in their life?
Why people have so many doubts about test tube babies?

Question 4.
Draw a neat labelled diagram of a human sperm. Fill the following table:

Name of the organ	Function
Tail
Mitochondria
Head
Middle piece

Answer:

Name of the part	Function
Tail	Helps in the movements of the sperm.
Mitochondria	Releases energy for the movements of the sperm cell and also provides energy for the penetration of sperm into Ovum.
Head	Helps in the process of fertilization.
Middle piece	It is the site of mitochondria which are the power houses of the cell.

Question 5.
Latha draw the following flow chart of Human reproduction like this. Is it correct or not.
If not, draw a correct one.

Answer:

Question 6.
What are the differences you observed in the reproduction of Amoeba and frog?
Answer:

Reproduction in Amoeba	Reproduction in Frog
1. Asexual mode of reproduction takes place in Amoeba.	1. Sexual mode of reproduction takes place ts Frog.
2. Fusion of garnets will not occur.	2. Fusion of garnets takes place.
3. Through binary fission or multiple fission (during unfavourable conditions) daughter amoebas are formed.	3. Through external fertilization, many offsprings will take birth in water.
4. Daughter amoebas formed during binary fission, completely resemble their mother.	4. Offsprings formed exhibit variations.
5. Mother Amoeba will extinct sifter binary fission.	5. Mother frog will not extinct after external fertilization.

Question 7.
Write the apparatus used in the project work of observing the life cycle of frog.
Answer:
To conduct this project we need,

Wide mouth transparent bottle/tube
Transparent glass
Dropper
Petridish
Some pebbles
Magnifying glass

Question 8.
Fill in the blanks with the correct terms.
The main male reproductive organs are A and in females are B. A and B release some cells respectively known as C and D. The fusion of C and D is known as E. E results in the formation of F. F gradually develops into a G and finally becomes a H.
Answer:
The main male reproductive organs are Testes and in females are Ovary. Testes and Ovary release some cells respectively known as Sperm cells and Ova. The fusion of Sperm cells and Ova is known as Fertilization. Fertilization results in the formation of Zygote. Zygote gradually develops into a Embryo and finally becomes a Foetus.

Question 9.
Observe this diagram and answer the questions.

a) The diagram belongs to which system?
b) Write the A, B, C in the above diagram.
c) What produces the “B” part?
d) What is function of Part “A”?
Answer:
a) Male reproductive system in human beings.
b) A – Seminal ducts B – Testes C – Penis
c) Sperm cells
d) This duct connects each testes to the urethra. They transport sperms into the penis.

Question 10.
Look at the picture and answer the questions.

a) Which system does the picture belong to?
b) What is “A”? What is produced here?
c) What is “B”?
d) What happens when fallopian tubes are closed?
Answer:
a) Female reproductive system in human beings.
b) Ovary
c) Uterus
d) Fertilization will not take place.

Question 11.
What happens to a female body when ovulation stops forever?
Answer:
a) If ovulation stops in females, menstrual changes do not occur in female body. This stage is known as ‘menopause’.
b) As ovum is not released, there is no chance of pregnancy in this stage.

Question 12.
Label the parts of female reproductive system and answer the following questions.

a) In which part of the female reproductive system does fertilization occur?
b) What is the function of the part labelled as 2?
Answer:
Labelled parts :

Fallopian Tube
Ovary
Uterus
Fallopian tube
Releasing ova

AP 9th Class Maths Important Questions Chapter 10 Surface Areas and Volumes

March 30, 2023July 29, 2021 by Srinivas

These AP 9th Class Maths Important Questions 10th Lesson Surface Areas and Volumes will help students prepare well for the exams.

AP State Syllabus 9th Class Maths 10th Lesson Important Questions and Answers Surface Areas and Volumes

Question 1.
If the surface area of sphere is 616 cm². Find it’s radius.
Solution:
Surface area of sphere = 4πr²
4πr² = 616
4 × \(\frac{22}{7}\) × r² = 616
r² = \(\frac{616 \times 7}{4 \times 22}\) = 49 ⇒ r = 7
∴ Radius = 7 cm.

Question 2.
A sphere and a hemisphere have same radius. What is the ratio of volume of the sphere to volume of the hemisphere?
Solution:
Radius of sphere = r = radius of hemisphere
Volume of sphere = \(\frac{4}{3}\) πr³
Volume of hemisphere = \(\frac{2}{3}\) πr³
∴ Ratios of volumes = \(\frac{4}{3}\) πr³ : \(\frac{2}{3}\) πr³
= 4 : 2 = 2 : 1

Question 3.
Two right circular cones of equal volumes have their heights in the ratio 1 : 2 then prove that the ratio of their radii is √2 : 1.
Solution:
Let Radii of two right circular cones be r₁ and r₂
Ratio of heights = h₁ and h₂
Ratio of volumes = V₁ and V₂
V₁ = V₂ so,

Question 4.
The surface area of a sphere of radius 5 cm is five times the area of CSA of a cone of radius 5 cm. Then find the volume of the cone.
Solution:
Radius of sphere = r = 5 cm
Surface area of sphere = 4πr²
= 4π(5)² = 100π
Radius of cone = 5 cm
Surface area of cone = πr(r + 1)
= π5(5 + l)
= 5π(5 + l)
Surface area of sphere = 5 x Surface area of cone
100π = 5 x 5π(5 + l)
100π = 25π(5 + l)
5 + l = \(\frac{100 \pi}{25 \pi}\) = 4
l = 4 – 5 = -1 cm
∴ Slant height of a cone is ’-1, so problem solving is not possible.

Question 5.
The volume of a cylinder is 308 cm³. The radius of its base is 3.5 cm. Find its curved surface area and total surface area.
Solution:
Volume of cylinder = πr²h
πr²h = 308
⇒ \(\frac{22}{7}\) × 3.5 × 3.5 × h = 308
⇒ 38.5 × h = 308
⇒ h = \(\frac{308}{38.5}\) = 8cm
CSA of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 8
= 2 × 11 × 8 = 196 cm
TSA of cylinder = 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 3.5 (3.5 + 8)
= 2 × 11 (11.5)
= 22 × 11.5 = 253 cm²

Question 6.
Sai, Raju, Latha and Sravani are studying 9th class. In SA-I, Sai got 60 marks out of 80. The average marks of all four students is 48. If the average marks of Sai and Raju is 54 and the average marks of Sai, Raju and Latha is 42 then find the marks obtained by four students individually.
Solution:
Let marks of Sai, Raju, Latha and Sravani be a, b, c, d.
\(\frac{a+b+c+d}{4}\) = 48
a + b + c + d = 192 …………… (1)
= \(\frac{a+b}{2}\) 54 ⇒ a + b = 108 …………….. (2)
\(\frac{a+b+c}{3}\) = 42 ⇒ a + b + c = 126 ……………… (3)
From (2) and (3)
(a + b + c) – (a + b) = 126 – 108 = 18
c = 18
From (1) and (2)
(a + b + c + d) – (a + b + c) = 192 – 126
d = 66
Given that a = 60°
∴ Sai = 60 Marks
Raju = 48 Marks
Latha = 18 Marks
Sravani = 66 Marks

AP 9th Class Maths Important Questions Chapter 6 Linear Equation in Two Variables

March 30, 2023July 29, 2021 by Srinivas

These AP 9th Class Maths Important Questions 6th Lesson Linear Equation in Two Variables will help students prepare well for the exams.

AP State Syllabus 9th Class Maths 6th Lesson Important Questions and Answers Linear Equation in Two Variables

Question 1.
Express the following statement as a linear equation in two variables. “Neeraja and Girija of class IX students together contributed Rs. 300/- towards the C.M. relief fund”.
Solution:
Contribution of Neeraja = ₹x
Contribution of Girija = ₹y
Sum of contribution of Neeraja and Girija is 300/-
∴ x + y = 300.

Question 2.
Re state the following statement with appropriate conditions to make it true statement.
“For every real number x, x2 ≥ x”.
Solution:
If x ≤ 0 or x ≥ 1 then x² ≥ x.
For every integer x, x² ≥ x.

Question 3.
Write the statement given below as a linear equation two in variables. “The sum of two numbers x and y is 75”.
Solution:
The sum of two numbers x and y is 75.
∴ x + y = 75

Question 4.
If x = 2 – α and y = 2 + α in a solution of 5x + 3y – 7 = 0 and x = 2β + 1 and y = β – 1 in a solution of 3x – 2y + 6 = 0 then find the value of α + β.
Solution:
Given equation = 5x + 3y – 7 = 0
x = 2 – α and y = 2 + α
∴ 5(2 – α) + 3(2 + α) – 7 = 0
⇒ 10 – 5α+ 6 + 3α – 7 = 0
⇒ 9 – 2α = 0
⇒ 2α = 9 ⇒ α = \(\frac{9}{2}\) = 4.5
Given equation 3x – 2y + 6 = 0
x = 2β + 1 and y = β – 1
∴ 3(2β + 1)- 2(β – 1) + 6 = 0
⇒ 6β + 3 – 2β + 2 + 6 = 0
⇒ 4β + 11 = 0 ⇒ β = \(\frac{-11}{4}\)
∴ α + β = \(\frac{9}{2}\) – \(\frac{11}{4}\) = \(\frac{18-11}{4}\) = \(\frac{7}{4}\)

Question 5.
Draw the graph of the equation 2x + 3y = 6. Find the coordinates of the points where the graph cuts the coordinate axes.
Solution:
Given equation 2x + 3y = 6

Coordinates of the points are (3, 0) (0, 2)

Question 6.
Draw the graph of the linear equations 2x + 3y = 12. At what points, the graph of the equation cuts the X-axis and Y-axis.
Solution:
2x + 3y = 12

The graph of the equation cuts the x – axis = (6, 0)
y – axis = (0, 4)

AP 9th Class Physical Science Important Questions Chapter 7 Reflection of Light at Curved Surfaces

March 30, 2023July 29, 2021 by Srinivas

These AP 9th Class Physical Science Important Questions 7th Lesson Reflection of Light at Curved Surfaces will help students prepare well for the exams.

AP State Syllabus 9th Class Physical Science 7th Lesson Important Questions and Answers Reflection of Light at Curved Surfaces

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Which mirror is used as rear-view mirror in the vehicles?
Answer:
Convex mirror is used as rear view mirror in the vehicles.

Question 2.
What is the relation between focal length (f) and radius of curvature (R)?
Answer:
The radius of curvature of a spherical mirror is twice to its focal length.
⇒ R = 2f (or) f = \(\frac{R}{2}\).

Question 3.
Can a virtual image be photographed by a camera?
Answer:
Yes, virtual image can be photographed by a camera.

Question 4.
Complete the diagram and draw the image.

Answer:

Question 5.
Predict and write the reason, why the value of the distance of the object (u) is always negative in the mirror equation.
Answer:
i) Direction of the incident rays is taken as positive (+ve).
ii) Object distance is measured from the pole to the object in the opposite direction of incident rays.

Question 6.
Which property of concave mirror is used in making the solar cooker?
Answer:
Rays coming parallel to the principal axis of a concave mirror is focused at focal point. Based on this property solar cooker is made.

Question 7.
Draw the ray diagram to show the formation of image for the object of height 1 cm. placed at 5 cm. distance, in front of a convex mirror having the radius of curvature R = 5 cm.
Answer:

Question 8.
What is reflection?
Answer:
The light rays falling on a surface are returned into the original medium. This phenomenon is called reflection.

Question 9.
What is the relation between focal length and radius of curvature?
Answer:
Radius of curvature = 2 x focal length
∴ R = 2f (or) f = \(\frac{R}{2}\)

Question 10.
What is the mirror formula for spherical mirrors?
Answer:
The mirror formula is \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
f = focal length of mirror ; u = object distance ; v = image distance

Question 11.
What is a real image? What is a virtual image?
Answer:
Real image :
The image formed due to convergence of light rays. The real image can be caught on the screen.

Virtual image :
The image that we get by extending the rays backwards is called a virtual image. A virtual image cannot be caught on the screen.

Question 12.
What is focal length?
Answer:
The distance between focus and vertex.

Question 13.
What is radius of curvature?
Answer:
The distance between vertex and centre of curvature.

Question 14.
What is magnification?
Answer:
The ratio of size of image to size of object is called magnification.

(OR)

The ratio of image distance to object distance is called magnification.

Question 15.
Why are concave and convex mirrors called spherical mirrors?
Anwer:
The reflecting surface of convex and concave mirror is considered to form a part of the surface of a sphere. So they are called spherical mirrors.

Question 16.
What is a reflecting surface?
Answer:
The surface used for reflection is called reflecting surface.

Question 17.
What is principal axis?
Answer:
The horizontal line which passes through the centre of curvature is called principal axis.

Question 18.
What is meant by converging of light rays?
Answer:
If light rays after reflection meet at a point, then we say the light rays are converging.

Question 19.
When do you say light rays are diverging?
Answer:
If light rays appear as if they are coming from a point after reflection, then we say light rays are diverging.

Question 20.
When does a ray reflect in the same path from a concave mirror?
Answer:
When it passes through centre of curvature.

Question 21.
When a light ray travelling from parallel to principal axis falls on concave mirror, then what is the path of reflected ray?
Answer:
The reflected ray passes through focal point.

Question 22.
Where do you place the vessel in solar cooker?
Answer:
We place the vessel in solar cooker at the focal point.

Question 23.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object.

Question 24.
Can a convex mirror burn a paper? If not, why?
Answer:
The rays coming parallel to principal axis after reflection diverge from the mirror. So we cannot burn a paper by using a convex mirror.

Question 25.
Which mirror has wider field of view?
Answer:
A convex mirror has wider field of view, that’s why they are used as rear view mirrors in vehicles.

Question 26.
Why does our image appear thin or bulged?
Answer:
Due to converging or diverging of light rays from the mirror.

Question 27.
Why is angle of incidence equal to angle of reflection when a light ray reflects from a surface?
Answer:
Because light selects the path that takes least time to cover a distance.

Question 28.
Are angle of reflection and angle of incidence also equal for curved surface?
Answer:
Yes, it is equal for curved surfaces like spherical mirrors.

Question 29.
What is a spherical mirror? Give different types of spherical mirrors.
Answer:
If the reflecting surface of mirror is considered to form a part of the surface of sphere, then it is called spherical mirror. Spherical mirrors are of two types :

Concave mirror
Convex mirror

Question 30.
Write about various distances related to mirrors.
Answer:
The various distances related to mirrors are
1) Focal length (f) :
The distance between vertex and focus is called focal length.

2) Radius of curvature (R) :
The distance between vertex and centre of curvature is called radius of curvature.

Question 31.
We wish to obtain an erect image of an object using a concave mirror of focal length of 15 cm. What should be range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object?
Answer:
The range of distance of object is between 0 and 15 cm.
The image is virtual and erect.
The image is larger than the object.

Question 32.
Name some apparatus which can work on the principle of reflection of light.
Answer:
Plane mirror, spherical mirrors, periscope, kaleidoscope.

Question 33.
If you want to get parallel beam by using concave mirror, then where do you keep the source?
Answer:
The object should be kept at focus because the light rays coming from focus after reflection from mirror travel parallel to principal axis.

Question 34.
If you want to form the image of an object at infinity, then where do you keep the object?
Answer:
The object should be kept at focus; then the image would be formed at infinity.

Question 35.
How do you get a virtual image with a concave mirror?
Answer:
When we place the object between vertex and focus then we will get a virtual image.

Question 36.
Why do dentists use concave mirror?
Answer:
If the object is between mirror and its focus we get enlarged virtual and straight image by using concave mirror. So dentists use this principle to see inner parts of mouth.

Question 37.
A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
> 10 cm.

Question 38.
What is your opinion on elevating buildings with mirrors?
Answer:
The mirrors used in elevating buildings are reinforced, tough and laminated glasses. These mirrors provide safety and make the buildings attractive.

Question 39.
Identify the mirror having focal length +15 cm.
Answer:
Convex mirror (since the focal length of convex mirror is taken as positive).

Question 40.
If the focal length of mirror is 10 cm, what is that mirror?
Answer:
The mirror is concave (since the focal length of concave mirror is taken as negative).

Question 41.
Can we focus a sunlight at a point using a mirror instead of magnifying glass?
Answer:
Yes, by using concave mirror we can focus sunlight at a point.

Question 42.
To reduce glaze of surroundings the windows of some department stores, rather than being vertical, slant inward at the bottom. How does this reduce glaze?
Answer:
This slant reflects the sunlight further down towards the ground, then it would happen as if they are vertical.

Question 43.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:
It gives erect and small image and covers large distance.

Question 44.
An object is placed at a distance 8 cm from a concave mirror of radius of curvature 16 cm. What are the characteristics of image?
Answer:
The image is real, inverted, and same size.

Question 45.
What happens when light falls on an opaque object?
Answer:
Some part of light is reflected back and remaining part is absorbed.

Question 46.
What happens when light is reflected from transparent object?
Answer:
Some part of light is reflected and remaining part is partly transmitted or partly absorbed.

Question 47.
Which objects at your home act as spherical mirrors?
Answer:
Objects at home that act as spherical mirrors are :

Spoons
Spectacles
Sink
Cooking vessel

Question 48.
Complete the following ray diagram.

Answer:

The light ray passing through centre of curvature falls normal to the concave mirror. So it retraces the same path.

Question 49.
If focal length is 20 cm, then what is radius of curvature of mirror?
Answer:
f = 20 cm
R = 2f = 2 × 20 = 40 cm.

Question 50.
The radius of curvature of a spherical mirror is 20 cm. What is the focal length?
Answer:
Radius of curvature (R) = 20 cm
R 20
Focal length (f) = \(\frac{\mathrm{R}}{2}=\frac{20}{2}\) = 10 cm.

Question 51.
The focal length of convex mirror is 16 cm. What is its radius of curvature?
Answer:
f = 16 cm
R = 2f = 2 × 16 = 32 cm

Question 52.
Write any two uses of concave mirror in our daily life.
Answer:
Uses of concave mirror :

Concave mirrors are used by dentists to see enlarged image of tooth.
Concave mirrors are used in car head lights.

Question 53.
Write any two uses of convex mirror in our daily life.
Answer:
Uses of convex mirror :

Convex mirrors are used as rear view mirrors in vehicles because convex mirrors increase field of view.
Convex mirrors are used in street light reflectors as they spread light over greater

Question 54.
Suggest a new use with a spherical mirror.
Answer:
Spherical mirrors are newly adapted in ATMs.

Question 55.
Focal length of a concave mirror is x. Find the sum of focal length and radius of curvature.
Answer:
Focal length = x; Radius of curvature = 2 x focal length = 2x.
The sum of focal length and radius of curvature = x + 2x = 3x.

Question 56.
If the angle between the mirror and incident ray is 40°, then find the angle of reflection.
Answer:
Given that angle between incident ray and mirror = 40°.
Suppose angle of incidence = x.
∴ 40 + x = 90
x = 90 – 40 = 50°.

But we know angle of incidence = angle of reflection
∴ Angle of reflection = 50°.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
Your friend has a doubt that whether a concave mirror or a convex mirror is used as a rear view mirror in the vehicles. What questions will you ask to clarify his doubts?
Answer:

Is the image in a rear-view mirror smaller or larger when compared to real object?
Which mirror forms smaller image than the object in the given mirrors?

Question 2.
The focal length of a huge concave mirror is 120 cm. A man is standing in front of it at a distance of 40 cm. What are the characteristics of his image in that mirror?
Answer:
i) Image form in the mirror
ii) Virtual image
iii)Erected image
iv) Enlarged image

Question 3.
How can you find out the focal length of concave mirror experimentally when there is no sunlight?
Answer:
Place the object / candle in front of the mirror and adjust the screen to get image on it. Measure the object distance, image distance. Substitute the values (as per sign connection) in mirror formula \(\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)\). We get the focal length of mirror.
(OR)
Place the object / candle and the screen at same point in front of the mirror. Adjust this set of material to get sharp image on the screen.

Measure the distance from mirror to object/screen. This distance is the radius of curvature and make it half, it gives focal length of the mirror.

Question 4.
The magnification of the image by the concave mirror is – 1. Mention the four char-acteristics of image from the above information.
Answer:

Image will be formed at the centre of curvature (C).
Image size is equal to that of the object size. ‘
Inverted image.
Real image.

Question 5.
Write about different points related to mirrors.
Answer:
The different points related to mirrors are
1) Vertex (P) :
The point where the central axis touches the mirror is called vertex.

2) Focus or focal point (F):
The light rays coming from distinct object appear to meet at point in case of concave mirror and tend to meet at point when drawn backward in case of convex mirror. That point is known as focus or focal point.

3) Centre of curvature (C) :
It is centre of the sphere to which the mirror belongs.

Question 6.
What happens if light rays parallel to principal axis fall on the concave mirror, and draw ray diagram?
Answer:

The light rays that are parallel to the principal axis get reflected such that they pass through the focal point of the mirror. R₁ is such ray in figure.

Question 7.
What happens to a ray which passes through focal point and falls on the concave mirror, and also draw the ray diagram?
Answer:

The light ray which goes through the focal point of the mirror travels parallel to principal axis. R₂ is such ray in figure.

Question 8.
How does an image form due to convex mirror? Draw the ray diagram.
Answer:

The parallel rays coming from distance object tend to diverge after reflection.
If we extend the reflected rays backwards they meet at ‘F’, i.e. focal point of the convex mirror.

Question 9.
Which light ray after reflection will travel along the same path in opposite direction? What can be such a ray for a spherical mirror? Draw the ray diagram.
Answer:

Any ray that is normal to the surface, on reflection, will travel along the same path but in opposite direction.
The line drawn from the centre of curvature of mirror is perpendicular to the tangent at the point, the line meets the curve.
So if we draw a ray starting from the tip of the object going through the centre of curvature to meet the mirror, it will get reflected along the same line. This ray is shown as R₃ in the figure.

Question 10.
What happens if an object is placed at centre of curvature of a mirror? Draw the ray diagram.
Answer:

From the ray diagram we conclude that the image of the object will be formed at the same distance as the object and it will be inverted and of the same size. The image is real because it forms on a screen.

Question 11.
Draw the ray diagrams with convex mirror and write rules of ray diagram of convex mirror.
Answer:
Rule -1 :

A ray running parallel to main axis, on meeting the convex mirror will get reflected so as to appear as if it is coming from the focal point.

Rule – 2 :

This is converse of rule 1. A ray going in the direction of focal point after reflection will become parallel to main axis.

Rule – 3 :

A ray going in the direction of the centre of curvature will get reflected back in opposite direction, and looks like that is coming from the centre of curvature.

Question 12.
Why do we use parabolic mirror instead of concave mirror?
Answer:

We use parabolic mirror instead of concave mirror because with the concave mirror all the rays coming parallel in it may not be focused at focal point (F).
Those rays which are very nearer to principal axis will only be focused at focal point.
It is very effective to make the mirror parabolic in order to make all the rays to converge at focus.

Question 13.
The magnification of mirror is given as – 3. What is the inference do you get from this information?
Answer:
Magnification – ve indicates it is an inverted image. So it is a real image.
Magnification 3 indicates the image size is three times the object size. So the image is enlarged. Since it is forming real image the mirror is concave.

Question 14.
Why are we able to see various objects around us?
Answer:
We are able to see various objects around us due to the diffused light reflected from these objects reaches to our eye which gives sense of vision to those objects.

Question 15.
Which type of mirror is used as a reflector in street lamp?
Answer:
The reflectors of the street lamp are made in convex in shape so that reflected rays diverge over the larger area on ground. Therefore convex mirror acts as a reflector in street lamp.

Question 16.
Which type of mirror is used in doctor’s head lamp?
Answer:
Doctors use head lamp to examine nose, throat, teeth, etc. of patients. In this lamp a parallel beam of light is allowed to fall on the concave mirror. The reflected light concentrates on focus on the mirror on a smaller area to be examined. So the concave mirror is used in doctor head lamp.

Question 17.
Would you able to burn a paper using concave mirror?
Answer:

Concave mirror focuses the parallel sun rays at focal point of the mirror.
So with a small concave mirror we can heat up and burn a paper.

Question 18.
How do you find the focal length of concave mirror?
Answer:

Hold a concave mirror such that sunlight falls on it.
Take a small paper and slowly move it in front of the mirror until we will get smallest and brightest spot of the Sun.
Find the distance between mirror and image of the Sun that will give the focal length of mirror.

Question 19.
See the table and identify the mirrors in each case.

Mirror	Magnification
X	– 1
Y	+ 1
Z	+ 0.5

Answer:
Magnification negative indicates that it is inverted image and also real. Magnification -1 means same size. So the mirror which gives real image of same size is concave mirror. So ‘X’ is concave.
The magnification +1 means the image is virtual, erect and same size. So the mirror Y is plane.
The magnification + 0.5 means the image is virtual, erect and diminished. So the mirror Z is convex.

Question 20.
An object is placed at various positions in front of concave mirror of focal length 10 cm. Complete the table by using given information without actually doing the problem.

Answer:

Question 21.
Draw a normal at any point of a concave mirror.
Answer:

Question 22.
Identify the following in a ray diagram showing the reflection of light in a concave mirror.
a) Pole of the mirror
b) Principal axis
c) Centre of curvature
d) Focal point
e) Focal length
Answer:

Question 23.
Complete the ray diagram.

Answer:
First we have to draw normal at point of contact to the concave mirror and then we have to use laws of reflection to draw the reflecting ray.

Question 24.
Figure shows two parallel light rays falling on a convex mirror. Complete the ray diagram.

Answer:

Question 25.
See the belog figure and complete the ray diagram.

Answer:

Question 26.
Assume that an object is kept at a distance of 20 cm in front of a concave mirror. If its focal length is 30 cm, then
a) what is the image distance?
b) what the magnification of mirror in this case?
Answer:
Object distance = u = 20 cm

Question 27.
There is an object in front of convex mirror at a distance of 5 cm. If its focal length is 10 cm, then
a) what is the image distance?
b) what is its magnification?
Answer:

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Sudheer wants to find focal length of a concave mirror experimentally.
a) What apparatus does he need?
b) Is the screen required or not? Explain.
c) Draw the table required to tabulate the values found in his experiment.
d) What is the formula used by him to find focal length?
Answer:
a) Apparatus required to Sudheer are

Concave mirror,
White paper or screen,
Scale,
V – stand,
Candle.

b) Yes, screen is required.
To catch and measure the image distance screen is required.

c) Table for observation and calculation of ‘f’.

d) Focal length \(\Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \text { (or) } \mathrm{f}=\frac{\mathrm{uv}}{\mathrm{u}+\mathrm{v}}\)
This is the formula used by him to find a focal length.

Question 2.
Show the formation of image with a ray diagram when an object is placed on the principal axis of a Concave mirror between focus and centre of curvature of the mirror.
Answer:

Question 3.
An object of 6 cm height is placed at a distance of 30 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror, will the image be formed? What are the characteristics of the image?
Answer:

Question 4.
List the materials required for conducting an experiment to find the focal length of a concave mirror. Explain the experimental process also.
Answer:
a) Material required for conducting an experiment to find the focal length of a concave mirror are

concave mirror
a piece of paper
meter scale.

b) Procedure of the experiment:

Hold a concave mirror such that sunlight falls on it.
Take a small paper and slowly move it in front of the mirror and find out the point where we get the smallest and brightest spot, which will be the image of the sun.
Measure the distance of this spot from the pole of the mirror.
This distance is the focal length (f) of the mirror.

Another Experiment:
a) Material required :
A candle, paper / screen, concave mirror, V-stand, measuring tape or meter scale.

b) Procedure :
1) Place the concave mirror on V-stand, a candle and meter scale as shown in the figure.

2) Keep the candle at different length from the mirror (10 cm to 80 cm) along the axis and by moving the paper find the position where the sharp image is got on the paper.

3) Measure the image distance (o) and note in the given table.

4) Find the average of focal lengths (f) obtained in the experiment.

5) The average T is the focal length of the given mirror.

Question 5.
An object of height 5 cm is placed at 30 cm distance on the principal axis in front of a concave mirror of focal length 20 cm. Find the image distance and size of the image.
Answer:
Object distance (u) = – 30 cm, Focal length (f) = – 20 cm, Height of object (h_o) = 5 cm, Image distance (v) = ?, Height of image (h_i) = ?

∴ The image is real, inverted with a height of 10 cm.

Question 6.
A student conducted an experiment to observe characteristics of images formed by spherical mirrors and recorded his observations as follows. Observe the table and answer the questions.

i) Above said information belongs to which spherical mirror?
ii) In which situation, magnification is less than 1.
iii) An object of height 8 cm placed at centre of curvature on principal axis, then where do you get the image and what is its height?
iv) “All real images are inverted”. Justify the statement by using above table.
Answer:
i) It is a Concave mirror.
ii) When object is kept beyond ‘C’ then magnification is less than 1.
iii) Image formed at ‘C’. The height of the image is 8 cm.
iv) According to the table if the image is erected image, it is a real image. In all the other cases every real image is virtual image.

Question 7.
In the following cases calculate the magnification values for a concave mirror. Give reason.
a) When the object is at the focal point of the mirror.
b) When the object is between focal point and the pole.
Answer:
In the case of concave mirror
a) When the object is at the focal point of the mirror, then its magnification value is -1.

Reason :
In this case size of the image is large, compared with the object. It is called virtual image. Image is formed behind the mirror so magnification has negative sign.
Nature of the object: It is real, inverted, enlarged and forms at infinity.

b) When the object is between focal point (F) and the pole (P) of the mirror, then its magnification value is +1.

Reason :
In this case image is formed on the same side of the object and it is also virtual image so, the sign of the magnification is positive.

Nature of the object:
It is virtual, erect, enlarged and on the same side of the object.

Question 8.
Write the derivation of mirror formula.
(OR)
Derive \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\).
(OR)
A student wants to find the image distance for a given object distance of a mirror. Then derive a formula for the mirror.
Answer:
Derivation of mirror formula :
In the figure P = pole, C = centre of curvature and F= focus of the concave miror. Object AB is placed beyond C. Image AB’ is formed in between F and C.
From the diagram triangles A’B’C and ABC are similar triangles.
\(\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}}\) ………………… (1)

Question 9.
Where is the base of the candle going to be in the image when the object is placed on the axis of the mirror beyond ‘C’?
Answer:

Any ray starting from a point on the axis and travelling along the axis will reflect on the axis itself.
So the base of the image is going to be the axis.
If the object is placed vertically on the axis, the image is going to be vertical.
Draw perpendicular from point A to axis.
The intersection point is the point where the base of the image of the candle is going to be formed

Question 10.
What happens if an object placed at a distance less than the focal length of the concave mirror? Draw the ray diagram.
(OR)
When do you get a virtual image by using a concave mirror and draw the ray diagram?
Answer:

The candle object (0) is placed at a distance less than the focal length of the mirror.
The first ray (R₁) will start from tip of the object and run parallel to axis to get reflected so as to pass through focal length.
The second ray (R₂) is the ray starting from the tip of the object and going through the focal point but it is not possible as such a ray will not meet the mirror.
The third ray (R₃), starting from the tip of the object goes to the centre of curvature but that also seem not to be possible.
Now consider a ray (R₄) that starts from the tip and goes in such a direction that it would go through the centre of curvature if extended backwards.
This ray is normal to surface and so will be reflected along the same line in opposite direction and will go through centre of curvature.
The two reflected rays diverge and will not meet.
When we extend these rays backward they appear to be coming from one point.
As seen from the figure (2) the image will be erect and enlarged and virtual.

Question 11.
A person in a dark room looking through a window can clearly see a person outside in the daylight, whereas the person outside cannot see the person inside. Why?
Answer:

There is usually some reflection that occurs at an interface between the two materials but most often of light passing through.
Imagine you are inside in the dark. A person outside in bright sunlight is sending out (reflection) lots of light, most of which would come through the window to you, so you see them clearly.
Since it is so bright outside, there is also a good amount of light which reflects back towards them.
This can distract them from little bit of light from you that is going towards them, so they have much harder time seeing you.

Question 12.
What is magnification? Derive an expression for magnification.
Answer:
Magnification :
The ratio of height of image to height of object is called magnification.

Question 13.
What are the rules to be followed while drawing ray diagrams?
Answer:
Various rules to be followed for drawing ray diagrams.
1) A ray parallel to the principal axis, after reflection will pass through the principal focus in case of a concave mirror or appear to diverge from the principal focus in case of convex mirror.

2) A ray passing through the principal focus of a concave mirror or a ray which is directed towards the principal focus of convex mirror, after reflection will emerge parallel to the principal axis.

3) A ray passing through the centre of curvature of a concave mirror or directed in the direction of centre of curvature of a convex mirror, after reflection, is reflected back along the same path.

4) A ray incident obliquely to the principal axis, towards a point P on the concave mirror or a convex mirror, is reflected obliquely. The incident ray and reflected rays follow laws of reflection.

Question 14.
How do you make a parallel beam with an experiment?
Answer:
Aim :
Making a beam of parallel lines.

Material used :
Two pins, thermocol block, candle.

Procedure:

Stick two pins on a thermocol block.
The pins are exactly parallel to each other.
As we can see in the figure, when a source of light is kept very near, we see the shadows diverging (from the base of the pins).
As we move the source away from the pins, the divergent angle starts reducing.
If we move the source far away, we will get parallel shadows. Thus we get a beam of parallel lines.

Question 15.
Write a table which shows the image formed by a concave mirror for different positions and also give size and nature of image.
Answer:

Question 16.
Ray diagrams of concave mirror.
Answer:
Object is placed ‘Infinitely’ :

Object is placed between ‘P’ and ‘F :

Question 17.
Complete the following ray diagrams and give reasons.

Answer:
a)

Reason :
The light appears to be passing through centre of curvature after reflection from convex mirror retraces the same path.

b)

Reason :
The light ray making certain angle of incidence ‘q’ with principal axis follows laws of reflection and reflects with same angle q.

c)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror and if we extended the ray backwards it passes through principal focus.

d)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror when the light ray drawn backwards passes through focus and second light ray follows laws of reflection (i.e., ∠i = ∠r).

These extended backward light rays meet and form a virtual, diminished and erect image between pole and focus inside the mirror.

Question 18.
Complete the following diagram to obtain image of object AB.

Answer:

The light ray which appears to coming from focus after reflection from convex mirror travels parallel to principal axis.
The light ray which is incident with certain angle ’x’ at pole ‘P’ reflects with same angle from convex mirror.
These two extended light rays meet at B. So AB’ is the image of the object AB.

Question 19.
An object 4 cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of image.
Answer:
Given that f = – 15 cm ; u = – 25 cm ; h0 = 4 cm ; v = ?

So the image is enlarged and inverted.

Question 20.
Focal length of a concave mirror is f. The distance from its focal point to the object is P. Find the ratio of heights of image.
Answer:
Concave mirror is a part of spherical mirror.

Question 21.
When do we get a blurred image from a distant object by using concave mirror?
Answer:

The diagram shows a few rays starting from the tip of the flame.
The reflected rays intersect A. So the reflected image of tip of flame will be at intersection point A.
If we hold the paper at any point before or beyond point A (for example at point B), we see that the rays will meet the paper at different points,
So the image of the tip of the flame will be formed at different points due to these rays.
If we draw more rays emanating from the same tip we will see that point A they will meet but at point ‘B’ they won’t.
So the image of the tip of the flame will be sharp if we hold the paper at A and become blurred (due to mixing of multiple images) when we move the paper slightly in any direction (forward or backward).

Question 22.
An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed? Find size and nature of the image.
Answer:
Given, h₀ = 7 cm ; u = – 27 cm ; f = – 18 cm ; v = ?

Question 23.
An object 3 cm high is placed at a distance of 15 cm from a concave mirror, the radius curvature is 20 cm. Find the nature, position and size of the image. (V = -30 cm, m = -2, h₂ = -6 cm)
Answer:
h₀ = 3 cm; u = -15 cm; r = -20 cm; f = \(\frac{r}{2}\) = – 10 cm

Question 24.
Focal length of a concave mirror is 15 cm. An object of length 5 cm is placed in front of this mirror. Draw neat diagrams to find the length and position of image when object is at (I) 5 cm, 0i) 12 cm, (iif) 20 cm, (jv) 35 cm away from the mirror.
Answer:
i) An object is kept at a distance of 5 cm from the mirror.

AP 10th Class Biology Important Questions Chapter 4 Excretion

March 29, 2023July 29, 2021 by Mahesh

These AP 10th Class Biology Important Questions and Answers 4th Lesson Excretion will help students prepare well for the exams.

AP State Syllabus 10th Class Biology 4th Lesson Important Questions and Answers Excretion

10th Class Biology 4th Lesson Excretion 1 Mark Important Questions and Answers

Question 1.
What is meant by excretion?
Answer:
Excretion is a biological process involved in separation and removal of wastes from body.

Question 2.
When you are on a field trip, your friend collected a sticky substance oozed out by a plant called gum. What are the plants you observe which give gum ?
Answer:
Acacia, Neem, Drumstick, Eukalyptus and Sapota are the gum yielding plants present in our surroundings.

Question 3.
What are primary metabolites?
Answer:
The substances that are directly involved in normal growth and development and reproduction of a plant are called primary metabolites, e.g.: Carbohydrates, fats and proteins.

Question 4.
What are secondary metabolites?
Answer:
The inorganic substances that are not directly involved in the normal growth, development and reproduction of an organism are called secondary metabolites, e.g.: Alkaloids, tannins, resins, gums and latex, etc.

Question 5.
Write any two substances present both in blood and urine.
Answer:
Glucose, Sodium, Potassium, Chlorides, Urea, Creatinine, Uric Acid, Calcium, Phosphorous.

Question 6.
When you are on a field trip, you might have collected some plants which contain alkaloids. Name the alkaloids which are harmful to us.
Answer:
Nicotine, Morphine, Cocaine.

Question 7.
Write the names of any two excretory organs in human beings.
Answer:
Kidney, Skin, Lungs, Liver, Large Intestine.

Question 8.
Which plants in your surroundings are useful for the production of medicines?
Answer:
Neem, Datura, Tulasi, Calotropis.

Question 9.
The body of a person is filled with extra water and waste products. His hands and feet were swollen. What do we call this condition? Failure of which system causes this condition?
Answer:
This condition is called Uremia. This condition is caused due to failure of excretory system (Kidneys).

Question 10.
How do plants get rid of their waste materials?
Answer:

Plants can get rid of excess water by transpiration and guttation.
The other nitrogenous waste products are stored in leaves, bark and fruits when these dead leaves, bark and ripe fruits fall off from the tree then the waste products in them are removed.

Question 11.
Give an example of a case where organ donation can save one’s life.
Answer:

Kidney of a brain dead person can be donated to save the life of renal failure persons.
Cornea of a dead person can be donated to the blind.
Bone marrow can be donated to save life of persons suffering from blood related genetic disorders.

Question 12.
Why is urine yellow in color?
Answer:
Because of urochrome, urine is yellow in colour. It forms in the liver from dead RBC.

Question 13.
Write two slogans to popularize the awareness on “Organ Donation”.
Answer:
Slogans:

Organ donation saves lives.
Donate organs today for better tomorrow.
Donate organs – Live after death.

Question 14.
Write two healthy habits which you practice to protect your kidneys from diseases.
Answer:

Drink plenty of water.
Eat low salt diet that saves kidney life.
Drink more fruit juices.

Question 15.
What precautions you have to take in the observation of internal structure of mammalian kidney?
Answer:

Wash the kidney thorougly with water.
Ensure that blood is completely drained from it.
Wash your hands with antibacterial lotion after completing the dissection.

Question 16.
Write two secondary metabolites, which you use in your daily life.
Answer:
Gum, Rubber, Coffee are the examples for secondary metabolites which we use in our daily life.

Question 17.
Why is vasopressin not secreted when a person drinks a lot of water?
Answer:
Vasopressin is secreted only when concentrated urine is to be passed out. When a person drinks a lot of water, there will be no need to secrete concentrated urine. The excess water taken by the person will be sent out in the form of dilute urine. Hence, vasopressin is not secreted.

Question 18.
In urine excretory system much water is reabsorbed. What happens if it doesn’t occur?
Answer:
If water is not reabsorbed it leads to excessive repeated dilute urination called diabetes insipidus. If water does not reabsorbed we would dry up in few hours.

Question 19.
A substance given below consists of other three substances. What is that substance? Where is it produced? Uric Acid, Sodium, Oxalate, Urine.
Answer:
Urine contains all the other three substances. It is produced in kidney.

Question 20.
Name the hospitals where the organ transplantation facility is available in Hyderabad.
Answer:
Organ transplantation facility is available only in two government hospitals. They are NIMS (Nizam Institute of Medical Sciences) and Osmania hospital. This facility is also available in some corporate hospitals.

Question 21.
What is anabolism?
Answer:
The synthesis of complex molecules in living organisms from simpler ones together with the storage of energy.

Question 22.
What is catabolism?
Answer:
The breakdown of complex molecules in living organisms to form simpler ones, together with the release of energy.

Question 23.
What is the meaning of the word excretion in Latin?
Answer:
In Latin ex means out, crenere means shift.

Question 24.
What are the wastes produced during metabolic activities?
Answer:
Carbon dioxide, water, nitrogenous compounds like ammonia, urea, uric acid, bile pigments, excess salts, etc., are the wastes produced during metabolic activities.

Question 25.
What is the most poisonous of all waste products?
Answer:
Ammonia is the most poisonous of all waste products.

Question 26.
What are the substances present in blood?
Answer:
Substances present in blood are glucose, sodium, potassium chloride, urea, creatinine, uric acid, cholesterol, triglycerides, calcium, phosphorous, bilirubin, proteins, etc.

Question 27.
What are the substances present in urine?
Answer:
The substances present in urine are protein, creatinine, calcium, phosphorous, uric acid, etc.

Question 28.
What are the substances present in both blood and urine in common?
Answer:
The substances present in the blood and urine in common are sodium, potassium, glucose, chlorides, urea, proteins, creatinine, calcium, phosphorous and uric acid.

Question 29.
What are the substances that need to be removed from body?
Answer:
Creatinine, uric acid, urea, cholesterol and calcium.

Question 30.
In human beings excretory system consists of?
Answer:
The excretory system in human beings consists of a pair of kidneys, a pair of ureters, urinary bladder and urethra.

Question 31.
Where are the kidneys present in human body?
Answer:
In human body kidneys are present in the abdominal cavity attached to dorsal body wall one on either side of backbone.

Question 32.
What is the shape and colour of the kidney?
Answer:
The kidney is bean shaped and reddish brown in colour.

Question 33.
What is the size of the kidney?
Answer:
The size of the kidney is 10 cm in length, 5 – 6 cm in breadth and 4 cm in thickness.

Question 34.
Why is the position of right kidney lower than the left kidney?
Answer:
The position of right kidney is lower than left kidney due to the presence of liver above it.

Question 35.
Which artery brings oxygenated blood to kidney?
Answer:
Renal artery brings oxygenated blood to kidney.

Question 36.
What are the two distinct regions present inside the kidney?
Answer:
The two distinct regions present inside the kidney are

dark coloured outer zone called the cortex and
pale coloured inner zone called medulla.

Question 37.
Each kidney is made up of how many nephrons?
Answer:
Each kidney is made up of about 1.3 to 1.8 million nephrons.

Question 38.
What is the other name of Nephron?
Answer:
The other name of nephron is uriniferous tubule.

Question 39.
What are the two basic parts of nephron?
Answer:
The two basic parts of nephron are malphigian body and renal tubule.

Question 40.
What is meant by Glomerulus?
Answer:
Malphigian body consists of a blind cupshaped broader end of nephron called Bowman’s capsule and bunch of blood capillaries called Glomerulus.

Question 41.
Which blood vessel forms glomerulus in Bowman’s capsule?
Answer:
Afferent arteriole forms glomerulus in Bowman’s capsule.

Question 42.
Renal tubule consists of how many parts?
Answer:
Renal tubule consists three parts. They are Proximal convoluted tubule (PCT), Loop of Henle, and Distal Convoluted Tubule (DCT).

Question 43.
What is the major function of proximal convoluted tubule?
Answer:
Proximal convoluted tubule reabsorbs useful substances like glucose, amino acids, phosphate, potassium, urea and other organic solutes from the filtrate.

Question 44.
What are podocyte cells?
Answer:
Glomerulus is lined by a single layer of squamous epithelial cells called podocyte cells.

Question 45.
What is the function of loop of Henle?
Answer:

In the descending loop of Henle reabsorption of water from the filtrate takes place.
Ascending loop of Henle is impermeable to water and only ions diffuse out into the surrounding cells.

Question 46.
What is the function of Distal convoluted tubule?
Answer:
Distal convoluted tubule maintains proper concentration and pH of the urine.

Question 47.
How many stages are involved in formation of urine? What are they?
Answer:
Formation of urine involves four stages. They are

Glomerular filtration,
Tubular reabsorption,
Tubular secretion and
Concentration of urine.

Question 48.
The amount of water reabsorption in the tubule depends on?
Answer:
The amount of water reabsorption in the tubule depends on amount of excess water present in the body and the amount of dissolved wastes to be excreted.

Question 49.
In which region seventy five percent of water content of the nephric filtrate is reabsorbed?
Answer:
Seventy five percent of water content of the nephric filtrate is reabsorbed in the region of proximal convoluted tubule.

Question 50.
Which hormone is responsible to pass concentrated urine?
Answer:
Vasopressin is responsible to pass concentrated urine.

Question 51.
What is micturition?
Answer:
Micturition is the process of discharge of urine from the urinary bladder.

Question 52.
What are the composition of various substances in urine?
Answer:
The composition of various substances in urine are 96% of water, 2.5% of organic substances and 1.5% of inorganic solutes.

Question 53.
How much amount of urine is excreted per day?
Answer:
Total amount of urine excreted per day is about 1.6 -1.8 litres.

Question 54.
What is the storage capacity of urinary bladder?
Answer:
The storage capacity of urinary bladder is 300 – 800 ml.

Question 55.
What is uremia?
Answer:
If kidneys stop working completely, our body is filled with extra water and waste products. This condition is called uremia.

Question 56.
What is haemodialysis?
Answer:
Artificial filtering of blood to remove wastes by using dialysis machine is known as haemodialysis.

Question 57.
What is the time required for each dialysis session?
Answer:
The time required for each dialysis session is 3 to 6 hours.

Question 58.
What are the organs that can be transplanted from brain dead patients?
Answer:
The organs that can be transplanted from brain dead patients are kidney, liver, heart, lungs, pancreas, skin, bone, intestines and eyes (cornea).

Question 59.
Where is the transplanted kidney fixed in the body of kidney failure patient?
Answer:
The transplanted kidney is fixed in the body of patient inside the lower abdomen.

Question 60.
What is cadaver transplantation?
Answer:
The process of transplantation of organs from brain dead patients to another is called cadaver transplantation.

Question 61.
What are the other excretory organs present in human body in addition to kidney?
Answer:
The other excretory organs present in human body in addition to kidney are lungs, skin, liver, large intestine, salivary glands and lacrimal glands.

Question 62.
What are the waste products excreted by lungs?
Answer:
The waste products excreted by lungs are carbon dioxide and water.

Question 63.
Sebum of sebaceous glands in skin contains wastes like?
Answer:
Sebum of sebaceous glands in skin contains wastes like waxes, sterols, hydro carbons and fatty acids.

Question 64.
What are the metabolic wastes of haemoglobin of red blood cells in liver?
Answer:
The metabolic wastes of haemoglobin of red blood cells in liver are bile pigments like bilirubin, biliverdin and urochrome.

Question 65.
How is urea produced in liver?
Answer:
Urea is produced in liver by the deamination of proteins.

Question 66.
What are the wastes excreted by intestine?
Answer:
The wastes excreted by intestine are excess salts of calcium, magnesium and iron.

Question 67.
How do unicellular organisms remove waste products?
Answer:
Unicellular organisms remove waste products by diffusion from the body surface to the surrounding water.

Question 68.
What is the osmoregulatory organelle in amoeba and paramoecium?
Answer:
The osmoregulatory organelle in amoeba and paramoecium is contractile vacuole.

Question 69.
In which animal phyla water bathes almost all their cells in body of organisms?
Answer:
In porifera and coelenterates water bathes almost all their cells in body of organisms.

Question 70.
Through which processes plants get rid of excess water?
Answer:
Through transpiration and guttation plants get rid of excess water.

Question 71.
What are Raphides?
Answer:
Waste gets stored in the fruits in the form of solid bodies called Raphides.

Question 72.
What are alkaloids?
Answer:
The poisonous nitrogenous by products produced by plants are alkaloids.

Question 73.
Which alkaloid is used as antimalarial drug?
Answer:
Quinine is the alkaloid used as antimalarial drug.

Question 74.
Which flowers and fruits are used to extract sedative scopalamine?
Answer:
The sedative scopalamine is extracted from the flowers and fruits of Datura stramonium.

Question 75.
What are the alkaloids produced by plants?
Answer:
The alkaloids produced by plants are nimbin, nicotine, morphine, codeine quinine, reserpine, caffeine/scopolamine, etc.

Question 76.
What are tannins?
Answer:
Tannins are alkaloids which are carbon compounds.

Question 77.
Which group of plants secretes resin?
Answer:
Resin secretes by a group of plants Gymnosperms.

Question 78.
What is latex?
Answer:
Latex is a sticky, milky white substance secreted by plants.

Question 79.
Which plant of latex is used to prepare rubber?
Answer:
Rubber is prepared from the latex of Hevea braziliensis.

Question 80.
Bio-diesel is obtained from the seeds of?
Answer:
Bio-diesel is obtained from the seeds of Jatropa.

Question 81.
Which person’s kidney is used for a successful kidney transplantation?
Answer:
From a close relative kidney can be used for successful transplantation.

Question 82.
What may be the feeling of the patients who are waiting for suitable organ donations due to failure of vital organs?
Answer:
The patients who are waiting may think that any donor with a kind heart can give them life by donating their organs.

Question 83.
What may be the feeling of a doctor while transplanting the organs?
Answer:
Doctor is also a human being, he predicts that the transplantation becomes a success.

Question 84.
How can you appreciate the father of Yaswanth Kumar for donating the organs of his brain dead son? (From annexure)
Answer:
We can appreciate the father of Yaswanth Kumar for his humanity where no one can accept to take or think such a delicate decision of donating the organs of his brain dead son.

Question 85.
What might be the feeling of the father H.V. Shiva Kumar father of Yaswanth Kumar for donating the organs of his son? (From annexure)
Answer:
The father might have thought that his son can live even after his death in some other people who are in need for the suitable organs. We can appreciate father’s humanity.

Question 86.
What does the society need about organ donation?
Answer:
The society needs much awareness in organ donation so that we can jsave many lives who are in need of different organs from donors for their survival.

Question 87.
Which substances are present above the normal limits both in the blood and urine in the sample given?
Answer:
Substances present above the normal limits in the blood are: creatinine, uric acid and cholesterol. Substances present above the normal limits in the urine are creatinine, calcium, uric acid and urea.

Question 88.
When is more urine excreted?
Answer:
A large intake of liquids or water rich food increases the volume of water in the blood, hence more urine is excreted.

Question 89.
What are the uses of Tannins?
Answer:
Tannins are used in tanning of leather and in medicines.

Question 90.
What are the economic importance of gums?
Answer:
Gums are valuable being used as adhesives and binding agents in the preparation of the medicines, food, etc.

Question 91.
Why do we get peculiar smell when you shift the potted plant?
Answer:
Plants release some peculiar chemical substances into their surroundings into soil through roots. These substances attract useful microorganisms to the plant. Hence we get peculiar smell when we shift the potted plant.

Question 92.
What is chewing gum ? How is it made of?
Answer:
Chewing gum is a type of gum for chewing made dates back 5000 years. Modern chewing gum originally made of chicle, natural latex from plant.

Question 93.
How do pollen grains affect our health?
Answer:
Whenever pollen grains enter in our body they cause allergy due to the presence of nitrogenous substances. These allergens cause skin allergy and asthma. Ex: Parthenium.

Question 94.
What are the defensive mechanism developed by plants of your village to protect themselves from the herbivores ? Give two examples.
Answer:
Plants produce nitrogenous compounds to protect themselves from the herbivores. These compounds are called as alkaloids. For example,
Datura – It produces alkaloid known as scopolamine.
Cactus – Develop spines on the body.

10th Class Biology 4th Lesson Excretion 2 Marks Important Questions and Answers

Question 1.
Fill in the table on the basis of your obervations in field trip.

S.No	Name of the plant	Secondary metabolites	Uses
A.	Neem
B.	Datura

Answer:

S.No	Name of the plant	Secondary metabolites	Uses
A.	Neem	Nimbin	Antiseptic
B.	Datura	Scopola mine	Sedative

Question 2.
Two kidneys are present in human beings as excretory organs. Haritha, whose age is 23 years, donated one kidney to her father. Now she has one kidney only. She gave birth to a female child.
a) How many kidneys are there in Haritha’s daughter?
Answer:
Harita’s daughter has 2 kidneys,

b) Support your answer.
Answer:
Bodily changes are not inherited, so Harita’s daughter has 2 kidneys.

Question 3.
What questions do you ask a nephrologist to know more about kidney related diseases?
Answer:

How are stones formed in the kidney?
What is dialysis?
What is the effect of smoking and alcoholism on kidneys?
What is ESRD?

Question 4.
Read the following lines and answer the questions.

Liver: It produces bile pigments (bilirubin, biliverdin and urochrome) which are metabolic wastes of haemoglobin of dead R.B.Cs. Urochrome, which is responsible for the amber colour of the urine, is eliminated through urine. Biliverdin and bilirubin are stored in gall bladder and later excreted through bile along with cholesterol and derivatives of steroid hormones, extra drug, vitamins and alkaline salts. Liver is also involved in urea formation.

i) Which substance adds colour to urine?
ii) The site of synthesis of urea in the body is …………..
Answer:
i) Urochrome
ii) Liver

Question 5.
Look at the picture and answer the questions, (label bottle on left as X and bottle on right as Y)
The above is a procedure of haemodialysis in a hospital.
i) What are the liquids X and Y, in the picture?
ii) What is the purpose of this procedure and for whom is it needed?
Answer:
i) X – Dialyzing fluid; Y – Urine
ii) It is a process of removal of poisonous nitrogenous wastes from blood. This process is needed by uremic or kidney failure patients.

Question 6.
Prepare four questions to find the reasons for obstructions in excretory system.
Answer:

What do we call the complete and irreversible kidney failure?
What happen if kidney stops working completely?
What is uremia?
Is there any solution to this problem?

Question 7.
Name the secondary metabolites which are useful in leather and rubber industry. From which plants we obtain them?
Answer:

Latex used in rubber and Tannins used in Leather Industry.
Latex – Hevea brasiliensis (Rubber Plant) Tannins – Cassia, Acacia.

Question 8.
Prepare four questions you will ask a nephrologist about Kidney failure.
Answer:

When does kidney fail?
What are the symptoms of kidney failure?
What precautions can we take to prevent failure of kidney?
Which alternative method can we adopt if kidney fails?

Question 9.
Observe the following table.

Name of the phylum / organism	Excretory system / organ
Protozoa	Diffusion
Porifera	Water bathes all their cells
Platyhelmirthes	Flame cells
Annelida	Nephridia
Arthropoda	Green glands
Reptiles, Aves and Mammals	Kidney

On the basis of above table, write answers to the following questions.

i) In above table, which living organisms contains kidneys as excretory organs like human beings?
Answer:
Reptiles and Aves.

ii) Write the excretory organs present in Earthworm and Cockroach.
Answer:
Earthworm – Nephridia Cockroach – Green glands

Question 10.
Observe the following table and answer the questions given below.

Alkaloid	Name of the plant	Uses
Quinine	Cinchona	Anti malarial drug
Nicotine	Tobacco	Pesticides
Morphine	Opium	Pain killer
Pyrethroids	Chrysanthemum species	Insecticides

i) Which alkaloid we get from the fruit, is used as pain killer?
Answer:
Morphine

ii) From which part of the plant do we get Quinine?
Answer:
Bark of Cinchona

Question 11.
What is called structural and functional unit of kidney? Why?
Answer:

Each kidney is made up of more than one million microscopic and thin tubular units called nephrons or uriniferous tubules. Hence nephron is known as structural and functional unit of kidney.
Nephron’s chief function is to regulate the concentration of water and soluble substances, reabsorbing what is needed and excreting the rest as urine. Hence it is known as functional unit of kidney.

Question 12.
Blood is filtered in Bowman’s capsule of nephron. For the filtration of blood some pressure is needed. How does the pressure arises in Bowman’s capsule?
Answer:

Blood flows inside the glomerulus of Bowman’s capsule under the influence of pressure due to the large diameter of afferent arteriole.
This increases or rise the blood pressure in the glomerulus capillaries leading to ultrafiltration of the blood in the Bowman’s capsule.

Question 13.
Classify the substances given below.
Ptyaline, Leptin, Morphine, Riboflavin, Testosterone, Thyamin, Niacine, Sucrase, Nicotine, Amylase, Retinol, Quinine, Calciferol, Adrenaline, Tripsin.
Answer:
The above substances can be classified into Enzymes, Hormones, Alkaloids and Vitamins.
Enzymes: Ptyaline, Sucrase, Amylase, Tripsin
Hormones: Testosterone, Adrenaline, Leptin
Alkaloids: Morphine, Nicotine, Quinine
Vitamins: Riboflavin, Thyamin, Niacine, Retinol, Calciferol

Question 14.
How does excretion take place in phylum protozoa?
Answer:

Specific excretory organs are absent in unicellular organisms. These organisms remove waste products by simple diffusion from the body surface into the surrounding water.
Excess water is sent out through contractile vacuoles in fresh water protozoans. Ex: Amoeba surface.
The major portion of excess water is eliminated by diffusion through body surface.

Question 15.
What are the functions of excretory system?
Answer:

Excretory system in animals performs more than one function.
The primary function of excretory system is to maintain ionic and osmotic balance in the animal body.
Excretory system helps to maintain appropriate concentration of salts and correct amount of water in the body.
It also helps in maintaining the body volume.
The secondary function of excretory system is excretion of nitrogenous wastes and foreign substances such as drugs.

Question 16.
Mention the groups of animals where excretion fakes place through diffusion from body surface.
Answer:

In protozoans wastes are excreted by diffusion through body surface.
Contractile vacuoles present in some protozoans are mainly concerned with the maintenance of osmotic and ionic balance. Excretion is mainly by diffusion.
In sponges, coelenterates like Hydra, excretion takes place by diffusion.

Question 17.
People in cold countries get very less / no sweat. What changes occur in their skin and in other excretory organs?
Answer:

Usually human skin has a unique system keep in equilibrium state with the surrounding temperature.
During winter season the temperature is low and the body temperature should be heated to balance the tolerance.
So the water content in our body is sent out in the form of urine.
Hence, people in cold countries get less sweat or no sweat.

10th Class Biology 4th Lesson Excretion 4 Marks Important Questions and Answers

Question 1.
Analyse the following information and answer the questions.

Alkaloid	Part of the plant	Uses
Quinine	Bark	Anti-malarial drug
Pyrethroids	Leaves	Insecticide
Reserpine	Roots	Medicine for snake bite
Caffeine	Seeds	Central nervous system stimulant
Nimbin	Seeds, Barks, Leaves	Antiseptic

i) Name the alkaloid which is used to cure malaria.
Answer:
Quininie

ii) Name the alkaloids used as insecticides.
Answer:
Pyrethroids

iii) Which system is stimulated by the alkaloid caffeine?
Answer:
Central nervous system

iv) Which parts of which plant is used as medicine for snake bite?
Answer:
The roots of Rauwolfia Serpentina (Snake root)

Question 2.
Which diagram do you draw to label these parts?
Answer:

Bowman’s capsule
Uriniferous tubule.
Collecting tubule.

Draw the diagram and label the parts.
Answer:

Question 3.
Fill the following table related to Secondary Metabolites in plants.

S.No.	Metabolite	Name of the plant	Use
1.		Pinus
2.	Gums
3.			Rubber preparation
4.	Nimbin

Answer:

S.No.	Metabolite	Name of the plant	Use
1.	Resin	Pinus	Used in varnishes
2.	Gums	Neem, Acacia	Used as adhesives and binding agents
3.	Latex	Hevea braziliensis (Rubber plant)	Rubber preparation
4.	Nimbin	Azadirachta indica (Neem)	Antiseptic

Question 4.
The given parts belong to which system? Draw a neat labelled diagram of the system.
a) Kidneys b) Ureters c) Urinary bladder
Answer:
The given parts belong to human excretory system.

Question 5.
Read the following passage and answer the questions.

Plants can get rid of excess water by a process like transpiration and guttation. Waste products may be stored in leaves, bark, and fruits. When these dead leaves, bark, and ripe fruits fall off from the tree then waste products in them are removed. Waste gets stored in the fruits in the form of solid bodies called ‘raphides’. However several compounds are synthesized by the plants for their own use especially for defence. Many plants synthesize chemicals and store them in roots, leaves, seeds, etc. for protection against herbivores. Most of the chemicals are unpleasant to taste. Hence herbivores usually do not prefer to eat such plants. Some of the chemicals are toxic and may even kill the animals that eat them.

i) What are raphides?
ii) How do plants protect themselves against herbivores?
iii) How do plants get rid of wastes?
iv) Name the processes by which plants lose excess of water.
Answer:
i) The wastes stored in some fruits in the form of solid bodies are called “raphides”.
ii) Many plants synthesize chemicals and store them in roots, leaves, seeds, etc. for protection against herbivores.
iii) Waste products may be stored in leaves, bark and fruits. When these dead leaves, bark and ripe fruits fall off from the tree. Then waste products in them are removed.
iv) Plants lose excess of water by transpiration and guttation.

Question 6.

S.No.	Test	Present level	Normal range
A. Blood Test
1.	Blood Pressure (BP)	160/90 mm/Hg	120/80 mm/Hg
2.	Glucose (Before food)	120- mg/dl	60-100 mg/dl
3.	Glucose (After food)	220 mg/dl	160-180 mg/dl
4.	Bilirubin	1.0 mg/dl	0.1-0.8 mg/dl
	B. Urine Test
1.	24 hrs protein	150 mg/d ay	100 mg/day
2.	Sodium	140 mmol/L	125-250 mmol/L.

A) Which test is required to know bilirubin?
Answer:
Blood test is required to know bilirubin.

B) How is the sugar disease confirmed?
Answer:
In blood test if the glucose levels in blood before and after food are more than the normal, the sugar disease is confirmed.

C) By observing the above report, what would be the other problems faced by that patient?
Answer:
The blood pressure of the patient 160/90 as the normal is 120/80 mm/Hg.
The Glucose levels in the blood of the patient shows more than the normal before and after food.
So he is suffering from Hypertension and diabetes.

D) What are the organs affected by these problems?
The organsaffected by these problems are heart and kidneys.

Question 7.
Explain the temporary and permanent methods to be adopted for Kidney failure (ESRD) persons.
Answer:
Temporary method for ESRD persons is Dialysis (Artificial kidney). Permanent method is transplantation of kidney.

Dialysis:

Blood is taken out from the main artery, mixed with an anticoagulant, such as heparin and then pumped in to dialyzer.
In dialyzer blood flows through cellophane tubes and these tubes are embeded in the dialysing fluid.
The membrane separates the blood flowing inside the tube and dialysing fluid (same as plasma without nitrogenous waste).

Kidney transplantation:

A functioning kidney is used in transplantation from a donor preferably a close relative.
The kidney that is received by a recipient must be a good match to his body, to minimise the chances of rejection by the immune system of the recipient.
Nowadays the process of organ donation helps a lot for kidney failure patients.

Question 8.
What are the accessory excretory organs in human body? How does the liver carry out excretion as a secondary function? (OR)
Write about the accessory excretory organs and their excretory substances in human beings.
Answer:

Accessory Excretory Organs	Excretory Materials
Lungs	CO₂ and Water
Skin	Sweat and Metabolic Waste (Sebum)
Liver	Bilirubin, Biliverdin, Urochrome
Large Intestine (Colon)	Excess of salts of calcium, magnesium and iron are excreted along with faecal matter.

Excretory functions of Liver:

Liver acts as detoxification centre of our body. Liver produces bile juice which contain bile pigments which are metabolic wastes of dead R.B.C.
Bilirubin, Biliverdin and urochrome are the pigments of bile juice.
Bilirubin, Biliverdin, cholesterol and derivatives of steroids, extra drugs, vitamins and alkaline salts are the wastes produced by liver.
Urochrome is eliminated through urine.
Liver also plays a role in the formation of concentrated urine.

Question 9.
Explain the formation of urine in a flow chart.
Answer:
Dissolved substances of blood like urea, glucose, amino acids, minerals, salts etc., are filtered out in Bowman’s capsule under high filtration pressure.
Glomerular filtrate is called primary urine.

Question 10.
Excreting wastes from the human body not only by kidneys but also by other organs helps you. How do you support it?
Answer:

In human body wastes are excreted not only by kidneys but also by other organs.
Kidney filters blood and eliminates nitrogenous wastes and other harmful things. Filters urea from blood.
Apart from kidney lungs, skin, liver, intestine, salivary glands and lacrymal glands.
Lungs remove carbon dioxide and water in respiration.
Skin excrete wastes in the form of sweat which contains water and certain salts.
Liver eliminates bile pigments bilirubin and biliverdin through urine.
Excess salts of calcium, magnesium and iron are excreted by epithellial cells of colon for elimination along with faeces by intestine.
Eccrine glands present on the forehead, the bottoms of the feet and the palms allow excess water to leave the body.
Salivary glands and lacrimal glands excrete small amounts of nitrogenous waste through saliva and tears.

Question 11.
Which plants can you get in your village? Among these by-products of which plants do you use in your real life?
Answer:

The plants grow or available in our village are Sapota, Coconut, Cassia, Mango, Guava, Borassus plantain, Tobacco, Rauwolfia, Coffee, Neem, Datura, Chrysanthe¬mum, Acacia, Pinus, Vallisneria, Teak, etc.,
Out of these plants alkaloids are available from the plants like Tobacco, Rauwolfia, Coffee, Neem, Datura and Chrysanthemum. The by-products from these plants are utilised in my real life.
Plant – Use
Tobacco – Insecticide
Rauwolfia serpentina – Medicine for snake bite
Coffee – Central nervous system stimulant
Neem – Antiseptic
Datura – Sedative
Chrysanthemum – Insecticides
Tannins are the by-products of cassia, acacia. These are used in tanning of leather and in medicines.
Resin the by-product of Pinus is used in varnishes.
Gums are extracted from neem and acacia.
They are used as adhesives and binding agents in the preparation of food, medicines.

Question 12.
Write an essay stating the advantages of by-products of plants in our real life.
(OR)
What are secondary metabolites? Briefly explain their uses.
(OR)
Not only the food of plants but also their wastes are useful to us. What evidences do you give for it?
Answer:

The materials which do not require for normal growth and development are called secondary metabolites. These are the by – products of plants, eg: Alkaloids, Tannins, Resins, Gums and Latex, etc. Though plants produce these chemicals for their own use man found the usage of these chemicals for own benefits. They are generally coloured and fragrant.

Alkaloids: These are nitrogenous by-products and poisonous. These are stored in different parts of the plants. Common alkaloids in plants and their uses are given in the table.

Alkaloid	Plant	Part	Uses
Quinine	Cinchona officinalis (Cinchona)	Bark	Antimalarial drug
Nicotine	Nicotiana tobacum (Tobacco)	Leaves	Insecticide
Morphine, Cocaine	Papaver somniferum (Opium)	Fruit	Pain killer
Reserpine	Rauwolfia serpentiana (Snake bite)	Root	Medicine for snake bite
Caffeine	Coffea Arabica (Coffee plant)	Seed	Central nervous system stimulant
Nimbin	Azadirachta indica (Neem)	Seeds, Barks, Leaves	Antiseptic
Scopolamine	Datura stramonium	Fruit, flower	Sedative
Pyrethroids	Chrysanthemum sps	Flower	Insecticides

Tannins: Tannins are carbon compounds. These are stored in different parts of the plant and are deep brown in colour. Tannins are used in tanning of leather and in medicines, e.g. Cassia, Acacia.
Resin: Occur mostly in Gymnosperms in specialized passages called resin passages. These are used in varnishes, e.g. Pinus.
Gums: Plants like Neem, Acacia oozes out a sticky substance called gum. When branches are cut. The gum swells by absorbing water and helps in the healing of damaged parts of a plant. Gums are economically valuable and used as adhesives and binding agents in the preparation of the medicines, food, etc.
Latex: Latex is a sticky, milky white substance secreted by plants. Latex is stored in latex cells or latex vessels. From the latex of Hevea braziliensis (Rubber plant) rubber is prepared. Latex from Jatropa is the source of bio-diesel.
Modern chewing gum originally made of chick natural latex from plant.

Question 13.
Blood is purified in kidneys. So many wastes are removed from the blood in nephron of the kidney. Which issue make you surprise in excretory system?
Answer:

Kidneys remove nitrogenous waste from our body. They also work towards balancing the amount of vitamins , minerals, fat and protein that are found in the blood. They do this so that our body can easily perform day to day activities.
Our intestine makes solid waste materials and is excreted through digestive tract.
Each day our body eliminates around 1.6 to 1.8 liters of urine which contains liquids, minerals and vitamins that are of no use to the body.
The bladder of a human body is nearly the same size as the average of human brain.
In one individuals life span the liver can produce around 184.275 kgs of bile (6500 ounces).
In our lifetime an individual could urinate close to 7,850,000,000 gallons of fluid.
A really extraordinary fact regarding the excretory system is that upto 400 ml of urine can be held in human bladder.
Urine contains a high amount of urea which can be used by plants as a source of nitrogen. Because of this diluted urine can be used in gardens and potted plants.
It is amazing to see that each kidney is made up of approximately more than one million microscopic tubular functional units called nephrons or uriniferous tubules.

Question 14.
What is the structural and functional unit of the excretory system? Draw a neat labelled diagram of that unit.
Answer:
Nephron is the structural and functional unit of the kidney.

Question 15.
Give an account of excretory system found In different phyla of animal kingdom.
Answer:

Different organisms have various excretory systems and organs.

Following are the excretory system found in various organisms.

Name of the phylum / organism	Excretory system / organ
Protozoa	Simple diffusion from the body surface into the surrounding water.
Porifera and coelenterates	Water bathes almost all their cells
Platyhelminthes	Flame cells
Nematoda	Renette cells
Annelids	Nephridia
Arthropoda	Green glands, Maiphigian tubules
Mollusca	Meta nephridia
Echinodermata	Water vascular system
Reptiles, Birds and Mammals	Kidneys

Question 16.
How do plants manage/send out waste products from their body?
Answer:

Plants can get rid of excess water by a process like transpiration and guttation.
Waste products may be stored in leaves, bark and fruits.
When these dead leaves, bark and ripe fruits fall off from the tree then waste products in them are got rid of.
Plants get rid of carbon dioxide and oxygen through diffusion.
Plants release some waste products through roots also.
Some waste products are deposited near bark as resins or gums.
In many plants waste products are stored in vacuoles of the cells. Plant cells have comparatively large vacuoles.

Question 17.
In recent days many people are coming forward to donate organs of brain dead people, who met with accidents. How will you appreciate the family members of organ donor?
Answer:

In recent days many families are willing to donate organs of brain dead person’s. This is truely significant change in attitude of people.
Most of the people are burning or burying body after death. Very a few people are would like to see their very dear one’s in other people by donating organs like heart, liver, kidneys, cornea, spleen and bone marrow etc.
The family members of brain dead are already in great sorrow. They need great courage and so much kindness toward needy patients.
With nobel decision of that family giving new life to 5 – 7 persons, who has no alternate treatment, other than organ transplantation. In society every one must appreciate their courage, kindness and sympathy. They stood role model to others.

Question 18.
How is the amount of urine produced regulated?
Answer:

The amount of urine produced largely depends on the amount of water reabsorbed in the renal tubule.
The amount of water reabsorbed by the renal tubule largely depends on
the amount of water present in excess in the body need to be removed. If water is abundantly present in the body tissues large quantities of dilute urine is send out of the body. When water is less in quantities in the body tissues a small quantity of concentrated urine is excreted.
When there is more quantity of dissolved wastes in the body more quantity of water is required to excrete them. So the amount of urine produced increases.
Deficiency of vasopressin causes excessive, repeated, dilute urination called diabetes insipidus.

Question 19.
Observe the below flow chart. Fill the boxes. Explain to which system this belongs to.

Answer:

Kidney
Loop of Henle
Pelvis
Urethra

This flow chart belongs to excretory system in human beings. It shows the way how the blood in the kidney moves filtered and urine is excreted.

Question 20.
Explain the external features of kidney in human beings.
Answer:

In human beings, there are a pair of bean-shaped, reddish-brown structures in the abdominal cavity attached to dorsal body wall one on either side of the backbone.
The size of the kidney is 10cm in length, 5 – 6 cm in breadth, and 4cm in thickness.
Each kidney is convex on the outer side and concave on the inner side.
The inner side of each kidney has a fissure or hilus for the entry of a renal artery, exit of a renal vein and an ureter.
Renal artery brings oxygenated blood loaded with waste products and renal vein carries deoxygenated blood.

Question 21.
Describe the excretory system of man.
Answer:

The excretory system of man consists of
a) a pair of kidneys
b) ureters and
c) urinary bladder and d) urethra.
Kidneys are bean shaped and are located in the abdominal region on either side of vertebral column.
From hilus of each kidney there are a pair of whitish, narrow tubular structures arise. They are known as ureters.
The ureter travels downwards and open, in the sac like structure called the urinary bladder, which stores urine.
Urethra is a tube that takes urine from urinary bladder to outside.
The opening of urinary bladder into urethra is guarded by a ring of muscles or sphincter.

Question 22.
Describe the internal structure of kidney with the help of diagrams. (OR) Describe the structure of nephron with the help of a diagram.
Answer:

L.S. of kidney show two distinct regions. Dark coloured outer zone called cortex and pale coloured inner zone called medulla.
Each kidney is made up of approximately more than one million microscopic and thin tubular functional units called nephrons or uniferous tubules.
Each nephron has basically two parts. One is malphigian body and other is renal tubule.
Malphigian body consists of a blind cup shaped broader end of nephron called Bowman s capsule and bunch of fine blood capillaries called glomerulus.
Glomerulus develops from afferent arteriole and it gives to rise to an efferent arteriole.
Glomerulus functions as a filtration unit.
Renal tubule has three parts. They are
1. Proximal Convoluted Tubule (PCT)
2. Loop of Henle which is ‘U’ shaped and
3. Distal Convoluted Tubule (DCT).
In tubule part reabsorption and secretion takes place. Urine is formed in the renal tubule part of nephron.
Distal convoluted tubules open into a collecting tube.
Collecting tube forms pyramids and calyces which open into the pelvis.
Pelvis leads into the ureter.

Question 23.
Describe the structure of renal tubule with neatly labelled diagram.
Answer:

Renal tubule is a specialised tubular structure made up of proximal convoluted tubule, a ‘U’ shaped tube called loop of Henle, and distal convoluted tubule.
The three tubular components are selectively permeable and only allow specific molecules to pass through them.
The renal tubule is surrounded by capillaries called peritubular capillaries that arise from the efferent arterioles.
The substances essential for the body are reabsorbed from the tubules into the peritubular capillaries and the unwanted or toxic molecules are secreted into the lumen of the renal tubule.