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AP State Board Syllabus AP SSC 10th Class Telugu Solutions 10th Class Telugu Grammar Chandassu ఛందస్సు Notes, Questions and Answers.

AP State Syllabus SSC 10th Class Telugu Grammar Chandassu ఛందస్సు

కవులు పద్యాలను, గేయాలను కొన్ని నియమాలకు లోబడి రాస్తారు. అందువల్లనే అవి రాగంతో పాడుకోడానికి వీలుగా ఉంటాయి.

1) లఘువు :
రెప్పపాటు కాలంలో లేదా చిటికె వేసే కాలంలో ఉచ్చరించే అక్షరాలు “లఘువులు.” ఇవి హ్రస్వాక్షరాలుగా మనం పిలుచుకొనే అక్షరాలు.

2) గురువు :
లఘువు ఉచ్చరించే సమయం కంటె, ఎక్కువ సమయం అవసరమయ్యే అక్షరాలు “గురువులు.”

గురులఘువుల గుర్తులు

లఘువు అని తెలుపడానికి గుర్తు : I ‘ల’
గురువు అని తెలుపడానికి గుర్తు : U ‘గ’

గురులఘువుల నిర్ణయం

ఎ) గురువుల లక్షణాలు – వాటిని గుర్తించే విధానము.

బి) లఘువుల లక్షణాలు – వాటిని గుర్తించే విధం:

గమనిక : గురువులు కాని అక్షరాలన్నీ లఘువులు

1) ఋ కారంతో కూడిన అక్షరం సంయుక్తాక్షరం కాదు. ‘ఋ’ అనేది అచ్చు. అందువల్ల అది లఘువు. దానికి ముందు అక్షరం కూడా లఘువే.

2) సంయుక్తాక్షరానికి ముందున్న అక్షరాన్ని ఊది పలికితేనే, అది గురువు అవుతుంది. లేకపోతే లఘువు అవుతుంది.

3) ఏకపదంలోనూ, సమాసంలోనూ సంయుక్తాక్షరానికి ముందున్న అక్షరం ఊది పలకబడుతుంది. కాబట్టి అది గురువు అవుతుంది.

గణ విభజన

1) ఒకే అక్షరం గణాలు : ఒకే అక్షరం గణంగా ఏర్పడితే, అది ఏకాక్షర గణం. ఈ గణంలో ఒక గురువు లేదా ఒక లఘువు ఒక్కొక్కటే గణంగా ఉంటాయి.

2) రెండక్షరాల గణాలు : రెండేసి అక్షరాలు కలిసి గణాలుగా ఏర్పడతాయి. వీటిలోనూ గురువులు, లఘువులు ఉంటాయి. ఇవి నాలుగు రకాలు.

అభ్యాసము : రెండక్షరాల గణాలు నాలుగు రకాలు ఉన్నాయి కదా ! ఒక్కొక్క దానికి 4 పదాల చొప్పున రాయండి.

మూడక్షరాల గణాలు

మూడక్షరాల గణాలు మొత్తం ఎనిమిది (8).

అ) మూడక్షరాల గణాలను గుర్తించే సులభ మార్గము :

య, మా, తా, రా, జ, భా, న, స అనే సూత్రాన్ని కంఠస్థం చేసి, పై విధంగా ఒక చక్రం గీయండి. మీకు కావలసిన గణం పేరు గల మొదటి అక్షరం ఎక్కడ ఉందో గుర్తించండి. ఆ అక్షరాన్నుండి కుడిగా ఉన్న మూడక్షరాలలోనూ గురు లఘువులు ఏ క్రమంగా ఉన్నాయో, మీకు కావలసిన గణానికి గురు లఘువులు ఆ క్రమంలో ఉంటాయి.

ఉదా : మీకు ‘య’ గణం యొక్క గురు లఘువుల క్రమం కావాలి అనుకోండి. అపుడు ‘య’ నుండి కుడివైపుగా ‘యమాతా’ అనే మూడక్షరాలను వేరుగా వ్రాయండి.

ఆ) మూడక్షరాల గణముల నిర్ణయంలో మరో పద్ధతి :

అని వ్రాసుకొని, దానికి గురు లఘువులు గుర్తించండి. మీకు కావలసిన గణం పేరు గల అక్షరంతో, ప్రక్క రెండు అక్షరాలూ కలిపి, దానిలోని గురు లఘువులు ఎలా ఉన్నాయో గమనిస్తే, ఏ గణానికి ఏ అక్షరాలు ఉంటాయో తెలుస్తుంది.
ఉదా :
1) య గణం = యమాతా = I U U = ఆది లఘువు
2) మ గణం = మాతారా = U U U = సర్వ గురువు
3) త గణం= తారాజ = U U I = అంత్య లఘువు
4) ర గణం = రాజభా = U I U = మధ్య లఘువు
5) జ గణం = జభాన = I U I = భాన మధ్య గురువు
6) భ గణం = భానస = U I I = ఆది గురువు
7) న గణం = నసల = I I I = సర్వ లఘువులు
8) లగము (లేక ‘వ’ గణం) = I U (లఘువు, గురువు)

నాలుగు అక్షరాల గణాలు

సూర్య గణాలు – ఇంద్ర గణాలు

యతి – ప్రాసలు

I. గమనిక : పద్యానికి సౌందర్యాన్ని చేకూర్చేవి యతి, ప్రాసలు.
1. యతి : పద్యపాదంలోని మొదటి అక్షరాన్ని ‘యతి’ అంటారు.
2. ప్రాస : పద్యపాదంలోని రెండవ అక్షరాన్ని ‘ప్రాస’ అంటారు.

II. గమనిక : నియమం చెప్పినచోట ‘యతి’, ‘ప్రాస’లు ప్రయోగించడం వల్ల చదవడానికి, వినడానికీ, జ్ఞాపకం పెట్టుకోడానికీ — సౌకర్యం కలుగుతుంది.

3. యతిమైత్రి :
పద్యపాదం యొక్క మొదటి అక్షరంతో, ఆ పద్యంలో నిర్ణయింపబడిన స్థానమందలి అక్షరం మైత్రి కలిగి ఉండడాన్ని యతిమైత్రి అంటారు. యతిమైత్రి యతిస్థానంలోని హల్లుకేకాక, అచ్చుతో కూడా మైత్రి ఉండాలి.

1. ఉత్పలమాల
కింది పద్యపాదాన్ని పరిశీలించండి.

గమనిక :
పై పాదాల్లో ‘భ, ర, న, భ, భ, ర, వ’ అనే గణాలు వరుసగా వచ్చాయి. ఇలా పద్యంలో నాలుగు పాదాల్లోనూ ఒకే రకమైన గణాలు ఒకే వరుసలో ఉన్న పద్యాన్ని ‘వృత్త పద్యం’ అంటారు.

యతి :
పద్య పాదంలో మొదటి అక్షరాన్ని ‘యతి’ అంటారు. ఈ యతి అక్షరం గానీ, దాని వర్ణమైత్రి అక్షరం గానీ ఆ పాదంలో మరొకచోట రావడాన్ని ‘యతిమైత్రి’ లేదా యతిస్థానం అంటారు.

పై పద్యపాదాల్లో పదవ అక్షరం (ఆ – అని జే – సి) యతి.

ప్రాస :
పై పద్యపాదాల్లో రెండవ అక్షరంగా నాలుగు పాదాల్లోనూ ‘య’ అనే అక్షరం వచ్చింది. ఈ పద్యాలలో రెండవ అక్షరంగా ఒకే గుణింతాక్షరం రావడాన్ని ‘ప్రాస’ నియమం అంటారు. పై లక్షణాలు గల పద్యాన్ని ‘ఉత్పలమాల’ పద్యం అంటారు.

ఉత్పలమాల పద్య లక్షణాలు :

1. ఇది వృత్తపద్యం.
2. ఈ పద్యంలో నాలుగు పాదాలుంటాయి.
3. ప్రతి పాదంలోనూ భ, ర, న, భ, భ, ర, వ అనే గణాలుంటాయి.
4. ప్రాస నియమం ఉంటుంది.
5. ప్రతి పాదంలోనూ 20 అక్షరాలుంటాయి.

2. చంపకమాల
కింది పద్య పాదాన్ని పరిశీలించండి.

చంపకమాల పద్య లక్షణాలు :

1. ఇది వృత్తపద్యం.
2. ఈ పద్యంలో నాలుగు పాదాలుంటాయి.
3. ప్రతి పాదంలో ‘న, జ, భ, జ, జ, జ, ర’ అనే గణాలుంటాయి.
4. ప్రతి పాదంలో 11వ అక్షరం యతిస్థానం (ఈ పాదంలో అ – య).
5. ప్రాస నియమం ఉంటుంది.
6. ప్రతి పాదంలోనూ 21 అక్షరాలుంటాయి.

3. శార్దూలం
కింది పద్యపాదాన్ని పరిశీలించండి.

శార్దూల పద్య లక్షణాలు :

1. ఇది వృత్తపద్యం.
2. ఈ పద్యంలో నాలుగు పాదాలుంటాయి.
3. ప్రతి పాదంలోనూ ‘మ, స, జ, స, త, త, గ’ అనే గణాలుంటాయి.
4. ప్రతి పాదంలో 13వ అక్షరం యతిస్థానం (ఈ పాదంలో ఆ – యం).
5. ప్రాస నియమం ఉంటుంది.
6. ప్రతి పాదంలోనూ 19 అక్షరాలుంటాయి.

4. మత్తేభం:
కింది పద్య పాదాన్ని పరిశీలించండి.

మత్తేభ పద్య లక్షణాలు :

1. ఇది వృత్తపద్యం.
2. ఈ పద్యంలో నాలుగు పాదాలుంటాయి.
3. ప్రతి పాదంలోనూ ‘స, భ, ర, న, మ, య, వ’ అనే గణాలుంటాయి.
4. ప్రతి పాదంలో 14వ అక్షరం యతిస్థానం (ఈ పాదంలో ప – పా).
5. ప్రాస నియమం ఉంటుంది.
6. ప్రతి పాదంలోనూ 20 అక్షరాలుంటాయి.

5. తేటగీతి

తేటగీతి పద్య లక్షణాలు :

1. ఇది ‘ఉపజాతి’ పద్యం.
2. ఈ పద్యానికి నాలుగు పాదాలుంటాయి.
3. ప్రతి పాదానికి ఒక సూర్యగణం, రెండు ఇంద్రగణాలు, రెండు సూర్యగణాలు వరుసగా ఉంటాయి.
4. నాలుగో గణం మొదటి అక్షరం యతి స్థానం.
5. ప్రాస యతి చెల్లుతుంది.
6. ప్రాస నియమం లేదు.

ఉదా :

పై పద్యంలో 1 సూర్య గణం, 2 ఇంద్ర గణాలు, 2 సూర్య గణాలు వరుసగా వచ్చాయి కాబట్టి ఇది తేటగీతి పద్యపాదం.
ఇక్కడ ‘ప్రాసయతి’ వాడబడింది.

6. ఆటవెలది

ఆటవెలది పద్య లక్షణాలు :
1) ఇది ‘ఉపజాతి’ పద్యం.
2) ఈ పద్యానికి నాలుగు పాదాలు ఉంటాయి.
3) 1, 3 పాదాల్లో వరుసగా మూడు సూర్య గణాలు, రెండు ఇంద్ర గణాల చొప్పున ఉంటాయి.
4) 2, 4 పాదాల్లో ఐదూ సూర్య గణాలే ఉంటాయి.
5) ప్రతి పాదంలోనూ నాల్గవ గణంలోని మొదటి అక్షరం యతి. యతిలేనిచోట ప్రాసయతి చెల్లుతుంది.
6) ప్రాస నియమం పాటింపనవసరం లేదు.
ఉదా :

7. సీసం : (ఉపజాతి పద్యాలంటే తేటగీతి, ఆటవెలది, సీసం)
సీసము పద్య లక్షణాలు :

1. సీస పద్యంలో నాలు పెద్ద పాదాలు ఉంటాయి. ఈ పెద్ద పాదం రెండు భాగాలుగా ఉంటుంది. ప్రతి భాగంలోనూ నాలుగేసి గణాల చొప్పున, ఒక్కొక్క పెద్ద పాదంలో ఎనిమిది గణాలు ఉంటాయి. ఈ 8 గణాల్లో మొదట ఆరు ఇంద్ర గణాలు, చివర రెండు సూర్య గణాలు ఉంటాయి.
2. సీస పద్యపాదంలోని రెండు భాగాల్లోనూ, ప్రతి భాగంలోనూ మూడవ గణం మొదటి అక్షరంతో యతిమైత్రి ఉండాలి. యతిలేని చోట ప్రాసయతి ఉండవచ్చు.
3. సీస పద్యంలో నాల్గు పెద్ద పాదాల తరువాత ఒక తేటగీతి గాని, ఆటవెలది గాని చేర్చాలి.

ఉదా :

8. ద్విపద
ద్విపద పద్య లక్షణాలు:

1. ‘ద్విపద’ పద్యంలో రెండు పాదాలు ఉంటాయి.
2. ప్రతి పాదంలోనూ వరుసగా మూడు ఇంద్ర గణాలు, ఒక సూర్య గణం ఉంటాయి.
3. మూడవ గణం మొదటి అక్షరంతో యతి.
4. ప్రాసయతి చెల్లుతుంది.
5. ప్రాస నియమం పాటింపబడుతుంది.

గమనిక :
ప్రాస నియమం లేని ద్విపదను ‘మంజరీ ద్విపద’ అంటారు.
ఉదా :

9. కందం
కందం పద్య లక్షణాలు :

1. ఈ ‘కందం’ పద్యములో గగ, భ, జ, స, నల అనే గణాలు ఉంటాయి.
2. మొదటి పాదం ‘లఘువు’తో మొదలయితే అన్ని పాదాల్లోనూ మొదటి అక్షరం లఘువుగానే ఉండాలి. మొదటి . పాదము ‘గురువు’తో మొదలయితే, అన్ని పాదాల్లోనూ మొదటి అక్షరం గురువుగానే ఉండాలి. 3) రెండవ, నాల్గవ పాదాల్లోని చివరి అక్షరం గురువుగా ఉండాలి.
3. 1, 2 పాదాలలో (3 + 5) 8 గణాలు; 3, 4 పాదాల్లో (3 + 5) = 8 గణాలు ఉంటాయి.
4. 1, 2 పాదాలు, 3, 4 పాదాలు కలిసిన మొత్తం 8 గణాల్లో 6వ గణం “నలము” గాని ‘జగణం’ కాని కావాలి.
5. బేసి గణం జగణం ఉండరాదు.
6. ప్రాస నియమం ఉండాలి.

ఉదా :

వృత్తాలు – లక్షణాలు – సులభంగా గుర్తు పట్టడం

పద్యపాదాలను గుర్తించి, గణ విభజన చేయడం

ఈ క్రింది పద్యపాదాలకు గణ విభజన చేసి, అవి ఏ పద్యపాదాలో గుర్తించి, యతిని పేర్కొనండి.
1) వెన్నెల వెల్లి పాల్కడలి వ్రేక దనంబున బేర్చి దిక్కులన్
జవాబు:

గమనిక : పై పద్యపాదంలో “భ, ర, న, భ, భ, ర, వ” అనే గణాలు వరుసగా వచ్చాయి. కాబట్టి ఇది ‘ఉత్పలమాల’ పద్య పాదం. యతిస్థానం 10వ అక్షరం (వె – వే).

2) దెసలను కొమ్మ లొయ్యనతి దీర్ఘములైన కరంబులన్ బ్రియం
జవాబు:

పై పద్యపాదంలో “న, జ, భ, జ, జ, జ, ర” అనే గణాలు వరుసగా వచ్చాయి. కాబట్టి ఇది చంపకమాల పద్యపాదం. యతి స్థానం 11వ అక్షరం (దె – దీ)

3) ఆ యేమీ యొక రాణి వాసమును బుణ్యవాసమున్ దెచ్చినా
జవాబు:

పై పద్యపాదంలో మ, స, జ, స, త, త, గ అనే గణాలు వరుసగా వచ్చాయి. కాబట్టి ఇది శార్దూల పద్యపాదం.. యతి స్థానము 13వ అక్షరం (ఆ – ణ్యా)

4) శివరాజంతట మేల్ము సుంగుఁదెరలో స్నిగ్దాంబుద చ్చాయలో
జవాబు:

పై పద్యపాదంలో “స, భ, ర, న, మ, య, వ” అనే గణాలు వరుసగా వచ్చాయి. కాబట్టి ఇది మత్తేభ పద్యపాదం. యతి స్థానం 14వ అక్షరం (శి – స్ని).

ఛందస్సుపై ప్రశ్నలు

1) ‘భ, ర, న, భ, భ, ర, వ’ అనే గణాలు, ఏ పద్యానికి చెందినవి ?
A) ఉత్పలమాల
B) చంపకమాల
C) శార్దూలము
D) మత్తేభము
జవాబు:
A) ఉత్పలమాల

2) ‘న జ భ జ జ జ ర’ గణాలు ఏ పద్యానికి చెందినవి?
A) శార్దూలము
B) మత్తేభము
C) ఉత్పలమాల
D) చంపకమాల
జవాబు:
D) చంపకమాల

3) ‘మసజసతతగ’ గణాలు ఏ పద్యానికి చెందినవి?
A) చంపకమాల
B) ఉత్పలమాల
C) మత్తేభం
D) శార్దూలం
జవాబు:
D) శార్దూలం

4) ‘సభరనమయవ’ గణాలు ఏ పద్యానికి చెందినవి?
A) ఉత్పలమాల
B) మత్తేభము
C) శార్దూలము
D) చంపకమాల
జవాబు:
B) మత్తేభము

5) 14వ అక్షరంతో యతి గల పద్యము
A) ఉత్పలమాల
B) చంపకమాల
C) శార్దూలము
D) మత్తేభము
జవాబు:
D) మత్తేభము

6) ఉత్పలమాల పద్యానికి ఈ అక్షరంతో యతి
A) 11వ అక్షరం
B) 10వ అక్షరం
C) 13వ అక్షరం
D) 14వ అక్షరం
జవాబు:
B) 10వ అక్షరం

7) చంపకమాల పద్యానికి ఈ అక్షరంతో యతి
A) 10వ అక్షరం
B) 13వ అక్షరం
C) 14వ అక్షరం
D) 11వ అక్షరం
జవాబు:
D) 11వ అక్షరం

8) 13వ అక్షరంతో యతి గల పద్యం ఇది
A) శార్దూలము
B) మత్తేభము
C) ఉత్పలమాల
D) చంపకమాల
జవాబు:
A) శార్దూలము

9) ‘అతడు’ – ఇది ఏ గణమో గర్తించండి?
A)త గణం
B) ర గణం
C) భ గణం
D) య గణం
జవాబు:
C) భ గణం

10) ‘మీయయ్య’ – ఇది ఏ గణమో గుర్తించండి?
A) ర గణం
B) త గణం
C) న గణం
D) మ గణం
జవాబు:
B) త గణం

11) ‘శ్రీరామా’ అనే పదం ఈ గణానికి చెందింది.
A) మ గణం
B) న గణం
C) ర గణం
D) స గణం
జవాబు:
A) మ గణం

12) ‘సీస పద్యం ‘ మీద చేరే పద్యాలలో ఇది ఒకటి
A) కందము
B) తేటగీతి
C) ఉత్పలమాల
D) ద్విపద
జవాబు:
B) తేటగీతి

13) ‘తేటగీతి’ పద్యంలో యతి ఏది?
A) 3వ గణాద్యక్షరం
B) రెండవ గణాద్యక్షరం
C) నాల్గవ గణాద్యక్షరం
D) ఐదవ గణాద్యక్షరం
జవాబు:
C) నాల్గవ గణాద్యక్షరం

14) పద్యంలో ప్రాసాక్షరము ఏది?
A) 4
B) 2
C) 3
D) 1
జవాబు:
B) 2

15) ‘యతి’ అంటే ఎన్నవ అక్షరము?
A) మూడవ
B) రెండవ
C) ఒకటవ
D) నాల్గువ
జవాబు:
C) ఒకటవ

16) ‘ఆటవెలది’ పద్యానికి గల పాదాలు
A) 2
B) 4
C) 5
D) 6
జవాబు:
B) 4

17) “అనయము దోషమే పరులయందు కనుంగొనునట్టియా’ – ఈ పాదంలో గురులఘువులు గుర్తించి, ఏ పద్యపాదమో పేర్కొనండి.
జవాబు:

ఇది చంపకమాల పద్యపాదము.

18 ‘పట్టుగ నీశ్వరుండు తన పాలిట నుండి పుడిచ్చినంతలో’ – ఈ పాదంలో గురులఘువులు గుర్తించి, ఏ పద్యపాదమో పేర్కొనండి.
జవాబు:

ఇది ఉత్పలమాల పద్యపాదము.

19) సురుచిర తారకా కుసుమ శోభి నభోంగణ భూమిఁ గాలమన్’ – ఈ పాదానికి గురులఘువులు గుర్తించి, ఏ పద్యపాదమో పేర్కొనండి. పేర్కొనండి.
జవాబు:

ఇది చంపకమాల పద్యపాదము.

AP State Syllabus AP Board 9th Class Biology Important Questions Chapter 1 Cell its Structure and Functions.

AP State Syllabus 9th Class Biology Important Questions 1st Lesson Cell its Structure and Functions

9th Class Biology 1st Lesson Cell its Structure and Functions 1 Mark Important Questions and Answers

Question 1.
What are the cell organelles present in a cell?
Answer:
The cell organelles present in a cell are endoplasmic reticulum, golgi apparatus, lysosomes, mitochondria, ribosomes, plastids and vacuoles.

Question 2.
What is the role played by cell wall in plant cells?
Answer:

• Cell wall excretes and inward wall pressure to resists the outward directed pres¬sure exerted by cell sap.
• So, the plant cells can withstand much greater changes in surrounding medium than animal cells.

Question 3.
What is cell theory?
Answer:

• Cell theory was proposed by Schleiden and Schwann.
• All living organisms are composed of cells and product of cell.
• All cells arise from pre-existing cells.

Question 4.
Write one precaution while observing nucleus in cheek cells.
Answer:

• Do not scrap the cheek too hard as it may injure the buccal mucosa.
• Excess stain should be drained off.

Question 5.
Name the colourless plastids in plants.
Answer:
Leucoplasts

Question 6.
Write the name of the plastids that are responsible for different colours in flowers and fruits.
Answer:
Chromoplasts

Question 7.
Mention the cell organell that is called “Protein factories”.
Answer:
Ribosomes

Question 8.
Name the cell organell that is known as “suicidal bags of the cell”.
Answer:
Lysosomes.

Question 9.
Write the main function of Smooth Endoplasmic Reticulum.
Answer:
The SER helps in the manufacture of fat molecules or lipids important for cell function.

Question 10.
What happens if Endoplasmic reticulum is destroyed in the cell?
Answer:
The transportation of substances from one part to another part of the cell will not occur.

Question 11.
What is the site of protein synthesis in the cell?
Answer:
Ribosomes

Question 12.
Write some examples for prokaryotic cells.
Answer:
Bacterium, Cyanobacteria and Blue green algae are examples for prokaryotic cells.

Question 13.
Write the unique feature seen in plant cells.
Answer:
Presence of cell wall is the unique feature seen in plant cells.

Question 14.
Name the selectively permeable membrane that covers the cell.
Answer:
Plasma membrane

Question 15.
What is an enzyme?
Answer:
An organic catalyst which catalyses a reaction within a cell.

Question 16.
Who coined the term ’Cytoblast’ and why?
Answer:
Schleiden called the nucleus as cytoblast, because he thought that new cells were created from the nucleus.

Question 17.
What is the site of cellular respiration?
Answer:
Mitochondria

Question 18.
What are vacuoles and write their function.
Answer:
Vacuoles are the fluid filled sac like structures present in the cytoplasm. They store solid or liquid contents.

9th Class Biology 1st Lesson Cell its Structure and Functions 2 Marks Important Questions and Answers

Question 1.
What is the difference between protoplasm and cytoplasm?
Answer:

• There is a fluid present inside the cell.
• For a long time it was believed that the essence of life was stored in the fluid.
• Hence the fluid was named as protoplasm, which means life fluid.
• When it became clear that the fluid is basically a medium in which various particles and membranes float, protoplasm was renamed as cytoplasm.

Question 2.
Take one grape fruit and place it in salt solution, Note the observations.
Answer:

• I take one grape fruit and placed it in salt solution.
• After sometime I observed that the fruit shrunk.
• This is because of loss of water inside the fruit, it comes out into salt solution.
• During this, the process of osmosis takes place.

Question 3.
Collect some parts of plants like, orange, beetroot, raddish, drumsticks, lady’s fin-ger, Jasmine, etc. and put a tick mark if you find the listed plastids present in them.
Answer:

Question 4.
Why animals depend upon plants for food?
Answer:

• Animal cells do not have chloroplasts.
• Chloroplast trap the energy from sunlight.
• It converts solar energy to chemical energy which takes place during photosynthesis.
• During the process of photosynthesis food materials are formed.
• Due to lack of chloroplasts animals are unable to prepare their own food.
• So, animals depend on plants for food.

Question 5.

Study the table and answer the questions.
1) Name the cell structure that helps in photosynthesis.
2) Name the cell organell that participates in intercellular digestion.
Answer:

1. Chloroplasts
2. Lysosomes

Question 6.
What happens if stomata are closed with paraffin wax?
Answer:

• Stomata helps in exchange of gases in leaf.
• If the stomata are closed with paraffin wax gaseous exchange will not takes place.

Question 7.
What questions do you pose to know more details about plasma membrane?
Answer:

1. How can you say that plasma membrane is also known as selectively permeable membrane?
2. Name the substances that can pass through the plasma membrane.
3. Give examples for selectively permeable membrane.

Question 8.
Write the main function of the cell wall.
Answer:

• The cell wall is tough but flexible porous layer that gives a definite shape to the cell.
• It provides protection to the cell from the external shocks.

Question 9.
Venu is asking his teacher about different functions of cell organells of Eukaryotic cell. What questions he would ask to his teacher ?
Answer:

1. Prokaryotic cells devoid of nucleus . Why?
2. Are the cell organells of same size in all the higher animals?
3. Why lysosomes are called suicidal bags of the cell?
4. What happens if cell wall is not present in plant cell?

Question 10.
Write about plastids.
Answer:

• Plastids are present only in plant cells.
• Plastids mainly of two types. 1. Chromoplasts (coloured) and leucoplasts (colourless).
• Chloroplasts are the type of chromoplasts present only in plants.
• The primary function of a chloroplast is to trap the energy of sunlight and transform it to chemical energy thus helping to carryout photosynthesis.
• Chromoplasts are responsible for having various colours of fruits, flowers and leaves.
• Leucoplasts are used to store starch, oil and proteins.

Question 11.
Write differences between plasma membrane and cell wall.
Answer:

Question 12.
What is the reason for colour change in tomatoes? (green – white – yellow – red)
Answer:
1) Plastids are responsible for colour change in tomatoes.

2) Plastids are of three types :
1. Chromoplasts (coloured)
2. Leucoplasts (colourless)
3. Chloroplasts (green)

3) All the three plastids have the capacity to change from one to another.

4) As the young tomatoes mature we see green, white, yellow and red coloured tomatoes.

Question 13.
Draw a neat and labelled diagram of nucleus.
Answer:

Question 14.
Draw a neat and labelled diagram showing L.S of mitochondria.
Answer:

Question 15.
Draw a neat and labelled diagram showing the structure of chloroplast.
Answer:

Question 16.
Draw a neat and labelled diagram of Endoplasmic Reticulum find in electron microscope.
Answer:

Question 17.
What is the function of a nucleus in a cell?
Answer:

• The nucleus plays a vital role in the cell.
• It controls all functions of the cell. It controls cell division.
• Nucleus contains chromosomes. These chromosomes contain DNA and proteins.

Question 18.
“Cell is the structural and functional unit of life” – How?
Answer:
a) A cell is capable of independently carrying out all necessary activities of life,
b) Hence, it is called the structural and functional unit of life.

Question 19.
What is Protoplasm? Who coined this term and when?
Answer:
a) The living fluid substance of the cell is called “Protoplasm”
b) Purkinje in 1839 coined the term protoplasm.

Question 20.
Name the smallest and largest known cells in this world?
Answer:
a) The smallest known cells is pneumonea cell. It is about 0.1 m in diameter.
b) An Ostrich egg cell is the largest known cell. It is 170 x 135 mm approximately.

Question 21.
What will happen to the size of the cell if it is placed in such solutions which vary in their concentrations.
a) When placed in Hypotonic solution?
b) When placed in Isotonic solution?
Answer:

• When a cell is placed in Hypotonic solution (dilute solution), water enters into the cell. Hence the cell swells up.
• When a cell is placed in Isotonic solution (same concentration), there is no movement of water. Hence the cell will stay the same size.

Question 22.
“A cell is a building unit of an organism”. Do you agree with this statement? If yes, explain why.
Answer:

• I agree with the above statement.
• A cell is a building unit of an organism because it is responsible for building the entire body of an organism.

Question 23.
What is Osmosis?
Answer:
Osmosis is the passage of water or any solvent from a region of its lower concentration to a higher concentration through a semi permeable membrane.

Question 24.
What are genes? What is their function?
Answer:
Genes are the segments of DNA present on the chromosomes. These are the hereditary units which are transmitted from one generation to another by chromosomes.

9th Class Biology 1st Lesson Cell its Structure and Functions 4 Marks Important Questions and Answers

Question 1.
Write a short note on the Golgi apparatus.
Answer:

• Camillo Golgi first observed the golgi bodies in 1898.
• This is made up of several membranes.
• These membranes create sac-like structure around which many fluid-filled vesicles abound.
• The proteins and other substances produced in the ribosomes reaches the golgi body through these vesicles.
• This organelle package various substances before Ciste they are transported to other parts of the cell. _
• The number of golgi bodies varies from cell to cell. Golgi apparatus
• They are large in number in those cells that secrete hormones and enzymes.

Question 2.
Write about mitochondria.
Answer:

• Mitochondria are small, spherical or cylindrical in shape.
• They are 2-8 micron long and about 0.5 micron wide.
• It is about 150 times smaller than the nucleus.
• 100 – 150 are present in each cell.
• They are made of a double-membrane
• The inner membrane of the wall protrudes into the interior in folds and forms cristae.
• The space between cristae is known as the matrix.
• They are responsible for cellular respiration.
• Energy generated and stored in mitochondria.
• Hence they are also called as cell’s power house.

Question 3.
Write a note on Cytoskeleton.
Answer:

• Endoplasmic Reticulum (ER) is one of the important cell organelle.
• It extends all over the cell, so it is also called as cytoskeleton.
• It is responsible for the transport of substances from one part of the cell to another.

• RER has ribosomes on its surface which are caused for protein manufacture.
• The SER helps in the manufacture of fat molecules.
• Invertebrate liver cells SER plays a crucial role in detoxifying many poisons and drugs.

Question 4.
How could you appreaciate the function of nucleus in a cell?
Answer:

• Nucleus is the most prominent one of all cell organelles.
• This is also known as cell’s control room.
• It was named by Robert Brown in 1831.
• All cells have nucleus except a few cells.
• In mammal red blood cells and phloem sieve tube in plants nucleus is absent.
• It regulates and controls all the functions of the cell.
• It determines the characteristics of the organism.
• It is the barrier of all genetic information.
• It involved in the process of cell division.
• If there is no nucleus in the cell, growth of organism not takes place.

Question 5.
Draw the diagram of nucleus and label it.
Answer:

Question 6.
Why the colouring fruits or vegetables changes? Support your answer.
Answer:

• In plants plastids are responsible for the colouring of fruits and vegetables.
• There are 3 types of plastids present in plants.
• Chloroplasts are responsible for green colour.
• Chromoplasts are responsible for different colours, i.e., orange, yellow, red etc.
• Leucoplasts are responsible for white colour.
• These plastids have the capacity to change from one form to another.
• E.g. : Young tomatoes are white in colour as they mature they turn to green and then to red in colour.

Question 7.
In what way different colours in flowers helpful to bio-diversity?
Answer:

• We can easily identify their species at a glance.
• By having different colours plants attracts insects for pollination, for their propagation.
• By having different colours plants appeal to the aesthetic sense of man, immense pleasure and happiness.
• This is useful in the propagation of plants by the human beings.
• Generally butterflies are known as the pollinators, but in fact they are the indicators of the health of an ecosystem.

Question 8.
Describe endoplasmic reticulum.
Answer:

• The network or membranes present in the cytoplasm for the transport of substances
  from the one part of the cell to another is known as endoplasmic reticulum.
• Endoplasmic reticulum is of two types.
  i) Rough endoplasmic reticulum and
  ii) Smooth endoplasmic reticulum.
• Endoplasmic reticulum having ribosomes on their surface is known as rough endoplasmic reticulum.
• The rough endoplasmic reticulum is the sites of protein synthesis.
• Ribosomes are absent in smooth endoplasmic reticulum and is involved in lipid synthesis.
• One function of the endoplasmic reticulum is to serve as channels for the transport of materials within the cytoplasm.
• It also function as a cytoplasmic framework providing a surface for some of the biochemical activities of the cell.

Question 9.
Observe the following slides under the microscope and draw their pictures. Write the cell organelles in them.
Answer:

Organelles present in Amoeba :
Nucleus, contractile vacuole, food vacuole etc.

Organelles present in Euglena :
Nucleus, chloroplasts, contractile vacuole, reservoir, paraflagellar body, endosome etc.

Cell organelles present in Paramoecium :
Anterior and posterior contractile vacuoles, micronucleus, macronucleus, cytostome, cytopyge, food vacuole etc.

Question 10.
Collect the names and photographs of scientists helped for the development of cell biology. Give brief note on them.
Answer:

• 1632 – 1723 Antonie Van Leeuwenhoek constructed simple microscope and draws protozoa, verticella from rain water and bacterium from his mouth.
• 1665 Robert Hooke discover cells in Cork, then in living plant tissues using an early compound microscope.
• In 1831 Robert Brown discovered nucleus. In 1839 Purkinje coined the term protoplasm.
• In 1839 Theodar Schwann and M.J. Schleiden proposed cell theory.
• 1855 – Rudolf Carl Virchow observed cell division.
• 1931 – Earnest Ruska built first transmission electron microscope.
• 1953 – Watson & Crick made their first announcement on the double helix structure of DNA.
• Albert Claude, father of cell biology awarded Nobel prize for Physiology (Medicine) in 1974.
• 1981 – Lynn Margulis published symbiosis in cell evolution detailing the endosymbiotic theory.

9th Class Biology 1st Lesson Cell its Structure and Functions Important Questions and Answers

Question 1.
What happens if there are no mitochondria in eukaryotic cell?
Answer:
If Mitochondria are absent in Eukaryotes, the energy required to perform all cellular activities will not be released. Hence, all the biological activities occurring in the cell will be stopped ultimately. This leads to the death of the eukaryotic cell.

Question 2.
What are the differences between protoplasm and cytoplasm?
Answer:

• Protoplasm is the content of the cell including the cell membrane, cytoplasm and the cell nucleus.
• Cytoplasm is the jelly like substance surrounding the nucleus within the cell membrane. The cytoplasm contains the cell organelles like mitochondria, ribosomes, etc.

Question 3.
a) Identify the figure and write the parts.
b) Write a short note on the above figure.

Answer:
a)

1. Matrix
2. Cristae
3. Inner membrane
4. Outer membrane

b)

1. The above shown cell organelle is mitochondria.
2. It performs cellular respiration and releases energy required for all cellular activities.
3. Mitochondria is also known as “Power house of the cell”.
4. Mitochondria are made up of a double membrane wall. The inner membrane of the wall protrudes into the interior in folds and forms structures called cristae.
5. The space between cristae is filled with a fluid known as the matrix.

Question 4.
Write the functions of the following cell organelles.

Answer:

• Mitochondria performs cellular respiration. They release energy for all biological activities of the cell.
• Chloroplasts trap the solar energy and helpful in photosynthesis through which plants derive their food.

Question 5.
a) Draw a neat labelled diagram of a plant cell.
b) Write the functions of endoplasmic reticulum.
Answer:
a)

b) Functions of Endoplasmic Reticulum :

1. The ER is to serve as channels for the transport of materials between various regions of the cytoplasm or between the cytoplasm and the nucleus.
2. It is the site of many bio – chemical activities in the cell.
3. It helps in the synthesis of proteins and lipids.
4. In vertebrate liver cells, SER plays a crucial role in detoxifying many poisons and drugs.

Question 6.
What does the modern cell theory propose?
Answer:
i) All living organisms are composed of cells and products of the cells.
ii) All cells arise from the pre existing cells.

Question 7.
What happens if lysosomes are absent in the cell?
Answer:
The materials that need to be destroyed are sent to lysosomes. They release the enzymes and digest them. If lysosomes are not present, the harmful substances which are dangerous to the cell would not be destroyed. The cell may die.

Question 8.
Name the chemical substance used in cheek cell lab activity.
Answer:
Methylene blue

Question 9.
Describe the nucleus of cell with the help of a well labelled diagram.

Answer:

• The membrane that encloses the nucleus and separates it from contents of cytoplasm is known as nuclear membrane.
• The entire genetic material of the cell is found in the nucleus.
• The nucleus has fluid like substance called nucleoplasm.
• In the centro of the nucleus, we can see a round shaped structure called nucleolus.
• Nucleus controls and regulates all the activities of the cell.
• Nucleus is closely involved in the process of cell division.

WorkBook Part

1. Write an activity to observe the nucleus in cheek cells.
2. Write a brief notes about mitochondria with help of the diagram.
3. Draw the diagram of Nucleus and label its parts.
4. Write name of the following figure and write its parts.
5. Write the name of the following figure and write its parts.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 3 Animal Tissues Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 3rd Lesson Animal Tissues

9th Class Biology 3rd Lesson Animal Tissues Textbook Questions and Answers

Improve Your Learning

Question 1.
What do you understand by the term tissue? (AS 1)
Answer:
Tissue is a group of cells similar in structure and function.
Eg : Nerve tissue, epithelial tissue, muscle tissue etc.

Question 2.
Show the difference between the three types of muscle fibres with diagrams. (AS 1)
Answer:

Question 3.
What is the specific function of the cardiac muscle? (AS 1)
Answer:
Specific function of the cardiac muscle :

1. Cardiac muscle present in the heart.
2. It is responsible for pumping of blood.

Question 4.
Differentiate between striated and unstriated muscles on the basis of their shape and site/location in the body. (AS 1)
Answer:

Question 5.
Draw a neatly labelled diagram of a neuron. (AS 3)
Answer:

Question 6.
Name the following. (AS 1)
a) Tissue that forms the inner lining of our mouth.
b) Tissue that connects muscle to bone in humans.
c) Tissue that transports food in animals.
d) Tissue that stores fat in our body.
e) Connective tissue present in the brain.
Answer:
a) Epithelial cells
b) Tendon
c) Connective tissue/blood
d) Adipose tissue
e) Areolar tissue

Question 7.
Identify the types of tissue in the following : Linings of the organs, skin, bone, internal lining of kidney tubule. (AS 1)
Answer:

Question 8.
If the platelets are not present in the blood, what happens? (AS 2)
Answer:

• If the platelets are not present in the blood, blood loss may be more from the injury.
• Whenever a blood vessel is injured, at the site of injury formation of a blood clot will not takes place.
• The wound will not be sealed by the clot.

Question 9.
If you touch at elbow, you get a shock like feeling. Why? (AS 7)
Answer:

• In human beings ulnar nerve runs from the shoulder to the hand.
• The ulnar nerve comes close to the surface near the elbow.
• Due to the superficial location it is not protected by muscle, fat or other soft tissues.
• Thinner skin layer around bone at elbow makes ulnar nerve more receptive for any small stimuli.
• That is the reason for getting a shock like feeling if we touch at elbow.

Question 10.
Why the blood is called a connective tissue?
Answer:
Connective tissue :
A loosely spaced tissue mainly carrying different materials to different parts of the body as well as rendering support, making connection between organs is called connective tissue.

Blood is considered as connective tissue because of the following reasons.

1. Blood connects different organs of our body together by carrying oxygen, nutrients, hormones and other signaling molecules and removing the waste.
2. It has all the three components of connective tissue i.e., cells, fibers and matrix.
3. Similar to other connective tissues, blood is rich in fibres like collagen fibers and blood clotting fibres.
4. Blood originates from the mesodermal layer of the embryo from which all other connective tissues also originate.

Question 11.
Write the procedure to identify your blood group with the help of kit. (As 3)
Answer:
Aim :
Identification of blood group.

Apparatus :
Blood identification kit, glass slide, wax pencil, disposable needle.

Materials used :
Cotton, 70/6 alcohol, toothpicks.

Procedure:

1. Take one porcelain plate, clean and dry it.
2. With a wax pencil, draw three circles on the plate to divide the surface into three parts and draw three circles.
3. Place one drop of the corresponding antiserum near the edge of the circles.
4. Clean the fingertip with an alcohol and let it dry.
5. Press on the bottom of the fingertip with the thumb and quickly prick the fingertip with the help, of a needle.
6. Quickly, let one drop of blood get into each circle, but not touching the anti-serum.
7. Apply gently pressure to the wound with cotton ball.
8. Use a toothpick to mix the blood and anti-serum and stir gently.
9. Watch to see if any of the samples show agglutination.

Result and Inference :
By using the following table determine the blood group.

If agglutination occurs in anti – RhD serum, the Rh factor is positive, and if it does not . the Rh factor is negative.

Note :
1. See the needle is sterile.
2. Usually choose left ring finger.
3. Don’t use same needle to other body.

Question 12.
Ramu felt weak. Ramu’s father took him to hospital. The doctor advised a blood test. The report says that he does not have the required levels of haemoglobin. What are its ill effects?
Answer:
Ill effects of haemoglobin :

• Blood is red in colour due to the presence of red coloured protein called haemoglobin.
• Haemoglobin helps in the transport of oxygen and carbon dioxide.
• Low haemoglobin is the main cause for anemia.
• If there is not enough haemoglobin in blood. The oxygen supply to various parts will be less, which causes shortness of breath.
• Low haemoglobin levels many aggravate extant heart problems.
• People with low haemoglobin levels get very tired as their cells do not get enough oxygen to perform their activities.

Question 13.
How blood test is useful to diagnose the disease? Explain with daily life situation. (AS 1)
Answer:

• Blood test is useful to diagnose diseases such as malaria, typhoid, cancer, HIV/AIDS, diabetes, anemia coronary heart disease, abnormalities in the functioning of kidney, liver, thyroid, etc.
• Abnormal red blood cell levels might be a sign of anemia. Dehydration, bleeding, and other disorder.
• Complete blood count with differential can measure the amounts of different types of white blood cells in our body.
• Abnormal white blood cell levels might be a sign of infection, blood cancer or an immune system disorder.
• Abnormal platelet levels might be a sign of a bleeding disorder or thrombotic disorder.
• Abnormal haemoglobin levels might be a sign of anemia, sickle cell anemia or thalassemia.
• Abnormal glucose levels in the blood might be a sign of diabetes.
• Abnormal calcium levels in the blood might suggest kidney problems, bone disease, thyroid disease, cancer, or malnutrition.
• Abnormal electrolyte levels might be a sign of dehydration, kidney disease, liver disease, heart failure or high B.P.
• Abnormal Blood Urea Nitrogen (BUN) and creatinine levels might suggest a kidney disease.
• High levels of enzymes like Troponin and creatine kinase is a sign of Heart attack.
• Abnormal cholestrol or triglyceride levels might be a sign of increased risk of coronary heart disease.
• Abnormal coagulation pannel test results might suggest risk of bleeding or developing clots in blood vessels.
• Existence of microorganisms or their antibodies in the blood suggest occurence of corresponding disease.
  E.g. : Plasmodium – Malaria, HIV – AIDS etc.

Question 14.
Collect the old blood reports of your friends/relatives and prepare a project report on the contents of the blood.
Answer:
On collection and observation of old blood reports I came to know that the contents of blood should present in definite proportions such as.

9th Class Biology 3rd Lesson Animal Tissues InText Questions and Answers

9th Class Biology Textbook Page No. 29

Question 1.
Why do old people shiver in winter when compared to youngsters? Is there any insulator like substance to prevent the escape of heat energy during winter?
Answer:

• Old people shivers in winter when compared to youngsters.
• They didn’t have enough fat storages below the skin.
• Fat storing adipose tissue is found below the skin and between internal organs.
• The cells of this tissue are filled with fat globules.
• Storage of fat also acts as insulator.

Question 2.
Which tissue gives definite shape to body of vertebrae?
Answer:

• Bone is one type of connective tissue.
• It forms the frame work that supports the body.
• It is a major component of the skeletal system of several vertebrae.

9th Class Biology Textbook Page No. 34

Question 3.
During winter, body shivers. Why?
Answer:

• When the body is exposed to cold air, we shiver.
• During shivering muscles contract and relax and produce large amount of heat.
• This keeps the body heat.
• It is one type of defensive mechanism of the body.

9th Class Biology Textbook Page No. 30

Question 4.
Blood is a type of connective tissue. Why is it called connective tissue?
Answer:
Blood is considered as connective tissue because of the following reasons.

1. Blood connects different organs of our body together by carrying oxygen, nutrients, hormones, and other signaling molecules and removing the waste.
2. It has all the three components of connective tissue i.e., cells, fibers, and matrix.
3. Similar to other connective tissues, blood is rich in fibres like collagen fibers and blood clotting fibres.
4. Blood originates from the mesodermal layer of the embryo from which ail other connective tissues also originate.

9th Class Biology 3rd Lesson Animal Tissues Activities

Lab Activity – 1

Question 1.
Aim:
Identification of tissue in collected sample.

Apparatus:
Microscope, slide, dilute HCl, forceps, brush.

Procedure:

1. Collect a small piece of chicken with bone from your nearby chicken centres or market.
2. Put it in dilute HCl for two hours.
3. Take the skin part of chicken piece.
4. Place the material with forceps or brush on the slide
5. Then keep the another slide on it and press both the slides gently.
6. Place a cover slip tap on it and observe under microscope.
7. Draw the diagram of what you have observed under microscope in your notebook.
8. Compare your diagram with the given picture.

Answer these questions.
1. Are all the cells similar?
Answer:
Yes. All the cells are similar.

2. How are they arranged?
Answer:
They are arranged in layers. Each cell is round and nucleated. Observed diagram

3. Are these cells tightly packed and formed as continuous sheath?
Answer:
Yes. The cells are tightly packed and formed as continuous sheath.

4. Is there any intercellular space?
Answer:
No. There is no intercellular space.

5. Think, why these cells look like continuous sheath.
Answer:
These cells are look like continuous sheath because there is no intercellular space and the cells are tightly packed.

6. Does this tissue covering protect inside and outside of the animal body?
Answer:
Yes. This tissue covering protect inside and outside of the animal body.

Question 2.
Aim:
Identification of tissue in collected sample.

Apparatus:
Microscope, slide, blood sample, syringe, cotton.

Procedure:

1. Take a sterilized syringe needle.
2. Collect one drop of blood from finger tip by pricking with syringe needle.
3. Take a slide. Keep the finger on the slide to collect one drop of blood.
4. Put another slide on it gently and press both :
5. Observe under microscope.
6. Draw the diagram of what you observe L microscope in your notebook. Compare diagram with the given picture.
   In this procedure we can identify red blood

Question 3.
Aim:
Identification of tissue in collected sample

Apparatus:
Microscope, slide, dilute HCl, vinegar, forceps.

Procedure:

1. Take a piece of muscle of chicken.
2. Put in diluted HCl or vinegar and leave it for two hours.
3. Next morning collect the piece of muscle on a slide with forceps.
4. Press gently with another slide, put few drops of water and place a cover slip on it.
5. Observe under microscope. Observed diagram
6. Draw the diagram what you have observed under microscope in your notebook. Compare your diagram with the above picture.

Answer these questions.

1. How are the cells arranged?
Answer:
Cells are arranged in layers one above the other.

2. Do you find any difference between skin cells and muscle cells?
Answer:
Muscle cells are long and nucleated.

3. If you want to observe the bone tissue in the chicken bone, settle it in vinegar or diluted HCl over night. Then only the bone becomes soft. Take a piece from it by using knife. Do you find any relation among these tissues?
Answer:
Usually muscle tissue is attached to bones.

4. Is this tissue useful for movements in our body?
Answer:
Yes. This tissue is useful for movements in our body.

Activity – 1

Question 4.
1. Collect the substance lining of mouth by using wooden spoon and observe this under microscope.
2. Draw the diagram that you observed in the microscope, in your notebook.
a) How are the cells arranged?
Answer:
Cells are extremely thin and flat and form a delicate lining.

b) Are there any intercellular spaces?
Answer:
No. Intercellular spaces are absent.

c) Think, why are the epithelial cells in skin are arranged in the form of layers?
Answer:

Because skin has to protect our body from cold, heat etc.

d) If you drink hot tea or chilled cool drink, how would you feel?
Answer:
Inner layers of our mouth cannot bear hot tea or chilled cool drink. We immediately spill hot or cold substances from our mouth.

e) If your skin burns or wounded, which tissue would effected ?
Answer:
Epithelial tissue.

Activity – 2

Question 5.
1. Take a permanent slide of cuboidal epithelium from your laboratory slide box and observe under microscope.
2. Draw the picture in your notebook.
3. How are the cells arranged?
Answer:
The cells are compactly arranged without intercellular spaces.

4. Conclusion :
These are the cuboidal epithelial cells which form the lining of organs or tubules or other parts and provide mainly mechanical support.

Activity – 3

Question 6.
1. Take a permanent slide of columnar epithelium from the slide box and observe under microscope.
2. Draw the figure that you observed under microscope. Observed diagram
3. How are the cells? Do you find any hair like projections on the outer surface of epithelial cells?

Answer:
a) The cells are long, compactly arranged without intercellular spaces.
b) Hair like projections are present on the outer surface of these cells.
c) These type of epithelial cells are present in the small intestine.

Activity – 4

Question 7.
1. Invite a scientist or doctor to your place.
2. Record an interview about blood structure and its functions.
3. It is important to make a questionnaire in order to conduct an interview.
4. After completion of interview, prepare a booklet about blood and display it on bulletin board or classroom library.

Booklet about blood.

1. Blood is a fluid connective tissue.
2. There are different types of cells in blood and each one has a different function.
3. All the cells in the blood cells float freely in the plasma.
4. Extracellular space is filled with fluid called plasma. There are no fibres in the blood.
5. Normal adult human beings have about 5 litres of blood. A chief component in plasma is water.
6. Besides water it also has several nutrients such as glucose, aminoacids, proteins, vitamins and hormones.
7. Plasma also contain factors responsible for blood clotting. Heparine helps to prevent blood clotting in blood vessels.
8. Cells present in blood are corpuscles. They are three types l.RBC, 2. WBC, 3. Blood platelets.
9. Red blood cells are also known as erythrocytes. They are red in colour due to the presence of haemoglobin.
10. haemoglobin helps in transport of oxygen and carbon dioxide.
11. When we are in mother’s womb, RBC are formed in the liver and spleen. After birth RBC are generated from the bone marrow of long bone.
12. RBC live for 120 days.
13. The second type of cells present in blood are white blood cells, which do not have haemoglobin. Hence they are called leucocytes.
14. There are two types of cells in WBC – granulocytes and agranulocytes.
15. There are three types of cells in the granulocytes – Neutrophils, Basophils and Esinophils.
16. These cells attack and destroy the microorganisms that enter the blood.
17. There are two types of agranulocytes – lymphocytes and monocytes.
18. Lymphocytes secret anti – bodies to guard against foreign material that enter into blood. So they are called microscopic policemen.
19. Monocytes move like amoeba and along with granulocytes. The foreign materials are destroyed inside these cells. They are called as ‘scavengers’.
20. Blood platelets are a separate group of cells which do not have a nucleus. They help in blood clotting.

Lab Activity – 2

Question 8.
Aim:
Identification of blood group.

Apparatus:
Blood identification kit, glass slide, wax pencil, disposable needle, cotton, tooth picks, 70% alcohol.

Kit components:

Procedure:

1. Take one porcelain plate, clean and dry it. The plate must be very clean so that it does not interfere with the reaction.
2. With a wax pencil, draw three circles on the plate to divide the surface into three parts and draw three circles, one in each part as shown in figure.
3. Place one drop of the corresponding antiserum near the edge but within each of the circles as shown in figure.
4. Choose a left ring finger clean it with alcohol in a cotton ball and let it dry. Keep the cotton ball nearby, as it is needed again. Dangle the hand down to increase the amount of blood in the fingers.
5. Press on the bottom of the finger tip with the thumb of the same hand and quickly prick the finger tip with the help of a needle.
6. Quickly, let one drop of blood get into each circle but not touching the anti-sera.
7. After putting three drops of blood, apply gentle pressure to the wound with cotton ball.
8. Use a toothpick to mix the blood and antiserum and stir gently. Do it for each of the circles using a fresh toothpick every time.
9. Watch to see if any of the samples show agglutination.

Result and inference :
Determine the blood type depending on the result. Following table can be used to determine the blood type :

If agglutination occurs in anti-RhD serum, the Rh factor is positive; and if it does not, the Rh factor is negative.

Result should be noted in the given table :

Activity – 5

Question 9.
Collect three types of muscle slides (Striated muscles, Non-striated muscles, Cardiac muscles) from slide box. Then observe these under microscope. Write your findings in the following table.

Activity – 6

Question 10.
Collect the slide of nerve cells from the slide box. Observe it under microscope. Write your findings.
Answer:

1. We can identify three distinct parts in nerve cells.
2. They are
   1. Cell body or cyton,
   2. Axon and
   3. Dendrites
3. Cell body or cyton has a large nucleus and cytoplasm. The cytoplasm contains granular structure called Nissal’s granules.
4. Projections arising from the cell body are called dendrites. They are sharp, branched, more in number.
5. One projection of the cyton is somewhat longer than remaining projections. This is called axon.
6. Nerve cell is covered with myeline sheath. Nodes of Ranvier are present in myelin sheath.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 9 Adaptations in Different Ecosystems Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 9th Lesson Adaptations in Different Ecosystems

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Textbook Questions and Answers

Improve Your Learning

Question 1.
What do you understand by adaptations in organisms and why do they adapt? (AS 1)
Answer:

• The ways and means that organisms adapt or develop over a certain period of time in different conditions for better survival are adaptation of organisms.
• Adaptation is a feature that is common in any population because it provides some improvement for better survival.

Question 2.
With the help of two examples, explain how these organisms have adapted themselves in the ecosystem? (AS 1)
Answer:

• Mangroves grow in a wet and salty place.
• They have evolved to have curious looking projections from their roots called pneu- matophores or knees.
• These pneumatophores develop from the lateral roots that are growing near the surface, and protrude upto 12 inches out of the soil.
• Pneumatophores aid the plants in maintaining adequate root respiration in a watery environment.
• We don’t find such structures in plants growing around us.
• Another example is in kaiabanda, the leaves are reduced to spines so that there is little transpiration loss and water is stored in the tissues of the stem (succulent stems)
• This helps the plant to live in conditions of water scarcity as we come across in deserts.
• With the above two examples, we can say that these organisms have adapted them-selves in the ecosystem.

Question 3.
Collect some aquatic plants- cut the leaves and stems. Observe them under microscope and record your observations like presence air /absence of air spaces etc. and answer the below. (AS 3)
a) Are there any other reasons for their floating?
Answer:
The bodies of aquatic plants are delicate with more than 80% of their weight consisting of water.

b) Draw a diagram of what you have observed under microscope.

Question 4.
What special adaptations can be seen in the following organisms? (AS 1)
a) mangrove trees
b) camel
c) fish
d) dolphins
e) planktons

a) Mangrove trees :
Answer:

1. Mangroves grow in a wet and salty place near the sea shore.
2. From their roots arise pneumatophores or knees.
3. These pneumatophores develop from the lateral roots that are growing near the surface and protrude upto 12 inches out of the soil.
4. Pneumatophores. aid the plants in maintaining adequate root respiration in a watery environment.

b) Camel:
Answer:

1. In camel hump stores fat fordater use.
2. Long eyelashes protects eye from sand.
3. Nostrils closes voluntarily to protect from blowing sand.
4. Long legs keeps the body away from hot ground.

c) Fish :
Answer:’

1. The body is covered by scales.
2. Fishes bear specialised structures to swim like fins.
3. Fishes have floaters in their body (special structures of their digestive canals) to be able to inhabit particu¬lar levels in the water body.
4. Fishes respire with gills.

d) Dolphins:
Answer:
Dolphins have adapted to their environment in the following ways :
Fins shape – A dolphins tail goes up and down to help it dive up to get the air. The shape of their fins also help to propel them through the water.

To help dolphins save oxygen while they dive under water, their heart beat slower during a dive and their blood is diverted from other parts of their body to their heart, lungs and brain. They also save oxygen via muscles, which have a protein called myo-globin which in turn stores oxygen.

They have a blubber or fat which provides insulation helping the dolphin stay warmer under cold water.

They have a body covering of skin. The upper most layer of skin produces an oil which forms a film that cover the dolphin’s body.

Being mammal dolphin breathe with lungs rather than gills. So they breathe from a blow hole which closes before the dolphin goes into the water. The long nose helps the dolphin to fight sharks and their teeth help them to catch fish.

They have well developed echo location by which they locate other animals and also communicate with each other.

e) Planktons:
Answer:
Microscopic photosynthetic organisms like planktons have droplets of oil in their cells that keeps them float.

Question 5.
If an animal of euphotic zone has to survive in abyssal zone, what adaptations are required to survive there? (AS 1)
Answer:
Adaptations required to survive in abyssal zone are :

1. The animals should have wide mouths and huge curved teeth which prevent escape of any prey.
2. Absence of skeleton, flattened bodies are required.
3. Animals should have special structures that produce light on their bellies, around their eyes, and at the sides of their bodies.
4. The animals should show bioluminescence in the dark waters.

Question 6.
Marine water fishes drink more water than fresh water fishes. Do you agree? Justify.
Answer:

• Yes, marine water fishes drink more water than fresh water fishes.
• Because several marine fishes have a lower internal salt concentration than that of the water they swim in.
• So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amount of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 7.
Visit a nearby pond or lake. Record the organisms you have observed and their adaptations. (AS 4)
Answer:

• Nearby pond or lake consists of three zones namely littoral zone, limnetic zone, and profundal zone.
• In the topmost littoral zone, the edge of a water body is home to snails, insects, several crustaceans, fishes and amphibians, and the eggs and larvae of dragonflies.
• Predators present are tortoise, snakes, and ducks.
• Adaptations : Several organisms have well developed sight, usually have dull and greyish bodies, and are fast swimmers.
• Limnetic zones contains fresh water fish, crustaceans like daphnia, cyclops, and small shrimps are present.
• Floating- plants like water hyacinth, wolfia, pistia along with algae are present.
• Adaptations seen in the plants this zone are presence of air space, leaves covered with wax, etc.
• In the profundal zone scavengers and predators for example crustaceans, crabs, fishes like eels and snails, turtles are present.
• They adapt themselves by feeding on dead animals that settle down.

Question 8.
Draw a lake showing different zones. Why are they called so? (AS 5)
Answer:
Zones of Lake :

1) Littoral zone :
The zone close to shore. They reaches all the way. Plants living in this zone perform photosynthesis.

2) Limnetic zone :
Sunlit part of the lake surrounded by the littoral zone. This zone extends at a depth where sunlight penetrates.

3) Profundal zone :
It is much colder and denser than previous zones.

Question 9.
Collect information of one lake from internet and prepare a table of organisms adapted at different zones.
Answer:
Different zones in lakes and types of organisms present:
1) The littoral zone :
a) The topmost and warmest zone at the edge of a water body is home to snails, clams, insects, several crustaceans, fishes and amphibians and eggs and larvae of dragonflies.
b) Plants like mosses, water lily, vallisneria, hydrilla etc. are found along with several types of algae.
c) Predators of this zone are tortoise, snakes and ducks.

2) The limnetic zone :
a) This zone contains variety of fresh water fish with bright shiny scales.
b) Transparent or whitish bodied crustaceans like daphnia, cyclops, small shrimps are also found in this zone.
c) There are different types of floating plants like water hyacinth wolfia, pistia along with a variety of algae.

3) The profundal zone :
a) Mostly heterotrophs are present.
b) Scavengers and predators like crustaceans, crabs, fishes like eels and glossogobius (isika dondu), snails, turtles etc are present.
c) Many kinds of bacteria are also present in this zone that help in decomposition.

Question 10.
Write the effect of temperature on the organisms adapted in a lake and pond in a tabular form. (AS 1)
Answer:

• In deeper lakes during summer only the surface water is heated up while the deeper layer remain cold. During summer the ponds dry up.
• In tropical regions water gets heated up and evaporates in lakes. During average temperatures the water in the pond heated up and evaporates.
• The requirements necessary to the organisms like oxygen and nutrients gets decreased in the lake.
• The salinity of the water increases, concentration of oxygen decreases and availability of food decreases in pond during average temperatures.
• In the cold regions upper layers of the lake gets frozen during winter and lower layers does not.
• The entire pond gets frozen during winter.
• Aquatic animals in tropics undergo aestivation or hybernation to overcome extreme cold or hot seasons.

Question 11.
Amphibians are wonderful creatures on the earth. How do you appreciate their adaptation? (AS 6)
Answer:

• Amphibian body has small waist, no neck. Streamlined body shape helps in swimming.
• Skin is thin and moist allows gaseous exchange in cutaneous respiration.
• Front legs used to keep the front portion of the body off the ground.
• Hind legs able to jump great distances and change direction quickly.
• Eyes are positioned on top of head gives the frog a wide angled visual field.
• Mouth is very large and broad can able to catch and eat large prey.
• Tongue attached at front of mouth enables it stick the prey when caught.
• Frogs start their lives as aquatic tadpoles with gills to breathe. As tadpole grows into frogs lungs replace the gills and allows frog to breathe on land.

Question 12.
Some animals and plants survive only in certain conditions. Nowadays human activities cause damage to these conditions. What do you think about this? (AS 7)
Answer:

• Human activities are causing lot of damage to biodiversity.
• Human activities such as deforestation, overgrazing, conversion of forest land to agricultural land, hunting and indiscriminate killing of animals for their products, and pollution can endanger the plant and animal species.
• If proper care is not taken plants and animals may disappear totally from the surface of the earth.

Question 13.
In the chapter on ecosystem, we had studied about the mangrove ecosystems. What kind of abiotic conditions did you study in them? (AS 1)
Answer:
Kinds of abiotic conditions in mangrove ecosystems are soil, pH, oxygen, nutrients, winds and currents, light, temperature, humidity, tides, salinity.

Question 15.
Are there any rivers meeting in the Bay of Bengal in the Coringa ecosystem? Collect information and make a note on them.
Answer:

• Coringa mangrove is situated South of Kakinada Bay and is about 150 km South of Visakhapatnam.
• Coringa is named after the river Coringa.
• Coringa mangroves receive fresh water from Coringa and Gaderu rivers, distributors of Gautami, Godavari rivers, and neritic waters from Kakinada Bay.
• Numerous creeks and canal traverse this coringa ecosystem.

Question 17.
The Murrel (Korramatta) and Rohu are fishes found in rivers. Will they be able to live in the coringa ecosystem ? Give reasons for your answer.
Answer:

• Yes, Murrel and Rohu be able to live in the coringa ecosystem.
• Because coringa ecosystem gets fresh water from rivers coringa, Gaderu and distributories of Gautami, Godavari rivers.
• If the salinity of the water in the coringa ecosystem increases, the water enters the body of fresh water fishes.
• The water can be excreted in the form of urine, but to maintain a suitable salt bal¬ance fresh water fish need to reabsorb salt through the kidneys and salt collecting cells in gills.

Question 18.
How the frogs got protected themselves from cold and heat?
Answer:

• Frogs are cold blooded animals so they can’t tolerate extreme cold or heat conditions.
• They protect themselves from extreme cold conditions by a process called hibernation (winter sleep) and from extreme heat conditions by Aestivation (summer sleep)
• During these processes they burrow deep in the ground and remain motionless until the conditions are favourable.
• During this period the rate of metabolic activities slow down and the animal goes into a nearly unconscious sleepy condition.

Question 19.
How do you appreciate the processing protection pebble plants from the enemies?
Answer:

• Pebble plants are also called living stones.
• They protect themselves from their enemies by adapting themselves to their habitat.
• They survive by living partly underground.
• They avoid being eaten by blending in with surrounding rocks.
• Leaves of these plants are not green as in almost all higher plants, but various shades of cream, grey and brown, patterned with darker windowed areas, dots and red lines.
• The markings on the top surface disguise the plant in its surroundings (camouflage)
• Thus, they adopt wonderfully to their habitats and protect themselves from their enemies.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems InText Questions and Answers

9th Class Biology Textbook Page No. 131

Question 1.
What is a habitat?
Answer:
Habitat is the immediate environment occupied by an organism or the living place of an organism.

Question 2.
Is a tree habitat only for a crow?
Answer:’
No. Tree is a habitat for variety of birds and insects.

Question 3.
In what way an ecosystem is different from habitat?
Answer:
In ecosystem biotic and abiotic components are present. Habitat is the place where organisms live in an ecosystem.

9th Class Biology Textbook Page No. 134

Question 4.
You may know animals that live in water. Do you find in them any suitable characters adapted to live in water? Write a note on them in your notebook.
Answer:

• Structural adaptations in the bodies like presence of special air spaces.
• Such air spaces help them to swim and float in water.
• The aquatic organisms bear specialized structures to swim like flippers as in turtles and fins in fishes.
• Fishes, dolphins have floaters in their body to be able to inhabit particular levels in the water body.

9th Class Biology Textbook Page No. 135

Question 5.
In what way flexible stem is useful to the aquatic plants?
Answer:

• In aquatic plants flexible stem contains a parenchymatous tissue known as arenchyma.
• Arenchyma consists of number of air filled spaces.
• These air spaces help the plant to float on water.

9th Class Biology Textbook Page No. 137

Question 6.
Observe the table and answer the following questions.
Answer:

a) How many zones can you see in the figure basis of light penetration? Name them.
Answer:
Three zones are present. They are eu- photic zone, bathyal zone and abyssal zone.

b) What types of abiotic conditions do you find as per the given table?
Answer:
Light, temperature and depth.

c) What will effect adaptation to marine life other than the conditions shown in the table and figure?
Answer:
Salinity, oxygen, rainfall, regular windflow, soil, pH, nutrients, humid-ity, oceanic currents effect adaptation to marine life.

d) What happens to the temperature and pressure as depth increases?
Answer:
As depth increases temperature decreases and pressure increases.

e) Which zone has more animals? Guess why.
Answer:

1. Bathyal zone has more animals. Because the conditions in this zone are suitable for the organisms to grow.
2. Red and brown kelps are the primary producers. They provide food to other organisms in that zone.

9th Class Biology Textbook Page No. 139

Question 7.
Does Pulikat lake of Nellore come under fresh water ecosystem or not? Why?
Answer:

• Pulikat lake of Nellore comes under marine or salt water ecosystem.
• Because the salinity of water in the lake is 3.5%.
• Main salts present in the Pulikat lake are sodium and potassium.

9th Class Biology Textbook Page No. 140

Question 8.
‘Think, how webbed feet helps ducks?
Answer:

1. Webbed feet of birds help them to adapt conditions on land as well as in water.
2. Webbed feet have enabled them to be good swimmers.

Question 9.
Why cranes have long legs and long beaks?
Answer:

1. Cranes have long, thin legs wander through the mud shallows searching for insects.
2. Long beak help them in searching of insects in the mud.

9th Class Biology Textbook Page No. 141

Question 10.
How are marine ecosystems different from fresh water ones?
Answer:

1. The saliny of water in marine ecosystem is 3.5% whereas it is 1.8% in fresh water.
2. Marine ecosystems are huge and they make up about three-fourths of the earths surface.
3. The number of organisms present in marine ecosystems are more when compared to fresh water ecosystem.

Question 11.
Write two types of adaptations you find in marine ecosystems, different from fresh water ecosystems.
Answer:

• Many marine animals have blubber fur insulation from the cold and some fish have an antifreeze like substance in their blood to keep it flowing.
• Marine animals must regulate the interaction of fresh water and salt water in their bodies.
• Specially developed kidneys, gills and body functions help to maintain salt concentrations across members through osmosis.

Question 12.
What are the similarities in adaptation on the basis of light penetration in the two aquatic ecosystems?
Answer:

• In both the aquatic ecosystems, light penetrates upto a depth of zoom only.
• The light intensity is sufficient to perform photosynthesis.
• In the low light intensities below 200 mts depth is sufficent to perform photosynthesis by some kelps.
• Due to the lack of light in abyssal and profundal zones, usually scavengers and predators exists.

9th Class Biology Textbook Page No. 142

Question 13.
Which zone do you think, when compared to marine ecosystems, is absent in fresh water ecosystem?
Answer:
Benthic zone is absent in fresh water ecosystem when compared to marine ecosystem.

Question 14.
What would be a major factor leading to different types of adaptations in marine, fresh water ecosystems?
Answer:
Light would be a major factor leading to different types of adaptation in marine, fresh water ecosystems.

Question 15.
Do all plants shed their leaves at same time in a year throughout the world?
Answer:

1. No. Some plants in temperate regions shed their leaves before the winter starts.
2. In tropical regions some plants shed their leaves before the start of summer.

9th Class Biology Textbook Page No. 143

Question 16.
Are thorny leaves also an adaptation to temperature?
Answer:

1. No. They are not adaptation to temperature.
2. They are adaptation to protect themselves from the animals who eat them.

Question 17.
If the trees have broad leaves at the time of snow fall season what will happen?
Answer:
If the trees have broad leaves at the time of snow fall season, the branches of tree can break due to the weight of snow gathered on each leaf and branch during snow fall.

Question 18.
Why polar bear has thick fur on its body?
Answer:

1. Polar bear has thick fur coat or hair covering on their bodies.
2. The fur act as insulator preventing heat loss from its body.

Question 19.
In what way thick skin helps the seal to protect from cold weather?
Answer:

1. In the thick layer of skin, fat is deposited in seals.
2. The thick layer of fat deposited under their act as insulators preventing heat loss from its body.
3. The fat not only insulates the body but helps in producing heat and energy.

9th Class Biology Textbook Page No. 132

Question 20.
Can you give some examples of fleshy leaf plants?
Answer:
Yes. Bryophyllum, Aloe, and Agave are the examples for fleshy leaved plants.

Question 21.
Why xerophytic plants do not have broad leaves?
Answer:
To prevent the excessive loss of water through respiration xerophytic plants do not have broad leaves.

Question 22.
You may see Kittanara, a xeric plant, grown as fence around crop fields in some areas in our state. Actually those places are not desert. How can they grow there?
Answer:
They grow there because this plant shows adaptations in that places.

9th Class Biology Textbook Page No. 133

Question 23.
Do all animals living in desert conditions show adaptations?
Answer:
Yes, all animals living in desert conditions show adaptations.

Question 24.
Why some animals have scales on their body?
Answer:

1. Scales mainly protect the animals from environment.
2. In desert animals scales allow them to retain moisture by preventing the evaporation of water through the skin.
3. This allows the animal to become dehydrated and animal requires small amount of water to survive.

Question 25.
Why the animals that lives in burrows usually comeout during night time only?
Answer:
To protect themselves from extreme hot conditions, animals that live in burrows usually comeout during night time only.

9th Class Biology Textbook Page No. 139

Question 26.
Which organism among jelly fishes and decomposers present in euphotic zone?
Answer:
Jelly fishes are present in euphotic zone.

Question 27.
What kinds of adaptations can be seen in the organisms of the euphotic zone?
Answer:

1. The organisms living in this zone are mostly floaten and swimmen.
2. Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface are transparent.
3. These usually have sharp vision.

Question 28.
What kind of adaptations can be seen in the organisms of abyssal zone?
Answer:

• The larger animals in abyssal zone have wide mouths and huge curved teeth which prevent escape of any prey.
• Absence of skeleton, flattened bodies are some other characteristics observed.
• Some animals also have special structures that produce light on their bellies, around their eyes and at the sides of their bodies.
• Some animals shows bioluminiscence in the dark waters.

Question 29.
What differences can you find in the animals of bathyal zone when compared to animals of euphotic and abyssal zones?
Answer:

• Most of the plants found in this zone are the red and brown kelps, sponges, corals even animals with tubular bodies like squids and large animals like whales, etc.
• Some of the animals in the bathyal zone have a flat body like the ray fishes.
• Big eyes sensitive to very dimlight may present in bathyal zone animals.

Question 30.
How organisms of different zones of marine ecosystem are adapted?
Answer:

• The animals of euphotic zone are mostly floaters and swimmers.
• Animals in this zone usually have shiny bodies reflecting light away to merge with shiny water surface.
• Animals of euphotic zone have very sharp vision.
• Some of the animals in bathyal zone have a flat body like the ray fishes.
• The animals may have big eyes sensitive to very dim light in bathyal zone.
• Absence of skeleton, flattened bodies are some adaptations found in animals of abyssal zone.
• Some animals in abyssal zone may have special structures that produce light on – their bellies, around their eyes and at the sides of their bodies.
• Some animals in abyssal zone shows bioluminiscence in the dark waters.

9th Class Biology Textbook Page No. 141

Question 31.
Organisms of the oceans have a lesser salt content in their bodies than the seawater around 3.5%. The fluid could drain out of the body of the organisms into the sea. This could be dangerous and fatal to the organism. How do they survive under such conditions?
Answer:

• Several marine species have a lower internal salt concentration than that of the water they swim in. So they tend to dehydrate as water is lost by osmosis.
• To compensate, they drink large amounts of water and excrete the salts both via their kidneys and through highly specialised cells in the gills.

Question 32.
Can fish in estuarine ecosystem survive in river as well as in sea?
Answer:

• Yes, fish in estuarine ecosystem survive in river as well as in sea.
• Two of the main challenges of estuarine life are the variability in salinity and sedimentation.
• Many species of fish living in estuarine have various methods of control to the salt shifts.
• They regulate the salt concentrations using osmoregulaters.

9th Class Biology 9th Lesson Adaptations in Different Ecosystems Activities

Activity – 1

Question 1.
i) Take a Kalabanda (Aloevera) and a Balsam plant in two separate pots.
ii) Water each of them with two tablespoons of water.
iii) Do not water them for a week.
iv) Observe the condition of the plants after a week.

Observations :
a) Which plant showed growth?
Answer:
Kalabanda plant showed growth.

b) Which plant dried first? Why?
Answer:
Balsam plant dried first. Because Balsam plants are not watered regularly. They need water to grow.

Activity – 2

Question 2.
i) Collect an aquatic plant out of a water body (e.g. Duck weed, Hydrilla, Vallisneria etc.) ii) Carry it back home and plant it in a pot and water it.
Observations :
a) From the above activity we see that some plants dry up without water very quickly, while other can grow even with very little water.
b) Each of these plants are adapted to the conditions in their surroundings on the basis of need of water.

Activity – 3

Question 3.
You know some of the animals that reside in and around lake or pond. Make a list of those animals and the characteristics of their body.
List of animals and reside in and around lake or pond :
Insects : Dragonfly, Damsefly, Mayfix, Stonefly, Dobsofly, Caddisfly, Cranefly, Water bugs, Beetles, etc.

Characteristics of the body of animais living in and around lake:

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 8 Challenges in Improving Agricultural Products Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 8th Lesson Challenges in Improving Agricultural Products

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products Textbook Questions and Answers

Improve Your Learning

Question 1.
Suggest some ways through which our country could increase the production of rice to meet at least global limits. (AS 1)
(OR)
Day by day population is increasing. But the cultivated land is very limited. To produce required quantity of food for the growing population, what are the poible solutions in your view?
Answer:

• Increasing area of cultivated land.
• Increasing production in the existing land.
• Developing high yielding rice varieties.
• Conserving the genetic diversity of rice so it can be used in the development of new varieties suited to different growing conditions.
• Developing rice crop management strategies that improve nutrient use efficiency.
• Management of crop protection and suitable irrigation methods.
• Organic manure to be used for rice crop.
• Alternating crops and mixed crop system to be followed.

Question 2.
How are biofertilizers more beneficial as compared to chemical fertilizers? (AS 1)
Answer:

• Biofertilizers add natural nutrients to soil.
• They increases soil organic matter and improves soil structures.
• Biofertilizers improves water holding capacity of the soil and reduces soil crusting problems.
• They reduces soil erosion from wind and water.
• Biofertilizers increases crop yield.
• Biofertilizers improves the percentage of humus and remained long time in the soil.

Question 3.
a) Find out the adverse effects of chemical fertilizers need for growing the high yielding varieties of crops. (AS 1)
Answer:

• Chemical fertilizers pollute lakes, rivers and streams.
• They destroy beneficial soil life including earthworms.
• By using chemical fertilizers, we can get high yielding for only 20 to 30 years.
• After that soil becomes reluctant to plant growth.
• Chemical fertilizers damage soil fertility.
• Make certain crops vulnerable to diseases.
• Prevent some plants from absorbing needed minerals.
• Food produced by using chemical fertilizers do not taste as good.

b) Can high yielding varieties be grown without them (Chemical Fertilizers) as well? How? (AS 1)
Answer:

• Yes, high yielding varieties be grown without chemical fertilizers.
• By using biofertilizers, instead of chemical fertilizers and synthetic pyrethroids we get higher yielding.

Question 4.
What threats to nature do chemical fertilizers, pesticides, insecticides and herbi-cides pose? (AS 6)
Answer:

• When we use insecticides to kill pests or weedicides/herbicides to destroy weeds, a large percentages of herbicides, chemical fertilizers, pesticides, insecticides remain in the soil.
• From the soil, these chemicals find their way into water sources.
• People spray these chemicals in fields are exposed to them and some of the chemi-cals enter their body.
• Insecticides destroy all insects in which some of them are useful in pollination.
• Extensive use of chemical fertilizers, pesticides, herbicides and weedicides makes the soil unsuitable to grow crops after sometime.

Question 5.
What are the adverse effects of using high yielding varieties of seeds? (AS 1)
Answer:
The adverse effect of using high yielding seeds is – they use more nutrients from the soil. Thus the soil can lose its fertility if they are used continuously.

Question 6.
What are the essential measures that a farmer needs to take before sowing the seeds of a crop? (AS 1)
Answer:

• Preparation of soil is done before sowing the seeds.
• The soil is ploughed to loosen and break the solid pieces of soil.
• The field is watered before sowing.
• Seed treatments against/soil-borne diseases to reduce the incidence of diseases.

Question 7.
Suppose you had a farm in a drought striken area of your state, what crops would you grow and how? (AS 1)
Answer:

• Sorghum, Pearlmillet, Red gram, Green gram, Horse gram can be grown in drought striken area.
• We can grow these crops by rain water harvesting building check dams, drip irrigation methods, watershed management and soil and water conservation methods.

Question 8.
What measures will you take to save your field from seasonal outburst of insects? (AS 1)
Answer:
Nowadays farmers use insecticides and other chemicals to save their crop field.

• I prefer catching the insects manually and removed from the field.
• I also use predatory insects to remove insects from the field.
• I place lighted bulb (Deepapu Teralu) so that insects could cluster around it.
• Insecticides are sprayed at regular intervals.

Question 9.
What basis would you adopt to explain to a farmer using chemical fertilizers switch over to organic fertilizers? (AS 4)
Answer:

• Organic fertilizers replenish the soil, keeps soil easily broken up into small pieces.
• Organic fertilizers promotes beneficial soil life.
• Organic fertilizers increase crop yield.
• They maintain a natural balance in the soil.
• They protect certain crops from diseases.
• Benefit the environment by recycling agricultural wastes into energy for local community.

Question 10.
A farmer had been using a particular insecticide for a long time. What consequences will it have on – a) insect population b) soil ecosystem? (AS 1)
Answer:
a) Insect population :

1. Insect develop immunity to the insecticide used.
2. And it has any effect on the insect it targets. Hence the number of insects increases,

b) Soil ecosystem :

1. A large percentage of insecticide chemicals remain in the soil.
2. These chemicals kill the worms in the soil which are useful to soil.
3. Hence soil ecosystem destroys by increasing the concentration of salts in the soil.

Question 11.
Venkatapuram village is in drought prone area. Somaiah wants to cultivate sugar-cane in his fields. Is it beneficial or not? What questions will you ask him to convey your belief? (AS 7)
Answer:

• It is not beneficial for Somaiah to cultivate sugarcane crop.
• Sugarcane grown in places where rich water resources are present.
• “Where do you get water to cultivate sugarcane crop?” I ask this question to Somaiah.
• I advise him to grow crops which needs less water (Aruthadi Pantalu) in drought prone area.

Question 12.
Draw a block diagram of water resources in your village. (AS 5)
Answer:

Question 13.
Ramaiah has soil testing done in his field. The percentages of nutrients are 34-20-45. Is it suitable for cultivating sugarcane crop ? Which crops can be cultivate without using pesticides in Ramaiah’s field? (AS 2)
Answer:

• Ramaiah’s field is not suitable for cultivating sugarcane crop.
• Because sugarcane needs 90% of nitrogen in the soil but Ramaiah’s field has only 34% of nitrogen.
• Maize and groundnut can be cultivated in Ramaiah’s field.

Question 14.
Organic manure is helpful to biodiversity. How do you support this statement? (AS 6)
Answer:

• Biological research on soil and soil organisms has proven beneficial to the system of organic farming.
• Varieties of bacteria and fungi breakdown chemicals, plant matter and animal waste into productive soil nutrients.
• In turn the producer benefits by heal their yields and more suitable soil for future crops.

Question 15.
Make a list of major weeds in your area (You have already conducted the project) Find out the different weeds that grow along with different crops in your area. (AS 4)

Answer:
Cynodondacylon, Digitaria longifolia, Dacty loctenium colonum, Setaria glauca, Cyperus rotundus, Cyperus difformis, Eichornia crassipes, Salvinia mollusta, Alternathera sps. Celosia argentea, Leucas aspera, Portulaca oleracea, cleome sps. Solanum nigum, Argemone mexicana, Abutilon indicum, Euphorbia sps. Vernonia Cinnera, Eichnochloa colonum, Commelina bengalensis, Avenafatua, Eichnochloa Crusgalli, Eleusine indica, Euphorbia hirta, Achyranthus, despera, Eclipta prostrata.

Question 16.
Spraying high dose of pesticides is hazardous to biodiversity and crop yielding. How can you support this statement?
Answer:

• When we use pesticides large percentage of it will remain in the soil. These kill the germs in the soil.
• From the soil pesticides find their way into water bodies affecting the aquatic animals.
• People who spray these pesticides in the fields are exposed to them and some of the chemicals enter their bodies causing health problems or some times the person dies.
• Pesticides destroy all the insects which are useful to the plants causing hazardous to biodiversity.

Question 17.
Natural pest controlling methods are useful to biodiversity. Comment it.
Answer:

• Some insects control the harmful insects and they are called friendly insects.
  E.g.: Spiders, Dragonfly, Krisopa etc.
• Trachoderma bacterium lives in the eggs of stemborer, tobacco caterpillar destroy the pests at the egg stage.
• Some bacteria like Bacillus Turengenisis destroy some pests.
• Mixed crops also control some pests and diseases.
• Hence natural pest control methods are useful to biodiversity because these methods destroys only the selected pests.

Question 18.
Observe the fields in your surroundings and collect the information from farmers about the process to remove weeds.
Answer:
Farmers use different methods to remove weeds. Some of them are
1) Manual method:
Many farmers still remove weeds by manually pulling them out of the field, making sure to include the roots that would otherwise allow them to resprout.

2) Stale seed bed method :
This method involves cultivating the soil, then leaving it follow for a week or so when the initial weeds sprout, the farmer lightly hoes them away before planting the desired crop.

3) Using Herbicides :
Selective herbicides kill certain targets while leaving the de¬sired crop relatively unharmed.

4) Biological control:
Vinegar kills the visible part of the weed. They will wrinkle and die next day.

5) Ploughing & Tilling :
Ploughing includes filling of soil, inter-cultural ploughing and summer ploughing. Ploughing up roots weeds causing them to die. Mechanical tilling can remove weeds around crop plants at various points in the growing process.
Crop rotation method also helps in controlling weeds.

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products InText Questions and Answers

9th Class Biology Textbook Page No. 110

Question 1.
Rate of growth of population and food grain production.

1. In which decade population growth is higher?
Answer:
Population growth is higher in 1961-1971.

2. In which decade food grain production is higher?
Answer:
Food grain production is higher in 1981-1991.

3. What major differences did you find in the table?
Answer:
The major differences find in the table are :
i) Foodgrain production is not increasing according to population growth.
ii) Ratio of FP/PG is in irregular order.

4. Is food grain production increasing according to population growth?
Answer:
No, last two decades food grain production is not increasing according to population growth.

5. In which decades production of food grains didn’t satisfy the needs of population? What will happen if the production is not sufficient?
Answer:
In 1991-2001 production of food grains didn’t satisfy the needs of population. If the food production is not sufficient then it leads to food crisis.

6. The decade 1991-2001 shows that rate of food production was nearly half as compared to population. What can you infer from the decade when population growth was highest?
Answer:
The reasons for the highest population growth :
i) Wide spread diseases are controlled.
ii) Health care programmes were made available in rural areas.
iii) So death rate declined.
iv) Therefore population growth become inevitable.

Question 2.

1. Find out from the graph the months in which the most water evaporates from plants.
Answer:
The months in which the most water evaporates from plants are May and June.

2. Are these the same months in monsoon season when the rainfall is heavy?
Answer:
No, these are not same.

3. So how does the availability of more water effect the plant?
Answer:
The availability of more water effects the plant with more evaporation.

9th Class Biology Textbook Page No. 110

Question 3.
When the weather is hot and the stomata dose, what effect would this have on the absorption of carbon dioxide by the plant?
Answer:
If the stomata closed, then the absorption of carbon dioxide by the plant is less.

Question 4.
What effect would a change in the amount of carbon dioxide absorbed have on the growth of the plant?
Answer:
If the absorption of carbon dioxide by the plant is less then the growth of the plant decreases.

Question 5.
If the plant does not get water at this time, what effect would this have on its growth? Discuss in your class and find out reasons.
Answer:
If the plant does not get water at this time then the growth of the plant will stopped.

Question 6.
What are the main water sources in your village for agriculture? How farmers utilise them?
Answer:

• Canals, Bore wells, ponds are the main water sources in our villages.
• Farmers utilise water from these sources to cultivate crops.

Question 7.
Make a list of crops which require less amount of water.
Answer:
Cotton, Jute, Bajra, Maize, Coconut, Black gram, Green gram etc.

9th Class Biology Textbook Page No. 111

Question 8.
If a field is cultivated for many years, what would happen to the nutrient content of the soil?
Answer:
If a field is cultivated for many years, then the nutrient content of the soil is decreased.

Question 9.
How does the soil get back or replenish these nutrients?
Answer:
The soil get back or replenish of these nurients by adding organic manure or chemical fertilizers.

9th Class Biology Textbook Page No. 112

Question 10.
A farmer cultivated sugar cane in his land for the last five years. Another farmer culti¬vated sugarcane in the first year and soya bean in the second year and sugarcane in third year.
– In which case do you think has the land lost most of its nutrients?
Answer:
The land lost most of its nutrients in the case of first farmer.

Question 11.
Have you ever seen two types of crops in the same field?
Answer:
Yes, I have seen two types of crops in the same field.

Question 12.
Which crops are grown in this way?
Answer:
In the fruit growing fields like Lemon, Pomegranate, Papaya, etc. pulses like Red gram, Black gram, Green gram, etc. are grown in this way.

Question 13.
What are the uses of cultivating mixed crops?
Answer:
The uses of cultivating mixed crops are :
1) The soil becomes fertile.
2) The nutrients which are used by one crop will be regained by cultivating another crop.

9th Class Biology Textbook Page No. 113

Question 14.
Is betel (Tamalapaku) a mixed crop? How can you justify your answer?
Answer:
Yes. Betel is mixed crop. Sorghum grown along with betel.

9th Class Biology Textbook Page No. 121

Question 15.
If we don’t use these chemicals, how can we get a good crop? How can we increase production? Is there an answer to this question? What coidd it be?
Answer:
Suppose we can use some other methods that do not give rise to these problems. For example, they say we can make use of the natural food chains to control pests. There are many insects that eat other insects. They are called predatory insects. We can make use of these insects. There are also birds that eat insects. We can use these birds to get rid of insects.

Question 16.
If insects that pollinate crops are killed, what effect will this have on crop production?
Answer:
If insects that pollinate crops are killed, the crop production will decrease.

Question 17.
In recent times, why farmers touch the flowers with handkerchiefs in sunflower fields?
Answer:
Farmers touch the flowers with handkerchiefs in sunflower fields to control the insects.

9th Class Biology Textbook Page No. 122

Question 18.
Do you know why Jatropa in cotton fields and marigold in mirchi fields are cultivated?
Answer:
Some mixed crops controls some diseases and pests. That’s why Jatropa in cotton field, marigold in mirchi fields are cultivated.

9th Class Biology Textbook Page No. 110

Question 19.
In what way this kind of water supply is useful to the crop as well as the farmer?
Answer:
To prevent water wastage and economically helpful to the farmer.

Question 20.
Water Shed is a process to improve ground water level. In what way it is related to irrigation? Support with your answer.
Answer:
If ground water level will be increased then it will help to irrigation.

9th Class Biology Textbook Page No. 117

Question 21.
In what way vermi compost is better than chemical fertilisers?
Answer:
After using vermi compost, investment on chemical fertilizers and other pesticides became reduced and the quantity of their agricultural products increased.

9th Class Biology 8th Lesson Challenges in Improving Agricultural Products Activities

Activity -1

Question 1.
Observe the Transpiration :

1. Take a polythene bag. Cover the bag on leaves and tie it.
2. Do this experiment during day time and night time separately.
3. Note the difference in your notebook.

Observations :

1. If we tie a plastic bag over a leaf, we will be able to see how much water a plant releases in the air.
2. It is estimated that a plant use only 0.1 percent of the water it absorbs to form carbohydrate.
3. The rate of transpiration is high during day time when compared to night time.

Question 2.
b) Draw the route map of Jawahar and Lai Bahadoor canals of Nagarjuna sagar in Andhra Pradesh & Telangana map.
Answer:

I. 1) Take one example from each of millets, cereals, vegetables, and fruits.
2) First you have to list out the known characters of the above and then list out the characters that you want to change or modify in them.
3) But you need to give your own reasons – why do you want to make such changes in them?

II. Red and yellow equal to rellow.

1) If you want to make your own hybrid flower you need to do the following. But it is time consuming process and patience job too.
2) For this you need red and yellow colour chandrakantha plants.
3) Select 5 or 6 red flowers on a plant.
4) Remove all the other flowers of that plant.
5) Take each flower, remove stamens carefully.
6) Take yellow flower and rub with that flower gently on the stigma of selected red flower for pollination. Do this process in the evening only.
7) Tie a tag with a thread loosely to the pollinated flower to avoid confusion in iden¬tifying these flowers for seeds in the next few days.
8) Within a week days you will get black seeds.
9) Keep them another two weeks to dry and sow them in a pot.
10) Take care to grow the plants until they flower.

Observations:
1) The colour of the flowers will be orange.

AP State Syllabus 9th Class Biology Solutions 5th Lesson Diversity in Living Organism

9th Class Biology 5th Lesson Diversity in Living Organism Textbook Questions and Answers

Improve Your Learning

Question 1.
Variations in organisms lead to diversity in living organisms. State reasons. (AS 1)
Answer:

• The presence of differences between organisms of the same species is called variation.
• Variation between different species is always greater than the animals within a species.
• The uniqueness of individual is the basis of the diversity that is shown by the living organisms.
• In our daily life, we see a variety of plants and animals in our locality.
• But if we go to some other places such as hills, forests or sea we entirely find different types of animals and plants.
• In fact, different parts of the world have their own typical kinds of living beings.
• Thus-we can say that variations in organisms lead to diversity in living organisms.

Question 2.
What was the basis for early classifications? (AS 1)
Answer:

• Living things are classified on the basis of their body structures.
• Living things are classified on the basis of dissimilarities and similarities.
• Charaka and Sushrut had classified plants on the basis of their medical importants.
• Parasar classified plants basing on the structure of flowers.
• Aristotle classified animals according to whether they lived on land, in water, or in the air.

Question 3.
What are the advantages of classifying organisms? (AS 1)
Answer:
Advantages of classifying organisms :

1. Classification makes the study of various organisms easy.
2. Classification helps to understand the inter relationships among different groups of organisms.
3. Classification helps in exploring the diversity of life forms.
4. Classification reveals evolution trends by showing simple body structures to complex body structures.
5. Geographical distribution of plants and animals is entirely dependent on the information given by classification.

Question 4.
What is the need of classification? What questions will you ask? (AS 2)
Answer:
Need of classification :

1. Classification gives better knowledge and better understanding of organisms that are studied.
2. It helps to study the organisms in a proper and systematic manner.
3. Classification helps to make comparison in an easier way.
4. It helps in understanding relationship among the organisms and their interdependence.
5. Classification makes our study more focused and helps us to handle huge population of organisms.
6. Classification gives us an idea of evolution.

Questions:
1. Who made the classification?
2. What are the advantages of classification?
3. What are the recent developments being done?

Question 5.
How do monocots differ from dicots? (AS 1)
Answer:

Question 6.
One day Kavita soaked seeds of green grams, wheat, maize, peas, and tamarind. After they became tender, she tried to split the seed. Name which would split, which would not, and identify them according to the characters. (AS 4)
Answer:

Question 7.
Make a flow chart of invertebrates in the kingdom Animalia, based upon their characteristic features. (AS 4)
Answer:

Question 8.
Write some common characters of Pisces, Reptilia, and Aves. (AS 1)
Answer:

1. Pisces, Reptilia, and Aves belong to vertebrate.
2. All these animals lay eggs.
3. All these animals possess vertebral column.

Question 9.
Name the kingdom to which these organisms belong according to Whittaker. (AS 1)

Answer:
a) Protista
b) Animalia
c) Fungi
d) Monera

Question 10.
Explain how animals in vertebrata are classified into further subgroups. (AS 1)
Answer:

• Vertebrata can be further classified into sub groups on the basis of simple to complex body structures and their functions.
• For example, fishes have two chambered hearts, amphibians have three chambered hearts while in birds and mammals have four chambered hearts to keep the oxygenated and deoxygenated blood separate.
• The following characteristic features are considered for classifying vertebrate into the further sub groups.

Question 11.
Platypus or Echidna is a group that forms a link between reptiles and mammals. Think and write about some characteristic features that these would have. (AS 4)
Answer:

• The platypus and echidna both belong to the group of animals known as monotremes.
• These two are characterised by the feature of egg laying mammals. Yet they are not birds or reptiles.
• Both creatures hatch their young from eggs, yet the mother of each species feeds her babies with milk from milk glands.
• These two are found in Australia and Tasmania.
• One of the characteristics of platypus is that it has an unusual duck like bill and does not have teeth an unusual characteristic for a mammal.
• Echidna, the spiny ant eater also does not have teeth. Tongue helps in feeding.
• Echidna and platypus young stay in burrow after they are hatched. Echidna develops a rudimentary pouch during breeding season.
• Both creatures have sharp claws for burrowing.
• Both the platypus and echidna like the water. Platypus hunt food in the water. Echidna regulates its temperature through swimming.

Question 12.
Sujata says Bat is not a bird but a mammal. How can you support Sujata’s statement?
Answer:

• Sujata’s statement that bat is not a bird but mammal is correct
• Like other mammals, including ourselves bats have hair or fur on their bodies.
• They are warm blooded animals.
• A baby bat that feed on its mother milk after it is born.
• Bats are the only mammals that can fly.

Question 13.
Which Phylum do I belong to? (AS 1)
a) My body is made of pores. I live in water. I do not have backbone also ……………….. .
b) I am an insect. 1 have jointed legs …………………. .
c) I am a marine living animal with spiny skin. My body is radially symmetrical ……………….. .
Answer:
a) Porifera
b) Arthropoda
c) Echinodermata

Question 14.
How can you appreciate the effort of scientists in classifying a wide range of organisms? (AS 6)
Answer:

• Classification makes the study of wide variety of organisms easy.
• It is essential to understand the inter-relationships among different groups of animals and plants.
• Classification gave us an idea of evolution of organisms from simple to complex ones.

9th Class Biology 5th Lesson Diversity in Living Organism InText Questions and Answers

9th Class Biology Textbook Page No. 59

Question 1.
Why do you think classification system has undergone changes over the years?
Answer:
Classification system has gone changes over the years due to the new discovered organisms, advancement in genetic science invention of powerful microscope.

Question 2.
If you were asked to classify organisms what would be your basis of classification?
Answer:
Our basis of classification would be whether the organism.

1. Has a prokaryotic or eukaryotic cell
2. Is unicellular or multicellular
3. Is autotrophic or heterotrophic
4. Mode of reproduction.
   Like that, I classify the organism in an orderly manner.

9th Class Biology 5th Lesson Diversity in Living Organism Tissues Activities

Question 1.
Collect leaves from different plants, observe them carefully and fill the table.

a) Could you find any two leaves which are similar with respect to any of the characters, size, shape, colour or any other as mentioned in the table?
Answer:
No. Every leaf has its size, shape and colour are same in many of the leaves.

b) Note down the differences you observed in the sample of leaves collected by you.
Write two such characters that differed most.
Answer:
1. Length and breadth of the leaves are different for each leaf.
2. Most of the leaves have reticulate venation but only in plantain parallel venation is present.

Question 2.
Collect 5 different plants from your surroundings and observe them carefully. Write your observations in the table given below.

a) Which characters given above varied most?
Answer:
Length of stem, Inter nodal distance, venation in leaves, and in type of root system.

b) Select a character mentioned above which shows minimum diversity.
Answer:
Flower shows least diversity arrangement of flowers in bunches.

c) Did you find any similarities? What were they?
Answer:
Yes. Similarity in venation, number of sepals and petals, and in type of root system.

d) Did you find patterns like plants with fibrous roots had flowers borne in groups? Note the other patterns that you observed.
Answer:
Yes. The flowers are borne in groups, reticulate venation.

e) Carefully observe the plants collected by you and note down some other characters not mentioned in the table.
Answer:
Spine are present in rose plants whereas they are absent in other plants.

f) Did you notice any two plants which were alike with regard to the above characteristics? If not, note down what differences you found?
Answer:
No. Venation, root system are the differences.

Question 3.
How do you observe the number of cotyledons in different seeds? Write your findings in the table.
Answer:
Observing cotyledons in seeds :

1. Collect seeds of plants from green gram, red gram, bengal gram, wheat, paddy, groundnut, maize and soak them for a day.
2. Take a maize seed and press it between fingers.
3. A small whitish structure come out of the maize seed.
4. Whitish structure is known as embryo/baby plant.
5. The portion left in our hand within the seed coat has a single cotyledon.
6. Repeat the activity with soaked seeds by pressing them with fingers.
7. Observe the pressed seeds with the help of a hand lens and fill the table.

Activity – 4

Question 4.
Collect the plants or pictures of the plants to complete the following table.
Answer:

Activity – 5

Question 5.
Collect housefly, mosquito, ant, dung beetle, butterfly, moth and cockroach from your surroundings. Observe them with magnifying glass and fill the table.
Answer:

a) What differences did you observe with regard to legs?
Answer:
In some insects legs are large in size whereas in some like Cockroach, Butterfly the legs are big in size.

b) What differences did you observe with regard to wings?
Answer:
In ant, housefly, mosquito and in dungbeetle a pair of wings are present whereas in butterfly, moth and cockroach 2 pairs of legs are present. In some like ant, housefly, mosquito the wings are small whereas in others they are big in size.

c) Is there any relationship between the number of wings and legs?
Answer:
As the size of the wings increases the length of legs decreased. The number of legs in all the insects are 3 pairs whereas the wings are one or two pairs.

Activity – 6

Question 6.
In order to observe diversity in animals select 10 children from your class and fill the following table with their data.
Answer:

a) Which character helps you to make a group with maximum individuals?
Answer:
Height helps us to a group with maximum individuals.

b) Which character helps you to have just a single individual in a group?
Answer:
Thumb impression.

c) Compare your group table with that of other groups and note down the differences you found.
Answer:
Student’s activity.

d) Did you find same observations of any two students in your class?
Answer:
No.

Activity – 7

Question 7.
Collect two small almost equal sized neem plants from your surroundings observe them and fill the table.

a) What differences did you find in the similar neem plants?
Answer:
The differences are in length of the stem and number of leaves.

b) Why do such differences present in nature?
Answer:
Every plant has got its own characteristics. The age of the plants also responsible for diverse characters.

Activity – 8

Question 8.
How do you observe moss plant through hand lens or dissection microscope? Draw the diagram and write the characteristics of moss plants.
Answer:

• Collect mosses from the greenish velvety growth on bricks during rainy season.
• Scrap a bit of the greenish velvety over a slide and observe under a dissection microscope.

Observations :

1. The structure that are seen in moss plants are not flowers but they are spores.
2. Spores are formed in Sporangium.
3. Spores contain very little amount of food.

Lab Activities

9th Class Biology Textbook Page No. 67

Question 1.
Observe slide of Hydra under a microscope. Draw the diagram and write your finding.
Observations :

a) Is the body made of single cell or a group of cells?
Answer:
The body of Hydra is made up of number of cells.

b) Did you find any hollow structure inside the body?
Answer:
The hollow structure found inside the body is called coelom or body cavity.

c) Did you find any other characters in it?
Answer:

• The proximal or aboral end is drawn into a slender stalk on the end of which is the basal disc for attachment.
• The free distal end or oral end bears the mouth which is situated on hypostome.
• The hypostome is encircled by 6 -10 tentacles.
• Bud is present at side with a mouth or tentacles like the parent.

Question 2.
Observe slide of tape worm (Taenea Solium) under microscope and write its external characters. Draw diagram of it.
Observations :

a) How does the body look like?
Answer:
1) The body of Taenia Solium (Tape worm) is long, dorso- ventrally, flattened, narrow, ribbon like.
2) Body consists of scolex or head, neck and strobila or body segments.

b) Did you see a body cavity in it?
Answer:
There is no true internal body cavity or coelom.

c) How does the anterior and posterior look like?
Answer:
The anterior (head) is smaller than the head of a pin. The posterior (tail) is very long and bigger than head and neck.

Question 3.
Observe slide of round worm (Ascaris) and write the characters of it by drawing the figure.
Observations :
a) Does the body look like the same as in the platyhelminthes?
Answer:
The body is round and cylindrical but not flat as in tape worm.

b) What are the differences you observed between tape worm and round worm?
Answer:

Pseudocoelom is present, whereas it is absent in Tape worm. The head and tail are tapering at the ends.

Question 4.
Observe the specimen of earthworm and draw the diagram. Write its characters you have observed.
Observations :

Answer:

1. The body of earthworm is bilaterally symmetrical and extensively segmented.
2. The anterior end is tapering while posterior end is more less blunt.

a) Touch the skin of the earthworm and say how do you feel?
Answer:
The skin of earthworm is moist.

b) What is the colour?
Answer:
The colour of earthworm is dark brown in colour.

c) Are there any differences you observed in its body colour and among the body parts?
Answer:
The dorsal surface is darker than the ventral surface.

d) How does it move?
Answer:
Earthworm moves by alternate contractions of circular and longitudinal muscles.

e) Are there any ring like structures seen in its body?
Answer:
Ring like segments are present in earthworm.

9th Class Biology Textbook Page No. 68

Question 5.
Observe the specimen of Cockroach. Draw its diagram and write its characteristics.
Observations :
a) How does the skin look like? Did you observe any hard layer on the skin?
Answer:

The entire body of cockroach is covered by a hard brown Coloured chitinous exoskeleton.

b) How many parts is the body divided into?
Answer:
The body is segmented and distinctly divisible into three parts

1. Head
2. Thorax
3. Abdomen.

c) Observe the legs and says how does it look like.
Answer:
Three pairs of legs are present. Each leg consists of five segmented. Jointed legs are present.

d) Name some more animals whose legs are jointed as seen in Cockroach.
Answer:
Prawn, scorpion, grasshopper, ant, mosquito have jointed legs.

Question 6.
Observe the snail by keeping inside a glass beaker and observe its characters.
Observations :

a) How does the outer body look like?
Answer:
The outer body is covered with shell.

b) Keep the snail unmoved for sometime and when it starts moving observe its body.
Answer:
The animal creeps by its ventral muscular foot. The movement is gliding movement.

c) Is the body soft or hard?
Answer:
The body of the snail is soft.

d) Did you find any antennae like structure in it?
Answer:
Yes, Tentacles are present.

Question 7.
Observe specimen of starfish and write your observations.
This specimen belongs to Phylum echinodermata.
Observations :
a) What do you find on the skin of the starfish?
Answer:

Spines are present on the skin of the starfish.

b) Are there any arms and ray shaped structure in it?
Answer:
Most of them are pentamemal, it means they have five fold symmetry with rays of arms in fives.

c) Do you find a small hole in the middle of the starfish?
Answer:
In the middle of starfish a small whole is present which is the mouth of it.

Question 8.
Collect a fish from a fish monger and observe its external characters.
External characters of fish :

a) Observe the skin of the fish. How does it look like?
Answer:
Body is covered with scales.

b) Write the body parts of the fish where scales are not present.
Answer:
On the head, on the tail, fins and on the lower side of the fish.

c) Open the mouth of the fish. What do you seen in it ?
Answer:
Teeth are present in the mouth. Tongue is also present.

Activity – 9

Question 9.
Try to find out the scientific names of at least 10 organisms that you see around you.
Answer:
Scientific names of plants around us :

Scientific names of animals around us :

AP Board 9th Class Biology Solutions

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 7 Animal Behaviour Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 7th Lesson Animal Behaviour

9th Class Biology 7th Lesson Animal Behaviour Textbook Questions and Answers

Improve Your Learning

Question 1.
What is the advantage of reflex action? (AS 1)
(a) It has to be learned
(b) It happens differently each time
(c) It does not have to be learned
(d) None of them
Answer:
(c) It does not have to be learned.

Question 2.
If a rat is given a mild electric shock when it goes to a certain part of its cage, it eventually avoids going there. This is because of- (AS 1)
(a) Imitation
(b) Conditioning
(c) Instinct
(d) Imprinting
Answer:
(b) Conditioning

Question 3.
Describe all types of behaviour discussed in the lesson with appropriate examples. (AS 1)
(OR)
Describe different types of animal behaviours discussed in the classroom with suitable examples.
Answer:
Scientists categorize animal behaviour into different categories like instinct, imprinting, conditioning, imitation.
1) Instinct:
A) Instinctual behaviours are behaviours that need not be learned.
B) They are inborn behaviours and can be complex like making nest by birds, choose mates and forming into groups for protection.
C) Reflexes are also a type of instinct behaviour.

2. Imprinting:
A) Chickens and ducklings are able to walk almost immediately after hatching from the egg.
B) Duckling can even swim after a few days.
C) They recognise their mother because of a behaviour type called imprinting.
D) Imprinting lets young animals recognise their mother from a young age. They can follow her for food and protection.

3. Conditioning:
A) Conditioning is a type of behaviour involving a response to a stimulus that is different from the natural one.
B) It is a type of learned behaviour.
C) If we take ringing of school bell as an example, student shows different types of conditioning to a school bell as per the time.

4. Imitation:
A) It is a type of behaviour where one animal copies another animal.
B) Kohler conducted experiments on imitation in chimpanzees.
C) One chimpanzee tried to take a fruit from a tree. But it failed in reaching the fruit.
D) Later chimpanzee used sticks to reach the fruit. This time it succeded in reaching the fruit.
E) Chimpanzee used sticks to spear juicy grubs to eat.
F) Other chimpanzees copy this behaviour.

Question 4.
Differentiate between (AS 1)
Answer:
a) Imitation and Imprinting.

Question 5.
How human behaviour is different from behaviour of other animals? Explain with an example. (AS 1)
Answer:

• Humans show many of the same types of behaviour as other animals.
• But human behaviour is often more complex because we are more intelligent and aware of ourselves.
• For example, hungry persons might want to start eating immediately when they sit down at the dining table.
• But the humans have learned that good manners mean they should wait until everyone is seated and ready to eat.
• But animals eat food whenever they find it.

Question 6.
Observe ants going on a line. Ask your teacher how they communicate and write a note on this. (AS 4)
Answer:

• Ants talk to each other using chemical signals called pheromones that they detect with antennae.
• Ants use their antennas to pick up smells.
• For example, when ants find food they have a pheromone trail from the food soure to the colony.
• Other ants follow this trial. As the other ants follow the trial, the pheromone scent becomes stronger.
• The paired antennas of ants provide information about the direction and intensity of scents.
• Since most ants live on the ground, they use the soil surface to leave pheromene trail that may be followed by other ants.
• Some ants produce sounds using gaster segments and their mandibles.
• Sounds may be used to communicate with colony members or with other species.

Question 7.
“Understanding of animal behaviour creates positive attitude towards animals”. How do you support this statement? Explain with suitable examples. (AS 6)
Answer:

• I support the above statement that understanding of animal behaviour creates positive attitude towards animals.
• Animals usually make sounds depending upon their needs. They show different facial expressions.
• For example, cattle make sounds whenever they need food and water.
• After giving food and water by the master they calmdown.
• When a crow dies, all the other crows come around making sounds ‘kaww, kaww’ shows their sadness, we have to understand nature of the crows in this situation.
• We have to identify the unity and integrity among the ants when they go in line in search of food.
• When dogs bark during nights, we should understand that they are doing that for our safe.
• We have to show positive attitude towards animals who are useful in our daily life. Live and let live should be our motto.

Question 8.
Look at this picture. How do you feel about sibbiling care nature of animals. Have you ever seen such kind of situations in your surroundings? Explain in your own words. (AS 7)

Answer:

• Sibbiling care nature of animals are animals with instinct behaviour.
• Every animal take care of their young ones until they are grown adults.
• I observed sibbiling care situations in my surroundings.
• The newly hatched chickens are taken by their mother to surrounding places to feed them.
• Whenever the chicks faces danger mother brings them under her wings.
• When the eagle tries to take away the chicks the mother attacks the eagle to save its chickens.
• Chicken feeds and protect young ones until they are able to collect their own food.
• A new born kitten is born blind. Its eyes normally do not open until it is 10 to 12 days.
• Finding the milk source is accomplished with help from mom, who encourages young kittens to feed a few minutes after birth.
• Cat often changes its living place by transfering young kitten. It does so to protect kittens from enemies.

9th Class Biology 7th Lesson Animal Behaviour Activities

Lab Activity

Question 1.
Behaviour of Cockroach : For this we need a choice box and calcium chloride.
Answer:
Making of Choice box :

• Take a box, and divide it into four chambers with the help of a card board.
• Make tiny holes in any two chambers of one side so that light can pass through these holes into the chambers.
• Let other two chambers as it is (dark).
• Now create humid environment with help of moist cotton wool in one of the lightened and one of the dark chambers.
• Create dry atmosphere with help of calcium chloride in one of the lightened and one of the dark chambers.

• So, the box has been divided into four chambers with different conditions i.e., light and dry, light and humid, dark and dry, dark and humid.
• Make four groups in class. Each group will put several cockroaches into a choice of chamber with four different conditions.
• Cover the box and leave the setup for 15-20 minutes.
• Count the number of cockroaches in each chamber.

Observations :

1. Cockroaches prefer dark and damp conditions.
2. The quarter of the choice chamber with these conditions contains most or all of the cockroaches.

Activity – 1

Question 2.
Observe the following behaviours of different animals. Identify their instinct, imprinting, conditioning or imitation.
a) Our pet dog barks only on strangers. If is not stopped, how would it behave?
Answer:
Conditioning.

b) Ants which usually go in a line reach sweet kept in tin. How do they know the way to reach the tin?
Answer:
Conditioning.

c) Mosquitoes, cockroaches come out of their places only when it is dark. How do they know the difference between light and dark?
Answer:
Instinct.

d) Bats and owl move and search for food during night only. How could they know what is a day what is a night?
Answer:
Instinct.

e) When you untie the neck of your bull at the time of ploughing, it moves towards plough without any instructions. In the same way, it moves towards tub at the time of feeding. How does the bull respond differently?
Answer:
Conditioning.

f) Birds collect material which is soft, strong to build its nest. How do they know the quality of material?
Answer:
Instinct.

g) Puppies, kitten fight each other when they saw a piece of cloth. They try to tare it off why?
Answer:
Imitation.

h) In a particular season some birds in our surroundings migrate from long distances. . How do they know their way?
Answer:
Instinct.

Activity – 2

Question 3.
Select any one of the animals in your surroundings. Observe it how it behaves in the following situation.
Answer:
1) Name of the animal:
Crow (corvus species)

2) Place where it lives :
They live in nests build on trees. Usually, they build nest where they feel safe from predators.

3) How it builds its place :
Crow builds its nest using tree branches, small sticks, hay etc.

4) Way of collecting food/prey:
Crows go around places where food is available. Crows are omnivorous and they eat almost everything.

5) External characters :
Crows are usually black in colour or black with little white plumage.

6) Expressions :
A) Crows make a wide variety of calls or vocalizations.
B) In many species the pattern and number of numerical vocalizations have been observed in response to events in the surroundings like arrival or departure of crows.
C) Crows show their happiness, Jadness, fear, threat by making sounds like ‘KOWWS’.

7) Group behaviour:
A) If one crow finds food it call others to join.
B) If one crow dies, all the other crows make ‘KOWWS’ continuously without interference.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 2 Plant Tissues Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 2nd Lesson Plant Tissues

9th Class Biology 2nd Lesson Plant Tissues Textbook Questions and Answers

Improve Your Learning

Question 1.
Define the terms (AS 1)
A) Tissue
B) Meristematic tissue
C) Dermal tissue.
Answer:
A) Tissue :
Tissue is a group of cells similar in structure and performing similar functions.

B) Meristematic tissue :
Meristematic tissue is the dividing tissue present in the growing regions of the plant.

C) Dermal tissue :
Tissues that form outer coverings are called as dermal tissues. It gives protection to the plant.

Question 2.
Differentiate the following. (AS 1)
A) Meristematic tissue and Ground tissue :

B) Apical Meristem and Lateral Meristem :

C) Parenchyma and Collenchyma :

D) Sclerenchyma and Parenchyma :

E) Xylem and Phloem :

F) Epidermis and Bark :

Question 3.
Name the following. (AS 1)
A) Growing tissue, which causes growth in the length of the plant.
Answer:
Apical Meristem

B) Growing tissue, which causes growth in the girth (diameter) of the plant.
Answer:
Lateral Meristem.

C) Large air cavities in the aquatic plants.
Answer:
Arenchyma.

D) Food material in parenchyma.
Answer:
Storage tissue

E) Pores essential for gaseous exchange and transpiration.
Answer:
Stomata

Question 4.
Compare and contrast the following. Xylem and Phloem. (AS 1)
Answer:

Meristematic tissue and Dermal tissue :
Answer:

Question 5.
Give reasons to the following. (AS 1)
A) Xylem is a conductive tissue.
Answer:

1. Xylem conducts water and mineral salts from the soil to the apical parts of the plants.
2. It transports materials away from the root.
3. Conduction of water by xylem is unidirectional i.e., from roots to apical parts of the plants.

B) Epidermis gives protection.
Answer:

1. Epidermis usually consists of a single layer of cells.
2. The walls of the cells of epidermis are thicker.
3. The epidermis protects the plants from loss of water, mechanical damage, and invasion by parasitic and disease causing organisms.

Question 6.
“Bark cells are impervious to gases and water”. What experiment will you perform to prove this? (AS 3)
Answer:

• In big trees dermal tissue forms several layers above the epidermis. It is called Bark.
• The several layers of bark does not allow any things like gases and water to pass through it.
• Hence bark cells are impervious to gases and water.

Question 7.
Though Chlorenchyma, Arenchyma, and storage tissues are parenchyma in nature, why do they have different (specific) names? (AS 1)
Answer:

• Chlorenchyma, Arenchyma and storage tissue are parenchymatous tissues,
• These three parenchymatous tissues are modified to perform various functions.
• The parenchyma which contains chloroplasts is called chlorenchyma. It performs photosynthesis.
• The parenchyma which contain large air cavities or spaces is called Arenchyma. It helps the plant to float.
• The parenchyma which stores water or food or waste products is called storage tissue.

Question 8.
Draw and label the diagram of the T.S of stem. (AS 5)
Answer:

Question 9.
Describe the functions of Meristem, Xylem and Phloem. (AS 1)
Answer:
1. Functions of Meristem:

1. It is a dividing meristematic tissue. It divides continuously. The cells formed from meristems later they differentiated as components of other tissues.
2. It brings about overall growth and repair.

2. Functions of xylem :

1. It conducts water and mineral salts from the root to apical parts of the plant like stems and leaves.
2. It gives mechanical support to the plant.

3. Functions of phloem :
Phloem conducts food material from the photosynthetic parts of the plants to other parts.

Question 10.
While observing internal parts of plants, how do you feel about its structure and functions? (AS 6)
Answer:

• While observing the internal parts of plants I felt that there are different types of tissues to perform various functions.
• For examples xylem and phloem of stem and stomata present on the outer layers of leaf are mend for performing different functions.
• Hence 1 felt that cells are organised from tissues and tissues are organised to perform various functions making the plant alive.

Question 11.
If you want to know more about tissues in plants, what questions you are going to ask? (AS 2)
Answer:
i) Which plant tissue provides both mechanical strength and flexibility?
ii) Which structure protects the plant body against the invasion of disease causing organisms?
iii) What will happen if apical meristem is destroyed or cut?
iv) What is the tissue present in the husk of coconut?
v) Why plants need different types of tissues?

Question 12.
Collect information about dermal tissues of plants in what way they help to them? Display it on wall magazine. (AS 4)
Answer:

• Dermal tissue usually consists of a single layer of tissues showing variations in the types of cells.
• On the basis of their location and function dermal tissues are studied as three different types epidermis (outer layer), mesodermis (middle layer) and endodermis (inner- most layer).
• Small pores are seen in the epidermis of the leaf called stomata.
• Cells of the roots have long hair like parts, called root hairs.
• Gum is secreted from the dermal layer of gum tree.
• The dermal layer protects the plants from loss of water, mechanical damage and invasion by parasitic and disease causing organisms.

9th Class Biology 2nd Lesson Plant Tissues Activities

Activity – 1

Question 1.
Parts of the plants and their functions.
Fill in the table.

Answer:

Activity – 2

Question 2.
How do you observe the cells in onion peel under microscope? Draw and label the diagram. Write your observations.
Answer:
Observing cells in onion peel:
Take a piece of onion peel. Place it on the slide.

Put a drop of water and then a drop of glycerine on it. Gently cover it with a cover slip. Observe it under microscope.

Observations:

1. All the cells are similar in shape and structure.
2. Intercellular spaces are present.
3. Cells are arranged in rows.
4. Each cell has cell wall, nucleus and cytoplasm.

Activity – 3

Question 3.
Observe the Cells in a leaf peel.
Answer:

• Take a betel leaf or a Tradescantia leaf.
• Tear it with a single stroke. So that a thin edge be seen at the torn end.
• Observe the thin edge where the leaf has been torn under the microscope in the” same manner as you had observed the onion peel.

Draw a diagram what you have observed and compare it with figure.

Observations:
We may have observed that the cells are present in groups with certain arrangement. With the help of following activities, we shall try to find out whether these arrangements have special roles to play in the plant body.

a) Are all the cells similar?
Answer:
All the cells are not similar in shape and size.

b) Is there any difference in their arrangement?
Answer:
They are arranged compactly living the small gaps called stomata. Stomata are surrounded by bean shaped cells (Guard cells).

c) What can we infer from the above activities?
Answer:
From the above activities it infers that some of the cells may modify or arranged in a different way to perform specific functions.

d) Have you noticed that the cells are in groups in both the activities?
Answer:
The cells are arranged in groups.

e) Compare and write a note on the arrangements of the cells that you have observed in both of the activities.
Answer:
In the second activity cells are almost of same size and shape but in this activity cells shape and size varies from one another depending on their function.

Activity – 4

Question 4.
How do you observe root tips of onion? Draw the diagram you observe under microscope. Write your observations.
Answer:

1. Take a transparent bottle filled with water. Take the onion bulb slightly larger than the mouth of the bottle.
2. Put the onion bulb on the mouth of the bottle.
3. Observe the growth of roots for few days till they grow to nearly an inch.
4. Take the onion out and cut some of the root tips.
5. Take an onion root tip. Place it on the slide. Put a drop of water and then a drop of glycerin on it.
6. Cover it with a cover slip. Put the 2-3 layer of filter paper on the cover slip.
7. Tap the cover slip gently press with the blunt end of the needle or brush to spread the material.
8. Observe under the microscope.

Observations:

1. All the cells are not similar in shape and structure.
2. Cells are arranged in rows.
3. Meristems are present below the root cap.

Activity – 5

Question 5.
Observe the roots of onion which had been cut off? Write your findings.
Answer:

• Take the onion and cut the end of the roots. Mark the cut end of roots with a permanent marker.
• Put the onion bulb on the mouth of the bottle.
• Leave the set up aside at least four to five days.
• Take care that there is enough water in the glass so that the roots are submerged.

Observations :

1. By removing the tip of the onion root, having a particular arrangement of cells, the growth of the root in length is stopped.
2. Cells are present in groups.

Activity – 6

Question 6.
Write down the arrangement of cells in the given table :
Answer:

Activity – 7

Question 7.
Observe temporary mount of T.S of Dicot stem under microscope and draw, label the diagram. Write your findings.
Answer:
Observing mount of T.S of Dicot Stem : Prepare a temporary mount of the T.S of dicot stem observe it under microscope.

Findings :

1. In the T.S of dicot stem meristematic tissue, vascular tissue, dermal tissue and ground tissue are present.
2. All the cells are not similar in shape and structure.

Activity – 8

Question 8.
Observe Rheo leaf peel under microscope. Draw and label the parts. Write your findings.
Answer:

1. Take a fresh leaf of Rheo leaf.
2. Tear it with a single stroke, so that a thin whitish edge can be seen at torn end.
3. Slowly remove it and observe the peel under microscope.

Findings:

1. Cells are structurally similar. They are compactly arranged without intercellular.
2. It is the dermal tissue of the plant.
3. It has an stomatal opening.

Activity – 9

Question 9.
Observe some tissues.
Answer:
Take permanent slides of Chlorenchyma, Arenchyma, Storage Tissue from your laboratory and observe them under the microscope. Find out the characteristics and differences and record them in your notebook.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 6 Sense Organs Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 6th Lesson Sense Organs

9th Class Biology 6th Lesson Sense Organs Textbook Questions and Answers

Improve Your Learning

I. Give reasons for :

Question 1.
We usually do not see bright colours in dim light.
Answer:

• Retina contains cells called rods and cones.
• Nearly 125 million tiny rods are present In our eye which contain pigment ‘rhodopsin’.
• Rods detect low intensity of light at night.
• Rods cannot make the fine distinctions of bright colours in dim light.

Question 2.
Removal of wax layer too often will raise incidence of ear infection.
Answer:

• External ear or pinna has wax producing ceruminous glands and oil producing sebaceous glands.
• These glands help to keep the ear canal lubricated prevent the dust and other par¬ticles from entering into the ear canal called Auditory Meatus.
• If we remove wax layer ear diseases like formation of pus, infection of eardrum etc. may be caused by bacterial and fungal infections.

Question 3.
During severe cough and cold we lose taste of food.
Answer:
When we suffering from severe cold and cough our nasal passages are completely blocked, food becomes tasteless as we are unable to smell the foods delicious aromas.

Question 4.
While cutting onions our tears start flowing.
Answer:

• The cells of onion contains amino acids like mithionine and enzyme cystine. They are kept separate in the onion cell.
• When we cut the onion, enzymes start mixing and produce propanethiol. Sulpher oxide, which is a volatile compound that starts moving towards our eyes.
• The gas that is emitted reacts with the water of our eyes and forms sulphuric acid.
• The sulphuric acid thus produced causes burning sensation in our eyes and this, in turn, leads to the tear glands secreting tears.
• Thus we end up with watery eyes every time we cut onions at home.

II. Find out the false statements and rewrite them as correct ones.

1. The rationale behind seeing is just the impression of the image in the retina.
Answer:
True

2. Ear functions only to hear.
Answer:
False
Correct statement: Ear helps in hearing and also in maintaining the equilibrium of our body.

3. Iris patterns are like fingerprints used in identifying individuals.
Answer:
True.

4. Saliva helps the taste buds in taste sensation.
Answer:
False
Correct statement: Taste receptors help the taste buds in taste sensation.

5. We are not able to adapt to sensations.
Answer:
False
Correct statement: We are able to adapt to sensations.

III. State the differences between the two.

1. Rods and cones.
Answer:

2. Iris and pupil.
Answer:

3. Pinna and tympanum.
Answer:

4. Nasal cavity and ear canal.
Answer:

IV. How do the following processes occur?

Question 1.
When we see an object, a real inverted image is formed on the retina.
Answer:

• When we see an object the eye gathers light through a convex lens, focusses it and forms an image in the retina at the back of the eye.
• The lens turns the image left to right and upside down.

Question 2.
The sound waves, collected by the pinna are changed as vibrations.
Answer:

• External ear or pinna collects the sound waves.
• They enter into the auditory meatus or ear canal. Then they strike the tympanum.
• The vibrations from the tympanum reach the malleus, incus and stapes in the middle ear.
• They magnify the intensity of the sound vibrations and send them to the membrane of oval window in middle ear.

Question 3.
We move our hand away from a hot object.
Answer:

• Moving our hand from a hot object is an unconditional reflex.
• Unconditional reflexes are inherited and shown from birth.
• Unconditional reflexes are present in all individuals and are basically same.

Question 4.
A pungent odour, makes us close our nose.
Answer:

• The olfactory receptors sense pungent odour, that information is sent to brain in the form of nerve impulses through sensory nerve.
• The brain interprets the information and identifies it as pungent odour.
• It sends message to our hand to close the nose immediately.
• The brain also sends information in the form of nerve impulses through motor nerves as we are unable to smell the food’s delicious aromas.

V. Fill in the blanks with suitable words. Then give reasons why the words are suitable.

1. Choroid layer provides ………………… to the eye.
2. The relationship between the tongue and ……………………. is more.
3. Iris pattern is used for individual …………………… .
4. Area where optic nerve leaves the eye is called the ……………………..
5. The ear drum is the …………………….
Answer:

1. Protection
2. Nose
3. Identification
4. Blind spots
5. Vibrating membrane

VI. Choose the correct option.

1. This vitamin is essential for the health of eye.
a) Vitamin ‘A’
b) Vitamin ‘B’
c) Vitamin ‘C’
d) Vitamin ‘D’
Answer:
a) Vitamin ‘A’

2. Sensation is a complex pathway involving
a) Sense organs
b) Sense organs and nerve impulses
c) Sense organs, nerve impulses, brain
d)Brain and nerve impulses
Answer:
c) Sense organs, nerve impulses, brain

3. The sound waves if not focused by external pinna and ear cannal will result in
a) Hearing several types of sound loudly
b) Not hearing anything
c) Slight hearing
d) Not being able to make out the type and origin of sound
Answer:
b) Not hearing anything

4. The muscles of the eyeball of a person becomes non functional, the invariable effect would be
a) The person fails to close eyes
b) Fails to move eye and see colours clearly
c) Feels pain in the eye
d) The nerves reaching the muscles become non-functional.
Answer:
b) Fails to move eye and see colours clearly

5. The tongue of a person is exposed to a high salty taste then:
a) The person learns to taste salty things better
b) Loves tasting salty things
c) Hates tasting salty things
d) Fails to taste a less salty thing just after the exposure.
Answer:
d) Fails to taste a less salty thing just after the exposure.

VII. Draw and label the diagrams, showing the structure of the
1. Eye

2. Ear

3. Tongue

VIII. How would you pay concern towards disabled people who is lacking sensory organs?
Answer:

• 1 will show utmost sympathy forwards disabled people who is lacking sensory organs.
• I will give my full cooperation in leading normal lives.
• With my deeds I will bring confidence among the disabled people.
• If the disabled people are blind everyday I will show the way to school and from school to home.
• I will see that the disabled people get the government help in a proper way.
• I will give my support to deaf people by giving symbols and signals to understand things.
• I will join the disabled people who are in the school age in the school mend for them.

IX. How do you appreciate the functions of sensory organs which helps us to enjoy the beauty of the nature?
Answer:

• Sense organs help us to enjoy the beauty of the nature.
• We enjoy the beauty of nature with our eyes, the melodious music with our ears, the taste of food with our tongue and feel the cool breeze on our skin.
• All these situations show just how our senses pick up information and react to them.
• Our sense organs are not just parts of us because nothing that we experience in our life, from the most important to the most boring, would be possible without the complicated power of our sense organs.
• Nothing in the entire universe of scientific exploration can even come close to match-ing the ability of our brain to use information sensed by our eyes, ears, skin, tongue and nose to produce a rich sensory experience in a matter of milli seconds.

X. Form a group with five students in your class and collect eye diseases and its char-acteristics by talking with ophthalmic assistant.
Answer:

XI. What happens if our skin loses its sensory nature?
Answer:

• The skin contains numerous sensory receptors which receive information from the outside environment.
• The sensory receptors of the skin are concerned with at least five different senses: pain, heat, cold, touch and pressure.
• The five are usually grouped together as the single sense of touch in the classification of the five senses of the whole human body.
• If the skin loss its sensory nature we cannot experience the pain, heat, cold, touch and pressure.

XII. Sagar is not able to listen things properly. Guess what would happen to him. What suggestions would you like to give him?
Answer:

• Sagar may be exposed to too much loud noise. This condition is noise induced hearing loss.
• Some times loud noise can cause a ringing, hissing or roaring sound in the ears called Tinnitus.
• Hearing problems may also be caused by a virus or bacteria.
• Hearing impairment happens when there is a problem with one or more partsflof the ear.
• So, I suggests Sagar to identify the reason for not listening things properly.
• I also advise him to consult a specialist called audiologist in ear problems.

9th Class Biology 6th Lesson Sense Organs InText Questions and Answers

9th Class Biology Textbook Page No. 76

Question 1.
Do you think our sense organs work together? Why, why not?
Answer:

• Yes, our sense organs work together.
• Every single function of the body is managed and controlled by the brain, including our organs and senses.
• Otherwise, we would have different interpretations of a stimulus, resulting confusion.
• But to be clear it is not the organs working together it is the brain constantly receiving stimuli from different senses.
• The brain is responsible for assimilating information and filling the pieces together.

9th Class Biology Textbook Page No. 85

Question 2.
If we do not have our external ear what will happen to us?
Answer:

• If we have no external ears, sound waves may not be collected by it.
• So we cannot hear anything and it leads to deafness.

9th Class Biology Textbook Page No. 87

Question 3.
If you are suffering from cold do you smell things in the natural way?
Answer:

• No, we cannot smell things in the natural way.
• When we have cold, we will notice foods seem tasteless because your nasal pas¬sages are blocked.

Question 4.
Do you find any relation between smell and taste?
Answer:

• Like smell, taste is also a sense based on identifying chemicals in food and the texture of it.
• The sense of taste and smell have a close and cooperative working relationship.

9th Class Biology Textbook Page No. 90

Question 5.
How sensitive is our skin?
Answer:

• Different parts of our body have different sensitivity and the skin around the neck and finger tips is more sensitive than skin on the palm, knee and arm.
• This is because sensitivity of our skin depends on
  a) thickness of our skin.
  b) the number of sensory receptors.

9th Class Biology Textbook Page No. 82

Question 6.
What will happen if we have no eyelashes?
Answer:

• The purpose of eyelashes is to keep moisture like sweat from setting into our eyes.
• Eyelashes protect the eye from debris and they are sensitive to being touched.
• If lashes are absent moisture like sweat will not be setting into our eyes.
• There will not be protection from debris to our eyes.

Question 7.
Are tears good for us?
Answer:

• Yes, tears are good for us.
• Whenever unwanted substances come in contact with the conjunctiva the lachry¬mal glands are stimulated to produce tears to wash the substances out of the eye.

9th Class Biology Textbook Page No. 89

Question 8.
Why are we suggested not to take too cool or too hot food material?
Answer:
To cool or too hot food material causes damage the sense of smell. It also damage the taste buds. If we take too cool or too hot food materials we will loose sense of smell and taste.

Question 9.
If you are suffering from fever that time you are not able to enjoy the taste of food why?
Answer:

• During fever, the temperature of our body increases from the normal body temperature of 98.6°F to high temperature sometimes to 105°F.
• At this temperature, the function of the enzymes in our tastebuds stop since they can work efficiently only in the temperature range of 77 to 98.6°F.
• The cells in the tastebuds cannot send messages to the nerve centres in the brain.
• That is why during fever we are not able to enjoy the taste of food.

9th Class Biology 6th Lesson Sense Organs Activities

Activity – 1

Question 1.
Note down a few lines of any text in your book. Write about the stimuli and re¬sponses and the sensory and motor functions with respect to the sense organs in¬volved.
Answer:
(Stimuli from the environment around are received by our body through some sense organs. As we already know, they are the eyes, ears, nose, tongue and skin. Let’s try to understand the path of receiving a stimulus to expressing a response (sensation))

1. The sense organs involved in writing the para above are eyes and skin.
2. Asking to write few lines in textbook is stimulus and writing the lines is response.
3. The sensory nerve in the eye take the information about writing lines to brain in the form of nerve impulses.
4. The brain interprets the signals and send the message through motor nerves to write the lines.
5. The skin in the palm helps to hold the pen in writing the lines.

Activity – 2

Question 2.
Testing tastes with tongue.
Answer:

• Dissolve a pinch of sugar in a glass of water.
• Drink a little of this. It doesn’t taste good.
• Try this for different concentrations of sugar adding by proper quantification that is weighing and preparing solutions to find out taste.
• You could take l/4th teaspoonful sugar each time which would be nearly 2 grams.
• After adding three spoons of sugar the taste of water becomes sugary.

Activity – 3

Question 3.
1. Observe the external structure of your friend’s eye, draw the diagram and lable it.
Answer:

2. Observe the eye ball of your friend in normal light. Then focus a beam of torch light on youj^friend’s eye.
Answer:

• Our eye contains eyelids, eyelashes, eyebrows and lachrymal glands.
• In normal light my friend’s eye ball is normal as usual, but after focusing a beam of torch light he immediately closed his eyes.
• After closing the eyes for two minutes, the black portion of eye is somewhat big in size.
• After opening forcibly when we throw the beam of torch light, the size of the small dark portion decrease.
• The small black portion in the centre of the eye expands in dark whereas in bright light it decreases in size.

Activity – 4

Question 4.
Testing the visual system.
Answer:

• Hold the text at arm’s length, close your right eye, and fix your left eye straight on the figure.
  
• Keep your right eye closed and bring the book slowly closer.
• When it is about 8 to 10 inches away the gap disappears as it is on the blind spot of your left eye.
• But you will not see a ‘hole’ in your visual field.
• Instead, your visual system “fills in” the missing area with information from the blue line on either side.

Activity – 5

Question 5.
Observe the iris and its surrounds of your friend’s eye. Can you find the pupil? Observe the colours and patterns in the iris of your friend’s eyes. Is there any difference from one another?
Answer:

• I found pupil in my friend’s eye.
• The colour of iris in my friend’s eyes are different.
• The colour of iris are blue in some, green in some and grey or brown in some of my friend’s eyes.
• The colour of iris are different but their shape is round in all my friend’s eyes.

Activity – 6

Question 6.
Enter into a dark room from a very bright place. What happens? Sit in a dark room for sometime. Then go into a bright light room. What happens?
Answer:

• If we enter into a dark room from a very bright place first we cannot see anything in the room.
• Because the pupil is very small in size and very less amount of light enters into the eyes.
• As the time progresses the pupil becomes large in size and we will be able to see things properly.
• If we go into a bright light room from a dark room, at first we cannot see anything in light from because the pupil does not bear the intensity of light.
• The pupil gradually lessens its size and we are able to see things properly.

Activity – 7

Question 7.
Eye and illusions.
Answer:

1. Take two pieces of white papers with same size.
2. Draw the picture of a cage on one pa¬per and the parrot on the other.
3. Then insert a stick and attach the blank sides of the papers with gum, see the figure.
4. Let it dry then twist the stick rapidly.
5. When we twist the stick rapidly, we see the parrot in the cage. What we are experi¬encing is an illusion.

Activity – 8

Question 8.
Testing of sound.
Answer:

• Take a plastic or iron funnel.
• Stretch a piece of rubber balloon and cover the wide part of the funnel with it.
• Tie it with rubber band.
• Ask your friend to shout ‘Oh’ at the narrow opening of the funnel.
• Observe the movements of the rubber sheet while he is shouting.
• Observe the rice grains also.

Observations:

1. Due to the vibrations in the rubber balloon the rice grains move up and down when we shout ‘Oh’ at the narrow opening of the funnel.
2. When we put the narrow end at the opening of our eye we hear the sound of heart as lub dub, lub dub …………….

Activity – 9

Question 9.
Touching test.
Answer:

• Blindfold your friend and ask him/her to identify different things by smell like lemon, tea, coffee, potato, tomato, tamarind, spinach, curd, brinjal etc.
• Keep as many things but be careful in choosing them.
• They should not be in powdered form.
• Don’t allow your friend to touch them.

Observations:

1. Biologically, the sense of smell or olfaction, begins with chemical events in the nose.
2. Their odours interact with receptor proteins associated with specialized nerve cells.
3. These cells incidentally are the body’s only nerve cells that come in direct contact with the outside environment.
4. Receptors present at the base of the skin lining the inner walls of the nose are highly sensitive to odour chemicals.
5. These odour chemicals can be complex and varied.

Activity -10

Question 10.
Tongue test
Answer:

• Close the eyes of your friend with a piece of cloth.
• Give her/him a piece of ginger, garlic, tamarind, banana and jaggery one by one.
• Ask her/him to taste by just taking these one at a time on the tongue.
• Remember that your friend needs to rinse his/her mouth between each test.
• Could your friend tell the taste by just putting the substances on the tongue? Yes, my friend told the taste.
• Now repeat the above experiment by asking your friend to take a bite and press the food on the plate.
• As food enters our mouth, we bite and chew it and press it against the palate with our tongue.
• This releases chemicals in food that trigger off our taste buds to act and carry stimulus to the brain to be processed for recognition of taste.
• The same taste bud is capable of producing different signals corresponding to the different chemicals in food.

Activity -11

Question 11.
Observe your tongue by standing in front of the mirror by sticking your tongue out. See how many different kinds of structures you can see on your tongue. Compare with the given diagram.
Answer:

• We can clearly see flake like structures that are filiform papillae.
• The roundish structures are fungi¬form papillae.
• There are large roundish ones at the back of the tongue which are circumvallate papillae.
• On the sides of the tongue, the bump like structures are foliate papillae.
• Taste buds are present on all of these except the filiform papillae that are not the sites of taste sensation.

Activity -12

Question 12.
Smell test
Answer:

• Blindfold your friend and ask him/her to close his or her nose as well.
• Give a few cumin seeds to your friend and ask him/her to chew.
• Ask your friend to identify what you have gave.
• You could try this with a small piece of potato as well.
• My friend has identified the cumin seeds and piece of potato.

Activity – 13

Question 13.
1) Make bundles of three toothpicks.
2) See to it that their pointed ends are at the same level.
3) Now ask your friend to make an outline of one of her/his palm.
4) Ask your friend to close her/his eyes. Now starting from the tip of the thumb keep pricking lightly with your toothpick bundle all over the plam.
5) Ask your friend how many points she/he could identify each time.
6) Repeat this with some of your friends.
Answer the following questions.
a) Where do you find maximum sensation on the palm?
Answer:
In the centre of the palm we find maximum sensation.

b) Where do you find minimum sensation?
Answer:
We find minimum sensation on the beginning of the palm.

c) Are palm sense patterns same for all your friends?
Answer:
Yes, palm sense patterns are same.

Activity – 14

Question 14.
Press your thumb gently on the tip of a sharpened pencil. Later press it on the blunt end of the pencil. How do you feel? Why?
Answer:

• When we press our thumb gently on the tip of a sharpened pencil it makes us feel pain.
• The sensory receptors sense the sharpened pencil’s press and the message is sent to brain through sensory nerve.
• Then the brain interprets the message and sends signals to feel pain through motor nerves.
• When we press our thumb with the blunt end of the pencil we do not feel pain because the sensory receptors (touch receptors) sends the message to brain through sensory nerve.
• Brain interprets the message and sends signals to our thumb to feel soft through motor nerves.

AP State Syllabus AP Board 9th Class Biology Solutions Chapter 10 Soil Pollution Textbook Questions and Answers.

AP State Syllabus 9th Class Biology Solutions 10th Lesson Soil Pollution

9th Class Biology 10th Lesson Soil Pollution Textbook Questions and Answers

Improve Your Learning

Question 1.
Define soil pollution. (AS 1)
Answer:
Soil or land pollution can be defined as the buildup in soils of persistent toxic compounds, chemicals, salts, radioactive materials, or disease causing agents, which have adverse effects on plant growth and animal health.

Question 2.
Why are plastic bags a big environmental nuisance? (AS 6)
Answer:

• Plastics are so versatile in use that their impact on environment are extremely wide ranging.
• Careless disposal of plastic bags chokes drains, blocks the porosity of the soil, and causes problems for ground water recharge.
• Plastic disturbs the soil microbe activity, and once ingested can kill animals.
• Plastic bags can also contaminate food stuffs due to leaching of toxic dyes and transfer of pathogens.
• Plastic bags remains strewn on the ground, or in unmanaged garbage dumps.
• Though small percentage lies strewn, it is this portion that is of concern as it causes extensive damage to the environment.

Question 3.
Describe an environmental friendly method to profitably dispose of human waste and cattle waste. (AS 1)
Answer:

• In recent years, an alternate and better method is used to obtain energy from not only from cattle waste but also from human waste.
• This is by anaerobic fermentation of the wastes to produce a gas which can be used as fuel.
• As this gas is produced from biological waste, this is called biogas.
• Biogas is a mixture of several gases : methane, carbondioxide, and small amounts of hydrogen, nitrogen, and hydrogen sulphide.

Question 4.
Chemical fertilizers are useful to crops. In which way they cause environmental pollution? (AS 1)
(OR)
We often read in newspapers that environmentalists often show deep concern of threats posed by chemical fertilisers and pesticides, etc. What are those threats to environment?
Answer:

• Fertilisers contaminate the soil with impurities, which come from the raw materials used for their manufacture.
• Due to excessive use of phosphate fertilizers soil becomes an indestructible poison for crops.
• Excessive use of fertilizers can endup polluting lakes, rivers and streams.
• This leads to promote the growth of algae in water bodies and is called eutropication.
• This abundant uncontrolled growth of plants blocks the flow of water and reduces oxygen content in the water.
• Other organisms living in the water do not get sufficient water, oxygen and ultimately die.
• Nitrogen fertilizer contribute to air pollution when it enters the atmosphere as ammonia and nitrogen oxide.
• This inturn cause acid rain and city smog associated health and environmental problems such as respiratory illness.

Question 5.
What steps can be taken to reduce pollution due to particulate matter from industries?
Answer:

• Industrial wastes can be treated physically, chemically and biologically until they are less hazardous.
• Acidic and alkaline wastes should be first neutralized; the insoluble material if biodegradable should be allowed to degrade under controlled conditions before being disposed.
• Electrostatic precipitators are used to reduce the particulate matter in the factory smoke.

Question 6.
What is a medical waste? Why it is called hazardous waste? What is the safe way to dispose medical waste? (AS 1)
Answer:

• Waste that is produced from hospitals is known as medical waste.
• Medical waste include needles, syringes, saline bottles, instruments used in surgeries, bandages soaked with blood and pus, used medicines, human excreta etc.
• Medical waste is called as hazardous waste because it containing toxic substances.
• Burying the medical waste in locations situated away from residential areas is the simplest method to dispose medical waste.

Question 7.
Prepare a flow chart to describe soil pollution, causes and methods of control. (AS 5)
(OR)
Prepare a pamphlet of your own to create awareness on soil pollution among the people in your area.
Answer:

Question 8.
What soil problems do you find in your area? Prepare a list of those problems and suggest a method for each of them to control those problems. (AS 7)
Answer:
The soil problems identified by me in our area :

Question 9.
What farm practices impact soil? Do they impact soil in a positive or a negative way?
Answer:

• Indiscriminate use of fertilizers, pesticides, insecticides, herbicides, no-till farming and growing same crop in all seasons are the farm practices impact soil.
• These farm practices may show positive or negative impact on the soil.
• By using chemical fertilizers we can get high yielding for only 20 to 30 years.
• After that soil becomes reluctant to plant growth. These chemicals damage fertility.
• Due to the extensive use of pesticides, insecticides, herbicides the salinity of the soil increases and it is not suitable for growing crops.
• Notill farming is a way of growing crops without disturbing the soil through tillage.
• Tillage activity can lead to compaction of soil, loss of organic matter in soil, loss of native vegetation, and death of the organisms in the soil.
• Growing the same crop in all seasons decreases the fertility.

Question 10.
Rank the negative impact practices in your area in the order in which you think they should be eliminated. (AS 1)
Answer:
Negative impact practices in our area :

1. Using chemical fertilizers
2. Using pesticides
3. Using insecticides
4. Using herbicides
5. Till farming
6. Deforestation
7. Using weedicides
8. Growing same crop in all seasons
9. Using locally prepared seeds

Question 11.
Rank the positive impact practices in order in which you think they should be used for the most benefit on your farm. (AS 1)
Answer:

1. Hybridised seeds
2. Organic manures
3. Organic weedicides
4. Predatory insects
5. No-till farming
6. Maintaining suitable pH value
7. Crop rotation
8. Salinity management
9. Soil organisms

Question 12.
Ravi said soil health is important. How can you support him? (AS 7)
Answer:

1. I support Ravi’s statement.
2. Healthy soil is fundamental to the quality of food it produces and to the health of those who eat the food produced from it.
3. When the soil components are present in appropriate percentage, the productivity is high.

Question 13.
How would soil texture affect the nutrients in soil? What would be its impact on crop production? (AS 2)
Answer:

• Soil with loose pores will allow water to collect and roots to expand. Loose soil is better than hard compact soil.
• Finer particles like clay increase surface area of the soil which allow nutrients to stay in the soil.
• Very porous soil, such as sand will allow nutrients to be leached more easily which can make less nutrients available to plants.
• Generally, a loose, airy soil structure is best for most plants.
• This can be accomplished by digging the bed and mixing together coarse and finer textures such as tilling compost into clay soil.

Question 14.
What are the three main physical properties of soil? What effects do this have on the plants? (AS 1)
Answer:

1. Colour texture, structure and porosity are the three main physical properties of soil.
2. These properties regulate and affect air and water movement in the soil and thus, soil ability to function.

Question 15.
What is pH? What is its range? What are the negative impacts if the pH of soil is too low or too high? (AS 1)
Answer:

1. The term pH is used to indicate the level of acidity or alkalinity of a soil.
2. The range of pH values of a good soil live from 5.5 to 7.5.
3. Below pH 7 the soils are termed as acidic and above pH 7 alkaline.

Negative impacts of low pH value :

1. The concentration of soluble metals especially aluminium and manganese may be toxic.
2. Calcium may be deficient.
3. Soil organisms responsible to transform N, S and P to plant available forms may be reduced.
4. Symbiotic nitrogen fixation in legume crops is greatly impaired.
5. Soils will be having low organic matter.
6. The availability of mineral elements to plants may be effected.

Negative impacts of high pH value :

1. If the pH is beyond 7, nutrient absorption and microbial activity will be affected which can be poisonous to plants.
2. pH extremes are unhealthy for most plants because they close or open membranes of plant cells too much.
3. This affects plant structure and their ability to uptake nutrients.
4. pH extremes make minerals and nutrients either too available or not available enough.

Question 16.
What is soil fertility? What are the sources of soil fertility? (AS 1)
Answer:

• Fertility of soil is closely associated with the properties of soil and is defined by its capacity to hold water and nutrients and supply them to plants when they need them, independent of direct application of nutrients.
• Soil organisms contribute to buildup soil organic matter, including humus, the soils most important nutrient reservoir.
• A major part of the soil microbial biomass is composed of fungi.
• Soil fertility is a complex process that involves the constant cycling of nutrients between organic and inorganic forms.
• As plant material and animal wastes decompose they release nutrients to the soil solution.
• Soil pH, its acidity or alkalinity is highly relevant to how readily nutrients become available in the soil.

Question 17.
Name 5 living things that live in soil. What do these things do to affect the soil?
Answer:

1. Viruses, earthworms, rats, ground squirrels, bacteria, fungi, algae, protozoa, dung beetle and different types of worms live in the soil.
2. These organisms feed on plant residues burrow the soil and help in aeration and percolation of water.
3. Soild microbes convert organic forms of elements to their inorganic forms.
4. Soil bacteria also control the forms of ions in which these nutrients occurs.

Question 18.
What is organic matter? Why it is important to plants? (AS 1)
Answer:

• Organic matter is the organic component of soil which includes the residues of dead plants and animals.
• Organic matter consists of nutrients necessary for plant growth such as nitrogen, phosphorus, and potassium.
• Soils which contain 30% or more organic matter are considered organic soil, all other soils are identified as mineral soils.
• Organic matter in soil improves water in filteration, decreases evaporation, and increases the water holding capacity.
• And also where there is organic matter, there will be numerous organisms present helping to convert it back to nutrients and these organisms help to create small pieces of nutrients, ideal for cultivation.

Question 19.
What are the factors affecting organic matter levels in soil? How this level of organic matter can be increased? (AS 1)
Answer:
1. Temperature, rainfall, natural vegetation, texture, drainage, cropping and tillage and crop rotation are the factors affecting organic matter levels in soil.

2. Temperature :
The decomposition of organic matter is accelerated in warm climates as compared to cooler climates.

3. For each 10°C decline in mean annual temperature the total organic matter and nutrients increases by two to three times.

4. Rainfall:
There is an increase in organic matter with an increase in rainfall.

5. Natural vegetation :
The total organic matter is higher in soils developed under grasslands than those under forests.

6. Texture :
Fine textured soils are generally higher in organic matter than coarse textured soils.

7. Drainage:
Poorly drained soils because of their high moisture content and relatively poor aeration are much higher in organic matter and nutrients than well drained soils.

8. Cropping and Tillage :
The cropped lands have much low nutrients and organic matter than comparable virgin soils.

9. Crop rotation :
Crop rotation of cereals with legumes results in higher soil organic matter.

Question 20.
What is solid waste? Explain best practices for solid waste management. (AS 1)
Answer:
Solid waste:
Solid waste may be defined as the organic and inorganic waste produced by various activities of the society which have lost their value to the first user.

Best practices for solid waste management:

1. By practicising four R’s : Reduce, Reuse, Recycle and Recover we would get less solid waste.
2. Materials such as glass containers, plastic bags, paper, cloth, etc., can be reused at domestic levels rather than being disposed, reducing solid waste pollution.
3. Solid waste management involves activities including collection, transfer, and transport to suitable sites and safe disposal of wastes by methods which are environmentally friendly methods.
4. Burying the waste in locations situated away from residential areas is the simplest and most widely used technique of solid waste management.
5. Solid waste management can also be done by methods such as sanitary landfill, composting and incineration, etc.

Question 21.
What is bioremediation? How it helps in controlling soil pollution? (AS 1)
Answer:

1. Bioremediation means to use a biological remedy to reduce or clean up contamination.
2. Microbes are often used to remedy environ¬mental problems found in soil, water and sediments.
3. Plants have also been used to assist bio¬remediation processes. This is called phytoremediation.
4. Biological processes have been used for some inorganic materials, like metals to lower radioactivity and to remediate organic contaminants.

Question 22.
Why soil conservation is important to us? What will happen if no preventive measures would be taken? (AS 2)
Answer:

• Soil conservation is important to us because it forms the basis for habitats and plants, which act as source of food to both humans and animals.
• Soil conservation is also important because with the erosion of the top soil layer, valuable nutrients are lost and crop yield diminish, which means very less food is produced per acre.
• We have to conserve soil because it has organic material that is good for plant growth.
• If no preventive measures are taken for soil conservation, soil erosion takes place.
• And also soil will be over used and it has more chemicals leading to unproductive soil.
• Amount of nutrients present in the soil decreases.

Question 23.
Look at the following symbol, what does it mean?

Answer:
1. It is the symbol of bioremediation.
2. Plants have been used to assist bioremediation.

9th Class Biology 10th Lesson Soil Pollution InText Questions and Answers

9th Class Biology Textbook Page No. 155

Question 1.
Today what are the pollutants produced from your school. How many of these are non-degradables?
Answer:
Wastes produced from our school :
Peels of fruits, vegetables, rice, glass materials, pens, polythene bags, biscuit and chocolate covers, icecream sticks, rubber, plastic tea glasses, paper leaves twigs etc.

Non-degradable pollutants :
Glass materials, pens, polythene bags, biscuit and chocolate wrappers, rubber, plastic glasses.

9th Class Biology 10th Lesson Soil Pollution Activities

Activity – 1

Question 1.
Answer:
1. During interval time Venu was eating a fruit.
2. He was about to throw the peel in corner of verandah.
3. His friend Ramu stopped him.
4. Ramu said you should not throw waste in the verandah. Drop it in the bin/basket given.
5. Prepare a list of waste materials we throw out in a day from morning to evening.
6. Classifying them as wet wastes and dry wastes with the help of the example given in the table.

7. Weigh the wet wastes, which you have listed in the table for one day.
8. Divide the weight by number of people in your home.
9. The result will be the per capita wet waste we are producing in one day.
10. Suppose if a family containing four members throws 400 gms of wet wastes per day,

Multiply it by 30 = 100 × 30 = 3000 gms per month
Multiply it by 365 = 3000 × 365 = 10,95,000 gms = 1095 kgs per year.

Activity – 2

Question 2.
Dumping and decomposing,
Answer:
1. Take a polythene bag / plastic bucket / or any container.
2. Fill half of it with soil.
3. Keep wet wastes and other wastes in it.
4. Wastes should include vegetable peels, rubber, plastic etc.
5. Add some more soil and sprinkle water regularly on it.
6. Dig it and observe in 15 days intervals.
7. Note your observations in the table.

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 14 Carbon and its Compounds Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 14th Lesson Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds Textbook Questions and Answers

Improve Your Learning

Question 1.
Name the simplest hydrocarbon. (AS1)
Answer:
The simplest hydrocarbon is alkane called Methane (CH₄). It’s an aliphatic, saturated compound of Hydrogen and Carbon.

Question 2.
What are the general molecular formulae of alkanes, alkenes and alkynes? (AS1)
Answer:
General molecular formula of alkane is C_nH_2n+2.
General molecular formula of alkene is C_nH_2n.
General molecular formula of alkyne is C_nH_2n-2.

Question 3.
Name the carboxylic acid used as a preservative. (AS1)
Answer:
Vinegar with chemical formula CH₃COOH is used as preservative. 5 – 8% of solution of acetic acid or ethanoic acid in water is called vinegar and it is used widely as preservative in pickles.

Question 4.
Name the product other than water formed on burning of ethanol in air. (AS1)
Answer:
C₂H₃OH + 3O₂ → 2CO₂ + 3H₂O + Energy
So, the product other than water formed on burning of ethanol in air is carbon dioxide (CO₂).

Question 5.
Give the IUPAC name of the following compounds. If more than one compound is possible, name all of them. (AS1)
i) An aldehyde derived from ethane.
ii) A ketone derived from butane.
iii) A chloride derived from propane.
iv) An alcohol derived from pentane.
Answer:
i) An aldehyde derived from ethane is ethanal. Its formula is CH₃CHO.
ii) A ketone derived from butane. Its IUPAC name is Butanone.
Its chemical formula is CH₃COCH₂CH₃
It is also known as methyl ethyl ketone. (Its general name)

iii) A chloride derived from propane.
A) 1-Chloro propane. Its formula is CH₃CH₂CH₂Cl.

iv) An alcohol derived from pentane :
A) 1-Pentanol. Its formula is CH₃CH₂CH₂CH₂CH₂OH.
B) 2-Pentanol. Its formula is CH₃CHOH CH₂CH₂CH₃
C) 3-Pentanol. Its formula is CH₃CH₂ CHOH CH₂CH₃

Question 6.
A mixture of oxygen and ethyne is burnt for welding ; can you tell why a mixture of ethyne and air is not used? (AS1)
Answer:

• Ethyne when burnt in the presence of oxygen gives enough heat that can be used for welding.
• Whereas if it is burnt in air which contains nitrogen, CO₂ and other inactive gaseous contents, sufficient oxygen is not available for burning ethyne to give the required heat.

Question 7.
Explain with the help of a chemical equation, how an addition reaction is used in vegetable ghee industry. (AS1)
Answer:

• The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation. The process of hydrogenation takes place in the presence of nickel or palladium metals as catalyst.
• The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.
• Vegetable oils are unsaturated fats having double bonds between some of their carbon atoms.
• When a vegetable oil (like groundnut oil) is heated with hydrogen in the presence of finely divided nickel as catalyst, a saturated oil called vegetable ghee (or vanaspati ghee) is formed. This a reaction is called hydrogenation of oils and it can be represented as follows.

Here vegetable oil is a liquid whereas vegetable ghee is a solid (or a semi solid).

Question 8.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆O? (AS1)
b) Give the IUPAC names of the above possible compounds and represent them in structures. (AS1)
c) What is the similarity in these compounds? (AS1)
Answer:
a) They are CH₃COCH₃and CH₃ CH₂ CHO

b) i) The IUPAC name of CH₃COCH₃ is propanone.
ii) The IUPAC name of CH₃ CH₂ CHO is propanal.

Question 9.
Name the simplest ketone apfl write its molecular formula. (AS1)
Answer:
Acetone is the simplest ketone. Its molecular formula is CH₃COCH₃ Its IUPAC name is propanone.

Question 10.
What do we call the Self linking property of carbon? (AS1)
Answer:
The property of self combination (or linking) of carbon atoms to form long chains is useful to us because it gives rise to an extremely large number of carbon compounds (or organic compounds). This is known as catenation.

Question 11.
Name the compound formed by heating ethanol at 443 K with excess of cone. H₂SO₄. (AS1)
(OR)
What is the compound formed when ethyhalcohol (Ethanol) is dehydrated ? Write the chemical equation of the reaction.
Answer:
1. When ethanol is heated with excess of cone. H₂SO₄ at 443 K (170° C), it gets dehydrated to form ethene (which is an unsaturated hydrocarbon).

2. During dehydration of ethanol molecules (CH₃ – CH₂OH), H from the CH₃ group and OH from CH₂OH group are removed in the form of a water molecule (H₂O) regulating in the formation of this molecule (CH₂ = CH₂).
3. In this reaction concentrated sulphuric acid acts as a dehydrating agent.

Question 12.
Give an example for esterification reaction. (AS1)
Answer:
The reaction between carboxylic acid and an alcohol in the presence of cone. H₂SO₄ to form a sweet odoured substance, ester with the functional group

is called esterification.

Ex: Ethanoic acid (carboxylic acid) reacts with Ethanol (alcohol) and forms ethyl acetate.

Question 13.
Name the product obtained when ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate. (AS1)
(OR)
If the ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate, what is the product obtained from them?
Answer:
Ethanol (Ethyl alcohol) undergoes oxidation to form the product of Acetaldehyde and finally Acetic acid.

Question 14.
Write the chemical equation representing the reaction of preparation of ethanol from ethane. (AS1)
Answer:
1. Ethane in the absence of air on heating forms ethene

2. Then Ethanol is prepared on large scale from ethene by the addition of water vapour to it in the presence of catalyst like P₂O₅, Tungsten oxide at high pressure and temperature.

Question 15.
Write the IUPAC name of the next homologous of CH₃OHCH₂CH₃. (AS1)
Answer:
The IUPAC name of the next homologous of CH₃OHCH₂CH₃ is HO-CH₃CH₂CH₂CH₃ 1 – butanol.

Question 16.
Define homologous series of carbon compounds. Mention any two characteristics of homologous series. (AS1)
Answer:
1. The series of carbon compounds in which two successive compounds differ by – CH₂ unit is called homologous series.
Ex : 1) CH₄, C₂H₆, C₃H₈, ………………..
2) CH₃OH, C₂H₅OH, C₃H₇OH, ………………..

2. If we observe above series of compounds, we will notice that each compound in the series differs by – CH₂ unit by its successive compound.

3. Characteristics of homologous series :
i) They have one general formula.
Ex : alkanes (C_nH_2n+2), alkynes (C_nH_2n-2), alcohols (C_nH_2n+1) OH, etc.
ii) Successive compounds in the series possess a difference of (-CH₂) unit.
iii) They have similar chemical properties.

Question 17.
Give the names of functional groups
(i) – CHO
(ii) – C = O. (AS1)
(OR)
Write the names of the given functional groups
(i) – CHO
(ii) – C = O
Answer:
i) – CHO → aldehyde
ii) – C = O → ketone

Question 18.
Why does carbon form compounds mainly by covalent bonding? (AS1)
Answer:
Since carbon atoms can achieve the inert gas electron arrangements only by the sharings of electrons, therefore, carbon always forms covalent bonds.

Question 19.
Allotropy is a property shown by which class substance: elements, compounds or mixtures? Explain allotropy with suitable examples. (AS1)
Answer:

1. Allotropy is a property shown by the elements.

2. The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy.

3. The different forms of the element are called allotropes and are formed due to the difference in the arrangement of atoms.

4. Example for allotropes : Allotropes of carbon.

Allotropes of carbon are classified into two types. They are
1) Amorphous forms,
2) Crystalline forms.

5) Amorphous forms of carbon:
Coal, coke, wood, charcoal, animal charcoal, lampblack, gas carbon, petroleum coke, sugar charcoal.

6) Crystalline forms of carbon :
Diamond, graphite and buckminsterfullerene.

Question 20.
Explain how sodium ethoxide is obtained from ethanol. Give chemical equations. (AS1)
Answer:
As ethanol is similar to water molecule (H₂O) with C₂H₅ group in place of hydrogen, it reacts with metallic sodium to liberate hydrogen and form sodium ethoxide.

Question 21.
Describe with chemical equation how ethanoic acid may be obtained from ethanol. (AS1)
Answer:
Ethyl alcohol (Ethanol) undergoes oxidation to form the product Acetaldehyde and finally acetic acid (Ethanoic acid).

Question 22.
Explain the cleansing action of soap. (AS1)
Answer:
When a dirty cloth is put in water containing dissolved soap, the hydrocarbon ends of the soap molecules in the micelle attach to the oil or grease particles present on the surface of dirty clothes.

Question 24.
Explain the structure of graphite in terms of bonding and give one property based on this structure. (AS1)
(OR)
Why does graphite act as lubricant?
Answer:

• Graphite forms a two dimensional layer structure with C – C bonds within the layers.
• There are relatively weak interactions between the layers.
• In the layer structure, the carbon atoms are in a trigonal planar environment.
• This is consistent with each carbon atom in sp² hybridisation.
• Interactions between the sp² orbitals (overlaps) lead to the formation of C – C bonds.
• Each carbon atom is with one unhybridised ‘p’ orbital.
• The unhybridised ‘p’ orbitals interact to form a π system that is delocalised over the whole layer.
• The interactions known as London dispersion forces between the layers which are separated by a distance of 3.35 A° are weakened by the presence of water molecules so that it is easy to cleave graphite.
• For this reason graphite is used as lubricant and as the lead in pencils.

Question 25.
Name the acid present in vinegar. (AS1)
Answer:
1) The acid present in vinegar is Ethenoic acid or acetic acid (CH₃COOH).
2) 5 – 8% solution of acetic acid in water is called vinegar.

Question 26.
What happens when a small piece of sodium is dropped into ethanol? (AS2)
Answer:
Ethanol reacts with sodium to liberate hydrogen and form sodium ethoxide.

Question 27.
Two carbon compounds A and B have molecular formula C₃H₈ and C₃H₆ respectively. Which one of the two is most likely to show addition? Justify your answer. (AS2)
Answer:

• It is a saturated hydrocarbon. It shows substitution reaction.

• This is an unsaturated hydrocarbon. Hence it shows addition to become saturated. During the reactions, addition of reagent takes place at the double bonded carbon atoms.

Justification :
In the following, C₃H₆ undergoes addition reaction.

Question 28.
Suggest a test to find the hardness of water and explain the procedure. (AS3)
(OR)
How do you test whether a given water sample is soft or hard?
Answer:

• Take about 10 ml hard water (well water or hand pump water) in a test tube.
• Add five drops of soap solution to it.
• Shake the test tube vigorously.
• We see that no lather is formed at first.
• Only a dirty white curd like scum is formed on the surface of water.
• From this, we conclude that soap does not form lather easily with hard water.
• We have to add much more soap to obtain lather with hard water.

Question 29.
Suggest a chemical test to distinguish between ethanol and ethanoic acid and explain the procedure. (AS3)
Answer:

1. Take ethanol and ethanoic acid in two different test tubes.
2. Add nearly 18 g of sodium bicarbonate (NaHCO₃) to each test tube.
3. Lots and lots of bubbles and foam will be observed from the test tube containing ethanoic acid. This is due to release of CO₂.
   NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
4. Ethanol will not react with sodium bicarbonate and thus we won’t observe any change in the test tube containing ethanol.
   Thus we can separate ethanol from ethanoic acid.

Question 30.
An organic compound ‘X’ with a molecular formula C₂H₆O undergoes oxidation with alkaline KMnO₄ and forms the compound ‘Y’, that has molecular formula C₂H₄O₂. (AS3)
i) Identify ‘X’ and ‘Y’.
Answer:
X is Ethanol is CH₃CH₂OH and T is Ethanoic acid, i.e., CH₃COOH.

ii) Write your observation regarding the product when the compound X is made to react with compound IT which is used as a preservative for pickles.
Answer:
Ethyl alcohol undergoes oxidation to form the product Acetaldehyde and finally Acetic acid.

Here CH₃COOH is used as preservative for pickles.

When X reacts with Y it forms ethyl acetate and water which is called esterification reaction.
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 31.
Prepare models of methane, ethane, ethene and ethyne molecules using clay balls and matchsticks. (AS4)
Answer:
Stick and ball model :
1) Methane (CH₄) :

2) Ethane (C2H₆):

3) Ethene (C₂H₄):

4) Ethyne (C₂H₂)

Question 32.
Collect information about artificial ripening of fruits by ethylene. (AS4)
Answer:

• Seasonal fruits like mango, banana, papaya, sapota and custard apple are often harvested in nature. But due to unripe condition they are subsequently allowed to ripen by natural release of ripening harmone (ethylene) from the fruit.
• However, natural ripening in some fruits is a slow process, which leads to high weight loss, desiccation of fruits and under ripening. With the rapid development of fruit trade, artificial ripening has become essential and the methods practised earlier by small traders are smoking and calcium carbide treatment.
• Fruits ripened with calcium carbide though seem attractive and colourful are inferior in taste, flavour and spoil faster.
• Government of India has banned the use of calcium carbide for artificial ripening of fruits under PFA Act 8-44AA, 1954.
• Artificial ripening of fruits by using the above steps spoils the health of consumers, so we should not use such type of fruits.
• Government has to take serious action on the fruit sellers who are practising the above said methods.

Question 33.
Draw the electronic dot structure of ethane molecule (C₂H₆). (AS6)
Answer:
C₂H₆:

Question 34.
How do you appreciate the role of esters in everyday life? (AS6)
Answer:

• Esters are usually volatile liquids having sweet or pleasant smell.
• They are also said to have fruity smell.
• Esters are used in making artificial perfumes.
• This is because of the fact that most of the esters have a pleasant smell.
• Esters are also used as flavouring agents.
• This means that esters are used in making artificial flavours and essences used in ice-cream, sweets and cool drinks.
• The alkaline hydrolysis of esters is known as saponification (Soap making).
• That’s why we can appreciate the role of esters in everyday life.

Question 35.
How do you condemn the use of alcohol as a social practice? (AS7)
Answer:

• Consumption of alcohol in the form of beverages is harmful to health.
• It causes severe damage to blood circulation system.
• Addiction to alcohol drinking leads to heart diseases and damages the liver.
• It also causes ulcers in small intestines due to increased acidity and damages the digestive system.
• Alcohol which is consumed in raw form under the names liquor, gudumba which is more harmful to health due to adulteration.
• Alcohol mixed with pyridine is called denatured spirit. Consumption of denatured spirit causes blindness and death.
• Hence use of alcohol is a social evil which harms the society.

Question 36.
An organic compound with molecular formula C₂H₄O₂ produces brisk effervescence on addition of sodium carbonate/bicarbonate.
Answer the following :
a) Identify the organic compound. (AS1)
Answer:
The organic compound is Ethanoic acid (CH₃COOH).

b) Write the chemical equation for the above reaction. (AS1)
Answer:

c) Name the gas evolved. (AS2)
Answer:
CO₂

d) How will you test the gas evolved? (AS3)
Answer:
1) Pass the evolved gas through lime water in a test tube.
2) We will find that lime water turns milky.
3) Only CO₂ gas can turn lime water milky.

e) List two important uses of the above compound. (AS1)
Answer:
1) Dilute ethanoic acid (CH₃COOH) is used as a food preservative in the preparation of pickles and sauces.
2) Ethanoic acid is used for making cellulose acetate which is an important artificial fibre.

Question 37.
1 ml glacial acetic acid and 1 m/of ethanol are mixed together in, a test tube. Few drops of concentrate sulphuric acid is added in the mixture are warmed in a water bath for 5 min.
Answer the following:
a) Name the resultant compound formed.
b) Represent the above change by a chemical equation.
c) What term is given to such a reaction?
d) What are the special characteristics of the compound formed?
Answer:
a) Ethyl acetate.

c) Esterification
d) It has fruity smell or pleasant smell.

Fill In The Blanks

1. Carbon compounds containing double and triple bonds are called ………………….
2. A compound which is basic constituent of many cough syrups ………………………
3. Very dilute solution of ethanoic acid is ………………..
4. A sweet odour substance formed by the reaction of an alcohol and a carboxylic acid is ………………
5. When sodium metal is dropped in ethanol …………………. gas will be released.
6. The functional group present in methanol is …………………….
7. IUPAC name of alkene containing 3 carbon atoms is ………………….
8. The first member of homologous series among alkynes is ……………………
9. The product that is formed by dehydration of ethanol in cone, sulphuric acid is ………………….
10. Number of single covalent bonds in ammonia are ………………..
11. Type of reactions shown by alkanes is ……………….
Answer:

1. unsaturated compounds
2. ethanol
3. vinegar
4. ester
5. H₂
6. – OH (Alcohol)
7. propene
8. ethyne (C₂H₂)
9. ethene (C₂H₄)
10. 3
11. substitutional

Multiple Choice Questions

1. Which of the four test tubes containing the following chemicals shows the brisk effervescence when dilute acetic acid was added to them?
i) KOH
ii) NaHCO₃
iii) K₂CO₃
iv) NaCl
A) i & ii
B) ii & iii
C) i & iv
D) ii & iv
Answer:
B) ii & iii

2. Which of the following solution of acetic acid in water can be used as preservative?
A) 5-10%
B) 10-15%
C) 15-20%
D) 100%
Answer:
A) 5-10%

3. The suffix used for naming an aldehyde is
A) – ol
B) – al
C) – one
D) – ene
Answer:
B) – al

4. Acetic acid, when dissolved in water, it dissociates into ions reversibly because it is a
A) weak acid
B) strong acid
C) weak base
D) strong base
Answer:
A) weak acid

5. Which one of the following hydrocarbons can show isomerism?
A) C₂H₄
B) C₂H₆
C) C₃H₈
D) C₄H₁₀
Answer:
D) C₄H₁₀

6. Combustion of hydrocarbon is generally accompanied by the evolution of
A) Heat
B) Light
C) Both heat and light
D) Electric current
Answer:
C) Both heat and light

7. 2 ml of ethanoic acid was taken in each of the three test tubes A, B and C and 2 ml, 4 ml and 8 ml water was added to them respectively. A clear solution is obtained in:
A) Test tube A only
B) Test tubes A & B only
C) Test tubes B and C only
D) All the test tubes
Answer:
D) All the test tubes

8. If 2 ml of acetic acid was added slowly in drops to 5 ml of water then we will notice
A) The acid forms a separate layer on the top of water
B) Water forms a separate layer on the top of the acid
C) Formation of a clear and homogenous solution
D) Formation of a pink and clear solution
Answer:
C) Formation of a clear and homogenous solution

9. A few drops of ethanoic acid were added to solid sodium carbonate. The possible results of the reactions are
A) A hissing sound was evolved
B) Brown fumes evolved
C) Brisk effervescence occurred
D) A pungent smelling gas evolved
Answer:
C) Brisk effervescence occurred

10. When acetic acid reacts with ethyl alcohol, we add cone. H₂SO₄, it acts as and the process is called
A) Oxidizing agent, saponification
B) Dehydrating agent, esterification
C) Reducing agent, esterification
D) Acid and esterification
Answer:
B) Dehydrating agent, esterification

10th Class Chemistry 14th Lesson Carbon and its Compounds InText Questions and Answers

10th Class Chemistry Textbook Page No. 254

Question 1.
Can carbon get helium configuration by losing four electrons from the outer shell?
Answer:

• If carbon loses four electrons from the outer shell, it has to form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is also a remote possibility.
• Carbon has to satisfy its tetravalency by sharing electrons with other atoms.
• It has to form four covalent bonds either with its own atoms or atoms of other elements.

10th Class Chemistry Textbook Page No. 255

Question 2.
How do carbon atoms form bonds in so many different ways?
Answer:
As per valence bond theory, the four unpaired electrons in a carbon atom is main cause to form many bonds.

Question 3.
Explain the four unpaired electrons in carbon atom through excited state.
Answer:
Electronic configuration of carbon (ground state):

Electronic configuration of carbon (excited state):

10th Class Chemistry Textbook Page No. 256

Question 4.
Where does this energy to excite electron come from?
Answer:

• We have to understand that free carbon atom would not be in excited state under normal conditions.
• When the carbon atom is ready to form bonds with other atoms, the energy required for excitation is taken up from bond energies, which are the liberated energies when bonds are formed between carbon atom and other atoms.

Question 5.
In methane (CH₄) molecule all four carbon – hydrogen bonds are identical and bond angle HCH is 109°28′. How can we explain this?
Answer:
In excited state, carbon atom has three unpaired electrons in p-orbitals and one electron in s-orbital. These four valence electrons are with different energies. These orbitals combine to form four identical orbitals. Four hydrogen atoms form four identical C -H bonds with bond angle 109° 28′. This is called hybridisation.

Question 6.
How do these energetically unequal valence electrons form four equivalent covalent bonds in methane molecule?
Answer:
1) When bonds are formed, energy is released and the system becomes more stable. If carbon forms four bonds rather than two, still more energy is released and so the resulting molecule becomes even more stable.

2) The energy difference between the 2s and 2p orbitals is very small. When carbon atom is ready to form bonds it gets a small amount of energy from bond energies and gets excited to promote an electron from the 2s to the empty 2p to give four unpaired electrons.

3) We have got four unpaired electrons ready for bonding, but these electrons are in two different kinds of orbitals and their energies are different.

4) We are not going to get four identical bonds unless these unpaired electrons are in four identical orbitals.

10th Class Chemistry Textbook Page No. 257

Question 7.
How to explain the four orbitals of carbon containing unpaired electrons as energetically equal?
Answer:
With hybridisation we explai n the four orbitals of carbon containing unpaired electrons are energetically equal.
Ex : Methane (CH₄).

10th Class Chemistry Textbook Page No. 258

Question 8.
How do you explain the ability of C – atom to form two single covalent bonds and one double bond?
Answer:
Ethylene (CH₂ = CH₂) explains the ability of carbon atom to form two single covalent bonds and one double bond.
Ex:

10th Class Chemistry Textbook Page No. 259

Question 9.
How do you explain the ability of carbon atom to form one single bond and one triple bond?
Answer:
Ethyne (HC \(\equiv\) CH) explains the ability of carbon atom to form one single bond between one hydrogen and carbon, and one triple bond between carbon and carbon.
Ex : H – C \(\equiv\) C – H.

10th Class Chemistry Textbook Page No. 260

Question 10.
What are bond angles H\(\widehat{\mathbf{C}}\)H in CH₄, C₂H₄ and C₂H₂ molecules?
Answer:

10th Class Chemistry Textbook Page No. 262

Question 11.
How do you understand the markings (writings) of a pencil on a paper?
Answer:

1. When we write with a pencil, the inter layer attractions breakdown and leave graphite layers on the paper.
2. It is easy to remove pencil marks from paper with an eraser because, the layers do not bind strongly to the paper.

10th Class Chemistry Textbook Page No. 265

Question 12.
Allotting completely one special branch in chemistry to compounds of only one element. Is it justified when there are so many elements and their compounds but not with any special branches?
Answer:

1. We understand that all molecules that make life possible carbohydrates, proteins, nucleic acids, lipids, hormones, and vitamins contain carbon.
2. The chemical reactions that take place in living systems are of carbon compounds.
3. Food that we get From nature, various medicines, cotton, silk and fuels like natural gas and petroleum almost all of them are carbon compounds.
4. Synthetic fabrics, plastics, synthetic rubber are also compounds of carbon.
5. Hence, carbon is a special element with the largest number of compounds:

10th Class Chemistry Textbook Page No. 266

Question 13.
What are hydrocarbons?
Answer:
The compounds containing only carbon and hydrogen in their molecules are called hydrocarbons.

Question 14.
Do all the compounds have equal number of C and H atoms?
Answer:
No. All the compounds do not have equal number of C and H atoms.

10th Class Chemistry Textbook Page No. 269

Question 15.
Observe the following two structures.
a) CH₃ – CH₂ – CH₂ – CH₃
b)

i) How about their structures? Are they same?
Answer:
No, they are not same compounds.

ii) How many carbon and hydrogen atoms are there in (a) and (b) structures?
Answer:
Carbon – 4 ; Hydrogen – 10.

iii) Write the condensed molecular formulae for (a) and (b), do they have same molecular formulae?
Answer:
C₄H₁₀; Yes.

Question 16.
Can carbon form bonds with the atoms of other elements?
Answer:
Carbon forms compounds not only with atoms of hydrogen but also with atoms of other elements like oxygen, nitrogen, sulphur, phosphorus, halogens, etc.

10th Class Chemistry Textbook Page No. 272

Question 17.
What do you mean by nomenclature of Organic componds?
Answer:
Nomenclature of organic chemistry is systematic method of naming organic compound.

Question 18.
What is the basis for nomenclature?
Answer:
The basic of the nomenclature is number of carbons in the parent chain in a compound.

10th Class Chemistry Textbook Page No. 273

Question 19.
What are the word – root and suffix?
Answer:
Word root:
Word root indicates the number of carbon atoms in the longest possible continuous carbon chain also known as parent chain.

Suffix :
Suffix is added immediately after the word root. It is two types

1) Primary Suffix :
It is used to indicate the degree of saturation or unsaturation of the main chain.

2) Secondary Suffix :
It is used to indicate the main functional group in the organic compound.

10th Class Chemistry Textbook Page No. 274

Question 20.
What do you mean by the term ‘alkyl’?
Answer:
Alkyl:
Alkyl is a substituent, that is attached to the molecular fragment.
General formula of alkyl is C_nH_{2n + 1}

10th Class Chemistry Textbook Page No. 278

Question 21.
Can we write the structure of a compound if the name of the compound is given?
Answer:
Yes, we can write the structure of a compound if the name of the compound is given.

10th Class Chemistry Textbook Page No. 279

Question 22.
Why do sometimes cooking vessels get blackened on a gas or kerosene stove?
Answer:
Because of the inlets of air getting closed, the fuel gases do not completely undergo combustion. Hence, it forms a sooty carbon form which gets coated over the vessels.

10th Class Chemistry Textbook Page No. 280

Question 23.
Do you know what is a catalyst?
Answer:
A catalyst is a substance which regulates the rate of a given reaction without itself finally undergoing any chemical change.

10th Class Chemistry Textbook Page No. 281

Question 24.
Do you know how the police detect whether suspected drivers have consumed alcohol or not?
Answer:

1. The police officer asks the suspect to blow air into a plastic bag through a mouth piece of the detecting instrument which contains crystals of potassium-di-chromate (K₂Cr₂O₇).
2. As K₂Cr₂O₇ is a good oxidizing agent, it oxidizes any ethanol in the driver’s breath to ethanal and ethanoic acid.
3. Orange Cr₂O₇^2- changes to bluish green Cr³⁺ during the process of the oxidation of alcohol.
4. The length of the tube that turned into green is the measure of the quantity of alcohol that had been drunk.
5. The police even use the IR Spectra to detect the bonds C – OH and C – H of CH₃ – CH₂OH.

10th Class Chemistry Textbook Page No. 283

Question 25.
What are esters?
Answer:

The compounds which contain the functional group and the general formula R – COO – R’, where R and R’ are alkyl groups or phenyl groups, are known as “Esters”.

10th Class Chemistry Textbook Page No. 284

Question 26.
What is a true solution?
Answer:
A true solution is that in which the solute particles dispersed in the solvent are less than 1 nm in diameter.

10th Class Chemistry Textbook Page No. 286

Question 27.
What is the action of soap particles on the greasy cloth?
Answer:

• Soaps and detergents make oil and dirt present on the cloth come out into water, thereby making the cloth clean.
• Soap has one polar end and one non-polar end.
• The polar end is hydrophilic in nature and this end is attracted towards water.
• The non-polar end is hydrophobic in nature and it is attracted towards grease or * . ; oil on the cloth, but not attracted towards water.
• When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it from the cloth.
• The hydrophobic end of the soap molecules move towards the dirt or grease particles. ’
• The hydrophobic ends attach to the dirt particle and try to pull out.
• The molecules of soap surround the dirt particle at the centre of the cluster and form a spherical structure called micelle.
• These micelles remain suspended in water like particles in a colloidal solution.
• The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.
• Thus, the dust particles remain trapped in micelles and are easily rinsed away with water.
• Hence, soap micelles remove dirt by dissolving it in water.

10th Class Chemistry Textbook Page No. 280

Question 28.
Why we are advised not to use animal fats for cooking?
Answer:

• Animal fats have recently been implicated as the cause of heart disease and obesity. So, we are advised not to use animal fats for cooking.
• Excess animal fat is stored in lipocytes, which expand in size until the fat is used for fuel.

Question 29.
Which oil is recommended for cooking? Why?
Answer:
Canola oil :

• A recent entrant into the Indian market Canola is flying off the shelves.
• Canola oil which is made from the crushed seeds of the Canola plant, is said to be amongst the healthiest of cooking oils.
• It has the lowest saturated fat content of any oil.
• It’s seen as a healthy alternative as it’s rich in monosaturated fats and is high in omega-3 and omega a fats.
• It has a medium smoking point and is an oil that works well for fruits, baking, sauteing, etc.

10th Class Chemistry 14th Lesson Carbon and its Compounds Activities

Activity – 1

Question 1.
Observe the structural formula of the following hydro carbons and write their names in your notebook.
Answer:
1) CH₃ – CH₂ – CH = CH₂
Sol. But-l-ene

Sol. 2-Methyl butane

3) CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – CH₃
Sol. Hexane

Sol. 3-Methyl, but-l-ene

5)

Sol. Prop-l-yne

Activity-2

Question 2.
Read the names of the following hydro carbons and draw their structures in your notebook.
1. 2,2-Dimethyl hexane
Sol.

2. But-l-yne
Sol. CH₃ – CH₂ – C = CH

3. 3-Methyl Pent-2-ene
Sol.

4. But-1.2-diene
Sol. CH₃ – CH₃ = c = CH₂

5. Hept-2 en, 4-yne
Sol.

Activity – 3

Question 3.
Write an activity to show esterification reactions.
Answer:
The compound formed is ester. The process is called esterification.

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
2. Warm it in a water bath or a beaker containing water for at least five minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
4. We will notice that the resulting mixture is sweet odoured subatance.
5. This substance is nothing but ethyl acetate, an ester.
6. This reaction is called esterification reaction.

Activity – 4

Question 4.
Write an activity to show soap solution separates oil from water.
Answer:

1. Take about 10 ml of water each in two test tubes.
2. Add a drop of oil to both the test tubes.
3. Label them as A and B.
4. Add a few drops of soap solution to test tube B.
5. Now shake both the test tubes vigorously for the same period of time.
6. We can see the oil and water layers separately in both the test tubes immediately after we stop shaking them.
7. Leave the test tubes undisturbed for sometime and observe.
8. The oil layer separates out first in which test tube we added drops of soap solution.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 11 Electric Current Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What do you mean by electric current?
(OR)
Define electric current.
Answer:
Electric current is defined as the amount of charge crossing any cross-section of the conductor in one second.

Question 2.
Which type of charge (positive or negative) flows through an electric wire when it is connected in an electric circuit?
Answer:
Negative type of charge flows through an electric wire when it is connected in an electric circuit.

Question 3.
Is there any evidence for the motion of charge in daily life situations?
Answer:
Yes, lightning is a live example.

Improve Your Learning

Question 1.
Explain how electron flow causes electric current with Lorentz – Drude theory of electrons. (AS1)
(OR)
How does electron flow cattle elfectric current with Lorentz – Drude theory of electrons? Explain.
Answer:
Lorentz – Drude theory :

1. Lorentz – Drude proposed that conductors like metals contain a large number of free electrons.
2. The positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.
3. The negative ions (electrons) move randomly in lattice in an open circuit.
4. When the lattice is closed the electrons are arranged in ordered motion.
5. When the electrons are in order motion, there will be a net charge (crossing through any cross section.
6. This order motion of electrons is called electric current.

Question 2.
How does a battery work? Explain. (AS1)
(OR)
How does a battery maintain a constant potential difference between its terminals?
Answer:
Working of a battery :

• A battery consists of two metal plates (positive electrode = anode and negative electrode = cathode) and a chemical (electrolyte).
• The electrolyte between the two metal plates consists of positive and negative ions which move in opposite directions.
• The electrolyte exerts a chemical force on these ions and makes them move in a specified direction.
• Depending upon the nature of the chemical, positive ions move towards one of the plates and accumulate on that plate.
• As a result of this accumulation of charges on this plate it becomes anode.
• Negative ions move in a direction opposite to the motion of positive ions and accumulate on the other plate.
• As a result of this the plate becomes negatively charged called cathode.
• This accumulation of different charges on respective plates continues till both plates are sufficiently charged.
• But the ions in motion experience electric force when sufficient number of charges are accumulated on the plates.
• The motion of ions continues towards their respective plates till the chemical force is equal to electric force.
• Thus the battery works.

Question 3.
Write the difference between potential difference and emf. (AS1)
Answer:
Potential Difference:
Work done by the electric force on unit charge is called potential difference.
\(\mathbf{V}=\frac{\mathbf{W}}{q}=\frac{\mathbf{F} l}{\mathbf{q}}\)

Electromotive force (emf):
The work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.
\(\varepsilon=\frac{W}{q}=\frac{F d}{q}\)

Question 4.
How can you verify that the resistance of a conductor is temperature dependent? (AS1)
(OR)
How do you prove increase in temperature affects the resistance with an activity?
Answer:
Resistance :
The resistance of a conductor is the obstruction offered to the flow of electrons in a conductor.

Resistance is temperature dependent:
Aim:
To show that the value of resistance of a conductor depends on temperature for constant voltage between the ends of the conductor.

Materials required :

1. A bulb
2. A battery
3. Key
4. Insulated wire
5. Multimeter

Procedure :

1. Take a bulb and measure the resistance when it is in open circuit using a multimeter.
2. Note the value of resistance in your notebook.
3. Connect a circuit with components as shown in figure.
4. Switch on the circuit. After few minutes, measure the resistance of the bulb again.
5. Note this value in your notebook.

Observation :

1. The value of resistance of the bulb in second instance is more than the resistance of the bulb in open circuit.
2. The bulb gets heated.

Result:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 5.
What do you mean by electric shock? Explain how it takes place. (AS1)
Answer:
Electric shock:
The electric shock is combined effect of potential difference, electric current, and resistance of the human body.

• An electric shock can be experienced when there exists a potential difference between one part of the body and another part.
• When current flows through human body, it chooses the path which offers low resistance.
• The resistance of a body is not uniform throughout it.
• As long as current flow continues inside the body, the current and resistance of human body go on changing inversely.
• This is called the electric shock.

Question 6.
Derive \(\mathbf{R}=\frac{\rho l}{\mathbf{A}}\). (AS1)
(OR)
What are laws of resistance and derive a formula for resistance.
Answer:
Resistance of a conductor is directly proportional to the length of the conductor,
i.e., R ∝ l ………………….. (1)
Resistance of a conductor is inversely proportional to the cross-section area of the conductor.
i.e., R ∝ \(\frac{1}{\mathrm{~A}}\) ………………….. (2)
From (1) and (2) R ∝ \(R \propto \frac{l}{A} \Rightarrow R=\frac{\rho l}{A}\)
where ρ is a constant,
ρ is called specific resistance or resistivity.

Question 7.
How do you verify that resistance of a conductor is proportional to the length of the conductor for constant cross-section area and temperature? (AS1)
Answer:

• Collect manganin wires of different lengths with the same cross-sectional areas.
• Make a circuit as shown in figure.
• Connect one of the manganin wires between the ends P and Q.
•  Measure the value of the current using the ammeter.
• Repeat the same for other lengths of the wires.
• Note the values of currents.
• We notice that the current decreases with increase in the length of the wire.
  ∴ R ∝ l (at constant temperature and cross-section area) …………… (1)
• Do the same with manganin wires with equal lengths but different cross-section area.
• We notice that the resistance was more when the cross-section area was less.
  ∴ R ∝ \(\frac{1}{\mathrm{~A}}\) ………………. (2)
  ∴ R ∝ \([latex]\frac{1}{\mathrm{~A}}\)[/latex]
  Thus we verify l and A.

Question 8.
Explain Kirchhoff’s laws with examples. (AS1)
(OR)
Write two examples of Kirchhoffs laws and explain it.
Answer:
Kirchhoff’s laws :
Two simple rules called Kirchhoff’s rules are applicable to any DC circuit containing batteries and resistors connected in any way.
The two laws are (i) Junction law and (ii) Loop law.

i) Junction law :

Here P is called junction point where conducting wires meet. The junction law states that, at any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents leaving the junction.
i.e., I₁ + I₄ + I₆ = I₂ + I₃ + I₅
This law is based on the conservation of charge.

ii) Loop law:

Loop law states that, the algebraic sum of the increases and decreases in potential difference (voltage) across various components of the circuit in a closed circuit loop must be zero.

This law is based on the conservation of energy.

Question 9.
What is the value of 1 KWH in Joules? (AS1)
Answer:
1 KWH = 1 KW x 1h
= 1000 W × 60 min = 1000 W × 60 × 60 s = 3.6 × 10₆ Ws = 3.6 × 10₆ J.
∴ 1 KWH = 3.6 × 10₆ J.

Question 10.
Explain overloading of household circuit. (AS1)
Answer:

• Electricity enters our homes through two wires called lines. These lines have low resistance and the potential difference between the wires is usually about 240 V.
• All electrical devices are connected in parallel in our home. Hence, the potential drop across each device is 240 V.
• Based on the resistance of each electric device, it draws some current from the supply. Total current drawn from the mains is equal to the sum of the currents passing through each device.
• If we add more devices to the household circuit the current drawn from the mains also increases.
• This leads to overheating and may cause a fire. This is called “overloading”.

Question 11.
Why do we use fuses in household circuits? (AS1)
(OR)
What is the use of fuses?
Answer:

• The fuse consists of a thin wire of low melting point.
• When the current in the fuse exceeds 20 A, the wire will heat up and melt.
• The circuit then becomes open and prevents the flow of current into the household circuit.
• Hence all the electric devices are saved from damage that could be caused by overload.
• Thus we can save the household wiring and devices by using fuses.

Question 12.
Deduce the expression for the equivalent resistance of three resistors connected in series. (AS1)
(OR)
Derive R = R₁ + R₂ + R₃
(OR)
The second end of a first resistor is connected to first end of second resistor. Then how are the resistors connected? Derive the expression for the resultant resistance of this connection.
Answer:
Series connection:
In series connection of resistors, there is only one path for the flow of current in the circuit. Hence, the current in the circuit is equal to I.
According to Ohm’s law,
∴ V₁ = IR₁ ; V₂ = IR₂ and V₃ = IR₃.
⇒ Let R be the equivalent resistance of the combination of resistors in series.

Also V = I R_eq
V = V₁ + V₂ + V₃
I R_eq = IR₁ + IR₂ + IR₃
⇒ I R_eq = I (R₁ + R₂ + R₃)
⇒ R_eq = R₁ + R₂ + R₃
∴ The sum of individual resistances is equal to their equivalent resistance when the resistors are connected in series.

Question 13.
Deduce the expression for the equivalent resistance of three resistors connected in parallel. (AS1)
(OR)
Derive : \(\frac{1}{\mathbf{R}}=\frac{1}{\mathbf{R}_{1}}+\frac{1}{\mathbf{R}_{2}}+\frac{1}{\mathbf{R}_{3}}\)
(OR)
Explain the expression for the equivalent resistance of three resistors which are connected in parallel.
(OR)
If all the first ends of resistors are connected to and second ends are connected to another point, then what type of connection is this? Derive the resultant resistance for this connection.
Answer:
Parallel Connection :
In parallel connection of resistors, there is same potential difference at the ends of the resistors. Hence the voltage in the circuit is equal to V.
Let Ip I2 and I3 be the currents flowing through R₁, R_2, and R₃ resistors respectively.
Hence, we can write I = I₁ + I₂ + I₃.
According to the Ohm’s law,

∴ The equivalent resistance of a parallel combination is less than the resistance of each of the resistors.

Question 14.
Silver is a better conductor of electricity than copper. Why do we use copper wire for conduction of electricity? (AS1)
Answer:
Silver is costlier than copper. So, we use copper wire for conduction of electricity even though silver is a better conductor of electricity.

Question 15.
Two bulbs have ratings 100 W, 220 V and 60 W, 220 V. Which one has the greater resistance? (AS1)
Answer:

∴ The second bulb possessing 60 W, 220 V has the greater resistance.

Question 16.
Why don’t we use series arrangement of electrical appliances like bulb, television, fan, and others in domestic circuits? (AS1)
Answer:

• If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases. To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series. They are connected in parallel.
• In series combination same current passes through all resistors. This is not suggestable for household appliances. Hence, they are connected in parallel.

Question 17.
A wire of length 1 m and radius 0.1 mm has a resistance of 100 Ω. Find the resistivity of the material. (AS1)
Answer:
1) Given l = 1 m, r = 0.1 mm = 10^-4 m, R = 100 Ω

Question 18.
Why do we consider tungsten as a suitable material for making the filament of a bulb? (AS2)
(OR)
What is the reason for using Tungsten as a filament in electric bulb?
Answer:
Tungsten has higher resistivity values and melting point. So, we consider tungsten as a suitable material for making the filament of a bulb.

Question 19.
Are the head lights of a car connected in series or parallel? Why? (AS2)
Answer:
The headlights of a car are connected in parallel.
Reason :

• When they are connected in parallel, same voltage (RD) will be maintained in the two lights.
• If one of the light damaged, the other will work without any disturbance.

Question 20.
Why should we connect electric appliances in parallel in a household circuit? What happens if they are connected in series?
Answer:

• The electric appliances are connected in parallel in a household circuit. Because in parallel wiring if any electric appliance is switched off, other appliances don’t get off.
• If one appliance, in a set of series combination breaks down, the circuit becomes open and the flow of current ceases.
• To avoid this the household appliances like bulb, T.V., fan, etc. are not connected in series.

Question 21.
Suppose that you have three resistors each of value 30Ω. How many resistors can you obtain by various combinations of these three resistors? Draw diagrams in support of your predictions. (AS2)
Answer:
Let R₁ = 30Ω, R₂ = 30Ω, R₃ = 30Ω
We get different resistors by different combinations as shown below.

Question 22.
State Ohm’s law. Suggest an experiment to verify it and explain the procedure. (AS3)
How do you prove experimentally the ratio V/l is a constant for a given conductor?
Answer:
Ohm’s law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Verification :
Aim :
To verify Ohm’s law or to show that \(\frac{\mathrm{V}}{\mathrm{I}}\) = constant for a conductor.

Materials required :
6V Battery eliminator, 0 to 1A Ammeter, 0 – 6V volt meter, copper wires, 50 cm manganin coil, Rheostat, switch and 3V LED, etc.

Procedure :

• Complete the circuit as shown in figure. Knob should be adjusted to 4.5V at battery eliminator.
• Using Rheostat change the potential difference between two ends of manganin wire from 0V to 4.5V (maximum).
• By using Rheostat adjust the potential difference 1V between two ends of manganin wire.
• Now observe the electric current through Ammeter in the circuit and note down in the following table.

• Using Rheostat change the potential difference with different values upto 4.5V and note down the current value (I) in the table.
• Take atleast five values of V and I and note down in the table.
• Find \(\frac{\mathrm{V}}{\mathrm{I}}\) for each set of values.
• We notice that \(\frac{\mathrm{V}}{\mathrm{I}}\) is a constant.
  V ∝ I ⇒ \(\frac{\mathrm{V}}{\mathrm{I}}\) = constant
  This constant is known as resistance of the conductor, denoted by R.
  ⇒ \(\frac{\mathrm{V}}{\mathrm{I}}\) = R
  ∴ Ohm’s law is verified.

How to Make Rheostat:

Make two holes at the two ends of 30cm Wooden scale. Through these holes fix two bolts with the help of nuts.Then take iron box filament i. e., nichrome wire and tie one end of thewire to the first bolt and wound wire with equal distance on the wooden scale to other end of the second bolt. Place this scale on the other scale perpendicularly as shown in the figure and stick them with glue. Now Rheostat is ready. Take support of your teacher to know the connection and functioning of Rheostat.

Question 23.
a) Take a battery and measure the potential difference. Make a circuit and measure the potential difference when the battery is connected in the circuit. Is there any difference in potential difference of battery? (AS4)
b) Measure the resistance of a bulb (filament) in open circuit with a multi-meter. Make a circuit with elements such as bulb, battery of 12 V and key in series. Close the key. Then again measure the resistance of the same bulb (filament:) for every 30 seconds. Record the observations in a proper table. What can you conclude from the above results? (AS4)
Answer:

a) When the battery is connected in a circuit, the voltage slowly decreases due to consumption of it. So, there is difference in voltage before using and after connecting.

b) After connecting battery (12 V), key in ammeter and bulb as shown in figure, we measure current (I) using the ammeter and voltage using multi-meter or voltmeter.

Note these values in the following table. Measure the resistance of the same bulb for every 30 seconds.

We conclude that the resistance is constant.

Question 24.
Draw a circuit diagram for a circuit in which two resistors A and B are connected in series with a battery and a voltmeter is connected to measure the potential difference across the resistor A. (AS5)
Answer:

V : Volt meter
A and B : Resistors
B : Battery
K: Key

Question 25.
How can you appreciate the role of a small fuse in house wiring circuit in preventing damage to various electrical appliances connected in the circuit? (AS7)
(OR)
We can save the household wiring and devices by using fuses. Write any four points by appreciating the role of fuse.
Answer:

• The fuse consists of a thin wire of low melting point. When the current in the fuse exceeds 20 A, the wire will heat up and melt.
• The circuit then becomes open and prevents the flow of current into the household circuit. So all the electric devices are saved from damage that could be caused by overload.
• Thus we can save the household wiring and devices by using fuses.
• In this way a small fuse prevents a great damage to costly electrical appliances in the circuit.

Question 26.
In the figure, the potential at A is………….. when the potential at B is zero. (AS7)

Answer:
Potential difference at A = V
Potential difference atB = V + 5 × 1 + 2 = 0 ⇒ V + 7V = 0
∴ V = – 7V

Question 27.
Observe the circuit and answer the questions given below. (AS7)

i) Are resistors C and D in series?
ii) Are resistors A and B in series?
iii) Is the battery in series with.any resistor?
iv) What is the potential drop across the resistor C?
v) What is the total emf in the circuit if the potential drop across resistor A is 6 V?
Answer:
The given circuit is written / drawn as
i) Yes, resistors ‘C’ and ‘D’ are connected in series. (Because, passing of the current is same to those resistors)
ii) No, resistors A’ and ‘B’ are not in series. (Because, different currents are passing through A and B. i.e., I₁ and I₂)
iii) The battery is in series with the resistor ‘A’. (Because, same current is passing through battery and resistor ‘A’, i.e., I)

iv) Potential drop across the resistor ‘C’
V₂ = V₃ + V₄
14V = V₃ + 8V
V₃ = 6V
Potential drop = 6V

v) Total emf
emf of combination of V₃ and V₄ = 14V ……………….. (1)
emf of combination of (1) and V₂ = 14 V ………………. (2)
emf of combination of (2) and V₁ = 6V + 14V = 20V
(Given, emf of ‘A’ = 6V)
Total emf = 20V

Question 28.
If the resistance of your body is 100000 Cl, what would be the current that flows in your body when you touch the terminals of a 12 V battery? (AS7)
Answer:
We know that, \(I=\frac{V}{R}\); here V = 12 V, R = 1,00,000Ω.
∴ The current passing through our body \(I=\frac{12 \mathrm{~V}}{100000 \Omega}\) = 0.00012 Ampere.

Question 29.
A uniform wire of resistance 100 Ω is melted and recast into wire of length double that of the original. What would be the resistance of the new wire formed? (AS7)
Answer:
Given R = 100 Ω
When ‘l = l’, R = 100 Ω.
When’l = 2l’, A’ = A / 2.

∴ Resistance is increased by four times.
∴ R = 4 × 1ooΩ = 400Ω.

Question 30.
A house has 3 tube lights, two fans and a Television. Each tube light draws 40 W. The fan draws 80 W and the Television draws 60 W. On the average, all the tube lights are kept on for five hours, two fans for 12 hours and the television for five hours every day. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWh. (AS7)
Answer:
Given 3 tube lights, two fans and a television.
Power consumed by 1 tube light = 40 W
∴ Power consumed by 3 tube lights = 3 × 40W = 120W
3 tube lights are kept on for five hours. So, consumption of power by 3 tube lights
= 5 × 120 W = 600 W ……………. (1)
Power consumed by 1 fan = 80 W
∴ Power consumed by 2 fans = 2x80W=160W
2 fans are kept on for 12 hours. So, consumption of power by 2 fans
= 12 × 160 W = 1920 W ……………. (2)
Power drawn by TV = 60 W
TV is kept on for 5 hours = 5 x 60 W = 300 W ………………. (3)
∴ Consumption of power in one day = (1) + (2) + (3)
= 600W+ 1920 W + 300 W = 2820 W = 2.820 KW
∴ Total consumption of power in 30 days at Rs. 3 per KW
= 2.820 × 30 × 3 = Rs. 253.80/-

Fill in The Blanks

1. The kilowatt hour is the unit of …………………..
2. A thick wire has ………………….. resistance than a thin wire.
3. An unknown circuit draws a current of 2 A from a 12 V battery. Its equivalent resistance is …………………..
4. The SI unit of potential difference is …………………..
5. The SI unit of current is …………………..
6. Three resistors of values 2Ω, 4Ω, 6Ω are connected in series. The equivalent resistance of combination of resistors is ……………………
7. Three resistors of values 2Ω, 4Ω, 6Ω are connected in parallel. The equivalent resistance of combination of resistors is ……………………
8. The power delivered by a battery of emf, 10 V is 10 W. Then the current delivered by the battery is ……………………
Answer:

1. electrical energy
2. less
3. 6 Ω
4. volt
5. Ampere
6. 12 Ω
7. \(\frac{11}{12} \Omega\)
8. 1 ampere

Multiple Choice Questions

1. A uniform wire of resistance 50 Ω. is cut into five equal parts. These parts are now connected in parallel. Then the equivalent resistance of the combination is
A) 2 Ω
B) 12 Ω
C) 250 Ω
D) 6250 Ω
Answer:
A) 2 Ω

2. A charge is moved from a point A to a point B. The work done to move unit charge during this process is called
A) potential at A
B) potential at B
C) potential difference between A and B
D) current from A to B
Answer:
C) potential difference between A and B

3. Joule/ coulomb is the same as
A) 1 – watt
B) 1 – volt
C) 1- ampere
D) 1 – ohm
Answer:
B) 1 – volt

4. The current in the wire depends
A) only on the potential difference applied
B) only on the resistance of the wire
C) on potential difference and resistance
D) none of them
Answer:
C) on potential difference and resistance

5. Consider the following statements.
a) In series connection, the same current flows through each element.
b) In parallel connection, the same potential difference gets applied across each element
A) both a and b are correct
B) a is correct but b is wrong
C) a is wrong but b is correct
D) both a and b are wrong
Answer:
A) both a and b are correct

10th Class Physics 11th Lesson Electric Current InText Questions and Answers

10th Class Physics Textbook Page No. 179

Question 1.
Does motion of charge always lead to electric current?
Answer:
Yes, it does.

Question 2.
Take a bulb, a battery, a switch and few insulated copper wires to the terminals of the battery through the bulb and switch. Now switch on the circuit and observe the bulb. What do you notice?
Answer:
The bulb glows.

10th Class Physics Textbook Page No. 180

Question 3.
Can you predict the reason for the bulb not glowing in situations 2 and 3?
Answer:
Yes, in situation 2 – there is no charge to travel in the circuit as the battery is disconnected. So, the bulb isn’t glowing.

In situation 3, we replaced the copper wires with nylon wires. Nylon is not a conductor. So, the bulb isn’t glowing.

Question 4.
Why do all materials not act as conductors?
Answer:
In conductors the gap between the atoms is very less. So, the transfer of energy is easy. But in other materials the gap is more. So, the transfer of energy is not possible.

Question 5.
How does a conductor transfer energy from source to bulb
Answer:

• A source has chemical energy which transfers electrons to the conductor.
• The conductor carries the electrons to the bulb when connected.
• Thus, the conductor transfers energy from source to bulb.

Question 6.
What happens to the motion of electrons when the ends of the conductor are connected to the battery?
Answer:
The energy transfer takes place from battery to the bulb through conductor.

10th Class Physics Textbook Page No. 181

Question 7.
Why do electrons move in specified direction?
Answer:
The electrons move in specified direction when the ends of the conductpr are connected to the terminals of a battery.
A uniform electric field is set up throughout the conductor. This field makes the electrons move in a specified direction.

Question 8.
In which direction do the electrons move?
Answer:
In a direction opposite to the direction of the field.

Question 9.
Do the electrons accelerate continuously?
Answer:
No, they lose energy and are again accelerated by the electric field.

Question 10.
Do they move with a constant speed?
Answer:
Yes, they move with a constant average Speed.

Question 11.
Why does a bulb glow immediately when we switch on?
Answer:
When we switch on any electric circuit, irrespective of length of the conductor, an electric field is set up throughout the conductor instantaneously due to the voltage of the source connected to the circuit.

Question 12.
How can we decide the direction of electric current?
Answer:
By the signs of the charge and drift speed.

10th Class Physics Textbook Page No. 183

Question 13.
How can we measure electric current?
Answer:
An ammeter is used to measure electric current.

Question 14.
Where do the electrons get energy for their motion from?
Answer:
From an electric field set up throughout the conductor.

Question 15.
Can you find the work done by the electric force?
Answer:
Yes. With the help of the formula W = F_el, we can find the work done by the electric force.

Question 16.
What is the work done by the electric force on unit charge?
Answer:
Work done by the electric force on unit charge \(\mathrm{V}=\frac{\mathrm{W}}{\mathrm{q}}=\frac{\mathrm{F}_{\mathrm{e}} l}{\mathrm{q}}\). It is called Potential difference.

10th Class Physics Textbook Page No. 184

Question 17.
What is the direction of electric current in terms of potential difference?
Answer:
Electrons move from low potential to high potential.

Question 18.
Do positive charges move in a conductor? Can you give an example of this?
Answer:
No, they don’t move. They are fixed in the lattice.
Eg : battery.

Question 19.
How does a battery maintain a constant potential difference between its terminals?
Answer:
We know that a battery consists electric force (Fe) and chemical force (Fc). These two forces are balanced in a battery. Due to this reason a battery maintains a constant , potential difference between its terminals.

Question 20.
Why does the battery discharge when its positive and negative terminals are connected through a conductor?
Answer:
A conductor permits the charges to pass through it. Due to this the exhaustion of charges is created after completion of all charges. So, when a battery is connected with a conductor it discharges.

10th Class Physics Textbook Page No. 185

Question 21.
What happens when the battery is connected in a circuit?
Answer:
A potential difference is created between the ends of the conductor, when the battery is connected in a circuit.

10th Class Physics Textbook Page No. 186

Question 22.
How can we measure potential difference or emf?
Answer:
With the help of a voltmeter, we measure potential difference or emf.

10th Class Physics Textbook Page No. 187

Question 23.
Is there any relation between emf of battery and drift speed of electrons in the conductor connected to a battery?
Answer:
Yes, when emf of a battery is more the drift speed of electrons will be more.

10th Class Physics Textbook Page No. 189

Question 24.
Can you guess the reason why the ratio of V and I in case of LED is not constant?
Answer:
This is due to forward voltage and maximum continuous current rating characters of LEDs.

Question 25.
Do all materials obey Ohm’s law?
Answer:
No, some materials don’t obey Ohm’s law.

Question 26.
Can we classify the materials based on Ohm’s law?
Answer:
Yes, the materials which obey Ohm’s law are conductors and others are same conductors or non-conductors.

Question 27.
What is resistance?
Answer:
The obstruction offered to the flow of electrons in a conductor is called the resistance.

Question 28.
Is the value of resistance the same for all materials?
Answer:
Yes, it varies.

Question 29.
Is there any application of Ohm’s law in daily life?
Answer:
Yes, this law is used in wiring.

Question 30.
What causes electric shock in the human body – current or voltage?
Answer:
Current with sufficient voltage.

10th Class Physics Textbook Page No. 190

Question 31.
Do you know the voltage of mains that we use in our household circuits?
Answer:
Yes, I know the voltage of mains that we use in our household circuits is 120 V.

Question 32.
What happens to our body if we touch live wire of 240 V?
Answer:
240 V current disturbs the functioning of organs inside the body. It is called electric shock. If the current flow continues further, it damages the tissues of the body which leads to decrease in resistance of the body. When this current flows for a longer time, damage to the tissues increases and thereby the resistance of human body decreases further. Hence, the current through the human body will increase. If this current reaches 0.07 A, it effects the functioning of the heart and if this much current passes through the heart for more than one second it could be fatal.

If this current flows for a longer time, the person in electric shock will be killed.

10th Class Physics Textbook Page No. 191

Question 33.
Why doesn’t a bird get a shock when it stands on a high voltage wire?
Answer:
There are two parallel lines carrying 240 V current. The voltage current will pass through the body if both the wires are touched at the same time. But, when the bird stands on only one wire, there is no potential difference between the legs. So, no current passes through the bird. Hence, it doesn’t feel any electric shock.

10th Class Physics Textbook Page No. 192

Question 34.
What could be the reason for increase in the resistance of the bulb when current flows through it?
Answer:
The increase in temperature of the filament in the bulb is responsible for increase in resistance of the bulb.

Question 35.
What happens to the resistance of a conductor if we increase its length?
Answer:
The resistance of a conductor increases with the increase of its length.

10th Class Physics Textbook Page No. 193

Question 36.
Does the thickness of a conductor influence its resistance?
Answer:
Yes, as the thickness of the conductor increases the resistance decreases.

10th Class Physics Textbook Page No. 195

Question 37.
How are electric devices connected in circuits?
Answer:
Electric devices are connected either in series or parallel in circuits.

Question 38.
When bulbs are connected (resistors) in series, what do you notice
Answer:
We notice that, the sum of the voltages of the bulbs (resistors) is equal to voltage across the combination of the resistors.

10th Class Physics Textbook Page No. 196

Question 39.
What do you notice when bulbs (resistors) are connected in series to the current?
Answer:
The current is not changing

Question 40.
What do you mean by equivalent resistance?
Answer:
If the current drawn by a resistor is equal to the current drawn by the combination of resistors, then the resistor is called equivalent resistor.

Question 41.
What happens when one of the resistors in series breaks down?
Answer:
The circuit becomes open and flow of current will be broken down.

Question 42.
Can you guess in what way household wiring has been done?
Answer:
Parallel connection.

10th Class Physics Textbook Page No. 197

Question 43.
How much current is drawn from the battery if the resistors are connected in parallel?
Is it equal to individual currents drawn by the resistors?
Answer:
It is the sum of currents flowing through each resistor. No, it is the sum of individual currents drawn by the resistors.

10th Class Physics Textbook Page No. 199

Question 44.
How could the sign convention be taken in a circuit?
Answer:
The potential difference across the resistor is taken as negative when we move along the direction of electric current through the resistor, and it is taken as positive when we move against the direction of electric current through the resistor.

10th Class Physics Textbook Page No. 201

Question 45.
You might have heard the sentences like “this month we have consumed 100 units of current”. What does ‘unit’ mean?
Answer:
Unit (or) kilo watt hour is the consumption of electric power in one hour by our electric appliances.

Question 46.
A bulb is marked 60 W and 120 V. What do these values indicate?
Answer:
It means, the resistance of the bulb is

Question 47.
What is the energy lost by the charge in 1 sec.?
Answer:
It is equal to \(\frac{\mathrm{W}}{\mathrm{t}}\).

10th Class Physics Textbook Page No. 202

Question 48.
What do you mean by overload?
Answer:
When a high current flows through the wire which is beyond the rating of wire then heating of wire takes place. This phenomenon is called overloading.

Question 49.
Why does it (overloading) cause damage to electric appliances?
Answer:
Due to overload the heat increases in the circuit and this melts the parts of the appliances. Thus overload causdt damage to the electric appliances.

10th Class Physics Textbook Page No. 203

Question 50.
What happens when this current (overloading) increases greatly to the household circuit?
Answer:
It causes fire.

Question 51.
How can we prevent damage due to overloading?
Answer:
To prevent damages due to overloading we connect an electric fuse to the household circuit.

10th Class Physics Textbook Page No. 203

Question 52.
What do you mean by short circuit?
Answer:

• The line wires that are entering the meter have a voltage of 240 V.
•  The minimum and maximum limit of current that can be drawn from the mains is 5 to 20 A.
• Thus, the maximum current that we can draw from the mains is 20 A.
• When the current drawn from the mains is more than 20 A, overheating occurs and may cause a fire. This is called overloading.
• A short circuit is an electrical circuit that allows a current to travel along an unintended path often where essentially no electrical impedance is encountered.

Question 52.
Why does a short circuit damage electric wiring and devices connected to it?
Answer:
In a short circuit the current drawn from the main exceeds the maximum limit 20 A. This will lead to overloading which can damage the electrical appliances.

10th Class Physics 11th Lesson Electric Current Activities

Activity – 1

Question 1.
Write an activity to check when a bulb glows in a circuit.
(OR)
How do you prove a source of energy is required to glow a bulb in a circuit?
Answer:
Aim :
To check when a bulb glows in a circuit.

Materials required:

1. A bulb
2. a battery
3. a switch
4. few insulated copper wire

Procedure (1) :

1. Take a bulb, a battery, a switch and few insulated copper wires.
2. Connect the ends of the copper wires to the terminals of the battery through the bulb and switch.
3. Now switch on the circuit.
   Observation (1) : The bulb glows.

Procedure (2) :

1. Remove the battery from the circuit and connect the remaining components to make a complete circuit.
2. Again switch on the circuit and observe the bulb.

Observation (2): The bulb does not glow.

Procedure (3) :
Replace the copper wires with nylon wires and connect the nylon wires to the terminals of the battery through a bulb and switch. Now switch on the circuit. We observe that the bulb does not glow. Because the wires are not conductors.

Observation (3) : The bulb does not glow.

Result:
The battery contains charges which glow the bulb.

Activity – 3

Question 2.
Write an activity to show that the values of current are different for different wires for a constant voltage.
(OR)
The resistance of a conductor depends on the material of the conductor. Prove this through an activity.
(OR)
List out the material required in the experiment to show that the electric resistance depends upon the nature of the material and write experimental procedure.
Answer:
Aim:
To show that the values of current are different for different wires for a constant voltage. Materials required : (wires of the same length and some cross-sectional area).

• Copper rod
• Nichrome rod
• Battery
• Ammeter
• Key
• Manganin Wire

Procedure :

1. Make a circuit as shown in figure.
2. Connect one of the wires between the ends P and Q.
3. Switch on the circuit. Measure the electric current for a fixed voltage, using the ammeter connected to the circuit. Note it in your notebook.
4. Repeat this experiment with other wires and note the current in your notebook.

Observation :
The values of current are different for different wires for a constant voltage.

Conclusion:
The resistance of a conductor depends on the material of the conductor.

Activity – 5

Question 3.
Write an activity to show that resistance is inversely proportional to the c section area of the conductor.
(OR)
What happens to resistance if the area of a cross-section of conductor is increased? Explain with an activity.
Aim :
To prove that resistance is inversely proportional to the cross-section area of the conductor.

Materials required :

1. A Battery
2. Mangnin Wires
3. Ammeter
4. Key
5. Manganin wires with different cross-section areas (lengths are same).

Procedure:

1. Make the circuit as given figure.
2. Connect one of the wires between points P and Q.
3. Switch on me circuit. Note the ammeter reading in your notebook.
4. Continue the experiment with different wires of same length but different cross-section areas. Note the ammeter readings in your notebook.

Observation :
As the cross-section area of the rods increases, the current increases.

Result (Conclusion) :
Resistance is inversely proportional to cross-section area of the conductor.

Activity – 6

Question 4.
Write an activity to prove that the sum of the potential differences of the bulb is equal to voltage across the combination of the resistors. (OR)
Prove that during series connection potential difference is distributed among the resistors experimentally.
Answer:
Aim:
To prove that the sum of the potential differences of the bulbs is equal to potential difference across the combination of the resistors.

Materials required :

1. Bulbs
2. Voltmeters
3. Insulated wires
4. Ammeter
5. Key

Procedure :

1. Take different bulbs. Using a multimeter measure their resistances. Note them as R,, R2 and Rv
2. Connect them as shown in figure.
3. Measure the voltage between terminals of the battery connected to the circuit.
4. Measure the voltages between the ends of each bulb and note them as Vj, V2 and V3 from voltmeters in your notebook.
5. Compare them.

Observation :
We notice that the’sum of the voltages of the bulbs is equal to voltages across the combination of the resistors.

Activity – 7

Question 5.
Write an activity to prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
(OR)
Prove that during parallel connection the current is distributed among the resistances by using an experimental activity.
Answer:
Aim:
To prove that the current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.
Materials required :

1. Bulbs
2. Ammeters
3. Buttery
4. Key
5. Wires

Procedure :

1. Connect the bulbs in parallel connection as shown in the given circuit.
2. Measure the voltage across each bulb using a voltmeter or multimeter.
3. Note these values in your notebook.

Observation :

1. The voltage at the ends of each bulb is the same.
2. Measure electric currents flowing through each bulb using ammeters. Note these values.
3. Measure the current (I) drawn from the battery using the ammeter 1.

Result (Conclusion) :
The current drawn from the battery is equal to the sum of individual currents drawn by the bulbs.