Bhagya

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 6 जड़ की मुसकान Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 6 जड़ की मुसकान

Hindi Guide for Class 10 PSEB जड़ की मुसकान Textbook Questions and Answers

(क) विषय-बोध

I. निम्नलिखित प्रश्नों के उत्तर एक-दो पंक्तियों में दीजिए

प्रश्न 1.
एक दिन तने ने जड़ को क्या कहा?
उत्तर:
तने ने जड़ से कहा कि वह तो निर्जीव है तथा सदा जीवन से डरी रहती है।

प्रश्न 2.
जड़ का इतिहास क्या है ?
उत्तर:
जड़ का इतिहास यह है कि वह सदा ज़मीन के अंदर मुँह गड़ा कर पड़ी रहती है। वह तो मिट्टी से कभी बाहर ही नहीं निकलती।

प्रश्न 3.
डाली तने को हीन क्यों समझती है?
उत्तर:
डाली तने को हीन इसलिए समझती है क्योंकि उसे जहाँ बैठा दिया जाता है वहीं बैठा रहता है, कभी भी गतिशील नहीं होता जबकि वे लहराती रहती हैं।

प्रश्न 4.
पत्तियाँ डाल की किस कमी की ओर संकेत करती हैं?
उत्तर:
पत्तियाँ डाल की इस कमी की ओर संकेत करती हैं कि वे इस ध्वनि प्रधान संसार में एक शब्द भी नहीं बोल सकती हैं। वह तो अपने भावों-विचारों को भी व्यक्त नहीं कर पाती।

प्रश्न 5.
फूलों ने पत्तियों की चंचलता का आधार क्या बताया?
उत्तर:
फूलों के अनुसार पत्तियों की चंचलता का आधार डाली है।

प्रश्न 6.
सबकी बातें सुनकर जड़ क्यों मुसकराई?
उत्तर:
सब की बातें सुनकर जड़ इसलिए मुसकराई थी कि केवल वह ही जानती थी कि यदि वह न होती तो वृक्ष का अस्तित्व ही न होता। न तना, न शाखाएं, न पत्ते और न ही फूल होते। जड़ ही उन सब का पोषण करती है। उन सभी का अस्तित्व उसी के कारण ही होता है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या करें

प्रश्न 1.
एक दिन तने ने भी कहा था,
जड़?
जड़ तो जड़ ही है
जीवन से सदा डरी रही है,
और यही है उसका सारा इतिहास
कि ज़मीन से मुँह गड़ाए पड़ी रही है;
उत्तर:
कवि लिखता है कि एक दिन तने ने जड़ के संबंध में कहा कि जड़ तो बिलकुल जड़ अर्थात् निर्जीव है। वह सदा जीवन से भयभीत रहती है। उसका सारा इतिहास यही है कि वह सदा ज़मीन में ही मुँह गड़ा कर पड़ी रहती है, वह संसार से मुँह छिपा कर जमीन के अंदर छिपी रहती है, किंतु तना स्वयं को जमीन से ऊपर उठा कर बढ़ता हुआ बताता है, वह मज़बूत बना हुआ है इसलिए तना कहलाता है, वह तना हुआ अर्थात् अकड़ा हुआ, घमंडी है।

प्रश्न 2.
एक दिन फूलों ने भी कहा था, पत्तियाँ?
पत्तियों ने क्या किया?
संख्या के बल पर बस डालों को छाप लिया,
डालों के बल पर ही चल-चपल रही हैं,
हवाओं के बल पर ही मचल रही हैं
लेकिन हम अपने से खुले, खिले फूले हैं
रंग लिए, रस लिए, पराग लिए
हमारी यश-गंध दूर-दूर-दूर फैली है,
भ्रमरों ने आकर हमारे गुन गाए हैं,
हम पर बौराए हैं।
उत्तर:
डालियों की बातें सुन कर एक दिन पत्तियों ने भी कहा कि डाल में भी भला कोई विशेषता है ? हम मानती हैं कि वे झूमती हैं, झुकती हैं, हिलती हैं परंतु इस आवाज़ वाली दुनिया में क्या कभी उन्होंने एक शब्द भी बोला है ? अर्थात् वे बिलकुल नहीं बोलती हैं। इसके विपरीत पत्तियाँ कहती हैं कि वे सदा हर-हर का शब्द बोलती रहती हैं। उनके आपस में टकराने से वातावरण उनकी खड़खड़ाहट की आवाज़ से भर जाता है। हर वर्ष वे नया स्वरूप प्राप्त कर लेती हैं और पतझड़ में झड़ जाती हैं। वसंत के आने पर वे फिर निकल आती हैं और डालियों पर छा जाती हैं। वे थके हुए मन वाले पथिकों की परेशानियों तथा गर्मी को दूर कर उन्हें शांति प्रदान करती हैं।

(ख) भाषा-बोध

I. निम्नलिखित शब्दों के विपरीत शब्द लिखें:

जीवन = ————-
जड़ = ————-
मज़बूत = ————-
ऊपर। = ————-
उत्तर:
शब्द = विपरीत शब्द
जीवन = मृत्यु
जड़ = चेतन
मजबूत = कमज़ोर
ऊपर = नीचे।

II. निम्नलिखित शब्दों के विशेषण शब्द बनाइए

इतिहास = ————
दिन = ————
वर्ष = ————
रंग = ————
रस। = ————
उत्तर:
शब्द = विशेषण
इतिहास = ऐतिहासिक
दिन = दैनिक
वर्ष = वार्षिक
रंग = रंगीला/रंगीन
रस। = रसीला।

III. निम्नलिखित शब्दों के दो-दो पर्यायवाची शब्द लिखें

प्रगति = ————
हवा = ————
ध्वनि = ————
फूल = ————
भ्रमर। = ————
उत्तर:
शब्द = पर्यायवाची शब्द
प्रगति = उन्नति, बढ़ोत्तरी, बढ़ती, वृद्धि।
हवा = वायु, पवन, वात, समीर।
ध्वनि = शब्द, आवाज़, गूंज, स्वर।
फूल = पुष्प, सुमन, प्रसून, कुसुम।
भ्रमर = भौंरा, भँवरा, सारंग, षट्पद।

IV. निम्नलिखित के अनेकार्थी शब्द लिखें

जड़ = ————
तना = ————
डाल = ————
डोली = ————
बोली। = ————
उत्तर:
जड़ = मूर्ख, स्थिर, वनस्पतियों की जड़, आधार।
तना = अकड़ा हुआ, पेड़-पौधों का जड़ से जुड़ा हिस्सा, कसा हुआ।
डाल = शाखा, डालना/उड़ेलना।
डोली = कांपी, मिट्टी का छोटा बर्तन, विवाह के पश्चात् लड़की का ससुराल गमन हेतु पालकी, शिविका।
बोली = वचन, भाषा, बोलचाल, नीलाम की आवाज़, व्यंग्य, कटाक्ष।

(ग) पाठ्येतर सक्रियता

प्रश्न 1.
रामवृक्ष बेनीपुरी का निबंध ‘नींव की ईंट’ पढ़िए और जड़ के महत्व पर अपने सहपाठियों के साथ चर्चा कीजिए।
उत्तर:
(विद्यार्थी स्वयं करें)

प्रश्न 2.
‘वही देश मज़बूत होता है, जिसकी संस्कृति मज़बूत जड़ के समान होती है।’ इस विषय पर कक्षा में भाषण प्रतियोगिता का आयोजन कीजिए।
उत्तर:
(विद्यार्थी स्वयं करें)

प्रश्न 3.
प्रस्तुत कविता को आगे बढ़ाइए। जड़, तना, पत्ते और फूल के साथ फल को भी शामिल कीजिए। कविता को आगे बढ़ाते निम्न पंक्तियों को पूरा करें-
एक दिन फलों ने भी कहा था,
……………….?
……………… ने क्या किया ?
वृथा ही फूलते हैं
आज फूले हैं
कल ……………. जाएंगे
हमें देखो, हम पशु, पक्षी और …………… का।
…………. भरते हैं
उन्हें ज़िन्दा रखने को
अपना ……………. करते हैं।
उत्तर:
एक दिन फलों ने भी कहा था फूल?
फूल ?
फूलों ने क्या किया?
वृथा ही फूलते हैं
आज फूले हैं
कल मुरझा जाएंगे
हमें देखो,
हम पशु, पक्षी और मनुष्य का
पेट भरते हैं
उन्हें ज़िन्दा रखने को
अपना जीवन बलिदान करते हैं।

(घ) ज्ञान-विस्तार

जिस प्रकार जड़ एक वृक्ष को बनाती है, उसी प्रकार परिवार की जड़ परिवार के पुरखे दादा-दादी, माता-पिता होते हैं, जो अपनी संतान के द्वारा परिवार को रक्षा, खुशहाली और वृद्धि प्रदान करते हैं। इसी प्रकार से जैसे माता-पिता अपनी संतान का पालन-पोषण करते हैं उसी प्रकार अध्यापक बच्चों को ज्ञान का प्रकाश देकर भावी नागरिक बनाते हैं। देश को स्वतंत्र कराने वाले देशभक्तों का बलिदान हमें देश की रक्षा की प्रेरणा देता है। इन सबके प्रति हमें कृतज्ञ होकर इनके दिखाये पथ पर चलना चाहिए।

PSEB 10th Class Hindi Guide जड़ की मुसकान Important Questions and Answers

प्रश्न 1.
प्रगति करने पर लोग क्या भूल जाते हैं?
उत्तर:
प्रगति करने पर लोग अक्सर अपने मूलभूत आधार को भूल जाते हैं तथा अपने जीवन की सारी प्रगति का श्रेय स्वयं को देते हैं, जबकि उनकी प्रगति के पीछे उन सब का हाथ होता है जिनके कारण वास्तव में उनकी प्रगति हुई।

प्रश्न 2.
भवन मज़बूत कब होता है?
उत्तर:
भवन तब मज़बूत होता है, जब उसकी नींव मज़बूत होती है।

प्रश्न 3.
तना किसके कारण मज़बूत बना है?
उत्तर:
तना जड़ के कारण मज़बूत बना है।

प्रश्न 4.
डालियाँ कहाँ से फूटती हैं?
उत्तर:
डालियाँ तने से फूटती हैं।

प्रश्न 5.
पत्तियाँ स्वर कैसे उत्पन्न करती हैं?
उत्तर:
पत्तियाँ हर-हर स्वर करती हैं जो उनकी आपसी रगड़ से उत्पन्न होता है।

प्रश्न 6.
फूलों पर कौन बौराएं हैं?
उत्तर:
फूलों पर भँवरे बौराएं हैं।

प्रश्न 7.
तने को अपने आप पर क्या घमंड था?
उत्तर:
तने को अपने आप पर यह घमंड था कि वह अपने साहस और शक्ति के कारण बढ़ सका। वह अपने बलबूते पर ज़मीन से ऊपर उठ सका। मिट्टी की गहराइयों से वह स्वयं बाहर निकला। उसने स्वयं अपनी शक्ति के कारण अपना विकास किया। वह जड़ की अपेक्षा अधिक सक्षम था।

प्रश्न 8.
फूलों ने अभिमान से भर कर क्या कहा था?
उत्तर:
फूलों ने अभिमान से भर कर कहा था कि उनके विकास में पत्तियों का कोई योगदान नहीं था। वे तो स्वयं खुले थे; फले थे और फूले थे। उन्होंने स्वयं ही रंग-रस पाया था। उनके यश को भँवरों ने गाया था और उनकी सुगंध दूर-दूर तक फैली थी।

एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
जड़ को जड़ किसने कहा?
उत्तर:
जड़ को जड़ तने ने कहा।

प्रश्न 2.
प्रगतिशील जगती में कौन तिल भर भी नहीं डोला?
उत्तर:
प्रगतिशील जगती में तना तिल भर भी नहीं डोला।

प्रश्न 3.
थके हुए पथिकों के मन का शाप-ताप कौन हरती हैं?
उत्तर:
पत्तियाँ थके हुए पथिकों के मन का शाप-ताप हरती हैं।

प्रश्न 4.
भ्रमरों ने किन के गुन गाए हैं?
उत्तर:
भ्रमरों ने फूलों के गुण गाए हैं।

बहुवैकल्पिक प्रश्नोत्तरनिम्नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
फूलों के गुन किसने गाए हैं
(क) भ्रमर
(ख) कोयल
(ग) मोर
(घ) पपीहा।
उत्तर:
(क) भ्रमर

प्रश्न 2.
हर वर्ष नूतन कौन होती है
(क) जड़
(ख) डाल
(ग) पत्ती
(घ) गंध।
उत्तर:
(ग) पत्ती

प्रश्न 3.
जीवन से सदा कौन डरी रही
(क) जड़
(ख) डाल
(ग) पत्ती
(घ) गंध।
उत्तर:
(क) जड़

एक शब्द/हाँ-नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
कौन खाया है, मोटाया है, सहलाया चोला है? (एक शब्द में उत्तर दें)
उत्तर:
तना

प्रश्न 2.
जड़ तो जड़ नहीं है। (सही या गलत में उत्तर लिखें)
उत्तर:
गलत

प्रश्न 3.
पत्तियाँ हर-हर स्वर नहीं करती हैं। (सही या गलत में उत्तर दें)
उत्तर:
गलत

प्रश्न 4.
सबकी सुनकर भी जड़ नहीं मुसकराई। (हाँ या नहीं में उत्तर दें)
उत्तर:
नहीं

प्रश्न 5.
डाल तने से फूटती है। (हाँ या नहीं में उत्तर दें)
उत्तर:
हाँ

प्रश्न 6.
लेकिन मैं ……….. से ऊपर उठा।
उत्तर:
ज़मीन

प्रश्न 7.
मर्मर स्वर …………….. भरती है।
उत्तर:
मर्मभरा

प्रश्न 8.
हमारी …………… दूर-दूर फैली है।
उत्तर:
यश-गंध।

जड़ की मुसकान पद्यांशों की सप्रसंग व्याख्या

1. एक दिन तने ने भी कहा था
जड़?
जड़ तो जड़ ही है।
जीवन से सदा डरी रही है,
और यही है उसका सारा इतिहास
कि ज़मीन से मुँह गड़ाए पड़ी रही है
लेकिन मैं ज़मीन से ऊपर उठा,
बाहर निकला,
बढ़ा हूँ
मजबूत बना हूँ,
इसी से तो तना हूँ।

शब्दार्थ:
जड़ = वृक्ष का मूल, जड़, मिर्जीव। मुँह गड़ाए पड़ी रही = सबसे छिप कर रहना। तना हूँ = वृक्ष का तना होना, अकड़ना, घमंड करना।

प्रसंग:
प्रस्तुत पंक्तियाँ हरिवंशराय बच्चन द्वारा रचित कविता ‘जड़ की मुसकान’ से ली गई हैं, जिसमें कवि ने वृक्ष के अस्तित्व में जड़ का महत्त्व स्पष्ट किया है।

व्याख्या:
कवि लिखता है कि एक दिन तने ने जड़ के संबंध में कहा कि जड़ तो बिलकुल जड़ अर्थात् निर्जीव है। वह सदा जीवन से भयभीत रहती है। उसका सारा इतिहास यही है कि वह सदा ज़मीन में ही मुँह गड़ा कर पड़ी रहती है, वह संसार से मुँह छिपा कर जमीन के अंदर छिपी रहती है, किंतु तना स्वयं को जमीन से ऊपर उठा कर बढ़ता हुआ बताता है, वह मज़बूत बना हुआ है इसलिए तना कहलाता है, वह तना हुआ अर्थात् अकड़ा हुआ, घमंडी है।

विशेष:

1. तना स्वयं को जड़ से श्रेष्ठ बताता है तथा जड़ को सदा मुँह छिपा कर धरती में छिप कर रहने वाली कहता है।
2. भाषा सहज, सरल है।
3. मानवीकरण और अनुप्रास अलंकार हैं।

2. एक दिन डालों ने भी कहा था,
तना?
किस बात पर है तना?
जहां बिठाल दिया गया था वहीं पर है बना;
प्रगतिशील जगती में तिल भर नहीं डोला है,
खाया है, मोटापा है, सहलाया चोला है;
लेकिन हम तने से फूटी,
दिशा-दिशा में गईं
ऊपर उठीं,
नीचे आईं
हर हवा के लिए दोल बनीं, लहराईं,
इसी से तो डाल कहलाईं।

शब्दार्थ:
बिठाल = बैठाया गया। प्रगतिशील = आगे बढ़ते रहने का भाव। जगती = दुनिया, संसार। तिल भर = ज़रा-सा, थोड़ा-सा। डोला = हिला, गतिशील, आगे बढ़ा। सहलाया चोला = सुविधा भोगी शरीर। दोल = हिलना। डाल = शाखा ; टहनी।

प्रसंग:
यह काव्यांश हरिवंशराय बच्चन द्वारा रचित कविता ‘जड़ की मुसकान’ से लिया गया है। इसमें कवि ने वृक्ष के अस्तित्व में जड़ का महत्त्व स्पष्ट बताया है।

व्याख्या:
कवि ने लिखा है कि एक दिन तने की बातें सुनकर पेड़ की डालियाँ भी बोलने लगीं। डाल ने कहा कि तना किस बात पर घमंड कर रहा है, क्योंकि उसे तो जहाँ बैठा दिया गया है, अभी भी वहीं पर बैठा है। इस निरंतर गतिमान रहने वाली दुनिया में वह ज़रा भी गतिशील नहीं हुआ है। वह खूब खा-खा कर मोटा हो गया है। उसका शरीर जड़ की मुसकान सुविधा भोगी बन गया है किंतु हम डालियाँ उसी तने से निकल कर अनेक दिशाओं में फैल गई हैं। डालियाँ ऊपर उठती हैं, नीचे भी आती हैं। हवा की हर लहर के साथ झूलती हुई लहराती हैं, इसी से हम डाली कहलाती हैं।

विशेष:

1. डाल स्वयं को गतिशील तथा तने को सुविधा भोगी शरीर वाला तथा एक ही स्थान पर बैठा रहने वाला बताती है।
2. भाषा सहज, सरल, भावपूर्ण है।
3. अनुप्रास, मानवीकरण और पुनरुक्ति प्रकाश अलंकार हैं।

3. एक दिन पत्तियों ने भी कहा था,
डाल?
डाल में क्या है कमाल?
माना वह झूमी, झुकी, डोली है
ध्वनि-प्रधान दुनिया में
एक शब्द भी वह कभी बोली है?
लेकिन हम हर-हर स्वर करती हैं
मर्मर स्वर मर्मभरा भरती हैं,
नूतन हर वर्ष हुई,
पतझर में झर
बहार-फूट फिर छहरती हैं,
विथकित-चित पंथी का
शाप-ताप हरती हैं।

शब्दार्थ:
कमाल = विशेषता, खास बात, खासियत। ध्वनि = आवाज़। हर-हर स्वर = पत्तियों के आपस में टकराने से उत्पन्न आवाज़। मर्मर = हरे पत्तों की खड़खड़ाहट। नूतन = नई। विथकित = थका हुआ। चित्त = मन । पंथी = मुसाफिर। शाप-ताप = परेशानी और गर्मी।

प्रसंग:
प्रस्तुत पंक्तियाँ हरिवंशराय बच्चन द्वारा रचित कविता ‘जड़ की मुसकान’ से ली गई हैं, जिसमें कवि ने वृक्ष के लिए जड़ के महत्त्व पर प्रकाश डाला है।

व्याख्या:
डालियों की बातें सुन कर एक दिन पत्तियों ने भी कहा कि डाल में भी भला कोई विशेषता है ? हम मानती हैं कि वे झूमती हैं, झुकती हैं, हिलती हैं परंतु इस आवाज़ वाली दुनिया में क्या कभी उन्होंने एक शब्द भी बोला है ? अर्थात् वे बिलकुल नहीं बोलती हैं। इसके विपरीत पत्तियाँ कहती हैं कि वे सदा हर-हर का शब्द बोलती रहती हैं। उनके आपस में टकराने से वातावरण उनकी खड़खड़ाहट की आवाज़ से भर जाता है। हर वर्ष वे नया स्वरूप प्राप्त कर लेती हैं और पतझड़ में झड़ जाती हैं। वसंत के आने पर वे फिर निकल आती हैं और डालियों पर छा जाती हैं। वे थके हुए मन वाले पथिकों की परेशानियों तथा गर्मी को दूर कर उन्हें शांति प्रदान करती हैं।

विशेष:

1. पत्तियां स्वयं को डाल से श्रेष्ठ सिद्ध कर अपनी छाया से पथिकों को विश्राम प्रदान करने वाली भी मानती हैं।
2. भाषा सरल तथा भावपूर्ण है।
3. मानवीकरण, अनुप्रास तथा पुनरुक्ति प्रकाश अलंकार हैं।

4. एक दिन फूलों ने भी कहा था,
पत्तियाँ?
पत्तियों ने क्या किया?
संख्या के बल पर बस डालों को छाप लिया,
डालों के बल पर ही चल-चपल रही हैं,
हवाओं के बल पर ही मचल रही हैं
लेकिन हम अपने से खुले, खिले, फूले हैं
रंग लिए, रस लिए, पराग लिए
हमारी यश-गंध दूर-दूर फैली है,
भ्रमरों ने आकर हमारे गुन गाए हैं,
हम पर बौराए हैं।
‘सबकी सुन पाई है,
जड़ मुसकराई है !

शब्दार्थ:
छाप लिया = ढक लिया। चल-चपल = चंचल, हिलती-डुलती। यश-गंध = प्रशंसा रूपी सुगंध। बौराए = पागल होना; होश खो देना।।

प्रसंग:
यह काव्यांश हरिवंशराय बच्चन द्वारा रचित कविता ‘जड़ की मुसकान’ नामक कविता से अवतरित है। इसमें कवि ने वृक्ष के अस्तित्व को जड़ पर निर्भर बताया है।

व्याख्या:
पत्तियों की बातें सुनकर एक दिन फूल भी बोला कि पत्तियों ने भला किया ही क्या है? अर्थात् पत्तियों में तो कोई विशेषता ही नहीं है। पत्तियों ने तो केवल अपनी अधिक संख्या होने के कारण केवल डालों को ही ढक लिया है। वे डालों के कारण ही हिल-डुल रही हैं और हवाओं के कारण ही मचल रहीं हैं। किंतु हम फूल स्वयं ही खिले हैं और हमारे यश की सुगंध दूर-दूर तक फैली हुई है। भँवरे भी आकर हमारे गुणों का गान करते हैं। वे हमारे ऊपर मंडराते रहते हैं; वे हम पर पागल-से हो रहे हैं। इन सबकी बातों को सुनकर जड़ केवल मुसकराती है क्योंकि वह जानती है कि यदि वह न होती तो तना, डाल, पत्तियाँ और फूल भी न होते। इन सब का अस्तित्व जड़ के कारण ही है।

विशेष:

1. किसी भी भवन को बनाने में जैसे नींव का महत्त्व होता है वैसे ही वृक्ष का महत्त्व उसकी जड़ से है, यदि जड़ सलामत तो वृक्ष भी रहेगा।
2. भाषा सहज, सरल तथा भावपूर्ण है।
3. मानवीकरण, अनुप्रास तथा पुनरुक्ति प्रकाश अलंकार है।

जड़ की मुसकान Summary

जड़ की मुसकान कवि परिचय

श्री हरिवंशराय बच्चन हिंदी-कविता में हालावाद के प्रवर्तक कवि माने जाते हैं। इनका जन्म 21 नवंबर, सन् 1907 ई० को उत्तर प्रदेश के प्रयाग (इलाहाबाद) के कायस्थ परिवार में हुआ था। उनकी प्रारंभिक शिक्षा म्युनिसिपल स्कूल, कायस्थ पाठशाला तथा गवर्नमेंट स्कूल में हुई। सन् 1938 ई० में इन्होंने इलाहाबाद विश्वविद्यालय से अंग्रेज़ी में एम० ए० किया तक तथा सन् 1942 से सन् 1952 तक इलाहाबाद विश्वविद्यालय में प्राध्यापक के पद पर कार्यरत रहे। इसके बाद ये इंग्लैंड चले गये। वहां इन्होंने सन् 1952 से 1954 तक रहकर कैंब्रिज विश्वविद्यालय से पीएच० डी० की उपाधि प्राप्त की थी। सन् 1955 ई० में भारत सरकार ने इन्हें विदेश मंत्रालय में हिंदी विशेषज्ञ के पद पर नियुक्त किया। ये राज्यसभा के सदस्य भी रहे। इन्हें सोवियतलैंड तथा साहित्य अकादमी पुरस्कार से सम्मानित किया गया। ‘दशद्वार से सोपान तक’ रचना पर इन्हें सरस्वती सम्मान दिया गया। इनकी प्रतिभा और साहित्य सेवा को देखकर भारत सरकार ने इन्हें ‘पद्मभूषण’ की उपाधि से अलंकृत किया। 18 जनवरी, सन् 2003 में ये इस संसार को छोड़कर चिरनिद्रा में लीन हो गए।

रचनाएँ- हरिवंशराय बच्चन जी बहुमुखी प्रतिभा संपन्न साहित्यकार थे। उन्होंने अनेक विधाओं पर सफल लेखनी चलाई है। उनकी प्रमुख रचनाएँ निम्नलिखित हैं-

1. काव्य संग्रह-मधुशाला, मधुबाला, मधुकलश, निशा निमंत्रण, एकांत संगीत, आकुल-अंतर, मिलन यामिनी, सतरंगिणी, आरती और अंगारे, नए पुराने झरोखे, टूटी-फूटी कड़ियाँ, बुद्ध और नाचघर।
2. आत्मकथा चार खंड-क्या भूलूँ क्या याद करूँ, नीड़ का निर्माण फिर, बसेरे से दूर, दशद्वार से सोपान तक।
3. अनुवाद-हैमलेट, जनगीता, मैकबेथ।
4. डायरी-प्रवास की डायरी।

जड़ की मुसकान कविता का सार

‘जड़ की मुसकान’ शीर्षक कविता में बच्चन जी ने स्पष्ट किया है कि किस प्रकार लोग अपने मूल आधार को भूल कर स्वयं को ही महत्व देने लगते हैं। एक वृक्ष का तना जड़ को सदा जीवन से भयभीत ज़मीन में गड़ी रहने वाली कह कर उसका उपहास करता है तथा स्वयं को उससे अधिक श्रेष्ठ बताता है। डालियाँ स्वयं को तने से भी अधिक अच्छा बताती हैं क्योंकि तना तो एक ही स्थान पर खड़ा रहता है जबकि वे दिशा-दिशा में जाकर हवा में डोलती रहती हैं। पत्तियाँ स्वयं को डाल से अधिक महत्त्वपूर्ण इसलिए मानती हैं क्योंकि वे मर-मर्र स्वर में बोल भी सकती हैं। फूल सर्वत्र सुगंध फैलाने वाले, भँवरों को आकर्षित करने वाले, अपनी सुंदरता से सबको अपनी ओर खींचने वाले जानकर स्वयं को सर्वश्रेष्ठ मानते हैं । इन सब की बातों को सुन कर जड़ केवल मुसकराती है, क्योंकि वह जानती है कि इन सब का अस्तित्व उसके कारण ही है। यदि वह सलामत है तो वृक्ष भी है। जैसे मज़बूत नींव मज़बूत भवन बनाती है, वैसे हो मज़बूत जड़ वृक्ष को भी हरा-भरा रखती है।

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 12 Electricity Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 12 Electricity

PSEB 10th Class Science Guide Light Reflection and Refraction Textbook Questions and Answers

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is
(a) 1/25
(b) 1/5
(c) 5
(d) 25.
Answer:
On cutting wire of resistance R into five equal parts, the resistance of each part is \(\frac{R}{5}\). Let the total resistance be R’ of 5 resistors each of resistance R/5 connected in parallel.

Hence answer (d) 25 is correct.

Question 2.
Which of the following terms does not represent electrical power in the circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
All three (a), (c) and (d) represent power. Therefore, only (b) IR² does not represent power.

Question 3.
An electric bulb is rated 220 V and 100 W. When operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W.
Answer:

Hence (d) 25 W is correct answer.

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of heat produced in series and parallel combination would be
(a) 1 : 2
(b) 2 :1
(c) 1: 4
(d) 4 : 1
Answer:
Since the wires are of the same material and are equal in lengths and diameters, therefore, if R is resistance of each. Let total resistance in series and parallel be Rs and R respectively, when connected in series ;
R_s = R + R
= 2R
When connected to a source V, then
Heat produced in series Hs = \(\frac{\mathrm{V}^{2} t}{\mathrm{R}_{s}}\)
when connected is parallel, \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\)
\(\frac{2}{R}\)
or
R_p = R/2
Heat produced in parallel H_p = \(\frac{\mathrm{V}^{2} t}{\mathrm{R}_{p}}\)

Hence, (c) 1 : 4 is correct answer.

Question 5.
How is a voltmeter connected in the circuit to measure potential difference between two points?
Answer:
Voltmeter is always connected in parallel between two points across which the potential difference is required to be measured.

Question 6.
A copper wire has diameter 0.5 mm and resistivity 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Here, Diameter (d) = 0.5 mm
= 0.5 × 10^-3 m
= 5 × 10^-4 m
Resistivity
ρ = 1.6 × 10^-8 Ω m
l = 1

l = 122.76 m
If the diameter of wire is doubled, resistance will become l/4th of the original resistance, since R ∝ 1/A ,
or R ∝ l²d²
Now Resistance \(\frac{R}{4}=\frac{10}{4}\).
= 2.5 Ω
Decrease in resistance will be = 10 – 10/4
= 10 – 2.5
= 7.5 ohm

Question 7.
The values of the current I flowing in a given resistor for corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
Graph between I and V is as shown in Figure It is almost a straight line.

Resistance is the slope of I-V graph.
R = Slope = \(\frac{B C}{A C}\)
or R = \(\frac{10.0-1.6}{3-0.5}\)
= \(\frac{8.4}{2.5}\)
= \(\frac{84}{25}\) = 3.36 Ω
∴ R = 3,36 Ω

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of resistor?
Answer:
Here, I = 2:5 mA = 2.5 × 10^-3 A; V = 12 volts ; R =?
We know from ohm’s law
V = IR

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively. How much current will flow through a 12 Ω resistor?
Answer:
Resistors are connected in series
∴ Total Resistance, R8 = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω
Potential Difference V = 9 V
Resistance, R = \(\frac{V}{I}\)
or I = \(\frac{V}{R}\)
= \(\frac{9}{13.4}\)
or I = 0.67 A
When in series, the current passing through each resistor is equal to the current in the circuit.
∴ Current through 12 Ω = 0.67 A

Question 10.
How many 176 Ω resistors in parallel are required to carry 5A on a 220 V line?
Answer:
Here R = 176 Ω
V= 220V
I = 5A
Resistance of parallel combination R_p = \(\frac{V}{I}\)
R_p = \(\frac{220}{5}\)

If n resistors each of resistance R are connected in parallel, then net resistance R_p is

or n = 4
4 resistors each 176 Ω connected in parallel will result in net resistance of 44 Ω causing a current of 5A to flow through when connected to 220 V.

Question 11.
How will you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω ; (ii) 4 Ω?
Answer:
(i) When two resistors each of 6 Ω are connected in parallel give total resistance Rp

∴ \(\frac{1}{6}+\frac{1}{6}\)
= \(\frac{1+1}{6}\)
= \(\frac{1}{3}\)
or R_p = 3 Ω
When this combination is connected in series with third resistor of 6 Ω, it gives a total resistance
(R) = R_p + 6 Ω
= 3Ω + 6Ω
= 9 Ω

(ii) When two resistors each of 6 Ω are connected in series, it gives rise to
R_s = 6 + 6 = 12 Ω

This 12 Ω when connected in parallel with 6 Ω gives total resistance R_p’
\(\frac{1}{R_{p}^{\prime}}=\frac{1}{12}+\frac{1}{6}\)
= \(\frac{1+2}{12}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)
R_p’ = 4Ω

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Given voltage V = 220 V
Current I = 5A
P = 10 W
∴ Resistance of each bulb, R = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}\)
= \(\frac{(220)^{2}}{10}\)
= \(\frac{220 \times 220}{10}\)
= 4840 Ω

Let ‘n’ be the number of bulbs of 4840 Ω resistance each connected in parallel. Their equivalent resistance R_p is given by ;

∴ n = 110
So, 110 bulbs can be connected in parallel (with each óther across two wires).

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series or in parallel. What are currents in three cases?
Answer:
1. When two coils A and B each of 24 Ω are used separately
R = 24 Ω and V = 220V
∴ I_sp = \(\frac{220}{24}\) = 9.17 A

2. When coils A and B are connected in series :
Resultant Resistance R_s = 24 Ω + 24 Ω = 48 Ω
∴Current I_sr = \(\frac{220}{48}\) = 4.58 A

3. When two coils A and B each of 24 Ω are connected in parallel

Question 14.
Compare the power used in 2 Ω resistor in each of the following circuits :
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors and
Answer:
Here R₁ = 1 Ω ; R₂ = 2 Ω
when connected in series the resultant resistance, R = R₁ + R₂
= 1 Ω + 2 Ω.
= 3 Ω
∴ Current through each resistor and also total current
I = \(\frac{\mathrm{V}}{\mathrm{R}_{s}}\)
= \(\frac{6}{3}\)
= 2A

Power used in 2 Ω resistor P₁ = I² × R
= (2)² × (2)
= 4 × 2
= 8 W
i.e Power P₁ = 8W

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
Let 2 Ω and 12 Ω in parallel combination give resultant resistance R


Current divides among its parallel resistances in the inverse ratio of resistances.

It is seen that P₁ = P₂ = 8W .
The Power consumed by 2 Ω resistor in each case is 8 W.

Question 15.
Two lamps, one rated loo W at 220 V and other 60 w at 220 V, are connected in parallel to electric main supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:
Given Power, P = V²/R :
or R = V²/P
Resistance of loo W lamp R₁ = \(\frac{(220)^{2}}{100}\)
= 484 Ω

Resistance of 60 W lamp R₂ = \(\frac{(220)^{2}}{60}\)
= 806.7 Ω

Let Rp be the resultant resistance when two lamps are connected in parallel

R_p = 302.5 Ω
Current drawn from the line, I = \(\frac{\mathrm{V}}{\mathrm{R}_{p}}\)
= \(\frac{220}{302.5}\)
= 0.727A

Question 16.
Which uses more energy, a 250 W TV set for 1 hour or a 1,200 W toaster for 10 minutes?
Answer:
(i) Energy consumed when TV set of 250W is used for 1 hr. .
E₁ = P₁ × t²
= 250 W × l h
∴ E₁ = 250 Wh

(ii) Energy consumed when 1200 W toaster is used for 10 minutes.
= 1,200 W × \(\frac{10}{60}\)hr
E₂ = 200 Wh.
From (i) and (ii) E₁ > E₂
Therefore, 250 W TV set consumes more energy than a toaster of 1200 W.

Question 17.
An electric heater of resistance 8 Q draws 15 A from service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Given R = 8 Ω, I = 15A and Time t = 2 hrs.
Rate at which heat is developed in the heater, its electric power,
P = I² × R
= 152 × 8
= 225 × 8
i.e P = 1,800 W or Js^-1
Thus, 1,800 joule of heat is developed per second

Question 18.
Explain the following :
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
Answer:
Melting point of tungsten is very high and also its resistance is high. Hence tungsten can be heated to very high temperature. Due to its high resistance, large quantity of heat is produced which makes it glow.

(b) Why are the conductors of electric heating devices, such as bread toasters and electric irons, m^e of an alloy rather than pure metal?
Answer:
Resistivity of alloys is very large as compared to pure metals. Moreover, alloys donot/Oxidise (burn out) easily as compared to pure metals. Therefore, conductors of electric heating devices are made of alloys.

(c) Why is series arrangement not used in domestic circuits?
Answer:
When electric appliances are connected in series,’ the equivalent resistance will be very large as
R_eq = R₁ + R₂ + R₃ + R₄ + ………. Due to high equivalent resistance huge quantity of heat is produced in the domestic circuit which may result in fire. This is why series arrangement is not used in domestic circuits.

(d) How does the resistance of a wire vary with its area of cross-section?
Answer:
Resistance of a conductor is inversely proportional to its area cross-section.

(e) Why are copper and aluminium usually employed for electricity transmission?
Answer:
Copper and aluminium have low resistivity and as such there will be small loss of energy when a certain current flows through them. Therefore, copper and aluminium wires are used for electricity transmission.

Science Guide for Class 10 PSEB Electricity InText Questions and Answers

Question 1.
What is electric circuit?
Answer:
Electric Circuit: The continuous or closed path of electric current from electric source flowing through different components is called electirc circuit.

Components of Electric Circuit: The main components of electric circuit are :

• Electric source (i.e. battery or cell)
• Conductor
• Switch (or key)
• Any other instrument connected in the circuit.

Question 2.
Define uint of electric current.
Answer:
S.I unit of electric current is ‘ampere’ which is denoted by ‘A’.

Ampere: When 1 coulomb of charge flows through a conductor in 1 second, then the electric current flowing through conductor is 1 ampere.
∴ 1A = \(\frac{1 \mathrm{C}}{\mathrm{ls}}\)

Question 3.
Calculate the number of electrons that constitute 1 coulomb of charge.
Answer:
We know, the charge present on 1 electron = 1.6 × 10^-19 C
Let ‘n be the number of electrons that constitute 1 coulomb of charge.
∴ n × 1.6 × 10^-19 = 1 C
n = \(\frac{1}{1.6 \times 10^{-19}}\)
= \(\frac{1 \times 10^{19}}{1.6}\)
= \(\frac{10}{16}\) × 10¹⁹
= 0. 625 × 10¹⁹
= 6.25 × 10¹⁸

Question 4.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Electric cell or battery is a device that helps to maintain a potential difference across the ends of a conductor. Chemical action within the cell generates the potential difference across the terminals of the cell.

Question 5.
What is meant by saying that the potential difference betwen two points is IV?
Answer:
The potential difference between two points in a current carrying conductor is said to be 1 V if 1 joule of work is done in moving a charge of one coulomb (1 C) from one point to the other.
i.e 1V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\)
or 1V = 1 JC^-1

Question 6.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Energy to be given to 1 C of charge to pass through T V potential difference = 1 J.
Energy to be given to 1C of charge to pass through 6 V battery = 6 J or W = VQ
= 6 volt x 1 coulomb
= 6\(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\) × 1C
= 6 J.

Question 7.
On what factors does the resistance of a conductor depend?
Or
What is the resistance of a conductor? State the factors on which resistance of a conductor depends?
Answer:
Resistance of a conductor is the opposition offered by it to the flow of electricity through it.

Factors on which resistance of a conductor depends :
Resistance R of a conductor depends upon :

• length of the conductor. It has been experimentally observed that :
  R ∝ l, the length of the wire
• Area of a cross-section of the conductor.
  i.e. R ∝ 1/A, where A is the area cross-section of the wire
  Combining above, we have :
  R ∝\(\frac{l}{\mathrm{~A}}\)
• Nature of material of the conductor.
  or R = ρ \(\frac{l}{\mathrm{~A}}\)
  where p is a constant called specific resistance or electric resistivity of the material of the conductor.

If l = 1, A = 1, ρ = R. .
i. e., specific resistance (or electric resistivity) of a conductor is the resistance of a wire of unit length and a unit area cross-section. It may be defined as :

Specific resistance of a conductor is the resistance of unit cube of the conductor.

Unit of Specific Resistance ρ : ρ = \(\frac{\mathrm{RA}}{l}\)
= \(\frac{\mathrm{ohm} \mathrm{m}^{2}}{\mathrm{~m}}\) = ohm m.

Question 8.
Will the current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?
Answer:
We know that the resistance of a wire is inversely proportional to the area. Therefore, the thicker the wire, more is the area and lesser the resistance. Thus, current will flow more easily through a thicker wire than through thin wire of the same material.

Question 9.
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half the former value. What change will occur in the current through it?
Answer:

• According to Ohm’s law, physical conditions remaining constant, V ∝ I where V is the potential difference across the ends of conductor and I, the current flowing through it.
• When potential difference is halved, the current will also be halved since resistance remain constant.

Question 10.
Why are coils of electric toasters and electric irons made of an alloy rather than that of a pure metal?
Answer:
Resistivity (specific resistance) of alloys is much higher than pure metals. Moreover, alloys also do not oxidise readily at very high temperatures. The alloys are, therefore, commonly used in making coils of electric heating devices like electric iron, electric toasters etc.

Question 11.
(a) Which among iron and mercury a better conductor?
Answer:
Resistivity of iron is 10 × 10^-8 ohm-m and that of mercury 94 × 10^-8 ohm-m, therefore, iron is a better conductor as compared to mercury.

(b) Which material is best conductor?
Answer:
As silver has low resistivity of 1.60 × 10^-8 ohm-m, so silver is the best conductor.

Question 12.
Draw schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor; an 8 Ω resistor and 12 Ω resistor and a plug key, all connected in series.
Answer:
Schematic diagram of the circuit.

Question 13.
Redraw the circuit in above question putting an ammeter to measure the current through the resistors and voltmeter to measure the potential difference across 12 Ω resistor. What would be the reading in the ammeter and the voltmeter?
Answer:
Modified circuit is as shown in Figure Since 5 Ω, 8 Ω and 12 Ω are connected in series, therefore, total resistance in series : (R_s) = R₁ + R₂ + R₃
= 5 + 8 +12 = 25 Ω

Now, Total Current through the circuit, I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{6}{25}\) = 0.24 A
∴ Reading of ammeter = 0.24 A
P.D. across 12 Ω i.e., reading of voltmeter
V = I × R
= 0.24 × 12.
V = 2.88 Ω

Question 14.
Judge the equivalent resistance when the following are connected in parallel (a) 1 Ω and 106 Ω ; (b) 1 Ω and 103 Ω and 106 Ω.
Answer:
(a) Here, R₁ = 1 Ω and R₂ = 106 Ω
When resistances are connected in parallel.
∴ \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
= \(\frac{1}{1}+\frac{1}{10^{6}}\)

(b) Here, R₁ = 1 Ω ; R₂ = 10³ Ω ; R₃ = 10⁶ Ω
In parallel combination of resistances, \(\frac{1}{\mathrm{R}_{p}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)

Question 15.
An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of the electric iron connected to the same source that takes as much current as all the three appliances and what is the current through it?
Answer:
Combined resistance of 100 Ω, 50 Ω and 500 Ω in parallel i.e., R_p is given by :
\(\frac{1}{\mathrm{R}_{p}}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}\)
= \(\frac{5+10+1}{500}\)
= \(\frac{16}{500}\)
R_p = \(\frac{500}{16}\)
= 31.25 Ω.
∴ Resistance of electric iron, R_p= 31.25 Ω

Current through electric iron, I = \(\frac{V}{R}\)
= \(\frac{220}{31.25}\)
∴ J = 7.04 A

Question 16.
What are the advantages of connecting electric devices in parallel with the battery instead of connecting them in series?
Answer:
When connected in parallel, all the electric devices are connected to the same potential difference but they draw different current. Moreover, in parallel arrangement if one electric device fails to work, the other devices are not affected. But when connected in series, the potential is divided between the different devices. In case, if one device is switched off or slops working, the other devices also do not work.

Question 4.
How can three resistors of resistance 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω ; (6) 1 Ω?
Answer:
(a) When all the three resistors are connected in series, the total resistance will be equal to the sum of all the three resistances.
i.e. R_s = 2Ω + 3Ω + 6Ω
= 11Ω this arrangement is rejected.

(a)
Therefore, when 3 Ω and 6 Ω are in parallel

(b)
R_p’ = 2 Ω. It is in series with 2 Ω.
∴ Total resistance when 3 Ω, 6 Ω in parallel are in series with 2 Ω is = R_p’ + 2 = 2 + 2 = 4Qas given in Figure (b)

(c)
If all the three resistances are in parallel, then
\(\frac{1}{\mathrm{R}_{p}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
= \(\frac{3+2+1}{6}\)
= \(\frac{6}{6}\)
R_p = 1 Ω

Question 17.
What is the (a) highest; (b) lowest total resistance that can be secured by combination of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) Total resistance will be maximum when all the four resistances are connected in series.
Let R_s be the total resistance in series combination
∴ R_s =R₁+ R₂ + R_{3 +} R₄
= 4Ω + 8Ω + 12Ω + 24Ω
∴ Highest resistance = 48 Ω

(b) Total resistance will be lowest when all the four given resistance are connected in parallel.
Let Rp be the total resistance in parallel combination,

Question 18.
Why does the cord carrying an electric current does not glow while heating element does?
Answer:
Heating effect = 12R where I is the current flowing and R is the resistance.
Current I is the same both in cord and in heating element.

Resistance R of the cord is negligible since it is made of copper (which has very less resistivity) while heating element is made of nichrome whose resistivity is 6,000 times more than copper. Nichrome gets heated up much more than copper. Therefore, heating element glows.

Question 19.
Compute the heat generated while transferring 96,000 C of charge in one hour through a potential difference of 50 V.
Answer:
Here, Quantity of charge (q) = 96000 C transferred in 1 hr.
Potential difference = 50 V
Work done to transfer Q coulomb of charge through potential difference W = VQ joule
W = 50 × 96,000
= 48,00,000 J
W = 4.8 × 10⁶ J
∴ Heat produced = Work done in transferring the charge = 4.8 × 10⁶ J

Question 20.
An electric iron of resistance 20 Ω takes a current of 5A. Calculate the heat developed in 30 s.
Answer:
Here, Resistance, R = 20 Ω ; I = 5 A ; t = 30 s.
We know, heat produced H = I2 Rt
= 52 × 20 × 30
H = 15,000 J

Question 21.
What determines the rate at which the energy is delivered by a current?
Answer:
Rate at which energy is delivered by the current is called power. It is measured in watt or in kilowatt.

Question 22.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, current. (I) = 5A
Potential difference (V) = 220 V
Time (t) =2h
= 2 × 60 × 60 s
But Power (P) = V × I .
= 220 × 5
= 1,100 W

Energy consumed in 2 h (= 2 × 60 × 60 s), E = P × t
E = (1,100) (2 × 60 × 60)
∴ E = 7,920,000 J

Punjab State Board PSEB 10th Class Science Important Questions Chapter 11 The Human Eye and The Colourful World Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 11 The Human Eye and The Colourful World

Long Answer Type Questions

Question 1.
With the help of a well labelled diagram, explain the construction and working of the human eye.
Answer:
Human eye is the most remarkable and most delicate natural optical instrument. The main parts of the eye are given below :

Structure of The Eye

The human eye consists of nearly spherical ball of about 2.5 cm in diameter.
1. Sclerotic: Its outermost coating is made of a tough and opaque white substance known as SCLEROTIC. It holds the eye ball in position and protects it from external injuries.

2. Cornea: Front portion of sclerotic is transparent and is known as CORNEA. It consists of a transparent substance. The outer surface of cornea is convex in shape. It allows the light to enter eye.

3. Choroid: There is a layer of black tissues, below sclerotic, called CHOROID. It serves to absorb any stray light and thus avoids blurring of the image by reflection from the eye-ball.

4. Iris: In front of eye, choroid merges into a coloured diaphragm known as iris with a hole in the middle called PUPIL. The iris corresponds to shutter in the camera. By means of involuntary muscle control, it regulates the amount of light entering the eye.

5. Eye Lens: It is a double convex lens made of transparent refracting tissues. The lens is held in position with the help of CILIARY MUSCLES. The ciliary muscles adjust the curvature of eye lens and hence its focal length to focus the images of all objects on retina.

The lens divides the eye-ball into two chambers—(i) the front chamber called anterior chamber and (ii) other between lens and the retina called posterior chamber. Anterior chamber is filled with a fluid, called AQUEOUS HUMOUR while the posterior chamber is filled with a jelly-like substance called VITREOUS HUMOUR.

6. Retina: The innermost coating of the eye, covering the rear of inner surface, is a very delicate membrane called the RETINA. It behaves like a screen as photographic film does in a camera on which image of object is formed.
The sensation of vision in the retina is carried to the brain by a nerve called OPTICAL NERVES.

7. Yellow Spot: The most sensitive part of retina is known as the YELLOW SPOT.

8. Blind Spot: The point where the optical nerve enters the eye is totally insensitive to light and is known as the BLIND SPOT.

9. Eye-lids: Eye Lids are provided to control the amount of light falling on the eye. Eye-lids also protect the eyes from dust etc.

Question 2.
What are the defects of human eye? How can they be corrected? Explain with diagrams.
Answer:
Defects of Human Eye: A normal healthy eye adjusts its focal length so as to form images of ail objects lying at different distances on the retina. Sometimes its power of accomodation decreases as a result of which the image is not formed on the retina resulting in two main defects viz. long sightedness and short sightedness.

In addition to these, presbyopia, colour blindness and astigmatism are also common defects.
1. Myopia or Short Sightedness: A person with myopic eye can see the near objects clearly but cannot see far off objects. The person suffering from myopia or short-sightedness has far point nearer than infinity. In normal eye, the far point is at infinity. The rays coming from distant object (at infinity) get focussed on retina [Fig. 11.5 (a)].

Causes of Defect: The defect myopia arises due to either :

• the length of eyeball is elongated (becomes longer than normal)
• the focal length of eye lens has decreased.

Due to either or both the causes, the eye is not able to focus the rays from distant object at retina [Figure (d)]. Focussing is there at a point 0 in front of retina. Therefore, the image formed on retina is blurred.

The defective eye is however able to focus the object upto its far point [Figure (c)].

Figure (a) Normal eye. Far point at infinity. Rays from distant object meet at retina.

Figure (b) Defective eye (eye-ball is enlarged), or focal length decreased, cannot focus rays from infinity at retina.

Defective Eye. Due to eye ball getting short or an increase in focal length of eye lens, the rays do not focus on retina.

Figure (c) Defective Eye is able to form image at the retina when object is moved from N to N’ the near point of defective eye.

Figure (d) Corrected Eye. A convex lens of suitable focal length converges the rays to match those coming from hf.
Normal eye is able to focus on retina the rays emerging out from N [Figure (a)].

However, the defective eye is not able to focus the rays from near point of normal eye i.e. N [Figure (b)].
It can focus the rays from near point of defective eye i.e., N’ [Figure (c)].

From Figure (b) and (c), we conclude that more inclined rays [Figure (6)] are not focussed on retina whereas less inclined rays from N’ get focussed on retina.

Correction of hypermetropia: This defect is corrected by placing a convex or a converging lens of suitable focal length before the eye so that the rays diverging from N after refraction, appear to come from the near point N’.

3. Presbyopia: This is an age related defect. Almost all persons above 40 years suffer from this defect. The flexibility of eye lens decreases with age and the ciliary musclles are not able to change the focal length of eye lens and the person cannot see distinctly. Because of mixed defect of myopia and hypermetropia a person needs either two spectacles one each mounted with convex lens and concave lens or a bifocal lens to correct this defect.

4. Colour Blindness: This defect is due to biological reasons. It is a genetic disorder. The patient having this defect can not distinguish colours because in retina of eye, there are insufficient number of cones. These are cells present in the eye which recognise red, blue and green colours. The person suffering from this defect cannot recognise specific colour due to insufficient cone like structures in his eye. This defect cannot be corrected. The person having this defect can see all the things but can not recognise some colours.

5. Astigmatism: The person suffering from this defect can not focus in both the horizontal and vertical axis clearly. This defect is caused due to varying curvature in lens in two axis i.e. the lens is not completely spherical. The person cannot focus in vertical direction. This defect can be corrected by wearing spectacles fitted with cylinderical lens.

Question 3.
What is prism? Explain deviation in glass prism by drawing a ray diagram.
Answer:
Prism: A portion of transparent medium bound by two plain refracting surfaces at an angle to each other is called a prism. The surfaces of the prism from which refraction occurs are called refracting surfaces and the angle between them is called angle of prism.

Deviation of Light by a Prism: Let PQR be the principal section of glass prism. Suppose a ray of light BC is incident at C on the surface PR of the prism. After refraction from this surface it goes bending toward normal along CD.

Now ray CD is incident at D on the refracting surface QR and after refraction it goes away from the normal drawn at D emerging along the direction DE. Therefore, prism deviates the ray of light coming along BC into direction DE. In this way prism produces angular deviation in the direction of light. If incident ray BC is produced in the forward direction and emergent ray DE in the backward direction, then they meet each other at G. The LFGD formed between these two is called angle of deviation. It is denoted by δ (delta).

The value of angle of deviation depends on (i) the material of the prism and (ii) angle of incidence. If we increase the value of angle of incidence (∠i) then the value of angle of deviation (δ) decreases. For a certain value of angle of incidence, the angle of deviation becomes minimum. This minimum value of angle of deviation is called minimum angle of deviation. If for a prism, A is the angle of prism and δ_m is the minimum angle of deviation, then, refractive index of material of the prism, μ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{m}}{2}\right)}{\sin \frac{\mathrm{A}}{2}}\)

Question 4.
What is meant by dispersion of light. Explain with the help of diagram and give the cause of dispersion.
Or
When a ray of light passes through a glass prism then a spectrum is obtained on the screen.
(а) Draw a diagram showing a spectrum of white light.
(b) Name the seven colours of spectrum in a serial order.
(c) Which colour of the spectrum suffers most deviation and which colour the least deviation?
Answer:
Dispersion of Light: The process of splitting of white light into its seven constituent colours is called dispersion or dispersion of light.
(а) Spectrum obtained by Dispersion through Prism. When a ray of white light coming from the sun passes through a prism then due to refraction it deviates from its path and bends towards the base of the prism and splits into its seven constituent colours and each light colour bends through a different angle forming a band of seven colours called spectrum.

(b) Generally we see the seven colours of specturm as a group. The order of seven colurs of spectrum are in the following order starting from the base of prism Violet, Indigo, Blue, Green, Yellow, Orange and Red. This order of colours can be easily remembered by a word “VIBGYOR.”

(c) The spectrum has red colour at its one end and violet colour at the other end. Red light travelling the fastest bends the least and violet light travelling the slowest bends the maximum.

Cause of Dispersion of Light: The refractive index of material (say transparent medium glass) depends on the colour of light. Refractive index of light of red colour is minimum and light of violet colour is maximum.

White light is composed of seven different colours of light (VIBGYOR) each colour having different colour due to different wavelength. Red colour has the longest wavelength and the violet has the shortest. The frequency of light is the same for colours. In air or vacuum, the speed of light is same for all colours. But in different media, the speed of light is different for different colours. So each light bends by different angle. Red colour in any medium bends the least while violet colour bends the most. So when light is passed through glass prism each light colour bends through different angle forming spectrum.

In this way, on passing of white light through a glass prism dispersion of light occurs.

Dispersion of white light does not takes place when it passes through a glass slab because there does not take place deviation of rays of light but lateral displacement occurs. The incident ray and emergent ray become parallel in glass slab.

Question 5.
Describe an experiment to show that different colours of white light can be recombined to form white light.
Answer:
To show that colours are not produced by the prism but are present in white light itself and the prism only separates these colours, Newton isolated a particular colour say green. He placed another prism in the path of green beam. No further splitting of colour took place. The light only deviated further. It clearly showed the prism just separates, a large number of colours coming together as white [Figure (b)].

Figure (a) Dispersion of white light by a prism.

Figure (b) Second pris,iz unable to split green bewn.

Figure (c) If a similar inverted second prism is placed. seven colours recombine spectrum colours into white light.
If a second prism of exactly of same angle with its refract ing edge opposite to the first is placed as shown, it is found that it results in white light again [Figure (c)]. The second prism deviates the rays in opposite direction. Thus dispersion produced by one prism is cancelled by the second similar prirn placed in the opposite direction.

Figure (d) When second prism is placed in sqnw way, the coloured beams are further spread.

If the second prism is placed in; the same way as the first, the coloured rays are again obtained on the screen but there are more spread [Figure (d)].

Short Answer Type Questions

Question 1.
What is name of defect of eyes due to loss of elasticity of eye-lens? How is it corrected?
Answer:
It is called presbyopia. It cad be corrected by using two separate spectacles one for near vision and the other for distant vision.

Question 2.
What is the function of ciliary muscles in the eye?
Answer:
Ciliary muscles pull/push the lens and thereby change its focal length in order to focus objects lying at different distances from the eye.

Question 3.
What happens when elasticity of the eye lens is reduced to zero?
Answer:
Decrease of Elasticity of Lens. In normal eye, the change in the power of the eye lens for seeing far and near points is very large. As the person grows older, the power of accommodation gradually decreases. A stage may even reach when ciliary muscles lose their power and crystalline lens become much less elastic, so that the power of acommodation is almost zero.

Question 4.
Why is eye considered the best gift of God?
Answer:
It is said that world exists only if there are healthy eyes. Human being can see with the god-gifted eye i.e. he can identify different objects properly, can distinguish, i-ecognise colours and can differentiate between small and big even without touching, can read and write and can see all wonders of the world. That is why eye is considered to be one of the best gifts that god has given.

Question 5.
When we enter some dark room then for some time we are not able to see anything and remaining there for long, if suddenly strong light is switched on then our eyes can not gee anything. Why?
Answer:
Behind cornea there is iris which regulates the size of pupil. With it the intensity of light entering the eye is controlled. When we enter some dark room then for image to be formed on the retina more light is needed. For allowing more light to enter the eye is wide opened and it takes some time for this. During that time we do not see anything. Similarly while sitting itt the dark, iris spreads so that more light may enter the eye. And if suddenly strong light appears then to reduce its size some time is required during which we cannot see anything.

Question 6.
What type of lens is present in front of eyeball? What is its main function?
Answer:
In human eye the convex lens is present in front of eye ball. It consists of fibrous jelly like substance. Its curvature is controlled by ciliary muscles. The most of light rays are refracted by cornea and aqueous humour. The crystalline lens ensures focal length. So that the image of the object may be formed on the retina.

Question 7.
What is the function of retina in human eye?
Answer:
When light falls on rejina, it excites rods and cones. The electric pulses produced are conveyed to brain through optic nerve.

Question 8.
Why do we experience difficulty when we read from too close?
Answer:
Because of its capacity and properties the eye lens can change its focal length to some limit but not below that. If any object is too close then the focal length of the eyelens does not change that much that it niay help in seeing that properly. Therefore, we experience difficulty in reading from a close distance. In doing so pressure is exerted on eye and we cannot see distinctly.

Question 9.
Why do aged persons need spectacles for reading?
Answer:
Approximately at the age of 60 years the near point of eye becomes 20 cm which was 25 cm for normal eye of a young person. Due to this the aged person faces difficulty while reading and, therefore, needs spectacles.

Question 10.
What is cataract? How is it corrected?
Answer:
Sometimes, the crystalline lens of aged person gets covered with a membrane which obstructs the passage of light rays through a transparent lens. At times the lens becomes completely opaque or cloudy. This condition is called cataract. This defect can be corrected either by using contact lens of suitable focal length or by surgery.

Question 11.
What is the necessity of eye donation? Explain.
Answer:
We know eye is the wonderful and priceless gift of God and 65% of people in the world are blind of which 45 lakh suffer with corneal blindness. They can be cured only by cornea transplantation. Therefore, after death we and our kins should donate eyes so that others who have become blind due to corneal defect may also see the world.

Question 12.
What things we should take into account while donating eyes?
Answer:

• After death eyes must be removed within 4 to 6 hours.
• The nearest eye bank should be informed immediately after death. The team of the eye bank removes eyes of the dead person either at his residence or in the nearest hospital in 10-15 minutes.
• Eye removal is a simple procedure and does not lead to any disfigurement.

Question 13.
What is meant by least distance of distinct vision?
Answer:
Least Distance of Distinct Vision. If an object is very close to the eye then it is not seen clearly. Therefore, that shortest distance where if an object placed is seen very distinctly is called the least distance of distinct vision. For normal eye this distance is 25 cm.

Question 14.
A 14 years old boy cannot read question written on the black-board lying 5 m away from him.
(i) Name the eye defect he is suffering from.
Answer:
He is suffering from Myopia.

(ii) Show with the help of a labelled diagram as to how this defect can be corrected.
Answer:
For correction of this defect a concave lens of suitable focal length is used.

Figure (a) Myopic Eye (b) Correction of Myopia by a concave lens of suitable focal length.

Question 15.
Why we see a rainbow just after rains?
Answer:
Rainbow is caused by dispersion of white sun light by tiny water droplets present in the atmosphere. Water droplets act as tiny prisms. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again when it comes out of raindrop. Due to dispersion of light and internal refraction, different colours reach the eye of the observer. Rainbow is always formed in a direction opposite to that of the sun.

Rainbow formation

Question 16.
How refraction of light elongates the length of day?
Or
Why does day appear longer than actually what it is due to refraction of light?
Answer:
The sun is visible a few minutes earlier than it actually rises above horizon. It happens because as we go up from the earth, the density of air layers decrease. The rays from sun S keep on bending towards normal till it enters the eye. Therefore, the sun appears to be at S’ (above horizon) although it is at S (below horizon). Thus the sun appears about two minutes earlier than actually when it should be.

For the same reason, the sun appears to set two minutes later than the actual. Hence the day appears to be about 4 minutes longer than what it is.

For the same reason explained above, the stars appear higher than their actual position as shown in Figure (b) given below :

Figure (a) The sun remains visible even after it has actually passed below the geometrical horizon.

Figure (b) A star in the direction S is seen in the direction S’ as a result of atmospheric refraction.

Question 17.
A star appears on the horizon. What is the true position of the star? Explain with diagram.
Answer:
True position of star is below the horizon. Incident rays from star, travel through earth’s atmosphere and reach observer’s eye. These incident rays travel from rarer to denser atmopshere so that they bend towards the normal. Thus, they appear to come fiom a different position slightly higher than the true position.

Question 18.
What is meant by scattering of light? Explain Tyndall effect. Give a few illustrations of scattering of light.
Answer:
The path of beam of- light becomes visible when it passes through space containing smoke, tiny water droplets, suspended dust particles. The path of light is visible when it pass through ak medium the size of whose particles are comparable to wavelength of light.

An arrangement for observing scattering of light in colloidal solution Colour of scattered light depends upon the size of scattering particles. Very small particles scatter blue light while larger particles scatter light of longer wavelength.

Question 19.
Why does the colour of the sky appear blue? [P.B., March 2019 (Set-A)\ Also tell how it would appear in the absence of earth’s atmosphere?
Answer:
The molecules of air/gases and other fine particles in the atmosphere have smaller size than the wavelength of visible light. These particles are therefore, more effective in scattering light of shorter wavelengths at blue and than light of longer wavelength at the red end. Thus, blue colour of the sky is due to scattering of sunlight by fine particles. In the absence of earth s atmosphere no scattering of light can occur. So, the sky appears black in that case.

Question 20.
What is short-sightedness? For a person suffering from this defect, where is the image of an object formed and by what type of spectacles this defect can be corrected?
Or
What is the cause of Myopia? How can it be corrected? Explain with a labelled diagram. (P.S.E.B. March 2017, Set-I)
Answer:
Myopia or Short Sightedness: A person with myopic eye can see the near objects clearly but cannot see far off objects. The person suffering from myopia or short-sightedness has far point nearer than infinity. In normal eye, the far point is at infinity. The rays coming from distant object (at infinity) get focussed on retina [Fig. 11.5 (a)].

Causes of Defect: The defect myopia arises due to either :

• the length of eyeball is elongated (becomes longer than normal)
• the focal length of eye lens has decreased.

Due to either or both the causes, the eye is not able to focus the rays from distant object at retina [Figure (d)]. Focussing is there at a point 0 in front of retina. Therefore, the image formed on retina is blurred.

The defective eye is however able to focus the object upto its far point [Figure (c)].

Figure (a) Normal eye. Far point at infinity. Rays from distant object meet at retina.

Figure (b) Defective eye (eye-ball is enlarged), or focal length decreased, cannot focus rays from infinity at retina.

Defective Eye. Due to eye ball getting short or an increase in focal length of eye lens, the rays do not focus on retina.

Figure (c) Defective Eye is able to form image at the retina when object is moved from N to N’ the near point of defective eye.

Figure (d) Corrected Eye. A convex lens of suitable focal length converges the rays to match those coming from hf.
Normal eye is able to focus on retina the rays emerging out from N [Figure (a)].

However, the defective eye is not able to focus the rays from near point of normal eye i.e. N [Figure (b)].
It can focus the rays from near point of defective eye i.e., N’ [Figure (c)].
From Figure (b) and (c), we conclude that more inclined rays [Figure (6)] are not focussed on retina whereas less inclined rays from N’ get focussed on retina.

Question 21.
What is the cause of long sightedness (Hypermetropia)? How can it be corrected? Explain with a labelled diagram?
Answer:
Correction of hypermetropia: This defect is corrected by placing a convex or a converging lens of suitable focal length before the eye so that the rays diverging from N after refraction, appear to come from the near point N’.

Question 22.
What is meant by Presbyopia and Colour blindness?
Answer:
Presbyopia. Some people suffer from myopia and hypermetropia both simultaneously, it is called presbyopia. Such people require bifocal lenses, the upper part of which is concave lens and the lower part a convex lens. The upper part is used to see distant objects while the lower part to see nearby objects (for reading).

Colour Blindness: This defect is caused due to decrease of cone like cells in the eyes of human beings. Due to this loss, eye is not sensitive for some particular colours. This is genetic defect and has no remedy. The persons having this defect can no doubt see objects but are unable to identify colours. This is called colour blindness.

Question 23.
Define the following :
Power of Accomodation, Far Point, Near Point, Least Distance of Distinct Vision, Persistence of Vision.
Answer:
Power of Accomodation Human ejre can see all nearby objects and distant objects. This ability of eyelens to adjust its focal length enabling it to see object lying at different distances is called power of accomodation.

Far Point. The point at the maximum distance from the eye where if an object is placed can be seen distinctly is called far point. For normal eye the far point is at infinity.
Near Point: The point at the leat distance from the eye where the object if lying can be seen distinctly is called near point.
Least Distance of Distinct Vision. It is a point in between the far point and near point and at minimum distance from eye where the object lying can be seen distinctly. For normal eye the least distance of distinct vision is 25 cm.
Persistence of Vision. The image of an object is formed on the retina of the eye but even if the object is removed still its image remains and does not fade away. This is called persistence of vision.

Question 24.
A person wears spectacles of power – 2.5 D. Name the defect of vision he is suffering from. Draw the ray diagram for (i) the defective eye, (ii) its correction after using a suitable lens.
Answer:
Since the power of the lens is negative, therefore, the lens used in spectacles is concave lens.

The defect of the eye is Myopia (or short sightedness)

Figure (a) Defective eye (eye-ball is enlarged), or focal length decrea.’ed, cannot focus rays from infinity at focus.

Figure (b) Corrected eye. Concate (or a divergent) lens diverges the parallel rays from infinity to an extent that they appear to diverge from F. They get focussed at retina.

Question 25.
Distinguish between simple microscope and compound miéroscope.
Answer:
Differences between Simple Microscope and Compound Microscope.

Numerical Problems

Question 1.
A person can not see clearly objects beyond a distance of 1.2 m. Name the defect of vision he is suffering from. What would be the power of correcting lens used to restore proper vision ?
Answer:
Since a person can see clearly only near objects (upto a distance of 1.2 m only) whereas normal human eye can see even distant objects (upto infinity), therefore, the defect of vision is myopia.
Here v = 1.2 m
u = ∞

∴ Power of correcting lens, P = \(\frac{1}{f(\text { in metres })}\)
= \(\frac{1}{-1.2}\)
= – 0.83 D

Question 2.
The near point of a person suffering from hypermetropia is 75 cm. Calculate the focal length and power of the lens required to enable him to read the newspaper which is kept at 25 cm from the eye.
Answer:
Here u = -25 cm
v = – 75 cm

∴ Focal length of the required lens
Now P = \(\frac{1}{f(\text { in metres })}\)
= \(\frac{1}{375}{\frac{1000}}\)
= \(\frac{1000}{375}\)
= 2.6 D

Question 3.
The near point of a myopic eye is 1 m. Find the power of the lens required to correct this defect. Assume that near point of the normal eye is 25 cm.
Answer:
Here u = – 25 cm [Normal near point]
V = – 1 m
= – 100 cm
f =?
Using lens Formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Now power of the required lens P = \(\frac{1}{f(\text { in metres })}\)
= \(\frac{100}{f(\text { in metres })}\)
= \(\frac{1}{\frac{100}{3}}\)
= \(\frac{100 \times 3}{100}\)
= + 3 D
Positive sign (+) indicates that the lens is convex.

Question 4.
A person cannot see objects beyond 1.5 m distinctly. What type of lens should be used to restore proper vision?
Answer:
Here u = – ∞ ; u = -1.5 m ; f =?

He should use concave or divergent lens say convexo concave lens of Focal length -1.5 m and power – 0.67 D.

Very Short Answer Type Questions

Question 1.
What do you mean by optical instruments? Name any two.
Answer:
Optical instruments are the devices based on various phenomena of optics. The’names are (i) Microscope (ii) Telescope.

Question 2.
What is least distance of distinct vision?
Answer:
Distance upto which a person can see clearly. It is 25 cm for normal eye.

Question 3.
What is far of point?
Answer:
This is most distant point upto which the eye can see clearly with unaided eyes.

Question 4.
Write a function of each of :
(i) Retina
Answer:
Retina: Retina is third layer of eye and acts as a screen for the image of the objects.

(ii) Sclerotic
Answer:
Sclerotic: Sclerotic is to maintain a solid shape of the eye and protects the internal soft parts from external injuries.

(iii) Ciliary muscles in human eye.
Answer:
Ciliary muscles in human eye. Eye lens is held in its position by ciliary muscles. Ciliary muscles help the eye to change the focal length by adjusting its curvature.

Question 5.
What is the function of sclerotic in human eye?
Answer:
Sclerotic is to provide a solid shape to eye and protect it from external injuries.

Question 6.
What is the function of ciliary muscles in human eye?
Answer:
Ciliary muscles help the eye to change its focal length by adjusting its curvature.

Question 7.
What is the function of rods on the retina?
Answer:
Rods are sensitive to light. More the intensity of light, more are they excited.

Question 8.
What are cones?
Answer:
Cones on retina are sensitive to different colours. If cones are absent or insufficient, the person is colour blind.

Question 9.
Why chickens come out late in the morning and return early in the evening?
Answer:
Chickens have very few rods on the retina, hence they are able to see only in bright light and not in dim-light.

Question 10.
Why cats / bats are able to see at night?
Answer:
They have very large number of rods on retina. Hence they are able to see even if there is very small amount of light.

Question 11.
Colour of eyes depend upon colour of which part of eye?
Answer:
It depends upon the colour of iris.

Question 12.
What is basic cause of colour blindness?
Answer:
It is due to no or insufficient number of cones on the retina.
Seeing sun or seeing towards it during solar eclipse may cause colour blindness.

Question 13.
Which phenomenon of light is shown in fig. below :

Answer:
Dispersion of light.

Question 14.
In the given diagram which defect of the human eye is being corrected using a concave lens?

Answer:
Myopia (Short-Sightedness).

Question 15.
Which defect of human eye is being corrected in the figure given below?

Answer:
Hypermetropia.

Question 16.
Which defect of the eye is shown in the figure given below?

Answer:
Hypermetropia (or Long-sightedness).

Multiple Choice Questions :

Question 1.
The approximate least distance of distinct vision of normal eye is
(A) 35 in
(3) 3.5 m
(C) 25 cm
(D) 2.5 cm.
Answer:
(C) 25 cm

Question 2.
The feal length of objective is changed by:
(A) Pupil
(B) Retina
(C) Ciliary muscles
(D) Iris.
Answer:
(C) Ciliary muscles

Question 3.
A person suffering from short sightedness cannot see objects beyond 1.2 m. For distinct vision he would use :
(A) Concave lens
(B) Cylinderical lens
(C) Convex lens
(D) None of these.
Answer:
(A) Concave tens

Question 4.
The far point of normal human eyes .
(A) At 25 cm
(B) At 25 mm
(C) At 25 m
(D) At infinity.
Answer:
(D) At. infinity.

Question 5.
In human eye, the image of object is formed at:
(A) Pupil
(B) Retina
(C) Cornea
(D) Eye ball.
Answer:
(B) Retina

Question 6.
Light entering the eye is mostly refracted by :
(A) Crystalline lens
(B) Outer surface of cornea
(C) Pupil
(D) Iri’.
Answer:
(B) Outer surfaceof cornea

Question 7.
Most insensitive part nf the eye is called:
(A) biqek spt
(B) vehow spot
(C) cornea
(D) blue Spot.
Answer:
(A) black spot

Question 8.
Focal length of the eye lens can be adjusted by action of:
(A) ciliarv muscles
(B) choroid
(C) optical nerves
(D) retina.
Answer:
(A) ciliary muscles

Question 9.
When light rays enter the eye most of the refraction occurs at:
(A) crystalline lens
(B) iris
(C) outer surface of cornea
(D) pupil.
Answer:
(A) crystalline lens

Question 10.
Distance of distinct vision of a normal eye is:
(A 25 m
(B) 2.5 m
(C) 25 cm
(D) 2.5 cm.
Answer:
(C) 25 cm

Question 11.
Accommodation of normal eyes is
(A) 5 cm to 15 cm
(B) 15 cm to m
(C) 1 m to 3 m
(D) 20 cm to infinity.
Answer:
(D) 20 cm to infinity.

Question 12.
When an object is placed beyond centre of curvature of a cancave mirror, the image is formed:
(A) beyond C;
(B) between C and F;
(C) at F;
(D) at infinity.
Answer:
(B) between C and F;

Fill in the blanks:

Question 1.
The least distance of distinct vision is __________
Answer:
25 cm.

Question 2.
Far point of normal human eye is __________
Answer:
infinity.

Question 3.
In human eye, image of an object is formed at __________
Answer:
Retina.

Question 4.
The ability of eye-lens to adjust its focal length is called __________
Answer:
Accommodation.

Question 5.
A shortsighted person cannot see the objects distinctly.
Answer:
distant.

Question 6.
The __________ shaped cells present in retina respond to the intensity of light.
Answer:
rod.

Question 7.
__________ helps in regulating the amount of light entering the eye.
Answer:
pupil.

Question 8.
The splitting of white light into its component colours is called __________
Answer:
dispersion.

Question 9.
A concave lens is used to rectify __________
Answer:
Myopia.

Question 10.
When the light is bright, the pupil becomes __________
Answer:
very small.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 11 The Human Eye and The Colourful World Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 11 The Human Eye and The Colourful World

PSEB 10th Class Science Guide The Human Eye and The Colourful World Textbook Questions and Answers

Question 1.
The human eye can focus object at different distances by adjusting the focal length of the eye lens. This is due to :
(a) presbyopia
(b) accommodation
(c) near-sightedness
(d) far-sightedness.
Answer:
(b) accommodation.

Question 2.
The human eye forms the image of an object at its :
(a) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d) retina.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about:
(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m.
Answer:
(c) 25 cm.

Question 4.
The change in focal length of an eye lens is caused by the action of the :
(a) pupil
(b) retina
(c) ciliary muscles
(d) iris.
Answer:
(c) ciliary muscles.

Question 5.
A person needs a lens of power – 5.5 D for correcting his vision. For correcting his near vision, he needs a lens of power + 1.5 D. What is the focal length of lens required for correcting
(i) distant vision and
Answer:
(i) For distant vision, f1 = \(\frac{1}{\mathrm{P}_{1}}\)
= \(\frac{1}{-5.5}\)
= – 0.182 m
⇒ f₁ = – 18.2 cm

(ii) near vision?
Answer:
For near vision, f₂ = \(\frac{1}{\mathrm{P}_{2}}\)
= \(\frac{1}{1.5}\)
= + 0.667 m
f₂ = + 66.7 cm

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
Here
Distance of far point u = – ∞
Focal length of the lens (v) = – 80 cm;

Using Less formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{(-80)}-\frac{1}{(-\infty)}\) [∵\(\frac{1}{\propto}\) = 0]
\(-\frac{1}{80}\)
Focal Length of the Lens (f) = – 80 cm = – 0.8 m
Negative sign (-) indicates that the nature of required lens is concave

Now,
P = \(\frac{1}{f}\)
= \(\frac{1}{-0.8}\)
P = – 1.25 D

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point/ of hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of normal eye is 25 cm.
Answer:
u = – 25 cm,
v = – 1 m
= – 100 cm
Using lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)


∴ Power = + 3 D
Convex (convergent) lens of power + 3 D

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
For seeing near objects, the ciliary muscles contract to make the lens thicker so as to reduce the focal length of eye lens, in order to form the image on the retina. Ciliary muscles cannot be contracted beyond certain limit and hence we cannot see clearly the objects lying closer than 25 cm, called the least distance of distinct vision.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
For a fixed focal length of the lens, the distance of image from the lens (i.e., v) decreases as the distance of the object from eye lens (i.e., u) is increased.

Since v cannot be decreased (as distance between eye lens and retina is fixed), f is decreased by action of ciliary muscles so as to satisfy lens formula in accordance with increased value of u.

Question 10.
Explain atmospheric refraction. Why do stars twinkle?
Answer:
Twinkling of the star is due to refraction of light from stars through atmosphere. The stars emit their own light. The light coming from the stars on entering the earth’s atmosphere undergoes refraction continuously at each layer of atmosphere having different density, before it reaches the earth. The stars are very much distant objects and may be considered as point sources. The refractive index of the air changes from time to time due to change of density of air. Due to change of the optical density of earth’s atmosphere, the path of rays from stars continuously changes. The apparent positions of the stars continuously change due to change in refractive index of the atmosphere. Thus the stars appear to twinkle.

Question 11.
Why planets do not appear twinkling?
Answer:
Since planets are quite close to the earth in comparison with the stars, they do not act as point sources but behave like extended sources. Planets may be considered as collection of a large number of point-sized sources of light. The total net deviation in amount of light entering our eyes from different point sources is zero. Therefore there is no net twinkling of planets.

Question 12.
Why does the sun appear reddish in the morning (as well as in evening)?
Or
Why does rising star appears red in colour?
Answer:
Earth is surrounded by envelope of gases called atmosphere. At the time of sunrise (or at sunset), light has to travel greater distance [AB at sunrise and BC at sunset] through the atmosphere to reach us than what it covers, when sun is over-head at noon [It has to travel only DB],

The wavelength of blue colour is about half that of red, blue light is scattered nearly 24 =16 times more than red colour of sunlight. As a result, the sun appears red at sunrise or at sunset due to negligible scattering of red colour of light.

Question 13.
Why does the sky appears dark instead of blue to an astronaut?
Answer:
The atmosphere exists only upto some height. At high altitudes, there is no atmosphere to scatter sun light. The sky, therefore, appears to be perfectly black to astronauts.

Science Guide for Class 10 PSEB The Human Eye and The Colourful World InText Questions and Answers

Question 1.
What is meant by power of accommodation of the eye?
Answer:
Power of accomodation of eye. It is the ability of the eye lens to adjust its focal length to see various objects lying at different distances. The minimum distance upto which a normal eye can see clearly is called near point. For normal eye, near point is about 20 cm. The maximum distance upto which the eye can see clearly is called far point. The distance between near point and far point is also called accommodation.

Question 2.
A person with myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
Here u = – ∞; v = – 1.2 ; f =?

Negative sign (-) shows that the corrective lens is concave lens.
But P = \(\frac{1}{f}\)
= \(-\frac{1}{1.2}\) = – 0.825 D
The person should wear spects mounted with concave lens having of power – 0.825 D

Question 3.
What is the far point and near point of human eye with normal vision?
Answer:
Far point: It is the distance of most distant point upto which a person can see object distinctly. It is infinity for normal vision.
Near point: It is distance of nearest point beyond which a person can see distinctly. It is 25 cm for normal vision.

Question 4.
A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How could it be corrected?
Answer:
Student is suffering from myopia or shortsightedness. This defect can be corrected by using spectacles having diverging (say concave or convexo-concave) lens of suitable focal length.

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 5 गाता खग Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 5 गाता खग

Hindi Guide for Class 10 PSEB गाता खग Textbook Questions and Answers

(क) विषय-बोध

I. निम्नलिखित प्रश्नों के उत्तर एक या दो पंक्तियों में दीजिए:

प्रश्न 1.
पक्षी प्रातः उठकर क्या गाता है?
उत्तर:
पक्षी प्रात: उठकर संसार के प्राणियों के सुखी तथा समृद्ध जीवन के गीत गाता है।

प्रश्न 2.
तारों की पंक्तियों की आँखों का अनुभव क्या है?
उत्तर:
तारों की पंक्तियों की आँखों को देखकर यह प्रतीत होता है, जैसे वे कह रही हों कि सारा जीवन करुणा और दुःख से भरा हुआ है।

प्रश्न 3.
फूल हमें क्या संदेश देते हैं?
उत्तर:
खिले हुए फूल मनुष्य को सदा मुस्कराते रहने का संदेश देते हैं।

प्रश्न 4.
लहरें किस उमंग में आगे बढ़ती जाती हैं?
उत्तर:
लहरें इस उमंग में आगे बढ़ती जाती हैं कि कभी न कभी तो उन्हें अपनी मंजिल मिल ही जाएगी।

प्रश्न 5.
बुलबुला विलीन होकर क्या पा जाता है?
उत्तर:
बुलबुला विलीन होकर अपने जीवन का अंतिम लक्ष्य पा लेता है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए

(1) हँसमुख प्रसून सिखलाते
पल भर है, जो हँस पाओ,
अपने उर की सौरभ से
जग का आँगन भर जाओ।
उत्तर:
कवि कहता है कि प्रभातकाल में खिले हुए पुष्प अपनी कोमलता, मनोहरता और सौरभ से वातावरण को पूरी तरह भरते हुए मानव को यह प्रेरणा देते हैं कि इस नाशवान और छोटे-से जीवन को अनेक प्रकार की विषमताओं और समस्याओं ने नीरस और उदास बना रखा है। यदि हो सके तो संसार में अपना छोटा-सा जीवन प्रसन्नता और आनंद है।

(2) उठ-उठ लहरें कहतीं यह
हम कूल विलोक न पाएँ,
पर इस उमंग में बह-बह
नित आगे बढ़ती जाएँ।
उत्तर:
कवि कहता है कि अनंत आकाश में छाए हुए अंधकार में निरंतर चमकने वाली ताराओं की पंक्तियाँ देखकर प्रतीत हो रहा है जैसे वह कह रही हों कि संपूर्ण जीवन करुणा तथा दुःख से भरा हुआ है। जिस प्रकार निरंतर दुःख सहते हुए किसी की आँखों में से आँसू बह जाते हैं और आँसुओं से भरी आँखों को देखकर दूसरे में सहानुभूति के कारण करुणा का संचार हो जाता है, उसी प्रकार विश्वव्यापी दुःख, अवसाद, विपन्नता और विषमता को देखकर अपलक ताराओं की करुणा से भरी आँखों से आँसू छलक पड़ते हैं। आँखों की भाषा नीरव और मौन होती है। केवल आँसुओं के माध्यम से ही प्रकट होती है।

(ख) भाषा-बोध

I. निम्नलिखित शब्दों के दो-दो पर्यायवाची शब्द लिखें

खग = ————
प्रसून = ————
उर = ————
किनारा। = ————
उत्तर:
शब्द – पर्यायवाची
खग- पक्षी, नभचर, विहग, पतंग।
प्रसून – कुसुम, सुमन, पुष्प, फूल।
उर – हृदय, छाती, वक्षस्थल, चित्त।
किनारा – तट, तीर, कूल, पुलिन।

II. निम्नलिखित शब्दों की भाववाचक संज्ञा बनाएं

सुंदर = ————
अपना = ————
हँसना = ————
नीरव। = ————
उत्तर:
शब्द – भाववाचक संज्ञा
सुंदर = सुंदरता
अपना = अपनापन
हँसना = हँसी
नीरव = नीरवता।

(ग) पाठ्येतर सक्रियता

1. कविता कंठस्थ करके सस्वर वाचन करें।
2. सुमित्रानंदन पंत ने प्रकृति के विभिन्न चित्र अपनी कविताओं के माध्यम से प्रस्तुत किए हैं। इस कविता में कवि ने लहरों, फूल, पक्षी, चमकते सितारों की पंक्तियाँ, नदी का किनारा आदि का वर्णन किया है। आप सुमित्रानंदन पंत की प्रकृति चित्रण से संबंधित कोई अन्य कविता याद कीजिए और कक्षा में सुनाइए।
उत्तर:
सुमित्रानंदन पंत की एक कविता “वर्षा गीत” यहां दी जा रही है:
नीलांजन नयना
उन्मद सिंधु सुता वर्षा यह
चातक प्रिय बयना!
नभ में श्यामल कुंतल छहरा
क्षिति में चल हरितांचल फहरा,
लेटी क्षितिज तले, अर्थोत्थित
शैल माल जघना!
इच्छाएँ करती उर मंथन
चिर अतृप्ति भरता गुरु गर्जन,
मुक्त विहँसती मत्त यौवना
स्फुरित तड़ित दशना !
रजत बिंदु चल नूपुर झंकृत
मंद भुरज रव नव धन घोषित
मुग्ध नृत्य करत बहँस्मित
कल बलाक रसना !
बकुल मुकुल से कबरी गुंफित
श्वास केतकी रज से सुरभित,
भू नभ को बाँहों में बांधे
इंद्रधनुष वसना!

(घ) ज्ञान-विस्तार

सुमित्रानंदन पंत की प्रकृति चित्रण से संबंधित कविताएँ निश्चित रूप से हिंदी साहित्य की अनमोल रचनाएँ हैं। उन्हीं के समान छायावाद के जयशंकर प्रसाद, महादेवी वर्मा, सूर्यकांत त्रिपाठी ‘निराला’, राम कुमार वर्मा आदि ने भी प्रकृतिचित्रण से संबंधित अमर रचनाओं का प्रणयन किया है; जैसे-
‘मैं नीर भरी दुःख की बदली-
“रूपसि तेरा घन केश पास
नभ गंगा की रजत धार में धो लाई क्या इन्हें रात!”

PSEB 10th Class Hindi Guide गाता खग Important Questions and Answers

प्रश्न 1.
संध्या के समय पक्षी क्या गाता है?
उत्तर:
संध्या के समय पक्षी मानव जीवन के कल्याण और मधुर जीवन के गीत गाता है।

प्रश्न 2.
तारों की पंक्तियाँ मानव के दुःख से दुःखी होकर कैसे आँसू बहाती हैं?
उत्तर:
तारों की पंक्तियाँ मानव के दुःख को देखकर ओस के रूप में आँसू बहाती हैं।

प्रश्न 3.
मानव जीवन में आशा और विश्वास का संदेश कौन भरते हैं?
उत्तर:
मानव जीवन में आशा और विश्वास का संदेश खिले हुए फूल भरते हैं।

प्रश्न 4.
मनुष्य को निरंतर आगे बढ़ते रहने का संदेश किन से मिलता है?
उत्तर:
मनुष्य को निरंतर आगे बढ़ते रहने का संदेश लहरों से मिलता है।

प्रश्न 5.
बुलबुले को अपना मकसद कब मिल जाता है ?
उत्तर:
जब बुलबुला पानी में समा जाता है तब उसे अपना मकसद मिल जाता है क्योंकि वह पहले पानी से अलग था परंतु पानी से मिलकर वह अपना स्थान प्राप्त कर लेता है।

प्रश्न 6.
कवि ने प्रकृति के माध्यम से क्या कामना की है ?
उत्तर:
कवि ने प्रकृति के माध्यम से मानवतावाद की स्थापना की कामना की है। उसने संसार के सभी लोगों के समृद्ध जीवन की इच्छा व्यक्त की है। वह चाहता है कि मानव सदा प्रसन्नता से भरा हुआ गाता रहे। मानव जीवन छोटा है इसीलिए वह संसार को सब प्रकार की खुशियाँ बाँट कर संसार के आंगन को खुशियों से भर दे।

प्रश्न 7.
‘गाता खग’ कविता की अंतिम पंक्तियों में कवि की क्या विशिष्टता दिखाई देती है ? स्पष्ट कीजिए।
उत्तर:
‘गाता खग’ कविता की अंतिम पंक्तियों में रहस्यवाद की झलक दिखाई देती है। जिस प्रकार नदी या समुद्र की लहरों से हवा टकरा कर बुलबुले को जन्म देती है पर वह बुलबुला कोशिश करके भी कभी किनारे को प्राप्त नहीं कर पाता। वह लहरों में ही कहीं इधर-उधर भटक कर बिखर जाता है और उसी जल में समा जाता है। जीव भी उसी प्रकार जीवन भर भटकने के पश्चात् मृत्यु प्राप्त करके परमात्मा में ही विलीन हो जाता है।

एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
संध्या के समय पक्षियों का चहचहाना क्या व्यक्त करता है?
उत्तर:
संध्या के समय पक्षियों का चहचहाना संसार के जीवन के मंगलमय रूप माधुर्य को व्यक्त करता है।

प्रश्न 2.
तारों की पंक्तियाँ टिमटिमा कर क्या कहती हैं?
उत्तर:
तारों की पंक्तियाँ टिमटिमा कर मानव जीवन में व्याप्त अवसाद पर दुःख व्यक्त करती हैं।

प्रश्न 3.
मनुष्य को जीवन में सदा खुशियाँ बिखेरने का संदेश कौन देते हैं ?
उत्तर:
खिले हुए पुष्प मनुष्य को जीवन में खुशियाँ बिखेरने का संदेश देते हैं।

प्रश्न 4.
लहरें मनुष्य को क्या संदेश दे रही हैं ?
उत्तर:
लहरें मनुष्य को निरंतर गतिमान रहने का संदेश दे रही हैं।

बहुवैकल्पिक प्रश्नोत्तरनिम्नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
सारा आशय कौन ‘विलीन’ होकर पा जाता है
(क) नदी
(ख) लहर
(ग) बुद् बुद्
(घ) कूल।
उत्तर:
(ग) बुद् बुद्

प्रश्न 2.
किनारे को कौन नहीं देख पाता
(क) नदी
(ख) लहर
(ग) बुद् बुद्
(घ) पक्षी।
उत्तर:
(ख) लहर

प्रश्न 3.
अपलक तारावलि किसका अनुभव कहती है
(क) आँख
(ख) कान
(ग) नाक
(घ) मुँह।
उत्तर:
(क) आँख

एक शब्द/हाँ-नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
सुंदर सुखमय जीवन के गीत प्रातः उठकर कौन गाता है? (एक शब्द में उत्तर दें)
उत्तर:
खग

प्रश्न 2.
फूल खिल कर संसार को रोना सिखाते हैं। (सही या गलत लिख कर उत्तर दें)
उत्तर:
गलत

प्रश्न 3.
तारावलि अपलक निहारती है। (सही या गलत लिखकर उत्तर दें)
उत्तर:
सही

प्रश्न 4.
किनारा न देखकर भी लहरें आगे बढ़ती जाती हैं। (हाँ या नहीं में उत्तर दें)
उत्तर:
हाँ

प्रश्न 5.
मनुष्य को अपने सद्गुणों से संसार को सुखद नहीं बनाना चाहिए। (हाँ या नहीं में उत्तर दें)
उत्तर:
नहीं

प्रश्न 6.
गाता खग संध्या ………. पर मंगल, ………… जग जीवन।
उत्तर:
तट, मधुमय

प्रश्न 7.
अपने उर की ……….. से जग का ………. भर जाओ।
उत्तर:
सौरभ, आँगन

प्रश्न 8.
पर इस …………. में बह बह नित आगे ………… जाएँ।
उत्तर:
उमंग, बढ़ती।

गाता खग पद्यांशों की सप्रसंग व्याख्या

1. गाता खग प्रातः उठकर
सुंदर, सुखमय जग-जीवन !
गाता खग संध्या – तट पर
मंगल, मधुमय जग-जीवन।

शब्दार्थ:
खग = पक्षी। तट = किनारा। मंगल = कल्याण। मधुमय = आनंदपूर्ण।

प्रसंग:
यह काव्यांश ‘सुमित्रानंदन पंत’ द्वारा रचित ‘गाता खग’ से लिया गया है। इसमें कवि पक्षियों के माध्यम से मानव के सुखद, सुंदर तथा मंगलमय जीवन की कामना कर रहा है।

व्याख्या:
कवि कहता है कि प्रभातकाल में आकाश में स्वतंत्र उड़ने वाले पक्षियों का कलरव गान यही उपदेश देता है कि यह जीवन सौंदर्य और सुख का भंडार है। जिस प्रकार प्रभात वेला में संपूर्ण प्रकृति अपनी सुंदरता, सरसता तथा शीतव्रता से जड़-चेतन को जीवन प्रदान कर देती है, उसी प्रकार पक्षियों का चहचहाना संगीत को भी माधुर्य प्रदान करता है, जिससे संपूर्ण विश्व सुखमय प्रतीत होता है। संध्या के समय जब संपूर्ण संसार विश्राम की ओर अग्रसर होता है, उस समय किसी शून्य, एकांत नदी अथवा सरोवर के किनारों पर एकत्रित होकर पक्षी पुनः चहकते हुए एकत्रित हो जाते हैं। उस समय उनका चहचहाना जीवन के मंगलमय रूप माधुर्य को व्यक्त करता है। कवि को पक्षियों के स्वर में जीवनसंगीत सुनाई पड़ता है।

विशेष:

1. कवि प्रभात तथा संध्या के समय होने वाले पक्षियों के कलरव को मानव जीवन के लिए समृद्धि एवं कल्याण का संदेश देने वाला मानता है।
2. भाषा तत्सम प्रधान है। मानवीकरण तथा अनुप्रास अलंकार हैं।

2. कहती अपलक तारावलि
अपनी आँखों का अनुभव,
अवलोक आँख आँसू की
भर आती आँखें नीरव !

शब्दार्थ:
अपलक = एकटक। तारावलि = तारों की पंक्ति। अवलोक = देखकर। नीरव = मौन, खामोश, चुपचाप।

प्रसंग:
प्रस्तुत पंक्तियाँ सुमित्रानंदन पंत द्वारा रचित कविता ‘गाता खग’ से ली गई हैं। कवि आकाश में चमकने वाले तारों के माध्यम से मानव जीवन में व्याप्त करुणा और संवेदना को व्यक्त कर रहा है।

व्याख्या:
कवि कहता है कि अनंत आकाश में छाए हुए अंधकार में निरंतर चमकने वाली ताराओं की पंक्तियाँ देखकर प्रतीत हो रहा है जैसे वह कह रही हों कि संपूर्ण जीवन करुणा तथा दुःख से भरा हुआ है। जिस प्रकार निरंतर दुःख सहते हुए किसी की आँखों में से आँसू बह जाते हैं और आँसुओं से भरी आँखों को देखकर दूसरे में सहानुभूति के कारण करुणा का संचार हो जाता है, उसी प्रकार विश्वव्यापी दुःख, अवसाद, विपन्नता और विषमता को देखकर अपलक ताराओं की करुणा से भरी आँखों से आँसू छलक पड़ते हैं। आँखों की भाषा नीरव और मौन होती है। केवल आँसुओं के माध्यम से ही प्रकट होती है।

विशेष:

1. कवि का मानना है कि तारों की पंक्तियाँ टिमटिमाकर मानव के दुःख और आँसू देखकर ओस के रूप में स्वयं भी आँसू बहाती है।
2. भाषा तत्सम प्रधान है। मानवीकरण तथा अनुप्रास अलंकार हैं।

3. हँसमुख प्रसून सिखलाते
पल भर है, जो हँस पाओ,
अपने उर की सौरभ से
जग का आँगन भर जाओ!

शब्दार्थ:
हँसमुख = प्रसन्न, खिले हुए। प्रसून = पुष्प, फूल। उर = हृदय। सौरभ = सुगंध। जग = संसार।

प्रसंग:
प्रस्तुत पंक्तियाँ सुमित्रानंदन पंत द्वारा रचित कविता ‘गाता खग’ से ली गई हैं। कवि मानव को सदा खिले हुए फूलों की तरह मुस्कराते रहने का संदेश दे रहा है।

व्याख्या:
कवि कहता है कि प्रभातकाल में खिले हुए पुष्प अपनी कोमलता, मनोहरता और सौरभ से वातावरण को पूरी तरह भरते हुए मानव को यह प्रेरणा देते हैं कि इस नाशवान और छोटे-से जीवन को अनेक प्रकार की विषमताओं और समस्याओं ने नीरस और उदास बना रखा है। यदि हो सके तो संसार में अपना छोटा-सा जीवन प्रसन्नता और आनंद है।

विशेष:

1. कवि की मान्यता है कि लहरें आपस में टकरा कर अथवा किनारे से टकरा कर बिखर जाती हैं तथा किनारे से दूर चली जाती हैं परंतु उनमें से निकला हुआ बुलबुला जल में विलीन हो कर अपने जीवन का उद्देश्य प्राप्त कर लेता है।
2. भाषा तत्सम प्रधान है। मानवीकरण तथा पुनरुक्ति प्रकाश अलंकार हैं।

गाता खग Summary

गाता खग कवि परिचय

छायावादी हिंदी-काव्य के उन्नायकों में श्री सुमित्रानंदन पंत का नाम प्रमुख है। इनका बचपन का नाम गुसाई दत्त था। अल्मोड़ा के निकट कौसानी नामक ग्राम में 20 मई, सन् 1900 ई० को उत्पन्न पंत जी को अपनी माँ का प्रेम नहीं मिल पाया था क्योंकि इन्हें जन्म देते ही वह सदा के लिए इस संसार को छोड़ गई थी। इनका जीवन प्रकृति की गोद में बीता था। इन्होंने अपनी प्रारंभिक शिक्षा गाँव के स्कूल से ही प्राप्त की थी। केवल नौ वर्ष की आयु में ही संस्कृत का ज्ञान प्राप्त कर लेने के पश्चात् इन्होंने सन् 1919 में म्योर सेंट्रल कॉलेज, इलाहाबाद में शिक्षा प्राप्ति के दौरान अंग्रेजी और बंगला की काव्य-कृतियों से परिचय प्राप्त किया। अंग्रेज़ी के शैली, वर्ड्सवर्थ, टैनीसन, कॉलरिज तथा बंगला के रविंद्र नाथ टैगोर की सौंदर्यात्मक रचनाओं ने इन्हें बहुत प्रभावित किया।

सन् 1921 ई० में गाँधी जी के असहयोग आंदोलन के प्रभावस्वरूप इन्होंने अपनी पढ़ाई छोड़ दी थी लेकिन राजनीति में सक्रिय भाग न लेकर अपनी काव्य-साधना जारी रखी थी। सन् 1930 ई० में ये कलाकांकर के महाराज के स्नेहभाजन बन कर रहने के लिए उनके पास ही चले गए और दस वर्ष तक इन्होंने ‘रूपाभ’ नामक पत्रिका का प्रकाशन आरम्भ किया था पर किन्हीं कारणों से यह एक वर्ष से अधिक न चल सकी। कुछ वर्ष तक रेडियो से संबंधित रहने के पश्चात् इन्होंने स्वतंत्र रूप से लेखन कार्य किया। इन्होंने रूस, जर्मनी, फ्रांस आदि देशों की यात्राएँ भी कीं। इन्हें कार्ल मार्क्स, स्वामी विवेकानंद, महात्मा गाँधी, अरविंद जैसी महान् विभूतियों के दर्शन ने अत्यन्त प्रभावित किया। सन् 1977 ई० में इनका देहांत हो गया था।

रचनाएँ-उच्छवास, ग्रंथि, वीणा, पल्लव, गुंजन, युगांत, युगवाणी, ग्राम्या, स्वर्ण-किरण, स्वर्ण-धूलि, युगपथ, उत्तरा, शिल्पी, वाणी, कला और बूढ़ा चांद, लोकायतन आदि। पंत जी के काल में प्रकृति-चित्रण के अतिरिक्त सौंदर्य-चित्रण, मानव प्रेम, शोषितों की पीड़ा, गांधीवादी विचारधारा आदि का प्रभाव भी दिखाई देता है। इसके काव्य में गीति-तत्व की प्रमुखता है। भाषा इनकी तत्सम प्रधान होते हुए भी तद्भव, देशज, विदेशी आदि शब्दों से युक्त है।

गाता खग कविता का सार

‘गाता खग’ कविता में सुमित्रानंदन पंत ने प्रकृति के माध्यम से मानव-जीवन, उसकी कामनाओं, कार्य-व्यापारों, नश्वरता आदि का वर्णन किया है। प्रभातकालीन आकाश में विचरण करते पक्षियों का कलरव, जीवन के सौंदर्य और सुख का संदेश देता है। अनंत आकाश में छाए अंधकार में टिमटिमाते तारे मानव जीवन की करुणा और दुःख का संदेश देते प्रतीत होते हैं। सुबह सवेरे खिले फूल वातावरण को सुगंधित करते हुए मानव को अपना जीवन प्रसन्नता और आनंद से व्यतीत करने की प्रेरणा देते हैं। सागर अथवा नदी में उठने वाली लहरों से मानव को निरंतर गतिमान रहते हुए साधना करने के लिए प्रेरित किया है। लहरों की सिहरन, किनारे से टकरा कर चूर-चूर हो कर सागर में समा जाना आसीम का असीम में समा जाना है।

Punjab State Board PSEB 10th Class Hindi Book Solutions Chapter 4 हम राज्य लिए मरते हैं Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Hindi Chapter 4 हम राज्य लिए मरते हैं

Hindi Guide for Class 10 PSEB हम राज्य लिए मरते हैं Textbook Questions and Answers

(क) विषय-बोध

I. निम्नलिखित प्रश्नों के उत्तर एक-दो पंक्तियों में दीजिए

प्रश्न 1.
प्रस्तुत गीत में उर्मिला किस की प्रशंसा कर रही है?
उत्तर:
प्रस्तुत गीत में उर्मिला किसानों के सरल एवं शांतिपूर्ण जीवन की प्रशंसा कर रही है।

प्रश्न 2.
किसान संसार को समृद्ध कैसे बनाते हैं?
उत्तर:
किसान अन्न पैदा कर के संसार को समृद्ध करते हैं।

प्रश्न 3.
किसान किस प्रकार परिश्रम रूपी समुद्र को धीरज से तैर कर पार करते हैं?
उत्तर:
किसान सहनशील हैं। वे परिश्रम रूपी समुद्र को अपने परिश्रम और धैर्य से तैर कर पार करते हैं।

प्रश्न 4.
किसान का अपने पर गर्व करना कैसे उचित है?
उत्तर:
किसान का अपने पर गर्व करना इसलिए उचित है क्योंकि वह समस्त संसार का अन्नदाता होता है।

प्रश्न 5.
किसान व्यर्थ के वाद-विवाद को छोड़कर किस धर्म का पालन करते हैं?
उत्तर:
किसान व्यर्थ के वाद-विवाद को छोड़कर धर्म की मूल बात को समझकर उसका पालन करते हैं।

प्रश्न 6.
‘हम राज्य लिए मरते हैं’ में उर्मिला राज्य के कारण होने वाले किस कलह की बात कहना चाहती है?
उत्तर:
उर्मिला राज्य के लिए श्री राम को वनवास दिए जाने तथा भरत को राज्य देने से उत्पन्न कलह की बात कहना चाहती है।

II. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या करें:

1. यदि वे करें, उचित है गर्व,
बात बात में उत्सव पर्व, हम से प्रहरी रक्षक जिनके,
वे किससे डरते हैं ?
हम राज्य लिए मरते हैं।
उत्तर:
उर्मिला कहती है कि यदि वे किसान अपने ऊपर गर्व करते हैं तो उनका ऐसा करना बिलकुल ठीक भी है। वे हर अवसर पर समारोह करते हैं तथा त्योहार मनाते हैं। जब हमारे जैसे पहरेदार उनके रक्षक हों तो भला वे किसी से क्यों डरेंगे ? वे निडरतापूर्वक अपने समारोह तथा पर्व मनाते हैं। इसके विपरीत हम लोग तो सदा राज्य के लिए ही मरते रहते हैं।

2. करके मीन मेख सब ओर,
किया करें बुध वाद कठोर,
शाखामयी बुद्धि तजकर वे मूल धर्म धरते हैं।
हम राज्य लिए मरते हैं।
उत्तर:
उर्मिला कहती है कि विद्वान् लोग हर बात में दोष निकाल कर व्यर्थ में बहस करते रहते हैं, चाहे उस से कुछ प्राप्त हो या न हो परंतु किसान इन व्यर्थ की बातों को त्यागकर सहज धर्म को अपनाते हैं। वे विद्वानों के चक्कर में न पड़कर धर्म के वास्तविक स्वरूप को सहज रूप से अपनाते हैं जबकि हम राज्य के लिए आपस में ही लड़ते-मरते रहते हैं।

3. होते कहीं वही हम लोग,
कौन भोगता फिर ये भोग?
उन्हीं अन्नदाताओं के सुख आज दुःख हरते हैं।
हम राज्य लिए मरते हैं।
उत्तर:
उर्मिला कहती है कि यदि कहीं हम भी किसान होते तो फिर राज्य की गृह-कलह के कारण उत्पन्न कष्टों को कौन सहन करता? यदि हम भी किसान होते तो राज्य की उलझनों को सहज करने वाला भी तो कोई होना चाहिए। उन्हीं अन्नदाता किसानों के सुखों को देखकर ही आज हमारे दुःख दूर हो रहे हैं फिर भी हम राज्य के लिए लड़ते-मरते रहते हैं।

(ख) भाषा-बोध

I. निम्नलिखित शब्दों के विपरीत शब्द लिखें:

संपन्न = ————
सुलभ = ————
धनी = ————
उचित = ————
उदार = ————
कठोर = ————
रक्षक = ————
धर्म। = ————
उत्तर:
शब्द – विपरीत शब्द
संपन्न – विपन्न
सुलभ – दुर्लभ
धनी – निर्धन
उचित – अनुचित
उदार – अनुदार
कठोर – कोमल
रक्षक – भक्षक
धर्म – अधर्म

II. निम्नलिखित शब्दों के दो-दो पयार्यवाची शब्द लिखें:

पत्नी = ————
कर्षक = ————
सागर = ————
उत्सव। = ————
उत्तर:
शब्द – पर्यायवाची शब्द
पत्नी – भार्या, अर्धांगिनी, गृहिणी, वधू।
कर्षक – किसान, कृषक, खेतीहर, हलवाहा, कृषिजीवी।
सागर – समुद्र, सिंधु, जलधि, रत्नाकर।
उत्सव – समारोह, पर्व, त्योहार, धूमधाम।

III. निम्नलिखित भिन्नार्थक शब्दों के अर्थ लिखकर वाक्य बनाइएँ:

अन्न = ————
अन्य = ————
उदार = ————
उधार। = ————
उत्तर:
अन्न – अनाज-किसान अन्न पैदा करने के कारण अन्नदाता कहलाता है।
अन्य – दूसरा-इस पर नहीं किसी अन्य विषय पर बात करें।
उदार – दानी-कर्ण बहुत ही उदार राजा था।
उधार – ऋण-क्या आप मुझे एक सौ रुपए उधार दे सकते हैं?

(ग) पाठ्येतर सक्रियता

प्रश्न 1.
किसान की दिनचर्या की जो बातें आपको अच्छी लगती हैं उनकी सूची बनाएँ।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 2.
पंजाब के किसान के जीवन से संबंधित बैशाखी’ त्योहार के कुछ चित्र संकलित करके अपने स्कूल की भित्त पत्रिका पर लगाएँ। उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 3.
किसान को अन्नदाता कहा जाता है। हरित क्रांति में किसानों के योगदान के विषय में जानकारी हासिल करें।
उत्तर:
(विद्यार्थी स्वयं करें।)

प्रश्न 4.
कविता को कंठस्थ करके उसका सस्वर वाचन करें।
उत्तर:
(विद्यार्थी स्वयं करें)

(घ) ज्ञान-विस्तार

द्विवेदी युगीन कवियों में मैथिलीशरण गुप्त द्वारा रचित ‘साकेत’ महाकाव्य खड़ी बोली हिंदी का श्रेष्ठ महाकाव्य है। इसका कथानक उर्मिला के माध्यम से रामकथा को प्रस्तुत करना है। उर्मिला लक्ष्मण की पत्नी थी। श्रीराम के साथ लक्ष्मण वन चले जाते हैं तो उनके विरह में व्याकुल उर्मिला की दशा का वर्णन भी साकेत के नौवें सर्ग में किया गया है। ‘साकेत’ अयोध्या का दूसरा नाम है। ‘साकेत’ का रेडियो रूपांतरण डॉ० प्रेम जनमेजय ने किया था, जो इंटरनेट पर भी मिलता है। किसान के जीवन पर अन्य कवियों की रचनाएँ भी मिलती हैं, जैसे-
फसल क्या है
लाख-लाख कोटि-कोटि हाथों के स्पर्श की गरिमा:
एक की नहीं, दो की नहीं,
हज़ार-हज़ार खेतों की मिट्टी का गुण धर्मः
उत्तर:
विद्यार्थी स्वयं अन्य कवियों की रचनाओं का संग्रह करें।

PSEB 10th Class Hindi Guide हम राज्य लिए मरते हैं Important Questions and Answers

प्रश्न 1.
‘हम राज्य लिए मरते हैं’ कविता कवि की किस मूल रचना और सर्ग से ली गई है?
उत्तर:
‘हम राज्य लिए मरते हैं’ शीर्षक कविता मैथिलीशरण गुप्त के द्वारा रचित ‘साकेत’ के नवम् सर्ग से ली गई है

प्रश्न 2.
‘हम राज्य लिए मरते हैं’ कविता का मूलभाव क्या है?
उत्तर:
इस कविता का मूल भाव यह है कि राजा तो गृह-कलह से दु:खी रहता है परंतु किसान अपने सरल, सहज, शांतिपूर्ण तथा परिश्रमी जीवन से सदा सुखी रहता है।

प्रश्न 3.
‘किसानों के पास गोधन है’-यहाँ गोधन से तात्पर्य है?
उत्तर:
गोधन से तात्पर्य गायों रूपी धन है। किसान के पास गायें होती हैं, वे ही उनका धन हैं।

प्रश्न 4.
राज्य में उत्पन्न उलझनों से कौन निपटता है?
उत्तर:
राज्य में उत्पन्न उलझनों से राज्य को ही निपटना पड़ता है।

प्रश्न 5.
राज्य का दुःख कैसे दूर हो सकता है?
उत्तर:
राज्य का दु:ख अन्नदाता किसान को देखकर दूर हो सकता है।

प्रश्न 6.
‘साकेत’ में किस की विरह-पीड़ा का सजीव चित्रण किया है?
उत्तर:
‘साकेत’ के ‘नवम् सर्ग’ में लक्ष्मण की पत्नी उर्मिला की विरह-पीड़ा का सजीव चित्रण किया गया है क्योंकि वह अपने पति के साथ वनवास के लिए नहीं जा सकी थी और अपने महल में रहकर निरंतर चौदह वर्ष तक विरह-वियोग में आँसू बहाती रही थी।

एक पंक्ति में उत्तरात्मक प्रश्न

प्रश्न 1.
कवि के अनुसार सच्चा राज्य कौन करते हैं ?
उत्तर:
सच्चा राज्य किसान करते हैं।

प्रश्न 2.
किसान के पास कौन-सा धन है ?
उत्तर:
किसान के पास गाय रूपी धन है।

प्रश्न 3.
किसानों के रक्षक कौन हैं ?
उत्तर:
अयोध्या नरेश किसानों के रक्षक हैं।

प्रश्न 4.
विद्वान् क्या करते हैं?
उत्तर:
विद्वान् हर बात में दोष निकाल कर व्यर्थ में तर्क-वितर्क करते हैं।

बहुवैकल्पिक प्रश्नोत्तरनिम्नलिखित प्रश्नों के उत्तर एक सही विकल्प चुनकर लिखें

प्रश्न 1.
‘सहनशीलता के आगार’ कौन हैं
(क) विद्वान्।
(ख) किसान
(ग) राजा
(घ) प्रहरी।
उत्तर:
(ख) किसान

प्रश्न 2.
‘शाखामयी बुद्धि’ कैसी बातें होती हैं
(क) तर्क पूर्ण
(ख) आध्यात्मिक
(ग) व्यर्थ की
(घ) पारलौकिक।
उत्तर:
(ग) व्यर्थ की

प्रश्न 3.
राजघराने की अपेक्षा किसका जीवन सहज है
(क) प्रहरी का
(ख) विद्वान् का
(ग) किसान का
(घ) न्यायाधीश का।
उत्तर:
(ग) किसान का

एक शब्द/हाँ-नहीं/सही-गलत/रिक्त स्थानों की पूर्ति के प्रश्न

प्रश्न 1.
हर अवसर पर समारोह तथा उत्सव कौन मनाते हैं? (एक शब्द में उत्तर दें)
उत्तर:
किसान

प्रश्न 2.
किसानों को अमृत के समान गाय का दूध सुलभ नहीं है। (हाँ या नहीं में उत्तर)
उत्तर:
नहीं

प्रश्न 3.
राज्य का दुःख अन्नदाता किसान को देख कर दूर हो जाता है। (हाँ या नहीं में उत्तर दें)
उत्तर:
हाँ

प्रश्न 4.
राजघराने के लोग राज्य के लिए नहीं मरते हैं। (सही या गलत में उत्तर दें)
उत्तर:
गलत

प्रश्न 5.
किसान निरंतर श्रम करते रहते हैं। (सही या गलत में उत्तर दें)
उत्तर:
सही

प्रश्न 6.
पत्नी सहित ……….. हैं वे, भव ………. भरते हैं।
उत्तर:
विचरते, वैभव

प्रश्न 7.
किया करें …………. वाद कठोर।
उत्तर:
बुध

प्रश्न 8.
उन्हीं ……….. के सुख …….. दुःख हरते हैं।
उत्तर:
अन्नदाताओं, आज।

हम राज्य लिए मरते हैं पद्यांशों की सप्रसंग व्याख्या

1. हम राज्य लिए मरते हैं।
सच्चा राज्य परन्तु हमारे कर्षक ही करते हैं।
जिनके खेतों में है अन्न,
कौन अधिक उनसे सम्पन्न ?
पत्नी-सहित विचरते हैं वे, भव वैभव भरते हैं,
हम राज्य लिए मरते हैं।

शब्दार्थ:
मरते हैं = झगड़ते हैं, दुःखी होते हैं। कर्षक = कृषक, किसान। सम्पन्न = धनी। विचरते = चलतेफिरते। भव = संसार। वैभव = धन-संपत्ति, ऐश्वर्य।

प्रसंग:
प्रस्तुत पंक्तियाँ मैथिलीशरण गुप्त द्वारा रचित कविता ‘हम राज्य लिए मरते हैं’ से ली गई हैं, जिसमें उर्मिला राज्य में व्याप्त गृह-कलह से दुःखी हो कर किसानों के शांतिपूर्ण जीवन की प्रशंसा कर रही है। .

व्याख्या:
उर्मिला कहती है कि हम तो राज्य के कारण गृह कलह से दुःखी हैं जबकि सच्चा राज्य तो हमारे राज्य के वे किसान करते हैं जिन के खेतों में अन्न पैदा होता है। उन से अधिक धनवान भला और कौन हो सकता है? अर्थात् सबसे दुखी तथा सम्पन्न अन्न उत्पन्न करने वाले किसान ही हैं। वे अपनी पत्नी के साथ घूमते-फिरते हुए संसार के ऐश्वर्य का उपभोग करते हैं, जबकि हम राज्य के लिए आपसी कलह से मर रहे हैं।

विशेष:

1. किसानों को सबसे सुखी माना है क्योंकि वे स्वयं अन्न पैदा कर स्वतंत्रतापूर्वक अपना जीवन व्यतीत करते हैं जबकि राजवंशी राज्य के कलह से ही त्रस्त हैं।
2. भाषा तत्सम प्रधान, भावपूर्ण है।

2. वे गोधन के धनी उदार,
उनको सुलभ सुधा की धार,
सहनशीलता के आगर वे श्रम सागर तरते हैं।
हम राज्य लिए मरते हैं।

शब्दार्थ:
गोधन = गायों रूपी धन। उदार = सरल, दानी। सुधा = अमृत। आगर = भंडार, खजाना। श्रम = मेहनत, परिश्रम। सागर = समुद्र।

प्रसंग:
प्रस्तुत पंक्तियाँ मैथिलीशरण गुप्त द्वारा रचित कविता ‘हम राज्य लिए मरते हैं’ से ली गई हैं, जिसमें उर्मिला राज्य के लिए गृह कलह से दुःखी होकर किसानों के संपन्न जीवन की प्रशंसा करती है।

व्याख्या:
उर्मिला किसानों की प्रशंसा करते हुए कहती हैं कि किसान के पास गायों रूपी धन है, जिसके कारण वे धनवान और उदार बने रहते हैं। उन्हें अमृत की धारा के समान गाय के दूध की धारा आसानी से मिल जाती है। वे सहनशीलता के भंडार हैं तथा निरंतर परिश्रम रूपी सागर में तैरते रहते हैं अर्थात् वे सदा मेहनत करते रहते हैं, जबकि हम लोग राज्य के लिए परस्पर लड़ते-मरते रहते हैं।

विशेष:

1. किसान गायों रूपी धन से अमीर हैं तथा निरंतर परिश्रम करते रहते हैं।
2. भाषा तत्सम प्रधान, अनुप्रास तथा रूपक अलंकार हैं।

3. यदि वे करें, उचित है गर्व,
बात बात में उत्सव-पर्व,
हम से प्रहरी रक्षक जिनके, वे किससे डरते हैं?
हम राज्य लिए मरते हैं।

शब्दार्थ:
उचित = सही है। गर्व = घमंड। उत्सव = समारोह। पर्व = त्योहार। प्रहरी = पहरेदार। रक्षक = रखवाले।

प्रसंग:
यह काव्यांश कविवर ‘मैथिलीशरण गुप्त द्वारा रचित ‘हम राज्य लिए मरते हैं’ नामक कविता से लिया गया है। इसमें उर्मिला ने किसानों के जीवन को अपने राज्य के लिए कलह ग्रस्त जीवन से श्रेष्ठ माना है।

व्याख्या:
उर्मिला कहती है कि यदि वे किसान अपने ऊपर गर्व करते हैं तो उनका ऐसा करना बिलकुल ठीक भी है। वे हर अवसर पर समारोह करते हैं तथा त्योहार मनाते हैं। जब हमारे जैसे पहरेदार उनके रक्षक हों तो भला वे किसी से क्यों डरेंगे ? वे निडरतापूर्वक अपने समारोह तथा पर्व मनाते हैं। इसके विपरीत हम लोग तो सदा राज्य के लिए ही मरते रहते हैं।

विशेष:

1. किसानों के आनंद एवं उल्लासमय जीवन की झलक प्रस्तुत की गई है।
2. भाषा तत्सम प्रधान, पुनरुक्ति प्रकाश अलंकार है।

4. करके मीन मेख सब ओर,
किया करें बुध वाद कठोर,
शाखामयी बुद्धि तजकर वे मूल धर्म धरते हैं।
हम राज्य लिए मरते हैं।

शब्दार्थ:
मीन मेख = दोष निकालना, तर्क-वितर्क करना, बहस करना। बुधं = बुद्धिमान, विद्वान्। वाद = वादविवाद, बहस। शाखामयी बुद्धि = व्यर्थ की बातें। तजकर = त्यागकर, छोड़कर। मूल धर्म = वास्तविक धर्म, सहज धर्म। धरते = धारण करना, पालन करना।

प्रसंग:
प्रस्तुत पंक्तियाँ मैथिलीशरण गुप्त द्वारा रचित कविता ‘हम राज्य लिए मरते हैं’ से ली गई हैं, जिसमें उर्मिला ने किसानों के सहज जीवन की प्रशंसा करते हुए राज्य के लिए व्यर्थ ही लड़ने-मरने वालों के जीवन को व्यर्थ माना है।

व्याख्या:
उर्मिला कहती है कि विद्वान् लोग हर बात में दोष निकाल कर व्यर्थ में बहस करते रहते हैं, चाहे उस से कुछ प्राप्त हो या न हो परंतु किसान इन व्यर्थ की बातों को त्यागकर सहज धर्म को अपनाते हैं। वे विद्वानों के चक्कर में न पड़कर धर्म के वास्तविक स्वरूप को सहज रूप से अपनाते हैं जबकि हम राज्य के लिए आपस में ही लड़ते-मरते रहते हैं।

विशेष:

1. आडंबरपूर्ण धर्म के स्थान पर धर्म के सहज स्वरूप को अपनाना श्रेष्ठ है।
2. भाषा तत्सम प्रधान और अनुप्रास अलंकार है।

5. होते कहीं वही हम लोग,
कौन भोगता फिर ये भोग?
उन्हीं अन्नदाताओं के सुख आज दुःख हरते हैं।
हम राज्य लिए मरते हैं।

शब्दार्थ:
भोगता = उपभोग करना। भोग = सुख। अन्नदाताओं = किसानों। हरते हैं = दूर करते हैं।

प्रसंग:
प्रस्तुत पंक्तियाँ मैथिलीशरण गुप्त द्वारा रचित कविता ‘हम राज्य लिए मरते हैं’ से ली गई हैं, जिसमें उर्मिला के राज्य के लिए गृह कलह करने वाले राज घराने से किसान के सहज जीवन को श्रेष्ठ माना है।

व्याख्या:
उर्मिला कहती है कि यदि कहीं हम भी किसान होते तो फिर राज्य की गृह-कलह के कारण उत्पन्न कष्टों को कौन सहन करता? यदि हम भी किसान होते तो राज्य की उलझनों को सहज करने वाला भी तो कोई होना चाहिए। उन्हीं अन्नदाता किसानों के सुखों को देखकर ही आज हमारे दुःख दूर हो रहे हैं फिर भी हम राज्य के लिए लड़ते-मरते रहते हैं।

विशेष:

1. वास्तविक सुख किसान के सरल जीवन में है, यह न समझ कर हम इस अभिमान में लड़-मर रहे हैं कि राज्य हमारा है।
2. भाषा तत्सम प्रधान और भावपूर्ण है।

हम राज्य लिए मरते हैं Summary

हम राज्य लिए मरते हैं कवि परिचय

राष्ट्रकवि मैथिलीशरण गुप्त भारतीयता के अमर गायक के रूप में प्रसिद्ध हैं। उनका जन्म सन् 1886 ई० में चिरगाँव जिला झांसी में हुआ था। उनके पिता श्री रामचरण गुप्त भगवान् राम के परम भक्त थे। माता दयालु स्वभाव की थीं। इनके गुणों का प्रभाव उन पर भी पड़ना स्वाभाविक था। वे भी राम के अनन्य उपासक बन गए। उनकी प्रारंभिक शिक्षा गाँव में ही हुई। बाद में इन्होंने झांसी के मेकडॉनल स्कूल से शिक्षा प्राप्त की थी। उन्होंने संस्कृत, हिंदी और बंगला के साहित्य का अध्ययन किया था और आचार्य महावीर प्रसाद द्विवेदी की प्रेरणा से इन्होंने काव्य-रचनाएँ की थीं जो ‘सरस्वती’ पत्रिका में छपने लगी थीं। गुप्त जी पर गाँधी जी के व्यक्तित्व का भी प्रभाव था। वे अपने जीवन काल में बारह वर्षों तक राज्यसभा के सदस्य भी रहे। आगरा विश्वविद्यालय ने डी० लिट् की उपाधि से तथा भारत सरकार ने पद्मभूषण अलंकार से इन्हें सम्मानित किया था। सन् 1964 ई० में गुप्त जी का देहावसान हुआ।

रचनाएँ-भारत भारती, रंग में भंग, नहुष, साकेत, यशोधरा, द्वापर, जयद्रथ वध, जय भारत, सिद्धराज, पंचवटी आदि इनकी प्रमुख रचनाएँ हैं। ‘झंकार’ उनके गीतों का संग्रह है। ‘साकेत’ नामक महाकाव्य पर गुप्त जी को हिंदी-साहित्य सम्मेलन, प्रयाग का मंगला प्रसाद पारितोषिक प्राप्त हुआ था। मानवतावादी दृष्टिकोण, राष्ट्रीयता की भावना, प्रकृति के विविध रूपों का चित्रण, भारतीय नारी का त्याग-भावना, संस्कृति प्रेम तथा नवीनता के प्रति आस्था आदि गुप्त जी के काव्य की प्रमुख विशेषताएँ हैं। अपनी प्रारंभिक रचना ‘भारत भारती’ में उन्होंने राष्ट्र के अतीत, वर्तमान और भविष्य पर अपने विचार प्रकट किए हैं।

हम राज्य लिए मरते हैं कविता का सार

‘हम राज्य लिए मरते हैं’ शीर्षक कविता श्री मैथिलीशरण गुप्त की रचना ‘साकेत’ के नवम् सर्ग से ली गई है, जिसमें उर्मिला राज्य के कारण उत्पन्न गृह-कलह से दुःखी होकर किसानों के सुखमय जीवन की प्रशंसा करती है। उर्मिला के अनुसार हम लोग तो राज्य के लिए लड़ते-झगड़ते रहते हैं जबकि सच्चा राज्य तो किसान करते हैं। उनके खेतों में अन्न है, वे संपन्न हैं, पत्नी सहित घूमते-फिरते हैं और संसार का सुख भोगते हैं। उनके पास गाय, उदारता, सहनशीलता, परिश्रम करने की शक्ति है। उनका अपने ऊपर गर्व करना, उत्सव-त्योहार मनाना, निडरतापूर्वक विचरण करना अत्यंत सहज कार्य है। विद्वान् चाहे व्यर्थ में धर्म पर तर्क-वितर्क करते रहें परन्तु किसान तर्क-वितर्क छोड़ कर धर्म के मूल को समझते हैं। उर्मिला सोचती है कि यदि हम किसान होते तो राज्य की उलझनों से उत्पन्न मुसीबतों को कौन सहन करता? उन्हीं किसानों के सुख देखकर आज हमारे दुःख दूर हो सकते हैं हम राज्य के लिए ही परस्पर लड़-मर रहे हैं।

Punjab State Board PSEB 10th Class Science Important Questions Chapter 10 Light Reflection and Refraction Important Questions and Answers.

PSEB 10th Class Science Important Questions Chapter 10 Light Reflection and Refraction

Long Answer Type Questions

Question 1.
(a) Form the image in case an object is moved from infinity to the concave mirror.
(b) Describe with the help of diagram the nature, size and position of the image formed when an object placed at centre of curvature of a concave mirror.
Answer:
(a) Images formed by concave mirror when it is moved from infinity to the mirror
1. Object is at Infinity: Two cases arise :

• When mirror is in parallel plane to the object: In such a case, rays from infinity come parallel to principal axis. After reflection they pass through principal focus F. Image is extremely small, real, inverted and it is formed at the principal focus [Figure (a)].
• When the rays strike the mirror obliquely. The ray AB passing through F after reflection goes parallel to principal axis towards BA’. Another ray DE through C striking normally to the mirror at E is reflected back. The two rays form an image at A’. Image is real, inverted, extremely diminished and is formed at F [Figure (b)].

2. Object beyond C. A ray of light AD from A parallel to principal axis after reflection passes through F.
Another ray from A passes through C, and is reflected back along the same path, forming real, diminished, inverted image A’B’, between F and C. [Figure (c)].

(c) Object beyoid C. a real, inverted, diminished image hetiveen F and C. inverted at C and is of saine size as that of object.

3. Object at centre of curvature, C i.e., at 2f. A ray AD from A parallel to principal axis after reflection from mirror passes through F.
Another ray AD’ from A through F. goes, parallel to principal axis i.e., towards, D A forming real, inverted image of AB at C i.e., at 2f. The image is of the same size as the object [Figure (d)].

(d) Object at C. Image is also at C. It is real.

4. Object between F and C between [f and 2f]
A ray AD from object going parallel to principal axis is reflected towards F. Another ray AE as of coming through C is reflected back forming image of A at A’. Similarly image of B is formed at B’. Image is real, inverted, enlarged and formed beyond C (2f) i.e., as shown in Figure (e).

(e) Object between F and C. Image is beyond C. It is real, inverted and enlarged.

5. Object at F. A ray AD parallel to principal axis passes through F. Another ray AE strikes the mirror normally at E is reflected back as it appears to pass through C. This reflected ray becomes parallel to reflected ray DY and, therefore, forms an image of object AB at infinity. The image is very much enlarged and is real and inverted [Figure (f)].

(f). Object at F. Real, inverted, extremely enlarged image is formed at infinity.

6. Object between F and P (Principal Focus and Pole).
A ray AD from A goes parallel to principal axis after reflection passes through F. Another ray AE striking the mirror normally through C is reflected back. These to reflected rays do not actually meet but meet when produced backward, therefore, they form virtual image of the object behind the mirror. The image is erect and enlarged and formed behind the mirror [Figure (g)].

(g). Object between F and P. An erect, enlarged, virtual image is formed behind the mirror.
(b) Please see Figure(d) and its description given on previous page.

Question 2.
What is refraction of light? with the help of diagram explain refraction of light by rectangular glass slab and show that emergent ray and incident ray are parallel.
Answer:
Refraction of Light. When a ray of light enters from one medium to another medium then it changes its path at the meeting plane of two media. This phenomenon of light is called refraction of light.

Refraction by glass slab.

Refraction by rectangular glass slab

A rectangular glass slab PQRS is placed in air. When? ray of ligxu, in air enters glass slab then AO, OO’ and O’B are incident ray, refracted ray and emergent ray respectively

By applying snell’s Law at ‘O’ \(\frac{\sin i_{1}}{\sin i_{1}}\) = ^aµ_b ….(i)

Now ray of hgnt is uavelling from glass (denser medium) to air (rarer medium).
Again using sneirs Law at O
^bµ_a = \(\frac{\sin i_{2}}{\sin r_{2}}=\frac{\sin r_{1}}{\sin r_{2}}\) ……….(iii) [∵∠i₂ = ∠r₁]

According to principle of reversibility of light
^bµ_a = \(\frac{1}{{ }^{a} \mu_{b}}\)

From (ii) and (iii)
^aµ_b = \(\frac{\sin r_{2}}{\sin r_{1}}\) ……..(iv)
Comparing (i) and (iv)
\(\frac{\sin i_{1}}{\sin r_{1}}=\frac{\sin r_{2}}{\sin r_{1}}\)
or sin i₁ = sin r₂
∴ ∠r₁ = ∠r₂
This means that angle of incidence is equal to emergent angle.

Therefore, when refraction of light takes place in a rectangular glass slab then emergent ray and incident ray are equal.

Question 3.
Draw the ray diagrams and find position, nature and size of image formed by a convex lens, when object is placed:
(i) Between F and 2F
Answer:
Object between F and 2F. When the object is between F and 2F, its real. inverted, magnified image is formed on the other side of the convex lens as shown in Figure (a).

(a) Object bet weer F and 2F real, inverted, magnified image is formed beyond 2F on the other side of lens.

(ii) Beyond 2F
Answer:
Object beyond 2F’. When the object is beyond 2F, a real, inverted, diminished image is formed between F’ and 2F [Fig (b)].

(b) Object beyond 2F, real, inverted, diminished image between Fund 2F

(iii) At F.
Answer:
Object at F. When object is placed at principal focus, a real, inverted and highly magnified image is formed at infinity [Fig. (c).]

(c) Object at F, highly magnified, real, inverted image is formed at infinity.

Question 4.
Give the nature, position and size of the image formed by a convex lens when the object lies at 2F and between focus F and optical centre O.
Answer:
1. Object at 2F. A ray AD from point A of the object going parallel to principal axis, passes through F on the other side of the lens goes towards DA’. Another ray from A through optical centre O goes towards OA’. The two meet at A’ forming image of A at A’. Image of object AB is formed at 2F on the other side of the lens. Image is real, inverted and of the same size as that of object [Figure (a)].

2. Object between F and O. When an object is placed between principal focus and optical centre of the lens, virtual, erect, magnified image is formed on the same side of the lens [Figure.(b)].

(b) Object between F and O, a virtual. erect, magnified image is formed on the same side.

Short Answer Type Questions

Question 1.
What is light? Give the nature of light.
Answer:
Light. Light is a form of energy which produces in us the sensation of sight. Light itself is not seen but it helps us to see the surrounding objects.

Nature of light. Light has a dual nature, a wave nature and a particle nature. To explain the phenomenon it was thought of as a wave but when it could not explain the treatment of interaction of light with matter then light was believed to have particle nature. The confusion about true nature of light was later explained when Model Quantum Theory emerged according to which “light is neither a wave nor a particle”. Modern Quantum Theory reconciles the particle properties of light with the wave nature

i. e. light is emitted as a particle but travels in the form of wave.

Question 2.
Write the characteristics of light.
Answer:

• It does not require medium for its propagation.
• It is in the form of electro-magnetic waves.
• Its speed depends upon the nature of medium through which it passes.

Question 3.
What are the various artificial sources of light? Give examples.
Answer:
Artificial Sources of Light: The main artificial sources of light are : fire, electricity gas and some chemical reactions.

Question 4.
What is Reflector?
Answer:
Reflector. Any smooth and polished surface that returns light into the same medium from which it is coming, is called reflector.

Question 5.
What is reflection of light? State the laws of reflection of light.
Answer:
Reflection of Light. When rays of light fall on a smooth and polished surface (incident) then they come back into the same medium in a particular direction. This phenomenon of coming back of light is known as reflection of light.

Laws of Reflection. There are two laws of reflection :

• The angle of incidence is equal to the angle of reflection i.e. ∠i = ∠r.
• The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
• In fig. AB is a reflecting surface (plane mirror), PQ is incident ray. QR is reflected ray and QN is normal at the point of incidence Q.

Figure shows that incident ray, reflected ray and the normal all lie in the same plane of paper.

Question 6.
What is the angle of incidence when incident ray falls normal to the mirror?
Answer:
When a ray of light is incident normal to the mirror then the angle of incidence is zero (i.e. ∠i= 0°).

Question 7.
What is the angle of reflection when a ray of light is incident normally to the mirror?
Answer:
When a ray of light is incident to the mirror normally (∠i = 0°) then after reflection it returns along the same path. In that case, angle of reflection will also be zero (∠r = 0°).

Question 8.
A ray of light falling normal to the mirror returns along the same path. Why?
Answer:
A ray of light falling normal to any mirror returns along the same path. In that case ∠i = 0° and according to the law of reflection ∠i = ∠r. So ∠r will be equal to zero and the reflected ray would be along the same path.

Question 9.
Define the terms :
(i) Spherical mirror
Answer:
Spherical Mirror: If the mirror is a part of some hollow sphere whose one surface is polished and the other surface is reflecting then such a mirror is called spherical mirror.

Spherical mirrors are of two types :
(a) Concave mirror
(b) Convex mirror.

(ii)Concave mirror
Answer:
Concave Mirror: A spherical mirror whose reflecting surface is towards the centre of that sphere of which the mirror is a part is called concave mirror. The outer surface of concave is polished and reflection occurs from the inner surface.

Concave mirror

(iii) Convex mirror
Answer:
Convex Mirror: A spherical mirror whose reflecting surface is away from the centre of that sphere of which the mirror is a part, is called convex mirror.
The inner surface of convex mirror is polished and reflection takes place from the outer surface.

Convex Mirror

(iv) Aperture.
Answer:
Aperture: That part of the mirror from which reflection takes place is called aperture of the mirror.
In figure (a) and (b) the length M1M2 is aperture

Aperture and Centre of Curvature

Question 10.
Define the terms :
(i) Centre of Curvature
Answer:
Centre of Curvature: It is the centre of that hollow sphere of which the mirror is a part. In the above figure (a) C is a centre of curvature of concave mirror and in figure (b) C is centre of curvature of convex mirror.

(ii) Pole
Answer:
Pole. The mid point of spherical mirror is called its pole or vertex. In the fig.l0.14.(a) and fig.10.14.(6) given below P is pole.

(iii) Principal Focus
Answer:
Principal Focus: The principal focus of the mirror is a point on the principal axis where parallel rays of light after reflection from the mirror actually meet at a point or appear to diverge.

(iv) Focal length.
Answer:
Focal Length: The distance between pole and focus is called focal length of the mirror. It is denoted by ‘f. In fig. 10.14. PF is focal length. In SI system the unit of focal length is metre (m).

Question 11.
What is the relation between focal length and radius of curvature of a concave mirror? What is focal length of a plane mirror?
Answer:
The focal length of a concave mirror is half the radius of curvature of that mirror. If ‘f is the focal length and ‘R‘ the radius of curvature of concave mirror, then
f = \(\frac{1}{2}\) × R
The focal length of plane mirror is infinite (∞).

Question 12.
When the image formed by a concave mirror is at infinity then what is the position of the object?
Answer:
When the object is placed at focus of a concave mirror then the image of the object is formed at infinity. This image is real, big in size and is bigger then the size of the object. In this position the rays of light coming from the object after suffering reflection from the mirror become parallel.

Reflected rays meet at Infinity.

Question 13.
Where should an object be placed so that its real and same size image is formed?
Answer:
To get real image of saine size being formed by a concave mirror, the object should be place at its centre of curvature. In this case the image will also be formed at the same position i.e at centre of curvature. This image will be real, inverted and of the same size of the object.

Question 14.
When is virtual and magnified image of an object formed in a concave mirror. Show with the help of a diagram.
Answer:
When an object is placed between pole and focus of a concave mirror then in that case a virtual, erect and magnified image is formed with respect to the size of the object.

Question 15.
Which mirror is shown in the figure? Where is object placed with respect to the mirror? Write one characteristic of the image formed.

Answer:
(a) Concave Mirror.
(b) Object is placed between Focus and Pole of the mirror.
(c) Characteristic of Image formed. The image is virtual, erect and magnified and is formed behind the mirror.

Question 16.
Which mirror is shown in the diagram? Where is object placed in relation to the mirror? Write characteristics of the image formed :

Answer:
In the diagram concave mirror is shown.
The object AB is placed beyond centre of curvature of the mirror i.e. it is at a distance more than radius of curvature (r = CP) from the pole of the mirror.

Characteristics of image formed
The image ¡VB’ of the object AB is :

• Real and inverted
• The image is smaller in size as compared to the size of the object.

Question 17.
Which mirror is used as a shaving mirror and why? Explain its working with the help of a ray diagram.
Answer:
Concave mirror is used as a shaving mirror because we place our face close to the mirror (in between Pole and Focus), the image formed is erect and magnified. This makes the fine hairs to appear quite big so that we can shave properly.

Question 18.
Which mirror always forms virtual, erect and smaller image?
Answer:
Convex mirror always forms a virtual, erect and reduced image of whatever may be the position of the object. The image is formed behind the mirror as is shown in figure.

Formation of Image in a Convex Mirror

Question 19.
Which mirror has wider filed of view?
Answer:
In a convex mirror the image formed is always virtual and smaller in size as compared to the size of the object. This image is formed behind the mirror. On moving the mirror away, the images of objects lying over the large area can be seen in the mirror. Therefore, its area of view becomes wide.

Question 20.
Which mirror is preferred as a driver’s mirror and why? Draw a ray diagram.
Answer:
Convex mirror is preferred as a driver’s mirror because the image formed by convex mirror is erect and smaller in size. Therefore, it enables the driver to see traffic coming behind over large area.

A convex mirror as a driver’s mirror

Question 21.
Where should an object be placed with respect to a concave mirror to get real and enlarged image? Show with the help of ray diagram.
Answer:

Object between F and C. Image ¿s beyond C. It is real, inverted and enlarged.

Question 22.
Write points of difference between convex mirror and concave mirror.
Answer:
Differences between Convex Mirror and Concave Mirror.

Question 23.
How will you distinguish between plane mirror, convex and concave mirror without touching?
Answer:
Look through all these mirrors turn by turn.

• In one of the mirrors, we shall observe that the image is erect, of the same size as the object and as far behind the mirror as the object is in front of it. It is plane mirror.
• In another mirror, the image is always diminished and erect. This is convex mirror.
• In third mirror, image will appear erect and very large when seen by placing eyes, close to the mirror. However, when the eyes are placed at a distance, image will be always inverted. This mirror is concave.

Question 24.
Define magnification of spherical mirror. What is the magnification produced in a plane mirror?
Or
What is magnification?
Answer:
Magnification: Magnification of spherical mirror is defined as the ratio of the size of the image formed by the mirror to the size of the object. It is denoted by ‘m’.
∴ Magnification (m) = \(\frac{\text { Size of the image (I) }\left(h_{2}\right)}{\text { Size of the object (O) ( } h_{1} \text { ) }}\)
m = \(\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{-v}{u}\)

Magnification in a Plane Mirror. Plane mirror can be considered to be a part of such a sphere whose radius of curvature is infinity.

Hence, the size of the object and the size of the image formed in a plane mirror are equal in size.

Question 25.
Write the characteristics of image formed in a plane mirror.
Answer:
Characteristics (or properties) of image in a Plane Mirror.

• It is virtual i.e. this image can not’ be obtained on a screen.
• It is erect.
• It is laterally inverted i.e. the right hand of the object appears left hand of the image and the left hand becomes the right hand.
• The size of the image formed in a plane mirror is equal to the size of the object.
• The image formed in a plane mirror is as far behind as the object is in front of it.

Question 26.
Write difference between real image and virtual image formed by a mirror.
Answer:
Differences between Real Image and Virtual Image.

Question 27.
Explain with the help of a diagram the formation of image formed by a plane mirror.
Or
How will you show that the image formed in a plane mirror is as far behind the plane mirror as the object is in front of it?
Answer:
Image formed by a Plane Mirror. Suppose MM’ is a plane mirror and 0 is a point object lying in front of it. From ‘O’ consider a ray of light ON falling normally to the mirror which retraces its path. Now consider two rays of light OA and OB starting from the object which are incident to the plane mirror. These rays after reflection from the plane mirror form two reflected rays AC and BD. These two reflected rays do not meet but appear to meet at ‘I’ where the image is formed. This ‘I’ is the virtual image of the object ‘O’.

On measurement, we find that NO = NI
i. e. Distance of the object from the mirror = Distance of the image from the mirror.
Therefore, the image formed in a plane mirror is as far behind the mirror as the object is in front of it.

Question 28.
Write uses of spherical mirrors.
Or
What are the uses of concave and convex mirrors?
Answer:
Uses of Spherical Mirrors. Spherical mirrors are of two types

1. Concave mirror
2. Convex mirror.

1. Uses of Concave Mirror:

• Large sized concave mirrors are used in solar cookers to focus solar energy on an object to be heated. Due to a limited reserve of oil and gas, the importance of solar cookers can no longer be ignored.
• Concave mirrors are used as reflectors in car head-lights, search-lights, torches and table lamps.
• Concave mirrors can be used as a shaving mirror or make up mirror. When the concave mirror is placed at a distance less than its focal length from the face, a magnified erect image is produced. Hence any cut or left over hair or minor flaw in the make up can easily be observed with the help of a concave mirror.
• Concave mirrors are used by E.N.T. doctors to concentrate hght on ear, nose, throat, etc.

2. Uses of Convex Mirror:

• Convex mirrors are used in automobiles (e.g., scooters, cars, trucks, buses, etc.) to see bulk of rear traffic. The advantage of using a convex mirror is that it gives a very larger field of view of the rear traffic than a plane mirror.
• The disadvantage of convex mirror is that since the image is always formed between P and F, we cannot accurately judge the distance and speed of the vehicles approaching from behind.
• Usually two mirrors, one plane and the other convex are used to see rear traffic whereas convex mirror on side of vehicle gives wider field of view, plane mirror in front of driver helps us in judging the distance and speed of the vehicles coming behind.

Question 29.
What are New Cartesian Sign Conventions used for reflection in spherical mirror? What is the mirror formula?
Answer:
New Cartesian Sign Conventions for Spherical Mirrors.
The modern cartesian sign conventions are :

• All the rays are drawn from left to right.
• All distances are measured from the pole of the spherical mirror.
• The distances measured in the direction of the incident light are taken as positive and those measured in a direction opposite to the direction of incident light are taken as negative.
• The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive and vice versa.

Since the object is always placed on the left of the mirror therfore, u is always negative.

Question 30.
What is the effect of density on ray of light when light enters from a rarer medium to a denser medium? Explain with diagram.
Or
What is the effect of density on bending of refracted ray during refraction?
Answer:
When a ray of light travelling from rarer medium enters into a denser medium then it bends towards the normal. The more the medium is denser, the more it will bend towards the normal. In the figures given below ray of light (a) from air to water (b) air to glass is shown. The bending of light is more in glass than in water because glass is more dense with respect to water.

Question 31.
When light enters from water to glass what is the change in its velocity?
Answer:
Water is a rarer medium with respect to glass. Therefore, when light enters from water to glass then velocity of light decreases. It suffers refraction and bends towards the normal. In this case angle of incidence (∠i) is greater than the angle of refraction (∠r).

Question 32.
If a ray of light travelling in glass enters into water will it bend towards or away from the normal?
Answer:
When light travels from glass medium (denser medium) enters into water (rarer medium), it suffers refraction and bends away from the normal. In this case angle of incidence (∠i) will be smaller than the angle of refraction (∠r) i.e. ∠i < ∠r and also the velocity of light in water will increase.

Question 33.
Find relation for refractive index in terms of real depth and apparent depth.
Answer:
Suppose an object is lying in the position A at the bottom of water tank. A ray of light AB starting from A strikes the surface of water normally and after refraction goes along BD. Another ray AC makes angle i with the normal at C and makes an angle of refraction r and reaches our eye. In this way A appears at A’. If g is the refractive index of water then
aμw = \(\frac{1}{{w} \mu_{a}}\) = μ

Question 34.
Why does a pencil immersed in water appear bent and short? Explain with the help of a ray diagram.
Answer:
A pencil half immersed in water appears to be bent and short due to refraction of light.
Consider a pencil AB such that its portion CB is immersed in water as shown in figure.

Consider rays of light starting from point B, the tip of the pencil. These rays travel from water (denser medium) and as they enter into air (rarer medium) bend away from normal. The refracted rays appear to come from point B’ slightly raised above. Thus, the immersed portion CB of the pencil appears bent at the surface of water as CB’.

Question 35.
Which phenomenon is shown in the figure? Give its definition and give laws of this phenomenon.

Answer:
The phenomenon of refraction of light is shown in the figure.

Refraction of light: When a ray of light travelling in one transparent medium enters into another transparent medium, it bends at the line of separation of two media. This change of path of hght in second medium is called refraction of light.

Laws of refraction are :

• Incident ray, normal and refracted rays all lie in the same plane.
• Ratio of sine of angle of incidence to the sine of the angle of refraction is constant
  i.e., \(\frac{\sin i}{\sin r}\) = µ, where µ is a constant called refractive index of second medium with respect to first medium .

Question 36.
Define Snell’s Law.
Answer:
Snell’s Law: The second law of refraction is known as Snell’s law. According to this law, ratio of sine of angle of incidence (sin i) to the sine of angle of refraction (sin r) is constant.

\(\frac{\sin i}{\sin r}\) = constant.

Question 37.
What is refractive index? Write its mathematical formula.
Answer:
Refractive Index: The ratio of velocity of light in air or vacuum to the velocity in a medium is called refractive index of medium.
∴ aµb = \(\frac{\text { velocity of light in vacuum }}{\text { velocity of light in medium }}\)
Refractive index has no unit. Since it is ratio of two similar quantities, therefore, it is expressed in number.

Question 38.
What is lens?
Answer:
Lens: A lens is a piece of transparent and refracting medium bounded by two surfaces either having both the surfaces spherical or one spherical and the other plane surface.

Types of lens. There are two main types of lens :

1. Convex lens (also called converging lens)
2. Concave lens (also called diverging lens)

Question 39.
Define the terms :
1. Optical centre
Answer:
Optical Centre: The mid point of lens is called optical centre of lens.

In figure (a) O is optical centre of convex lens and in figure (b) O’ is optical centre of concave lens. Any ray of light passing through this point goes undeviated.

2. Principal axis
Answer:
Principal Axis: The principal axis of a lens is an imaginary line which passes through its optical centre and is normal to the spherical surface of the lens. In figure (a) and figure (b) EF is principal axis of convex and concave lens respectively.

3. Principal focus of lens.
Answer:
Principal Focus of Lens: It is a point on the principal axis where, rays of light coming parallel to the principal axis after refraction meet in case of convex lens and appear to meet on producing backward in case of concave lens.

Question 40.
How is image formed by a concave lens ? Show by drawing diagram as to what will be the position and nature of the image formed by a concave lens?
Answer:
The ray diagram for image of an object placed in front of concave lens is shown in figure. The image formed by concave lens always lies between its optical centre and focus. This image is virtual, erect and small in size.

Question 41.
What is lens formula? Give its sign conventions and assumptions.
Answer:
Lens formula. It is the relationship between the distance of object (u), the distance of image (v) from optical centre of the lens, and focal length (f).
i.e. \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

The formula is equally applicable to convex, concave or any other type of lens (e.g., concavo-convex, plano-convex, convexo-concave, plano-concave etc.)

Sign Conventions.

• The object is always placed to the left of the lens.
• All distances parallel to principal axis are measured from optical centre of the lens.
• All distances measured to the right of optical centre are taken as positive while those measured to the left of the optical centre are taken as negative.
• Distances measured perpendicular to and above the principal axis are taken as positive.
• Distances measured perpendicular to and below the principal axis are taken as negative.

Assumptions.

• Object is taken on the principal axis.
• The lens is thin.

Question 42.
What is magnification of a lens?
Answer:

• Magnification of a lens is defined as the ratio of the height of the image formed by the lens to the actual height of object.
• If h is the height of the object and h’, the height of the image formed by lens, then magnification,
  m = \(\frac{h^{\prime}}{h}\)
• If u and v are the distances of object and image respectively from the optical centre, then
  m = \(\frac{h^{\prime}}{h}=-\frac{v}{u}\)

Question 43.
Compare convex and concave lenses.
Answer:
Differences between Convex Lens and Concave Lens.

Question 44.
What are the differences between reflection and refraction?
Answer:
Differences between Reflection and Refraction.

Question 45.
Define 1 dioptre of power of a lens.
Answer:
Power of a lens is reciprocal of focal length.
P = \(\frac{1}{f}\)

If f = 1 m, P = \(\frac{1}{f}=\frac{1}{1}\) =1 dioptre
Dioptre: Dioptre or 1 dioptre is the power of a lens of focal length 1 metre.

Numerical Problems

Question 1.
The radius of curvature of a concave mirror is 30 cm. What is its focal length?
Answer:
Here, Radius of curvature of concave mirror (R) = – 30 cm
Focal length of the mirror (f) =?
We know that R = 2f
or f = \(\frac{R}{2}\)
= \(\frac{-30}{2}\)
∴ Focal length of concave mirror (f) = – 15 cm

Question 2.
The radius of curvature of a convex mirror is 40 cm. What is its focal length?
Answer:
Here, Radius of curvature of convex mirror (R) = + 40 cm
Focal length of convex mirror (f) = ?
We know that R = 2f
or f = \(\frac{R}{2}\)
= \(\frac{+40}{2}\)
∴ Focal length of convex mirror (f) – + 20 cm

Question 3.
A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Find where will the image be formed?
Answer:
Here object distance (u) = – 10 cm
Magnification (m) = \(-\frac{v}{u}\) [Real image]
Image distance (v) = ?
We know m = \(-\frac{v}{u}\)
– 3 = \(-\frac{v}{-10}\)
– 3 = \(\frac{v}{10}\)
v = – 3 × 10
= – 30 cm
Negative sign shows that the image is formed at a distance of 30 cm in front of a concave mirror.

Question 4.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Sol. Here, focal length of convex mirror if) = + 15 cm
object distance from convex mirror (u)= – 10 cm
Image distance (v) =?

∴ Image distance v = + 6 cm
Positive sign shows that the image is formed behind the mirror (on the other side of the mirror) at a distance of 6 cm. This image is virtual and erect.

Question 5.
Light enters from air to diamond having refractive index 2.4. What is the speed of light in diamond? Given speed of light in vacuum = 3 × 10⁸ m s^-1.
Answer:
Speed of light in Vacuum, c = 3 × 10⁸ m s^-1 Refractive index of diamond, n = 2.4
Speed of light in diamond, v_d =?

Question 6.
Light travels from air to water of refractive index 1.33. Calculate the speed of light in water, if speed of light in air is 3.0 × 10⁸ m/s.
Answer:
Given : Speed of light in air (c) = 3 × 10⁸ m/s
Refractive index of water ^aµ_w – 1.33
Speed of light in water (v) =?

Question 7.
Refractive index of water w.r.t. air is \(\frac{4}{3}\) and that of glass w.r.t air is \(\frac{3}{2}\) What will be the refractive index of glass w.r.t water?
Answer:
Here, Refractive Index of water w.r.t. air (^aµ_w) = \(\frac{4}{3}\)
Refractive index of glass w.r.t air (^aµ_g) = \(\frac{3}{2}\)
Refractive index of glass w.r.t water (^wµ_g) =?
Refractive index of glass w.r.t water (^wµ_g) = ^wµ_a x ^aµ_g

Question 8.
An object is 2m away from a lens, which forms an erect image \(\frac{1}{4}\)th the size of the object. Determine the focal length of the lens. What type of the lens is this?
Answer:
Here, object distance (u) = – 2m

Negative sign indicates that the lens is concave lens.

Question 9.
A convex lens of focal length 20 cm is placed at a distance of 24 cm from the screen. How far from the lens should an object be placed so as to form a real image on the screen. Also find the nature and magnification of the image produced.
Answer:
Given : Focal length of the lens (f) = 20 cm
Image distance (v) = 24 cm
Object distance (u) =?
Magnification (m) =?

Negative sign shows that the image formed is real and inverted.

Question 10.
A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification produced by the lens.
Answer:
Given : Focal length of concave lens (f) = – 15 cm
Image distance (v) = – 10 cm
(∵ The image formed in a concave lens is virtual and erect and is formed on the same side of the lens, so it has negative (-) sign.)
Object distance (u) = ?

Negative sign suggest that the object should be placed at a distance of 30 cm to the left of the lens.
Now magnification (m) = \(-\frac{v}{u}\)
= \(-\frac{(-10 \mathrm{~cm})}{(-30 \mathrm{~cm})}\)
= \(\frac{1}{3}\)
= – 0.33
Negative sign shows that the image is virtual and erect and its size is \(\frac{1}{3}\)rd the size of the object.

Question 11.
A convex lens of power 4 D is placed at a distance of 40 cm from a wall. At what distance from the lens should a candle be placed so that its image is formed on the wall?
Answer:
Given : Power of convex lens(P) = + 4 D
We know P = \(\frac{100}{f(in cm)}\)
f = \(\frac{100}{P}\)
= \(\frac{100}{4}\) = 25 cm
Image distance (v) = 40cm

Very Short Answer Type Questions

Question 1.
Which mirror has a widest field of view?
Answer:
Convex mirror.

Question 2.
Define a spherical mirror.
Answer:
Spherical Mirror. A spherical mirror is a part of hollow sphere whose one surface is polished and the other is reflecting.

Question 3.
What is concave mirror?
Answer:
Concave Mirror. Concave mirror is one which is silvered from outer side and reflection takes place at inner surface.

Question 4.
What is convex mirror?
Answer:
Convex Mirror. Convex mirror is one which is silvered from inner side and reflection takes place from outer convex side.

Question 5.
Define the pole of a mirror.
Answer:
Pole. Pole is the mid-point of mirror.

Question 6.
Define principal focus of a mirror.
Answer:
Principal focus. It is a point on the principal axis at which the ray of light incident in a direction parallel to its axis actually meet or converge or appear to diverge after reflection from the mirror.

Question 7.
Define the focal length of mirror.
Answer:
Focal Length. It is the distance between pole and principal focus of the mirror.

Question 8.
What is light?
Answer:
Light. It is a form of energy which causes in us sensation of sight.

Question 9.
What is a real image?
Answer:
Image formed on a screen is called real image.

Question 10.
What is refraction of light?
Answer:
The phenomenon of change of path of light from one transparent medium to another is called refraction of light.

Question 11.
Define refractive index?
Answer:
It is the ratio of sine of the angle of incidence to the sine of the angle of refraction.

Question 12.
What is a lens?
Answer:
Lens. It is a portion of a transparent medium which is bounded by two surfaces.

Question 13.
Define power of a lens.
Answer:
Power of lens. It is the ability of a lens to bend the rays.

Question 14.
Define a ‘dioptre’.
Answer:
Dioptre. It is the power of a lens whose focal length is 1 m.

Question 15.
Define focal length of a lens.
Answer:
Focal length of Lens. The distance between optical centre and principal focus is called focal length.

Question 16.
What is the nature of the mirror having focal length – 15 cm?
Answer:
It is a concave mirror haying focal length – 15 cm and radius of curvature – 30 cm.

Question 17.
A mirror has magnification 0.4, what type of the mirror is and what type of the image is formed?
Answer:
It is a convex mirror since the magnification is positive as well as less than one. Image is diminished and erect.

Question 18.
What is curvature of a mirror? What is its value for plane mirror?
Answer:
Curvature is reciprocal of focal length.
Radius of curvature of a plane mirror is infinity and hence its curvature is = \(\frac{1}{\infty}\) = zero.

Question 19.
A ray strikes the mirror normally, what is the angle of incidence?
Answer:
Zero degree. Angle of incidence is the angle which the incident ray makes with the normal.

Multiple Choice Questions :

Question 1.
The power of a lens is 2 Dioptre. Its focal length will be ________

(A) 20cm
(B)4Ocm
(C) 10cm
(D)5Ocm.
Answer:
(D) 50 cm

Question 2.
________ form.s virtual and same size of image of an object.
(A) Concave mirror
(B) Convex mirror
(C) Plane mirror
(D) None of these.
Answer:
(C) Plane mirror

Question 3.
The image of an object formed by a convex mirror is always ______
(A) Real, inverted and smaller than the object
(B) Virtual, inverted and smaller than the object
(C) Virtual, erect and smaller than the object
(D) Virtual, erect and bigger than the object.
Answer:
(C) Virtual, erect and smaller than the object

Question 4.
________ is used in motor vehicles to take rearview.
(A) Concave mirror
(B) Plane mirror
(C) Convex mirror
(D) Any spherical mirror
Answer:
(C) Convex mirror

Question 5.
\(\frac{\sin i}{\sin r}\) relation was given by __________
(A) Newton
(B) Raman
(C) Snell
(D) Faraday.
Answer:
(C) Snell

Question 6.
The focal length of a lens is expressed by which of the following:
(A) \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
(B) \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
(C) \(\frac{1}{f}=\frac{1}{v}=\frac{1}{u}\)
(D) \(\frac{1}{f}=\frac{1}{u-frac{1}{v}\)
Answer:
(B) \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Question 7.
The relation between the radius of curvature (R) and focal length (f) of a concave mirror is __________
(A) f = R
(B) f = \(\frac{R}{2}\)
(C) R = \(\frac{f}{2}\)
(D) R = \(\frac{f}{4}\)
Answer:
(B) f = \(\frac{R}{2}\)

Question 8.
Where will the real and inverted image of an object placed at the centre of curvature of a concave mirror will be formed?
(A) At F
(B) At C
(C) Between C and F
(D) At infinity
Answer:
(B) At C

Question 9.
The mirror used by dentist is ________
(A) Convex mirror
(B) Concave mirror
(C) Plane mirror
(D) None of the above.
Answer:
(B) Concave mirror

Question 10.
The mirror used for getting real and enlarged image is ________
(A) Convex mirror
(B) Concave mirror
(C) Plane mirror
(D) None of these.
Answer:
(B) Concave mirror

Question 11.
Parallel rays incident on a mirror, after reflection converge at a point, then the mirror will be _________
(A) Plane
(B) Concave
(C) Convex
(D) None of these.
Answer:
(B) Concave

Question 12.
The Unit of power of lens is ____________
(A) Coulomb
(B) Watt
(C) Joule
(D) Dioptre
Answer:
(D) Dioptre

Question 13.
Power of a lens is – 5D, its focal length is :
(A) 20 cm
(B) – 20 cm
(C) – 0.2 m
(D) 5 cm.
Answer:
(C) – 0.2 m

Question 14.
Which of the following lenses would you prefer to use while reading small letters found in dictionary?
(A) A convex lens of focal length 50 cm
(B) A concave lens of focal length 50 cm
(C) A convex lens of focal length 5 cm
(D) A concave lens of focal length 5 cm
Answer:
(C)A convex lens of focal length 5 cm

Question 15.
Shaving mirrors are ________
(A) Convex mirrors
(B) Plane mirrors
(C) Concave mirrors
(D) Parabolic mirrors.
Answer:
(C) Concave mirrors

Question 16.
Which of the following is the property of light?
(A) Reflection
(B) Refraction
(C) Rectilinear propagation
(D) All of these.
Answer:
(D) All of these.

Question 17.
Twinkling of stars is due to atmospheric :
(A) Reflection of light
(B) Dispersion of light
(C) Interference of light
(D) Refraction of light.
Answer:
(D) Refraction of light.

Question 18.
Defect of vision that cannot be corrected by spectacles is :
(A) Myopia
(B) Presbyopia
(C) Cataract
(D) Hypermetropia.
Answer:
(C) Cataract

Question 19.
Which one of the following material cannot be used to make a lens?
(A) Clay
(B) Glass
(C) Water
(D) Plastic.
Answer:
(A) Clay

Question 20.
Where should object be placed in front of a convex lens to get real image of the size of object?
(A) At principal focus of lens
(B) At twice the focal length
(C) At infinity
(D) Between optical centre of lens and its principal focus. –
Answer:
(B) At twice the focal length

Question 21.
A spherical mirror and a thin spherical lens have each a focal length -15 cm. The mirror and lens are likely to be _____
(A) Both are convex
(B) Both are concave
(C) Mirror is concave but lens is convex
(D) Mirror is convex and lens is concave.
Answer:
(B) Both are concave

Question 22.
No matter how far you stand from a mirror, your image appear erect. The mirror is likely to be:
(A) Plane
(B) Concave
(C) Concavo-convex
(D) Convex.
Answer:
(A) Plane

Question 23.
The refractive index of diamond is:
(A) 2.42
(B) 2.43
(C) 2.45
(D) 2.40.
Answer:
(A) 2.42

Question 24.
A ray of light travelling in a glass merges into air, it will bend :
(A) Towards the normal
(B) Away from the normal
(C) Goes along the normal
(D) All of these.
Answer:
(B) Away from the normal.

Fill in the blanks :

Question 1.
A concave mirror forms image inverted and equal in size, when object is placed at ________
Answer:
centre of curvature (c).

Question 2.
A ________ mirror is used to see the rear view in cars.
Answer:
convex mirror.

Question 3.
The ratio of sine of angle of incidence to the sine of angle of refraction is called ________
Answer:
Refractive Index.

Question 4.
When a beam of light passes from optically rarer medium to optically denser medium, it bends ________ the normal.
Answer:
towards.

Question 5.
________ is the unit of Power of the lens.
Answer:
Dioptre.

Question 6.
A virtual and enlarged image is formed by a ________ mirror.
Answer:
concave.

Punjab State Board PSEB 10th Class Science Book Solutions Chapter 10 Light Reflection and Refraction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

PSEB 10th Class Science Guide Light Reflection and Refraction Textbook Questions and Answers

Question 1.
Which of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay.
Answer:
(d) Clay, which is opaque.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than object. Where should be the position of the object?
(а) between principal focus and the centre of curvature.
(b) at centre of curvature.
(c) beyond centre of curvature.
(d) between the pole of the mirror and its principal focus.
Answer:
(d) between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(а) at principal focus of the lens.
(b) at twice the focal length of lens.
(c) at infinity.
(d) between optical centre of the lens and its principal focus.
Answer:
(b) at twice the focal length of the lens.

Question 41.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and lens are likely to be :
(а) both are concave.
(b) both are convex.
(c) mirror is concave and lens is convex.
(d) mirror is convex but lens is concave.
Answer:
(a) both are concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be :
(a) plane only.
(b) concave only.
(c) convex only.
(d) either plane or convex.
Answer:
(d) Either plane or convex.

Question 6.
Which of the following lenses would you prefer to use while reading small letters in a dictionary?
(а) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
(c) A convex lens of focal length 5 cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than object? Draw a ray diagram to show the image formation in this case.
Answer:
The image formed in the concave mirror would be erect only when the object is placed between pole and principal focus of the concave mirror. Therefore, range of distance is greater than zero and less than focal length i.e., between 0-15 cm.

This image is erect, enlarged. virtual and behind the mirror.

Question 8.
Name the type of mirror used in the following situations :
(a) Head-light of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirrors are used for head-lights of car to give concentrated parallel beam of light.
(b) Side/rear-view mirror of a vehicle is mostly convex mirror because it forms an erect, virtual and diminished image and therefore, gives a wider field of view of the traffic behind.
However, for judging the distance and speed of the rear vehicle, a plane mirror fixed inside the car.
(c) Solar furnaces use concave mirror. It is because parallel sun rays (coming from infinity) after reflection from concave mirror are converged at focus to produce much heat.

Question 9.
One half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Explain your observation.
Answer:
If lower half of a convex lens is covered with black paper no doubt it will form complete image but intensity (brightness) of image with half lens will be less as that with complete lens exposed. The nature, size and location of the image will be the same since light from all parts of the object reaches the exposed part of the lens.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of image formed.
Answer:

Here, Length of the object (h) = 5 cm,
Image distance, v =?,
Object distance form the lens (u) = – 25 cm [Object distance is always negative]
h’ = ?; f = + 10cm [For convex lens, f is positive]


Negative sign shows that the image is inverted, real and diminished (3.3 cm) which is formed at 16.7 cm distance on the right side of lens.

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw ray diagram.
Answer:
Distance of the image from the lens (V) = – 10 cm [Virtual image]
Focal length of concave lens (f) = – 15 cm [Focal length of concave lens is taken negative];
Distance of the object from the lens (u) = ?


Negative sign shows that the object is placed to the left side of lens.

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length of 15 cm. Find the position and nature of the image.
Answer:
Distance of the object from the lens (u) = – 10 cm [u is always negative]
Focal length of the lens (f) = + 15 cm [convex mirror]
Distance of the image from the lens. (v) = ?

Negative sign shows that the image is virtual and erect and is formed towards left of the mirror.

Question 13.
The magnification produced by a plane mirror is + 1. What does this mean?
Answer:
It means that image produced by plane mirror is virtual, erect and of the same size as that of the object.

Question 14.
An object 5.0 cm of length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Here height of the object, h = 5.0 cm ;
u = – 20 cm [u is always negative]
Radius of curvature (r) = + 30 cm [convex mirror]
But f = \(\frac{r}{2}\)
= \(\frac{30}{2}\)
= + 15 cm

Height of the image h’ = ?

Image distance from the mirror v = \(\frac{60}{7}\) = 8.56 cm
Thus, the image is formed behind the mirror or to the right of the mirror.

Positive sign (+) indicates that the image is erect and formed above the principal axis.

Question 15 .
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should the screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Here h = 7.0 cm ; u = – 27 cm [u is always negative]
f = – 18 cm [concave mirror]
v =?; h’ =?

Using mirror formula ;

∴ Image distance from the mirror v = – 54 cm
i. e., at 54 cm on the same side of the mirror as that of the object.
Now, Magnification (m) = \(\frac{h^{\prime}}{h}=-\frac{v}{u}\)
\(\frac{h^{\prime}}{7.0}=-\frac{(-54)}{(-27)}\)
= -2
h’ = – 2 × 7
∴ h’ = – 14 cm
Image is of size 14 cm i.e. magnified. Negative sign (-) shows that the image is real and inverted.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Here, Power of lens (P) = – 2.0 D
Focal length of lens (f) =?
We have f = \(\frac{1}{P}\)
= \(\frac{1}{-2}\)
= – 0.5 m
∴ f = – 0.5 m = – 50 cm
Negative sign indicates that the lens is concave. Therefore, it is a divergent lens.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of lens. Is prescribed lens diverging or converging?
Answer:
Here power of lens (P) = + 1.5 D
Focal length of lens (f) =?
We know f = \(\frac{1}{P}\)
= \(\frac{1}{1.5}\)
∴ f = 0.67 m = 67 cm
The positive sign (+) shows that the lens is a convex lens which is diverging in nature.

Science Guide for Class 10 PSEB Light Reflection and Refraction InText Questions and Answers

Question 1.
Define the principal focus of concave mirror.
Answer:
Principal focus: It is a point on the principal axis of a concave mirror where the rays parallel to principal axis meet after reflection from the mirror.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Spherical mirror may be either convex mirror or a concave mirror.

For convex mirror,
Radius of curvature (R) = + 20 cm
Focal length (f) =?

We know, f = \(\frac{R}{2}\)
f = \(\frac{20}{2}\) = 10 cm

For concave mirrior,
Radius of curvature (R) = – 20 cm
Focal Length (f) = ?

We know, f = \(\frac{R}{2}\)
∴ f = \(\frac{-20}{2}\) = -10 cm

Question 3.
Name a mirror which can give an erect and enlarged image of an object.
Answer:
Concave mirror produces an erect and enlarged image when the object is placed between its pole and principal focus.

Question 4.
Why do we prefer a convex mirror as back view mirror in vehicles?
Answer:
Preference of convex mirror as rear view mirror in vehicles. We prefer to have convex mirror as a rear view mirror in vehicles because of the following reasons :

• It produces an erect image of the object (traffic behind).
• The size of the image formed is much smaller than the size of the object.
• It has a wider field of view.

Question 5.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Radius of curvature (R) = + 32 cm [for convex mirror, R is +ve]
Focal length (f) =?

We know, f = \(\frac{R}{2}\)

∴ f = \(\frac{32}{2}\) = 16 cm

Question 6.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Magnification (m) = – 3 [for real image m is taken as negative]
u = – 10 cm v =? ; [u is always negative]

We know, m = \(-\frac{v}{u}\)
-3 = \(-\frac{v}{(-10)}\)
∴ v = – 30 cm
Since, the value of v is negative, therefore, the image is formed at 30 cm on the same side (to the left) of the mirror as that of the object.

Question 7.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards normal or away from normal? Why?
Answer:

anw = \(\frac{\sin i}{\sin r}\)
Since anw > 1,
sin i > sin r
or i > r
or r < i
Hence, a ray of light travelling from optically rarer medium (air) to an optically denser medium (water), bends towards normal.

Question 8.
Light enters from air to glass having refractive index 1.50. What is speed of light in glass? Speed of light in vacuum is 3 × 10⁸ m s^-1.
Answer:
Refractive index of glass (µ) = 1.50
Speed of light in vacuum (c) = 3 × 10⁸ m s^-1

We know, Refractive index = \(\frac{\text { Speed of light in vacuum }(c)}{\text { Speed of light in glass }\left(v_{g}\right)}\)
1.50 = \(\frac{3 \times 10^{8}}{v_{g}}\)
or v_g = \(\frac{3 \times 10^{8}}{1.50}\)
v_g = 2 × 10⁸ m s^-1

Question 9.
Find out from table 10.3 of the text-book, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:

• Optical density of a medium depends upon its refractive index. The higher the refractive index, higher the optical density and vice versa.
• From the data available in the table it is apparent that diamond is having highest refractive index (µ = 2.42) and air is having lowest refractive index (µ = 1,0003).

Question 10.
Refractive indices of kerosene, turpentine and water are 1.44,1.47 and 1.33 respectively. In which material does the light travel fastest and why?
Answer:
Refractive indices of kerosene, turpentine and water are 1.44, 1.47 and 1.33 respectively. We know that velocity of light in a medium is inversely proportional to refractive index of the medium.

Therefore, light will travel fastest in water (having least refractive index i.e. µ = 1.33) and slowest in turpentine oil (having maximum refractive index i.e., µ = 1.47.)

Question 11.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
We know, Refractive index of diamond = \(\frac{\text { Velocity of light in vacuum }}{\text { Refractive index of diamond }}\)
or
Velocity of light m diamond = \(\frac{\text { Velocity of light in vacuum }}{\text { Refractive index of diamond }}\)
The statement means that velocity of light in diamond is \(\frac{1}{2.42}\) times the velocity of light in vacuum.
i.e., v_D = \(\frac{3 \times 10^{8}}{2.42}\) m s^-1
= 1.24 × 10⁸ m s^-1-1
In other words, light travels 2.42 times faster in vacuum than in diamond.

Question 12.
Define 1 dioptre of power of a lens.
Or
What is power of Lens? Give the commercial unit of power.
Answer:

• Power of Lens: It is defined as the ability of a lens to bend incident light rays.
• It is measured by taking reciprocal of focal length measured in metres.
• The power of a lens is said to be 1 dioptre if its focal length is one metre.

Question 13.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle be placed in front of the convex lens if the image is equal to size of the object? Also, find the power of the lens.
Answer:
Here u =?, f =?, P =?
v = + 50 cm [for real image u is taken as positive]
m = -1 [Real image is always inverted]

Question 14.
Find the power of a concave lens of focal length 2 metres.
Answer:
Here f = -2 m [for concave lens f is taken as negative]
P =?
Power of lens P = \(\frac{1}{f}\)
= \(\frac{1}{(-2)}\)
P = – 0.5 D

Punjab State Board PSEB 9th Class Science Book Solutions Chapter 6 Tissues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Science Chapter 6 Tissues

PSEB 9th Class Science Guide Tissues Textbook Questions and Answers

Question 1.
Define the term ’tissue’.
Answer:
Tissue: A group of cells that are similar in structure and/or work together a particular function is called tissue.

Question 2.
How many types of elements together make xylem tissue? Name them.
Answer:
Four types of elements make xylem tissue. They are:

1. Tracheids
2. Vessels
3. Xylem parenchyma
4. Xylem fibres

Question 3.
How are simple tissues different from complex tissues in plants?
Answer:
Differences between simple tissue and complex tissue in plants:

Complex tissue

Question 4.
Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell walls.
Answer:

Question 5.
What are functions of stomata?
Answer:
Functions of stomata:

• They are necessary for exchanging gases with the atmosphere.
• Transpiration also takes place through stomata.

Question 6.
Diagrammatically show the differences between three types of muscle fibres.
Answer:
Three Types of Muscle Fibres:

Question 7.
What is the specific function of cardiac muscles?
Answer:
Cardiac muscles undergo rhythmic contraction and relaxation. They are responsible for heart beat and thus plays a role in circulation (pumping) of blood in the body.

Question 8.
Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Answer:
Differences between Striated, Non-Striated and Cardiac Muscle Fibres:

Question 9.
Draw a labelled diagram of a neuron.
Answer:
Structure of neuron

Question 10.
Name the following:
(a) Tissue that forms inner lining of our mouth.
(b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants.
(d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix.
(f) Tissue present in the brain.
Answer:
(a) Squamous epithelium
(c) Phloem
(e) Blood (Vascular tissue)

Question 11.
Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Answer:

1. Skin
2. Bark of tree
3. Bone
4. Lining of kidney tubule
5. Vascular bundle

Question 12.
Name the regions in which parenchyma tissue is present.
Answer:
Pith and cortex of stem and root.

Question 13.
What is the role of epidermis in plants?
Answer:
Role of epidermis:

1. Protection of all parts of plant.
2. Secretion of waxy-water resistant layer on their outer surface.
3. Epidermis aids in protection against loss of water, mechanical injury and invasion by parasitic fungi.

Question 14.
How does cork act as a protective tissue?
Answer:
Cell of cork are dead and compactly arranged without intercellular spaces. They have a chemical called suberin in their walls that makes them impervious to gases and water. Thus cork acts as a protective tissue.

Question 15.
Complete the table:
Answer:

1. Parenchyma
2. Sclerenchyma
3. Phloem

Science Guide for Class 9 PSEB Tissues InText Questions and Answers

Question 1.
What is a tissue?
Answer:
Tissue: A group of cells that are similar in structure and/or work together a particular function is called tissue.

Question 2.
What is utility of tissues in multicellular organisms?
Answer:

• They provide protection and mechanical strength.
• Tissues provide highest possible efficiency of function.

Question 3.
Name types of simple tissues.
Answer:
Types of simple tissues:

1. Parenchyma
2. Collenchyma
3. Sclerenchyma.

Question 4.
Where is apical meristem found?
Answer:
Apical meristem is present at the growing tips of stems and roots of plants.

Question 5.
Which tissue makes up the husk of coconut?
Answer:
Sclerenchyma tissue.

Question 6.
What are constituents of phloem?
Answer:
Four types of elements constitute phloem:

1. Sieve tubes
2. Companion cells
3. Phloem fibers
4. Phloem parenchyma

Question 7.
Name the tissue responsible for the movement of our body.
Answer:
Muscular tissue.

Question 8.
What does a neuron look like?
Answer:
A neuron consists of cell body cyton with hair-like parts called dendrites and a long axon. Thus gives the appearance of a miniature tree.

Question 9.
Give three features of cardiac muscles.
Answer:

1. Heart muscles are cylindrical, branched, and uninucleate.
2. They are involuntary and undergo rhythmic contraction and relaxation.
3. Intercalated discs are present at the junction of two cells.

Question 10.
What are the functions of areolar tissue?
Answer:
Areolar tissue fills space inside the organ, supports internal organs, and helps in repairing the tissues.