AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 6 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत ।

अ) परोक्षे कार्यहन्तारं प्रत्यक्षे प्रियवादिनम् ।
वर्जयेत्तादृशं मित्रं विषकुम्भं पयोमुखम् ॥
समाधान:
पदच्छेद (Word Division) : परोक्षे, कार्यहन्तारं, प्रत्यक्षे, प्रियवादिनम्, वर्जयेत्, तादृशं, मित्रं, विषकुम्भं, पयोमुखम् ।
अन्वयक्रम : परोक्षे, कार्यहन्तारं प्रत्यक्षे, प्रियवादिनम्, तादृशं मित्रं, पयोमुखं, विषकुम्भं इव, वर्जयेत् ।
अर्था (Meanings) : परोक्षे = behind, कार्यहन्तारं = one who spoils our work, प्रत्यक्षे = in front of us, प्रियवादिनम् = who speaks sweetly, तादृशम= such, मित्रम् = friend, पयोमुखम = milk topped, with milk at rim, विषकुम्भम् = pitcher of poison, वर्जयेत् = should abandon.
भाव (Substance) : One should abandon such a friend who spoils our work at our back, but speaks sweetly in front of us as he is a vessel filled with poison but having milk at rim.

आ) प्रथमे नार्जिता विद्या द्वितीये नार्जितं धनम् ।
तृतीये नार्जितं पुण्यं चतुर्थे किं करिष्यति ॥
समाधान:
पदच्छेद (Word Division ) : प्रथमे न, आर्जितं, विद्या, द्वितीये, न, आर्जितं धनम्, तृतीये न, आर्जितं पुण्यं, चतुर्थे किं करिष्यति ।
अन्वयक्रम : प्रथमे, विद्या, न, आर्जितम्, द्वितीये, धनं, न आर्जितम्,
तृतीये, पुण्यं, न, आर्जितम्, चतुर्थे किं करिष्यति ।
अर्था (Meanings) : प्रथमे = during the first stage of life, in the first quarter; विद्या = education; न + आर्जितम् = not gained; द्वितीये = in the second stage, second quarter; धनम् = money, ricers; न `आर्जितम् = not gained; तृतीये: = in the third stage of life, in the third quarter; पुण्यं = merit; न + आर्जितम् not gained; चतुर्थे = in the last stage, in the last quarter; किं करिष्यति = what will he do?
(Substance): If a person does not acquire knowledge in the first stage of his life, money in the second stage, and merit in the third stage, what will he do in the last stage?

2. एकं निबन्धप्रश्नं समाधत्त ।

अ) दिलीपस्य धर्मनिष्ठामधिकृत्य लिखत ?
Answer:
Introduction: The lesson Dharmanishta is an extract from the 2nd canto of Raghuvamsa, written by Kalidasa. This lesson describes the moral and devotional character of king Dilipa.

A Lion attacks Nandini: The childless king Dilipa his wife Sudakshina were engaged in the service of Nandini, the cow of sage Vasishta. One day Nandini which went to the forest guarded by the king wished to test the devotion of the king, and entered a Himalayan cave. Thinking that the cow was unassailable to any wild animal, the king took his eyes off her to watch the beauty of the mountain. Then a lion attacked her. Alerted by her cry, the king became ready to kill the lion by shooting an arrow at it. But his hand got fixed on the shaft of the arrow. चित्रार्पितारम्भ इवावतस्थे ।

Then the lion told him that he was a servant of Lord Siva. He was appointed there to protect the Devadaru tree grown by Parvathi. He was made a lion, and any animal that arrived there when he was hungry would become his food.

Dilipa’s offer: Then the king said that Siva was to be respected, and at the same time, the property of the teacher was to be saved. He offered his own body as food to the lion requesting the release of the cow. But the lion said that the king was young and an emperor. He seemed to be imprudent in giving up more for little. By showing compassion to the cow, she alone would be saved. But if he lived, he could always save his people from calamities. The king said that a king should protect his subjects.

तद्विपरीतवृत्तेः, प्राणैः उपक्रोशमलीमसैर्वा । He could not offer other cows for Nandini, the daughter of Surabhi to his preceptor. He pleaded with the lion that how he could stand before his teacher when the cow was lost and himself was unhurt. He further said that people like him did not care for physical bodies. पिण्डेष्वनास्था खलु भौतिकेषु ।

The lion agreed and released the cow. But when the king bent his head before the lion, he received a shower of flowers from the Vidyadharas. The delighted cow, followed by the king returned safely to the hermitage.

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

आ) पाठ्यभागमनुसृत्य गीर्वाणवाण्याः सारांशं लिखत ?
Answer:
Introduction : The lesson Chitravimsati was written by Sri Jatavallabhula Purushottama Sastry. It is an extract from his work Chitra-satakam. There are two topics in this lesson Girvanavani and Chitraloka. The first part describes the greatness of Sanskrit language.

Girvanabhasha :
The nectar oozing Sanskrit language is spoken by the gods. It is the ocean of the gems of good sayings and the treasure trove of great literature, and is acquired only by the meritorious. संस्कृताख्या सुकृतैकलभ्या | It is the ladder to liberation and the staircase to the heaven: It is relished by people of different tastes. Just as the children inherit the virtues of the mother, so also all the languages inherit the qualities of the divine mother language. In Sanskrit as the words are formed from the roots, they are naturally pure.

A language is the mother because she spreads in the world, and also she protects the dependents like a mother. On this earth, such a mother tongue is Sanskrit only, how can other languages have the title of mother ? वाच्याः कथं मातृपदेन चान्या: ? Even now, those who are experts in that language make the non-learners of the language ashamed by reciting the ancient verses.

Lakhs of people speak it, and lakhs understand it. Lakhs of books in that language are sold every year. How can we live abandoning that language which is always in our body like life, and which is used during marriage ceremony, rituals of manes and gods, and in poetry recitals and philosophical discourses ? Our history is before us with life. How is it dead? In this world, beautiful poetry is removed from dharma and righteous poetry is not beautiful. Except in Sanskrit where else is poetry that is golden and fragrant ?

3. एकं निबन्धप्रश्नं समाधत्त ।

अ) भोजस्य सालभंजिकया उक्तं विक्रमस्य औदार्यं लिखत ।
Answer:
Introduction: The lesson Vikramasya Audaryam is taken from vikramarkacharitam written by Sivadasa. When Bhoja wants to ascend the throne of Vikrama the statue on the fourth step tells this story about the generosity of king Vikrama.

The childless Brahmin:
While Vikramaditya ruled Ujjaini, there was a learned Brahmin in that city, who was virtuous, but who had no children. Once, his wife said to him that there was no heaven for one who had no children सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना । The Brahmin worshipped Lord Siva. The god appeared in his dream and asked him to perform Pradoshavrata. The Brahmin did so and he begot a son, whom he named Devadatta. When his son grew up, the Brahmin performed his marriage, and went on a pilgrimage to Varanasi.

Devadatta’s help to Vikrama :
Once Devadatta went into the forest to collect fuel sticks for sacrifice. At that time, king Vikrama came to the forest for hunting. There he asked Devadatta the way to the city, even though he knew it.

Devadatta himself led him to the city. The king praised him, and appointed him in his court. One day the king praised Devadatta’s help in the assembly saying that he could never repay his debt.

Devadatta kidnaps the prince :
Devadatta wondered whether the king’s praise was true or false. He wanted to test the words of the king. He kidnapped the prince, and gave one of his ornaments to his servant and sent him to the market to sell it. The king’s men who were searching for the prince caught him, and brought him to the king. When the king asked him, the servant said that he was Devadatta’s servant. The king sent for Devadatta, who said that he killed the prince or money. The members of the court said that Devadatta should be sentenced to death. But the king said that Devadatta rescued him in the forest. One should not find faults in those dependent on him. fordy महतां गुणदोषचिन्ता | He said that his son died because of his previous deeds only. He honoured Devadatta and sent him away.

Devadatta returned with the prince, and told the king that he wanted to test the words of the king. He praised the generosity of the king, who said that one should not forget the help done to him. यः कृतमुपकारं विस्मरति स एंव पुरुषाधमः ।
Thus the statue told Bhoja about the generosity of king Vikrama.

आ) ‘भिषजो भैषज्यम्’ इति पाठ्यभागस्य सारांशं लिखत ।
Answer:
Introduction: The lesson Bhishajah Bhaishajyam was written by Prof. Pullela Sriramachandra. It is taken from his Sriramachandra-laghukavya-sangraha. This lesson describes the story of a selfish doctor, and the fruit he reaped for his selfishness.

The Villager’s Plea: One day some villagers came to Dr. Venkata Rao, and requested him to attend to a boy who was injured in an accident. Venkata Rao childed them for not coming in time. वैद्योऽपि मानव एव | He accused them of trying to get treatment done without paying fee. He insulted them saying the boy would be delicate as the baby of a donkey. When they left as the boy was serious क्षणे क्षणे किल परिक्षीयते बालस्य दशा, Venkata Rao thought nothing would happen if one puppy died.

The poor and intelligent Venkata Rao : Venkata Rao was the son of a poor farmer. He was very intelligent and secured a seat in medical college. His father sold their agricultural land for his education. A rich man married his daughter to Venkata Rao. Venkata Rao’s practice also picked up. Along with money, the three defects grew in him. They were considering others as insects, himself as god, and accepting others praise as the truth. When he spoke of his father as a beggar, his father left him and returned to their village.

Marriage was also a business affair for Venkata Rao. विवाहो नाम वणिग्व्यवहार एव | For him money was everything. He never loved his wife. His son Suresh alone became the object of his affection.

Desire for Power: Venkata Rao thought that money was useless without power. He became an MLA twice. But he could not become a minister. He was unhappy about that. He could not win the seat the third time. He blamed the people for that. His hatred for people grew.

Manjuhasini, the Lady Doctor: At that time, Dr. Manjuhasini joined the government hospital there. She was Venkata Rao’s classmate in medical college. She rejected Venkata Rao’s advances. Venkata Rao was hoping that she might have changed now as he became rich.

The death of his son: Venkata Rato received a phone call from Manjuhasini requesting his help in an emergency case. His driver tried to inform him that his son was not there at the school when he went there after getting the brake repaired. Venkata Rao cut him short saying that the boy would have reached home. But when he went to the hospital he saw the same villagers who came to him earlier in the day, and the body of his dead son.

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

4. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) भृगुवंशकेतुः कः ?
समाधान:
भृगुवंशकेतुः परशुरामः ।

आ) कीदृशं वस्तु तथैव तिष्ठति ?
समाधान:
हुतं दत्तं च वस्तु तथैव तिष्ठति ।

इ) केन द्रुपदः गर्वाधः जातः ?
समाधान:
द्रुपदः धनश्रिया गर्वान्धः जातः । एवम् अर्जुनः उक्तवान् ।

ई) कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय किं दत्तवती ?
समाधान:
कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय पिष्टमिश्रितजलं दत्तवती । तत् पीत्वा अश्वत्थामा तुष्टः अभवत् ।

उ) कालिदासश्लोके भोजेन प्राप्तं समाधानं किम् ?
समाधान:
कालिदासश्लोकेन भोजेन प्राप्तं समाधानम् एवम् अस्ति – द्वयोः सम्भाषणकर्त्रीः मध्ये सूचनां विना न प्रविशेत् इति ।

ऊ) भोजः कोशाध्यक्षं प्रति किं सूचितवान् ?
समाधान:
कविकुलगुरोः कालिदासस्य सदनं एकलक्षसुवर्णमुद्राः प्रेष्यन्ताम् इति भोजः कोशाध्यक्षं सूचितवान् ।

5. द्वयोः संदर्भ व्याख्यानं लिखत ।

अ) परेषां सहसावज्ञा न कर्तव्या कथञ्चन ।
समाधान:
परिचयः – एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् ।
एषः पाठः – रामायणस्य युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः – प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
भावः – परेषां बलानि अपरिमेयानि । तेषां सहसा अवज्ञा न कुर्यात् ।

आ) पिण्जेष्वनास्था खलु भौतिकेषु ।
समाधान:
परिचयः एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषः भागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः दिलीपः सिंहं प्रति एवं उक्तवान् ।
भावः पृथिव्यादि भूतविकारेषु शरीरेषु अनपेक्षा खलु ।
विवरणम्ः हे सिंह ! मम थशोरुप शरीरे दयालुः भव । देहं अशाश्वतम्, कीर्तिरेव शाश्वतम् ।

इ) ग्रामे यत्र गृहे समैक्यपरता सा धर्मभूमिर्मम |
समाधान:
परिचयः – एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् । अस्य कविः श्रीमान् दोर्बल प्रभाकर शर्मा ।
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः – यत्र गृहे माता भर्तुः सुतानां च इष्टवस्तूनि परिवेषति, पिता वृत्तिप्रदः, भ्रातरः स्नेहसम्पन्नाः ग्रामे शोभायात्राः गृहे एक्यता इत्यादीनि सन्ति, सा मम धर्मभूमिः इति कविः वदति ।

ई) भृत्या हि पक्त्री वद किं न सा स्त्री ।
समाधान:
परिचय : गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री ।
सन्दर्भ : लोके चित्रविषयान् वर्णयन् कविः एवं वदति । मम सुता महानसे न निरोधनीया, स्त्रियः पूज्याः इति चित्रं वदन्ति । तर्हि भूक्तिः कथं सिद्ध्यति इति कविः पृच्छति ।
भाव : या भृत्या पचति सा स्त्री नास्ति किम् ?
विवरणम् : मदीया कन्या महानसे नैव पाकं कराति ? स्त्रियः पूज्याः, इति वदन्ति, भृत्या पाचिका स्त्री कि ? वद ।

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

6. द्वयोः ससंदर्भ व्याख्यानं लिखत ।

अ) सतां सङ्गो हि भेषजम् ।
समाधान:
परिचयः – एतत-वाक्यं, “मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः – शोकाविष्टं कौण्डिन्यं पति कपिलः एवं अवदत् ।
भावः – सज्जनसांगत्यं एव सज्जनां भेषजम् ।
विवरणम्ः – सर्वे सज्जनसाङ्गत्यं करणीयम् तदा एव जीवनं सुखमयं भवेत् ।

आ) सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना ।
समाधान:
परिचय : इदं वाक्यं “विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास |
सन्दर्भ : ब्राह्मणस्य पत्नी स्व भर्तारं प्रति एवं अवदत् ।
भाव : सत्कुमारिण वशं राज्ञा वसुधा सूर्येण भुवनत्रयं प्रकाश्यन्ते ।
विवरणम् : वाणी व्याकरणेन, हंसमिधुनैः पद्यः पण्डितैः सभा सुपुत्रेण कुलं राज्ञा, भूमिः सूर्येण लोकत्रयं विराजन्ते ।

इ) क्षणे क्षणे किल परिक्षीयते बालस्य दशा ।
समाधान:
परिचयः – वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः एकः वृद्धः युवकं उद्दिश्य एवं अवदत् ।
भावः – क्षणे क्षणे परीशील्यमाने बलकस्य दशा क्षीयेत ।
विवरणम्ः – वृया अनेन कलहेन, बालकस्य परिस्थितिः सीयते ।

ई) अप्पय्यः साक्षात् परमशिवस्य अवतारः ।
समाधान:
परिचय : एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भ : केचन जनाः अन्तिमसमये अप्पय्यदीक्षितम् चिदम्बरे नटराज- प्रतिमायां लीयमानम् अपश्यन् । तस्मिन्नेव समये गृहे अप्पय्यः प्राणविहीनः अभवत् । तेन सः परमेश्वरस्य अवतारः इति विज्ञाः मन्यन्ते स्म ।
भाव : अप्पय्यः साक्षात् परमेश्वरस्य अपरः अवतारः एव ।

7. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) कदा प्रभृति अशुभानि निमित्तानि दृश्यन्ते ?
समाधान:
यदा प्रभृति सीता आगता तदा प्रभृति अशुभानि निमित्तानि दृश्यन्ते ।

आ) धर्मनिष्ठा इति पाठ्यांशः कस्मात् संगृहीतः ?
समाधान:
धर्मनिष्ठा, इति पाठ्यभागः, रघुवंशमहाकाव्ये तृतीयसर्गत् संगृहीतः ।

इ) पृथिव्यां त्रीणि रत्नानि कानि ?
समाधान:
पृथिव्यां त्रीणि रत्नानि जलं, अन्नं, सुभाषितम् ।

ई) देवः कुत्र सहायकृत् भवति ?
समाधान:
उद्यमं, साहस, धैर्यं, बुद्धिः, शक्तिः, पराक्रमः यत्र वर्तन्ते तत्र सहायकृत् भवति ।

उ) मानवानां प्रकृतिशुभगुणाः के ?
समाधान:
क्षान्तिः, संस्कारगन्धः, संतृप्तिः, मृदुवाक् इत्यादयः मानवानां प्रकृतिगुणाः ।

ऊ) लोके चित्रमतिः कः ?
समाधान:
ईश्वरे भक्तिः सर्वेषां अस्ति, ईश्वरप्रतिपादित वेदे भक्तिः नास्ति, एतत् लोके चित्रगतिः ।

8. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) हितोपदेशे कति भागाः सन्ति ? ते च के ?
समाधान:
हितोपदेशे चत्वारः भागाः सन्ति । ते
1. मित्रलाभः 2. मित्रभेदः 3. विग्रहः 4. सन्धिः

आ) ब्राह्मणं प्रति तत्प्रेयसी किं जगाद ?
समाधान:
ब्राह्मणं प्रति तत्प्रेयसी एवं जगाद – अपुत्रस्य गतिर्नास्ति स्वर्गो नैव च नैव च । सत्पुत्रेणैव कुलं प्रकाशते इति ।

इ) धर्मपालस्य कति पुत्राः ? ते च के ?
समाधान:
धर्मपालस्य त्रयः पुत्रः आसन् । ते च सुमन्त्रः, सुमित्रः कामपालः च ।

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

ई) श्रीधरः स्वमातुः मरणसमये कि कुर्वन्नासीत् ?
समाधान:
श्रीधरः स्वमातुः मरणसमये विश्वविद्यालये शोधग्रन्थसमर्पणकार्यं कुर्वन् आस्ते ।

उ) वेङ्कटरावस्य पिता कः ?
समाधान:
वेङ्कटरावस्य पिता सुब्बरायशास्त्री ।

ऊ) अप्पय्यदीक्षितः आन्ध्रभाषां कथं प्रशशंस ?
समाधान:
आन्ध्रत्वम् आन्ध्रभाषा च नाल्पस्य तपसः फलम् इति अप्पय्यः आन्ध्रभाषां प्रशसंश

9. एकेन वाक्येन समाधानं दत्त ।

अ) सुखधर्मनाशनं केन भवति ?
समाधान:
सुखधर्मनाशनं कोपेन भवति ।

आ) दिलीपस्य धर्मपत्नी का ?
समाधान:
सुदक्षिणा

इ) सर्वत्र का पूज्यते ?
समाधान:
विद्या

ई) चिरसम्प्रदाय रुचिराः के ?
समाधान:
चिरसम्प्रदायरुचिराः संस्काराः ।

उ) “चित्रविंशतिः” इति पाठ्यभागः कस्मात् स्वीकृत: ?
समाधान:
गीर्वाणवाणी – चित्रलोक:

10. एकेन वाक्येन समाधानं दत्त ।

अ) हितोपदेशः केन विरचितः ?
समाधान:
नारायणपण्डितेन

आ) देवदत्तः राजानं कुत्र अनीनयत् ?
समाधान:
नगरम्

इ) दशकुमारचरितं केन विरचितम् ?
समाधान:
दण्डिना

ई) श्रीधरस्य ज्येष्ठभ्रातृवत् कः वर्धते ?
समाधान:
श्रीधरस्य ज्येष्ठभ्रातृवत् तस्य मातुः कासः वर्धते ।

उ) भिषजः भैषज्यम् इति पाठ्यांशः केन विरचितः ?
समाधान:
पुल्लेल श्रीरामचन्द्रः

11. अधोनिर्दिष्टं कथां पठित्वा प्रश्नानां समाधानानि दत्त |

प्रश्नाः,
(1) वृक्षे प्रतिवसतः पक्षिणो नाम किम् ?
समाधान:
वृक्षे प्रतिवसतः पक्षिणो नाम सिन्धुकः ।

(2) कस्मिन् सुवर्णमुत्पद्यते ?
समाधान:
पुरीषे सुवर्णमुत्पद्यते ।

(3) पक्षिणमुद्दिश्य कः समाययौ ?
समाधान:
पक्षिणमुद्दिश्य व्याधः समाययौ ।

(4) स पक्षी कुत्र पुरीषमुत्ससर्ज ?
समाधान:
स पक्षी व्याधस्य अग्रत एव पुरीषमुत्ससर्ज ।

(5) व्याधः वृक्षे कं बबन्ध ?
समाधान:
व्याधः वृक्षे पाशं बबन्ध |

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त ।

अ) शरत् + चन्द्रः
समाधान:
शरच्चन्द्रः = श्चुत्व सन्धिः

आ) उत् + डयनम्
समाधान:
उड्डयनम् = ष्टुत्व सन्धिः

इ) तत् + मात्रम्
समाधान:
तन्मात्रम् = अनुनासिक सन्धिः

ई) षट् + आननः
समाधान:
षडाननः = जश्त्व सन्धिः

उ) सः + अपि
समाधान:
सोऽपि = विसर्ग सन्धिः

ऊ) हरिः + गच्छति
समाधान:
हरिर्गच्छति = विसर्गरेफादेश सन्धिः

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

13. नामनिर्देशपूर्वकं त्रीणि बिघटयत ।

अ) सज्जनः
समाधान:
सद् + जनः = श्चुत्व सन्धिः

आ) तट्टीका
समाधान:
तत् + टीका = ष्टुत्व·सन्धिः

इ) अजन्तः
समाधान:
अच् + अन्तः = जश्त्व सन्धिः

ई) षण्मुखः
समाधान:
षट् + मुखः = अनुनासिक सन्धिः

उ) कोऽपि
समाधान:
कः + अपि = विसर्ग सन्धि:

ऊ) पितुरिच्छा
समाधान:
पितुः + इच्छा = विसर्गरेफादेश सन्धिः

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत ।
अ) वाणिक
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions 2

आ) खज्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions 3

इ) जगत
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions 4

ई) यद् (न.पुं.)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions 5

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत ।

अ) वृक्षमूलम्
समाधान:
वृक्षस्य मूलम् = षष्ठीतत्पुरुष समासः

आ) अल्पवातः
समाधान:
अल्पश्च असौ वातश्च = विशेषणपूर्वपदकर्मधारय समासः

इ) त्रिलोकी
समाधान:
त्रयाणां लोकानां समाहारः = द्विगु समासः

AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions

ई) धर्मार्थकामाः
समाधान:
धर्मः च अर्थः च कामः च = द्वन्द्व समासः

उ) शाकप्रति
समाधान:
शाकस्य लेशः = अव्ययीभाव समासः

ऊ) उपदशाः
समाधान:
दशानां समीपे ये सन्ति ते = संख्योत्तरपदबहुव्रीहि समासः

16. अधो निर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखत |
AP Inter 2nd Year Sanskrit Model Paper Set 6 with Solutions 1
समाधान:
प्रश्नाः
1. वयं सत्यं पठिष्यामः
2. यूयं समाधानं लेखिष्यथ
3. ते ग्रामं गमिष्यन्ति
4. ताः पाकं करिष्यन्ति
5. दावाः सत्यं वदिष्यन्ति

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 5 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत

अ) तक्षकस्य विषं दन्ते मक्षिकायास्तु मस्तके ।
वृश्चिकस्य विषं पुच्छे सर्वाङ्गे दुर्जने विषम् ॥
समाधान:
समा. पदच्छेद (Word Division) : तक्षकस्य, विषं, दन्ते, मक्षिकायाः, मस्तके, वृश्चिकस्य, विषं, पुच्छे, सर्वाङ्गे, दुर्जने, विषम् ।
अन्वयक्रम : तक्षकस्य, विषं, दन्ते, मक्षिकायाः, मस्तके, वृश्चिकस्य, पुच्छे, विषं, दुर्जने, सर्वाङ्ग, विषम् ।
अर्थ (Meanings) : तक्षकस्य = to a serpent, विषम् = poison, दन्ते in fangs, मक्षिकायाः = to a bee or spider, मस्तके = in head (poison), वृश्चिकस्य = to a scorpion, पुच्छे = in tail, विषम् = poison, दुर्जने = to an evil person, सर्वांगे = throughout the body; विषम् = poison.
भाव (Substance) : A serpent has poison is its fangs, a bee has in its head, and a scorpion has poison in its tail. But an evil person has poison in his whole body.

आ) परोपदेशसमये जनाः सर्वेऽपि पण्डिताः ।
तदनुष्ठानसमये मुनयोऽपि न पण्डिताः ॥
समाधान:
पदच्छेद (Word Division) : परोपदेशसमये, जनाः, सर्वेः, अपि, पण्डिताः, तत्, अनुष्ठानसमये, मुनयः, अपि, न, पंडिताः ।
अन्वयक्रम : सर्वे जनाः, अपि, परोपदेशसमये, पण्डिताः, तत्, अनुष्ठानसमये, मुनयः, अपि, न पण्डिताः ।
अर्था (Meanings) : पर + उपदेशसमये = while advising others; सर्वे अपि = all the; जनाः = people; पंडिता: = are scholars; तत् + अनुष्ठानसमये = while putting into practice; मुनयः अपि = even the sages; न पण्डिताः = are not scholars.
(Substance): While giving advice to others, everyone acts as a scholar, but while putting it into practice, even the sages are not scholars.

2. एकं निबन्धप्रश्नं समाधत्त ।

अ) रावणाय विभीषणेन कृतमुपदेशं लिखत ।
Answer:
Introduction: The lesson Vibhishanopadesa is an extract from the Yuddhakanda of the Ramayana, written by Valmiki. Vibhishana advises Ravana to send Sita to Rama.

The Advice of Vibhishana: Vibhishana advised Ravana that one should use their might only when the three other methods namely sama, dana and bheda fail. Then also it would work again the weak and already unfortunate enemies. They should not underestimate the enemy. परेषां सहसावज्ञा न कर्तव्या कथञ्चन | They should protect lives. They need not have an unnecessary fight with one who followed Dharma. Sita should be sent to Rama.

Vibhishana advised Ravana to abandon his anger which would destroy his comfort and dharma. Ravana should follow dharma which would increase joy and fame.

The Inauspicious signs: Vibhishana mentioned the various inauspicious signs witnessed since the arrival of Sita. “The milk of the cows has decreased. The mighty elephants have lost their rut. The horses are neighing piteously. The donkeys, camels and mules are shedding tears losing their hair. Flocks of crows are cawing harshly on all sides. Cruel animals have assembled at the gates of the city, and are roaring loudly.”

He compared Sita with a serpent. She had the hood of bosom, poison of worry, fangs of smile, heads of five fingers and a great body. He warned that Ravana would not be spared with life by Rama. Even the Sun, Indra, Yama and other could not protect him.

Friends, who were like enemies : Vibhishana accused the Rakshasas saying that they were enemies in the guise of friends. He said that a minister should weight the strength, loss and gain on both sides, and give a beneficial advice. वदेत् क्षमं स्वामिहितं मन्त्री । He asked whether anyone could stand before the arrows of Rama. He advised that Sita should be sent to Rama along with precious gifts.

When Ravana said that he would not have spared if this speech was made by any other person, Vibhishana became angry. He flew in to the space, and said that as a brother Ravana could say anything, but he would not forgive him. He said that people who spoke sweetly could be found easily. But those who gave bitter but beneficial advice, and its listener were hard to be found. अप्रियस्य च पथ्यस्य वक्ता श्रोता च दुर्लभः | He wished him happiness, and left.

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

आ) मातृभूमेः वैभवं कविः कथं प्रशशंस ?
Answer:
Introduction: The lesson Sa Matrubhumirmama was written by Dorbala Prabhakara Sarma. The poet describes the great ness of India in this poem.

Vedabhumi, Jnanabhumi and Dharmabhumi: The poet praises the motherland as a land of Vedas, wisdom, languages, gods, dharma, action, yoga, tapas, rivers, holy places and great people. Here, the cattle of the Vedas show the right path.

The people here are cultured, intelligent, healthy and sweetly speaking. यत्रैताः हृदि जागृताः प्रकृतयः सा मातृभूमिर्मम | The land is prosperous, waters are pure, breeze is purifying and life is friendly. There are Rik mantras that praise the gods, Yajush mantras that are used in sacrifices, Atharva mantras that show the path of prosperity. This land is full of knowledge with the Vedas, Upanishads, Puranas, Itihasas, Kavyas etc. विद्या यत्र परोपकारविभवाः सा ज्ञानभूमिर्मम | There are different languages belonging to Vanga, Anga, Andhra, Kerala, Maharashtra, Sindhi, Gujarati, Karnataka etc.

Selfless People: The gods of rivers, villages, cities, towns, households, lands and others bestow boons. The lords of quarters such as Indra, Agni, Varuna, Vayu, Kubera, Yama etc. along with Vishnu, Uma and others grant welfare to the people. In a household, the mother takes care of the interests of her husband and children, father works, brothers give relief, teachers preach and the elders offer advice. ग्रमे यत्र गृहे समैक्यपरता सा धर्मभूमिर्मम | The people work selflessly, are engaged in meditation, prayers and contemplation. The wisdom of the sages ensures the welfare of the world, strengthens culture and enriches the knowledge.

Rivers and holy places: Rivers like Ganga, Yamuna, Saraswati, Godavari, Krishna, Narmada etc. remove sins, diseases and obstacles. Mountains such as Hemadri, Rajatadri, Aravali, Mahendra, Himalayas, Sahya etc. embellish this land. There are holy cities such as Kasi, Ayodhya, Puri, Kanchi, Madhura, Avanti, Tirupati, Ahobilam etc. Great sages and poets like Vyasa, Valmiki, Bhrigu, Sankara, Kalidasa, Bhavabhuti etc. belonged to this land.

3. एकं निबन्धप्रश्नं समाधत्त ।

अ) “मन्दविषसर्पकथा” इति पाठ्यभागस्य सारांशं लिखत ।
Answer:
Introduction: The lesson Mandavishasarpakatha is an extract from the Hitopadesa written by Narayana Pandita. This tells the story of an old serpent that devoured the frogs making friendship with them.

The old serpent: There was a serpent named Mandavisha in a ruined forest. Because of its old age, it became unable to search for food, and remained on the bank of a pond. A frog asked him why he did not take food. Then the serpent told his story.

Kapila’s advice: Once Mandavisha had bitten the son of Kaundinya, a Brahmin from Brahmapura. When the father was crying rolling on the ground, an educated Brahmin Kapila advised him not to weep like that. He said. “Death is certain for those who are born, and birth is certain for those dead. Death is certain, now or hundred years after मृत्युर्वै प्राणिनां ध्रुवः | Death approaches nearer and nearer day by day. One will not have permanent association with anything, even with his body. “On hearing that, Kaundinya decided to retire to the forests. But Kapila objected to it, saying that there would be evils even in a forest. To the detached one, the

house itself would be a hermitage. निवृत्तरागस्य गृहं तपोवनम् | He said that there was pain only, and not happiness in the world.
Kapila advised Kaundinya to renounce all the associations. If it was not possible, he should associate with the good only. Hi सङ्गो हि भेषजम् | Then the Brahmin cursed the serpent to become a vehicle of the frogs. Hence, Mandavisha arrived there.

Killing the frogs: When the frog informed this to their king, he came and rode on the back of the serpent for some time. The next day, the frog-king permitted the serpent to have the frogs as his food, as he was weak. Then the serpent killed all the frogs · one by one, and at last ate the frog-king too.

क्रमशः खादितवान् । मण्डूकनाथमपि भक्षितवान् ।
जातस्य हि ध्रुवो मृत्युः
थ्रुवं जन्म मृतस्य चः
अपि स्वेन शरीरेण
किमुतान्येन केननित् ।
निवृत्तरागस्य गृहं तपोवनम्
सतां सङ्गो हि भेषजाम्
स्कन्धेनापि कहेत् शत्रून्
कार्यमासाद्य बुहिमान्
जातस्य हि ध्रुवो मृत्युः
ध्रुवं जन्म मृतस्य च
संभोगो हि वियोगस्य
संच्वचयति सम्भवम्

आ) अप्पय्यदीक्षितेन्द्र इति पाठ्यभागसारांशं संग्रहेण विवृणुत ?
Ans. Introduction: The lesson Appayyadikshitendra was written by Sanka Usharani. This lesson narrates the story of the great scholar and devotee Appayya Dikshita.

Family of scholars: Appayya Dikshita was born in the family of scholars. His father was Rangarajadhwari was a genius. His grandfather was in the court of Acchana Dikshita. Appayya was born in 1554 in Virinnchipura in Tamilnadu.

Appayya was a poet, philosopher, scholar, devotee, Grammarian, Mimamsaka, Logician, Ritualist etc. He studied all the fourteen branches of knowledge at the feet of his teacher Nrisimha Ramaswamy.

Marriage and works: A scholar named Ratnakheta Srinivasa wanted to defeat Appayya with the grace of the goddess Kamakshi. However, the goddess advised him to give his daughter in marriage to Appayya. अप्पय्यः असाधारणः वादे जेतुमशक्तः | He did so. Appayyà had two daughters and three son.
Many students came to him to receive education. Bhattoji Dikshita came to him to learn Mimamsa and Vedanta.

Appayya wrote 104 works on many subjects like Vyakarana, Vedanta, Alankara etc. Among the Vedanta works Siddhanta leasangraha and Parimala vyakhya were famous. Sivakarna mritam and Sivarchanachandrika were his famous works devoted to Siva. Kuvalayananda and Chitramimamsa were Alankara works.
Appayya was invited by Vellore Chinabommanayaka to adorn his court. Appayya was epileptic. But when the king came to his house to observe his disease, Appayya transferred his disease on to his upper garment. Dikshita means one who performed sacrifices. Appayya performed more than 104 sacrifices including Somayaga, Vajapeya etc. He even performed Viswajit. Once while he was performing a sacrifice, the king arrived there and offered him clothes, ornaments etc. Which he immediately put in the fire. The fire god appeared wearing them.

Devotion of Siva: Appayya was a devotee of Siva. To test his devotion, he ate dattura leaves and became mad. In that state, he recited Unmattapanchasat. The king honoured him with a shower of gold coins when Appayya completed his Sivarkamanidipika. राजा चं कनकाभिषेकेण सत्कृतवान् | Appayya built a temple for Siva with that gold in his native village. Though Appayya was a Tamilian he used Telugu idioms and proverbs. He praised Telugu language thus:

Being born in Andhra, knowing the language of Telugu, studying Prabhakara’s Mimamsa work, and belonging to the Yajurveda School are the fruit of not less penance.

Though the householder Appayya was in the path of action, his mind was in the path of detachment only. He spent his last days in the presence of Nataraja of Chidambara. He became one with Siva at the age of 72 in the year of 16261 He was considered the incarnation of Siva Himself. अप्पय्य: साक्षात् परमशिवस्य अवतारः ।

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

4. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) कर्णभारं केन विरचितम् ?
समाधान:
कर्णभारं भासेन विरचितम् ।

आ) नृपश्रिय: कीदृशा: ?
समाधान:
नृपश्रियः भुजङ्गजिह्वाचपला ।

इ) बालानां कलहः कीदृश: ?
समाधान:
बालानां कलहः तत्कालीनः भवति । बालाः कलहायन्ते अनन्तरं क्रीडन्ते च ।

ई) द्रोणडुपदयोः सम्बन्धः कीदृश: ?
समाधान:
द्रोणद्रुपदयोः स्नेहबन्धः अच्छेद्यः अविभाज्यः च । सः पटुचिक्कणात्मकः अभवत् ।

उ) कालिदासश्लोके भोजेन प्राप्तं समाधानं किम ?
समाधान:
कालिदासश्लोकेन भोजेन प्राप्तं समाधानम् एवम् अस्ति द्वयोः सम्भाषणकर्त्रीः मध्ये सूचनां विना न प्रविशेत् इति ।

ऊ) कालिदासः महाराजं कथं सन्तोषितवान् ?
समाधान:
भोजस्य मनोगतस्य यथोचितम् उत्तरं दत्वा कालिदासः भोजं संतोषितवान् ।

5. द्वयोः ससंदर्भ व्याख्यानं लिखत ।

अ) वदेत् क्षमं स्वामिहितं मन्त्री ।
समाधान:
परिचयः एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् । एषः पाठः रामायणस्यं युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
भावः परबलं, स्वबलं तथा क्षयं वृद्धिं च बुद्ध्या समीक्ष्य मन्त्री स्वामिहितं वदेत् ।

आ) राज्येन किं तद्विपरीतवृत्तेः प्राणैः उपक्रोशमलीमसैर्वाः ।
समाधान:
परिचयः एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषःभागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः दिलीपः सिरं प्रति एवं वदति ।
भावः अक्षत्रविरुद्धवृत्तेः राज्येन किं प्रयोजनम् निन्दया मलिनयुतं प्राणेन किं प्रयोजनम् ।
‘विवरणम्ः क्षत्रियः राज्यं पालयति, स्वपराक्रमेन धर्म पालयति, आक्षितान् रक्षति ।

इ) विद्या यत्र परोपकारविभवाः सा ज्ञान भूमिर्मम ।
समाधान:
परिचयः – एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् । अस्य कविः श्रीमान् दोर्बल प्रभाकर शर्मा ।
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः, धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः – यत्र वेदाः उपनिषदः, पुराणेतिहासाः आयुर्योगधर्मसंगीतग्रन्थाः च परोपकारविभवाः भवन्ति सा ज्ञानभूमिः मम मातृभूमिः इति कविः वदति ।

ई) सा संस्कृताख्या सुकृतैकलभ्या ।
समाधान:
परिचय : गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री |
सन्दर्भ : गीर्वाणवाण्याः वैभवं वर्णयनन् कविः एवं वदति । संस्कृतभाषा सुधास्रवन्ती, सुरभाषिता, सूच्चारिता, सूक्तिरन्नवार्धिः च ।
भाव : संस्कृतभाषा पुण्यैफलेन एव लभ्या ।
विवरणम् : संस्कृतभाषा अतिप्राचीना, सुधास्रवन्ती च सूक्तिरत्नै सुशोभिता ।

6. द्वयोः ससंदर्भ व्याख्यानं लिखत ।

अ) मृत्युर्वै प्राणिनां ध्रुवः ।
समाधान:
परिचयः एतत-वाक्यं, “मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः पुत्रशोकं अनुभवन्तं कौण्डिन्यं कपिलः प्रति एवं अवदत् भावः सर्वेषां प्राणिनां मृत्युः निश्चयम्
विवरणम्ः जातस्य मरणं, मृतस्य जननं निश्चयम्, अतः सर्वेषां प्राणिनां मृत्युः ध्रुवम् ।

आ) यः कृतमुपकारं विस्मरति स एव पुरुषाधमः ।
समाधान:
परिचय : इदं वाक्यं “विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास ।
सन्दर्भ : राज्ञः वचनं क्षुत्वा एकः एवं अवदत् ।
भाव : यः कृतं उपकारं विस्मरति सः मानवः पुरुषाधमः भवति ।
विवरणम् : नारिकेल वृक्षः पीतं तोयं स्मृत्वा नरागां अमृततुल्यं जलं ददाति यः मानवः परैः कृतं उपकारं विस्मरति सः मानवः पुरुषाधमः भवति ।

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

इ) विवाहो नाम वणिग्व्यवहार एव ।
समाधान:
परिचयः वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः एकः वृद्धः युवकं उद्दिश्य एवं अवदत् ।
भावः परिणयः नाम केवलं वाणिज्यव्यवहारः ।
विवरणम्ः सर्वस्य धनं प्रधानं, सर्वेगुणाः धनं आश्रयन्ति । तद्दत् विवाहः अपि केवलं वाणिज्यमेव ।

ई) राजा तं कनकाभिषेकेण सत्कृतवान् ।
समाधान:
परिचय : एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भ : अप्पय्यदीक्षितः शिवार्कमणिदीपिकां अरचयत् । तदा राजा चिनबोकनायकः तं कनकाभिषेकेण सत्कृतवान् । तेन हिरण्येन अप्पय्यः स्वग्रामे शिवालयं निर्मितवान् ।
भाव : राजा अप्पय्यदीक्षितं कनकाभिषेकं कृत्वा सम्मानितवान् ।

7. त्रयाणां प्रश्नांनां समाधानानि लिखत ।

अ) नराः क इव रणे सीदन्ति ?
समाधान:
शूराः अपि नराः वालुकासेतवः इव रणे सीदन्ति ।

आ) सुदक्षिणादिलीपौ किमर्थं नन्दिनीधेनोः सेवाम् अकरुताम् ।
समाधान:
सुदक्षिमादिलीपौ सन्तानार्थं – नन्दिनीधेनोः सेवाम् अकुरुताम् ।

इ) व्याधितस्य मित्रं किम् ?
समाधान:
व्याधितस्य औषधं मित्रम् ।

ई) अस्पृशनेव क्तिानि कः परेभ्यः प्रयच्छति ?
समाधान:
अस्पृशन्नेव वित्तानि कृपणः परेभ्यः प्रयच्छति ।

उं) सर्वप्राणिनामुपकारकारि किम् ?
समाधान:
सर्वप्राणइनाम् उपकारकारि सुमहत् आर्षविज्ञानामृतम् ।

ऊ) संस्कृताव्या वाणी कीदृशी ?
समाधान:
संस्कृताख्या वाणी सुधासवन्ती, सूक्तिसुरत्नर्वार्धः, सुकाव्यसंदोहनिधिः ।

8. त्रयाणां प्रश्नानां समाधानं लिखत |

अ) कौण्डिन्येन मन्दविषः किम् इति शप्तः ?
समाधान:
कौण्डिन्येन मन्दविषः – “अद्य आरभ्य मण्ङ्गकानां वाहनं भविष्यसि” इति शप्तः ।

आ) ब्राह्मणः पुत्रप्राप्त्यर्थं कीदृशं व्रतमाचरितवान् ?
समाधान:
ब्राह्मणः पुत्रप्रास्यर्थं प्रदोषव्रतमाचरितवान् । मार्गशीर्षशुद्धत्रयोदश्यां मन्दवासरे कल्पोक्तप्रकारेण सः व्रतमाचरितवान् ।

इ) राजहंसः कः ? तस्य नगरी का ?
समाधान:
राजहंसः मगधदेशस्य राजा । तस्य नगरी पुष्पपुरी ।

ई) वैद्यः श्रीधरं प्रति किमुक्तवान् ?
समाधान:
अधिककार्यभारेण हृद्रोगः प्राप्तः कञ्चित्कालं विश्रान्तिरपेक्षिता ! नियतरूपेण औषधसेवनं करणीयम् – इति वैद्यः श्रीधरम् उक्तवान् |

उ) मञ्जुहासिनी का ?
समाधान:
मञ्जुहासिनी आत्मनः प्रेयसी ।

ऊ) रत्नखेटिश्रीनिवासस्य अभिलाषा का ?
समाधान:
रत्नखे श्रीनिवासस्य अभिलाषा – अप्पय्यः जेतव्यः, स्वपादयोः पातनीयः च इति ।

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

9. एकेन वाक्येन समाधानं दत्त ।

अ) केषां बलानि अमेयानि ?
समाधान:
परेषां बलानि अमेयानि ।

आ) रघुवंशमहाकाव्यं केन विरचितम् ?
समाधान:
कालिदासेन

इ) यावज्जीवं कः दहेत् ?
समाधान:
यावज्जीवं जडः दहेत् ।

ई) पापहरा नदी का ?
समाधान:
पापहरा नदी गङ्गा ।

उ) सर्वासु सुतासु कस्याः गुणाः भवन्ति ?
समाधान:
मातुः

10. एकेन वाक्येन समाधानं दत्त ।

अ) कौडिन्यस्य पुत्रः कः ?
समाधान:
सुशीलः

आ) विक्रमार्कचरितं केन विरचितम् ?
समाधान:
शिवदासेन

इ) राजहंसस्य पत्नी का ?
समाधान:
वसुमती

ई) अपर्याप्तः समयः इति पाठ्यभागः केन विरचितः ?
समाधान:
अपर्याप्तः समयः इति पाठ्यभागः राणी सदाशिवमूर्तिना विरचितः ।

उ) वेङ्कटरावस्य पत्नी का ?
समाधान:
सुशीला

11. अधो निर्दिष्ट कथां पठित्वा प्रश्नानां समाधानानि दत्त।

अस्ति मगधदेशे चम्पकवतीनाम अरण्यानी। तस्यां चिरात् महता स्नेहेन मृगकाको निवसतः। स च मृगः स्वेच्छया भ्राम्यन् हृष्टपुष्टाड़ः केनचित् शृगालेन अवलोकितः। तं दृष्ट्वा शृगालोऽचिन्तयत् – आह – ‘कथम् एतन्मासं सुललितं भक्षयामि ? भवतु। विश्वासं तावदुत्पादयामि” इत्यालोच्य उपसृत्य अब्रवीत् – “मित्र ! कुशलं ते” इति। मृगेणोक्तम् – “कस्त्वम्” ? इति । स बूले “क्षुद्रबुद्धिर्नाम जम्बुकोऽहम् । अत्र आरण्ये बन्धुहीने मृतवत् एकाकी निवसामि । इदानीं त्वां मित्रमासाद्य पुनः सबन्धुर्जीवलोकं प्रविष्टोऽस्मि। अघुना तवानुचरेण मयां भवितव्यम्”। मृगेणोक्तम् – “एवमस्तु” इति ।
प्रश्नाः
(1) चम्पकवतीनाम अरण्यानी कुत्र अस्ति ?
समाधान:
चम्पकवतीनाम अरण्यानी मगधदेशे अस्ति ।

(2) मृगकाकौ कथं निवसतः ?
समाधान:
मृगकाकौ महता स्नेहेन निवसतः ।

(3) मृगः केन अवलोकितः ?
समाधान:
मृगः केनचित् शृगालेन अवलोकितः ।

(4) जम्बुकस्य नाम किम् ?
समाधान:
जम्बुकस्य नाम क्षुद्रबुद्धिः

(5) शृगालः मृगं दृष्ट्वा किमचिन्तयत् ?
समाधान:
शृगालः मृगं दृष्ट्वा “कथं एतन्मासं सुललितं भक्षयामि ? भवतु । विश्वासं तावदुत्पादयामि” इति अचिन्तयत ।

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त ।

अ) तत् + टीका
समाधान:
तट्टीका = ष्टुत्व सन्धिः

आ) पेष् + टा
समाधान:
पेष्टा = ष्टुत्व सन्धिः

इ) अच् + अन्तः
समाधान:
अचन्तः = जश्त्व सन्धिः

ई) वाक् + मयम्
समाधान:
वाङ्मयम = अनुनासिक सन्धिः

उ) रामः + अपि
समाधान:
रामोऽपि = विसर्ग सन्धिः

ऊ) कविः + आयाति
समाधान:
कविरायाति = विसर्गरेफादेश सन्धिः

13. नामनिर्देशपूर्वकं त्रीणि विघटयत ।

अ) रामष्टीक
समाधान:
रामस् + टीकते = ष्टुत्व सन्धिः

आ) मनश्चलति
समाधान:
मनस् + चलति = श्चुत्व सन्धिः

इ) जगन्नाथः
समाधान:
जगत् + नायः = अनुनासिक सन्धिः

ई) षडाननः
समा. षट् + आननः = जश्त्व सन्धिः

उ) शिवोऽहम्
समाधान:
शिवः + अहम् = विसर्ग सन्धिः

ऊ) नृपतिर्जयति
समाधान:
नृपति + जयति = विसर्गरेफादेश सन्धिः

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत ।
अ) जलमुक्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions 2

आ) पचत्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions 3

इ) मनस्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions 4

ई) यद् (पुं)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions 5

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत ।

अ) ईश्वराधीनम्
समाधान:
ईश्वरे अधीनम् = सप्तमीतत्पुरुष समासः

आ) मयूरव्यंसकाः
समाधान:
व्यंसकाः च ते मयूराः च = विशेषणपूर्वपदकर्मधारय समासः

इ) नवरात्रम्
समाधान:
नवानां रात्रीनां समाहारः = द्विगु समासः

ई) रामलक्ष्मणौ
समाधान:
रामः च लक्ष्मणः च = द्वन्द्व समासः

उ) प्रत्यक्षम्
समाधान:
अक्ष्णोः समीपे = अव्ययीभाव समासः

ऊ) नीलोज्वलवपुः
समाधान:
नीलं उज्वलं वपुः यस्य सः = बहुपदबहुव्रीहि समासः

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions

16. अधो निर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखत ।

AP Inter 2nd Year Sanskrit Model Paper Set 5 with Solutions 1
अ) प्रश्नाः
समाधान:
1. रमेशः किं रकोति ?
2. गोविन्दः किमर्थं वसति ?
3. त्वं कुत्र पिबसि ?
4. रामः कदा गरछति ?
5. अहं किं पठामि ?.

AP Inter 1st Year Maths 1B Question Paper April 2022

Access to a variety of AP Inter 1st Year Maths 1B Model Papers and AP Inter 1st Year Maths 1B Question Paper April 2022 allows students to familiarize themselves with different question patterns.

AP Inter 1st Year Maths 1B Question Paper April 2022

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

  1. Answer all questions.
  2. Each question carries two marks.

1. Transform the equation 3x + 4y + 12 = 0 into

  1. Slope-intercept form
  2. intercept form

2. Find the value of p, if the straight lines x + P = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

3. If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of a tetrahedron, find the fourth vertex.

AP Inter 1st Year Maths 1B Question Paper April 2022

4. Find the equation of the plane whose intercepts on X, Y, Z- axes are 1, 2, 4 respectively.

5. Compute \(\lim _{x \rightarrow 0} \frac{e^x-1}{\sqrt{1+x}-1}\)

6. Compute \(\lim _{x \rightarrow \infty} \frac{8|x|+3 x}{3|x|-2 x}\)

7. If f(x) = sin (log x), (x > 0) find f(x).

8. Find the derivative of cos (log x + ex).

9. Find the slope of the normal to the curve x = a cos3θ, y = a sin3θ at θ = \(\frac{\pi}{4}\).

10. Find the intervals on which f(x) = x2 – 3x + 8 is increasing or decreasing.

Section – B

II. Short answer type questions :

  1. Attempt any five questions.
  2. Each question carries four marks.

11. If the distance from P to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of the locus of P.

12. A(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9.

13. A straight line with slope 1 passes through Q (-3, 5) and meets the straight line x + y – 6 = 0 at P, Find the distance PQ.

AP Inter 1st Year Maths 1B Question Paper April 2022

14. If f given by
AP Inter 1st Year Maths 1B Question Paper April 2022 1
is a continuous function on R, then find the values of k.

15. Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) for the function
AP Inter 1st Year Maths 1B Question Paper April 2022 2

16. Find the angle between the curve 2y = \(e^{\frac{-x}{2}}\) and Y-axis.

17. Show that \(\frac{x}{1+x}\) < In (1 + x)< x, ∀ x > 0.

Section – C
(5 × 7 = 35)

III. Long answer type questions :

  1. Attempt any five questions.
  2. Each question carries seven marks.

18. If p and q are the lengths of the perpendiculars from the origin to the straight lines x secα + y cosecα = a and x cosα – y sinα = a cos2α, prove that 4p2 + q2 = a2.

19. Show that the pairs of straight lines 6x2 – 5xy – 6y2 = 0 and 6x2 – 5xy – 6y2 + x + 5y – 1 = 0 form a square.

20. Find the values of k if the lines joining the origin to the points of intersection of the curve 2x2 – 2xy + 3y2 + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular.

21. Find the direction cosines of two lines which are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0.

AP Inter 1st Year Maths 1B Question Paper April 2022

22. If xy = yy then
AP Inter 1st Year Maths 1B Question Paper April 2022 3

23. Show that the tangent at P(x1, y1) on the curve
= \(\sqrt{x}+\sqrt{y}\) = \(\sqrt{a}\) is
\(y y_1{ }^{\frac{-1}{2}}\) = \(a^{\frac{1}{2}}\)

24. The profit function P(x) of a company, selling x items per day is given by P(x) = (150 – x) x – 1600. Find the number of items that the company should sell for maximum profit. Also find the maximum profit.

TS Inter 1st Year Chemistry Question Paper May 2017

Access to a variety of TS Inter 1st Year Chemistry Model Papers and TS Inter 1st Year Chemistry Question Paper May 2017 helps students overcome exam anxiety by fostering familiarity.

TS Inter 1st Year Chemistry Question Paper May 2017

Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions in Section ‘B’ and any two questions in Section ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
  5. Draw labelled diagrams wherever necessary for questions in Sections ‘3’ and ‘C’.

Section – A

Note : Answer all questions.

Question 1.
Why is gypsum added to cement?
Answer:
Gypsum is added to cement to slow down the process of setting of the cement and to get sufficiently hardened cement.

Question 2.
Name any two man-made silicates.
Answer:
Cement and glass are man-made silicates.

Question 3.
What is Lewis acid? Give one example.
Answer:
Lewis acid: The substance which accepts electron pair and forms dative bond is called Lewis acid. E.g. :AlCl3, FeCl3, Ag+ etc.

Question 4.
Calcualte kinetic energy of 2 moles of Nitrogen at 27°C.
Answer:
Kinetic energy (KE) = \(\frac{3}{2}\) nRT
= \(\frac{3}{2}\) × 2 × 2 × 300 K
K = 1800 Cal.

TS Inter 1st Year Chemistry Question Paper May 2017

Question 5.
Give an account of the biological importance of Na+ ions.
Answer:
Biological importance of Na+ :

  1. Na+ ions participate in the transmission of nerve signals.
  2. Na+ ions regulates the flow of water across cell membranes.
  3. Na+ ions responsible for transport of sugars and amino acids into cells.

Question 6.
Assign oxidation number to the underlined elements in
(a) NaHSO4
(b) KMnO4
Answer:
(a) NaHSO4
+ 1′ + 1 + x + 4(-2) = 0
2 + x – 8 = 0
x = + 6

(b) KMnO4
1 + x + 4(-2) = 0
1 + x – 8 = 0
x – 7 = 0
x = + 7

Question 7.
How does graphite function as a lubricant?
Answer:
Graphite is soft and it has layer lattice. In this structure one of the layer slided over another due to weak Vander Waal’s force of attraction. These layers are slippery. Hence it is greasy and function as lubricant. .

Question 8.
Give the possible BOD values of clean water and the polluted water.
Answer:
BOD value of pure water is 1 ppm. Fairly Pure water is 3 ppm and doubtful purity 5 ppm.
BOD of municipal sewage is 100 – 4000 ppm.

Question 9.
What is PAN? What effect is caused by it?
Answer:

  1. Peroxy Acetyl Nitrate is called as PAN.
  2. Peroxy Acetyl Nitrate is (PAN) is a powerful eye irritant.

TS Inter 1st Year Chemistry Question Paper May 2017

Question 10.
Give two examples for position isomerism.
Answer:
Position isomerism : Compounds having same molecular formula but they differ in position of substituent or functional group (or) multiple bond are called position isomers. This is position isomerism.
TS Inter 1st Year Chemistry Question Paper May 2017 - 1

Section – B

Question 11.
Explain the hybridization involved in SF6 molecule.
Answer:
In this hybridisation one ‘s’ orbital, three ‘p’ orbitals and two ‘d’ orbitals of the excited atom combine to form six equivalent sp3d2 hybrid orbitals, e.g. : SF6
TS Inter 1st Year Chemistry Question Paper May 2017 - 2

These six sp3d2 hybrid orbital six 2pz orbitals of fluoring atoms to form six σsp3d2 bonds. The directions of the bonds give an octa-hedral shape to the molecule. The bond angle is 90° or 180° & 90°.

Question 12.
Deduce (a) Boyle’s law and (b) Charle’s law from kinetic gas equation.
Answer:
a) Deduction of Boyle’s law :
Kinetic gas equation is
TS Inter 1st Year Chemistry Question Paper May 2017 - 3
Hence Boyle’s law proved from kinetic gas equation.

b) Deduction of Charle’s law :
Kinetic gas equation is PV = \(\frac{1}{3}\) mnu2
TS Inter 1st Year Chemistry Question Paper May 2017 - 4
Hence Charle’s law proved from kinetic gas equation.

Question 13.
A carbon compound contains 12.8% carbon, 2.1 % Hydrogen, 85.1% Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.
Answer:
TS Inter 1st Year Chemistry Question Paper May 2017 - 5
∴ Emperical formula of the compound = C1 H2 Br
Molecular formula = n (Emperical formula)
n = \(\frac{\text { Molecular wt }}{\text { Emperical wt }}=\frac{187.9}{94}=2\)
Given molecular wt = 187. 9
∴ Molecular formula = 2 (CH2Br)
= C2H4Br2
Given molecular wt = 187. 9
Emperical wt = 94 (CH4 Br)

TS Inter 1st Year Chemistry Question Paper May 2017

Question 14.
State and explain the Hess’ law of constant heat summation.
Answer:
Hess’s Law :
Hess’s law states that the total arr\ount of heat evolved or absorbed in a chemical reaction is always same whether the reaction is carried out in one step (or) in several steps.

Illustration :
This means that the heat of reaction depends only on the initial and final stages and not on the intermediate stages through which the reaction is carried out. Let us consider a reaction in which A gives D. The reaction is brought out in one step and let the heat of reaction be ΔH.
A → D; ΔH
Suppose the same reaction is brought out in three stages as follows
A → B : ΔH1
B → C : ΔH2
C → D : ΔH3
TS Inter 1st Year Chemistry Question Paper May 2017 - 6
The net heat of reaction is ΔH1 + ΔH2+ ΔH3.
According to Hess law ΔH = ΔH1 + ΔH2 + ΔH3.
Ex : Consider the formation of CO2 It can be prepared in two ways.
1) Direct method: By heating carbon in excess of O2.
C(s) + O2(g) → CO2(g); ΔH = – 393.5 kJ

2) Indirect method: Carbon can be converted into CO2 in the following two steps.
C(s) + \(\frac{1}{2}\) O2(g) → CO2(g); ΔH1 = – 110.5 kJ
CO (g) + \(\frac{1}{2}\) O2(g) → CO2(g); ΔH2 = – 283.02 kJ
Total ΔH = -393.52 kJ
The two ΔH values are same.

Question 15.
Discuss the application of Le-Chatelier’s principle for the industrial synthesis of Ammonia.
Answer:
Applications of Le Chatelier’s principle to synthesis of Ammonia by Haber’s process :
TS Inter 1st Year Chemistry Question Paper May 2017 - 7
Nitrogen and Hydrogen combine to form ammonia. The formation of ammonia is reversible and exothermic reaction. It is accompanied by decrease in volume.

Effect of Pressure :
1 volume of N2 combines with 3 volumes of H2 to form 2 volumes of NH3. There is decrease in volume in the forward reaction (4 volumes to 2 volumes). According to Le Chatelier’s principle increase of pressure favours the reaction where there is decrease in volume. So higher the pressure, greater the yield of ammonia. In practice 200 atmospheres are used in the manufacture of ammonia by Haber’s process. Low pressures favour the reverse reaction i.e., decomposition of NH3 already formed.

Effect of Temperature : The formation of ammonia (forward reaction) is exothermic reaction
N2 + 3 H2 → 2 NH3; ΔH = -92 kj

Low temperatures favour the forward reaction. But at low tempera¬tures the reaction is too slow. Therefore an optimum temperature (725 K – 775 K) is chosen in Haber’s process. To speed up the reaction, a catalyst, finely divided iron is used. To increase the activity of the catalyst molybdenum or a mixture of oxides of K and M is used as promoter.

The reverse reaction (i.e.,) decomposition of NH3 is an endo¬thermic reaction. High temperatures favour the decomposition of NH3. Therefore high temperatures are avoided in Haber’s process.
Thus the optimum conditions are

Pressure : 200 atm
Temperature : 725 – 775 k
Catalyst : Fe (Powered)
Promoter : Mo (or) (K2O + Al2O3)

Question 16.
Write a few lines on the utility of Hydrogen as a fuel.
Answer:
Hydrogen as a fuel :

  1. The heat of combustion of hydrogen is high i.e about 242kj/ mole. Hence hydrogen is used as industrial fuel.
  2. The energy released by the combustion of dihydrogen is more , than the petrol (3 times).
  3. Hydrogen is major constituent in fuel gases like coal gas and water gas.
  4. Hydrogen is also used in fuel cells for the generation of electric power.
  5. 5% dihydrogen is used in CNG for running four-wheeler vehicles.
  6. By hydrogen economy principle the storage and transportation of energy in the form of liquid (or) gaseous state.
  7. Here energy is transmitted in the form of dihydrogen and not as electric power.

TS Inter 1st Year Chemistry Question Paper May 2017

Question 17.
Explain borax bead test with a suitable example.
Answer:
Borax bead test: This test is useful in the identification of basic radicals in qualitative analysis. On heating borax swells into a white, opaque mass of anhydrous sodium tetra borate. When it is fused, bo¬rax glass is obtained. Borax glass is sodium meta borate and B2O3. The boric anhydride, B2O3, combined with metal oxides to form metal metaborates as coloured beads. The reactions are as follows :
TS Inter 1st Year Chemistry Question Paper May 2017 - 8

Question 18.
Explain the formation of co-ordinate covalent bond with one example.
Answer:
Co-ordinate covalent bond (dative bond) is a special type of co¬valent bond. It is proposed by Sidgwick. It is formed by the sharing of electrons between two atoms in which both the electrons of the shared electron pair are contributed by one atom and the other atom nearly participates in sharing. The bond is represented as (“→”) an arrow starting from the donar atom and directed towards the acceptor atom.

Exmples.
1) Ammonia – Boron trifluoride H3N : → BF3
Ammonia combines with boron trifluoride to give ammonium boron trifluoride.
TS Inter 1st Year Chemistry Question Paper May 2017 - 9

In ammonia nitrogen has a complete octet and also it has a lone pair of electrons. In BF3 the boron atom has a total of six electrons after sharing with fluorine. Nitrogen donates the electron pair to boron to form a co-ordinate covalent bond between ammonia and boron trifluoride.
TS Inter 1st Year Chemistry Question Paper May 2017 - 10

Properties of co-ordinate covalent bond :

  1. The bond do not ionise in water.
  2. The compounds are generally soluble in organic solvents and are sparingly soluble in water.
  3. These compounds exhibit space isomerism because the bond is rigid and directional.
  4. The bond is semipoiar in nature – so their volatility lies in between covalent and ionic bonds.

Section – C

Question 19.
How are the quantum numbers n, l and m arrived at? Explain the significance of these quantum numbers.
Answer:

  • In general a large no.of orbitals are possible in an atom.
  • These orbitals are distinguished by their size, shape and orientation.
  • An orbital of smaller size means there is more chance to find electron near the nucleus.
  • Atomic orbitals are precisely distinguished by quantum numbers. Each orbital is designated by three major quantum numbers.
    1. Principal quantum number (n)
    2. Azimuthal quantum number (l)
    3. Magnetic quantum number (m)

1) Principal quantum number :
The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the distance between the nucleus and the orbit also increases.
It is denoted by the letter ‘n’. It can have any simple integer value 1,2, 3, but not zero. These are also termed as K, L, M, N etc. The radius and energy of an orbit can be determined basing on “n” value.

The radius of n orbit is rn = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The radius of n orbit is En = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number: It was proposed by Sommerfeld. It is also known as angular momentum quantum number or subsidiary quantum number. It indicates the shapes of orbitals. It is denoted by T. The values of depend on the values of ‘n1, T has values ranging from ‘0’ to (n – 1) i.e., I = 0, 1, 2,……(n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10, 14 respectively.

Subshell l value Shape
s 1 = O spherically symmetric
P 1 = 1 dumb – bell
d l = 2 double dumb-bell
f l = 3 four fold dumb-bell

Energy levels and subshells

Principal quantum number(n) Azimuthal quantum number(1) Symbol Number of subshells
1 0 s 1 (1s)
2 0 s 2 (2s, 2p)
1 P
3 0 s 3 (3s, 3p, 3d)
1 p
2 d
4 0 s 4 (4s, 4p, 4d, 4f)
1 P
2 d
3 f

3) Magnetic quantum number:
It was proposed by Lande. It shows the orientation of the orbitals in space. ‘p’ orbital has three orientations. The orbital oriented along the x-axis is called px orbital, along the y-axis is called py – orbital and along the z-axis is called pz orbital. In a similar way d – orbital has five orientations. They are dxy , dyz, dzx, dx2-y2 and dz2 it is denoted by ‘m’. Its values depends on azimuthal quantum number, ‘m‘ can have all the integral values from -l to +1 including zero. The total number of ‘nT values are (21 + 1).

Subshell l value m value -l to 0 to +l No.orbitals = 2l + 1
s 0 0 1
p 1 -1, 0, +1 3
d 2 -2, -1,0, +1, +2 5
f 3 -3, -2, -1, 0, +1, +2, +3 7

TS Inter 1st Year Chemistry Question Paper May 2017 - 11

Question 20.
Define IE1 and IE2. Why is IE2 > IE1 for a given atom? Discuss four factors that effect IE of an element.
Answer:
1) Ionization energy is the amount of energy required to remove the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom.

It is denoted as I., and is expressed in electron volts per atom, kilo calories (or) kilo joules per mole.
M(g) + I1 → M(g)+ + e+
I1 is first ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I2.
M+(g) + l2 → M(g)2+ + e

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential.
For sodium, I1 is 5.1 eV and I2 is 47.3 eV.

I1 is 5.1 ev and I2 is 47.3 eV
I1 < I2 < I3 …… In

Factors affecting ionization potential :
1) Atomic radius:
As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2) Nuclear charge:
As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3) Screening effect or shielding effect:
In multi electron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. Jhe electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of’electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.

(Magnitude of screening effect) ∝ \(\frac{1}{\text { (Ionization enthalpy) }}\)

TREND IN A GROUP:
The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.

TREND IN A PERIOD:
In a period from left to right I.P. value increases as the size of the elements decreases along the period.

TS Inter 1st Year Chemistry Question Paper May 2017

Question 21.
How do we get Benzene from acetylene? Give the corresponding. equation. Explain the halogenation, alkylation and acylation of benzene with equations.
Answer:
Preparation of Benzene from acetylene : On passing acetylene gas through red hot iron tubes, it trimerises to give benzene.
TS Inter 1st Year Chemistry Question Paper May 2017 - 12

Chemical Properties:
(l)Halogenation: Benzene reacts with chlorine in the presence of FeCl3 or AlC3 to give chloro-benzene.
TS Inter 1st Year Chemistry Question Paper May 2017 - 13

(2) ‘Friedel’- Craft’s Alkylation: Benzene reacts with alkyl halides in the presence of AlCl3 to give alkyl benzene.
TS Inter 1st Year Chemistry Question Paper May 2017 - 14

(3) Friedel – Craft’s Acylation: Benzene reacts with acetyl ‘chloride in the presence of AIC13 to give acetophenone.
TS Inter 1st Year Chemistry Question Paper May 2017 - 15

(4) Nitration : Benzene when heated with a mixture of cone. H2SO4 and cone. HNO3 below 60°C to give Nitrobenzene.
TS Inter 1st Year Chemistry Question Paper May 2017 - 16

(5) Sulphonation : With fuming H2SO4, benzene, reacts to form benzene sulphonic acid
TS Inter 1st Year Chemistry Question Paper May 2017 - 17

TS Inter 1st Year Chemistry Question Paper March 2017

Access to a variety of TS Inter 1st Year Chemistry Model Papers and TS Inter 1st Year Chemistry Question Paper March 2017 helps students overcome exam anxiety by fostering familiarity.

TS Inter 1st Year Chemistry Question Paper March 2017

Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions in Section ‘B’ and any two questions in Section ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
  5. Draw labelled diagrams wherever necessary for questions in Sections ‘3’ and ‘C’.

Section – A

Note : Answer all questions.

Question 1.
Name two adverse effects caused by acid rains.
Answer:
Adverse effects caused by Acid rains :

  1. These cause respiratory ailments in human beings and animals.
  2. These damage the-old buildings and historical monuments like Tajmahal.
  3. These effect the plants and animal life in aquatic ecosystem.

Question 2.
What is Chemical Oxygen Demand (COD)?
Answer:
Chemical Oxygen Demand (COD) : The amount of oxygen required to oxidise the organic substances present in polluted water is called Chemical Oxygen Demand (COD).

TS Inter 1st Year Chemistry Question Paper March 2017

Question 3.
Write the functional isomers of organic compound C3H6O.
Answer: Functional isomers of organic compound C3H60 are
CH3 – CH2 – CHO 1-propana
CH3 – CO – CH3 propan 2-one

Question 4.
Define inert pair effect.
Answer:
Inert pair effect: The reluctance of ‘ns1 pair of electrons to take part in bond formation is called inertpair effect. Ex : ‘Tl’ exhibits +1, oxidation state instead of +3 due to inert pair effect.

Question 5.
Write the biological importance of Na+ ions.
Answer:
Biological importance of Na+ ions :

  1. Na+ ions participate in the transmission of nerve signals.
  2. Na+ ions responsible or transport of sugars and aminoacids into cells.
  3. Na+ ions regulates the flow of water accross cell on embrAnswer:

Question 6.
What is meant by ionic product of water? What is its value at room temperature?
Answer:
Ionic product of water : The product of H+ ion concentration and OH ion concentration in water is called ionic product of water (KW). The value of ionic product of water at room temperature is 1.008 × 10-14 mole2/lit2.

TS Inter 1st Year Chemistry Question Paper March 2017

Question 7.
Calculate the kinetic energy of 5 moles of nitrogen at 27°C.
Answer:
KE = \(\frac{3}{2}\) nRT
n = 5 moles
R = 2 cal
T = 27° C = 27 + 273 = 300 K
= \(\frac{3}{2}\) × 5 × 2 × 300
= 4500 cal.

Question 8.
Calculate the oxidation number in Cr2O72- ion on chromium.
Answer:
Cr2O72-
2x + 7 (-2) = -2
2x – 14 = -2
2x = 12
x = +6

Question 9.
What is Piaster of Paris?
Answer:
The hemi hydrate of calcium sulphate is called plaster of paris,
The formula of plaster of pans is CaSO4\(\frac{1}{2}\)H2O.

Question 10.
Give the formula of borazine. What is its common name?
Answer:
The formula of borazine is B3N3H6 (Borozole)
It’s common name is inorganic benzene. It is iso structural with benzene.
TS Inter 1st Year Chemistry Question Paper March 2017 - 1

Section – B

Question 11.
Write the postulates of Kinetic Molecular Theory of Gases.
Answer:
Assumptions :

  1. Gases are composed of minute particles called molecules. All the molecules of a gas are identical.
  2. Gaseous molecules are always, at a random movement. The molecules are moving in all possible directions in straight lines with very high velocities. They keep on colliding against each other and against the walls of the vessel at very small intervals of time.
  3. The actual volume occupied by the molecules is negligible when compared to the total volume occupied by the gas.
  4. There is no appreciable attraction or repulsion between the molecules.
  5. There is no loss of kinetic energy when the molecules collide with each other or with the wall of vessel. This is because the molecules are spherical and perfectly elastic in nature. .
  6. The pressure exerted by the gas is due to the bombardment of the molecules of the gas on the walls of the vessel.
  7. The average kinetic energy of the molecules of the gas is directly proportional to the absolute temperature, Average K.E. ∝ T.
  8. The force of gravity has no effect on the speed of gas molecules.

Boyle’s law :
According to kinetic theory of gases, the pressure of a gas is due to collisions of gas molecules on the walls of the vessel. At a particular temperature the molecules make definite number of collisions with the walls of the vessel; When the volume of the vessel is reduced the molecules have to travel lesser distance only before making collisions on the walls. As a result the number of collisions per unit increases. The pressure then increases, i.e., the pressure increases when the volume is reduced at constant temperature. This explains Boyle’s law.

Charles’ law :
According to kinetic theory of gases, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas.
K.E. ∝ T
but K.E. = \(\frac{1}{2}\) me2
As temperature increases, the velocity of the molecules also increases. As a result the molecules make more number of collisions against the walls of the vessel. This results in an increase of pressure if the volume is kept constant. If the volume is allowed to increase the number of collisions decrease due to the increased distance between the molecules and the walls of the vessel. The pressure then decreases. In other words, with rise of temperature, the volume should increase in order to keep the pressure constant. V ∝ T at constant pressure. This is Charles’ law.

TS Inter 1st Year Chemistry Question Paper March 2017

Question 12.
A carbon compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.06 gm. What are its empirical and molecular formulas?
Answer:
TS Inter 1st Year Chemistry Question Paper March 2017 - 2

Question 13.
a) State the third law of thermodynamics.
Answer:
The entropy of a pure and perfectly crystalline substance is zero at absolute zero.

b) Define entropy.
Answer:
The measure of randomness or disorder ness of a system is called entropy (s).

TS Inter 1st Year Chemistry Question Paper March 2017

Question 14.
Explain the concept of Bronsted-Lowry acid base theory with suitable examples.
Answer:
According to Bronsted theory a substance which can donate a proton to the other substance is known as acid. A substance which accept a proton from other substance is a base.
Ex : HCl + H2O ⇌ H3O+ + Cl2
NH3 + H2O ⇌ NH4+ + OH
Here HCl donates a proton to water and behaves as Bronsted Lowry acid. Similarly NH3 accepts a proton from H2O and acts as Bronsted Lowry base.

Above reaction is a reversible reaction so that H3O+ can donate proton to act as acid. Cl-1 can accept a proton to act as base. Thus each acid base reaction equilibrium involves two acids and two bases. Each pair differs by a proton, such acid base pair is called Conjugate acid base ‘pair.
TS Inter 1st Year Chemistry Question Paper March 2017 - 3

According to this theory strengths of acids and bases can be explained. An acid which show great tendency to donate protons is a strong acid and an acid which shows less tendency to donate proton is a weak acid. A base which shows great tendency to accept a proton is a strong base and a base which shows less tendency to accept a proton is a weak base.

Question 15.
a) Electron – deficient hydrides.
Answer:
These are the molecular hydrides in which the available no.of valency electrons is less than the number required for normal covalent bond formation.
(or)
These are the molecular hydrides in which the available valency electrons are lessthan the required for writting the Lewis structure of the molecule.
Eg : (AlH3)n, B2H6 etc.
These hydrides acts as Lewis acids i.e., electron pair acceptors. These forms dative bond with donors.

b) Electron – rich hydrides
Answer:
These are the molecular hydrides in which the valency electrons on the central atom are more than that are required for bond formation..
(or)
These are the molecular hydrides in which the available valency electrons are more than the required for writting the Lewis structure of the molecule.

  1. These hydrides contains lone pairs on central atoms.
    TS Inter 1st Year Chemistry Question Paper March 2017 - 4
  2. These hydrides have high boiling points than those of the hydrides of the subsequent members of group because of hydrogen bond formation.

Question 16.
Explain the differences in properties of diamond and graphite on the basis of their structures.
Answer:

Diamond Graphite
a) Each carbon is sp3 hybri- dised. a) Each carbon is sp2 hybri- dised.
b) Each carbon is bonded to 4 other carbons tetrahe- drally. b) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings. It has sheet like structure.
c) It has a 3 dimensional structure. c) It has a 2 dimensional structure.
d) C – C bond length is 1.54 A° and bond angle is 109° 28′. d) C – C bond length in hex- agonal rings is 1.42 A°and bond angle is 120°.
e) Carbon atoms are firmly held with strong covalent bonds. e) The distance between two adjacent layers is 3.35 A°. These layers are held by weak Vander Waal’s forces.
f) Diamond is very hard. f) Graphite is soft.
g) Density = 3.5 g/cc. g) Density – 2.2g/cc.
h) Diamond is an insulator due to the absence of free electrons. h) Graphite is a conductor due to the presence of free electrons.
i) It is transparent to light and X-rays. It has high refractive index (2.45). i) It has layer lattice. The layers are slippery. Hence it is greasy.

Question 17.
Explain Wurtz reaction and Friedel – Crafts alkylation with examples.
Answer:
Wurtz reactions:
Alkyl halides reacts with sodium metal in presence of dry ether to form alkanes.
TS Inter 1st Year Chemistry Question Paper March 2017 - 5

Friedal – Crafts alkylation:
Benzene reads with alkyl halides in presence of catalyst anhydrous AlC13 to from alkyl benzene.
TS Inter 1st Year Chemistry Question Paper March 2017 - 6

Question 18.
What is polymerization ? Explain with one example.
Answer:
Polymerization : A large molecular weight complex compound which is formed by the repeated combination of smaller units is called polymer. The process of formation of polymer is called polymerisation. Ex : Ethylene undergo polymerisation at 200°C and 1500 – 200 atm pressure gives polythene.
TS Inter 1st Year Chemistry Question Paper March 2017 - 7

Section – C

Question 19.
a) What are the postulates of Bohr’s model of hydrogen atom?
Answer:
Postulates :

  1. The electron in the hydrogen atom can revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits (or) stationary states. These circular orbits are concentric (having same center) around the nucleus.
  2. The energy of an electron in the orbit does not change with time. .
  3. When an electron moves from lower stationary state to higher stationary state absorption of energy takes place.
  4. When an electron moves from higher stationary state to lower . – stationary state emission of energy takes place.
  5. When an electronic transition takes place between two sta¬tionary states that differ in energy by ∆E is given by ∆E = E2 – E1 = hν)
    ∴ The frequency of radiation absorbed (or) emitted
    ν = \(\frac{E_2-E_1}{h}\) E1 and E2 are energies of lower, higher energy states respectively.
  6. The angular momentum of an electron is given by mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
    An electron revolve only in the orbits for which it’s angular momentum is integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

b) Explain the significance of ‘n’ and ‘l’ quantum numbers.
Answer:
1) Principal quantum number: The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the distance between the nucleus and the orbit also increases. It is denoted by the letter ‘n’. It can have any simple integer value 1, 2, 3, but not zero. These are also termed as K, L, M, N etc. The radius and energy of an orbit can be determined basing on “n” value.

The radius of nth orbit is rn = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The energy of nth orbit is En = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number :
It was proposed by Sommer- feld. It is also known as angular momentum quantum number or subsidiary quantum number. It indicates the shapes of orbitals. It is denoted by ‘l’. The values of ‘l’ depend on the values of ‘n’, T has values ranging from ‘O’ to (n – 1) i.e., l = 0, 1, 2,……(n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10, 14 respectively.

Subsheli l- value Shape
s l = 0 spherically symmetric
p l = 1 dumb – bell
d l = 2 double dumb-bell
f l = 3 four fold dumb-bell i

Energy levels and subshelis

Principal auanturn number (n) Azimuthal quantum number (l) Symbol Number of subshells
1 0 s 1 (1s)
2 0 s 2 (2s, 2p)
1 p
3 0 s 3 (3s, 3p, 3d)
1 p
2 d
4 0 s 4 (4s, 4p, 4d, 4f)
1 p
2 d
3 f

Question 20.
What is hybridization ? Explain sp, sp2 and sp3 hybridizations with one example each.
Answer:
Hybridisation is defined as the process of mixing of atomic orbitals of nearly equal energy of an atom to give the same number of new set of orbitals of equal energy and shapes.

Depending on the number and nature of orbitals involving hybridisation it is classified into different types. If ‘s’ and ‘p’ atomic orbitals are involved three types are possible namely sp3, sp2 and sp.

1. sp3 hybridisation :
In this hybridisation one’s and three ‘p‘ atomic orbitals of the excited atom combine to form four equivalent sp3 hybridised orbitals. This hybridisation is known as tetrahedral or tetragonal hybridisation.

Each sp3 hybridised orbital possess 25% ‘s’ nature and 75% of ‘p’ nature. The shape of the molecule is tetrahedral with a bond angle 109°28′, e.g. : Formation of Methane molecule :

  1. The central atom of methane is carbon.
  2. The electronic configuration of carbon in ground state is 1s2 2s2 2px2 2p20 2pz0 and on excitation it is 1s2 2s1 2px1 2py1 2pz1. During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2pz orbital.
  3. The 2s1,2px1 2py1 2pz1 undergo sp3 hybridisation giving four equivalent sp3 hybridised orbitals.
  4. Each sp3 hybrid orbital overlaps with the 1s orbitals of hydrogen forming σGsp3-s bond.
  5. In case of methane four σGsp3-s bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109°28′.

TS Inter 1st Year Chemistry Question Paper March 2017 - 8

2. sp2 hybridisation :
In this hybridisation one ‘s’ and two ‘p’ atomic orbitals of the excited atom combine to form three equivalent sp2 hybridised orbitals. This hybridisation is also known as trigonal hybridisation. In sp2 hybridisation each sp2 hybrid orbital has 33.33% ‘s’ nature and 66.66% ‘p’ nature. The shape of the molecule is trigonal with a bond angle 120°.

E.g. : Boron trichloride molecule formation :

  1. The electronic configuration of ‘B’ in the ground state is 1s2 2s2 2px1 2py0 2pz0.
  2. On excitation the configuration is 1s2 2s1 2px1 2py1 2pz1. Now there are three half filled orbitals are available for hybridisation.
  3. Now sp2 hybridisation takes place at boron atom giving three sp2c hybrid orbitals.
  4. Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom. The overlapping is σsp-p (CZ atom has the unpaired electron in 2pz orbital). In boron trichloride there are three ‘σ’ bonds.

TS Inter 1st Year Chemistry Question Paper March 2017 - 9

3. sp hybridisation :
In this hybridisation one ‘s’ and one ‘p’ atomic orbitals of the excited atom combine to form two equivalent sp hybridised orbitals. This hybridisation is also known as diagonal hybridisation, in sp hybridisation each sp hybrid orbital has 50% ‘s’ character and 50% ‘p’ character. The shape of the molecule is linear or diagonal with a bond angle 180°.

Ex. : Beryllium chloride molecule formation :

  1. Be atom has 1s2 2s2 2px0 2pb0 2pz0 electronic configuration.
  2. In ground state it has no half filled orbitals. On excitation the configuration becomes 1s2 2s1 2Px1 2py0 2pz0.
  3. Now sp hybridisation takes place at beryllium atom giving two sp hybrid orbitals. Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom.
  4. The overlaping is σsp-p (Cl atom has the unpaired electron in 2pz orbital). In beryllium chloride there are two ‘σ’ bonds.

TS Inter 1st Year Chemistry Question Paper March 2017 - 10

TS Inter 1st Year Chemistry Question Paper March 2017

Question 21.
Define IE1 and IE2. Why is IE2 > IE1 for a given atom? Discuss the factors that affect IE of an element.
Answer:
1) Ionization energy is the amount of energy required to remove the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom. It is denoted as I1 and is expressed in electron volts per atom, kilo calories (or) kilo joules per mole.
M(g) + I1 → M(g)+ + e
l1 is frist ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I2.
M+(g) + l2 → M(g)2+ + e

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential. For sodium, I1 is 5.1 eV and I2 is 47.3 eV.
I1 < I2 < I3 …… In

Factors affecting ionization potential :
1. Atomic radius:
As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2. Nuclear charge :
As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3. Screening effect or shielding effect:
In multielectron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. The electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.

(Magnitude of screening effect) ∝ \(\frac{1}{\text { (Ionization enthalpy) }}\)

TREND IN’A GROUP :
The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.

TREND IN A PERIOD :
In a period from left to right I.P. value increases as the size of the elements decreases along the period.

TS Inter 1st Year Chemistry Question Paper May 2016

Access to a variety of TS Inter 1st Year Chemistry Model Papers and TS Inter 1st Year Chemistry Question Paper May 2016 helps students overcome exam anxiety by fostering familiarity.

TS Inter 1st Year Chemistry Question Paper May 2016

Note : Read the following instructions carefully.

  1. Answer all questions of Section ‘A’. Answer any six questions in Section ‘B’ and any two questions in Section ‘C’.
  2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
  3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
  4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
  5. Draw labelled diagrams wherever necessary for questions in Sections ‘3’ and ‘C’.

Section – A

Note : Answer all questions.

Question 1.
Define receptor and sink.
Answer:
Receptor: The medium which is effected by a pollutant is called receptor.
Sink : The medium which retains and interacts with long lived pollutant is called the sink.
Ex : Oceans are important sinks for atmospheric COr

Question 2.
Give the hybridisation of carbon in garphite and diamond.
Answer:
a) In CO3-2 carbon atom undergoes sp2 hybridisation.
b) In Diamond carbon atom undergoes sp3 hybridisation.
c) In Graphite carbon atom undergoes sp2 hybridisation.
d) In Fullerenes carbon atom undergoes sp2 hybridisation.

TS Inter 1st Year Chemistry Question Paper May 2016

Question 3.
Calculate the oxidation numbers of oxygen in OF2 and O2 F2.
Answer:
OF2
x + 2(-1) = 0
x – 2 = 0
x = +2

O2 F2
2x + 2(-1) = 0
2x – 2 = 0
2x = 2
x = +1

Question 4.
Give any two biological importance of sodium and potassium ions.
Answer:
Biological importance of Na2 and K2 :
1. Na+ ions participate in transmission of nerve signals.
2. Na+ ions regulates the flow of water accross cell membrane.
3. K+ ions are useful in activating enzymes.
4. K+ ions participate in the oxidation of glucose to produce ATP.

Question 5.
Calculate the pH of 0.01 M HCl solution.
Answer: PH : The negative logarithm of H+ ion concentration base ten expressed in Moles/Lit.
PH = -log 10 [H+]

Problem:
Given 0.1 M HCI solution.
[H+] = 0.1 M
PH = – log[H+]
= -log 0.1
= – log 10-1
= 1
PH = 1

Question 6.
What is Chemical Oxygen Demand (COD)?
Answer:
Chemical oxygen demand (COD) : The amount of oxygen required to oxidise organic substance present in polluted water is called COD.

TS Inter 1st Year Chemistry Question Paper May 2016

Question 7.
Which of the gas diffuses faster among N2,O2 and CH4 and why?
Answer:
CHH gas diffuse faster among N2, O2 and CH4.
Reason : CH4 (16) has low molecular weight than N2 (28) and O2 (32).

Question 8.
Write the formula of Plaster of Paris.
Answer:
Plaster of paris is the hemi hydrate of CaSO4 with formula
CaSO4.\(\frac{1}{2}\)H2O.

Preparation:
Plaster of paris is obtained by heating gypsum at 393 K.
TS Inter 1st Year Chemistry Question Paper May 2016 - 1

  1.  If temperature is used greater than 393 K then an hydrous CaSO4 is formed which is called ‘dead burnt plaster.
  2. Plaster of paris has an important property of setting with water.
  3. It forms a hard solid in 5 to 15 min. When it is mixed with suitable quantity of water.
  4. It is majorly used in building industry and as well as plasters.
  5. It is used in the bone fractures (or) sprain conditions.
  6. It is used in dentistry.
  7. It is used in manufacturing status and busts.

Question 9.
Write the IUPAC names of the following :
TS Inter 1st Year Chemistry Question Paper May 2016 - 2
Answer:
TS Inter 1st Year Chemistry Question Paper May 2016 - 3

Question 10.
What is producer gas?
Answer:

  1. Producer gas is mixture of CO and N2.
  2. It is prepared by passing air over hot coke.

Section – B

Question 11.
State and explain the Hess’s law of constant heat summation with example.
Answer:
Hess’s law states that the total amount of heat evolved or absorbed in a chemical reaction is always same whether the reaction is carried out in one step (or) in several steps. Illustration : This means that the heat of reaction depends only on the initial and final stages and not on the intermediate stages through which the reaction is carried out. Let us consider a reaction in which A gives D. The reaction is brought out in one step and let the heat of reaction be ∆H.
TS Inter 1st Year Chemistry Question Paper May 2016 - 4
A → D; → ∆H
Suppose the same reaction is brought out in three stages as follows
A → B : ∆H1
B → C : ∆H2
C → D : ∆H3
The net heat of reaction is ∆H1 + ∆H2+ ∆H3.
According to Hess law ∆H = ∆H1 + ∆H1 + ∆H1.
Ex : Consider the formation of CO2. It can be prepared in two ways.
1) Direct method : By heating carbon in excess of O2

C(s) + O2(g) → CO2(g) ; ∆H = -393.5 kj
2) Indirect method : Carbon can be converted into CO2 in the following two steps.
C(s) + [/latex]\frac{1}{2}[/latex]O2(g) → CO2(g); ΔA1 = – 110.5 kj
CO(g) + [/latex]\frac{1}{2}[/latex]O2(g) → CO2(g); ΔA2 = – 283.02 kj
Total ΔH = -393.52 kj (∆H1 + ∆H2
The two ∆H values are same.

Question 12.
Explain the structure of Diborane.
Answer:
Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons. In diborane each boron atom undergoes sp3 hybridization out of the four hybrid orbitals one is vacant. Each borane forms two a – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their’1s’orbital.

The remaining hybrid orbitals of boran used for the forma-tion of B-H-B bridge bonds. In the formation of B-H-B bridge, half filled sp3 hybrid orbital of one borane atom and vacant sp3 hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

‘These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH2 units. Diborane contains two coplanar BH2 groups. The four hydro-gen atoms are called terminal hydrogen atoms and the remain-ing two hydrogens are called bridge hydrogen atoms.

TS Inter 1st Year Chemistry Question Paper May 2016 - 5

Bonding in diborane. Each B atom uses sp3 hybrids for bonding. Out of the four sp3 hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

TS Inter 1st Year Chemistry Question Paper May 2016

Question 13.
Define the terms hard water and soft water. How is temporary- hardness removed by Clark’s process?
Answer:
Hard water : Water does not give lather readily with soap is called hard water.

Hard water contains hardness. This hardness is due to presence of Ca, Mg soluble salts.
Presence of Ca, Mg – bicarbonates causes temporary hardness.
Presence of Ca, Mg – chlorides, sulphates causes permanent hardness.

Soft water : Water which give lather immediately with soap is called soft water.
(Or)
i) Ion – Exchange method :

  1. This method is useful to remove the permanent hardness of water.
  2. This method is also named as permutit (or) zeolite process.
  3. Permutit is the artificial zeolite, i.e. sodium aluminium orthosilicate. (Na2Al2Si2O2xH2O (or) NaAlSiO4)
  4. Permutit is written in short form as Naz.
  5. When permutit is added to hard water, the following ion- exchange reactions takes place.
    2Naz(s) + Ca(aq)+2 → Caz2(s) + 2Na(aq)+
    2Naz(s) + Mg(aq)+2 → Mg2(s) + 2Na(aq)+
  6. Caz2 and Mgz2 are called as exhausted permutit. These are regenerated to permutit by the treatment with brine solution. Caz(s) + 2Na(aq)+2 → 2Naz(s) + Ca(aq)+2

ii) Calgon process:

  1. Calgon is sodium hexametaphosphate. [Na6P6O18 (or) (NaPO3)6]
  2. Calgon doesnot precipitate the Ca (or) Mg – salts but removes Ca+2 and Mg+2 ions from water.
  3. The removal of Ca+2 (or) Mg+2 ions from water may takes place either by adsorption (or) by complex formation

Reactions:
Na6P6O18 → 2Na+ + Na4P6O18-2
TS Inter 1st Year Chemistry Question Paper May 2016 - 6

Question 14.
Define Le Chatelier principle and mention the conditions for the preparation of the ammonia by Haber’s process.
Answer:
Lechatelier Principle Statement: When a system at equilibrium is subjected to a change of temperature, pressure, concentration then the equilibrium shifts in such a direction in order to nallify the effect of change.

Haber’s process:
N2(g) + 3H2(g) ⇌ 2NH3(g) + 92 kj

  1. In the above equilibrium recution decrease in no. of moles observed. So according to lechatelier’s principle high pressure favours the formation of NH3. So an optimum pressure of 200 atm is used.
  2. The given reaction is exothermic so low temperature favours the formation of NH3. But in industries at low temperature reaction proceds slowly. So suitable temperature is 725-775 K.
  3. To increase the rate of reaction catalyst ‘Fe’ is used and to increase the activity of catalyst promotor ‘Mo’ is used.
    Pressure – 200 atm
    Temperature – 725 – 775 k
    Catalyst – Fe
    Promotor – Mo

TS Inter 1st Year Chemistry Question Paper May 2016

Question 15.
State and write chain and position isomerisms with examples.
Answer:
Chain Isomerism :
Compounds having same molecular formula but differ in the carbon chain or carbon skeleton are called chain isomer and this is called chain isomerism.
Eg:
TS Inter 1st Year Chemistry Question Paper May 2016 - 7

Position Isomerism :
Compounds having same molecular formula but differ in the position of substituent or functional group or multiple bond are called position isomers and this is called position isomerism.
Eg:
TS Inter 1st Year Chemistry Question Paper May 2016 - 8

Question 16.
Define molarity. Calculate the molarity of NaOH in the solution prepared by dissolving 4 gr in enough water to form 250 ml of the solution.
Answer:
Molarity : The No.of moles of solute present in one liter of solution is called molarity (M).
TS Inter 1st Year Chemistry Question Paper May 2016 - 9
Note that molarity of a solution depends upon temperature because’volume of a solution is temperature dependent.

Question 17.
Complete the following reaction and name the products A, B and C:
TS Inter 1st Year Chemistry Question Paper May 2016 - 18

Question 18.
Derive the Graham’s law of diffusion from kinetic gas equation.
Answer:
Derivation of Charle’s Law from Kinetic gas equation :
Kinetic gas equation is
TS Inter 1st Year Chemistry Question Paper May 2016 - 11

Hence Charle’s Law Proved from kinetic gas equation. Derivation of Graham’s Law of diffusion from Kinetic gas Equation. Knietic gas equation is
TS Inter 1st Year Chemistry Question Paper May 2016 - 12

Section – C

Question 19.
Explain the significance of four quantum numbers associated with an electron in an atom.
Answer:

  • In general a large no.of orbitals are possible in an atom.
  • These orbitals are distinguished by their size, shape and orientation.
  • An orbital of smaller size means there is more chance to find electron near the nucleus.
  • Atomic orbitals are precisely distinguished by quantum numbers. Each orbital is designated by three major quantum numbers.
    1. Principal quantum number (n)
    2. Azimuthal quantum number (l)
    3. Magnetic quantum number (m)

1) Principal quantum number:
The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the dis¬tance between the nucleus and the orbit also increases.
It is denoted by the letter ‘n’. It can have any simple integer value 1, 2, 3,…… but not zero. These are also termed as K, L, M, N etc.

The radius and energy of an orbit can be determined basing on “n” value.
The radius of nth orbit is rn = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The radius of nth orbit is En = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number:
It was proposed by Sommer feld. It is also known as angular momentum quantum number or subsidiary quantum number. it indicates the shapes of orbitals. It is denoted by T. The values of T depend on the values of ‘n1, T has values ranging from ‘O’ to (n – 1) i.e., ( l = 0, 1, 2 …… (n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10,14 respectively.

Subshell l – value Shape
s l = 0 spherically symmetric
p l = 1 dumb – bell
d l = 2 double dumb – bell
f l = 3 four fold dumb – bell

Energy levels and subshells

Principal quantum number (n) Azimuthal quantum number. (l) Symbol Number of subshells
1 0 s 1 (1s)
2 0 s 2 (2s, 2p)
1 P
3 0 s 3 (3s, 3p, 3d)
1 p
2 d
4 0 s 4 (4s, 4p, 4d, 4f)
1 P
2 d
3 f

3) Magnetic quantum number:
It was proposed by Lande. It shows the orientation of the orbitals in space. ‘p’ – orbital has three orientations. The orbital oriented along the x – axis is called px orbital, along the py – axis is called pz – orbital and along the z – axis is called pz orbital. In a similar way d – orbital has five orientations. They are dxy
AP Inter 1st Year Chemistry Question Paper May 2016 - 13

Question 20.
What is periodic property? How are the following properties vary in a group and period?
a) Atomic radius
b) Electron gain enthalpy
c) Electronegativity.
Answer:
Periodicity: The properties of elements which change gradually with change in electronic configuration and repeats in regular intervals are called periodic properties and this repetetion is called periodicity.

a) Atomic radius :
In groups and periods of the periodic table the ionization enthalpy values are periodically change depend upon the electronic configuration and size of elements. In a group of elements ionization energy decreases from top to bottom because atomic radius increases. In general, in a period the atomic size decreases. Because of this, the ionization energy increases across a period.

b) Electron gain enthalpy:
Electron gain enthalpy increases in a period from left to right because the size of the atom decreases and the nature of the element changes from metallic to non – metallic nature when we move from left to right in a period. Electron gain enthalpy decreases from top to bottom in a group because there is an increase in the atomic size. But the second element has greater electron gain enthalpy than the first element’.
e.g. : Chlorine has more electron affinity value ( – 348 kJ mol-1) than Fluorine (- 333 kJ mol-1). It is because fluorine atom is smaller in size than chlorine atom. There is repulsion between the incoming electron and electrons already present in fluorine atom i.e., due to stronger inter electronic repul, sions.

c) Electronegativity:
Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases. In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

TS Inter 1st Year Chemistry Question Paper May 2016

Question 21.
Define the hybridisation. Explain sp, sp2 and sp3 hybridisations each with one example.
Answer:
Hybridisation is defined as the process of mixing of atomic orbitals of nearly equal energy of an atom to give the same number of new set of orbitals of equal energy and shapes. Depending on the number and nature of orbitals involving hybridisation it is classified into different types. If ‘s’ and ‘p’ atomic orbitals are involved three types are possible namely sp3, sp2 and sp.

1. sp3 hybridisation:
In this hybridisation one’s and three ‘p’ atomic orbitals of the excited atom combine to form four equivalent sp3 hybridised orbitals.

This hybridisation is known as tetrahedral or tetragonal hybridisation. Each sp3 hybridised orbital possess 25% ‘s’ nature and 75% of ‘p’ nature. The shape of the molecule is tetrahedral with a bond angle 109°28′, e.g.: Formation of Methane molecule:
1) The central atom of methane is carbon.

2) The electronic configuration of carbon in ground state is 1s2 2s2 2p2x 2p1y 2p0z and on excitation it is 1s2 2s1 2p1x 2py1 2pz1. During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p2 orbital.

3) The 2s1 2p1x 2p1y 2p1z undergo sp3 hybridisation giving four equivalent sp3 hybridised orbitals.

4) Each sp3 hybrid orbital overlaps with the 1s orbitals of hydrogen forming σsp3-s bond.

5) In case of methane four σsp3-s bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109°28′.
TS Inter 1st Year Chemistry Question Paper May 2016 - 15
E.g.: Boron trichloride molecule formation :
1) The electronic configuration of ‘B’ in the ground state is 1s2 2s2 2p1x 2p0y 2p0z.
2) On excitation the configuration is 1s2 2s1 2p1x2p1y,2pz0 Now there are three half filled orbitals are available for hybridi-sation.
3) Now sp2 hybridisation takes place at boron atom giving three sp2 hybrid orbitals.

4) Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom. The overlapping is σsp2-p (Cl atom has the unpaired electron in 2pz orbital). In boron trichloride there are three ‘σ’ bonds.
TS Inter 1st Year Chemistry Question Paper May 2016 - 16

3. sp hybridisation:
In this hybridisation one ‘s’ and one ‘p’ atomic orbitals of the excited atom combine to form two equivalent sp hybridised orbitals.

This hybridisation is also known as diagonal hybridisation. In sp hybridisation each sp hybrid orbital has 50% ‘s’ character and 50% ‘p’ character. The shape of the molecule is linear or diagonal with a bond angle 180°.

Ex.: Beryllium chloride molecule formation :
1) Be atom has 1s1 2s1 2p0x 2p0y 2p0z electronic configuration.
2) In ground state it has no half filled orbitals. On excitation the configuration becomes 1s2 2s1 2p1x 2p0y2p0z

3) Now sp hybridisation takes place at beryllium atom giving two sp hybrid orbitals. Each of them with one unpaired electron forms a ‘o’ bond with one chlorine atom.

4) The overlaping is σsp-p (Cl atom has the unpaired electron in 2pz orbital). In beryllium chloride there are two ‘σ’ bonds.
TS Inter 1st Year Chemistry Question Paper May 2016 - 17

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 4 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत ।

अ) पृथिव्यां त्रीणि रत्नानि जलमन्नं सुभाषितम् ।
मूढैः पाषाणखण्डेषु रत्नसंज्ञा विधीयते ॥
समाधान:
पदच्छेद (Word Division) : पृथिव्यां त्रीणि, रत्नानि, जलं, अन्नं, सुभाषितम्, मूढैः, पाषाणखण्डेषु, रत्नसंज्ञा, विधीयते ।
अन्वयक्रमः पृथिव्यां, जलं, अन्नं, सुभाषितं, त्रीणि, रन्तानि, मूढैः, पाषाणखण्डैः, रन्तसंज्ञा, विथीयते ।
अर्था (Meanings) : पृथिव्याम् = On the earth, जलम् = water, अन्नम् = food, सुभाषितम् = and wise saying, त्रीणि = are the three, रत्नानि = gems, मूढैः = a fool, पाषाणखण्डेषु = to stones, रत्नसंज्ञा = the name of gem, विधीयते = gives.
भाव (Substance) : Water, food and wise saying these three are the gems on the earth. But, a fool gives the name of a gem to a piece of stone.

आ) अर्थानामार्जनं कार्यं वर्धनं रक्षणं तथा ।
भक्ष्यमाणो निरादायः सुमेरुरपि हीयते ॥
समाधान:
पदच्छेद (Word Division) : अर्थानां, आर्जनं, कार्यं, वर्थनं, रक्षणं, तथा, भक्ष्यमाणः निरादायः सुमेरुः, अपि, हीयते ।
अन्वयक्रम : अर्थानों, आर्जनं, वर्धनं, तथा रक्षणं, कार्यं, निरादायः, भक्ष्यमाणः, सुमेरुः, अपि, हीयते ।
अर्था (Meanings) : अर्थानाम् = of wealth, money; आर्जनम् = acquisition; वर्धनम् = increase, growth; तथा = and; रक्षणम् = protection; कार्यम् = is to be done; भक्ष्यमाणः = if eaten; निरादायः without income; सुमेरुः + अपि = even the golden Meru Mountain; हीयते = will shrink.
भाव (Substance): One should acquire, increase and protect wealth. If eaten, without augmenting, even the Meru Mountain will shrink.

2. एकं निबन्धप्रश्नं समाधत्त ।

अ) क्षितीशेन वसिष्ठ धेनुः कथं आराधिता । विशदयत ?
Answer:
Introduction: The lesson Dharmanishta is an extract from the 2nd canto of Raghuvamsa, written by Kalidasa. This lesson describes the moral and devotional character of king Dilipa.

A Lion attacks Nandini: The childless king Dilipa his wife Sudakshina were engaged in the service of Nandini, the cow of sage Vasishta. One day Nandini which went to the forest guarded by the king wished to test the devotion of the king, and entered a Himalayan cave. Thinking that the cow was unassailable to any wild animal, the king took his eyes off her to watch the beauty of the mountain. Then a lion attacked her. Alerted by her cry, the king became ready to kill the lion by shooting an arrow at it. But his hand got fixed on the shaft of the arrow. चित्रार्पितारम्भ इवावतस्थे । Then the lion told him that he was a servant of Lord Siva. He was appointed there to protect the Devadaru tree grown by Parvathi. He was made a lion, and any animal that arrived there when he was hungry would become his food.

Dilipa’s offer: Then the king said that Siva was to be respected, and at the same time, the property of the teacher was to be saved. He offered his own body as food to the lion requesting the release of the cow. But the lion said that the king was young and an emperor. He seemed to be imprudent in giving up more for little. By showing compassion to the cow, she alone would be saved. But if he lived, he could always save his people from calamities. The king said that a king should protect his subjects.

तद्विपरीतवृत्तेः, प्राणैः उपक्रोशमलीमसैर्वा | He could not offer other cows for Nandini, the daughter of Surabhi to his preceptor. He pleaded with the lion that how he could stand before his teacher when the cow was lost and himself was unhurt. He further said that people like him did not care for physical bodies. पिण्डेष्वनास्था खलु भौतिकेषु ।

The lion agreed and released the cow. But when the king bent his head before the lion, he received a shower of flowers from the Vidyadharas. The delighted cow, followed by the king returned safely to the hermitage.

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

आ) गीर्वाणवाण्याः वैभवं लिखत ।
Answer:
Introduction :
The lesson Chitravimsati was written by Sri Jatavallabhula Purushottama Sastry. It is an extract from his work Chitra-satakam. There are two topics in this lesson Girvanavani and Chitraloka. The first part describes the greatness of Sanskrit language.

Girvanabhasha :
The nectar oozing Sanskrit language is spoken by the gods. It is the ocean of the gems of good sayings and the treasure trove of great literature, and is acquired only by the meritorious. Eigen सुकृतैकलभ्या । It is the ladder to liberation and the staircase to the heaven. It is relished by people of different tastes. Just as the children inherit the virtues of the mother, so also all the languages inherit the qualities of the divine mother language. In Sanskrit as the words are formed from the roots, they are naturally pure.

A language is the mother because she spreads in the world, and also she protects the dependents like a mother. On this earth, such a mother tongue is Sanskrit only, how can other languages have the title of mother ? वाच्याः कथं मातृपदेन चान्या: ? Even now, those who are experts in that language make the non-learners of the language ashamed by reciting the ancient verses.

Lakhs of people speak it, and lakhs understand it. Lakhs of books in that language are sold every year. How can we live abandoning that language which is always in our body like life, and which is used during marriage ceremony, rituals of manes and gods, and in poetry recitals and philosophical discourses? Our history is before us with life. How is it dead? In this world, beautiful poetry is removed from dharma and righteous poetry is not beautiful. Except in Sanskrit where else is poetry that is golden and fragrant ?

3. एकं निबन्धप्रश्नं समाधत्त ।

अ) विक्रमस्य औदार्यमिति पाठ्यभागस्य सारांशं लिखत ।
Answer:
Introduction: The lesson Vikramasya Audaryam is taken from vikramarkacharitam written by Sivadasa. When Bhoja wants to ascend the throne of Vikrama the statue on the fourth step tells this story about the generosity of king Vikrama.

The childless Brahmin:
While Vikramaditya ruled Ujjaini, there was a learned Brahmin in that city, who was virtuous, but who had no children. Once, his wife said to him that there was no heaven for one who had no children सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना | The Brahmin worshipped Lord Siva. The god appeared in his dream and asked him to perform Pradoshavrata. The Brahmin did so and he begot a son, whom he named Devadatta. When his son grew up, the Brahmin performed his marriage, and went on a pilgrimage to Varanasi.

Devadatta’s help to Vikrama :
Once Devadatta went into the forest to collect fuel sticks for sacrifice. At that time, king Vikrama came to the forest for hunting. There he asked Devadatta the way to the city, even though he knew it.

Devadatta himself led him to the city. The king praised him, and appointed him in his court. One day the king praised Devadatta’s help in the assembly saying that he could never repay his debt.

Devadatta kidnaps the prince :
Devadatta wondered whether the king’s praise was true or false. He wanted to test the words of the king. He kidnapped the prince, and gave one of his ornaments to his servant and sent him to the market to sell it. The king’s men who were searching for the prince caught him, and brought him to the king. When the king asked him, the servant said that he was Devadatta’s servant. The king sent for Devadatta, who said that he killed the prince or money. The members of the court said that Devadatta should be sentenced to death. But the king said that Devadatta rescued him in the forest. One should not find faults in those dependent on him. ferdy महतां गुणदोषचिन्ता | He said that his son died because of his previous deeds only. He honoured Devadatta and sent him away.

Devadatta returned with the prince, and told the king that he wanted to test the words of the king. He praised the generosity of the king, who said that one should not forget the help done to him. यः कृतमुपकारं विस्मरति स एव पुरुषाधमः ।
Thus the statue told Bhoja about the generosity of king Vikrama.

आ) अप्पय्यदीक्षितस्य जीवितचरितं संक्षेपेण लिखत ।
Answer:
Introduction: The lesson Appayyadikshitendra was written by Sanka Usharani. This lesson narrates the story of the great scholar and devotee Appayya Dikshita.

Family of scholars: Appayya Dikshita was born in the family of scholars. His father was Rangarajadhwari was a genius. His grandfather was in the court of Acchana Dikshita. Appayya was born in 1554 in Virinnchipura in Tamilnadu.

Appayya was a poet, philosopher, scholar, devotee, Grammarian, Mimamsaka, Logician, Ritualist etc. He studied all the fourteen branches of knowledge at the feet of his teacher Nrisimha Ramaswamy.

Marriage and works: A scholar named Ratnakheta Srinivasa wanted to defeat Appayya with the grace of the goddess Kamakshi. However, the goddess advised him to give his daughter in marriage to Appayya. अप्पय्यः असाधारणः वादे जेतुमशक्तः । He did so. Appayya had two daughters and three son.
Many students came to him to receive education. Bhattoji Dikshita came to him to learn Mimamsa and Vedanta.

Appayya wrote 104 works on many subjects like Vyakarana, Vedanta, Alankara etc. Among the Vedanta works Siddhanta lesasangraha and Parimala vyakhya were famous. Sivakarna mritam and Sivarchanachandrika were his famous works devoted to Siva. Kuvalayananda and Chitramimamsa were Alankara works.
Appayya was invited by Vellore Chinabommanayaka to adorn his court. Appayya was epileptic. But when the king came to his house to observe his disease, Appayya transferred his disease on to his upper garment. Dikshita means one who performed sacrifices. Appayya performed more than 104 sacrifices including Somayaga, Vajapeya etc. He even performed Viswajit. Once while he was performing a sacrifice, the king arrived there and offered him clothes, ornaments etc. Which he immediately put in the fire. The fire god appeared wearing them.

Devotion of Siva: Appayya was a devotee of Siva. To test his devotion, he ate dattura leaves and became mad. In that state, he recited Unmattapanchasat. The king honoured him with a shower of gold coins when Appayya completed his Sivarkamanidipika. चं कनकाभिषेकेण सत्कृतवान् | Appayya built a temple for Siva with that gold in his native village. Though Appayya was a Tamilian he used Telugu idioms and proverbs. He praised Telugu language thus: Being born in Andhra, knowing the language of Telugu, study- ing Prabhakara’s Mimamsa work, and belonging to the Yajurveda School are the fruit of not less penance.

Though the householder Appayya was in the path of action, his mind was in the path of detachment only. He spent his last days in the presence of Nataraja of Chidambara. He became one with Siva at the age of 72 in the year of 16261 He was considered the incarnation of Siva Himself. अप्पय्यः साक्षात् परमशिवस्य अवतारः ।

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

4. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) नृपश्रियः कीदृशाः ?
समाधान:
नृपश्रियः भुजङ्गजिह्वाचपला ।

आ) कर्णभारं केन विरचितम् ?
समाधान:
कर्णभारं भासेन विरचितम् ।

इ) कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय किं दत्तवती ?
समाधान:
कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय पिष्टमिश्रितजलं दत्तवती । तत् पीत्वा अश्वत्थामा तुष्टः अभवत् ।

ई) केन द्रुपदः गर्वान्धः जातः ?
समाधान:
द्रुपदः धनश्रिया गर्वान्धः जातः । एवम् अर्जुनः उक्तवान् ।

उ) कालिदासः महाराजं कथं सन्तोषितवान् ?
समाधान:
भोजस्य मनोगतस्य यथोचितम् उत्तरं दत्वा कालिदासः भोजं संतोषितवान् ।

ऊ) कालिदासश्लोके भोजेन प्राप्तं समाधानं किम् ?
समाधान:
कालिदासश्लोकेन भोजेन प्राप्तं समाधानम् एवम् अस्ति – द्वयोः सम्भाषणकर्त्रीः मध्ये सूचनां विना न प्रविशेत् इति ।

5. द्वयोः संदर्भ व्याख्यानं लिखत ।

अ) अप्रियस्य च पथ्यस्य वक्ता श्रोता च दुर्लभः ।
समाधान:
परिचयः – एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् । एषः पाठः रामायणस्य युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः – प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
भावः – लोके प्रियवादिनः जनाः सुलभाः भवन्ति । परन्तु अप्रियस्य हितस्य वक्ता, श्रोता च दुर्लभः एव ।

आ) चित्रार्पितारम्भ इवाऽवतस्थे ।
समाधान:
परिचयः – एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषःभागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः – कविः दिलीपमहाराजस्य स्थितिं एवं वदति ।
भावः – चित्र लिखित शरीद्धरण उद्योग इव स्थितः ।
विवरणम्ः – सिंहं प्रहर्तुं दिलीपः उद्युक्तः । तस्य वामेतरः करः सायकपुंखे एव लग्मः अमवत् ।

इ) यत्रैताः हृदि जागृताः प्रकृतयः सा मातृभूमिर्मम ।
समाधान:
परिचयः – एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् । अस्य कविः श्रीमान् दोर्बल प्रभाकर शर्मा ।
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः – मम मातृभूमिः क्षेमकरी, तत्र जलं शुचि, जीवनं स्नेहात्मकं, मनः सुरचनं इति कविः वदति । यत्र प्रकृतयः हृदि अवबोधितः सा मम मातृभूमिः इति कविः वदति ।

ई) वाच्याः कथं मातृपदेन चान्याः ।
समाधान:
परिचयः – गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री ।
सन्दर्भः – गीर्वाणवाण्याः वैभवं वर्णयनन् कविः एवं वदति । या भाषा लोके प्रसारिता, माता इव रक्षति सा एव मातृभाषा इति कथ्यन्ते ।
भावः – कथं अन्यभाषाः मातृभाषा इति कथ्यन्ते ।
विवरणम्ः – संस्कृतभाषा सर्वासां भाषाणां जननी, सा भाषा मातेव रक्षति ।

6. द्वयोः ससंदर्भ व्याख्यानं लिखत ।

अ) निवृत्तरागस्य गृहं तपोवनम् ।
समाधान:
परिचयः एतत-वाक्यं, “मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः तपःकर्तुं बनाय प्रस्थितं कौण्डिन्यं प्रति कपिलः एवं अवदत् । भावः विरक्तस्य जनस्य गृहमेव तपोव तुल्यम्
विवरणम्ः सर्वे पञ्चेन्दियाणि वशेयुः, तदा एव तस्य शान्तिः सिध्यति, तस्मिन् समये गृहः एव तपोवनसदृशम् ।

आ) नैवाश्रितेषु महतां गुणदोषचिन्ता |
समाधान:
परिचयः – इदं वाक्यं “विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास ।
सन्दर्भः – समामध्ये राजा सचिवानां प्रति एवं अवदत् ।
भावः – शरणागतेषु महतां गुणदोष चिन्ता न कुर्तव्या ।
विवरणम्ः – चन्द्रः क्षथी, प्रकृतिवक्रतनु, कलडिए तथापि सः शकरं आश्रितवान् सः शंकरः चन्द्रस्य गुणदोषचिन्तनं ।

इ) वैद्योऽपि मानव एव ।
समाधान:
परिचयः वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः वेङ्कटरावः ग्रामीणं प्रति एवं अवदत् ।
भावः वैद्यः अपि चरः एव ।
विवरणम्ः अरे भूर्ख ! स्वसुखनिन्तैव युषमाकम् वैद्यः मानवमात्र एव ।

ई) अप्पय्यः असाधारणः वादे जेतुमशक्तः ।
समाधान:
परिचयः – एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भः – अप्पय्यं जेतुं, स्वपादावनतं कर्तुं च रत्नखेटश्रीनिवासः अभिलषति स्म । तदर्थं सः काञ्चीपुरे कामाक्षी प्रार्थितवान् । तदा देवी कामाक्षी अप्पय्यं जेतुं न शक्यते । तव पुत्रिकां तस्मै विवाहे प्रयच्छ इति उक्तवती ।
भावः – अप्पय्यः असामान्यः । वादे तं जेतुं न शक्यते ।

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

7. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) सीता कीदृशी ?
समाधान:
सीता महाहिसदृशी । सीतायाः चिन्ता एव विषम्, स्मित एव दंष्ट्राः स्तनौ एव भोगाः तथा अङ्गुल्यः एव शिरांसि ।

आ) दिलीपेन सेव्यमानं नन्दिनीधेनुं कः चकर्ष ?
समाधान:
दिलीपेन सेव्यमानं नन्दिनीधेनुः सिंहः चकर्ष ।

इ) सर्वेन्द्रियाणां प्रधानं किम् ?
समाधान:
सर्वेन्द्रियाणां नयनं प्रधानम् ।

ई) परोपदेशसमये जनाः सर्वेऽपि कीदृशाः भवन्ति ?
समाधान:
परोपदेशसमये जनाः सर्वेऽपि पण्डिताह भवन्ति ।

उ) देवगिरीणां पञ्च नामानि लिखत ?
समाधान:
हेमाद्रिः, रजताद्रिः, हिमाद्रिः, सह्याद्रिः, विन्ध्याद्रिः इति पञ्चदेवगिरयः ।

ऊ) अमर्त्यवाण्याः मातृता कथमागता ?
समाधान:
सर्वासुभाषासु अमर्त्य वाज्यः गुणाः सन्ति, अतः अमर्त्यवाण्याः मातृता आगता ।

8. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) मन्दविषो नाम सर्पः कुत्र वसति ?
समाधान:
मन्दविषः नाम सर्पः जीर्णोधाने वसति

आ) विक्रमेण पृष्टः भृत्यः किमुत्तरमदात् ?
समाधान:
विक्रमेण पृष्टः भृत्यः एवम् उत्तरम् अदात् – अहं देवदत्तस्य भृत्यः । एतदाभरणं विक्रीय धनमादाय इति प्रेषितः अस्मि । इति ।

इ) मानसारः कथं गदां प्राप्तवान् ?
समाधान:
मानसारः परमेश्वरं समाराध्य सन्तुष्टात् तस्मात् गदां प्राप्तवान् ।

ई) मार्गशीर्षमासः कीदृशः आसीत् ?
समाधान:
मार्गशीर्षमासः शीतवायुविजृम्भणेन युक्तः आसीत् ।

उ) प्रभुत्वचिकित्सालये नियोगेनागता भिषगङ्गना का ?.
समाधान:
प्रभुत्वचिकित्सालये नियोगेनागता भिषगाङ्गना आत्मनः सहाध्यायिनी भिषगङ्गला ।

ऊ) अप्पय्यदीक्षितः कुत्र जन्म लेभे ?
समाधान:
अप्पय्यदीक्षितः तमिळदेशे आरणिसमीपे अडयप्पलमिति ग्रामे जन्म लेभे ।

9. एकेन वाक्येन समाधानं दत्त ।

अ) रणे रामेण कः हतः ?
समाधान:
रणे रामेण खरः हतः ।

आ) मुनिहोमधेनुः का ?
समाधान:
नन्दिनी

इ) कैः पाषाणखण्डेषु रत्नसंज्ञा विधीयते ?
समाधान:
मूढैः

ई) मूर्खस्य कति चिह्नानि भवन्ति ?
समाधान:
मूर्खस्य पञ्च चिह्नानि भवन्ति ।

उ) देहे सदा प्राण इव का अस्ति ?
समाधान:
धिषणा

10. एकेन वाक्येन समाधानं दत्त ।

अ) मण्डूकनाथः कः ?
समाधान:
जालपादः

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

आ) कुमारमानीय राज्ञे क ददौ ?
समाधान:
देवदत्तः

इ) राजवाहनः कस्य पुत्रः ?
समाधान:
राजहंसस्य

ई) श्रीधरस्य मातुः नाम किम् ?
समाधान:
श्रीधरस्य मातुः नाम वेदवती ।

उ) वेङ्कटरावस्य पुत्रः कः ?
समाधान:
सुरेशः

11. अधो निर्दिष्टं कथां पठित्वा प्रश्नानां समाधानानि दत्त |

1. देवालयः कुत्र अस्ति ?
समाधान:
देवालयः चित्रकूटपर्वतनिकटतपोवनमध्ये अस्ति ।

2. जलधारा कस्मात् पतति ?
समाधान:
जलधारा अत्युदग्रात् शिखरात् पतति ।

3. प्रतिदिनं होमकुण्डे कः होमं करोति ?
समाधान:
प्रतिदिनं होमकुण्डे कश्चित् ब्राह्मण होमं करोति ।

4. जनः कीदृशं स्थानमद्राक्षीत् ?
समाधान:
जनः अतिविचितं स्यानं अहासीत ।

5. तस्य देहात् अतीव कलुषमुदकं निस्सरति ?
समाधान:
पापाचारस्य देहात् अतीव कलुषमुदकं निस्सरति ।

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त ।

अ) जगत् + जननी
समाधान:
जगज्जननी = श्चुत्व सन्धिः

आ) सद् + जनः
समाधान:
सज्जनः = ष्टुत्व सन्धिः

इ) षट् + आननः
समाधान:
षडाननः = जश्त्व सन्धिः

ई) वाक् + मयम्
समाधान:
वाङ्मयम् = अनुनासिक सन्धिः

उ) रामः + अपि
समाधान:
रामोऽपि = विसर्ग सन्धिः

ऊ) नृपतिः + जयति
समाधान:
नृपतिर्जयति = विसर्गरेफादेश सन्धिः

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

13. नामनिर्देशपूर्वकं त्रीणि विघटयत ।

अ) शरच्चन्द्रः
समाधान:
शरत् + चन्द्रः = श्चुत्व सन्धिः

आ) पेष्टा
समाधान:
पेष् + टा = ष्टुत्व सन्धिः

इ) तदपि
समाधान:
तत् + अपि = जश्त्व सन्धिः

ई) षण्मुखः
समाधान:
षट् + मुखः = अनुनासिक सन्धिः

उ) रामोऽपि
समाधान:
रामः + अपि = विसर्ग सन्धिः

ऊ) कविरायाति
समाधान:
कविः + आयाति = विसर्गरेफादेश सन्धिः

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत ।

अ) राजन्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions 2

आ) वाक्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions 3

इ) पचत्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions 4

ई) एतद् (न.पुं.)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions 5

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत ।

अ) कृष्णश्रितः
समाधान:
कृष्ण श्रितः = द्वितीयातत्पुरुष समासः

आ) पुरुषव्याघ्रः
समाधान:
पुरुषः व्याघ्रः इव = उपमानोत्तरपदकर्मधारय समासः

इ) सप्तर्षयः
समाधान:
सप्त च ते ऋषयश्च = द्विगु समासः

ई) शिवकेशवौ
समाधान:
शिवः च केशवः च = द्वन्द्व समासः

AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions

उ) यथाशक्ति
समाधान:
शक्तिम् अनतिक्रम्य अव्ययीभाव समासः

ऊ) द्वित्राः
समाधान:
द्वौ वा त्रयो वा = संख्योभयपद बहुव्रीहि समासः

16. अधो निर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखंत |
AP Inter 2nd Year Sanskrit Model Paper Set 4 with Solutions 1
समाधान:
प्रश्नाः
1. गणेशः सत्यं वदिष्यति ?
2. त्वं कदा गमिष्यसि ?
3. अहं पाठं पठिष्यामि ?
4. कृष्णः कदा करिष्यति ?
5. रामः फलं खादिष्यति ?

AP Inter 1st Year Maths 1B Question Paper May 2019

Access to a variety of AP Inter 1st Year Maths 1B Model Papers and AP Inter 1st Year Maths 1B Question Paper May 2019 allows students to familiarize themselves with different question patterns.

AP Inter 1st Year Maths 1B Question Paper May 2019

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

  1. Answer all questions.
  2. Each question carries two marks.

Question 1.
Find the ratio in which the straight line 3x + 4y = 6 divides the line joining the points (2, -1) and (1, 1). State whether the points lie on the same side or on either side of the straight line.
Solution:
Let L ≡ 3x + 4y – 6 = 0
L11 = 3(2) + 4(-1) – 6
= 6 – 4 – 6
= – 4
L22 = 3(1) + 4(1) – 6
= 3 + 4 – 6
= 1
∴ The straight line L = 0 divides the line joining the points (2, -1) and (1, 1) in the ratio -L11 : L22
= -(-4) : 1
= 4 : 1
L11 and L22 have the opposites sign, the two points lie on the opposite side of the given tine L = 0.

Question 2.
Find the value of p, if the straight lines 3x + 7y – 1 =0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given straight line equations are
3x + 7y – 1 = 0 ……. (1)
7x – py + 3 = 0 ……. (2)
Since (1), (2) are mutually perpendicular
∴ a1a2 + b1b2 = 0
⇒ 3(7) + 7(-p) = 0
⇒ 21 – 7p = 0
⇒ 7p = 21 ⇒ p = 3.

AP Inter 1st Year Maths 1B Question Paper May 2019

Question 3.
Show that the points A(3, 2, -4), B(5, 4, – 6) and C(9, 8, -10) are collinear and find the ratio in which B divides AC.
Solution:
Given A = (3, 2, – 4)
B = (5, 4, -6)
C = (9, 8, -10)
AP Inter 1st Year Maths 1B Question Paper May 2019 2
∴ A, B, C are collinear.
B divides AC in the ratio = AB: BC
= 2\(\sqrt{3}\) + 4\(\sqrt{3}\)
= 1 : 2

Question 4.
Find the equation to the plane parallel to the ZX – plane and passing through (0, 4, 4).
Solution:
The equation to the plane parallel to ZX – plane and passing through (x1, y1, z1) is y = y1
∴ The equation to the plane parallel to ZX -p lane and passing through (0, 4, 4) is y = 4.

Question 5.
Show that: \(\lim _{x \rightarrow 2} \frac{|x-2|}{x-2}\) = -1.
Solution:
AP Inter 1st Year Maths 1B Question Paper May 2019 3
AP Inter 1st Year Maths 1B Question Paper May 2019 4

Question 6.
Compute \(\lim _{x \rightarrow-\infty} \frac{5 x^3+4}{\sqrt{2 x^4+1}}\)
Solution:
AP Inter 1st Year Maths 1B Question Paper May 2019 5

Question 7.
If f(x) = cos [log (cot x)], then find f(x).
Solution:
Given f(x) = cos [log(cot x)]
Differentiating w.r.to ‘x’ on both sides
f(x) = – sin[log(cot x)]. \(\frac{d}{d x}\)[log(cot x)]
= -sin[log(cot x)]. \(\frac{1}{\cot x} \cdot \frac{d}{d x}\)(cot x)
= – sin[log(cot x)]. \(\frac{1}{\cot x}\)(- cosec2 x)
= sin[log(cot x)]. \(\frac{\sin x}{\cos x} \cdot \frac{1}{\sin ^2 x}\)
= \(\frac{2 \sin [\log (\cot x)]}{2 \sin x \cos x}\) = \(\frac{2 \sin [\log (\cot x)]}{\sin 2 x}\)

AP Inter 1st Year Maths 1B Question Paper May 2019

Question 8.
If f(x) = 2x2 + 3x – 5, then prove that f(0) – 3f'(-1) = 0.
Solution:
Given f(x) = 2x2 + 3x – 5
f (x) = 4x + 3
f'(0) = 4(0) + 3 = 3
f'(-1)= 4(-1) + 3 = – 4 + 3 = – 1
∴ f'(0) + 3f'(-1) = 3 + 3(-1) = 3 – 3 = 0
∴ f'(0) + 3f'(-1) = 0.

Question 9.
The time t, of a complete oscillation of a simple pendulum of length l is given by t = 2π\(\sqrt{\frac{l}{\mathrm{~g}}}\) where g is gravitational constant. Find the approximate percentage of error in t when the percen tage of error in l is 1%.
Solution:
AP Inter 1st Year Maths 1B Question Paper May 2019 6
AP Inter 1st Year Maths 1B Question Paper May 2019 7

Question 10.
Define stationary point with example.
Solution:
Stationary point: A point x = c in the domain of the function is said to be a stationary point of y = f(x) if f(c) = 0.

Section – B
(5 × 4 = 20)

II. Short answer type questions:

  1. Attempt ANY FIVE questions.
  2. Each question carries FOUR marks.

Question 11.
Find the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P.
Solution:
Let A = (2, 3), B = (-1, 5)
Let p(x1, y1) be any point on the locus.
Given geometric condition is PA2 + PB2 = AB2
AP Inter 1st Year Maths 1B Question Paper May 2019 8

Question 12.
When the origin is shifted to the point (2, 3), the transformed equation of a curve is : x2 + 3xy – 2y2 + 17x – 7y – 11 = 0. Find the original equation of the curve.
Solution:
Equations of transformation are x = x’ + h, y = y’ + k
x’ = x – h = x – 2, y’ = y – 3
Transformed equation is x2 + 3xy – 2y2 + 17x – 7y – 11 = 0
Original equation is (x – 2)2 + 3(x – 2) (y – 3) – 2 (y – 3)2 + 17(x – 2) – 7(y – 3) – 11 = 0
x2 – 4x + 4 + 3xy – 9x – 6y + 18 – 2y2 + 12y – 18 + 17x – 34 – 7y + 21 – 11 = 0
x2 + 3xy – 2y2 + 4x – y – 20 = 0
This is the required original equation.

Question 13.
A straight line passing through A(1, -2) makes an angle tan-1\(\left(\frac{4}{3}\right)\) with the positive direction of the X-axis in the anticlockwise sense. Find the points on the straight line whose distance from A is 5.
Solution:
The parametric equations of the line through A(1, -2) and whose slope is \(\frac{4}{3}\) (∵ tanθ = \(\frac{4}{3}\)) are
x = 1 + r cosθ and y = -2 + r sinθ
x = 1 + r(\(\frac{3}{5}\)) and y = -2 + r(\(\frac{4}{5}\))
The points on the above line at a distance of |r| = 5
⇒ r = ± 5
If r = 5 then x = 1 + 5(\(\frac{3}{5}\)) and y = -2 + 5(\(\frac{4}{5}\))
x = 4 and y = 2
If r = -5 then x = 1 – 5(\(\frac{3}{5}\)) and y = -2 – 5(\(\frac{4}{5}\))
x = – 2 and y = – 6
Hence required points are (4, 2) and (-2, – 6).

Question 14.
Check the continuity of the following function at 2:
AP Inter 1st Year Maths 1B Question Paper May 2019 1
Solution:
AP Inter 1st Year Maths 1B Question Paper May 2019 9

Question 15.
Find the derivative of f(x) = cot x using the first principle.
Solution:
Let f(x) = cot x
AP Inter 1st Year Maths 1B Question Paper May 2019 10
AP Inter 1st Year Maths 1B Question Paper May 2019 11

Question 16.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm, how fast is the enclosed area increases ?
Solution:
Let r, A be the radius, area of a circular ripple.
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5 cm/sec. dt
A = πr2
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = \(\pi \cdot 2 \mathrm{r} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}\)
When r = 8
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π.2(5).8
= 80 π cm2/sec.
∴ The enclosed area increases 80 π cm2/sec.

AP Inter 1st Year Maths 1B Question Paper May 2019

Question 17.
Find the angle between the curves y2 = 8x; 4x2 + y2 = 32.
Solution:
Given curve equations are y2 = 8x ….. (1)
4x2 + y2 = 32 …… (2)
From (1) and (2)
4x2 + 8x = 32
⇒ 4x2 + 8x – 32 = 0
⇒ x2 + 2x – 8 = 0
⇒ (x + 4) (x – 2) = 0
⇒ x = -4, x = 2
If x = 2 then y = ±4.
∴ The points of intersection of (1) and (2) are (2, 4) and (2, – 4)
AP Inter 1st Year Maths 1B Question Paper May 2019 12

Section – C

III. Long Answer type questions:

  1. Attempt ANY FIVE questions.
  2. Each question carries SEVEN marks.

Question 18.
Find the equations of the straight lines passing through the point (1, 2) and making an angle of 60° with the line \(\sqrt{3}\)x + y + 2 = 0.
Solution:
Let m be the slope of the required straight line Slope of the line \(\sqrt{3}\)x + y + 2 = 0 is –\(\sqrt{3}\)
Since the required straight line makes an angle of 60° with the line \(\sqrt{3}\)x + y + 2 = 0.
AP Inter 1st Year Maths 1B Question Paper May 2019 13
∴ The equation of the straight line passing through the point (1, 2) with slope \(\sqrt{3}\) is
y – 2 = \(\sqrt{3}\)(x – 1)
The equation of the straight line passing through the point (1, 2) with slope ‘o’ is
y – 2 = 0(x – 1) ⇒ y – 2 = 0
Hence required straight line equations are y – 2 = \(\sqrt{3}\)(x – 1) and y – 2 = 0.

Question 19.
Find the centroid and the area of the triangle formed by the lines 3x2 – 4xy + y2 = 0, 2x – y = 6.
Solution:
Given 3x2 – 4xy + y2 = 0
⇒ 3x2 – 3xy – xy + y2 = 0
⇒ 3x(x – y) – y(x – y) = 0
⇒ (x – y) (3x – y) = 0
⇒ x – y = 0 ………… (1)
3x – y = 0 ……… (2)
2x – y – 6 = 0 …… (3)
The point of intersection of (1) and (2) is 0(0, 0)
Solving (2) & (3)
AP Inter 1st Year Maths 1B Question Paper May 2019 14
AP Inter 1st Year Maths 1B Question Paper May 2019 15

AP Inter 1st Year Maths 1B Question Paper May 2019

Question 20.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is :
\(\frac{|c|}{\sqrt{(a-b)^2+4 b^2}}\)
Solution:
Let S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Let S = 0 represent lines be l1x + m1y + n1 = 0 …… (1)
l2x + m2y + n2 = 0 …… (2)
Then ax2 + 2hxy + by2 + 2gx + 2fy + c = (l1x + m1y + n1)(l2x + m2y + n2)
Comparing the co-efficients of like terms on both sides, we have
AP Inter 1st Year Maths 1B Question Paper May 2019 16
The perpendicular distance from origin to the line (1) = \(\frac{\left|\mathrm{n}_1\right|}{\sqrt{l_1^2+\mathrm{m}_1^2}}\)
The perpendicular distance from origin to the line (2) = \(\frac{\left|\mathrm{n}_2\right|}{\sqrt{l_2^2+\mathrm{m}_2^2}}\)
The product of the perpendicular distance from origin to the pair of straight lines
AP Inter 1st Year Maths 1B Question Paper May 2019 17

Question 21.
Find the angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0.
Solution:
Given 3l + m + 5n = 0 …… (1)
6mn – 2nl + 5lm = 0 ……. (2)
From (1), m = – (3l + 5n)
Substituting in (2)
-6n(3l + 5n) – 2nl – 5l (3l + 5n) = 0
-18ln – 30n2 – 2nl – 15l2– 25ln = 0
-15l2 – 45ln – 30n2 = 0
l2 + 3ln + 2n2 = 0
(l + 2n) (l + n) = 0
l + 2n = 0 or l + n = 0
AP Inter 1st Year Maths 1B Question Paper May 2019 18

Question 22.
Find the derivative of the function :
xx + (cot x)x.
Solution:
Let u = xx and v = (cot x)x
Let y = xx + (cot x)x
Then y = u + v
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d u}{d x}+\frac{d v}{d x}\) ……. (1)
Taking logarithms on both sides, we have
log u = x log x
Differentiating w.r.to (x) on both sides, we have
AP Inter 1st Year Maths 1B Question Paper May 2019 19
Taking logarithms on bothsides, we have
log v = x log (cot x)
Differentiating w.r.to ‘x’ on both sides, we have
AP Inter 1st Year Maths 1B Question Paper May 2019 20
AP Inter 1st Year Maths 1B Question Paper May 2019 21

AP Inter 1st Year Maths 1B Question Paper May 2019

Question 23.
Find the equations of the tangents to the curve y = 3x2 – x3, where it meets the X – axis.
Solution:
Given curve equation is y = 3x2 – x3
y = 0 ⇒ 3x2 – x3 = 0
⇒ x2 (3 – x) = 0
⇒ x = 0, 3
∴ The curve crosses the x – axis at 0(0, 0) and A(3, 0)
y = 3x2 – x3 ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6x – 3x2
\(\left(\frac{d y}{d x}\right)_{(0,0)}\) = 0 – 0 = 0
Equation of the tangent at 0(0, 0) is
y – 0 = 0(x – 0) ⇒ y = 0
y = 3x2 – x3 ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6x – 3x2
\(\left(\frac{d y}{d x}\right)_{(3,0)}\) = 18 – 27 = -9
Equation of the tangent at A(3, 0) is
y – 0 = – 9(x – 3)
y = – 9x + 27
⇒ 9x + y = 27
Hence required tangent equations are y = 0 and 9x + y = 27.

Question 24.
From a rectangular sheet of dimensions 30 cm × 80 cm, four equal squares of side x cm are removed at the corners, and the sides are then turned up so as to form an open rectangular box. Find the value of x, so that the volume of the box is the greatest.
Solution:
Length of the box = 80 – 2x = l
Breadth of the box = 30 – 2x = b
Height of the box = x = h
AP Inter 1st Year Maths 1B Question Paper May 2019 22
Volume = lbh
= (80 – 2x) (30 – 2x). x
= x(2400 – 220x + 4x2)
f(x) = 4x3 – 220x2 + 2400x
f(x) = 12x2 – 440x + 2400
= 4[3x2 – 110x + 600]
f(x) = 0 ⇒ 3x2 – 110x + 600 = 0
x = \(\frac{110 \pm \sqrt{12100-7200}}{6}\)
= \(\frac{110 \pm 70}{6}\) = \(\frac{180}{6}\) or \(\frac{40}{6}\) = \(\frac{30}{3}\) or \(\frac{20}{6}\)
If x = 30,
b = 30 – 2x
= 30 – 2(30)
= -30 < 0
⇒ x ≠ 30
∴ x = \(\frac{20}{3}\)
f'(x) = 24x – 440
When x = \(\frac{20}{3}\) f”(x) = 24.\(\frac{20}{3}\) – 440
= 160 – 440
= -280 < 0
f(x) is maximum when x = \(\frac{20}{3}\)
Volume of the box is maximum when x = \(\frac{20}{3}\) cm.

AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 3 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत ।

अ) दातृत्वं प्रियवक्तृत्वं धीरत्वमुचितज्ञता ।
अभ्यासेन न लभ्यन्ते चत्वारः सहजा गुणाः ॥
समाधान:
पदच्छेद (Word Division) : दातृत्वं, प्रियवक्तृत्वं, धीरत्वं, उचितज्ञता, अभ्यासेन, न, लभ्यन्ते, चत्वारः, सहजा, गुणाः ।
अन्वयक्रम : दातृत्वं, प्रियवक्तृत्वं, धीरत्वं, उचितज्ञता, अभ्यासेन, न, लभ्यन्ते, चत्वारः, गुणाः, सहजाः ।
अर्थ (Meanings) : दातृत्वम् = generosity, प्रियवक्तृत्वं = speaking sweetly, धीरत्वम् = courage, उचितज्ञता = doing things properly, अभ्यासेन = by practice, न लभ्यन्ते = are not acquired, चत्वारः = these four, गुणाः = qualities, सहजा: = (they are) natural.
भाव (Substance) : The four qualities namely generosity, sweet speech, courage and propriety are not acquired by practice, but are natural.

आ) प्रथमे नार्जिता विद्या द्वितीये नार्जितं धनम् ।
तृतीये नार्जितं पुण्यं चतुर्थे किं करिष्यति ॥
समाधान:
पदच्छेद (Word Division) : प्रथमे न, आर्जितं, विद्या, द्वितीये, न, आर्जितं धनम्, तृतीये न, आर्जितं पुण्यं चतुर्थे किं करिष्यति ।
अन्वयक्रम : प्रथमे, विद्या, नं, आर्जितम्, द्वितीये, धनं, न आर्जितम्, तृतीये, पुण्यं, न, आर्जितम्, चतुर्थे किं करिष्यति ।
अर्था (Meanings) : प्रथमे = during the first stage of life, in the first quarter; विद्या = education; न + आर्जितम् = not gained; द्वितीये = in the second stage, second quarter; धनम् = money, ricers; न आर्जितम् = not gained; तृतीये:= in the third stage of life, in the third quarter; पुण्यं = merit; न + आर्जितम् = not gained; चतुर्थे = in the last stage, in the last quarter; किं करिष्यति = what will he do?
भाव (Substance): If a person does not acquire knowledge in the first stage of his life, money in the second stage, and merit in the third stage, what will he do in the last stage?

2. एकं निबन्धप्रश्नं समाधत्त ।

अ) विभीषणोपदेशः इति पाठ्यभाग सारांशं लिखत ?
Answer:
Introduction: The lesson Vibhishanopadesa is an extract from the Yuddhakanda of the Ramayana, written by Valmiki. Vibhishana advises Ravana to send Sita to Rama.

The Advice of Vibhishana: Vibhishana advised Ravana that one should use their might only when the three other methods namely sama, dana and bheda fail. Then also it would work again the weak and already unfortunate enemies. They should not underestimate the enemy. परेषां सहसावज्ञा न कर्तव्या कथञ्चन । They should protect lives. They need not have an unnecessary fight with one who followed Dharma. Sita should be sent to Rama.

Vibhishana advised Ravana to abandon his anger which would destroy his comfort and dharma. Ravana should follow dharma which would increase joy and fame.

The Inauspicious signs: Vibhishana mentioned the various inauspicious signs witnessed since the arrival of Sita. “The milk of the cows has decreased. The mighty elephants have lost their rut. The horses are neighing piteously. The donkeys, camels and mules are shedding tears losing their hair. Flocks of crows are cawing harshly on all sides. Cruel animals have assembled at the gates of the city, and are roaring loudly.”

He compared Sita with a serpent. She had the hood of bosom, poison of worry, fangs of smile, heads of five fingers and a great body. He warned that Ravana would not be spared with life by Rama. Even the Sun, Indra, Yama and other could not protect him.

Friends, who were like enemies : Vibhishana accused the Rakshasas saying that they were enemies in the guise of friends. He said that a minister should weight the strength, loss and gain on both sides, and give a beneficial advice. वदेत् क्षमं स्वामिहितं मन्त्री। He asked whether anyone could stand before the arrows of Rama. He advised that Sita should be sent to Rama along with precious gifts. When Ravana said that he would not have spared if this speech was made by any other person, Vibhishana became angry. He flew in to the space, and said that as a brother Ravana could say anything, but he would not forgive him. He said that people who spoke sweetly could be found easily. But those who gave bitter but beneficial advice, and its listener were hard to be found. अप्रियस्य च पथ्यस्य वक्ता श्रोता च दुर्लभः | He wished him happiness, and left.

आ) सा मातृभूमिर्मम इति पाठ्यभाग सारांशं लिखत ?
Answer:
Introduction: The lesson Sa Matrubhumirmama was written by Dorbala Prabhakara Sarma. The poet describes the greatness of India in this poem.
Vedabhumi, Jnanabhumi and Dharmabhumi: The poet praises the motherland as a land of Vedas, wisdom, languages, gods, dharma, action, yoga, tapas, rivers, holy places and great people. Here, the cattle of the Vedas show the right path.

The people here are cultured, intelligent, healthy and sweetly speaking. यत्रैताः हृदि जागृताः प्रकृतयः सा मातृभूमिर्मम । The land is prosperous, waters are pure, breeze is purifying and life is friendly.

There are Rik mantras that praise the gods, Yajush mantras that are used in sacrifices, Atharva mantras that show the path of prosperity. This land is full of knowledge with the Vedas, Upanishads, Puranas, Itihasas, Kavyas etc. विद्या यत्र परोपकारविभवाः सा ज्ञानभूमिर्मम | There are different languages belonging to Vanga, Anga, Andhra, Kerala, Maharashtra, Sindhi, Gujarati, Karnataka etc.

Selfless People: The gods of rivers, villages, cities, towns, households, lands and others bestow boons. The lords of quarters such as Indra, Agni, Varuna, Vayu, Kubera, Yama etc. along with Vishnu, Uma and others grant welfare to the people. In a household, the mother takes care of the interests of her husband and children, father works, brothers give relief, teachers preach and the elders offer advice. ग्रमे यत्र गृहे समैक्यपरता सा धर्मभूमिर्मम | The people work selflessly, are engaged in meditation, prayers and contemplation. The wisdom of the sages ensures the welfare of the world, strengthens culture and enriches the knowledge.

Rivers and holy places: Rivers like Ganga, Yamuna, Saraswati, Godavari, Krishna, Narmada etc. remove sins, diseases and obstacles. Mountains such as Hemadri, Rajatadri, Aravali, Mahendra, Himalayas, Sahya etc. embellish this land. There are holy cities such as Kasi, Ayodhya, Puri, Kanchi, Madhura, Avanti, Tirupati, Ahobilam etc. Great sages and poets like Vyasa, Valmiki, Bhrigu, Sankara, Kalidasa, Bhavabhuti etc. belonged to this land.

3. एकं निबन्धप्रश्नं समाधत्त ।

अ) भोजराजाय सालभंजिकया उक्तं विक्रमस्य औदार्यं लिखत |
Answer:
Introduction: The lesson Vikramasya Audaryam is taken from vikramarkacharitam written by Sivadasa. When Bhoja wants to ascend the throne of Vikrama the statue on the fourth step tells this story about the generosity of king Vikrama.

The childless Brahmin:
While Vikramaditya ruled Ujjaini, there was a learned Brahmin in that city, who was virtuous, but who had no children. Once, his wife said to him that there was no heaven for one who had no children सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना | The Brahmin worshipped Lord Siva. The god appeared in his dream and asked him to perform Pradoshavrata. The Brahmin did so and he begot a son, whom he named Devadatta. When his son grew up, the Brahmin performed his marriage, and went on a pilgrimage to Varanasi.

Devadatta’s help to Vikrama:
Once Devadatta went into the forest to collect fuel sticks for sacrifice. At that time, king Vikrama came to the forest for hunting. There he asked Devadatta the way to the city, even though he knew it.

Devadatta himself led him to the city. The king praised him, and appointed him in his court. One day the king praised Devadatta’s help in the assembly saying that he could never repay his debt.

Devadatta kidnaps the prince:
Devadatta wondered whether the king’s praise was true or false. He wanted to test the words of the king. He kidnapped the prince, and gave one of his ornaments to his servant and sent him to the market to sell it. The king’s men who were searching for the prince caught him, and brought him to the king. When the king asked him, the servant said that he was Devadatta’s servant. The king sent for Devadatta, who said that he killed the prince or money. The members of the court said that Devadatta should be sentenced to death. But the king said that Devadatta rescued him in the forest. One should not find faults in those dependent on him. Hardy महतां गुणदोषचिन्ता | He said that his son died because of his previous deeds only. He honoured Devadatta and sent him away.

Devadatta returned with the prince, and told the king that he wanted to test the words of the king. He praised the generosity of the king, who said that one should not forget the help done to him. यः कृतमुपकारं विस्मरति स एव पुरुषाधमः ।
Thus the statue told Bhoja about the generosity of king Vikrama.

आ) वेङ्कटरावस्य स्वभावं संग्रहेन लिखत ।
Answer:
Introduction: The lesson Bhishajah Bhaishajyam was written by Prof. Pullela Sriramachandra. It is taken from his Sriramachandra-laghukavya-sangraha. This lesson describes the story of a selfish doctor, and the fruit he reaped for his selfishness.

The Villager’s Plea: One day some villagers came to Dr. Venkata Rao, and requested him to attend to a boy who was injured in an accident. Venkata Rao childed them for not coming in time. वैद्योऽपि मानव एवं | He accused them of trying to get treatment done without paying fee. He insulted them saying the boy would be delicate as the baby of a donkey. When they left as the boy was serious क्षणे क्षणे किल परिक्षीयते बालस्य दशा, Venkata Rao thought nothing would happen if one puppy died.

The poor and intelligent Venkata Rao: Venkata Rao was the son of a poor farmer. He was very intelligent and secured a seat in medical college. His father sold their agricultural land for his education. A rich man married his daughter to Venkata Rao. Venkata Rao’s practice also picked up. Along with money, the three defects grew in him. They were considering others as insects, himself as god, and accepting others praise as the truth. When he spoke of his father as a beggar, his father left him and returned to their village.

Marriage was also a business affair for Venkata Rao. विवाहो नाम वणिग्व्यवहार एव । For him money was everything. He never loved his wife. His son Suresh alone became the object of his affection.

Desire for Power: Venkata Rao thought that money was useless without power. He became an MLA twice. But he could not become a minister. He was unhappy about that. He could not win the seat the third time. He blamed the people for that. His hatred for people grew.

Manjuhasini, the Lady Doctor: At that time, Dr. Manjuhasini joined the government hospital there. She was Venkata Rao’s classmate in medical college. She rejected Venkata Rao’s advances. Venkata Rao was hoping that she might have changed now as he became rich.

The death of his son : Venkata Rao received a phone call from Manjuhasini requesting his help in an emergency case. His driver tried to inform him that his son was not there at the school when he went there after getting the brake repaired. Venkata Rao cut him short saying that the boy would have reached home. But when he went to the hospital he saw the same villagers who came to him earlier in the day, and the body of his dead son.

4. त्रयाणां प्रश्नानां समाधानानि लिखत |

अ) कीदृशं वस्तु तथैव तिष्ठति ?
समाधान:
हुतं दत्तं च वस्तु तथैव तिष्ठति ।

आ) नृपश्रियः कीदृशाः ?
समाधान:
नृपश्रियः भुजङ्गजिह्वाचपला ।

इ) द्रोणद्रुपदयोः सम्बन्धः कीदृश: ?
समाधान:
द्रोणद्रुपदयोः स्नेहबन्धः अच्छेद्यः अविभाज्यः च । सः पटुचिक्कणात्मकः अभवत् ।

ई) बालानां कलहः कीदृश: ?
समाधान:
बालानां कलहः तत्कालीनः भवति । बालाः कलहायन्ते अनन्तरं क्रीडन्ते च ।

उ) भोजः कोशाध्यक्षं प्रति किं सूचितवान् ?
समाधान:
कविकुलगुरोः कालिदासस्य सदनं एकलक्षसुवर्णमुद्राः प्रेष्यन्ताम् इति भोजः कोशाध्यक्षं सूचितवान् ।

ऊ) कालिदासेन उक्तस्य श्लोकस्य भावः कः ?
समाधान:
अहं खादन् न गच्छामि, हसन् न जल्पामि, गतं न शोचामि कृतं न स्मरामि, द्वयोः संभाषणमध्ये न प्रविशामि । तेन कथम् अहं मूर्खः भवामि ।

5. द्वयोः संदर्भ व्याख्यानं लिखत ।

अ) परेषां सहसावज्ञा न कर्तव्या कथञ्चन ।
समाधान:
परिचयः – एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् । एषः पाठः रामायणस्य युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
भावः – परेषां बलानि अपरिमेयानि । तेषां सहसा अवज्ञा न कुर्यात् ।

आ) राज्येन किं तद्विपरीतवृत्तेः प्राणैः उपक्रोशमलीमसैर्वाः ।
समाधान:
परिचयः एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषः भागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः दिलीपः सिरं प्रति एवं वदति ।
भावः अक्षत्रविरुद्धवृत्तेः राज्येन किं प्रयोजनम् निन्दया मलिनयुतं प्राणेन किं प्रयोजनम् ।
विवरणम्ः क्षत्रियः राज्यं पालयति, स्वपराक्रमेन धर्म पालयति, आक्षितान् रक्षति ।

इ) विद्या यत्र परोपकारविभवाः सा ज्ञानभूमिर्मम ।
समाधान:
परिचयः – एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् । अस्य कविः श्रीमान् दोर्बल प्रभाकरशर्मा |
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः, धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः – यत्र वेदाः उपनिषदः, पुराणेतिहासाः आयुर्योगधर्मसङ्गीतग्रन्थाः च परोपकारविभवाः भवन्ति सा ज्ञानभूमिः मम मातृभूमिः इति कविः वदति ।

ई) सा संस्कृताख्या सुकृतैकलभ्या ।
समाधान:
परिचय : गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री ।
सन्दर्भ : गीर्वाणवाण्याः वैभवं वर्णयनन् कविः एवं वदति । संस्कृतभाषा सुधास्रवन्ती, सुरभाषिता, सूच्चारिता, सूक्तिरन्नवार्धिः च ।
भाव : संस्कृतभाषा पुण्यैफलेन एव लभ्या ।
विवरणम् : संस्कृतभाषा अतिप्राचीना, सुधास्रवन्ती च, सूक्तिरत्नै सुशोभिता।

6. द्वयोः ससंदर्भ व्याख्यानं लिखत ।

अ) मृत्युर्वै प्रणिनां ध्रुवः ।
समाधान:
परिचयः एतत-वाक्यं, “मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः पुत्रशोकं अनुभवन्तं कौण्डिन्यं कपिलः प्रति एवं अवदत् भावः सर्वेषां प्राणिनां मृत्युः निश्चयम्
विवरणम्ः जातस्य मरणं, मृतस्य जननं निश्चयम्, अतः सर्वेषां प्राणिनां मृत्युः ध्रुवम् ।

आ) सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना ।
समाधान:
परिचय : इदं वाक्यं “विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास ।
सन्दर्भ : ब्राह्मणस्य पत्नी स्व भर्तारं प्रति एवं अवदत् ।
भाव : सत्कुमारिण वशं राज्ञा वसुधा सूर्येण भुवनत्रयं प्रकाश्यन्ते ।
विवरणम् : वाणी व्याकरणेन, हंसमिधुनैः पद्यः, पण्डितैः सभा सुपुत्रेण : कुलं राज्ञा, भूमिः सूर्येण लोकत्रयं विराजन्ते ।

इ) विवाहो नाम वणिग्व्यवहार एव ।
समाधान:
परिचयः वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः एकः वृद्धः युवकं उद्दिश्य एवं अवदत् ।
भावः परिणयः नाम केवलं वाणिज्यव्यवहारः ।
विवरणम्ः सर्वस्य धनं प्रधानं, सर्वेगुणाः धनं आश्रयन्ति । तद्दत् विवाहः अपि केवलं वाणिज्यमेव ।

ई) राजा तं कनकाभिषेकेण सत्कृतवान् ।
समाधान:
परिचय : एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भ : अप्पय्यदीक्षितः शिवार्कमणिदीपिकां अरचयत् । तदा राजा चिनबोकनायकः तं कनकाभिषेकेण सत्कृतवान् । तेन हिरण्येन अप्पय्यः स्वग्रामे शिवालयं निर्मितवान् ।
भाव : राजा अप्पय्यदीक्षितं कनकाभिषेकं कृत्वा सम्मानितवान् ।

7. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) नराः क इव रणे सीदन्ति ?
समाधान:
शूराः अपि नराः वालुकासेतवः इव रणे सीदन्ति ।

आ) सुदक्षिणादिलीपौ किमर्थं नन्दिनीधेनोः सेवाम् अकरुताम् ?
समाधान:
सुदक्षिमादिलीपौ सन्तानार्थं – नन्दिनीधेनोः सेवाम् अकुरुताम् ।

इ) पृथिव्यां त्रीणि रत्नानि कानि ?
समाधान:
पृथिव्यां त्रीणि रत्नानि जलं, अन्नं, सुभाषितम् ।

ई) अस्पृशन्नेव वित्तानि कः परेभ्यः प्रयच्छति ?
समाधान:
अस्पृशन्नेव वित्तानि कृपणः परेभ्यः प्रयच्छति ।

उ) सर्वप्राणिनामुपकारकारि किम् ?
समाधान:
सर्वप्राणइनाम् उपकारकारि सुमहत् आर्षविज्ञानामृतम् ।

ऊ) संस्कृताख्या वाणी कीदृशी ?
समाधान:
संस्कृताख्या वाणी सुधासवन्ती, सूक्तिसुरत्नर्वार्धः, सुकाव्यसंदोहनिधिः ।

8. त्रयाणां प्रश्नानां समाधानानि लिखत ।

अ) कौण्डिन्येन मन्दविषः किम् इति शप्तः ?
समाधान:
कौण्डिन्येन मन्दविषः ”अद्य आरभ्य मण्ङ्गकानां वाहनं भविष्यसि ” इति शप्तः ।

आ) ब्राह्मणं प्रति तत्प्रेयसी किं जगाद ?
समाधान:
ब्राह्मणं प्रति तत्प्रेयसी एवं जगाद – अपुत्रस्य गतिर्नास्ति स्वर्गो नैव च नैव च । सत्पुत्रेणैव कुलं प्रकाशते इति ।

इ) धर्मपालस्य कति पुत्राः ? ते च के ?
समाधान:
धर्मपालस्य त्रयः पुत्रः आसन् । ते च सुमन्त्रः, सुमित्रः कामपालः च ।

ई) वैद्यः श्रीधरं प्रति किमुक्तवान् ?
समाधान:
अधिककार्यभारेण हृद्रोगः प्राप्तः, कञ्चित्कालं विश्रान्तिरपेक्षिता । नियतरूपेण औषधसेवनं करणीयम् – इति वैद्यः श्रीधरम् उक्तवान् |

उ) वेङ्कटरावस्य पिता कः ?
समाधान:
वेङ्कटरावस्य पिता सुब्बरायशास्त्री ।

ऊ) रत्नखेटिश्रीनिवासस्य अभिलाषा का ?
समाधान:
रत्नखेटश्रीनिवासस्य अभिलाषा – अप्पय्यः जेतव्यः, स्वपादयोः पातनीयः चं इति ।

9. एकेन वाक्येन समाधानं दत्त ।

अ) सुखधर्मनाशनं केन भवति ?
समाधान:
सुखधर्मनाशनं कोपेन भवति ।

आ) रघुवंशमहाकाव्यं केन विरचितम् ?
समाधान:
कालिदासेन

इ) सर्वत्र का पूज्यते ?
समाधान:
विद्या

ई) केन सर्वे वशाः भवन्ति ?
समाधान:
द्रव्येण सर्वे वशाः भवन्ति ।

उ) सर्वासु सुतासु कस्याः गुणाः भवन्ति ?
समाधान:
मातुः

10. एकेन वाक्येन समाधानं दत्त ।

अ) हितोपदेशः केन विरचितः ?
समाधान:
नारायणपण्डितेन ।

आ) विक्रमार्कचरितं केन विरचितम् ?
समाधान:
शिवदासेन 1

इ) राजहंसस्य पत्नी का ?
समाधान:
वसुमती

ई) अपर्याप्तः समयः इति पाठ्यभागः केन विरचितः ?
समाधान:
अपर्याप्तः समयः इति पाठ्यभागः राणी सदाशिवमूर्तिना विरचितः ।

उ) भिषजः भैषज्यम् इति पाठ्यांशः केन विरचितः ?
समाधान:
पुल्लेल श्रीरामचन्द्रः

11. अधो निर्दिष्टकथां पठित्वा प्रश्नानां समाधानानि दत्त ।

प्रश्नाः
(1) सिंहस्य नाम किम् ?
समाधान:
सिंहस्य नाम करालकेसरी ।

(2) परिचारकः कः ?
समाधान:
परिचारकः धूसरको नाम सृगालः ।

(3) सृगालः येन सह युद्धं कृतवान् ?
समाधान:
सृगालः हस्तिना सह युद्धं कृतवान् ।

(4) धूसरकः सिंहं प्रति किमब्रवीत् ?
समाधान:
धूसरकः सिंहं प्रति ‘स्वामिन्! बुभुक्षया पीडितोऽहम् पदात्पदमपि चलितुं न शक्नोमि । तत्कथं ते शुश्रूषां करोमि ” इति अब्रवीत ।

(5) सिंहः सृगालं प्रति किमाह ?
समाधान:
सिंहः सृणालं प्रति – “भोः ! गच्छः अन्वेषय किंञ्चित् सत्त्वम् । येनेमाम्ं अवस्थां गतोऽपि व्यापादयामि” ।

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त ।

अ) जगद् + जननी
समाधान:
जगज्जननी = श्चुत्व सन्धिः

आ) पेस् + टा
समाधान:
पेष्टा = जश्त्व सन्धिः

इ) अच् + अन्तः
समाधान:
अजन्तः = जश्त्व सन्धिः

ई) जगत् + नाथः
समाधान:
जगन्नाथः = अनुनासिक सन्धिः

उ) कः + अपि
समाधान:
=
विसर्ग सन्धिः

ऊ) हरिः + गच्छति
समाधान:
हरिर्गच्छति = विसर्गरेकादेश सन्धिः

13. नामनिर्देशपूर्वकं त्रीणि विघटयत ।

अ) उड्डयनम्
समाधान:
उत् + डयनम् = ष्टुत्व सन्धिः

आ) मनश्चलति
समाधान:
मनस् + चलति = श्चुत्व सन्धिः

इ) षडाननः
समाधान:
षट् + आननः = जश्त्व सन्धिः

ई) तन्मात्रम्
समाधान:
तत् + मात्रम् = अनुनासिक सन्धिः

उ) सोऽपि
समाधान:
सः + अपि = विसर्ग सन्धिः

ऊ) पितुरिच्छा
समाधान:
पितुः + इच्छा = विसर्गरेफादेशः सन्धिः

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत ।

अ) भवत्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions 2

आ) सम्पद्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions 3

इ) महत्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions 4

ई) एतत् (पुं)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions 5

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत ।

अ) अहिंसा
समाधान:
न हिंसा = नञ् तत्पुरुष समासः

आ) लतातन्वी
समाधान:
लता इव तन्वी = उपमानपूर्वपदकर्मधारय समासः

इ) त्रिभुवनम्
समाधान:
त्रयाणां भुवनानां समाहारः = द्विगु समासः

ई) कंसकृष्णौ
समाधान:
कंसः च कृष्णः च = द्वन्द्व समासः

उ) शाकप्रति
समाधान:
शाकस्य लेशः = अव्ययीभाव समासः

ऊ) महाबलः
समाधान:
मह्त् बलम् यस्य सः = द्विपदाबहुव्रीहि समासः

16. अधो निर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखत ।
AP Inter 2nd Year Sanskrit Model Paper Set 3 with Solutions 1
प्रश्नाः
1. शिवः चित्रं पश्यति ।
2. शिवः देवं पूजयति ।
3. शिवः क्षीरं पिबति ।
4. सः चित्रं पश्यति ।
5. सः देवं पूजयति ।

AP Inter 1st Year Maths 1B Question Paper March 2019

Access to a variety of AP Inter 1st Year Maths 1B Model Papers and AP Inter 1st Year Maths 1B Question Paper March 2019 allows students to familiarize themselves with different question patterns.

AP Inter 1st Year Maths 1B Question Paper March 2019

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

  1. Answer all questions.
  2. Each question carries two marks.

Question 1.
Find the angle which the straight line y = x – 4 makes with the y-axis.
Solution:
Given straight line is y = \(\sqrt{3}\)x – 4
m = \(\sqrt{3}\) ⇒ tan θ = \(\sqrt{3}\)
⇒ θ = 60°
The straight line y = \(\sqrt{3}\)x – 4 makes angle with the Y – axis is 90° – 60° = 30°.

Question 2.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 =0.
Solution:
Given straight lines are 3x + 4y – 3 = 0
⇒ 6x + 8y – 6 = 0 ……. (1)
6x + 8y – 1 = 0 ……… (2)
∴ The distance between the parallel straight lines.
(1) and (2) is \(\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}\)
= \(\frac{|-1-(-6)|}{\sqrt{36+64}}\) = \(\frac{|-1+6|}{\sqrt{100}}\)
= \(\frac{5}{100}\) = \(\frac{1}{2}\)

AP Inter 1st Year Maths 1B Question Paper March 2019

Question 3.
Find ‘x’, if the distance between (5, -1, 7) and (x, 5, 1) is 9 units.
Solution:
Let A = (5, -1, 7)
B = (x, 5, 1)
Given AB = 9
⇒ \(\sqrt{(x-5)^2+(5+1)^2(1-7)^2}\) = 9
⇒ (x – 5)2 + 36 + 36 = 81
⇒ (x – 5)2 = 9
⇒ x – 5 = ±3
⇒ x – 5 = 3 (or) x – 5 = -3
⇒ x = 8 (or) x = 2.

Question 4.
Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form.
Solution:
Given plane equation is 4x – 4y + 2z + 5 = 0
⇒ 4x – 4y + 2z = – 5
⇒ \(\frac{4 x}{-5}-\frac{4 y}{-5}+\frac{2 z}{-5}\) = 1
⇒ \(\frac{x}{-5 / 4}+\frac{y}{5 / 4}+\frac{z}{-5 / 4}\)
x- intercept = \(\frac{-5}{4}\)
y – intercept = \(\frac{5}{4}\)
z – intercept = \(\frac{-5}{2}\)

Question 5.
Compute \(\lim _{x \rightarrow 0} \frac{e^{3+x}-e^3}{x}\)
Solution:
AP Inter 1st Year Maths 1B Question Paper March 2019 20

Question 6.
Compute \(\lim _{x \rightarrow 3} \frac{x^2+3 x+2}{x^2-6 x+9}\)
Solution:
AP Inter 1st Year Maths 1B Question Paper March 2019 2

Question 7.
Find the derivative of the function tan-1 (log x).
Solution:
Let y = tan-1 (logx)
differentiating with respect to ‘x’ on both sides, we have
AP Inter 1st Year Maths 1B Question Paper March 2019 37

Question 8.
If y = \(\frac{2 x+3}{4 x+5}\) then find y”.
Solution:
Given y = \(\frac{2 x+3}{4 x+5}\)
differentiating with respect to x on both sides, we have
AP Inter 1st Year Maths 1B Question Paper March 2019 21

Question 9.
Define relative error and percentage error of the variable y.
Solution:
Relative error in y = \(\frac{\Delta y}{y}\)
Percentage error in y = \(\frac{\Delta y}{y}\) × 100

AP Inter 1st Year Maths 1B Question Paper March 2019

Question 10.
Find the absolute extremum of f(x) = x2 defined on [-2, 2].
Solution:
Given f(x) = x2 defined on [-2, 2]
Clearly f is continuous on [-2, 2].
It can be shown that it has only local minimum and the point of local minimum is 0.
∴ The absolute maximum of f is the largest value of f(-2), f(0) and f(2)
f(-2) = (-2)2 = 4
f(0) = 02 = 0
f(2) = 22 = 4.
∴ The absolute maximum value is 4.
similarly the absolute minimum is the least value of 4, 0, 4.
∴ The absolute minimum value is 0.

Section – B

II. Short Answer Type Questions.

  1. Attempt any five questions.
  2. Each question carries four marks.

Question 11.
A (5, 3) and B (3, -2) are two fixed points. Find the equation of locus of P, so that the area of triangle PAB is 9.
Solution:
Given A = (5, 3)
B = (3, – 2)
Let P(x1, y1) be any point on the locus.
Given geometric condition is area of ∆PAB is 9.
⇒ \(\frac{1}{2}\left|\begin{array}{cccc}
x_1 & 5 & 3 & x_1 \\
y_1 & 3 & -2 & y_1
\end{array}\right|\) = 9
AP Inter 1st Year Maths 1B Question Paper March 2019 22

Question 12.
When the origin is shifted to the point (3, -4) and transformed equation is x + y2 = 4. Find the original equation.
Solution:
Here (h, k) = (3, -4)
we know x = x + h ⇒ x = x – h
⇒ x = x – 3
y = y + k ⇒ y = y – k
⇒ y = y – (-4)
⇒ y = y + 4
Given transformed equation is x2 + y2 = 4
⇒ (x – 3)2 + (y + 4)2 = 4
⇒ x2 – 6x + 9 + y2 + 8y + 16 = 4
⇒ x2 + y2 – 6x + 8y + 21 = 0.
∴ Required original equation is x2 + y2 – 6x + 8y + 21 = 0.

Question 13.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a3 + b3 + c3 = 3 abc.
Solution:
Given ax + bc + c = 0 ……… (1)
bx + cy + a = 0 ……… (2)
cx + ay + b = 0 ……… (3)
solving (1) and (2)
AP Inter 1st Year Maths 1B Question Paper March 2019 23

Question 14.
Compute \(\lim _{x \rightarrow a}\left(\frac{x \sin a-a \sin x}{x-a}\right)\)
Solution:
AP Inter 1st Year Maths 1B Question Paper March 2019 24
AP Inter 1st Year Maths 1B Question Paper March 2019 25

Question 15.
Find the derivative of the function cot x from the first principle.
Solution:
Let f'(x) = cot x
f (x + h) = cot (x + h)
By first principle
AP Inter 1st Year Maths 1B Question Paper March 2019 26
AP Inter 1st Year Maths 1B Question Paper March 2019 27

AP Inter 1st Year Maths 1B Question Paper March 2019

Question 16.
Find the approximate value of \(\sqrt[3]{999}\).
Solution:
Let f (x) = x1/3, x = 1000, ∆x = – 1
AP Inter 1st Year Maths 1B Question Paper March 2019 28

Question 17.
The distance – time formula for the motion of a particle along a straight line is s = t3 – 9t2 + 24t – 18. Find when and where the velocity is zero.
Solution:
Given s = t3 – 9t2 + 24t – 18.
Velocity v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3t2 – 18t + 24t.
v = 0 ⇒ 3t2 – 18t + 24t = 0
⇒ t2 – 6t + 8 = 0
⇒ t2 – 2t – 4t + 8 = 0
⇒ t (t – 2) – 4 (t – 2) = 0
⇒ (t – 2)(t – 4) = 0
t = 2 (or) t = 4.
When t = 2
s = 23 – 9 (2)2 + 24(2) – 18
= 8 – 36 + 48 – 18
= 56 – 54
= 2 units
When t = 4
s = 43 – 9 (4)2 + 24 (4) – 18
= 64 – 144 + 96 – 18
= 160 – 162 = 2 units.

Section – C

III. Long Answer Type Questions.

  1. Attempt any five questions.
  2. Each question carries seven marks.

Question 18.
If Q (h, k) is the image of the point P(x1, y1) with respect to the straight line ax + by + c = 0 then prove that \(\frac{h-x_1}{a}\) = \(\frac{k-y_1}{b}\) = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
Solution:
Let L ≡ ax + by + c = 0
Given Q (h, k) is the image of p(x1, y1)
with respect to the straight line L = 0.
AP Inter 1st Year Maths 1B Question Paper March 2019 29
⇒ a(h + x1) + b(k + y1) + 2c = 0
⇒ a(x1 + aλ + x1) + b(y1 + b + y1) + 2c = 0.
⇒ 2ax1 + a2λ + 2by1 + b2λ + 2c = 0.
⇒ (a2 + b2)λ = -2(ax1 + by1 + c)
⇒ λ = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
from (1)
∴ \(\frac{h-x 1}{a}\) = \(\frac{k-y_1}{b}\) = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)

Question 19.
If the equation S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of parallel straight lines then show that
(i) h2 = ab
(ii) af2 = bg2 and
(iii) the distance between the parallel lines
= \(\sqrt[2]{\frac{g^2-a c}{a(a+b)}}\) = \(\sqrt[2]{\frac{f^2-a c}{b(a+b)}}\)
Solution:
Let the two parallel straight lines represented by S = 0 be
lx + my + n1 = 0 ………… (1)
and lx + my + n2 = 0 ……. (2)
Then, S ≡ λ(lx + my + n1) (lx + my + n2), for some real λ ≠ 0.
From this we have
l2 = \(\frac{a}{\lambda}\), m2 = \(\frac{b}{\lambda}\), n1n2 = \(\frac{c}{\lambda}\), lm = \(\frac{h}{\lambda}\)
l(n1 + n2) = \(\frac{2 \mathrm{~g}}{\lambda}\) and m(n1 + n2) = \(\frac{2 f}{\lambda}\).
Now

(i) h2 = λ2l2m2 = (λl2)(λm2) = ab.

(ii) 4af2 = (λ2l2) (λ2m2) (n1 + n2)2
= (λm2) (λ2l2) (n1 + n2)2 = b(4g2)
so that af2 = bg2.

(iii) Distance between the parallel lines
AP Inter 1st Year Maths 1B Question Paper March 2019 30

AP Inter 1st Year Maths 1B Question Paper March 2019

Question 20.
Find the value of k, if the lines joining the origin to the points of intersection of the curves 2x2 – 2xy + 3y2 + 2x – y – 1 =0 and the line x + 2y = k are mutually perpendicular.
Solution:
AP Inter 1st Year Maths 1B Question Paper March 2019 31
Equation of the circle is x2 + y2 = a2 …… (1)
Equation of AB is lx + my = 1 ……….. (2)
Homogenising (1) with the help of (2)
Combined equation of OA, OB is
x2 + y2 = a2.12
x2 + y2 = a2 (lx + my)2
= a2(l2x2 + m2y2 + 2lmxy)
= a2l2x2 + a2m2y2 + 2a2lmxy
i.e., a2l2x2 + 2a2 lmxy + a2 m2y2 – x2 – y2 = 0
(a2l2 – 1) x2 + 2a2 lmxy + (a2m2 – 1) y2 = 0
Since OA, OB are perpendicular
Co-efficient of x2 + co-efficient of y2 = 0
a2l2 – 1 + a2m2 – 1 = 0
a2(l2 + m2) = 2
This is the required condition.

Question 21.
Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2 + m2 – n2 = 0.
Solution:
Given l + m + n = 0 …… (1)
l2 + m2 – n2 = 0 ……. (2)
From (1) l = – m – n …… (3)
Substituting in (2)
(- m – n)2 + (m2 – n2) = 0.
⇒ (m + n)2 + (m + n) (m – n) = 0.
⇒ (m + n) [m + n + m – n] = 0
⇒ 2m (m + n) = 0
⇒ m = 0, m + n = 0
When m = 0
From (3), l = -n
∴ \(\frac{l}{l}\) = \(\frac{m}{0}\) = \(\frac{n}{-l}\)
When m = – n
From (3) l = -(-n) – n
= n – n
= 0.
∴ \(\frac{l}{0}\) = \(\frac{m}{l}\) = \(\frac{n}{-l}\)
∴ The d.rs of the two lines are (1, 0, -1) and (0, 1, -1) let ‘θ’ be the angle between the lines.
AP Inter 1st Year Maths 1B Question Paper March 2019 32
AP Inter 1st Year Maths 1B Question Paper March 2019 33

Question 22.
If y = x \(\sqrt{a^2+x^2}\) + a2 log (x + \(\sqrt{a^2+x^2}\)) then prove that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(2 \sqrt{a^2+x^2}\)
Solution:
AP Inter 1st Year Maths 1B Question Paper March 2019 34

AP Inter 1st Year Maths 1B Question Paper March 2019

Question 23.
Show that the condition for the orthogonality of the curves ax2 + by2 = 1 and a1x2 + b1y2 = 1 is \(\frac{1}{a}-\frac{1}{b}\) = \(\frac{1}{a_1}-\frac{1}{b_1}\).
Solution:
Let P(x1, y1) be the point of intersection of the curves
Given curve equation is ax2 + by2 = 1
Differentiating (1) w.r. to y on both sides, we have
AP Inter 1st Year Maths 1B Question Paper March 2019 35
Since ‘P’ lies on ax2 + by2 = 1 and a1x2 + b1y2 = 1
Solving these two equations, we get
AP Inter 1st Year Maths 1B Question Paper March 2019 36

Question 24.
Find the points of local extrema and local extrema for the function f(x) = cos 4x defined on \(\left(0, \frac{\pi}{2}\right)\)
Solution:
The function f(n) = cos 4x defined on (0, \(\frac{\pi}{2}\)) …….. (1)
f'(x) = – 4 sin 4x ……… (2)
f”(x) = -16 cos 4x …….. (3)
f'(x) = 0 ⇒ – 4 sin 4x = 0
⇒ sin 4x = 0.
⇒ 4x = 0, π, 2π, 3π, 4π, ….
⇒ x = 0, \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\), π
Since x ∈ (0, \(\frac{\pi}{2}\))
∴ x = \(\frac{\pi}{4}\)
f”(\(\frac{\pi}{4}\)) = -16 cos 4 (\(\frac{\pi}{4}\))
= -16 cos π
= – 16 (-1)
= 16 > 0.
∴ The function f has local minimum at x = \(\frac{\pi}{4}\),
∴ Local minimum value = f(\(\frac{\pi}{4}\))
= cos 4(\(\frac{\pi}{4}\))
= cos π
= – 1.

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 2 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत । (1 × 8 = 8)

अ) तक्षकस्य विषं दन्ते मक्षिकायास्तु मस्तके |
वृश्चिकस्य विषं पुच्छे सर्वाङ्गे दुर्जने विषम् ||

पदच्छेद (Word Division) : तक्षकस्य, विषं, दन्ते, मक्षिकायाः, तु, मस्तके, वृश्चिकस्य, विषं, पुच्छे, सर्वाङ्गे, दुर्जने, विषम् ।
अन्वयक्रम : तक्षकस्य, विषं, दन्ते, मक्षिकायाः, मस्तके, वृश्चिकस्य, पुच्छे, विषं, दुर्जने, सर्वाङ्गे, विषम् ।
अर्था (Meanings) : तक्षकस्य = to a serpent, विषम् = poison, दन्ते in fangs, मक्षिकायाः = to a bee or spider, मस्तके = in head (poison), वृश्चिकस्य = to a scorpion, पुच्छे = in tail, विषम् = poison, दुर्जने = to an evil person, सर्वांगे = throughout the body, विषम् = poison..
भाव (Substance) : A serpent has poison is its fangs, a bee has in its head, and a scorpion has poison in its tail. But an evil person has poison in his whole body.

आ) उद्यमं साहसं धैर्यं बुद्धिः शक्तिः पराक्रमः |
षडेते यत्र वर्तन्ते तत्र देवः सहायकृत् ॥

पदच्छेद (Word Division) : उद्यमं, साहस, धैर्य, बुद्धिः, शक्तिः, पराक्रमः, षडेते, यत्र, वर्तन्ते, तत्र देवः, सहायकृतः ।
अन्वयक्रम : उद्यमं, साहस, धैर्यं, बुद्धिः, शक्तिः, पराक्रमः, षडेते, यत्र, वर्तन्ते, तत्र, देवः, सहायकृत् ।
अर्थ (Meanings) : उद्यमम् = effort; साहसं = adventure; धैर्यं = courage; बुद्धिः = intellect; शक्तिः = strength; पराक्रमः = valour; एतेषद् = these six qualities; यत्र = where; वर्तन्ते = exist; तत्र-देवः there the god; सहायकृत् = does help.
भाव (Substance): Where these six qualities namely effort, adventure, courage, intellect, strength and valour are there, god will be a help there.

2. एकं निबन्धप्रश्नं समाधत्त । (1 × 6 = 6)

अ) धर्मनिष्ठा इति पाठ्यांशस्य सारांशं संक्षेपेण लिखत ?
Answer:
Introduction: The lesson Dharmanishta is an extract from the 2nd canto of Raghuvamsa, written by Kalidasa. This lesson describes the moral and devotional character of king Dilipa.

A Lion attacks Nandini: The childless king Dilipa his wife Sudakshina were engaged in the service of Nandini, the cow of sage Vasishta. One day Nandini which went to the forest guarded by the king wished to test the devotion of the king, and entered a Himalayan cave. Thinking that the cow was unassailable to any wild animal, the king took his eyes off her to watch the beauty of the mountain. Then a lion attacked her. Alerted by her cry, the king became ready to kill the lion by shooting an arrow at it. But his hand got fixed on the shaft of the arrow. चित्रार्पितारम्भ इवावतस्थे ।

Then the lion told him that he was a servant of Lord Siva. He was appointed there to protect the Devadaru tree grown by Parvathi. He was made a lion, and any animal that arrived there when he was hungry would become his food.

Dilipa’s offer: Then the king said that Siva was to be respected, and at the same time, the property of the teacher was to be saved. He offered his own body as food to the lion requesting the release of the cow. But the lion said that the king was young and an emperor. He seemed to be imprudent in giving up more for little. By showing compassion to the cow, she alone would be saved. But if he lived, he could always save his people from calamities.

The king said that a king should protect his subjects. तद्विपरीतवृत्तेः, प्राणैः उपक्रोशमलीमसैर्वा । He could not offer other cows for Nandini, the daughter of Surabhi to his preceptor. He pleaded with the lion that how he could stand before his teacher when the cow was lost and himself was unhurt. He further said that people like him did not care for physical bodies. पिण्डेष्वनास्था खलु भौतिकेषु ।

The lion agreed and released the cow. But when the king bent his head before the lion, he received a shower of flowers from the Vidyadharas. The delighted cow, followed by the king returned safely to the hermitage.

आ) पाठ्यभागमनुसृत्य गीर्वाणवाण्याः सारांशं लिखत ।
Answer:
Introduction: The lesson Chitravimsati was written by Sri Jatavallabhula Purushottama Sastry. It is an extract from his work Chitra-satakam. There are two topics in this lesson Girvanavani and Chitraloka. The first part describes the greatness of Sanskrit language.

Girvanabhasha: The nectar oozing Sanskrit language is spoken by the gods. It is the ocean of the gems of good sayings and the treasure trove of great literature, and is acquired only by the meritorious. संस्कृताख्या सुकृतैकलभ्या | It is the ladder to liberation and the staircase to the heaven. It is relished by people of different tastes. Just as the children inherit the virtues of the mother, so also all the languages inherit the qualities of the divine mother language. In Sanskrit as the words are formed from the roots, they are naturally pure.

A language is the mother because she spreads in the world, and also she protects the dependents like a mother. On this earth, such a mother tongue is Sanskrit only, how can other languages have the title of mother ? वाच्याः कथं मातृपदेन चान्याः ? Even now, those who are experts in that language make the non-learners of the language ashamed by reciting the ancient verses. Lakhs of people speak it, and lakhs understand it. Lakhs of books in that language are sold every year. How can we live abandoning that language which is always in our body like life, and which is used during marriage ceremony, rituals of manes and gods, and in poetry recitals and philosophical discourses? Our history is before us with life. How is it dead? In this world, beautiful poetry is removed from dharma and righteous poetry is not beautiful. Except in Sanskrit where else is poetry that is golden and fragrant ?

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

3. एकं निबन्धप्रश्नं समाधत्त । (1 × 6 = 6)

अ) विक्रमस्य औदार्यमिति पाठ्यभागस्य सारांशं लिखत ?
Answer:
Introduction: The lesson Vikramasya Audaryam is taken from vikramarkacharitam written by Sivadasa. When Bhoja wants to ascend the throne of Vikrama the statue on the fourth step tells this story about the generosity of king Vikrama.

The childless Brahmin:
While Vikramaditya ruled Ujjaini, there was a learned Brahmin in that city, who was virtuous, but who had no children. Once, his wife said to him that there was no heaven for one who had no children सत्पुत्रेण कुलं नृपेण वसुधा लोकत्रयं भानुना । The Brahmin worshipped Lord Siva. The god appeared in his dream and asked him to perform Pradoshavrata. The Brahmin did so and he begot a son, whom he named Devadatta. When his son grew up, the Brahmin performed his marriage, and went on a pilgrimage to Varanasi.

Devadatta’s help to Vikrama:
Once Devadatta went into the forest to collect fuel sticks for sacrifice. At that time, king Vikrama came to the forest for hunting. There he asked Devadatta the way to the city, even though he knew it.

Devadatta himself led him to the city. The king praised him, and appointed him in his court. One day the king praised Devadatta’s help in the assembly saying that he could never repay his debt. Devadatta kidnaps the prince :

Devadatta wondered whether the king’s praise was true or false. He wanted to test the words of the king. He kidnapped the prince, and gave one of his ornaments to his servant and sent him to the market to sell it. The king’s men who were searching for the prince caught him, and brought him to the king. When the king asked him, the servant said that he was Devadatta’s servant. The king sent for Devadatta, who said that he killed the prince or money. The members of the court said that Devadatta should be sentenced to death. But the king said that Devadatta rescued him in the forest. One should not find faults in those dependent on him. adg महतां गुणदोषचिन्ता | He said that his son died because of his previous deeds only. He honoured Devadatta and sent him away.

Devadatta returned with the prince, and told the king that he wanted to test the words of the king. He praised the generosity of the king, who said that one should not forget the help done to him. यः कृतमुपकारं विस्मरति स एव पुरुषाधमः ।
Thus the statue told Bhoja about the generosity of king Vikrama.

आ) अप्पय्यदीक्षितेन्द्र इति पाठ्यभाग सारांशं संग्रहेण विवृणुत ?
Answer:
Introduction: The lesson Appayyadikshitendra was written by Sanka Usharani. This lesson narrates the story of the great scholar and devotee Appayya Dikshita.

Family of scholars: Appayya Dikshita was born in the family of scholars. His father was Rangarajadhwari was a genius. His grandfather was in the court of Acchana Dikshita. Appayya was born in 1554 in Virinnchipura in Tamilnadu.

Appayya was a poet, philosopher, scholar, devotee, Gram- marian, Mimamsaka, Logician, Ritualist etc. He studied all the four- teen branches of knowledge at the feet of his teacher Nrisimha Ramaswamy.

Marriage and works: A scholar named Ratnakheta Srinivasa wanted to defeat Appayya with the grace of the goddess Kamakshi. However, the goddess advised him to give his daughter in marriage to Appayya. अप्पय्यः असाधारणः वादे जेतुमशक्तः । He did so. Appayya had two daughters and three son.
Many students came to him to receive education. Bhattoji Dikshita came to him to learn Mimamsa and Vedanta.

Appayya wrote 104 works on many subjects like Vyakarana, Vedanta, Alankara etc. Among the Vedanta works Siddhanta lesasangraha and Parimala vyakhya were famous. Sivakarna mritam and Sivarchanachandrika were his famous works devoted to Siva. Kuvalayananda and Chitramimamsa were Alankara works.

Appayya was invited by Vellore Chinabommanayaka to adorn his court. Appayya was epileptic. But when the king came to his house to observe his disease, Appayya transferred his disease on to his upper garment. Dikshita means one who performed sacrifices. Appayya performed more than 104 sacrifices including Somayaga, Vajapeya etc. He even performed Viswajit. Once while he was performing a sacrifice, the king arrived there and offered him clothes, ornaments etc. Which he immediately put in the fire. The fire god appeared wearing them.

Devotion of Siva: Appayya was a devotee of Siva. To test his devotion, he ate dattura leaves and became mad. In that state, he recited Unmattapanchasat. The king honoured him with a shower of gold coins when Appayya completed his Sivarkamanidipika. राजा चं कनकाभिषेकेण सत्कृतवान् | Appayya built a temple for Siva with that gold in his native village. Though Appayya was a Tamilian he used Telugu idioms and proverbs. He praised Telugu language thus: Being born in Andhra, knowing the language of Telugu, study- ing Prabhakara’s Mimamsa work, and belonging to the Yajurveda School are the fruit of not less penance.

Though the householder Appayya was in the path of action, his mind was in the path of detachment only. He spent his last days in the presence of Nataraja of Chidambara. He became one with Siva at the age of 72 in the year of 16261 He was considered the incarnation of Siva Himself. अप्पय्यः साक्षात् परमशिवस्य अवतारः ।

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

4. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) कर्णभारं केन विरचितम् ?
समाधान:
कर्णभारं भासेन विरचितम् ।

आ) भृगुवंशकेतुः कः ?
समाधान:
भृगुवंशकेतुः परशुरामः ।

इ) द्रोणद्रुपदयोः सम्बन्धः कीदृश: ?
समाधान:
द्रोणद्रुपदयोः स्नेहबन्धः अच्छेद्यः अविभाज्यः च । सः पटुचिक्कणात्म अभवत् ।

ई) कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय किं दत्तवती ?
समाधान:
कृपी गोक्षीरं ददामीत्युक्त्वा पुत्राय पिष्टमिश्रितजलं दत्तवती । तत् पीत्वा अश्वत्थामा तुष्टः अभवत् ।

उ) कालिदासेन उक्तस्य श्लोकस्य भावः कः ?
समाधान:
अहं खादन् न गच्छामि, हसन् न जल्पामि, गतं न शोचामि, कृतं न स्मरामि, द्वयोः,संभाषणमध्ये न प्रविशामि । तेन कथम् अहं मूर्खः भवामि |

ऊ) कालिदासः महाराजं कथं सन्तोषितवान् ?
समाधान:
भोजस्य मनोगतस्य यथोचितम् उत्तरं दत्वा कालिदासः भोजं संतोषितवान् ।

5. द्वयोः ससंदर्भ व्याख्यानं लिखत । (2 × 3 = 6)

अ) वदेत् क्षमं स्वामिहितं मन्त्री ।
समाधान:
परिचयः – एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् । एषः पाठः रामायणस्य युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः – प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
वदेत् ।
भावः – परबलं, स्वबलं तथा क्षयं वृद्धिं च बुद्ध्या समीक्ष्य मन्त्री स्वामिहितं

आ) चित्रार्पितारम्भ इवाऽवतस्थे ।
समाधान:
परिचयः – एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषःभागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः – कविः दिलीपमहाराजस्य स्थितिं एवं वदति ।
भावः – चित्र लिखित शरीद्धरण उद्योग इव स्थितः ।
विवरणम्ः – सिंहं प्रहर्तुं दिलीपः उद्युक्तः । तस्य वामेतरः करः सायकपुंखे एव लग्मः अमवत् ।

इ) ग्रामे यत्र गृहे समैक्यपरता सा धर्मभूमिर्मम ।
समाधान:
परिचयः एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् ।
अस्य कविः श्रीमान् दोर्बल प्रभाकरशर्मा |
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः यत्र गृहे माता भर्तुः सुतानां च इष्टवस्तूनि परिवेषति, पिता वृत्तिप्रदः, भ्रातरः स्नेहसम्पन्नाः ग्रामे शोभायात्राः, गृहे एक्यता इत्यादीनि सन्ति, सा मम धर्मभूमिः इति कविः वदति ।

ई) वाच्याः कथं मातृपदेन चान्याः ।
समाधान:
परिचयः – गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री ।
सन्दर्भः – गीर्वाणवाण्याः वैभवं वर्णयनन् कविः एवं वदति । या भाषा लोके प्रसारिता, माता इव रक्षति सा एव मातृभाषा इति कथ्यन्ते ।
भावः – कथं अन्यभाषाः मातृभाषा इति कथ्यन्ते ।
विवरणम्ः – संस्कृतभाषा सर्वासां भाषाणां जननी, सा भाषा मातेव रक्षति ।

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

6. द्वयोः ससंदर्भ व्याख्यानं लिखत । (2 × 3 = 6)

अ) सतां सङ्गो हि भेषजम् ।
समाधान:
परिचयः – एतत-वाक्यं, “मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः – शोकाविष्टं कौण्डिन्यं पति कपिलः एवं अवदत् ।
भावः – सज्जनसांगत्यं एव सज्जनां भेषजम् ।
विवरणम्ः -सर्वे सज्जनसाङ्गत्यं करणीयम् तदा एव जीवनं सुखमयं भवेत् ।

आ) नैवाक्षितेषु महतां गुणदोषचिन्ता ।
समाधान:
परिचयः – इदं वाक्यं ‘“विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास ।
सन्दर्भः – समामध्ये राजा सचिवानां प्रति एवं अवदत् ।
भावः – शरणागतेषु महतां गुणदोष चिन्ता न कुर्तव्या ।
विवरणम्ः – चन्द्रः क्षथी, प्रकृतिवक्रतनु, कलडिए तथापि सः शकरं आश्रितवान् सः शंकरः चन्द्रस्य गुणदोषचिन्तनं ।

इ) क्षणे क्षणे किल परिक्षीयते बालस्य दशा ।
समाधान:
परिचयः – वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः – एकः वृद्धः युवकं उद्दिश्य एवं अवदत् ।
भावः – क्षणे क्षणे परीशील्यमाने बलकस्य दशा क्षीयेत ।
विवरणम्ः – वृया अनेन कलहेन, बालकस्य परिस्थितिः सीयते ।

ई) अप्पय्यः असाधारणः वादे जेतुमशक्तः ।
समाधान:
परिचयः – एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भः – अप्पय्यं जेतुं, स्वपादावनतं कर्तुं च रत्नखेटश्रीनिवासः अभिलषति
स्म । तदर्थं सः काञ्चीपुरे कामाक्षी प्रार्थितवान् । तदा देवी कामाक्षी अप्पय्यं जेतुं न शक्यते । तव पुत्रिकां तस्मै विवाहे प्रयच्छ इति उक्तवती ।
भावः – अप्पय्यः असामान्यः । वादे तं जेतुं न शक्यते ।

7. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) सीता कीदृशी ?
समाधान:
सीता महाहिसदृशी । सीतायाः चिन्ता एव विषम्, स्मित एव दंष्ट्राः स्तनौ एव भोगाः तथा अङ्गुल्यः एव शिरांसि ।

आ) धर्मनिष्ठा इति पाठ्यांशः कस्मात् संगृहीतः ?
समाधान:
धर्मनिष्ठा, इति पाठ्यभागः, रघुवंशमहाकाव्ये तृतीयसर्गत् संगृहीतः ।

इ) सर्वेन्द्रियाणां प्रधानं किम् ?
समाधान:
सर्वेन्द्रियाणां नयनं प्रधानम् ।

ई) परोपदेशसमये जनाः सर्वेऽपि कीदृशाः भवन्ति ?
समाधान:
परोपदेशसमये जनाः सर्वेऽपि पण्डिताह भवन्ति ।

उ) देवगिरीणां पञ्च नामानि लिखत ?
समाधान:
हेमाद्रिः, रजताद्रिः, हिमाद्रिः, सह्याद्रिः, विन्ध्याद्रिः इति पञ्चदेवगिरयः ।

ऊ) लोके चित्रगतिः कः ?
समाधान:
ईश्वरें भक्तिः सर्वेषां अस्ति, ईश्वरप्रतिपादित वेदे भक्तिः नास्ति, एतत् लोके चित्रगतिः ।

8. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) मन्दविषो नाम सर्पः कुत्र वसति ?
समाधान:
मन्दविषः नाम सर्पः जीर्णोधाने वसति ।

आ) ब्राह्मणः पुत्रप्राप्त्यर्थं कीदृशं व्रतमाचरितवान् ?
समाधान:
ब्राह्मणः पुत्रप्रास्यर्थं प्रदोषव्रतमाचरितवान् । मार्गशीर्षशुद्धत्रयोदश्यां मन्दवासरे कल्पोक्तप्रकारेण सः व्रतमाचरितवान् ।

इ) राजहंसः कः ? तस्य नगरी का ?
समाधान:
राजहंसः मगधदेशस्य राजा । तस्य नगरी पुष्पपुरी ।

ई) मार्गशीर्षमासः कीदृशः आसीत् ?
समाधान:
मार्गशीर्षमासः शीतवायुविजृम्भणेन युक्तः आसीत् ।

उ) मञ्जुहासिनी का ?
समाधान:
मञ्जुहासिनी आत्मनः प्रेयसी ।

ऊ) अप्पय्यदीक्षितः कुत्र जन्म लेभे ?
समाधान:
अप्पय्यदीक्षितः तमिळदेशे आरणिसमीपे अडयप्पलमिति ग्रामे जन्म लेभे ।

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

9. एकेन वाक्येन समाधानं दत्त । (5 × 1 = 5)

अ) रणे रामेण कः हतः ?
समाधान:
रणे रामेण खरः हतः ।

आ) मुनिहोमधेनुः का ?
समाधान:
नन्दिनी

इ) यावज्जीवं कः दहेत् ?
समाधान:
यावज्जीवं जडः दहेत् ।

ई) अनुच्छदनसमये मुनयोऽपि कीदृशाः भवन्ति ?
समाधान:
अनुच्छदनसमये मुनयः अपि पण्डिताः भवन्ति ।

उ) देहे सदा प्राण इव का अस्ति ?
समाधान:
धिषणा.

10. एकेन वाक्येन समाधानं दत्त । (5 × 1 = 5)

अ) मण्डूकनाथः कः ?
समाधान:
जालपादः

आ) देवदत्तः राजानं कुत्रं अनीनयत् ?
समाधान:
नगरम्

इ) दशकुमारचरितं केन विरचितम् ?
समाधान:
दण्डिना

ई) श्रीधरस्य ज्येष्ठभ्रातृवत् कः वर्धते ?
समाधान:
श्रीधरस्य ज्येष्ठभ्रातृवत् तस्य मातुः कासः वर्धते ।

उ) वेङ्कटरावस्य पुत्रः कः ?
समाधान:
सुरेशः

11. अधो निर्दिष्टं कथां पठित्वा प्रश्नानां समाधानानि दत्त | (5 × 1 = 5)

अस्ति कस्मिंश्चित् पर्वतैकदेशे महान् वृक्षः । तत्र च सिन्धुकनामा कोऽपि पक्षी प्रतिवसति स्म । तस्य पुरीषे सुवर्णमुत्पद्यते । अथ कदाचित् तमुद्दिश्य व्याधः कोऽपि समाययौ । स च पक्षी तदग्रत एव पुरीषमुत्ससर्ज । अथ पतनसमकालमेव तत्सुवर्णीभूतं दृष्ट्रा व्याधो विस्मयमगमत् । अहो मम शिशुकाला- दारभ्य शकुनि-बन्ध-व्यसनिनोऽशीति-वर्षाणि समभूवन् । न च कदाचित्पक्षिपुरीषे सुवर्णं दृष्टम् इति विचिन्त्य तत्र वृक्षे पाशं बबन्ध |

प्रश्नाः
(1) वृक्षे प्रतिवसतः पक्षिणो नाम किम् ?
समाधान:
वृक्षे प्रतिवसतः पक्षिणो नाम सिन्धुकः ।

(2) कस्मिन् सुवर्णमुत्पद्यते ?
समाधान:
पुरीषे सुवर्णमुत्पद्यते ।

(3) पक्षिणमुद्दिश्य कः समाययौ ?
समाधान:
पक्षिणमुद्दिश्य व्याधः समाययैौ ।

(4) स पक्षी कुत्र पुरीषमुत्ससर्ज ?
समाधान:
सं पक्षी व्याधस्य अग्रत एव पुरीषमुत्ससर्ज ।

(5) व्याधः वृक्षे कं बबन्ध ?
समाधान:
व्याधः वृक्षे पाशं बबन्ध ।

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त । (3 × 2 = 6)

अ) सत् + चिदानन्दः
समाधान:
सच्चिदानन्दः = श्चुत्व सन्धिः

आ) रामस् + टीकते
समाधान:
रामष्टीकते = ष्टुत्व सन्धिः

इ) वाक् + ईश:
समाधान:
वागीशः = जश्त्व सन्धिः

ई) वाक् + मयम
समाधान:
वाङ्मयम् = अनुनासिक सन्धिः

उ) सः + अपि
समाधान:
सोऽपि = विसर्ग सन्धिः

ऊ) कविः + आयाति
समाधान:
कविरायाति = विसर्गरेफादेश सन्धिः

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

13. नामनिर्देशपूर्वकं त्रीणि विघटयत । (3 × 2 = 6)

अ) सज्जनः
समाधान:
सद् + जनः = श्चुत्व सन्धिः

आ) तट्टीका
समाधान:
तत् + टीका = ष्टुत्व सन्धिः

इ) अजन्तः
समाधान:
अच + अन्तः = जश्त्व सन्धिः

ई) षण्मुखः
समाधान:
षट् + मुखः = अनुनासिक सन्धिः

उ) कोऽपि
समाधान:
कः + अपि = विसर्ग सन्धिः

ऊ) हरिर्गच्छति
समाधान:
हरिः + गच्छति = विसर्गरेफादेशः सन्धिः

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत । (2 × 6 = 12)

अ) जलमुक्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions 2

आ) स्रज्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions 3

इ) मनस्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions 4
ई) एतद् (स्त्री)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions 5

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत ।

अ) विद्यानिपुणः
समाधान:
विद्यया निपुणः = तृतीया तत्पुरुष समासः

आ) युक्तिरत्नानि
समाधान:
पुक्तयः राव रत्नानि = अवधारणापूर्वपदकर्मधारय समासः

इ) त्रिलोकी
समाधान:
त्रयाणां लोकानां समाहारः = द्विगु समासः

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions

ई) कंसकृष्णौ
समाधान:
कंसः च कृष्णः च = द्वन्द्व समासः

उ) यथाशक्ति
समाधान:
शक्तिम् अनतिक्रम्य = अव्ययीभाव समासः

ऊ) सहपुत्रः
समाधान:
पुत्रेण सह वर्तते इति = सहपूर्वपद बहुव्रीहि समासः

16. अधोनिर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखत । (5 × 1 = 5)

AP Inter 2nd Year Sanskrit Model Paper Set 2 with Solutions 1
समाधान:
अ) प्रश्नाः
1. रमेशः किं रकोति ?
2. गोविन्दः किमर्थं वसति ?
3. त्वं कुत्र पिबसि ?
4. रामः कदा गरछति ?
5. अहं किं पठामि ?

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

Access to a variety of AP Inter 2nd Year Sanskrit Model Papers Set 1 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

Time : 3 Hours
Max.Marks : 100

Note :

  • All questions should be attempted.
  • Question Nos. 1, 2 and 3 should be answered either in the medium of instructions of the candidate or in Sanskrit (Devanagari Script) only.
  • The remaining questions should be answered in Sanskrit (Devanagari Script) only.

सूचना :
प्रथम, द्वितीय, तृतीय प्रश्नान् विहाय अन्ये सर्वेऽपि प्रश्नाः संस्कृत भाषायामेव समाधातव्याः ।

1. एकस्य श्लोकस्य प्रतिपदार्थं भावं च लिखत । (1 × 8 = 8)

अ) परोक्षे कार्यहन्तारं प्रत्यक्षे प्रियवादिनम् ।
वर्जयेत्तादृशं मित्रं विषकुम्भं पयोमुखम् ||
समाधान:
पदच्छेद (Word Division) : परोक्षे, कार्यहन्तारं, प्रत्यक्षे, प्रियवादिनम्, वर्जयेत्, तादृशं, मित्रं, विषकुम्भं, पयोमुखम् ।
अन्वयक्रम : परोक्षे, कार्यहन्तारं प्रत्यक्षे, प्रियवादिनम्, तादृशं, मित्रं, पयोमुखं, विषकुम्भं इव, वर्जयेत् ।
अर्थ (Meanings) : परोक्षे = behind, कार्यहन्तारं = one who spoils , our work, प्रत्यक्षे = in front of us, प्रियवादिनम् = who speaks sweetly, तादृशम= such, मित्रम् = friend, पयोमुखम = milk topped, with milk at rim, विषकुम्भम् = pitcher of poison, वर्जयेत् = should abandon.
भाव (Substance) : One should abandon such a friend who spoils our work at our back, but speaks sweetly in front of us as he is a vessel filled with poison but having milk at rim.

आ) परोपदेशसमये जनाः सर्वेऽपि पण्डिताः ।
तदनुष्ठानसमये मुनयोऽपि न पण्डिताः ॥
समाधान:
पदच्छेद (Word Division) : परोपदेशसमये, जनाः, सर्वेः, अपि, पण्डिताः, तत्, अनुष्ठानसमये, मुनयः, अपि, न, पंडिताः ।
अन्वयक्रम : सर्वे, जनाः, अपि, परोपदेशसमये, पण्डिताः, तत्, अनुष्ठानसमये, मुनयः, अपि न पण्डिताः ।
अर्था (Meanings) : पर + उपदेशसमये = while advising others; सर्वे अपि = all the; जनाः = people; पंडिताः = are scholars; तत् + अनुष्ठानसमये = while putting into practice; मुनयः अपि = even the sages; न पण्डिताः= are not scholars.
भाव (Substance): While giving advice to others, everyone
acts as a scholar, but while putting it into practice, even the sages are not scholars.

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

2. एकं निबन्धप्रश्नं समाधत्त । (1 × 6 = 6)

अ) रावणाय विभीषणेन कृतमुपदेशं लिखत ।
Answer:
Introduction: The lesson Vibhishanopadesa is an extract from the Yuddhakanda of the Ramayana, written by Valmiki. Vibhishana advises Ravana to send Sita to Rama.

The Advice of Vibhishana: Vibhishana advised Ravana that one should use their might only when the three other methods namely sama, dana and bheda fail. Then also it would work again the weak and already unfortunate enemies. They should not underestimate the enemy. परेषां सहसावज्ञा न कर्तव्या कथञ्चन । They should protect lives. They need not have an unnecessary fight with one who followed Dharma. Sita should be sent to Rama.

Vibhishana advised Ravana to abandon his anger which would destroy his comfort and dharma. Ravana should follow dharma which would increase joy and fame.

The Inauspicious signs: Vibhishana mentioned the various inauspicious signs witnessed since the arrival of Sita. “The milk of the cows has decreased. The mighty elephants have lost their rut. The horses are neighing piteously. The donkeys, camels and mules are shedding tears losing their hair. Flocks of crows are cawing harshly on all sides. Cruel animals have assembled at the gates of the city, and are roaring loudly.”

He compared Sita with a serpent. She had the hood of bosom, poison of worry, fangs of smile, heads of five fingers and a great body. He warned that Ravana would not be spared with life by Rama. Even the Sun, Indra, Yama and other could not protect him.

Friends, who were like enemies: Vibhishana accused the Rakshasas saying that they were enemies in the guise of friends. He said that a minister should weight the strength, loss and gain on both sides, and give a beneficial advice. वदेत् क्षमं स्वामिहितं मन्त्री | He asked whether anyone could stand before the arrows of Rama. He advised that Sita should be sent to Rama along with precious gifts.

When Ravana said that he would not have spared if this speech was made by any other person, Vibhishana became angry. He flew in to the space, and said that as a brother Ravana could say anything, but he would not forgive him. He said that people who spoke sweetly could be found easily. But those who gave bitter but beneficial advice, and its listener were hard to be found. अप्रियस्य च पथ्यस्य वक्ता श्रोता च दुर्लभः | He wished him happiness, and left.

आ) मातृभूमेः वैभवं कविः कथं प्रशशंस ?
Answer:
Introduction: The lesson Sa Matrubhumirmama was written by Dorbala Prabhakara Sarma. The poet describes the greatness of India in this poem.

Vedabhumi, Jnanabhumi and Dharmabhumi: The poet praises the motherland as a land of Vedas, wisdom, languages, gods, dharma, action, yoga, tapas, rivers, holy places and great people. Here, the cattle of the Vedas show the right path:

The people here are cultured, intelligent, healthy and sweetly speaking. यत्रैताः हृदि जागृताः प्रकृतयः सा मातृभूमिर्मम । The land is prosperous, waters are pure, breeze is purifying and life is friendly. There are Rik mantras that praise the gods, Yajush mantras that are used in sacrifices, Atharva mantras that show the path of prosperity. This land is full of knowledge with the Vedas, Upanishads, Puranas, Itihasas, Kavyas etc. विद्या यत्र परोपकारविभवाः सा ज्ञानभूमिर्मम | There are different languages belonging to Vanga, Anga, Andhra, Kerala, Maharashtra, Sindhi, Gujarati, Karnataka etc.

Selfless People: The gods of rivers, villages, cities, towns, households, lands and others bestow boons. The lords of quarters such as Indra, Agni, Varuna, Vayu, Kubera, Yama etc. along with Vishnu, Uma and others grant welfare to the people. In a household, the mother takes care of the interests of her husband and children, father works, brothers give relief, teachers preach and the elders offer advice. ग्रमे यत्र गृहे समैक्यपरता सा धर्मभूमिर्मम । The people work selflessly, are engaged in meditation, prayers and contemplation. The wisdom of the sages ensures the welfare of the world, strengthens culture and enriches the knowledge.

Rivers and holy places: Rivers like Ganga, Yamuna, Saraswati, Godavari, Krishna, Narmada etc. remove sins, diseases and obstacles. Mountains such as Hemadri, Rajatadri, Aravali, Mahendra, Hima-layas, Sahya etc: embellish this land. There are holy cities such as Kasi, Ayodhya, Puri, Kanchi, Madhura, Avanti, Tirupati, Ahobilam etc. Great sages and poets like Vyasa, Valmiki, Bhrigu, Sankara, Kalidasa, Bhavabhuti etc. belonged to this land.

3. एकं निबन्धप्रश्नं समाधत्त । (1 × 6 = 6)

अ) मन्दविषसर्पकथां सङ्ग्रहेण लिखत ।
Answer:
Introduction: The lesson Mandavishasarpakatha is an extract from the Hitopadesa written by Narayana Pandita. This tells the story of an old serpent that devoured the frogs making friendship with them..

The old serpent: There was a serpent named Mandavisha in a ruined forest. Because of its old age, it became unable to search for food, and remained on the bank of a pond. A frog asked him why he did not take food. Then the serpent told his story
.
Kapila’s advice: Once Mandavisha had bitten the son of Kaundinya, a Brahmin from Brahmapura. When the father was crying rolling on the ground, an educated Brahmin Kapila advised him not to weep like that. He said. “Death is certain for those who are born, and birth is certain for those dead. Death is certain, now or hundred years after. मृत्युर्वै प्राणिनां ध्रुवः | Death approaches nearer and nearer day by day. One will not have permanent association with anything, even with his body. “On hearing that, Kaundinya decided to retire to the forests. But Kapila objected to it, saying that there would be evils even in a forest. To the detached one, the

house itself would be a hermitage. निवृत्तरागस्य गृहं तपोवनम् | He said that there was pain only, and not happiness in the world.

Kapila advised Kaundinya to renounce all the associations. If it was not possible, he should associate with the good only. Hi सङ्गो हि भेषजम् । Then the Brahmin cursed the serpent to become a vehicle of the frogs. Hence, Mandavisha arrived there.

Killing the frogs: When the frog informed this to their king, he came and rode on the back of the serpent for some time. The next day, the frog-king permitted the serpent to have the frogs as his food, as he was weak. Then the serpent killed all the frogs one by one, and at last ate the frog-king too.

क्रमशः खादितवान् । मण्डूकनाथमपि भक्षितवान् |
जातस्य हि ध्रुवो मृत्युः
थ्रुवं जन्म मृतस्य चः
अपि स्वेन शरीरेण
किमुतान्येन केननित् ।
निवृत्तरागस्य गृहं तपोवनम्
सतां सङ्गो हि भेषजाम्
स्कन्धेनापि कहेत् शतून्
कार्यमासाद्य बुहिमान्
जातस्य हि ध्रुवो मृत्युः
ध्रुवं जन्म मृतस्य च
संभोगो हि वियोगस्य
संच्वचयति सम्भवम्

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

आ) भिषजः भैषज्यम् इति पाठ्यभागस्य सारांशं संक्षेपेण लिखत ।
Answer:
Introduction: The lesson Bhishajah Bhaishajyam was written by Prof. Pullela Sriramachandra. It is taken from his Sriramachandra laghukavya sangraha. This lesson describes the story of a selfish doctor, and the fruit he reaped for his selfishness.

The Villager’s Plea: One day some villagers came to Dr. Venkata Rao, and requested him to attend to a boy who was in- jured in an accident. Venkata Rao childed them for not coming in time. वैद्योऽपि मानव एव। He accused them of trying to get treatment done without paying fee. He insulted them saying the boy would be delicate as the baby of a donkey. When they left as the boy was serious क्षणे क्षणे किल परिक्षीयते बालस्य दशा, Venkata Rao thought nothing would happen if one puppy died.

The poor and intelligent Venkata Rao: Venkata Rao was the son of a poor farmer. He was very intelligent and secured a seat in medical college. His father sold their agricultural land for his edu- cation. A rich man married his daughter to Venkata Rao. Venkata Rao’s practice also picked up. Along with money, the three defects grew in him. They were considering others as insects, himself as god, and accepting others praise as the truth. When he spoke of his father as a beggar, his father left him and returned to their village.

Marriage was also a business affair for Venkata Rao. विवाहो नाम वणिग्व्यवहार एव । For him money was everything. He never loved his wife. His son Suresh alone became the object of his affection.

Desire for Power: Venkata Rao thought that money was useless without power. He became an MLA twice. But he could not become a minister. He was unhappy about that. He could not win the seat the third time. He blamed the people for that. His hatred for people grew.

Manjuhasini, the Lady Doctor : At that time, Dr. Manjuhasini joined the government hospital there. She was Venkata Rao’s class- mate in medical college. She rejected Venkata Rao’s advances. Venkata Rao was hoping that she might have changed now as he became rich.

The death of his son : Venkata Rao received a phone call from Manjuhasini requesting his help in an emergency case. His driver tried to inform him that his son was not there at the school when he went there after getting the brake repaired. Venkata Rao cut him short saying that the boy would have reached home. But when he went to the hospital he saw the same villagers who came to him earlier in the day, and the body of his dead son.

4. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) कीदृशं वस्तु तथैव तिष्ठति ?
समाधान:
हुतं दत्तं च वस्तु तथैव तिष्ठति ।

आ) कर्णभारं केन विरचितम् ?
समाधान:
कर्णभारं भासेन विरचितम् ।

इ) केन द्रुपदः गर्वान्धः जातः ?
समाधान:
द्रुपदः धनश्रिया गर्वान्धः जातः । एवम् अर्जुनः उक्तवानथ् ।

ई) बालानां कलहः कीदृश: ?
समाधान:
बालानां कलहः तत्कालीनः भवति । बालाः कलहायन्ते अनन्तरं क्रीडन्ते च ।

उ) कालिदासश्लोके भोजेन प्राप्तं समाधानं किम् ?
समाधान:
कालिदासश्लोकेन भोजेन प्राप्तं समाधानम् एवम् अस्ति – द्वयोः सम्भाषणकर्त्रीः मध्ये सूचनां विना न प्रविशेत् इति ।

ऊ) भोजः कोशाध्यक्षं प्रति किं सूचितवान् ?
समाधान:
कविकुलगुरोः कालिदासस्य सदनं एकलक्षसुवर्णमुद्राः प्रेष्यन्ताम् इति भोजः कोशाध्यक्षं सूचितवान् ।

5. द्वयोः ससंदर्भ व्याख्यानं लिखत | (2 × 3 = 6)

अ) अप्रियस्य च पथ्यस्य वक्ता श्रोता च दुर्लभः |
समाधान:
परिचयः – एतत् वाक्यं विभीषणोपदेशः इति पाठ्यभागात् स्वीकृतम् । एषः पाठः रामायणस्य युद्धकाण्डात् गृहीतः । अस्य कविः वाल्मीकिः ।
सन्दर्भः – प्रदीयतां दाशरथाय मैथिली इति रावणं प्रति उपदिशन् विभीषणः एवं वदति ।
भावः – लोके प्रियवादिनः जनाः सुलभाः भवन्ति । परन्तु अप्रियस्य हितस्य वक्ता, श्रोता च दुर्लभः एव ।

आ) पिण्डेष्वनास्था खलु भौतिकेषु ।
समाधान:
परिचयः – एतत् वाक्यं ‘धर्मनिष्ठा’ इति पाठ्यभागात् स्वीकृतम्, एषःभागः कालिदासस्य रघुवंश महाकाव्ये पञ्चमसर्गात् स्वीकृतः ।
सन्दर्भः – दिलीपः सिंहं प्रति एवं उक्तवान् |
भावः – पृथिव्यादि भूतविकारेषु शरीरेषु अनपेक्षा खलु ।
विवरणम्ः – हे सिंह ! मम थशोरुप शरीरे दयालुः भव । देहं अशाश्वतम्, कीर्तिरेव शाश्वतम् ।

इ) यत्रैताः हृदि जागृताः प्रकृतयः सा मातृभूमिर्मम |
समाधान:
परिचयः – एतत् वाक्यं सा मातृभूमिर्मम इति पाठ्यभागात् स्वीकृतम् । अस्य कविः श्रीमान् दोर्बल प्रभाकरशर्मा |
सन्दर्भः – मातृभूमेः वैशिष्ट्यं वर्णयन् मम मातृभूमिः ज्ञानभूमिः धर्मभूमिः इति वदन् कविः एवं वर्णयति ।
भावः – मम मातृभूमिः क्षेमकरी, तत्र जलं शुचि, जीवनं स्नेहात्मकं, मनः सुरचनं इति कविः वदति । यत्र प्रकृतयः हृदि अवबोधितः सा मम मातृभूमिः इति कविः वदति ।

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

ई) भृत्या हि पक्त्री वद किं न सा स्त्री ।
समाधान:
परिचय : गतत् वाक्यं ‘नित्रविंशतिः’ इति पाठ्यभागात् स्वीकृतम्’ अस्य पाठ्यभागस्य रचयिता जटावल्लभपुरुषोत्तम शास्त्री ।
सन्दर्भ : लोके चित्रविषयान् वर्णयन् कविः एवं वदति । मम सुता महानसे न निरोधनीया, स्त्रियः पूज्याः इति चित्रं वदन्ति । तर्हि भूक्तिः कथं सिद्ध्यति इति कविः पृच्छति ।
भाव : या भृत्या पचति सा स्त्री नास्ति किम् ?
विवरणम् : मदीया कन्या महानसे नैव पाकं कराति ? स्त्रियः पूज्याः, इति वदन्ति, भृत्या पाचिका स्त्री कि ? वद ।

6. द्वयोः ससंदर्भ व्याख्यानं लिखत | (2 × 3 = 6)

अ) निवृत्तरागस्य गृहं तपोवनम् ।
समाधान:
परिचयः एतत-वाक्यं, ”मन्दविषसर्पकथा” इति पाठ्यभागात् स्वीकृतम् अस्य पाठ्यभागस्य रचयिता नारायणपण्डितः ।
सन्दर्भः तपःकर्तुं बनाय प्रस्थितं कौण्डिन्यं प्रति कपिलः एवं अवदत् । भावः विरक्तस्य जनस्य गृहमेव तपोवल तुल्यम् ।
विवरणम्ः सर्वे पञ्चेन्दियाणि वशेयुः, तदा एव तस्य शान्तिः सिध्यति, तस्मिन् समये गृहः एव तपोवनसदृशम् ।

आ) यः कृतमुकारं विस्मरति स एव पुरुषाधमः ।
समाधान:
परिचय : इदं वाक्यं “विक्रमस्य औदार्यम्” इति पाठ्यभागात् स्वीकृतम् । अस्य पाठस्य रचयित शिवदास ।
सन्दर्भ : राज्ञः वचनं क्षुत्वा एकः एवं अवदत् ।
भाव : यः कृतं उपकारं विस्मरति सः मानवः पुरुषाधभः भवति ।
विवरणम् : नारिकेल वृक्षः पीतं तोयं स्मृत्वा नरागां अमृततुल्यं जलं ददाति यः मानवः परैः कृतं उपकारं विस्मरति सः मानवः पुरुषाधमः भवति ।

इ) वैद्योऽपि मानव एव ।
समाधान:
परिचयः वेङ्कटरावस्य भिषजो भैषज्यम् इति पाठ असित अस्य पाठस्य रचयित, श्री पुल्लेल श्रीरामचन्द्रः ।
सन्दर्भः वेङ्कटरावः ग्रामीणं प्रति एवं अवदत् ।
भावः वैद्यः अपि चरः एव ।
विवरणम्ः अरे भूर्ख ! स्वसुखनिन्तैव युषमाकम् वैद्यः मानवमात्र एव ।

ई) अप्पय्यः साक्षात् परमशिवस्य अवतारः ।
समाधान:
परिचय : एतत् वाक्यम् अप्पय्यदीक्षितेन्द्र इति पाठ्यभागात् स्वीकृतम् । अस्य रचयित्री सङ्का उषाराणी ।
सन्दर्भ : केचन जनाः अन्तिमसमये अप्पय्यदीक्षितम् चिदम्बरे नटराज- प्रतिमायां लीयमानम् अपश्यन् । तस्मिन्नेव समये गृहे अप्पय्यः प्राणविहीनः अभवत् । तेन सः परमेश्वरस्य अवतारः इति विज्ञाः मन्यन्ते स्म ।
भाव : अप्पय्यः साक्षात् परमेश्वरस्य अपरः अवतारः एव ।

7. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) कदा प्रभृति अशुभानि निमित्तानि दृश्यन्ते ?
समाधान:
यदा प्रभृति सीता आगता तदा प्रभृति अशुभानि निमित्तानि दृश्यन्ते ।

आ) दिलीपेन सेव्यमानं नन्दिनीधेनुं कः चकर्ष ?
समाधान:
दिलीपेन सेव्यमानं नन्दिनीधेनुः सिंहः चकर्ष ।

इ) व्याधितस्य मित्रं किम् ?
समाधान:
व्याधितस्य औषधं मित्रम् ।

ई) देवः कुत्र सहायकृते भवति ?
समाधान:
उद्यमं, साहस, धैर्यं, बुद्धिः, शक्तिः, पराक्रमः यत्र वर्तन्ते तत्र सहायकृत् भवति ।

उ) मानवानां प्रकृतिशुभगुणाः के ?
समाधान:
क्षान्तिः, संस्कारगन्धः, संतृप्तिः, मृदुवाक् इत्यादयः मानवानां प्रकृतिगुणाः ।

ऊ) अमर्त्यवाण्याः मातृता कथम् आगता ?
समा.
सर्वासुभाषासु अमर्त्य वाज्यः गुणाः सन्ति, अतः अमर्त्यवाण्याः मातृता आगता ।

8. त्रयाणां प्रश्नानां समाधानानि लिखत । (3 × 2 = 6)

अ) हितोपदेशे कति भागाः सन्ति ? ते च के ?
समाधान:
हितोपदेशे चत्वारः भागाः सन्ति । ते
1. मित्रलाभः 2. मित्रभेदः 3. विग्रहः 4. सन्धिः

आ) विक्रमेण पृष्टः भृत्यः किम् उत्तरमदात् ? –
समाधान:
विक्रमेण पृष्टः भृत्यः एवम् उत्तरम् अदात् – अहं देवदत्तस्य भृत्यः । एतदाभरणं विक्रीय धनमादाय इति प्रेषितः अस्मि । इति ।

इ) मानसारः कथं गदां प्राप्तवान् ?
समाधान:
मानसारः परमेश्वरं समाराध्य सन्तुष्टात् तस्मात् गदां प्राप्तवान् ।

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

ई) श्रीधरः स्वमातुः मरणसमये किं कुर्वन्ना सीत् ?
समाधान:
श्रीधरः स्वमातुः मरणसमये विश्वविद्यालये शोधग्रन्थसमर्पणकार्यं कुर्वन् आस्ते ।
उ) प्रभुत्वचिकित्सालये नियोगेनागता भिषगङ्गना का ?
समाधान:
प्रभुत्वचिकित्सालये नियोगेनागता भिषगाङ्गना आत्मनः सहाध्यायिनी भिषगङ्गला ।

ऊ) अप्पय्यदीक्षितः आन्ध्रभाषां कथं प्रशशंस ?
समाधान:
आन्ध्रत्वम् आन्ध्रभाषा च नाल्पस्य तपसः फलम् इति अप्पय्यः आन्ध्रभाषां प्रशसंश

9. एकेन वाक्येन समाधानं दत्त । (5 × 1 = 5)

अ) केषां बलानि अमेयानि ?
समाधान:
परेषां बलानि अमेयानि ।

आ) दिलीपस्य धर्मपत्नी का ?
समाधान:
सुदक्षिणा

इ) कैः पाषाणखण्डेषु रत्नसंज्ञा विधीयते ?
समाधान:
मूढैः

ई) मूर्खस्य कति चिह्नानि भवन्ति ?
समाधान:
मूर्खस्य पञ्च चिह्नानि भवन्ति ।

उ) चित्रविंशतिः इति पाठ्यभागः कस्मात् स्वीकृतः ?
समाधान:
गीर्वाणवाणी – चित्रलोकः

10. एकेन वाक्येन समाधानं दत्त । (5 × 1 = 5)

अ) कौण्डिन्यस्य पुत्रः कः ?
समाधान:
सुशीलः

आ) कुमारमानीय राज्ञे कः ददौ ?
समाधान:
देवदत्तः

इ) राजवाहनः कस्य पुत्रः ?
समाधान:
राजहंसस्य

ई) श्रीधरस्य मातुः नाम किम् ?
समाधान:
श्रीधरस्य मातुः नाम वेदवती ।

उ) वेङ्कटरावस्य पत्नी का ?
समाधान:
सुशीला

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

11. अधो निर्दिष्टं कथां पठित्वा प्रश्नानां समाधानानि दत्त | (5 × 1 = 5)

अस्ति भागीरथीतीरे गृध्रकूटनाम्नि पर्वते महान् पर्पटीवृक्षः । तस्य कोटरे दैव-दृर्विपाकात् गलितनखनयनो जरद्गवनामा गृध्रः प्रतिक्सति । अथ कृपया तज्जीवनाय तद् वृक्षवासिनः पक्षिणः स्वाहारात् किञ्चित् उद्धृत्य तस्मै ददति । तेन असौ जीवति । तेषां शाबकरक्षां च करोति । अथ कदाचित् दीर्घकर्णनामा मार्जारः पक्षिशाबकान् भक्षयितुं तत्र आगतः । ततस्तम् आयान्तं दृष्ट्वा पक्षिशाबकैः भयार्तैः कोलाहलः कृतः । तच्छ्रुत्वा जरद्गवेण उक्तम् “कोऽयमायाति” ? इति । दीर्घकर्णो गृध्रमवलोक्य सभयमाह ” हा हतोऽस्मि । यतोऽयं मां व्यापादयिष्यति” इति ।

प्रश्नाः
1. पर्पटीवृक्षः कुत्रास्ति ?
समाधान:
पर्पटीवृक्षः भागीरथीतीरे गृध्रकूटनाम्नि पर्वते अस्ति ।

2. वृक्षस्य कोटरे कः प्रतिवसति ?
समाधान:
वृक्षस्य कोटरे जरद्गवनाम गृध्रः प्रतिवसति ।

3. मार्जारः किमर्थं तत्र आगतः ?
समाधान:
मार्जारः पक्षिशाबकान् भक्षयितुं तत्र आगतः ।

4. कोलाहलः कैः कृतः ?
समाधान:
कोलाहलः पक्षिशाबकैः कृतः ।

5. दीर्घकर्णोः गृधमवलोक्य किमाह ?
समाधान:
दीर्घकर्णो गृध्रमवलोक्य “हा हतोऽस्मि । यतोऽयं मां व्यापादयिष्यति” इति ।

12. नामनिर्देशपूर्वकं त्रीणि सन्धत्त । (3 × 2 = 6)

अ) शरत् + चन्द्रः
समाधान:
शरच्चन्द्रः = श्चुत्व सन्धिः

आ) तत् + टीका
समाधान:
तट्टीका = ष्टुत्व सन्धिः

इ) षट् + आननः
समाधान:
षडाननः = जश्त्व सन्धिः

ई) तत् + मात्रम्
समाधान:
तन्मात्रम् = अनुनासिक सन्धिः

उ) शिवः + अहम्
समाधान:
शिवोऽहम् = विसर्ग सन्धिः

ऊ) पितुः + इच्छा
समाधान:
पितुरिच्छा = विसर्गरेफादेश सन्धिः

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

13. नामनिर्देशपूर्वकं त्रीणि विघटयत । (3 × 2 = 6)

अ) मनश्चलति
समाधान:
मनस् + चलति = श्चुत्व सन्धिः

आ) रामष्टीकते
समाधान:
रामस् + टीकते = ष्टुत्व सन्धिः

इ) वागीशः
समाधान:
वाक् + ईशः = जश्त्व सन्धिः

ई) जगन्नाथः
समाधान:
जगत् + नाथः = अनुनासिक सन्धिः

उ) सोऽपि
समाधान:
सः + अपि = विसर्ग सन्धिः

ऊ) नृपतिर्जयति
समाधान:
नृपतिः + जयति = विसर्ग सन्धिः

14. द्वयोः शब्दयोः सर्वविभक्तिरूपाणि लिखत । (2 × 6 = 12)

अ) वणिक्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions 2

आ) सरित्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions 3

इ) महत्
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions 4

ई) यद् (पुं)
समाधान:
AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions 5

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions

15. समासनामनिर्देशपूर्वकं त्रयाणां विग्रहवाक्यानि लिखत । (3 × 2 = 6)

अ) व्याघ्रभीतः
समाधान:
व्याघ्रात् भीतः = पंचमी तत्पुरुष समासः

आ) लतातन्वी
समाधान:
लता इव तन्वी = उपमान पूर्वपद कर्मधारय समासः

इ) नवरात्रम्
समाधान:
नवानां रात्रीणां समाहारः = द्विगु समासः

ई) रामलक्ष्मणौ
समाधान:
रामः च लक्ष्मणः च = द्वन्द्व समासः

उ) प्रत्यक्षम्
समाधान:
अक्षणोः समीपे = अव्ययीभाव समासः

ऊ) उपदशाः
समाधान:
दशानां समीपे ये सन्ति ते = संख्योत्तरपद बहुव्रीहि समासः

16. अधो निर्दिष्ट पट्टिकामाधारीकृत्य पञ्चसाधुवाक्यानि लिखत । (5 × 1 = 5)

AP Inter 2nd Year Sanskrit Model Paper Set 1 with Solutions 1
समाधान:
1. कविता पाठं अपठत् ।
2. राजेशः देवं अनमत् ।
3. त्वं क्षीरं अपिब: ।
4. रामः देवं अनमत् ।
5. अहं पाठं अलिखम् ।