Prasanna

By going through these CBSE Class 11 Maths Notes Chapter 11 Conic Sections Class 11 Notes, students can recall all the concepts quickly.

Conic Sections Notes Class 11 Maths Chapter 11

Conic : Let T be a fixed line and F be a fixed point (not on line T).
Let P be any point in the plane of the line T and the point F. Then, the set of all points P, such that | FP | = e | PM |, where M is the foot’ of perpendicular from P on the line l and e > 0 is . a fixed real number is called a conic.

Notes :

1. The fixed point F is called focus of the conic and the fixed line T is called directrix of the conic associated with F.
2. The positive real number e is called eccentricity of the conic.
3. A line through the focus F and perpendicular to the directrix l is called axis of the conic.
4. The point of intersection of the conic and its axis is called
   vertex.
5. A conic with eccentricity ‘e’ is called
   (i) a parabola, if e = 1. (ii) an ellipse, if e < 1. (iii) a hyperbola if e > 1..

Parabola : Let T be a fixed line and F be a fixed point (not on 1). Let P be any point in the plane of line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = | PM | is called a parabola.

Equation of the Parabola (Standard Form): The equation of the parabola in the standard form is y² = 4ax.

Latus Rectum : The latus rectum of conic is the chord passing through the focus and perpendicular to its axis.

Length of Latus Rectum : The length of the latus rectum of the parabola y² = 4ax is 4a.

Focal Distance : The distance of any point on parabola from the focus is called focal distance of that point.

Focal Chord : Any chord that passes through the focus of the parabola is called a focal chord of the parabola.

The focal distance of a point P(x₁, y₁ of the parabola y² = 4ax is a + x₁

Main Facts about Four Standard Forms of Parabola

Ellipse : Let l be a fixed line, F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F.

Let M be the foot of perpendicular from P on the line l.
Then, the set of all points P such that | FP | = e | PM | (0 < e < 1) is called an ellipse.

Notes : 1. The fixed line is called directrix of the ellipse.
2. The fixed point F is called focus of the ellipse.
3. The positive real number e < 1 is called eccentricity of the ellipse.

Standard form of an Ellipse : The equation of the ellipse in the standard form is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1, where b² = a²( 1 – e²) = a² – a²e².
Putting ae = c. Therefore, c² = a² – b².

Note: The ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 has two foci, namely (ae,o) and (- ae, 0) and their respective directrices are x – \(\frac{a}{e}\) = 0 and x + \(\frac{a}{e}\) = 0.

Main Facts about two Standard forms of Ellipse

Notes : 1. The focal distances of any point (x₁, y₁) of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 are a ± ex₁
2. Sum of the focal distances of a point of the ellipse is equal to length of the major axis.

Hyperbola: Let l be a fixed line and F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = e | PM | (e > 1) is called a hyperbola.

2. Sum of the focal distances of a length of the major axis.

Hyperbola: Let l be a fixed line and F be a fixed point (not on l). Let P be any point in the plane of the line l and the point F. Let M be the foot of perpendicular from P on the line l. Then, the set of all points P such that | FP | = e | PM | (e > 1) is called a hyperbola.

Standard form of Hyperbola : The equation of a hyperbola in the standard form is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, where b² = a²(e² – 1). Putting ae -c, b² – c² – a² or c² – a² + b².

Notes: 1. The hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1 has two foci namely (- ae, 0) and (ae, 0) with two respective directrices as x + \(\frac{a}{e}\) = 0 and x – \(-\frac{a}{e}\) = 0

2. | ex₁ – a | and | ex₁ + a | are the focal distances of any point (x₁, y₁) on the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1.

Main facts about two standard forms of hyperbola

Circle : The locus of a point which moves in a plane such that its distance from a fixed point, in the plane, is always a constant, is called a circle. The fixed point is called the centre and the constant distance is called the Radius of the circle.

Equation of the Circle : The equation of a circle with centre (h, k) and radius V’ is given by (x – h)² + (y – k)² = r².
If the centre of a circle is origin and radius is ‘r’ then its equation is x² + y² = r².

General Equation of a Circle : The general equation of a circle is x² + y² + 2gx + 2fy + c = 0.

Its centre is (-g,-f) and radius = \(\sqrt{g^{2}+f^{2}-c}\)

Note : 1. If g² + f² – c = 0, then radius = 0.
In this case, x² +y² + 2gx + 2fy + c = 0 represents a circle whose radius is zero. Such a circle is called a Point Circle, [i.e., it represents the point (-g, -f)].

2. If g² + f² – c < 0, then radius is non-real. In this case, x2 +y2 + 2gx + 2fy + c = 0 represents the empty set. 3. If g2 + f2 – c > 0, then radius is real and therefore,
x² + y² + 2gx + 2fy + c = 0 represents a circle whose centre is at the point (- g,- f) and radius = \(\sqrt{g^{2}+f^{2}-c}\). (where g² + f² – c > 0).

Conditions for an Equation to Represent a Circle :

(i) It should be a second degree equation in x and y.
(ii) Coefficient of x² = Coefficient of y² ≠ 0.
(iii) It should not contain any term of the product xy.

Method to find the centre, radius of a circle :

(i) Write the equation of the circle with the coefficient of x² and y² being unity.
(ii) Compare this with x² + y² + 2gx + 2fy + c = 0 and find the value of g,f and c.
(iii) Then, (-g,-f) is the centre and \(\sqrt{g^{2}+f^{2}-c}\) is the radius of the circle.

By going through these CBSE Class 9 Sanskrit Notes Chapter 5 सूक्तिमौक्तिकम् Summary, Notes, word meanings, translation in Hindi, students can recall all the concepts quickly.

Class 9 Sanskrit Chapter 5 सूक्तिमौक्तिकम् Summary Notes

सूक्तिमौक्तिकम् Summary

नीति-ग्रंथों की दृष्टि से संस्कृत साहित्य काफी समृद्ध है। इन ग्रंथों में सरल और सारगर्भित भाषा में नैतिक शिक्षाएँ दी गई हैं। इनके द्वारा मनुष्य अपना जीवन सफल और समृद्ध बना सकता है। ऐसे ही मूल्यवान कुछ सुभाषित इस पाठ में संकलित हैं, जिनका सार इस प्रकार है

• मनुष्य को अपने आचरण की रक्षा करनी चाहिए। धन नश्वर है। वह कभी आता है तो कभी चला
• जैसा व्यवहार स्वयं को अच्छा न लगे, वैसा व्यवहार दूसरों के साथ नहीं करना चाहिए।
• मीठे बोल सभी को प्रिय लगते हैं। अतः मीठा बोलना चाहिए। मनुष्य को बोलने में कंजूसी नहीं करनी चाहिए।
  महापुरुष अपने लिए कुछ नहीं करते हैं। वे सदा परोपकार करते रहते हैं। कारण कि महापुरुषों का पृथ्वी पर आगमन परोपकार के लिए ही होता है।
• मनुष्य को गुणों के लिए यत्न करना चाहिए। गुणों के द्वारा वह महान बनता है। . सज्जन लोगों की मित्रता स्थायी होती है, जबकि दुर्जन लोगों की मित्रता अस्थायी।
• हंस तालाब की शोभा होते हैं। यदि किसी तालाब में हंस नहीं हैं तो यह उस तालाब के लिए हानिकर है।
• गुणज्ञ व्यक्ति को पाकर गुण गुण बन जाते हैं, परंतु निर्गुण को प्राप्त करके वे ही गुण दोष बन जाते हैं।

सूक्तिमौक्तिकम् Word Meanings Translation in Hindi

1. वृत्तं यत्नेन संरक्षेद् वित्तमेति च याति च।
अक्षीणो वित्ततः क्षीणो वृत्ततस्तु हतो हतः॥ -मनुस्मृतिः

शब्दार्थाः-
वित्तम् – धन, ऐश्वर्य,
वृत्तम् – आचरण, चरित्र
अक्षीणः – नष्ट नहीं होता
क्षीणः – नष्ट होना
वृत्ततः – आचरण से
हतः – नष्ट हो जाना
एति – आता है
याति – जाता है
संरक्षेत् – रक्षा करनी चाहिए
यत्नेन – प्रयत्नपूर्वक।

अर्थ- हमें अपने आचरण की प्रयत्नपूर्वक रक्षा करनी चाहिए, क्योंकि धन तो आता है और चला जाता है, धन के नष्ट हो जाने पर मनुष्य नष्ट नहीं होता है। परंतु चरित्र या आचरण के नष्ट हो जाने पर मनुष्य भी नष्ट हो जाता है।

अव्ययानां वाक्येषु प्रयोगः –
पदानि – वाक्येषु प्रयोगः
च (और) – वित्तं आयाति याति च।
तु (तो) – वृत्ततः क्षीणः तु हतो हतः।

विलोमपदानि –
पदानि – विलोमपदानि
एति – याति
अक्षीणः – क्षीणः

2. श्रूयतां धर्मसर्वस्वं श्रुत्वा चैवावधार्यताम्।।
आत्मनः प्रतिकूलानि परेषां न समाचरेत्॥ -विदुरनीतिः

शब्दार्थाः –
श्रूयता – सुनो
धर्मसर्वस्वं – धर्म के तत्व को
श्रुत्वा – सुनकर
अवधार्यताम् – ग्रहण करो, पालन करो,
प्रतिकूलानि – विपरीत
परेषाम् – दूसरों के प्रति
समाचरेत् – आचरण नहीं करना चाहिए
आत्मन: – अपने।

अर्थ-
धर्म के तत्व को सुनो और सुनकर उसको ग्रहण करो, उसका पालन करो। अपने से प्रतिकूल व्यवहार का आचरण दूसरों के प्रति कभी नहीं करना चाहिए अर्थात् जो व्यवहार आपको अपने लिए पसंद नहीं है, वैसा आचरण दूसरों के साथ नहीं करना चाहिए।

अव्ययानां वाक्येषु प्रयोगः –
पदानि – वाक्येषु प्रयोगः
न (नहीं) – आत्मनः परेषां न समाचरेत्।

पर्यायपदानि –
पदानि – पर्यायपदानि
धर्मसर्वस्वम् – कर्त्तव्यसारः
समाचारेत् – आचरणं कर्त्तव्यम्
प्रतिकूलानि – विपरीतानि

विपर्ययपदानि
श्रूयताम् – वदताम्
प्रतिकूलानि – अनुकूलानि

3. प्रियवाक्यप्रदानेन सर्वे तुष्यन्ति जन्तवः।
तस्माद् तदेव वक्तव्यं वचने का दरिद्रता। -चाणक्यनीतिः

शब्दार्था:-
प्रियवाक्यप्रदानेन – प्रिय वाक्य बोलने से
तुष्यन्ति – प्रसन्न होते हैं
जन्तवः – प्राणी
वक्तव्यम् – कहने चाहिए
वचने – बोलने में
दरिद्रता – गरीबी, कंजूसी।

अर्थ- प्रिय वचन बोलने से सब प्राणी प्रसन्न होते हैं तो हमें हमेशा मीठा ही बोलना चाहिए। मीठे वचन बोलने में कंजूसी नहीं करनी चाहिए।

विशेषण – विशेष्य – चयनम् –
विशेष्यः – विशेषणम्
तत् – वक्तव्य
का – दरिद्रता
सर्वे – जन्तवः

पर्यायपदानि –
पदानि – विलोमपदानि
तुष्यन्ति – प्रसन्नाः भवन्त
प्रियं – मधुरं
वक्तव्यम् – कथनीयम्

विलोमपदानि –
पदानि – विलोमपदानि
वक्तव्यम् – श्रवणीयम्
तुष्यन्ति – रोदन्ति
प्रियं – कट

4. पिबन्ति नद्यः स्वयमेव नाम्भः।
स्वयं न खादन्ति फलानि वृक्षाः।
नादन्ति सस्यं खल वारिवाहाः
परोपकाराय सतां विभूतयः॥ – सुभाषितरत्नभाण्डागारम्

शब्दार्था:-
नद्यः – नदियाँ
अम्भः – पानी
पिबन्ति – पीती हैं
वृक्षाः – पेड़
खादन्ति – खाते हैं
खलु – निश्चित ही
वारिवाहा: – बादल
सस्य – अनाज
अदन्ति – खाते हैं
सतां – सज्जनों की
विभूतयः – धन – संपत्ति
परोपकाराय – दूसरों की भलाई के लिए।

अर्थ- नदियाँ अपना पानी स्वयं नहीं पीतीं। पेड़ अपने फल स्वयं नहीं खाते, निश्चित ही बादल अनाज (फसल) को नहीं खाते (इसी प्रकार) सज्जनों (श्रेष्ठ लोगों) की धन-सम्पत्तियाँ दूसरों के लिए ही होती हैं।

अव्ययानां वाक्येषु प्रयोगः –
पदानि – वाक्येषु प्रयोगः
न – नद्यः जलं न पिबन्ति।
खलु (निश्चित ही) – खलु वारिवाहाः सस्यं न खादन्ति।
स्वयं (अपना) – स्वकार्यं स्वयं कुरु।

पर्यायपदानि –
पदानि – पर्यायपदानि
वारिवाहाः – मेघाः
अम्भः – जलं, वारि
अदन्ति – खादन्ति
विभूतयः – समृद्धयः
वृक्षाः – तरवः, महीरुहाः

5. गुणेष्वेव हि कर्तव्यः प्रयत्नः पुरुषैः सदा।
गुणयुक्तो दरिद्रोऽपि नेश्वरैरगुणैः समः॥ – मृच्छकटिकम्

शब्दार्था:-
गुणेषु – गुणों में
प्रयत्नः – कोशिश
पुरुषैः – पुरुषों के द्वारा
कर्तव्यः – करना चाहिए
हि – निश्चित ही
सदा – हमेशा
गुणयुक्तः – गुणवान्
दरिद्रः – गरीब
अपि – भी
ईश्वरैः – ऐश्वर्यशाली
समः – समान
न – नहीं
गुणैः – गुणों से

अर्थ – मनुष्य को सदा गुणों को ही प्राप्त करने का प्रयत्न करना चाहिए। गरीब होता हुआ भी वह गुणवान व्यक्ति ऐश्वर्यशाली गुणहीन के समान नहीं हो सकता (अर्थात् वह उससे कहीं अधिक श्रेष्ठ होता है।)

अव्ययानां वाक्येषु प्रयोगः –
वाक्येषु – वाक्येषु प्रयोगः
एव (ही) – ईश्वरः सर्वत्र एव अस्ति।
हि (निश्चित ही) – त्वम् ह्यः हि मम गृहम् आगतवान्।
समः (समान) – श्रेष्ठः जनः ईश्वरेण समः भवति।

पर्यायपदानि –
पदानि – पर्यायपदानि
दरिद्रः – निर्धनः
गुणयुक्तः – गुणसम्पन्नः
समः – समानः

विलोमपदानि –
पदानि – पर्यायपदानि
अगुणैः – सगुणैः
सदा – कदाचित्
गुणयुक्तः – गुणहीनः
समः – असमः

6. आरम्भगुर्वी क्षयिणी क्रमेण
लघ्वी पुरा वृद्धिमती च पश्चात्।
दिनस्य पूर्वार्द्धपरार्द्धभिन्ना
छायेव मैत्री खलसज्जनानाम॥ – नीतिशतकम

शब्दार्थाः-
आरम्भगुर्वी – आरंभ में लंबी
क्रमेण – धीरे-धीरे
क्षयिणी – घटते स्वभाव वाली
पुरा – पहले
लघ्वी – छोटी
वद्धिमती – लंबी होती हुई
पूर्वार्द्ध – पूर्वाह्न
अपरार्द्ध – अपराह्न
छायेव – छाया के समान, भिन्न
‘खल – दुष्ट
सज्जनानाम् – सज्जनों की।

अर्थ – आरंभ में लंबी फिर धीरे-धीरे छोटी होने वाली तथा पहले छोटी फिर धीरे-धीरे बढ़ने वाली पूर्वाह्न तथा अपराह्न काल की छाया की तरह दुष्टों और सज्जनों की मित्रता अलग-अलग होती है।

विशेषण – विशेष्य चयनम् –
विशेषणम् – विशेष्यः
गुर्वी/लघ्वी – मैत्री

अव्ययानां वाक्येषु प्रयोगः –
पदानि – वाक्येषु प्रयोगः
पुरा (पहले) – पुरा भारतस्य नाम आर्यावर्तः आसीत्।
पश्चात् (बाद) – पश्चात् आर्यावर्तः नाम भारतम् अभवत्।

पर्यायपवानि –
पवानि – पर्यायपवानि
आरम्भगुर्वी – आद्यौ दीर्घा
वृद्धिमती – वृद्धिम् उपगता
खल – दुष्ट
पुरा – प्राचीनकाले
क्षयिणी – क्षयशीला
पूर्वार्द्धपरार्द्धभिन्ना – पूर्वार्द्धन परार्द्धन च पृथग्भूता
सज्जनानाम् – सुजनानाम्/श्रेष्ठ जनानाम्
क्रमेण – क्रमानुसारेण

7. यत्रापि कुत्रापि गता भवेयु
हंसा महीमण्डलमण्डनाय।
हानिस्तु तेषां हि सरोवराणां
येषां मरालैः सह विप्रयोगः॥ – भामिनीविलासः

शब्दार्थाः –
महीमण्डल – पृथ्वी
मण्डनाय – सुशोभित करने के लिए
गताः – चले जाने वाले
भवेयुः – होने चाहिए
मरालैः – हंसों से
हंसाः – हंस
सरोवराणां – तालाबों का
विप्रयोगः – अलग होना
हानिः – हानि
सह – साथ
तेषां – उनका
येषां – उनका।

अर्थ – पृथ्वी को सुशोभित करने वाले हंस भूमण्डल में (इस पृथ्वी पर) जहाँ कहीं (सर्वत्र) भी प्रवेश करने में समर्थ हैं, हानि तो उन सरोवरों की ही है, जिनका हंसों से वियोग (अलग होना) हो जाता है।

विशेषण – विशेष्य चयनम् –
विशेषणम् – विशेष्यः
गतः – हंसाः
तेषाम् – सरोवराणाम्

अव्ययानां वाक्येषु प्रयोगः –
पवानि – वाक्येषु प्रयोग
यत्र-कुत्र (जहाँ-कहाँ)- यत्र कृष्णः कुत्र पराजयः?
अपि (भी) – त्वम् अपि पठ।
(साथ) – रामेण सह सीता अगच्छत्। पर्यायपदानि

पर्यायपवानि –
पवानि पर्यायपदानि
महीमण्डल – पृथ्वीमण्डल
मण्डनाय – अलङ्करणाय
मरालैः – हंसाः
विप्रयोगः – वियोगः
सरोवराणाम् – तडागानाम्

8. गुणा गुणज्ञेषु गुणा भवन्ति
ते निर्गुणं प्राप्य भवन्ति दोषाः।
आस्वाधतोयाः प्रवहन्ति नद्यः
समुद्रमासाद्य भवन्त्यपेयाः॥ – हितोपदेशः

शब्दार्था:-
गुणज्ञेषु – गुणों को जानने वालों में (गुणवान),
निर्गुणं – गुणहीन
दोषा: – दुर्गु
आस्वाद्यतोया: – स्वादयुक्त, जलवाली,
आसाद्य – प्राप्त करके
प्रवहन्ति – बहती हैं
नद्यः – नदियाँ
अपेया: – न पीने योग्य
प्राप्य – प्राप्त करके
भवन्ति – हो जाती हैं।

अर्थ – गुणवान लोगों में रहने के कारण ही गुणों को सगुण कहा जाता है। गुणहीन को प्राप्त करके वे दुर्गुण (दोष) बन जाते हैं। जिस प्रकार नदियाँ स्वादयुक्त जलवाली होती हैं, परंतु समुद्र को प्राप्त करके न पीने योग्य अर्थात् (कुस्वादु या नमकीन) हो जाती हैं।

विशेषण-विशेष्य चयनम् –
आस्वाद्यतोयाः – नघ:
अपेयाः – नघ:

पर्यायपदानि
गुणज्ञेषु – गुणज्ञातृषु जनेषु
आसाद्य – प्राप्य
निर्गुणं – गुणहीनः
आस्वाद्यतोयाः – स्वादनीयजलसम्पन्नाः
अपेयाः – न पेयाः, न पानयोग्याः

Here we are providing Class 11 Biology Important Extra Questions and Answers Chapter 9 Biomolecules. Important Questions for Class 11 Biology are the best resource for students which helps in Class 11 board exams.

Class 11 Biology Chapter 9 Important Extra Questions Biomolecules

Biomolecules Important Extra Questions Very Short Answer Type

Question 1.
What is hydrolysis?
Answer:
During the digestion of carbohydrates, the glycosidic bond between sugar residues is broken by the addition of water and this is called hydrolysis.

Question 2.
Define fatty acid.
Answer:
Fatty acids are organic acids with a hydrocarbon chain ending in a carboxyl group.

Question 3.
What are iso-enzymes?
Answer:
The enzymes possessing slightly different molecular structures but similar in their bio-catalytic action.

Question 4.
Give the names of 2 non-polar organic solvents that are used for lipid extraction from cells.
Answer:
Chloroform, Ether.

Question 5.
Name one monosaccharide sugar that is found in the blood plasma of human beings.
Answer:
Glucose.

Question 6.
What is the function of calcium in the human body?
Answer:
Calcium in bones and teeth provides strength and rigidity to them.

Question 7.
Name the bonds uniting the monosaccharide subunits.
Answer:
Monosaccharide sub-units are joined together by glycoside bonds.

Question 8.
What are lygases?
Answer:
Lygases are the enzymes that join two substrate molecules.

Question 9.
Name the ending group of fatty acids which are organic acids with a hydrocarbon chain.
Answer:
Carboxyl.

Question 10.
Which is the most common form of sugar in fruits?
Answer:
Fructose.

Question 11.
How can we overcome the deficiency of iodine?
Answer:
By using iodized common salt.

Question 12.
Names the important food storage of carbohydrates.
Answer:
Starch and Glycogen.

Question 13.
Define allosteric modulation.
Answer:
It regulates the activity of some enzymes internally.

Question 14.
Mention two functions of the sodium and potassium ions in the body.
Answer:
Functions of the sodium and potassium ions in the body are:

1. To maintain the volumes of extracellular and intracellular fluids.
2. Transmission of nerve impulses.

Question 15.
Name the small molecules of the cell.
Answer:
The small molecules of the cell are:

1. Minerals
2. Water
3. Amino acids
4. Sugars
5. Lipids
6. Nucleotides.

Biomolecules Important Extra Questions Short Answer Type

Question 1.
What are monosaccharides? Give few examples.
Answer:
Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed into still smaller carbohydrates. The general formula is C_n H_2n O_n e.g. Ribose, Glucose, Fructose.

Question 2.
What is a disaccharide?
Answer:
A disaccharide is a sugar molecule composed of two monosaccharide sub-units e.g. a molecule of sucrose is formed from a molecule of glucose and a molecule of fructose by dehydration.

Question 3.
Why do fats release more energy than carbohydrates on oxidation?
Answer:
Like carbohydrates, fats are made up of C, H, and O but they contain fewer oxygen molecules than carbohydrates. On oxidation they consume more oxygen releasing more energy.

Question 4.
What is the function of calcium in our body? In what form is calcium deposited in the middle lamella?
Answer:
Calcium is impregnated in bones and teeth. It provides them with strength and rigidity.
Calcium is deposited in the middle lamella in the form of calcium pectate.

Question 5.
Define cellular pool. What are the characteristics of a small molecule in the cellular pool?
Answer:
The collection of various types of molecules in a cell is termed a cellular pool.

The characteristics of small molecules in the cellular pool are

1. Low molecular weight
2. Simple molecular conformation.
3. Higher solubility.

Question 6.
What are lipids or fats? State their characteristic. What are the functions of subcutaneous fat in our body?
Answer:
Fat or lipids are esters of glycerol and fatty acids. They are made up of C, H, and O but include proportionately less oxygen as compared to carbohydrates. They are insoluble in water and soluble in non-polar organic solvents.

Functions of subcutaneous fat are

• Storage of food (chemical form of energy).
• Shock absorption.
• Insulation.

Question 7.
How are amino acids linked to form a peptide chain?
Answer:
Amino acids are condensed together to form a peptide chain. The bond is formed between the carboxyl group of one amino acid and the amino group of adjacent amino acid. This is called a peptide bond and it is formed by dehydration.

Question 8.
What are phospholipids?
Answer:
Phospholipids are lipids containing phosphate groups e.g. phosphoglyceride. They have a hydrophilic polar head and a hydrophobic non-polar tail.

Question 9.
What are macromolecules? Give examples.
Answer:
Simple molecules assemble and form large and complex molecules called macromolecules e.g. proteins, lipids, nucleic acid, and carbohydrates.

Question 10.
Give two examples of storage polysaccharides.
Answer:
Two advantages of storing carbohydrates in the form of polysaccharides:

1. Food storage polysaccharides are starch and glycogen. Starch is found in rice, wheat, etc. Glycogen is stored in the liver. During their formation, many molecules of water are removed from monosaccharides.
2. If necessary, polysaccharides are broken down by enzymes for the release of energy.

Question 11.
What is chitin?
Answer:
Chitin: Chitin is similar to cellulose in many ways except its basic units are not glucose, but a similar molecule that contains nitrogen (N-acetyl glucosamine) and is soft as well as leathery.

Question 12.
Explain how glycosidic bonds are formed?
Answer:
Formation of Glycosidic Bonds: The aldehyde or ketone group of a monosaccharide can react and bind with an alcoholic group of another

Glycosidic bond

organic compound to join the two compounds together. This bond is known as a glycosidic bond. This bond may be hydrolyzed to give the original compounds. Monosaccharides by uniting together through glycosidic bonds give rise to compound carbohydrates.

Question 13.
Describe functions of polysaccharides in living organisms.
Answer:
Food storage polysaccharides: As the name suggests saccharides which perform the function of storing food. Examples are starch, glycogen.

Starch: It is formed as a result of photosynthesis. It is found in large quantities in rice, wheat, cereals, legumes, potato, tapioca, and bananas. It is an energy-giving substance as it stores energy.

Glycogen: It is found in the muscles and liver of mammals and stores energy. There are distinct advantages of storing carbohydrates in the form of polysaccharides

1. During the formation, many molecules of water get removed and bulk reduced
2. Polysaccharides are relatively easy to store and get broken down easily by enzymes to release energy.

Structural Polysaccharides: Examples of these are cellulose and chitin. These take part in the formation of the organism.

Cellulose: It is a plant product. It is perhaps the most abundant material found in the living world. If forms cell walls. It is a fibrous polysaccharide that has high tensile strength. Wood and cotton have large quantities of cellulose.

Chitin: Chitin is similar to cellulose in many ways except that its basic unit is not glucose, but a similar molecule that contains nitrogen (N-acetyl glucosamine). Chitin is soft and leathery, it becomes hard when it gets impregnated with calcium carbonate or certain proteins. The insolubility of these polysaccharides in water helps in retaining the particular form it also helps in strengthening the structure of organisms.

Question 14.
What is meant by the tertiary structure of proteins?
Answer:
The tertiary structure of proteins: Many amino acid units form polypeptides. The peptide bonds holding the amino acids together in a particular way constitute the primary structure of the protein. A functional protein contains one or more polypeptide chains.

Through the formation of hydrogen bonds, peptide chains assume a secondary structure. Secondary protein may be in the form of a twisted helix or pleated sheet.

When the individual peptide chains of the secondary structure of the protein are further extensively coiled and folded into sphere-like shapes with the hydrogen bonds between the amino and carboxyl group and various other kinds of bonds cross-linking on-chain to another they form tertiary structure.

The ability of proteins to carry out specific reactions is the result of their primary, secondary, and tertiary structure.

The tertiary structure (myoglobin)

Question 15.
Explain the composition of triglycerides.
Answer:
Fat is esters of fatty acids with glycerol. Each molecule of glycerol can react with 3molecules of fatty acid. On the basis of fatty acids that are attached to the glycerol molecule, the esters are called either mono, di, or triglycerides.

Triglyceride-Tripalmitin

Question 16.
Mention the difference between saturated and unsaturated fat?
Answer:
Differences between saturated and unsaturated fat:

Question 17.
Describe the structure of phospholipid. How are they arranged in the cell membrane?
Answer:
Structure of phospholipids: Phospholipids are a class of lipids that serve as the structural component of the cell membrane. Phospholipids have only 2 fatty acids attached to the glycerol while the 3rd glycerol binding site holds a phosphate group. This phosphate is bound to alcohol.

These lipids have both hydrophilic and hydrophobic pans due to a charge on the phosphoric acid/alcohol head of a molecule and lack a charge of the long tail of the molecule (made by fatty acids). When exposed to an aqueous solution the charged heads are attracted to the water phase and the non-polar tails are repelled from the water phase.

Phospholipids
When two layers of polar lipids come together to make a double layer, the outer hydrophilic face of every single layer will orient itself towards the solution, and the hydrophobic part will become immersed in the core of the bilayer.

Water acts as a solvent for polar molecules and the arrangement of phospholipids in the lipid bilayer of membranes is dependent on water.

Question 18.
Write short notes on
(i) Steroids
Answer:
Steroids: Steroids are complex compounds mostly found in animal hormones and cell membranes. The best example of steroids is cholesterol. The cell membrane of fungi contains ergosterol. The prostaglandins are fatty acid derivatives. They occur in minute amounts and function in blood clotting, smooth muscle contraction, and allergic reactions, etc.

(ii) Wax.
Answer:
Wax: Waxes are lipids. They are esters formed by the combination of a saturated long-chain fatty acid with long-chain alcohol. They play an important role in protection as tires form a water-proof covering over the root hairs and parts of the body in some organisms. Wax is soft and pliable. The paraffin is hard when cold. Fruits, feathers, leaves, the skin of man, and the exoskeleton of insects are waterproofed by the coating of wax. The bacteria causing TB and leprosy produce wax (Wax D.).

Question 19.
Describe the structure and function of ATP.
Answer:
ATP: It is a primary and universal carrier of chemical energy in the cell. Living cells capture, store and transport energy in a chemical form, largely as ATP and it is the ATP that is the carrier and intermediate source of chemical energy to those reactions in the cell which do not occur simultaneously.

These reactions can take place only if the chemical energy is released. It was Fritze Lipmann in 1941 who postulated this unifying concept and proposed the ATP cycle as given in the figure below.

Adenosine triphosphate

Question 20.
How are amino acids bonded together? Describe how these bonds are formed?
Answer:
Proteins are also formed from amino acids but they have small peptides. The two amino acids are linked by the formation of a peptide bond. Successive amino acids can be linked by peptide bonds to form a linear chain of many amino acids.

When a few amino acids are joined together, the molecule is called a peptide. Proteins are macromolecules formed from a large number of amino acids.

Peptide band

Question 21.
Draw the structure of amino acid.

(A) Acidic amino acid
Answer:
Amino acids: These are small molecules made of carbon, hydrogen, oxygen, and nitrogen, in certain cases sulfur is also found. Amino acids may be monocarboxylic or dicarboxylic acids bearing one or two amino groups.

The a-carbon is next to the carboxyl – C. The four valencies of the a-carbon of an amino acid hold respectively an amino (NH₂) group, a carboxyl (COOH) group, a hydrogen atom, and side-chain fig (B) which may be polar or non-polar.

A free amino group is basic, a free carboxyl group is acidic. Lysine and alanine are basic amino acids because they have two amino groups and one carboxyl group. Glutamic acid and aspartic acid contain one amino and two carboxyl groups each is classified as acidic amino acids.

Alanine, glycine, valine, and phenylalanine are neutral amino acids because these contain one carboxyl group. Two amino acids can be linked by the formation Of a bond called a peptide bond. With the help of peptide bonds, many amino acids form a linear chain of many amino acids.

(B) Non-polar-side chain

Question 22.
Describe the primary structure of the protein.
Answer:
Primary Structure of Protein: Proteins are made of amino acids which have carboxyl group (COOH) and amino (group) (-NH₂). The COOH end of an amino acid is joined to the -NH₂ end of the other amino acid. Many amino acids are joined by peptide bonds which held them together in a particular sequence and constitute the primary structure of proteins. The structure does not make a protein functional.

It is a linear sequence of amino acids.

Question 23.
Name different types of RNA.
Answer:
There are three types of RNA.

1. Messenger RNA (mRNA).
2. Ribosomal RNA (rRNA)
3. Transfer RNA (tRNA).

Question 24.
List the differences between DNA and RNA.
Answer:
Differences between DNA and RNA:

Question 25.
Distinguish between prosthetic group and co-factors.
Answer:
Differences between coenzyme, cofactor, and prosthetic groups:

Question 26.
Explain the structure of an enzyme.
Answer:
Structure of an Enzyme: The enzymes are chemical substances. They catalyze the chemical reactions in the cells. They are secreted and synthesized by living cells. Most all the enzymes are proteinous in nature. Some enzymes contain a nonprotein part called the prosthetic group. Some p esthetic groups are metal compounds. NAD is a coenzyme. All enzymes have active sites.

Question 27.
Distinguish between Unsaturated fatty acids and Saturated fatty acid
Answer:

Question 28.
Distinguish between aldose sugar and Ketose sugar
Answer:

Question 29.
Distinguish between Oil and Fat.
Answer:

Biomolecules Important Extra Questions Long Answer Type

Question 1.
Enlist the functions of small carbohydrates?
Answer:

1. Monosaccharides are formed during the photosynthetic pathway. They are stored in plants and are utilized by other living organisms depending on them.
2. Glucose is the blood sugar of many animals and on oxidation, it provides energy for all vital activities.
3. Nucleotides and nucleosides contain pentose sugar in the form of ribose and deoxyribose sugars. They form a part of nucleic acids.
4. Lactose of milk is formed from glucose and galactose and mammary glands of mammals.
5. Glucose is used for the synthesis of fats and amino acids.
6. Structural polysaccharides like cellulose and oligosaccharides are derived from mono-saccharides.
7. Food storage polysaccharides like starch and glycogen are derived from monosaccharides.

Question 2.
Enumerate the functions of Lipids.
Answer:

1. Lipids are storage products in plants as well as animals.
   (a) In plants, fats are stored in cotyledons or endosperm to provide nourishment to the developing embryo.
   (b) In animals fats are stored in adipocytes to be used whenever required by the body.
2. In animals, subcutaneous fats act as an insulation layer and shock \ absorber.
3. They form structural components of membranes, phospholipids, glycolipids, and sterols.
4. They take part in the synthesis of steroid hormones, vitamin D, and bile salts.
5. Act as a solvent for fat-soluble vitamins i.e., vitamin A, D, E, and K.
6. The neutral fats form a concentrated fuel producing more than twice as much energy per gram as do the carbohydrates. They thus, represent an economical food reserve in the body.
7. The wax lipids form a waterproof protective coating on animal furs, plant stem, leaves, and fruits.

Question 3.
How does water help in maintaining the constancy of the internal environment of an organism?
Answer:
Some substances, capable of neutralizing acids or bases, remain in solution in the cytoplasm as extracellular fluids, e.g., bicarbonate (HCO₃), carbonic acid, dibasic phosphate (HPO₄^-2). Acids and bases mix in the body fluids with these substances and are neutralized by them. Because of its solvent action water aids in keeping a constant pH.

Water also helps in maintaining constant body temperature by eliminating excess heat through the evaporation of sweat. Elimination of waste products through urine also helps in maintaining the constancy of the internal environment of an organism.

Question 4.
What are peroxisomes and phagosomes?
Answer:
Peroxisomes: These were for the first time observed in the kidney of rodents. They are found both in plants and animals. Their size varies from 0.5 to lp in diameter. They are delimited by a single membrane and contain a finely granular matrix. They often possess a central core called nucleoid which may consist of parallel tubules or twisted with strands. Peroxisomes are generally observed in close association with the endoplas¬mic reticulum.

Peroxisomes in different plant and animal cells differ con¬siderably in their enzymatic make-up, but they contain some peroxide-producing enzymes like urate, oxidase, D-amino acid oxidase, B-hydroxy acid oxidase, and catalase. Peroxisomes are somehow associated with some metabolic processes like photorespiration and lipid metabolism in animal cells.

Sphaerosomes: There are cell organelles bounded by a single membrane. They contain enzymes and are visible under the light microscope. These show some affinities for fat stains, including Sudan stain and sodium tetroxide.

These organelles originate from E.R. by budding. They contain enzymatic proteins which help in synthesizing oils and fats. Further devel¬opment of phagosomes takes place through an increase in the lipid content with a concomitant decrease in protein.

Question 5.
Enumerate the importance of Energy carriers.
Answer:
Energy carriers consist of nucleotides having one or two additional phosphate groups linked up at their phosphate end forming diphosphates and triphosphates. Linkage of additional phosphate groups occurs at the cost of a large amount of energy. This energy is provided by the oxidation of food mainly glucose and by photosynthesis.

Separation of the additional phosphate groups from the nucleotides by enzymatic hydrolysis releases a correspondingly large amount of energy.

Thus, ADP and ATP provide ready energy for biological activities.

The bonds joining the additional phosphate groups to the nucleotides are called high energy or energy-rich bonds, as they carry a great deal of energy. The nucleotides having more than one phosphate group are called higher nucleotides.

The energy of energy carriers, when set free is utilized for driving energy-dependent reactions in the cell and is biologically useful energy. ATP is the most common energy carrier in cells and is often called the energy currency of the cell.

Question 6.
Explain the functions of amino acids.
Answer:

1. Amino acids are the building blocks for proteins.
2. The amino acid Tyrosine takes part in the formation of the skin pigment melanin as well as hormones thyroxine and adrenaline.
3. Glycine is important for the formation of heme.
4. Tryptophan takes part in the formation of the vitamin nicotinamide.
5. In plants, tryptophan forms the growth hormone indole-3- acetic acid.
6. Amino acids are converted into glucose by deamination.
7. Histamine and other biogenic amines are formed by the removal of carboxyl groups from amino acids.

Question 7.
Give reasons for following
(i) Salts dissolve in water but oil does not
Answer:
Water molecules are hydrogen-bonded to form short-lived macromolecular aggregates. To dissolve in water, a solute molecule must form hydrogen bonds with water molecules. Salts are polar compounds, their hydrophilic polar groups form hydrogen bonds with water molecules. So they dissolve oils having hydrophobic non-polar groups that cannot join the lattice structure of water. Thus non-polar molecules of oil do not dissolve in water.

(ii) Amino acid can be basic
Answer:
A free amino group is basic and a free carboxyl group is acidic. Amino acids can be basic because they may carry two amino groups and one carboxyl group e.g., Arginine. One free amino group causes amino acids to be basic.

(iii) Phospholipids form a thin layer on the surface of an aqueous medium.
Answer:
Phospholipids form a thin layer on the surface of an aqueous medium due to the simultaneous presence of both polar and non-polar groups in the molecule. As a result, the phospholipid molecules may arrange themselves in a double-layered membrane in aqueous media.

Question 8.
Illustrate lock and key hypothesis of enzyme action?
Answer:
Mechanism of Enzyme action: The working of enzymes is a complex one. All enzymes first of all combine with the reactions they catalyze. In other words, enzymes with substrates form an intermediate complex before decomposition of the substrate can occur.

This two-way reaction can be represented as follows.
1st step: Enzyme substrate complex = Enzyme + Product.
Formation of the enzyme-substrate complex during enzyme action.

From the above, it is clear that the enzymes must combine first with substrate molecules in order to act. In order to explain the mode of action of an enzyme. Fischer proposed a lock and key theory. According to him if the right key fits in the right lock. The lock can be opened, otherwise not.

Model of enzyme activity

To explain the above in context with the enzyme action it is believed that molecules have specific configurations into which other molecules can fit. The molecules which are acted upon by the enzymes are called substrates of the enzymes. Under the above assumption, only those substrate molecules with the proper geometric shape can fit into the active site of the enzymes.

If this happens, the above molecules may compete with the substrate, and the reaction may either slow down or stop. Substances are called competitive inhibitors because they act to prevent the production of a substance.

An induced-fit model of enzyme action was given by Koshland (1959). Buttressing and catalytic are two groups of the active site of the enzyme. Their site when the substrate attaches to its bonds is broken.

Question 9.
What is the structure of DNA?
Answer:
The nucleic acids are among the largest of all molecules found in living beings. They contain three types of molecules (a) 5 carbon sugar, (b) Phosphoric acid (usually called phosphates when in chemical combi¬nation), and nitrogen-containing bases (Purines and Pyrimidines). The three join together to form a nucleotide i.e., sugar+ base + phosphate = Nucleotide. Only a few nucleotides are possible. They differ only in the kind of purines or pyrimidine (nitrogen-containing bases).

In 1953 J.D. Watson and F.H.C. Crick working in Cambridge Uni¬versity, England prepared a model of DNA molecule elucidating the struc¬ture of DNA molecule. They were awarded the Nobel Prize for this outstanding work.

Structure of DNA

Watson and Crick model of DNA: According to Watson and Crick, the DNA molecule consisted of two strands twisted around each other in the form of a helix. Each strand is made of polynucleotides, each polynucleotide consisting of many nucleotides which remain united with its complimentary’ chain with the help of bases.

Adenine always unites with thymine and cytosine with guanine. It means that one polynucleotide chain of DNA molecule is complementary to the other.

The distance between two chains of the helix is about 20 A and the helix turns over every 34 A. Each mm of the chain consists of about 10 nucleotides.

Structure of DNA

Question 10.
How does the substrate concentration affect the velocity of enzyme reaction?
Answer:
Michaelis constant or more appropriately Michaelis-Menten constant (Km) is a mathematical derivation given by Leonor Michaelis and Monde Menten in 1913 with the help of which velocity of reaction can be calculated for any substrate concentration.

Effect of substrate concentration on enzyme action

Km or Michaelis constant is the substrate concentration at which the chemical reaction attains half its maximum velocity. The constant is an inverse measure of the affinity of an enzyme for its substrate, that is the smaller the Km the greater the substrate affinity and vice versa. The value usually lies between 10⁴ – 10⁵ M

By going through these CBSE Class 11 Maths Notes Chapter 10 Straight lines Class 11 Notes, students can recall all the concepts quickly.

Straight lines Notes Class 11 Maths Chapter 10

Co-ordinate Geometry: The branch of Mathematics in which geometrical problem are solved through algebra by using the co-ordinate system, is known as co-ordinate geometry or analytical geometry.

Algebrical Representation of a Point: It is done by means of two numbers defined in a particular way called co-ordinates of the point.

Co-ordinate Axes : Let X’OX and Y’OY, be two mutually perpendicular lines taken as axes whose positive directions are shown by arrows on the axes. Let these axes intersect at O. Then, point O is called origin and the axes taken are called co¬ordinate axes.

(i) X’OX is called axis of x or x-axis.
(ii) Y’OY is called axis of y or y-axis.
Since the axes taken are mutually at right angles, they are called rectangular axes.

Quadrants : The co-ordinate axes divide the plane into four parts called quadrants.

XOY, YOX’, XOY’ and Y’OX are called first, second, third and fourth quadrants respectively. The names of the quadrants as I, II, III and IV are fixed once for all.

Co-ordinates : Let P be any point in the plane of axes. From P, draw PL, PM ⊥ on y-axis and x-axis respectively. Then, length LP is called the x-coordinate or the abscissa of point P and MP is called the y-co-ordinate or the ordinate of point P. Point whose abscissa is x and ordinate is y, is called the point (x, y).

Signs of Co-ordinates : The co-ordinates of point lying in the first, second, third and fourth quadrants are as (+, +), (-, +), (-, -) and (+, -) respectively.

Note :

1. Co-ordinates of orgin are (0, 0).
2. Ordinate of every point on the x-axis is zero.
3. Abscissa of every point on the y-axis is zero.

Distance Between Two Points : The distance between two
points P(x₁, y₁) and Q(x₂,y₂) is given by PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Note : 1. When the line PQ is parallel to they-axis, the abscissa of points P and Q will be equal, i.e., x₁ = x₂. Therefore, PQ = |y₂ – y₁|

2. When the line PQ is parallel to the x-axis, the ordinate of points P and Q wll be equal i.e.,y1= y2. Therefore, PQ = |x₂ – x₁ |.

3. The distance of a point P(x, y) from the origin (0, 0) is given
by OP = \(\sqrt{x^{2}+y^{2}}\)

Tests : In questions relating to geometrical figures such as triangle, quadrilateral, etc., take the given vertices in the given order.

(i) For an isosceles triangle : At least two sides are equal.
(ii) For an equilateral triangle : Three sides are equal.
(iii) For a right angled triangle : Sum of the squares of two sides is equal to the square of the third side.
(iv) For collinear points : Sum of the distances between two point-pairs is equal to the distance between the third point pair.
(v) For a square : Four sides are equal, two diagonals are
equal.
(vi) For a rhombus : Four sides are equal and there is no need to prove that two diagonals are unequal as a square is also a rhombus.
(vii) For a rectangle : Opposite sides are equal and two diagonals are equal.
(viii) For a parallelogram: Opposite sides are equal and there is no need to prove that two diagonals are unequal as a rectangle is also a parallelogram.

Internal Division : A point R between the points P and Q on the line segment PQ is said to divide PQ internally in the ratio m₁ : m₂,

Here, in the internal division, sense of line segments PR and RQ is same.
∴ m₁ and m₂ are both positive.

External Division : A point R on the part of the line segment PQ produced or QP produced is said to divide PQ externally in the ratio m₁ : m₂

Hence, if a point R divides PQ externally in ratio m₁ : m₂, then we can say that R divides internally in the ratio – m₁ : m₂ or m₁: – m₂.

Section Formula (Internal division) : The co-ordinates of the point R(x, y) which divides internally the straight line segment joining points P(x₁, y₁) and Q(x₂, y₂) in the ratio m₁: m₂ is given by
\(\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right)\)

How to remember: To find the x-co-ordinate of R, multiply m₁ with the x-coordinate of the point remote from m₁ and m₂ by the x co-ordinate of the point remote from m₂ and add these product, and divide the sum by m₁ + m₂. Similarly, findy.

Note : 1. The co-ordinates of the mid-point of the line segment joining( the points P(x₁ y₁) and Q(x₂, y₂) are given by
\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

2. If a point R divides the line segment joining the points P(x₁ y₁) and Q(x₂, y₂) in the ratio k : 1, then the co-ordinates of R are given by \(\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)\)

3. If R divides PQ externally in the ratio m₁ : m₂
divides PQ internally in the ratio m₁: -m₂.
∴ The co-ordinates of R are given by
\(\left(\frac{m_{1} x_{2}-m_{2} x_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}\right)\)

Area of a Triangle : The area of the triangle having vertices
as (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by \(\frac { 1 }{ 2 }\) [(x₁y₂ – x₂y₁ + (x₂y₃ – x₃y₂ + (x₃y₁ – x₁y₃)]
or
\(\frac { 1 }{ 2 }\) [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)]

How to remember : Result of area can be written down in a simpler way as in the figure given :

Numbers to be multiplied are shown by arrows. All the products with arrow downwards are of positive sign and are added to the products with arrow upwards with negative sign.

Condition of Collinearity of Three Points : Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) will be collinear, if and only if the area of the triangle ABC is zero. That is,
(x₁y₂ – x₂y₁) + (x₂y₃ – x₃y₂) + (x₃y₁ – x₁y₃) = 0.
or x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) = 0.

Slope of a Line : The slope (or gradient) of a line is the tangent of the angle which the part of the line above the x-axis makes with the positive direction of the x-axis.

Note : 1. The slope of the line is independent of the sense of line. Consider sense AB. Then slope = tan θ and on cosidering sense BA, the slope = tan (π + θ) = tan θ. Thus, it is same for both senses of the line.

2. The slope of a line is denoted by m, i.e., m = tan θ, where θ is the angle which the line makes with the positive direction ofx-axis.

3. Slope of a line is not defined, when θ = 90°, as tan 90° is not defined.

4. The slope of a line equally inclined to axes is ± 1.
[∵ tan 45° = 1, tan 135° = – 1]

5. If a line is parallel to the x-axis, θ = 0 and its slope m = tan 0° = 0.

Slope of a Line Joining Two Points : The slope of the line joining points (x₁,y₁) and (x₂ , y₂) is given by \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Condition of Parallelism and Perpendicularity :
(i) The two lines are parallel, if their slopes are equal, i.e., m₁ = m₂.
(ii) The two lines are perpendicular, if the product of their slopes = – 1, i.e., m₁m₂ = -1
Or the slope of one line is the negative reciprocal of the other.

Intercepts : Let a straight line AB meet the axes in A and B. Then,
(i) OA is called the intercept of the line on x-axis or x-intercept.
(ii) OB is called the intercept of the line ony-axis or y-intercept.
(iii) AB is called the portion of the line intercepted between the axes.

Rule for Signs of Intercepts: (i) The intercept on the x-axis is positive, if measured to the right of the origin and negative, if measured to the left.
(ii) The intercept on the y-axis is positive, if it is measured above the origin, and negative, if measured below.

Locus of a point: Locus of a point is the path traced by a moving point when it moves under some given geometrical conditions.

Method to Find Locus of Moving Point:

Step 1 : Suppose the co-ordinates of moving point as (x, y).
Step 2 : Write down the given condition under which the point moves.
Step 3 : Express the geometrical condition in step 2 in terms of x and y, i.e., co-ordinates of the moving point.
Step 4 : Simplify, if required, the result in step 3. Equation so obtained, will be the equation of the locus.

Equation of Straight Lines Parallel to Co-ordinate Axes
(i) Equation of a straight line parallel to x-axis and at a distance of ‘b’ from it is y = b.
(ii) Equation of x-axis is y = 0.
(iii) Equation of a straight line parallel to y-axis and at a distance of ‘a’ from it is x = a.
(iv) Equation of y-axis is x – 0.

A Line Through The Origin : The equation of a straight line passing through the origin and making an angle θ with the positive direction of the x-axis is y = mx, where m = tanθ.

Slope-Intercept Form: (i) The equation of straight line which cuts off a given intercept ‘c’ on they-axis and is inclined at a given angle ‘θ’ to the x-axis is y = mx + c, where m = tan θ.
(ii) The equation of a line whose slope is m and the x-intercept is d, which is y = m(x – d).

Intercept Form : The equation of a straight line which cuts off given intercepts a and b from the axes is \(\frac{x}{b}+\frac{y}{b}\) = 1.

Normal Form : The equation of a straight line in terms of the length of the perpendicular p from the origin upon it and the angle a which this perpendicular makes with the x-axis is x cos α+y sin α = P

Point-Slope Form : The equation of a straight line drawn through a given point (x₁,y₁) making an angle 0 with x-axis is y – y₁ = m(x – x₁), where m = tan θ.

Two-Point Form : The equation of a straight line passing through, two given points (x₁, y₁) and (x₂, y₂) is y – y₁ =\(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x-x₁).

Distance Form : The equation of a straight line in the form x – x y – y
\(\frac{x-x_{1}}{\cos \theta}=\frac{y-y_{1}}{\sin \theta}\) = r. where (x₁, y₁) is a point on the line, r the distance between the points (x, y) and (x₁, y₁), θ is the angle which the line makes with the x-axis.

General Equation of a Straight Line : Any equation of the first degree in x and y represents a straight line, such as
Ax + By + C = 0 …(1)

Some Cases :

(i) If A = 0, B ≠ 0, then (1) reduces to By + C = 0
⇒ y = \(-\frac{\mathrm{C}}{\mathrm{B}}\) which is the equation of horizontal line ii.e. parallel to x axis)
(ii) If A = C = 0, B ≠ 0, then (1) reduces to y = 0, which is the equation of x-axis.
(iii) If A ≠ 0, B = 0, then (1) reduces to Ax + C = 0 ⇒ x = \(-\frac{\mathrm{C}}{\mathrm{A}}\) , which is the equation of a vertical line (i.e., parallel to y-axis).
(iv) If B = C = 0, A ≠ 0, then (1) reduces x = 0, which is the equation of y-axis.
(v) If A ≠ 0, B ≠ 0, C = 0, then (1) reduces to Ax + By = 0 ⇒ y = \(-\frac{\mathrm{A}}{\mathrm{B}} x\) , which is the equation of a straight line passing through the origin.

Rule to Find the Intercepts of a Line on the Axes : In the equation of the line. Puty = 0, and find x. This gives the intercept on the x-axis. Again, put x = 0, and findy. This is intercept on the v- axis.

Rule to Reduce the General Equation to the Normal Form:
In the given equation :
Divide both sides by \(\sqrt{(\text { coeff. of } x)^{2}+\left(\text { coeff. of } y^{2}\right)^{2}}\)
Transpose the constant term to R.H.S. and make it positive (by changing the signs throughout, if necessary).

Identical Lines : If equations ax + by + c = C and a’x + b’y + c’ = 0 represent the same straight line, then \(\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}\)

Point of Intersection of Two Lines : Let equation of the lines be
a₁x + b₁y + c₁= 0
and a₂c + b₂y + c₂ = 0

The point of intersection can be obtained by solving these equations by cross-multiplication.

Test for Concurrence of Three Lines : If the equations of three lines are a₁x + b₁y + c₁= 0, a₂c + b₂y + c₂ = 0, a₃x + b₃y + c₃= 0 and if three constants l, m, n can be found such that l(a₁x + b₁y + c₁) + m(a₂x + b₂y + c₂) + n(a₁x + b₃y + c₃ = 0 identically (i.e., = 0, for all values of x and y), then the three straight lines meet in a point.

Angle between Two Lines: (i) The angle θ between two lines y = m₁x + c₁ and y = m₂x + c₂ is given by tan θ = ±\(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) .

Condition of Parallelism : If lines are parallel, then θ, the angle between them = 0.
⇒ tan θ = tan 0° = 0
⇒ \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)= 0
⇒ m₁ – m₂ = 0
⇒ m₁ = m₂.
Thus, the two lines are parallel, if their slopes are equal.

Condition of perpendicularity : If lines are perpendicular to each other, then 0, the angle between them = 90°.
⇒ tan θ = tan 90° = ∞
⇒ \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) = ∞
⇒ The denominator, 1 + m₁m₂ = 0.
⇒ m₁ m₂ = – 1 or m₂ = \(-\frac{1}{m_{1}}\)

Hence, the two lines are perpendicular to each other, if the product of their slopes = – 1, or the slopes of one line is the negative reciprocal of the other.

Note : 1. Angle between the two lines means the acutal angle between the lines Hence, for finding the angle between the two lines, we shall be rejecting the negative values of tan 0.
2. The complete angle formula is used, when the angle between the lines is given.

(ii) The angle θ between the lines Ax + By + C = 0 and A’x + By + C’ = 0 is given by tan θ = .
\(\frac{\mathbf{A}^{\prime} \mathbf{B}-\mathbf{A B}^{\prime}}{\mathbf{A A}^{\prime}+\mathbf{B B}^{\prime}}\)

Condition of parallelism : If the lines are parallel, then θ =
⇒ tan θ = tan 0° = 0.
⇒ \(\frac{\mathrm{A}^{\prime} \mathrm{B}-\mathrm{AB}^{\prime}}{\mathrm{AA}^{\prime}+\mathrm{BB}^{\prime}}\) = 0
⇒ A’B-AB’ = 0
⇒ \(\frac{\mathrm{A}}{\mathrm{A}^{\prime}}=\frac{\mathrm{B}}{\mathrm{B}^{\prime}}\)

Thus, if two lines (equations in the general form) are parallel, then the ratio of the coeff. of x = ratio of the coeff. of y.
Condition of perpendicularity : If lines are perpendicular to each other, then θ = 90°.
tan θ = tan 90° = ∞
⇒ \(\frac{A^{\prime} B-A B^{\prime}}{A A^{\prime}+B B^{\prime}}\) = ∞
⇒ AA’ + BB’ = 0.

Hence, if two lines (equation in the general form) are perpendicular to each other, then the product of the coeffs. of x + product of the coeffs. of y = 0.

Perpendicular Distance Formula : (i) The perpendicular distance of the point (x1, yx) from the line xcos a + ysin a = p is ± (x₁cos α+ y₁sin α – p).
(ii) The perpendicular distance of the point (x₁ y₁) from the line ax + by + c = 0is ± \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)

Distance Between the Parallel Lines: The distance between parallel lines ax + by + c = 0 and ax + by + c is \(\frac{c-c^{\prime}}{\sqrt{a^{2}+b^{2}}}\)

We have given detailed NCERT Solutions for Class 10 Sanskrit आदर्शप्रश्नपत्रम् Solved Questions and Answers come in handy for quickly completing your homework.

CBSE Class 10 Sanskrit आदर्शप्रश्नपत्रम् Solved

समयः होरात्रयम्
सम्पूर्णाङ्काः – 80

सामान्यनिर्देशाः

• अस्मिन् प्रश्नपत्रे चत्वारः खण्डाः सन्ति।
• प्रत्येकं खण्डम् अधिकृत्य एकस्मिन् स्थाने क्रमेण उत्तराणि लेखनीयानि।
• सर्वेषां प्रश्नानाम् उत्तराणि संस्कृतेन लेखनीयानि।
• प्रश्नपत्रानुसारं प्रश्नसंख्या अवश्यमेव लेखनीया।

खण्ड – ‘क’
(अपठितः अनुच्छेदः – 10 अङ्काः)

प्रश्न 1.
अधोलिखितम् अनुच्छेदं पठित्वा यथानिर्देशं प्रश्नान् उत्तरत्-

बाल्यावस्था जीवनस्य महत्वपूर्णः कालः भवति। अस्मिन् समये यदि वयं परिश्रमं कुर्मः तदा जीवन सुखमयं भवति। विद्यार्थी बाल्यकाले विद्याध्ययने प्रवृत्तः भवति चेत् तस्य परीक्षाफलं शोभनं भवति, विषयस्यापि ज्ञानं सरलतया भवति। सः स्वस्वप्नं पूरयितुं समर्थो भवति। एवमेव चरित्रनिर्माणे चापि बाल्यकालः महत्वपूर्ण स्थानम् आदधाति। बाल्यकाले यादृशाः संस्काराः लभ्यन्ते तादृशः एव आचारः व्यवहारः च आजीवनम् अस्माभिः सह तिष्ठतः। अत एव अस्माभिः बाल्यकाले अवधानपूर्वकं गुणाधानस्य प्रयास: कर्त्तव्यः। अस्मिन् समये अध्ययनप्राप्तये अपि सावधानमनसा प्रयत्नः करणीयः। शरीरस्वास्थ्यरक्षायै चापि बाल्यकालादेव पौष्टिकाहारः ग्रहीतव्यः व्यायामः चापि करणीयः।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव) (1 × 2 = 2)
(क) बाल्यकाले अवधानपूर्वकं कस्य प्रयासः कर्त्तव्यः?
(ख) किमर्थं पौष्टिकाहारः ग्रहीतव्यः?
(ग) बाल्यकालः कस्मिन् महत्त्वपूर्ण स्थानम् आदधाति?
उत्तरम्:
(क) गुणाधानस्य
(ख) शरीरस्वास्थ्यरक्षायै
(ग) चरित्रनिर्माणे

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नद्वयमेव) (2 × 2 = 4)
(क) जीवनं कदा सुखमयं भवति?
(ख) कौ अस्माभिः सह सदैव तिष्ठतः?
(ग) शरीरस्वास्थ्यरक्षायै किं कर्तव्यम्?
उत्तरम्:
(क) यदा बाल्यकाले वयं परिश्रमं कुर्मः तदा जीवनं सुखमयं भवति।
(ख) आचाराः व्यवहारः च अस्माभिः सह सदैव तिष्ठतः।
(ग) शरीरस्वास्थ्यरक्षायै चापि बाल्यकालादेव पौष्टिकाहारः ग्रहीतव्यः व्यायामः चापि करणीयः।

(iii) निर्देशानुसारम् उत्तरत (केवलं प्रश्नत्रयमेव) (1 × 3 = 3)
(क) ‘अनवधानमनसा’ इत्यस्य विलोमपदं गद्यांशात् चित्वा लिखत।
(i) ध्यानमनसा
(ii) सावधानमनसा
(ii) मनसा
(iv) सावधानेन
उत्तरम्:
(ii) सावधानमनसा

(ख) ‘महत्त्वपूर्णः कालः’ इत्यनयोः किं विशेषणपदम्?
(i) पूर्णः
(ii) महत्त्वम्
(iii) काल:
(iv) महत्त्वपूर्णः
उत्तरम्:
(iv) महत्त्वपूर्णः

(ग) ‘बाल्यावस्था जीवनस्य महत्त्वपूर्णः कालः भवति’ इति वाक्ये किं क्रियापदम्?
(i) कालः
(ii) भवति
(iii) जीवनस्य
(iv) बाल्यावस्था
उत्तरम्:
(ii) भवति

(घ) अनुच्छेदात् ‘समयः’ इति पदस्य पर्यायं चित्वा लिखत।
(i) आचारः
(ii) संस्काराः
(iii) प्रयत्नः
(iv) काल:
उत्तरम्:
(iv) कालः

(iv) प्रदत्तगद्यांशस्य कृते समुचितं शीर्षकं लिखत। (1 × 1 = 1)
उत्तरम्:
बाल्यावस्थायाः महत्त्वम्

खण्ड – ‘ख’
(रचनात्मक कार्यम् – 15 अङ्काः)

प्रश्न 2.
भवान् (अनिमेषः) शैक्षिकयात्रार्थं शिक्षकैः सह अरुणाचलप्रदेशं गन्तुमिच्छति। तदर्थम् अनुमति प्राप्तये यात्राव्ययार्थं च रूप्यकाणि लब्धं पितरं प्रति लिखितं पत्रमिदं मञ्जूषाप्रदत्तपदानां सहायतया पूरयत- (½ × 10 = 5)

मञ्जूषा – अनुमतिः, मातृचरणयोः, रूप्यकाणि, यात्राव्ययार्थम्, व्यस्तः, शैक्षिकयात्रार्थम्, छात्राः, कुशली, यथाशीघ्रम्, भवतः

छात्रावासतः
दिनाङ्कः _______

पूज्य पितृचरणाः
प्रणतयः।
अत्र अहं सकुशलः स्वाध्ययने (i) _______ अस्मि। आशासे भवान् अपि (ii) _______ अस्ति। मम विद्यालयात् दशमकक्षायाः अनेके (iii) _______ शिक्षकै सह (iv) _______ अरुणाचल-प्रदेश गन्तुम् इच्छन्ति। यदि भवतः (v) _______ यात्, अहमपि तैः सह गन्तुकामोऽस्मि। (vi) _______ सहस्ररूप्यकाणि अपि आवश्यकानि। (vii) _______ स्वमन्तव्यं सूचयतु। यदि (viii) _______ सम्मतिः अस्ति (ix) _______ अपि प्रेषयतु। (x) _______ मम प्रणतिः निवेदनीया।

भवतः पुत्रः
अनिमेषः

उत्तरम्:
(i) व्यस्तः
(ii) कुशली
(iii) छात्राः
(iv) शैक्षिकयात्रार्थम्
(v) अनुमतिः
(vi) यात्राव्ययार्थम्
(vii) यथाशीघ्रम्
(viii) भवतः
(ix) रूप्यकाणि
(x) मातृचरणयोः।

प्रश्न 3.
अधः प्रदत्तं चित्रं दृष्ट्वा तदाधारितानि पञ्चवाक्यानि लिखत- (5)

अथवा
‘वर्धमानं प्रदूषणम्’ इति विषयमधिकृत्य मञ्जूषातः पदानि चित्वा पञ्चवाक्यात्मकम् अनुच्छेदं लिखत।

मञ्जूषा – वैश्विक-उष्णता, यानानां व्यवहारः, विषाक्तवायुः, अवकरक्षेपणम्, जलप्रदूषणम्, वायुप्रदूषणम्, ध्वनिप्रदूषणम्, कोलाहलः, शुचिपर्यावरणम्, मलिनम्, यन्त्रागारम्

उत्तरम्:
1. इदं चित्रं विजयदशमी पर्वणः अस्ति।
2. चित्रे श्रीरामः रावणं बाणेन हन्ति)
3. रावणः पापस्य प्रतीकः अस्ति।
4. सः रामस्य भार्या सीताम् अहरत्।
5. श्रीरामः तं हत्वा सीताम् अमोचयत्।
अथवा
1. वर्धमानं प्रदूषणं मानवजीवनस्य नाशकः अस्ति।
2. अनेन वैश्विक उष्णता विषाक्त वायुः प्रदूषणं च वर्धन्ते।
3. प्रदूषणं तु जलप्रदूषणं, वायु प्रदूषणं ध्वनिप्रदूषणं इति त्रिधा भवति।
4. नगराणां कोलाहलः विषाक्त वायुः चापि प्रदूषणं वर्धयतः।
5. अतः प्रदूषणस्य निवारणाय यन्त्रागारं दूरं निर्माय शुचिपर्यावरणं कर्तव्यम्।

प्रश्न 4.
अधोलिखित-वाक्यानां संस्कृतेन अनुवादं कुरुत (केवलं पञ्च एव)
(i) छात्र विद्यालय जाता है। A student goes to school.
(ii) मैं परिश्रम करता हूँ। I work hard.
(iii) तुम क्या करते हो? What do you do?
(iv) पिता के साथ पुत्र घूमता है। Son walk with his father.
(v) शिक्षक पाठ पढ़ाते हैं। Teacher teaches the lesson.
(vi) वे सब पढ़ने में कुशल बनें। They all are become pefect in reading.
उत्तरम्:
(i) छात्रः विद्यालयं गच्छति।
(ii) अहम् परिश्रमं करोमि।
(iii) त्वम् किं करोषि।
(iv) पित्रा सह पुत्रः भ्रमति।
(v) शिक्षकाः पाठं पाठयति।
(vi) ते पठने कुशलाः भवेयुः/भवन्तु।

खण्ड – ‘ग’
(अनुप्रयुक्त-व्याकरणम् – 25 अङ्काः)

प्रश्न 5.
अधोलिखित-वाक्येषु रोखाङ्कितपदानां सन्धि विच्छेदं वा कुरुत (केवलं चत्वारि एव) (1 × 4 = 4)

(i) पिताऽस्य किं तपस्तेपे इत्युक्तिः तत् कृतज्ञता।
(ii) सेवितव्यो महावृक्षः फलच्छायासमन्वितः।
(iii) विचित्रे खलु संसारे नास्ति किञ्चिन्निरर्थकम्।
(iv) स एवं वह्निः + दहते शरीरम्।
(v) ओम् जय जगत् + ईश हरे।
उत्तरम्:
(i) तपः + तेपे
(ii) फल + छायासमन्वितः
(iii) किञ्चित् + निरर्थकम्
(iv) वह्निर्दहते
(v) जगदीश

प्रश्न 6.
अधोलिखित-वाक्येषु रेखाङ्कितपदानां समस्तपदं लिखत (केवलं चत्वारि एव) (1 × 4 = 4)

(i) यदि अहं कृष्णः वर्णः यस्य सः तर्हि श्रीरामस्य वर्णः कीदृशः?
(क) कृष्णवर्णा
(ख) कृष्णवर्णम्
(ग) कृष्णवर्णः
(घ) कृष्णवर्णाः
उत्तरम्:
(ग) कृष्णवर्णः

(ii) सर्वेषां महत्त्वं विद्यते समयम् अनतिक्रम्य।
(क) यथासमयम्
(ख) यथासमयः
(ग) यथासमय
(घ) यथासमयाय
उत्तरम्:
(क) यथासमयम्

(iii) मिलित्वा प्रकृतेः सौन्दर्यम् रक्षणीयम्।
(क) प्रकृति सौन्दर्य
(ख) प्रकृतिसौन्दर्यः
(ग) प्राकृति सौन्दर्यः
(घ) प्रकृतिसौन्दर्यम्
उत्तरम्:
(घ) प्रकृतिसौन्दर्यम्

(iv) माता च पिता च वन्दनीयौ।
(क) मातापितौ
(ख) पितरौ / मातापितरौ
(ग) पितौ
(घ) मातरौ
उत्तरम्:
(ख) पितरौ / मातापितरौ

(v) महात्मा सर्वत्र पूज्यते।
(क) महान् आत्मा यस्य सः
(ख) महान् आत्मा (ग) महान् आत्मा यस्यः सा
(घ) महान्तः आत्मायस्य सः
उत्तरम्:
(क) महान् आत्मा यस्य सः

प्रश्न 7.
अधोलिखित-वाक्येषु रेखाङ्कितपदेषु प्रकृतिप्रत्ययौ संयोज्य विभज्य वा रिक्तस्थानानि पूरयत (केवलं चत्वारि मव) (1 × 4 = 4)

(i) श्रमक्लमपिपासोष्णशीतादीनां सहिष्णु + तल् _______।
(क) सहिष्णुतल्
(ख) सहिष्णुता
(ग) सहिष्णुताम्
(घ) सहिष्णुतया
उत्तरम्:
(ख) सहिष्णुता

(ii) वैज्ञानिकाः ___________ कथयन्ति यत् पाषाणशिलानां संघर्षणेन कम्पनं जायते।
(क) वैज्ञान + ठक्
(ख) विज्ञान + ठक्
(ग) विज्ञान + इक
(घ) विज्ञानी + अक्
उत्तरम्:
(ख) विज्ञान + ठक्

(iii) बुद्धि + मतुप ___________ सर्वत्र पूज्यते।
(क) बुद्धिमत्
(ख) बुद्धिमन्
(ग) बुद्धिवान्
(घ) बुद्धिमान्
उत्तरम्:
(घ) बुद्धिमान्

(iv) राजसिहंस्य पत्नी बुद्धिमती ___________ आसीत्।
(क) बुद्धिमत् + डीप्
(ख) बुद्धि + मती
(ग) बुद्धि + मतुप्
(घ) बुद्धि + मतु
उत्तरम्:
(क) बुद्धिमत् + ङीप्

(v) मानवत्वम् _____________ एव सर्वत्र पूज्यते।
(क) मानव + तव
(ख) मानव + त्वम्
(ग) मानव + त्व
(घ) मानव + त्वाम्
उत्तरम्:
(ग) मानव + त्व

प्रश्न 8.
प्रदत्तेभ्यः विकल्पेभ्यः समुचितं। विकल्पं चित्वा वाच्यानुसारं रिक्तस्थानानि पूरयत (केवलं त्रीणि एव) (1 × 3 = 3)

(i) _______ सत्यं कथ्यते। (त्वम् / त्वाम् / त्वया)
(ii) काकः पिकस्य _______ पालयति। (सन्तति: / सन्ततिम् / सन्तत्या)
(iii) पित्रा पुत्राय विद्याधनं _______। (ददाति / दीयते / यच्छति)
(iv) पुत्रः अपि तस्मै धन्यवादं _______। (कुर्वन्ति / करोति / करोषि)
उत्तरम्:
(i) त्वया
(ii) सन्ततिम्
(iii) दीयते
(iv) करोति

प्रश्न 9.
कोष्ठके प्रदत्त-समयवाचकान् अङ्कान् संस्कृतेन लिखत (केवलं चत्वारि एव) (1 × 4 = 4)

(i) बालकः _______ वादने उत्तिष्ठति। (6:00)
(ii) सः _______ वादने विद्यालयं गच्छति। (7:15)
(iii) _______ वादने विद्यालये अर्धावकाशः भवति। (11:30)
(iv) सः _______ वादने विद्यालयात् गृहम् गच्छति। (1:45)
(v) पुनः बालक: सायं _______ वादने पठनाय उपविशति (5:15)
उत्तरम्:
(i) षड्
(ii) सपादसप्त
(iii) सार्ध-एकादश
(iv) पादोन द्वि
(v) सपादपञ्च

प्रश्न 10.
मञ्जूषाप्रदत्त-अव्ययपदैः रिक्तस्थानानि पूरयत- (½ × 6 = 3)

मञ्जूषा – यत्र, अद्य, सदा, यदि, तत्र, तर्हि

(i) ___________ परिश्रमं कुर्मः ___________ सफलाः भवामः।
(ii) ___________ समयस्य सदुपयोगः करणीयः।
(iii) जीवनम् ___________ दुर्वहं जातमस्ति ।
(iv) ___________ हरीतिमा ___________ पर्यावरणं शुद्धम्।
उत्तरम्:
(i) यदि, तर्हि
(ii) सदा
(iii) अद्य
(iv) यत्र, तत्र

प्रश्न 11.
रेखाङ्कितपदं संशोध्य वाक्यानि लिखत (केवलं त्रीणि एव) (1 × 3 = 3)

(i) ते नार्यः गल्पं कुर्वन्ति।
(क) सा
(ख) तौ
(ग) ताः
(घ) सः
उत्तरम्:
ताः नार्यः गल्पं कुर्वन्ति।

(ii) अहं परिश्रमं करोति।
(क) सः
(ख) तौ
(ग) ते
(घ) त्वम्
उत्तरम्:
सः परिश्रमं करोति।

(iii) बालकः कार्यं कुर्वन्ति।
(क) बालकौ
(ख) बालकाः
(ग) बालिके
(घ) बालिका
उत्तरम्:
बालकाः कार्यं कुर्वन्ति।

(iv) त्रयः बालिकाः गीतं गायन्ति।
(क) त्रीणि
(ख) त्रि
(ग) तिस्रः
(घ) चतुरः
उत्तरम्:
तिस्रः बालिकाः गीतं गायन्ति।

खण्ड – ‘घ’
(पठित-अवबोधनम् – 30 अङ्काः)

प्रश्न 12.
अधोलिखितं गद्यांशं पठित्वा यथानिर्देशं प्रश्नान् उत्तरत-

कश्चन निर्धनो जनः भूरि परिश्रम्य किञ्चिद् वित्तमुपार्जितवान्। तेन वित्तेन स्वपुत्रम् एकस्मिन् महाविद्यालये प्रवेशं दापयितुं सफलो जातः। तत्तनयः तत्रैव छात्रावासे निवसन् अध्ययने संलग्नः समभूत्। एकदा पिता तनूजस्य रुग्णतामाकर्ण्य व्याकुलो जातः, पुत्रं च द्रष्टुं प्रस्थितः। परमर्थकार्येन पीडितः सः बसयानं विहाय पदातिरेव प्राचलत्।
(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव) (½ × 2 = 2)
(क) निर्धनो जनः भूरि परिश्रम्य किम् उपार्जितवान्?
(ख) अर्थकार्येन पीडितः सः केन प्राचलत्?
(ग) जनः कीदृशो आसीत्?
उत्तरम्:
(क) वित्तम्
(ख) पदातिः
(ग) निर्धनः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नएकमेव) (1 × 1 = 1)
(क) निर्धनः जनः वित्तेन किं कृतवान्?
(ख) तस्य तनयः छात्रावासे कस्मिन् संलग्नः समभूत्?
उत्तरम्:
(क) निर्धनः जनः वित्तेन स्वपुत्रम् एकस्मिन् महाविद्यालये प्रवेशं दापयितुं सफलः भूतवान्।
(ख) तस्य तनयः छात्रावासे अध्ययने संलग्नः समभूत्।

(iii) निर्देशानुसारम् उत्तरत (केवलं प्रश्नत्रयमेव) (1 × 3 = 3)
(क) ‘धनम्’ इत्यस्य समानार्थकपदं गद्यांशात् चित्वा लिखत।
(ख) ‘निर्धनः जनः’ इत्यनयोः किं विशेषणपदम्?
(ग) ‘परमर्थकार्येन पीडितः सः बसयानं विहाय पदातिरेव प्राचलत्’ इति वाक्ये ‘सः’ इति सर्वनामपदं कस्मै प्रयुक्तम्?
(घ) अनुच्छेदे ‘गृहीत्वा’ पदस्य कः विपर्ययः आगतः?
उत्तरम्:
(क) वित्तम्
(ख) निर्धनः
(ग) निर्धनजनाय / निर्धनपित्रे
(घ) विहाय

प्रश्न 13.
अधोलिखितं पद्याशं पठित्वा यथानिर्देशं प्रश्नान् उत्तरत-

उदीरितोऽर्थः पशुनापि गृह्यते हयाश्च नागाश्च वहन्ति बोधिताः।
अनुक्तमप्यूहति पण्डितो जनः परेङ्गितज्ञानफला हि बुद्धयः।।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव) (½ × 2 = 1)
(क) केन उदीरितः अर्थः गृह्यते?
(ख) का: परेङ्गितज्ञानफलाः भवन्ति?
(ग) अनुक्तम् अपि कीदृशः जनः ऊहति?
उत्तरम्:
(क) पशुना
(ख) बुद्धयः
(ग) पण्डितः

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नएकमेव) (1 × 1 = 1)
(क) पण्डितः जनः किम् ऊहति?
(ख) के बोधिताः बहन्ति?
उत्तरम्:
(क) पडितः जनः अनुक्तम् अपि ऊहति।
(ख) हयाः च नागाः च बोधिता: वहन्ति।

(iii) निर्देशानुसारम् उत्तरत (केवलं प्रश्नत्रयमेव) (1 × 3 = 3)
(क) ‘अश्वाः ‘ इति अर्थे पद्ये किं पदं प्रयुक्तम्?
(ख) ‘मूर्खः’ इत्यस्य विलोमपदं श्लोकात् चित्वा लिखत।
(ग) ‘हयाश्च नागाश्च वहन्ति बोधिताः’ इति वाक्ये किं क्रियापदम्?
(घ) श्लोके ‘गृह्यते’ इति क्रियायाः कर्तृपदं किम्?
उत्तरम्:
(क) हयाः
(ख) पण्डितः
(ग) वहन्ति
(घ) पशुना

प्रश्न 14.
अधोलिखितं नाट्यांशं पठित्वा यथानिर्देशं प्रश्नान् उत्तरत-

चाणक्यः – भो श्रेष्ठिन्! प्रीताभ्यः प्रकृतिभ्यः प्रतिप्रियमिच्छन्ति राजानः।
चन्दनदासः – आज्ञापयतु आर्यः, किं कियत् च अस्मज्जनादिष्यते इति।
चाणक्यः – भो श्रेष्ठिन्! चन्द्रगुप्तराज्यमिदं न नन्दराज्यम्। नन्दस्यैव अर्थसम्बन्धः प्रीतिमुत्पादयति। चन्द्रगुप्तस्य तु भवतामपरिक्लेश एव।
चन्दनदासः – (सहर्षम्) आर्य! अनुगृहीतोऽस्मि।
चाणक्यः – भो श्रेष्ठिन्! स चापरिक्लेशः कथमाविर्भवति इति ननु भवता प्रष्टव्याः स्मः।

(i) एकपदेन उत्तरत (केवलं प्रश्नद्वयमेव) (½ × 2 = 1)
(क) प्रीताभ्यः प्रकृतिभ्यः राजानः किम् इच्छन्ति?
(ख) कस्य अर्थसम्बन्धः प्रीतिमुत्पादयति?
(ग) कस्य इदं राज्यम् अस्ति?
उत्तरम्:
(क) प्रतिप्रियम्
(ख) नन्दस्य
(ग) चन्द्रगुप्तस्य

(ii) पूर्णवाक्येन उत्तरत (केवलं प्रश्नएकमेव) (1 × 1 = 1)
(क) चन्दनदासेन (भवता) किं प्रष्टव्यम्?
(ख) राजानः किम् इच्छन्ति?
उत्तरम्:
(क) चन्दनदासेन चाणक्यस्य आदेशः प्रष्टव्यः।
(ख) प्रीताभ्यः प्रकृतिभ्यः प्रतिप्रियमिच्छन्ति राजानः।

(iii) निर्देशानुसारम् उत्तरत (केवलं प्रश्नत्रयमेव) (1 × 3 = 3)
(क) ‘दुःखाभावः’ इति अर्थे नाट्यांशे किं पदं प्रयुक्तम्?
(ख) ‘आर्य अनुगृहीतोऽस्मि’ अत्र ‘अस्मि’ इति क्रियापदस्य कर्तृपदम् किम्?
(ग) ‘किं कियत् च अस्मज्जनादिष्यते’ इति वाक्ये किं क्रियापदम्?
(घ) नाट्यांशे ‘प्रेम’ इति पदस्य कः पर्यायः अस्ति?
उत्तरम्:
(क) अपरिक्लेशः
(ख) अहम्
(ग) आदिष्यते
(घ) प्रीतिम्

प्रश्न 15.
अधोलिखितवाक्येषु रेखाङ्कितपदमाधृत्य प्रश्ननिर्माणं कुरुत (केवलं प्रश्नचतुर्णामेव) (1 × 4 = 4)

(i) विद्वांस एव लोकेऽस्मिन् चक्षुष्मन्तः प्रकीर्तिताः।
(क) के
(ख) कः
(ग) कथम्
(घ) काः
उत्तरम्:
(क) के

(ii) समीपे एका नदी प्रवहति।
(क) कीदृशी
(ख) काः
(ग) का
(घ) कुत्र
उत्तरम्:
(ग) का

(iii) अतिदाक्षिण्येन अलम्।
(क) कथम्
(ख) केन
(ग) कः
(घ) कया
उत्तरम्:
(ख) केन

(iv) गहनकानने सा व्याघ्रं ददर्श।
(क) के
(ख) कस्मिन्
(ग) कदा
(घ) कुत्र
उत्तरम्:
(घ) कुत्र

(v) बुद्धिमती व्याघ्रभयात् मुक्ता जाता।
(क) का
(ख) काः
(ग) कीदृशी
(घ) के
उत्तरम्:
(ग) कीदृशी

प्रश्न 16.
अधोलिखितयोः श्लोकयोः अन्वये रिक्तस्थानानि पूरयत- (½ × 8 = 4)

(i) त्यक्त्वा धर्मप्रदां वाचं परुषां योऽभ्युदीरयेत्।
परित्यज्य फलं पक्वं भुङ्क्तेऽपक्वं विमूढधीः।।

अन्वयः – यः ___________ वाचं त्यक्त्वा ___________ अभ्युदीरयेत् (स:) ___________ पक्वं फलं परित्यज्य अपक्वं ___________।

(ii) अपत्येषु च सर्वेषु जननी तुल्यवत्सला।
पुत्रे दीने तु सा माता कृपार्द्रहृदया भवेत्।।

अन्वयः – सर्वेषु च ___________ जननी ___________ भवति। परं माता दीने ___________ तु ___________ भवेत्।

अथवा

अधोलिखितस्य श्लोकस्य भावार्थे रिक्तस्थानपूर्ति मञ्जूषाप्रदत्तपदानां सहायतया कुरुत- (1 × 4 = 4)

यः इच्छत्यात्मनः श्रेयः प्रभूतानि सुखानि च।
न कुर्यादहितं कर्म सः परेभ्यः कदापि च।।

भावार्थ: – य: नरः आत्मनः ___________ इच्छति, तस्य ___________ अपि इच्छा अस्ति, सः ___________ कृते कदापि अकल्याणकर ___________ न कुर्यात् इति अवधातव्यम्।

मञ्जूषा – कार्यम् सुखप्राप्तेः कल्याणम्, अन्येषाम्

उत्तरम्:
(i) धर्मप्रदां, परुषाम्, विमूढधीः, भुङ्क्ते।
(ii) अपत्येषु, तुल्यवत्सला, पुत्रे, कृपार्द्रहृदया
अथवा
कल्याणम्, सुखप्राप्तेः, अन्येषाम्, कार्यम्

प्रश्न 17.
अधोलिखितवाक्यानि घटनाक्रमानुसारं लिखत- (½ × 8 = 4)

(i) बुद्धिमती व्याघ्रभयात् मुक्ता जाता।
(ii) ग्रामे राजसिंहः नाम राजपुत्रः वसति स्म।
(iii) गलबद्धशृगालक: व्याघ्रः सहसा पलायित:।
(iv) व्याघ्रः शृगालं निजगले बद्ध्वा काननं गतवान्।
(v) तस्य भार्या पुत्रद्वयोपेता पितुर्गृहं प्रति चलिता।
(vi) जम्बुक: व्याघ्र पुनः तत्र गन्तुं प्रेरितवान्।
(vii) इयं व्याघ्रमारी इति विचिन्त्य भयेन व्याघ्रः पलायितः।
(viii) सा पुत्रौ उक्तवती-एकं व्याघ्रं विभज्य खादतम्।
उत्तरम्:
(i) ग्रामे राजसिंहः नाम राजपुत्रः वसति स्म।
(ii) तस्य भार्या पुत्रद्वयोपेता पितुर्गृहं प्रति चलिता।
(iii) सा पुत्रौ उक्तवती-एकं व्याघ्रं विभज्य खादतम्।
(iv) इयं व्याघ्रमारी इति विचिन्त्य भयेन व्याघ्रः पलायितः।
(v) जम्बुक: व्याघ्रं पुनः तत्र गन्तुं प्रेरितवान्।
(vi) व्याघ्रः शृगालं निजगले बद्ध्वा काननं गतवान्।
(vii) गलबद्धशृगालकः व्याघ्रः सहसा पलायितः।
(viii) बुद्धिमती व्याघ्रभयात् मुक्ता जाता।

प्रश्न 18.
अधोलिखितवाक्येषु रेखाङ्कितपदानां प्रसङ्गानुसारं शुद्धम् उत्तरं विकल्पेभ्यः चित्वा लिखत (केवलं प्रश्नत्रयमेव) (1 × 3 = 3)

(i) अत्र जीवनं दुर्वहं जातम्।
सरलम् / कठिनम् / जटिलम्

(ii) संव्यवहाराणां वृद्धिलाभाः प्रचीयन्ते।
संस्काराणाम् / संलापानाम् / व्यापाराणाम्

(iii) तौयैः अल्पैः अपि तरोः पुष्टिः भवति।
जलै: / त्वग्भि / दुग्धैः।

(iv) अवक्रता यथा चित्रे भवेत्।
सरलता / कठोरता / मृदुता
उत्तरम्:
(i) कठिनम्
(ii) संस्काराणाम्
(iii) जलै:
(iv) सरलता