Class 10

Students must go through these JAC Class 10 Science Notes Chapter 1 Chemical Reactions and Equations to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 1 Chemical Reactions and Equations

→ Chemical reactions are observed in daily life such as spoiling of milk in summer, rusting of iron, formation of glucose by photosynthesis, digestion of food, etc.

→ In balanced chemical equation, the number of atoms of each element taking part in the reaction remains the same before and after the reaction.

→ In combination reaction, two or more substances combine to form a single new substance.

→ In decomposition reaction, a single substance breaks down to form two or more substances.

→ A reaction in which heat is absorbed during the formation of product is called an endothermic reaction.

→ A reaction in which heat is released during the formation of product is called an exothermic reaction.

→ In displacement reaction, more reactive element or a substance displaces the less reactive element or a substance from its solution.

→ In double displacement reaction, two different atoms or ions get exchanged.

→ Salt obtained in precipitation reaction is insoluble in water.

→ In an oxidation reaction, a substance gains oxygen or loses hydrogen.

→ In a reduction reaction, a substance gains hydrogen or loses oxygen.

→ Oxidising agent gives oxygen or receives hydrogen.

→ Reducing agent gives hydrogen or receives oxygen.

→ Oxidising agent undergoes reduction and reducing agent undergoes oxidation.

→ Metal corrosion: A reaction in which a surface of metal is affected by oxygen, water, chemical and atmospheric gases is called corrosion of metal or corrosion.

→ Rancidity: The food materials containing oils and fats when exposed to air for a long time, undergo oxidation changing their smell and taste. This is called rancidity.

Students should go through these JAC Class 10 Maths Notes Chapter 9 त्रिकोणमिति का अनुप्रयोग will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 9 त्रिकोणमिति का अनुप्रयोग

भूमिका :
इस अध्याय में हम उन विधियों के विषय में पढ़ेंगे जिनमें त्रिकोणमिति का प्रयोग हमारे आसपास के जीवन से जुड़ा होता है। त्रिकोणमिति की आवश्यकता रवगोलकी में पृथ्वी से ग्रहों और तारों की दूरियाँ परिकलित करने में होती थी। त्रिकोणमिती का प्रयोग भूगोल और नौचालन, मानचित्र बनाने और देशांतर और अक्षांश के सापेक्ष एक द्वीप की स्थिति ज्ञात करने में की जाती है।

इस अध्याय में हम अध्ययन करेंगे कि वास्तविक माप के बिना, त्रिकोणमिति का प्रयोग विभिन्न वस्तुओं की ऊँचाइयाँ और दूरियाँ ज्ञात करने में किया जाता है।

ऊंचाइयाँ और दूरियाँ :
दृष्टि रेखा (Line of sight) : प्रेक्षक की आँख से प्रेक्षक द्वारा देखी गई वस्तु के बिन्दु को मिलाने वाली रेखा को दृष्टि रेखा कहते हैं, अथवा जब हम किसी वस्तु (object) को देखते हैं, तो हमारी आँख और वस्तु को जोड़ने वाली रेखा को दृष्टि रेखा कहते हैं।

चित्र में आँख बिन्दु पर है और वस्तु की स्थिति P है। अत: OP दृष्टि रेखा होगी।

पूरक कोण (Complimentary angles) : यदि दो कोणों का योग 90° हो, तो ये कोण पूरक कोण कहलाते हैं।

आन्तरिक एकान्तर कोण (Alternate interior angles) : यदि दो रेखाओं को एक तिर्यक रेखा काटती है तो दोनों रेखाओं के अन्दर तिर्यक रेखा के विपरीत दिशा में बने कोण एकान्तर कोण कहलाते हैं, यदि दोनों रेखाएँ परस्पर समान्तर हैं, तो बने आन्तरिक एकान्तर कोण बराबर होते हैं।

उन्नयन कोण (Angle of elevation): जब कोई वस्तु, आँख से ऊपर हो, तो दृष्टि रेखा, क्षैतिज के साथ जो कोण बनता है उसे उन्नयन या उन्नतांश या उन्नति कोण कहते हैं।

चित्र में आँख बिन्दु पर है और वस्तु (object) की स्थिति P है। अतः OP दृष्टि रेखा है जो क्षैतिज रेखा OX से कोण ∠XOP बनाती है। अतः उन्नयन कोण = ∠XOP

अवनमन कोण (Angle of depression) : जब कोई वस्तु, आँख से नीचे हो, तो दृष्टि रेखा, क्षैतिज के साथ जो कोण बनता है उसे अवनमन या अवनति कोण कहते हैं।

चित्र में आँख बिन्दु पर और वस्तु (object) की स्थिति P है अतः OP दृष्टि रेखा है जो क्षैतिज रेखा OX’ से कोण X’OP बनाती है। अतः अवनमन कोण = X’OP
ऊँचाई एवं दूरी की समस्याओं को हल करते समय निम्नलिखित बिन्दुओं को ध्यान में रखना चाहिए :
(i) सर्वप्रथम प्रश्न को ध्यानपूर्वक पढ़ने के उपरान्त चित्र बनाकर समकोण त्रिभुज का निर्माण करते हैं।
(ii) समकोण त्रिभुज में ज्ञात कोण के त्रिकोणमितीय अनुपातों (sine, cosine, tangent) आदि को ज्ञात भुजा के पदों में व्यक्त करते हैं।
(iii) चित्र में स्पष्ट है कि O का P के सापेक्ष उन्नयन कोण = P का O के सापेक्ष अवनमन कोण।

चित्र में वस्तुओं द्वारा प्रेक्षक की आंख पर अन्तरित अवनमन कोणों के उदाहरण :

Students must go through these JAC Class 10 Science Notes Chapter 15 Our Environment to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 15 Our Environment

→ Environment: It is the sum total of all external conditions and influences that affect the life and development of an organism, i.e., the environment includes all physical or abiotic and biological factors. All the factors of environment are affecting the organisms.

→ Ecosystem: A system formed by an interaction between biotic environment is components and the physical called an ecosystem.

→ Food chain and food web: Living organisms of an ecosystem depend on each other for their food requirement and form a chain. This is termed as food chain.

• A food chain describes unidirectional flow of energy and cyclic flow of materials in an ecosystem.
• Each ecosystem has its own specific food chain. The individuals involved in one food chain are linked with food chains of other ecosystem and forms a complex net which is termed as a food web.

→ Trophic levels: Each step in the food chain constitutes a trophic level.

• There are generally 3-4 trophic levels in food chain.
• Sun is the energy source for any food chain.
• Producers constitute 1st trophic level in a food chain and they are the entrance for solar energy in an ecosystem.

→ Biological magnification : A progressive increase in the concentration of non- biodegrable substances in organisms at each successive trophic level in known as biological magnification.

→ Ozone layer: Ozone at the higher levels of the atmosphere (stratosphere) is a product of UV radiation acting on oxygen (Oa) molecule.

• Ozone forms shield surrounding the earth that protect against harmful UV radiation from the sun.

→ Ozone depletion : An event of dropping sharply in the amount of ozone in the atmosphere is known as ozone depletion.

• CFCs – Chlorofluorocarbons used in refrige-rators and air-conditioners is responsible for ozone depletion.

→ Waste: Unwanted, unusable items, remains and other products of household garbage are called waste.

The disposal of the waste generated by us is causing serious environmental problems.

→ Household waste management: Each person in urban areas produces 500grams of garbage everyday.

• Such wastes can be a source of extra income when properly managed. Use of reusable and rechargeable items, proper disposal of wastes and prevents the use of plastic, polythene.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 वृतों से संबंधित क्षेत्रफल Ex 12.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 वृतों से संबंधित क्षेत्रफल Exercise 12.1

(जब तक अन्यथा न कहा जाए, का प्रयोग कीजिए ।)

प्रश्न 1.
दो वृत्तों की त्रिज्याएँ क्रमशः 19 सेमी और 9 सेमी हैं। उस वृत्त की त्रिज्या ज्ञात कीजिए जिसकी परिधि इन दोनों वृत्तों की परिधियों के योग के बराबर है।
हल :
दिया है पहले वृत्त की त्रिज्या (r₁) = 19 सेमी
∴ पहले वृत्त की परिधि = 2πr₁
= 2πr₁
= 2π × 19 = 38π सेमी
और दूसरे वृत्त की त्रिज्या (r₂) = 9 सेमी
∴ दूसरे वृत्त की परिधि = 2πr₂
= 2π × 9 = 18π सेमी
∴ दोनों वृत्तों की परिधियों का योग
= (38π + 18π)
= 56π सेमी
माना कि तीसरे वाँछित वृत्त की त्रिज्या r सेमी है।
प्रश्नानुसार,
वांछित वृत्त की परिधि = दोनों वृत्तों की परिधियों का योग
2πr = 56π
∴ r = \(\frac {56π}{2π}\) = 28 सेमी
अतः अभीष्ट वृत्त की त्रिज्या = 28 सेमी

प्रश्न 2.
दो वृत्तों की त्रिज्याएँ क्रमशः 8 सेमी और 6 सेमी हैं। उस वृत्त की त्रिज्या ज्ञात कीजिए जिसका क्षेत्रफल इन दोनों वृत्तों के क्षेत्रफलों के योग के बराबर है।
हल :
दिया है
पहले वृत्त की त्रिज्या (r₁) = 8 सेमी
∴ पहले वृत्त का क्षेत्रफल = πr₁²
= π × 8 × 8
= 64π वर्ग सेमी
और दूसरे वृत्त की त्रिज्या (r₂) = 6 सेमी
दूसरे वृत्त का क्षेत्रफल = πr₁²
∴ दोनों वृत्तों के क्षेत्रफलों का योग
= (64π + 36π)
= 100π वर्ग सेमी
माना कि तीसरे वाँछित वृत्त की त्रिज्या r सेमी है।
प्रश्नानुसार,
वाँछित वृत्त का क्षेत्रफल = दोनों वृत्तों का क्षेत्रफल
πr² = 100π
⇒ r² = 100
∴ r = \(\sqrt{100}\)
⇒ r = 10 सेमी
अतः अभीष्ट वृत्त की त्रिज्या = 10 सेमी

प्रश्न 3.
दी गई आकृति एक तीरंदाजी लक्ष्य को दर्शाती है, जिसमें केन्द्र से बाहर की ओर पाँच क्षेत्र GOLD, RED, BLUE, BLACK और WHITE चिह्नित हैं, जिनसे अंक अर्जित किए जा सकते हैं। GOLD अंक वाले क्षेत्र का व्यास 21 सेमी है तथा प्रत्येक अन्य पट्टी 10.5 सेमी चौड़ी है। अंक प्राप्त कराने वाले इन पाँचों क्षेत्रों में से प्रत्येक का क्षेत्रफल ज्ञात कीजिए।

हल :
सबसे पहले GOLD क्षेत्र का व्यास = 21 सेमी
GOLD क्षेत्र की त्रिज्या (r₁) = \(\frac {21}{2}\) = 10.5 सेमी
∴ GOLD क्षेत्र का क्षेत्रफल = πr₁²
= \(\frac {22}{7}\) × 10.5 × 10.5 = 346.5 सेमी²
और अगली प्रत्येक पट्टी की चौड़ाई = 10.5 सेमी
तीरंदाजी के पाँच क्षेत्रों का क्रम = GOLD, RED, BLUE, BLACK, WHITE
∴ RED क्षेत्र की भीतरी त्रिज्या (r₁) = 10.5 सेमी
तब बाहरी त्रिज्या r₂ = r₁ + 10.5
= 10.5 + 10.5 = 21.0 सेमी
तब BLUE क्षेत्र की भीतरी त्रिज्या (r₂) = 21.0 सेमी
तथा बाहरी त्रिज्या r₃ = r₂ + 10.5 सेमी
= 21.0 + 10.5 = 31.5 सेमी
तब BLACK क्षेत्र की भीतरी त्रिज्या (r₃)
= 31.5 सेमी
तथा बाहरी त्रिज्या = r₄ = r₃ + 10.5 सेमी
= 31.5 + 10.5 = 42.0 सेमी
तब WHITE क्षेत्र की भीतरी त्रिज्या (r₄)
= 42.0 सेमी
तथा बाहरी त्रिज्या r₅ = r₄ + 10.5 सेमी
= 42.0 + 10.5 = 32.5 सेमी
∵ क्षेत्र वलयाकार हैं :
∴ RED क्षेत्र का क्षेत्रफल = πr₂² – πr₁²
= π[r²₂ – r₁²]
= π[(21)² – (10.5)²]
= \(\frac {22}{7}\) [441 – 110.25]
= \(\frac {22}{7}\) × 330.75 = \(\frac {7276.5}{7}\)
= 1039.5 सेमी²
BLUE क्षेत्र का क्षेत्रफल = πr₃² – πr₂²
= π[r₃² – r₂²]
= π[(31.5)² – (21)²]
= π(992.25 – 441)
= \(\frac {22}{7}\) × 551.25
= \(\frac {12127.5}{7}\)
= 1732.5 सेमी²
BLACK क्षेत्र का क्षेत्रफल = πr₄² – πr₃²
= π [(42)² – (31.5)²]
= π [1764 – 992.25]
= \(\frac {22}{7}\) × 771.75
= \(\frac {16978.5}{7}\)
= 2425.5 सेमी²
WHITE क्षेत्र का क्षेत्रफल = πr₅² – πr₄²
= π (r₄² – r₃²)
= π[(r₅)² – (r₄)²)
= π [(52.5)² – (42)²]
= \(\frac {22}{7}\) [2756.25 – 1764]
= \(\frac {22}{7}\) × 992.25
= \(\frac {21829.5}{7}\)
= 3118.5 सेमी²

प्रश्न 4.
किसी कार के प्रत्येक पहिए का व्यास 80 सेमी है। यदि यह कार 66 किमी प्रति घण्टे की चाल से चल रही है, तो 10 मिनट में प्रत्येक पहिया कितने चक्कर लगाता है ?
हल :
दिया है : कार के पहिए का व्यास = 80 सेमी
कार के पहिए की परिधि = π × व्यास
= \(\frac {22}{7}\) × 80
= \(\frac {1760}{7}\) सेमी
कार की चाल = 66 किमी / घण्टा
= 66 × \(\frac {1000}{60}\) मीटर / मिनट
= 66 × \(\frac {1000}{60}\) × 100 सेमी / मिनट
= 110000 सेमी / मिनट
∴ 10 मिनट में तय की गई दूरी = 110000 × 10
= 1100000 सेमी
∴ चक्करों की संख्या

अतः कार के प्रत्येक पहिए द्वारा 10 मिनट में लगाए गए चक्करों की संख्या = 4375

प्रश्न 5.
निम्नलिखित में सही उत्तर चुनिए तथा अपने उत्तर का औचित्य दीजिए:
यदि एक वृत्त का परिमाप और क्षेत्रफल संख्यात्मक रूप से बराबर हैं, तो उस वृत्त की त्रिज्या है :
(A) 2 मात्रक
(B) मात्रक
(C) 4 मात्रक
(D) 7 मात्रक
हल :
वृत्त की परिधि = वृत्त का क्षेत्रफल
2πR = πR²
⇒ 2R = R²
∴ R = 2 मात्रक
अतः सही विकल्प A है

Students should go through these JAC Class 10 Maths Notes Chapter 6 त्रिभुज will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 6 त्रिभुज

भूमिका :
सर्वांगसम आकृति के बारे में हम पिछली कक्षा IX में पढ़ चुके हैं। ऐसी दो ज्यामितीय आकृतियाँ जिनके आकार व रूप बिल्कुल समान हों एवं परस्पर अध्यारोपण पर एक-दूसरे को पूरा-पूरा ढक लेती हैं, सर्वांगसम आकृतियाँ कहलाती हैं।

इस अध्याय में हम ऐसी ही आकृतियों का अध्ययन करेंगे जिनका रूप या आकृतियाँ (Shape) बिल्कुल समान हों किन्तु आकार में भिन्नता हो, समरूप आकृतियाँ कहलाती हैं।
दो सर्वांगसम आकृतियाँ भी समरूप होती हैं। किन्तु इसका विलोम सर्वदा सत्य नहीं होता अर्थात् समरूप आकृतियाँ सदैव सर्वांगसम नहीं होती हैं।
→ बहुभुज (Polygon) : रेखाखण्डों से बनी साधारण वक्र बन्द आकृति को बहुभुज कहते हैं।

→ त्रिभुज (Triangle) : तीन भुजाओं वाले बहुभुज को त्रिभुज कहते हैं।

→ विषमबाहु त्रिभुज (Scalene Triangle) : एक त्रिभुज जिसकी तीनों भुजाएँ असमान हों, विषमबाहु त्रिभुज कहलाता है।

→ समद्विबाहु त्रिभुज (Isosceles Triangle) : एक त्रिभुज जिसकी कोई सी दो भुजाएँ समान हों, समद्विबाहु त्रिभुज कहलाता है।

→ समबाहु त्रिभुज (Equilateral Triangle) : एक त्रिभुज जिसकी तीनों भुजाएँ समान हों, समबाहु त्रिभुज कहलाता है।

→ न्यूनकोण त्रिभुज (Acute angled Triangle) : एक त्रिभुज जिसके तीनों कोण न्यून कोण ( less than 90°) हो, न्यूनकोण त्रिभुज कहलाता है।

→ अधिक कोण त्रिभुज (Obtuse angled Triangle) : एक त्रिभुज जिसका एक कोण अधिक कोण (greater than 90°) हो, अधिक कोण त्रिभुज कहलाता है।

→ समकोण त्रिभुज (Right angled Triangle) : एक त्रिभुज जिसका एक कोण समकोण है, समकोण त्रिभुज कहलाता है।

→ त्रिभुज का परिमाप (Perimeter of a Triangle) : त्रिभुज की तीनों भुजाओं का योग त्रिभुज का परिमाप कहलाता है।

→ त्रिभुज की माध्यिका (Median of a Triangle) : त्रिभुज के शीर्ष से इसके सम्मुख भुजा के मध्य बिन्दु को मिलाने वाली रेखा को त्रिभुज की माध्यिका कहते हैं।

→ त्रिभुज का शीर्षलम्ब (Altitude of a Triangle) : त्रिभुज के एक शीर्ष से सम्मुख भुजा पर खींची गयी लम्ब रेखा को त्रिभुज का शीर्षलम्ब कहते हैं।

→ त्रिभुज का कोण समद्विभाजक (Bisector of angle of a Triangle) : त्रिभुज के एक शीर्ष कोण को समद्विभाजित करने वाली रेखा को त्रिभुज का कोण समद्विभाजक कहते है।

→ समकोणीय त्रिभुज (Equiangular Triangle) : यदि दो त्रिभुजों के संगत कोण बराबर हों, तो वे समकोणीय त्रिभुज कहलाते हैं।

समरूप आकृतियाँ :
ऐसी आकृतियाँ जिनका आकार तो समान है, परन्तु माप भिन्न है, समरूप आकृतियाँ कहलाती हैं।

उदाहरण : चित्र (A) में दो भवन, चित्र (B) में चार त्रिभुज, चित्र (C) में चार वृत्त, चित्र (D) में तीन पंचभुज और चित्र (E) में चार वर्ग को देखने पर इनका आकार समान एवं माप भिन्न-भिन्न है अर्थात् सभी समान संख्या की भुजाओं के समबहुभुज हैं जैसे: समबाहु त्रिभुज, वर्ग, वृत्त, समपंचभुज इत्यादि समरूप होते हैं।

समरूप बहुभुज :
दो बहुभुज समरूप होते हैं, यदि उनके संगत कोण समान हों एवं उनकी संगत भुजाएँ समानुपाती हों।

चित्र में, दो बहुभुज ABCDEF एवं PQRSTU समरूप हों, तो संगत कोण समान होंगे, अर्थात्
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S, ∠E = ∠T एवं ∠F = ∠U
एवं संगत भुजाएँ समानुपाती होंगी, अर्थात्
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C D}{R S}=\frac{D E}{S T}=\frac{E F}{T U}=\frac{F A}{U P}\)
टिप्पणी: यदि एक बहुभुज दूसरे बहुभुज के समरूप हो और दूसरा बहुभुज, तीसरे बहुभुज के समरूप हो, तो पहला बहुभुज, तीसरे बहुभुज के भी समरूप होता है।

दो बहुभुजों के समरूप होने के लिए भुजाओं का समानुपाती होना ही पर्याप्त नहीं है, जैसे कि चित्र में, ABCD एक वर्ग है और PQRS एक समचतुर्भुज है। वर्ग ABCD की भुजाएँ समचतुर्भुज PQRS की भुजाओं की समानुपाती हैं परन्तु वर्ग ABCD के कोण समचतुर्भुज PQRS के कोणों के समान नहीं हैं। अतः वर्ग एवं समचतुर्भुज की भुजाएँ समानुपाती होते हुए भी दोनों समरूप नहीं हैं।
भुजाओं की समान संख्या वाले दो बहुभुज समरूप होते हैं, यदि
(i) सभी संगत कोण बराबर हों।
(ii) सभी संगत भुजाएँ एक ही अनुपात में (या समानुपाती हों।

Jharkhand Board JAC Class 10 Science Important Questions Chapter 3 Metals and Non-metals Important Questions and Answers.

JAC Board Class 10 Science Important Questions Chapter 3 Metals and Non-metals

Additional Questions and Answers

Question 1.
Give scientific reasons for the following statements:
(1) Hydrogen gas is not evolved when Al reacts with nitric acid.
Answer:
When Al reacts with nitric acid, hydrogen gas is not evolved because HNO₃ is a strong oxidising agent. It oxidises the H₂ produced with water and itself is reduced to any of the nitrogen oxide.

(2) Melting points and boiling points of ionic compounds are high.
Answer:
Melting and boiling points : A considerable large amount of energy is required to break the strong inter-ionic attraction. Hence, ionic compounds have high melting and boiling points.
For example,

Ionic compound	Melting point (K)	Boiling point (K)
NaCl	1074	1686
LiCl	887	1600
CaCl2	1045	1900
CaO	2850	3120
MgCl2	981	1685

(3) Ionic compounds are non-conductor in the solid state, but they conduct electricity in the molten state or in an aqueous solution.
Answer:
Conduction of electricity : The conduction of electricity through a solution involves the movement of charged particles. A solution of an ionic compound in water contains ions, which move to the opposite electrodes, when electricity is passed through the solution.

Ionic compounds in the solid state do not conduct electricity because the movement of ions in the solid is not possible due to their rigid structure. But, ionic compounds conduct electricity in the molten state. Since the electrostatic forces of attraction between the oppositely charged ions are overcome and free ions are obtained due to heat. As a result, these ions move freely and conduct electricity.

Question 2.
Explain : Roasting and Calcination
Answer:
(1) Roasting: In this method, the concen-trated sulphide ore is heated in presence of excess < of air for a long time. So metal sulphides ore oxidised to metal oxides and SO₂. This method is known as roasting.

(2) Calcination : The ore containing metal carbonate or metal hydroxide is heated strongly in absence of air to convert it into metal oxide after the removal of volatile impurities and hydrated water. This process is known as calcination.
For example,

Question 3.
Explain : Alloying of gold
Answer:
Pure gold (24 carats) is very soft and s so the jewellery or ornaments made from it do not resist much pressure. Even a little pressure s can change their shape. Moreover they cannot resist much of wear and tear caused by friction. So, gold is alloyed with other metals like copper and silver to make it hard. 22 carat gold means it contains 22 parts of gold and 2 parts copper or silver in 24 parts by weight of an alloy.

Question 4.
Distinguish between :
(1) Metals and non-metals on the basis of their physical properties.
Answer:

Metals	Non-metals
1. Metals are in solid form. (Exception: mercury and gallium)	1. Non-metals are in the solid, or gaseous form.
2. They are heavy in weight.	2. They are light in weight.
3. They	3. They are bad conductors of heat and electricity.
4. Metals have lustre.	4. Non-metals do not have lustre. (Exception: Graphite, Silicon, Phosphorus)
5. Generally, they have high melting points.	5. Generally, they have low melting points.
6. Many metals produce ringing sound.	6. Non-metals do not produce ringing sound.
7. They can be hammered into thin sheets and drawn into wires.	7. Non-metals cannot be hammered into sheets nor can be drawn into wires.

(2) Calcination and roasting
Answer:

Calcination	Roasting
1. The process in which carbonate ores are changed into oxides by heating strongly in limited air is called calcination.	1. The process in which sulphide ores are converted into oxides by heating strongly in the presence of excess air is called roasting.
2. CO2 gas is evolved.	2. SO2gas is evolved.
	3. 2ZnS + 3O2 → 2ZnO + 2SO2

Question 5.
Explain :
(a) Why copper is used to make taps, hot water tanks and not any other metal?
(b) What will happen, if iron nails are S’ kept in a solution containing copper sulphate?
(c) Write the balanced chemical equations S> for the following and balance it:
(i) Ca + H₂O →
(ii) Al + HCl →
(iii) Fe + H₂O →
Answer:
(a) Copper does not react with cold and hot water, nor It reacts with the steam of water. Moreover, it is cheap and easily available, hence, S it is used to make taps and hot water tanks.

(b) Wlien iron nails are kept in a copper sulphate solution, the blue colour of the solution c fades due to displacement of copper by iron.

(c)

• Ca + 2H₂O → Ca(OH)₂ + H₂(g)
• 2Al + 6HCl → 2AlCl₃ + 3H₂(g)
• 3Fe + 4H₂O → Fe₃O₄ + 4H₂(g)

Question 6.
How will you prove that zinc is more reactive than copper?
Answer:
Take strips of zinc and copper and two test tubes with copper sulphate and zinc sulphate S solution. Add zinc strip in copper sulphate solution and copper strip in zinc sulphate solution.

Observation : In the test tube with zinc strip in copper sulphate solution shows that blue colour of copper sulphate solution fades. The other test tube will not show any change. This proves that zinc is more reactive than copper.

Question 7.
5 mL each of concentrated HCl and concentrated HNO₃ are taken in test tubes labelled as A and B while a mixture of concentrated HCl (15 mL) and concentrated HNO₃ (5 mL) is taken in test tube labelled as C. A small piece of metal is placed in each test tube. No change is observed in test tubes A and B, but metal got dissolved in test tube C. What would be the metal?
Answer:
Metal would be silver, gold or platinum.

Question 8.
The electronic configuration of three elements X, Y and Z are as follows:
X → 2, 8 Y → 2, 8, 6 Z → 2, 8, 1
Identify the metal and non-metal.
Answer:
From electronic configuration, X has 10 electrons, Y has 16 electrons and Z has 11 electrons. Hence these elements are
X = ₁₀Ne, Y = ₁₆S and Z = ₁₁Na.
Hence, X and Y are non-metals while Z is a metal.

Question 9.
Arrange the elements Au, Fe, Cu, Mg, Ca, Zn, Ag and K in descending order of their reactivity.
Answer:
Descending order of reactivity:
K > Ca > Mg > Zn > Fe > Cu > Ag > Au.

Objective Questions and Answers

Question 1.
Answer the following questions in short:
(1) What is called metallic lustre?
Answer:
Metals possesses a shining surface in their pure form. This property is called metallic lustre.

(2) State the full form of PVC.
Answer:
The full form of PVC : PolyVinyl Chloride

(3) Give the example of metal and non¬metal existing in liquid state.
Answer:
Liquid metal: Mercury (Hg) and Gallium
(Ga); Liquid non-metal: Bromine (Br)

(4) Which metals will melt, if you keep them on your palm?
Answer:
Gallium (Ga) and Caesium (Cs).

(5) What is an amphoteric oxide?
Answer:
Metal oxide which reacts with both acid as well as base to produce salt and water is S known as an amphoteric oxide.

(6) Which metals react violently with cold water?
Answer:
Sodium and potassium metals react > violently with cold water.

(7) What is an aqua regia?
Answer:
A freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio of 3 : 1 by volume is called an aqua regia.

(8) State one use of anhydrous calcium chloride.
Answer:
Anhydrous calcium chloride is used as drying agent, because it absorbs moisture from the air.

(9) What is meant by activity or reactivity series?
Answer:
When different metals are arranged in decreasing order of their activities or reactivities then it constitutes a series known as activity or reactivity series.

(10) Why does the noble gases possess little chemical activity?
Answer:
Noble gases possesses completely filled valence shell with 8 electrons. Hence they are chemically inert.

(11) State the reason for conduction of electricity through a solution.
Answer:
The conduction of electricity through a solution is due to the presence of ions which migrate to opposite electrodes and carry electricity.

(12) In which forms are the metals obtained from the earth’s crust?
Answer:
Metals are obtained in the form of oxide, carbonate and sulphide ores from the earth’s crust.

(13) On what factors does the process (method) used to remove the gangue from ores depend?
Answer:
The process (method) used for removing the gangue from the ores depends on the differences between the physical or chemical properties of the gangue and ores.

(14) What type of metals are used as reducing agent in displacement reactions? Mention the examples.
Answer:
Highly reactive metals are used as reducing agent in displacement reactions. For example, sodium, calcium, aluminium, etc.

(15) Name the two allotropes of carbon.
Answer:
Diamond and graphite.

Question 2.
Define:
(1) Displacement reaction
Answer:
A reaction in which more reactive metal displaces less reactive metal from its compound in solution or in molten form is known as displacement reaction.

(2) Electrovalent compounds
Answer:
The compounds formed by the transfer of electrons from a metal to a non-metal are known as electrovalent compounds. They are ionic compounds.

(3) Ore
Answer:
A mineral containing high percentage of a particular metal and from which metal can be profitably extracted is called an ore.

(4) Roasting
Answer:
The process of strongly heating sulphide ores in the presence of an excess of air and converting them into oxides is called roasting.

(5) Calcination
Answer:
The process of strongly heating carbonate ores in limited air, so that the volatile impurities are removed converting ores into oxides is called calcination.

(6) Anode-mud
Answer:
In electrolytic refining, insoluble impurities such as gold, silver, platinum are settled at the bottom of the anode are known as an anode-mud.

(7) Corrosion
Answer:
Metal gets rusted, when it is exposed to water, air and moisture for long time. This process of rusting is known as metal corrosion.

(8) Alloy
Answer:
The homogenous mixture of two or more metals, or a metal and a non-metal is called an alloy.

Question 3.
Fill in the blanks :

1. Iodine is a ………………. but it possesses
2. Diamond and graphite are allotropes of ……………….
3. Metals forms ………………. oxide.
4. The colour of copper (II) oxide is ……………….
5. Solder is an alloy of ………………. and ……………….
6. The electrical conductivity and melting points of alloys are ………………. than the pure metals.
7. The metals at the bottom of the reactivity series are ………………. reactive.
8. Ores are usually contaminated with large amount of impurities such as soil, sand, etc. which are called ……………….
9. The process of obtaining metal from its compound is ……………….
10. Iron (III) oxide is heated with aluminium powder forms iron in the molten state. This process is known as ……………….

Answer:

1. non-metal, lustre
2. carbon
3. basic or amphoteric
4. black
5. lead, tin
6. less
7. less
8. gangue
9. reduction
10. Thermit process

Question 4.
State whether the following statements are true or false:

1. All metals have similar hardness.
2. Silver and copper are good conductors of heat.
3. Bromine is a gas.
4. Graphite is a non-conductor of heat.
5. Li, Na and K can be cut with the knife.
6. Non-metallic elements dissolve in water to form acidic oxide.
7. Magnesium burns with dazzling blue flame.
8. Magnesium is less reactive than sodium.
9. Magnesium does not react with cold and hot s water, but it reacts with steam to form metal > oxide and hydrogen gas.
10. Gold and platinum do not dissolve in aqua regia.
11. Reactivity decreases in order of Mg > Al > Zn.
12. Copper reacts with dilute HCl.
13. Magnesium chloride possesses two ionic bonds.
14. The compounds formed by transfer of electrons are known as ionic compounds; moreover, they are known as electrovalent compounds.
15. Electrovalent compounds are kerosene and petrol.

Answer:

1. False
2. True
3. False
4. False
5. True
6. True
7. False
8. True
9. True
10. False
11. True
12. False
13. True
14. True
15. False

Question 5.
Choose the correct option from those given below each question:
1. Which of the following is an alloy?
A. Brass
B. Bronze
C. Steel
D. All of the given
Answer:
D. All of the given

2. Height and weight of iron pillar near the s Qutub Minar in Delhi are respectively…
A. 8 metre, 6 tonnes
B. 6 metre, 8 tonnes
C. 6 metre, 6 tonnes
D. 8 metre, 8 tonnes
Answer:
A. 8 metre, 6 tonnes

3. Which of the following metals is highly malleable?
A. Gold
B. Silver
C. A and B both
D. None of these
Answer:
C. A and B both

4. Which of the following is not useful as a good conductor of heat?
A. Silver
B. Copper
C. Lead
D. All of the given
Answer:
C. Lead

5. Which coating is applied on electric wire?
A. DDT
B. PVC
C. PTFE
D. PAN
Answer:
B. PVC

6. Which metal exists as liquid at room temperature?
A. Mercury
B. Bromine
C. Sodium
D. Calcium
Answer:
A. Mercury

7. Which allotrope of carbon is known as the hardest natural substance?
A. Graphite
B. Diamond
C. Coke
D. Carbon black
Answer:
B. Diamond

8. Most of the metal oxides in water are …
A. Soluble
B. Insoluble
C. Partial soluble
D. Highly soluble
Answer:
B. Insoluble

9. Which layer is formed on Al, when it is exposed to air?
A. Al₃O₂
B. Al₂O₃
C. AlO
D. AIN
Answer:
B. Al₂O₃

10. HNO₃ is reduced in …
A. NO
B. NO₂
C. N₂O
D. All of the given
Answer:
D. All of the given

11. Which of the following reactions is called roasting?

Answer:

12. During which reaction is dihydrogen gas not produced under normal conditions?
A. Metal + dilute sulphuric acid
B. Metal + dilute hydrochloric acid
C. Metal + dilute nitric acid
D. Metal + water
Answer:
D. Metal + water

13. In which of the following, displacement reaction is possible?
A. Solution of NaCl + coin of copper
B. Solution of MgCl₂ + coin of aluminium
C. Solution of FeSO₄ + coin of silver
D. Solution of AgNO₃ + coin of copper
Answer:
D. Solution of AgNO₃ + coin of copper

14. Which of the following reactions is not possible?
A. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
B. Zn(s) + FeSO₄(aq) → ZnSO₄(aq) + Fe(s)
C. Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
D. Cu(s) + FeSO₄(aq) → CuSO₄(aq) + Fe(s)
Answer:
D. Cu(s) + FeSO₄(aq) → CuSO₄(aq) + Fe(s)

15. Which of the following statements is incorrect?
A. Corrosion of copper takes place by contact with air and water.
B. The melting points and boiling points of metals are low.
C. The method to convert carbonate containing ore to metal oxide is called calcination.
D. The displacement of less active metals from their solution takes place by more active metal.
Answer:
B. The melting points and boiling points of metals are low.

Question 6.
Answer the following questions in one word :

1. Metals can be hammered into thin sheets. What does this property called?
2. How long thin wire can be drawn from one gram gold?
3. Name two metals which are poor conductors of heat.
4. How many electrons are present in the outermost shell of fluorine?
5. The electron of which shell is removed when sodium cation is formed from sodium atom?
6. By which force the anions and cations are held together in ionic compounds?
7. Are the melting points and boiling points of ionic compounds high or low?
8. Which soluble salts are present in sea-water?
9. Which metals are available in free state?
10. State the molecular formula of cinnabar.
11. Name the metal which can be cut with a knife.
12. Name the solution which is used for dissolving gold.

Answer:

1. Malleability
2. 2 km
3. Lead and mercury
4. 7 electrons
5. M-Shell
6. Electrostatic attraction forces
7. High
8. Sodium chloride and magnesium chloride
9. Gold, silver, platinum and copper
10. HgS
11. Sodium (Na)
12. Aqua regia

Question 7.
Mention the formulae, names and physical states of the products in the following reactions :
( 1 ) 2Cu(s) + O₂(g) →
( 2 ) 4Al(s) + 3O₂(g) →
( 3 ) Al₂O₃(s) + 6HCl(aq) →
( 4 ) Al₂O₃(s) + 2NaOH(aq) →
( 5 ) K₂O(s) + H₂O(l) →
( 6 ) Ca(s) + 2H₂O(l) →

( 13 ) 3Fe(s) + 4H₂O(g) →
( 14 ) CO(g) + H₂O(g) →
( 15 ) 3NO₂(g) + H₂O(l) →
Answer:

Value Based Questions With Answers

Question 1.
Arvind delivered a speech in the school assembly on “How to minimise the use of heavy metals?” He told how mercury thermometers when broken and thrown away in the garbage leads to pollution of soil and underground water pollution. He also showed how cadmium and lead also causes dangerous health problems.

1. Name two heavy metals which are present in the mobile batteries.
2. Name the disease caused due to mercury entering into our food-chain.
3. What value of Arvind is seen in the act?

Answer:

1. Lead and cadmium are used in the mobile batteries.
2. Mercury causes minamata disease.
3. Arvind showed the value of concern for nature, aware citizen and responsible behaviour.

Question 2.
A jeweller made jewellery (ornaments) of 22 carat gold and also charged his customers for the rate of 24 carat gold. His business developed due to this act.

1. Why can’t we make ornaments of 24 carat gold?
2. Name two metals that can be added to make jewellery.
3. What value of the jeweller is reflected in this act?

Answer:

1. 24 carat gold is soft and pure; and it cannot be moulded in the shape and design given to it while making the jewellery with strength. Hence, it is not advisable to make ornaments of 24 carat gold.
2. Copper and silver can be added to gold to make it strong.
3. Jeweller has shown the value of dishonesty, unfaithfulness and greediness towards his customers.

Question 3.
Naman saw his friend riding a cycle which was completely rusted at the edge of the pedal. He advised his friend to weld it and apply a
coating of oil paint over it.

1. Why do iron rust?
2. Give two remedies to avoid rusting.
3. What value of Naman is seen in the above act?

Answer:

1. Iron, when exposed to air and moisture, undergoes chemical change to form iron oxide which is reddish brown powder called rust.
2. Remedies to avoid rusting are : (i) Apply paint on the surface of iron. (ii) Apply grease or oil coating on iron.
3. Naman showed the value of awareness, concern and helpful nature.

Practical Skill Based Questions With Answers

Question 1.
Mention three safety measures essential while performing the reactivity series experiment in s the laboratory.
Answer:
Essential safety measures while performing the experiment of reactivity series are as follows :

• The metal or any other chemicals should not come in contact with skin.
• Do not inhale any fumes or gas released during the experiment.
• One should not taste any chemical or material.
• One should wear goggles, hand gloves and lab coat during the usage of the chemicals.

Question 2.
How will you identify the copper sulphate, iron sulphate and barium sulphate in the laboratory?
Answer:
Above given substances can be identified by their colours. For example, the colour of iron (II) sulphate is green, the colour of iron (III) sulphate is yellowish brown. The s colour of copper (II) sulphate is blue, and the white colour salt is barium sulphate.

Question 3.
Write the name and molecular formula of any three metal salts which are white in colour.
Answer:

• Zinc sulphate (ZnSO₄)
• Magnesium sulphate (MgSO₄)
• Aluminium sulphate (Al₂(SO₄)₃)

Above metal salts are white in colour.

Question 4.
A student put the spatula of iron in the test tube containing copper sulphate while performing an experiment in the laboratory. What will he observe next day?
Answer:
The student will observe that the blue coloured solution of copper sulphate turns to green and the iron spatula gets the coating of brown colour metal on it, only on the surface which is dipped in the solution.

Question 5.
Three solutions of zinc sulphate, magnesium sulphate and aluminium sulphate are prepared in laboratory by a lab assistant. But he forgot to label the reagent bottles. How will you label these bottles by using the metal reactivity studies?
Answer:
Take three test tubes and add little amount of each solution from the reagent bottle separately. Drop a small piece of magnesium in it. The test tube in which there is no reaction contains magnesium sulphate. Now, to identify the remaining two solutions, take these two solutions in two different test tubes and add a small piece of aluminium metal to them. The test tube in which, there is no reaction contains aluminium sulphate and the other one is zinc sulphate.

Memory Map

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Exercise 8.2

प्रश्न 1.
निम्नलिखित के मान निकालिए :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°

हल:
(i) sin 60° cos 30° + sin 30° cos 60°

प्रश्न 2.
सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) =
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}\) =
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
(iii) sin 2A = 2 sin A तब सत्य होता है, जबकि A बराबर है :
(A) 0°
(B) 30° 2tan 30°
(C) 45°
(D) 60°
(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\) बराबर है:
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
हल:

अत: सही विकल्प (A) है।

(ii) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-1^2}{1+1^2}\)
= \(\frac{1-1}{1+1}=\frac{0}{2}\) = 0
अत: सही विकल्प (D) है।

(iii) sin 2A = 2 sin A
यदि A = 0 हो तो
बायाँ पक्ष = sin 2A = sin (2 × 0)
= sin 0° = 0
दायाँ पक्ष = 2 sin A = 2 sin 0° = 0
अत: सही विकल्प (A) है।

अत: सही विकल्प (C) है।

प्रश्न 3.
यदि tan (A + B) = \(\sqrt{3}\) और tan (A – B) = \(\frac{1}{\sqrt{3}}\); 0° < A + B ≤ 90°; A > B तो A और B का मान ज्ञात कीजिए।
हल:
tan (A + B) = \(\sqrt{3}\)
tan (A + B) = tan 60°
A + B = 60° ….(1)
और tan (A – B) = \(\frac{1}{\sqrt{3}}\)
tan (A – B) = tan 30°
A – B = 30° …(2)
समीकरण (1) और (2) को जोड़ने पर
A + B = 60°
A – B = 30°
2A = 90°
A = \(\frac{90^{\circ}}{2}\) = 45°
A का मान समीकरण (1) में रखने पर,
45° + B = 60°
⇒ B = 60° – 45°
∴ B = 15°
अतः A = 45° और B = 15°.

प्रश्न 4.
बताइए कि निम्नलिखित में कौन-कौन सत्य हैं या असत्य हैं। कारण सहित अपने उत्तर की पुष्टि कीजिए:
(i) sin (A + B) = sin A + sin B
(ii) θ में वृद्धि होने के साथ sin θ के मान में भी वृद्धि होती है।
(iii) 6 में वृद्धि होने के साथ cos θ के मान में भी वृद्धि होती है।
(iv) θ के सभी मानों पर sin θ = cos θ
(v) A = 0° पर cot A परिभाषित नहीं है।
हल:
(i) माना कि
A = 30° तथा B = 60°
तो sin (A + B) = sin (30° + 60°)
= sin 90°
= 1
और sin A + sin B = sin 30° + sin 60°
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}=1\)
अतः sin (A + B) ≠ sin A + sin B
∴ दिया गया कथन असत्य है।

(ii) ∵ θ के मान 0°, 30°, 45°, 60°, 90° लेने पर,
sin 0° = 0, sin 30° = \(\frac{1}{2}\)
sin 45° = \(\frac{1}{\sqrt{2}}\)
sin 60° = \(\frac{\sqrt{3}}{2}\)
sin 90° = 1
अतः θ का मान बढ़ने पर sin θ का मान बढ़ता है। परन्तु यह θ = 90° तक ही सही है, आगे नहीं।
दिया गया कथन सत्य है।

(iii) ∵ cos 0° = 1 और cos 90° = 0
अतः θ का मान बढ़ाने पर cos θ के मान में वृद्धि नहीं होती।
∴ दिया गया कथन असत्य है।

(iv) ∵ sin θ = cos θ
θ = 30° लेने पर
sin 30° = \(\frac{1}{2}\)
cos 30° = \(\frac{\sqrt{3}}{2}\)
∴ sin 30° ≠ cos 30°
∴ दिया गया कथन असत्य है।

(v) tan 0° = 0
cot 0° = \(\frac{1}{\tan 0^{\circ}}\)
= \(\frac{1}{0}\) = अपरिभाषित
A = 0° पर cot A अपरिभाषित है।
∴ दिया गया कथन सत्य है।

Jharkhand Board JAC Class 10 Science Solutions Chapter 4 Carbon and Its Compounds Textbook Exercise Questions and Answers.

JAC Board Class 10 Science Solutions Chapter 4 Carbon and Its Compounds

Jharkhand Board Class 10 Science Carbon and Its Compounds Textbook Questions and Answers

Question 1.
Ethane, with the molecular formula C₂H₆ has …
(a) 6 covalent bonds.
(b) 7 covalent bonds.
(c) 8 covalent bonds.
(d) 9 covalent bonds.
Answer:
7 covalent bonds.

Question 2.
Butanone is a four carbon compound with the functional group ….
(a) carboxylic acid.
(b) aldehyde.
(c) ketone.
(d) alcohol.
Answer:
ketone.

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that…
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer:
the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The atomic numbers of carbon, hydrogen and chlorine are 6, 1 and 17 respectively. Hence, their electronic configuration may be written as follows :

Atom	K shell	L shell	M shell
Carbon	2	4	–
Hydrogen	1	–	–
Chlorine	2	8	7

• This indicates that, carbon requires 4 electrons to complete its octet; hydrogen requires 1 electron to complete its duplet and chlorine requires 1 electron to complete its octet.
• Therefore, carbon shares its four electrons, one electron each with three hydrogen atoms and one electron with chlorine atom to form four covalent bonds completing its octet.
  
• Thus, carbon attains the stable electronic configuration of nearest noble gas neon, hydrogen attains the electronic configuration of noble gas helium, while chlorine acquires the electronic configuration of noble gas argon.
• In short, chloromethane forms three C – H and one C – Cl covalent bonds.

Question 5.
Draw the electron dot structures for –
(a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂
Answer:

Question 6.
What is an homologous series ? Explain with an example.
Answer:
Definition : A series of compounds in which hydrogen atom / atoms remain in carbon chain are replaced by same functional group, then it constitute the homologous series.
OR
A series of organic compounds in succession which differ by a definite group (like – CH₂ -) is called homologous series.
Explanation:
CH₃ – OH Methanol
CH₃ – CH₂ – OH Ethanol
CH₃ – CH₂ – CH₂ – OH Propanol
CH₃ – CH₂ – CH₂ – CH₂ – OH Butanol

• It is a homologous series of alcohol where each member of the series possesses functional group – OH (Hydroxyl group).
• The difference in molecular formula between two successive members of the series is equal to -CH₂-, hence, difference in molecular masses of any two successive members of the series is 14 u.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Answer:
Difference in physical properties :

Physical property	Ethanol	Ethanoic acid
1. Smell	Pleasant (Sweet)	Vinegar-like strong
2. Taste	Burning taste	Sour
3. Melting point (K)	156	290
4. Boiling point (K)	351	391

Difference in chemical properties :

Reaction with	Ethanol	Ethanoic acid
1. Sodium bicarbonate	CO2 gas does not evolve.	CO2 gas is evolved.
2. Alkaline KMnO4	Pink colour of KMnO4 disappear.	Pink colour of KMnO4 does not disappear.

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
A soap molecule has two ends which possess different properties. One end of soap has a polar head which is called hydrophilic head, while the other end has a non-polar tail which is also called hydrophobic tail.

• Polar head has affinity towards water molecules, while non-polar tail being hydrophobic has no affinity towards water molecules.
• When soap is dissolved in water, the polar head dissolves in water but non-polar tail remains outside the water surface. As a result, spherical micelles are formed.
  
• Since, soap is soluble in ethanol, therefore micelle formation does not take place in ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
When carbon is burnt in the presence of air or oxygen, it forms carbon dioxide and water with a release of a large amount of heat and light.
C(s) + O₂(g) → CO₂(g) + Heat + Light

• When carbon and its compounds are burnt, additional heat energy is not required for consistent burning.
• Hence, carbon and its compounds are used as fuels for most applications.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hard water possesses calcium and magnesium ions which combine with soap molecules to form curdy white precipitate of calcium and magnesium salts, which is called scum.

Question 11.
What change will you observe if you test soap with litmus paper (red or blue)?
Answer:
Soap is basic (alkaline) in nature, it turns red litmus paper blue, while, it does not have any effect on blue litmus paper.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:
A reaction in which unsaturated hydrocarbons add hydrogen in the presence of catalysts such as nickel or palladium to form saturated hydrocarbons is known as hydrogenation reaction.

Industrial application : It is used to convert vegetable oil into vegetable ghee.

Question 13.
Which of the following hydrocarbons undergo addition reactions?
C₃H₆, C₃H₈, C₂H₂ and CH₄
Answer:
C₃H₆ and C₂H₂ are unsaturated hydrocarbons, hence, they undergo addition reaction.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Butter contains saturated fats while cooking oil contains unsaturated fats.

Unsaturated compound decolourise the pink colour of alkaline KMnO₄ solution.
When butter is treated with a few drops of alkaline KMnO₄ solution, the pink colour of KMnO₄ does not disappear, while, when cooking oil is treated with a few drops of alkaline KMnO₄ solution, its pink colour disappears.

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:
The molecules of soap are sodium or potassium salts of long chain carboxylic acids.

• Two ends of soap molecule possess different properties. One end is hydrophilic and it dissolves in water, while the other end is hydrophobic and it dissolves in hydrocarbons.
• When soap is at the surface of water, the hydrophobic ‘tail’ of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon ‘tail’ protruding out of water.
  
• Inside water, these molecules have a particular orientation that keeps the hydrocarbon portion out of the water.
• This happens due to the formation of clusters of molecules in which the hydrophobic tails are in the interior part of the cluster and the ionic ends are on the surface of the cluster.
• This formation is called a micelle.
• Soap in the form of a micelle collects the oily dirt in the centre of it and is able to clean.
• These micelles exist in solution as a colloid.
• Micelles do not cluster to precipitate because of ion-ion repulsion.
• Thus, the dirt suspended in the micelles is also easily rinsed away.
• The soap micelles are large enough to scatter light, hence a soap solution appears cloudy.

Jharkhand Board Class 10 Science Carbon and Its Compounds InText Questions and Answers

Question 1.
What would be the electron dot structure of carbon dioxide which has the s formula CO₂?
Answer:
The electronic configuration of carbon and oxygen are as follows:

Thus, carbon has 4 electrons and oxygen s has 6 electrons in the valence shell.

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made-up of eight atoms of sulphur? [Hint: The eight atoms of sulphur are joined together in the form of a ring.]
Answer:
The electronic configuration of ₁₆S is as follows :

Question 3.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
The conversion of ethanol to ethanoic acid is shown below:

In this reaction, a molecule of ethanol contains one oxygen atom while ethanoic acid contains two oxygen atoms. Hence, oxygen atom is added during the reaction; therefore, this conversion is an oxidation reaction.

Question 4.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
Ethyne is an unsaturated hydrocarbon.

• Combustion of ethyne in air produces a yellow flame with lot of black smoke.
• Due to this incomplete combustion, heat produced is very less than the heat required for welding; hence, the mixture of ethyne and air is not used for welding.
• When the combustion of ethyne is carried out with oxygen, due to complete combustion heat produced is very large which is usually needed for welding; hence the mixture of ethyne and oxygen is used for welding.
• 2[CH ≡ CH] + 5O₂ → 4CO₂ + 2H₂O + Heat and Light

Question 5.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:
An alcohol and a carboxylic acid can be distinguished experimentally by the following tests :
(1) Sodium hydrogencarbonate test: Take a small amount of each substance in two different test tubes and add to it an aqueous solution of sodium hydrogencarbonate.

• Test tube in which brisk effervescence of CO₂ gas appear contains carboxylic acid.
• Solution of ethanol does not produce brisk effervescence of CO₂ gas.

(2) Alkaline potassium permanganate test:
Take a small amount of each substance in two different test tubes, and add to it a few drops of alkaline potassium permanganate and warm the test tubes.

• The purple colour of potassium permanganate disappear in the test tube containing alcohol.
• Solution of carboxylic acid does not discharge the colour of potassium permanganate.

Question 6.
What are oxidising agents?
Answer:
Substances, which are capable of adding oxygen to other substances are known as oxidising agents. For example, alkaline KMnO₄, acidic K₂Cr₂O₇.

Question 7.
Would you be able to check if water is hard by using a detergent?
Answer:
No. Because detergent produces foam in both hard and soft water.

Question 8.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
Soap decreases the surface tension of water. When the soap is applied to any dirty clothes or on stains, the non-polar tail containing hydrocarbon part is attracted towards the molecule of dirt, oil or grease; while the polar tail part is attracted towards water. Thus, micelles are formed. Hence, to remove the micelles from the surface of the clothes, agaitation of clothes is necesary.

Question 9.
How many structural isomers can you draw for pentane?
Answer:
Structural isomers of pentane (C₅H₁₂) :

Question 10.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:
Carbon atom forms covalent bonds by sharing of electrons with other atoms to form numerous compounds. The number of carbon compounds known to chemists was recently estimated about three millions. Carbon has ability to form a large number of compounds. Two factors noticed in this case are as follows :

Catenation property of carbon : Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large number of molecules. This property of carbon is called catenation.

• Carbon compounds consist of long chains of carbon, branched chains of carbon or in the ring form or cyclic form.
  
• Carbon atom may be linked with other atoms by single, double or triple bonds.
• Compounds of carbon, which are linked by only single bonds between the carbon atoms are called saturated compoundsr
• Compounds of carbon, having double or triple c bonds between the carbon atoms are called unsaturated compounds.
• No other element exhibits the property of catenation to the extent seen in carbon compounds. Silicon forms compounds with hydrogen which have chains of upto seven or eight atoms, but these compounds are very reactive.
• The carbon-carbon bond is very strong and hence it is stable.
• Since carbon has a valency of four, it is capable of bonding with four other atoms of carbon or atoms of some other monovalent elements.
• Compounds of carbon are formed with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements giving rise to compounds with specific properties which depend on the elements other than carbon present in the molecule.
• The bond formed by carbon with most other elements are very strong, which makes compounds exceptionally stable.
• Carbon being small in size, it enables the nucleus to hold on to the shared pairs of electrons strongly. Hence, strong bonds are formed by carbon.

Question 11.
What will be the formula and electron dot structure of cyclopentane?
Answer:
The general formula of cycloalkane is C_nH_2n; hence, molecular formula of cyclopentane Is C₅H₁₀.
The electron dot structure of cyclopentane is shown below:

Question 12.
Draw the structures for the following compounds:
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal
Are structural isomers possible for bromo-pentane?
Answer:

Possible structural isomers of bromopentane are as follows :

Question 13.
How would you name the following compounds?

Answer:
The name 0f the given compounds are:

• Bromoethane
• Methanal
• Hex-l-yne or Hexyne.

Activity 4.1 [T.B. Pg. 58]

Aim : To demonstrate that most of the things we use in our day to day life consists of carbon.

Questions:

• Make a list of ten things you have used or consumed since morning.
• Compile this list with the lists made by your s classmates and then sort the items into the? adjascent table.

If there are items which are made-up of? more than one material, put them into both < the relevant columns.
Answer:
A list of ten things used or consumed s since morning is given below:
Toothbrush, toothpaste, bucket, water, soap, S detergent, cooking utensils, cup, milk, medicine, S’ newspaper, books.

Things made of metal	Things made of glass / clay	Others
Cooking utensils, bucket	Cup	Toothbrush, toothpaste, water, soap, detergent, milk, medicine, news-paper, books.

Activity 4.2 [T. B. Pg. 67]

Aim : To study the concept of homologous series.

Questions:

Question 1.
Calculate the difference in the formulae and molecular masses for (a) CH₃OH and C₂H₅OH, (b) C₂H₅OH and C₃H₇OH and (c) C₃H₇OH and C₄H₉OH
Answer:

Difference in formulae	Difference in molecular masses
(a) C2H5OH – CH3OH = CH2	[2(12) + 6(1) +1(16)] – [1(12) + 4(1) + 1(16)] = 46 – 32 = 14 u
(b) C3H7OH – C2H5OH = CH2	[3(12) + 8(1)+1(16)] – [2(12) + 6(1) + 1(16)] = 60 – 46 = 14 u
(c) C4H9OH- C3H7OH = CH2	[4(12) + 10(1) + 1(16)] – [3(12) + 8(1) + 1(16)] = 74 – 60 = 14 u

Question 2.
Is there any similarity in these three compounds mentioned above?
Answer:
Similarity observed in above three compounds are as follows :

• Difference in molecular formulae of these compounds is CH₂.
• Difference in molecular masses of these compounds is 14 u.
• In the nomenclature of these compounds, a common suffix -ol is written.
• These compounds are represented by the general formula C_nH_2n+1OH.
• Chemical reactions of these compounds are identical.
• All the three compounds are alcohols.

Question 3.
Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
Answer:
Alcohols in the increasing order of carbon atoms are as follows:
CH₃OH, C₂H₅OH, C₃H₇OH, C₄H₉OH
Yes, we call this family a homologous series.

Question 4.
Generate the homologous series for compounds containing up to four carbons for the other functional groups given in Q. 30.
Answer:
Homologous series for halogen group : [C_nH_2n+1X]
CH₃Cl, C₂H₅Cl, C₃H₇Cl, C₄H₉Cl
→ Homologous series for aldehyde group : [C_nH_2nO]
HCHO, CH₃CHO, CH₃CH₂CHO, CH₃CH₂CH₂CHO

→ Homologous series for ketone group : [C_nH_2nO]
CH₃COCH₃, CH₃COC₂H₅, CH₃COC₃H₇, CH₃COC₄H₉

→ Homologous series for carboxylic acid group : [C_nH_2nO₂]
HCOOH, CH₃COOH, CH₃CH₂COOH, CH₃CH₂CH₂COOH

Activity 4.3 [T. B. Pg. 69]

Aim : To study the flame of burning of carbon compounds.

Caution:
This activity needs the teacher’s assistance.

Activity:

• Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula t and burn them in the flame.
• Observe the nature of the flame and note s whether smoke is produced.
• Place a metal plate above the flame. Is there a deposition on the plate in case of any of the compounds?

Observation :

• Camphor and alcohol burn with a blue non-luminous flame and there is no sooty deposit on the metal plate.
• Naphthalene burns with a luminous yellow flame and there is sooty deposit on the metal plate.

Inference :
Camphor and alcohol are saturated hydrocarbons, while, naphthalene is an unsaturated hydrocarbon.

Activity 4.4 [T. B. Pg. 69]

Aim: To study the flame formed by complete and incomplete combustion.

Activity:
Light a bunsen burner and adjust the air hole at the base to get different types of flames / presence of smoke.

Questions:

Question 1.
When do you get a yellow flame?
Answer:
When the holes of the bunsen burner are not opened, enough supply of oxygen-rich air is not available, hence, incomplete combustion of hydrocarbon produces a yellow flame.

Question 2.
When do you get a blue flame?
Answer:
When the holes of the bunsen burner are opened, enough supply of oxygen-rich air is available, hence, complete combustion of hydrocarbon takes place and blue flame is formed.

Activity 4.5 [T. B. Pg. 70]

Aim: To study the oxidation of alcohols with alkaline KMnO₄.

Activity:

• Take about 3 mL of ethanol in a test tube and warm it gently in a water bath.
• Add a 5 % solution of alkaline potassium permanganate dropwise to this solution.

Questions :

Question 1.
Does the colour of potassium permanganate persist when it is added initially?
Answer:
When KMnO₄ is added to alcohol, the colour of KMnO₄ disappears initially, because, it oxidises ethanol to ethanoic acid and itself S is reduced to MnO₂.
[Note: This reaction is known as Bayer’s test.]

Question 2.
Why does the colour of potassium perman-ganate not disappear when excess is added?
Answer:
KMnO₄ solution when added more than required, then some of the KMnO₄ remains unused and hence its colour does not disappear.

Activity 4.6 [T. B. Pg. 72]

Aim: To study the reaction of sodium metal with ethanol.

Activity:

• Teacher’s demonstration
• Drop a small piece of sodium, about the size j of a couple of grains of rice into ethanol? (absolute alcohol).

Questions:

Question 1.
What do you observe?
Answer:
Hydrogen gas is evolved due to reaction between ethanol and sodium metal.

Question 2.
How will you test the gas evolved?
Answer:
When a burning matchstick is brought near? the hydrogen gas, it burns with a popping sound.

Activity 4.7 [T. B. Pg. 73]

Aim : To study the strength of dilute acetic acid and dilute hydrochloric acid.

Activity:
Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.

Questions:

Question 1.
Are both acids indicated by the litmus test?
Answer:
Both acids turn blue litmus to red, thus, both the acids are indicated by the litmus test.

Question 2.
Does the universal indicator show them as equally strong acids?
Answer:
The pH values of both acids of same concentration as obtained by using universal indicator are different. The pH value of dilute acetic acid is 4, while that of dilute hydrochloric acid is 2, which indicates that acetic acid is a weak acid than hydrochloric acid.

Activity 4.8 [T. B. Pg. 73]

Aim: To study the preparation and properties of an ester.

Activity:

• Take 1 mL ethanol (absolute alcohol) and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
• Warm in a water-bath for at least five minutes as shown in figure 4.4.
• Pour into a beaker containing 20 – 50 mL of water and smell the resulting mixture.
  
• When the mixture of absolute alcohol and glacial acetic acid is heated in presence of few drops of concentrated sulphuric acid, it forms a sweet smelling ester.

Activity 4.9 [T. B. Pg. 74]

Aim: To show the evolution of carbon dioxide gas by the reaction of carboxylic acid with sodium carbonate or sodium hydrogencarbonate.

• Set up the apparatus as shown in Chapter 2, Activity 2.5.
• Take a spatula full of sodium carbonate in a test tube and add 2 mL of dilute ethanoic acid.

Questions:

Question 1.
What do you observe?
Answer:
When sodium carbonate is added to 2 mL dilute solution of ethanoic acid, a brisk effervescence appear due to evolution of carbon dioxide gas.

Question 2.
Pass the gas produced through freshly prepared lime water. What do you observe?
Answer:
When the evolved gas is passed through freshly prepared lime water, it turns milky due to the formation of insoluble calcium hydroxide.

Question 3.
Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test?
Answer:
Yes. Carbon dioxide gas produced by the reaction can be identified by this test.
Repeat this activity with sodium hydrogen- carbonate instead of sodium carbonate.

Activity 4.10 [T. B. Pg. 74]

Aim: To study the solubility of oil in soap solution.

Activity:

• Take about 10 mL of water in two separate test tubes.
• Add a drop of oil (cooking oil) to both the test tubes and label them as A and B.
• To test tube B, add a few drops of soap solution.
• Now, shake both the test tubes vigorously s for the same period of time.

Questions:

Question 1.
Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them?
Answer:
In test tube A, two separate layers are seen, while in test tube B separate layers are not seen.

Question 2.
Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out? In which test tube does this happen first?
Answer:
In test tube A, two separate layers of oil and water are formed, while, in test tube B separate layers are not formed.

Activity 4.11 [T. B. Pg. 76]

Aim : To study the reaction of soap with hard and soft water.

Activity:

• Take about 10 mL of distilled water (or rain water) and 10 mL of hard water (from a tube) well or hand pump) in separate test tubes.
• Add a couple of drops of soap solution to both.
• Shake the test tubes vigorously for an equal period of time and observe the amount of foam formed.

Questions:

Question 1.
In which test tube do you get more foam?
Answer:
The test tube which contains soft water, i.e., distilled water or rain water, produces foam easily.

Question 2.
In which test tube do you observe a curdy white precipitate?
Answer:
The test tube which contains hard water, i.e., tube well water, produces curdy white precipitates.

Note for the teacher:
If hard water is not available in your locality, prepare Some hard water by dissolving hydrogen-carbonates/sulphates, chlorides of calcium or magnesium in water.

Activity 4.12 [T. B. Pg. 76]

Aim : To study the solubility of soap and detergent in hard water.

Activity:

• Take two test tubes with about 10 mL of hard water in each.
• Add five drops of soap solution to one and five drops of detergent solution to the other.
• Shake both test tubes for the same period.

Questions:

Question 1.
Do both test tubes have the same amount of foam?
Answer:
Both the test tubes do not contain the same amount of foam. The test tube with soap solution contains little foam in it.

Question 2.
In which test tube is a curdy white solid formed?
Answer:
The test tube to which soap solution was added produced a curdy white solid substance.

Students must go through these JAC Class 10 Science Notes Chapter 14 Sources of Energy to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 14 Sources of Energy

→ Energy can neither be created nor destroyed.

→ Energy is the capacity of a system to perform work.

→ Source of energy: The source which can provide energy conveniently is called source of energy.

→ A good source of energy : A source should be easily accessible, easy to store and transport and able to do large amount of work.

→ Fossil fuel : Coal, petroleum products and natural gas, etc. are fossil fuels.

→ The growing demand for energy is largely met by fossil fuels which are non-renewable sources of energy.

→ The thermal power plants produce heat energy from fuels and converts it into electrical energy.

→ A quarter of energy requirement in India is met by Hydropower plants.

→ Biomass : The matter obtained from living organisms is called biomass.

→ Fire wood, cow-dung cake. etc. form fuels which are plant and animal products. These sources are called biomass, which are traditionally used as fuels.

→ Charcoal: When wood is burnt in a limited supply of oxygen charcol is left behind as the residue.

→ Biogas: It is produced by decomposition of cow-dung, agriculture wastes, vegetable waste and sewage in absence of oxygen.

• Biogas is an excellent fuel as it contains upto 75 % methane.
• Biogas plant provides fuel gas as well as manure.

→ Wind energy: The kinetic energy of the wind can be used to get electricity on a commercial scale. Wind energy is an environment friendly and efficient source of renewable energy.

→ Solar energy: Solar energy is the root of all energy.

• India receives solar energy for greater part of the year.
• Solar cooker, solar water heater, solar cell panel, etc. are widely used now-a-days.

→ Ocean energy : Tidal energy, wave energy and ocean thermal energy are considered as ocean energy.

• Due to high and low tides, the difference in sea level gives tidal energy.
• Wave energy would be a viable proposition only where waves are very strong.
• Temperature difference between the water at the surface and water at depths is exploited to obtain ocean thermal energy.

→ Geothermal energy : Heat energy can be exploited from the steam trapped in rocks near the hot springs. This is called geothermal energy.

→ Nuclear energy: It is obtained by nuclear reactions, i.e.,

• Nuclear fission – The fission of an atom of uranium.
• Nuclear fusion – Joining of lighter nuclei (hydrogen isotopes) to create a heavier nucleus (helium).

→ CNG (Compressed Natural Gas) : CNG is considered as the cleaner alternative to petrol or diesel.

→ Renewable energy is available in our natural environment, in the form of repetitive current of energy.

Students should go through these JAC Class 10 Maths Notes Chapter 8 त्रिकोणमिति का परिचय will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 10 Maths Notes Chapter 8 त्रिकोणमिति का परिचय

भूमिका :
त्रिकोणमिति के अंग्रेजी शब्द Trigonometry की व्युत्पत्ति ग्रीक शब्दों ‘tri’ (जिसका अर्थ है, तीन), ‘gon’ (जिसका अर्थ है, भुजा) और ‘metron’ (जिसका अर्थ है, माप) से हुई है। वस्तुत: त्रिकोणमिति में एक त्रिभुज की भुजाओं और कोणों के बीच के सम्बन्धों का अध्ययन करते हैं।

पाइथागोरस प्रमेय (Pythagoras Theorem) : “किसी समकोण त्रिभुज में समकोण बनाने वाली भुजाओं के वर्गों का योग त्रिभुज के कर्ण के वर्ग के बराबर होता है।”
(कर्ण)² = (समकोण बनाने वाली एक भुजा)² + (समकोण बनाने वाली दूसरी भुजा)²
चित्र में ΔABC समकोण त्रिभुज है, जिसमें ∠B = 90°, कर्ण = CA तथा समकोण बनाने वाली भुजाएँ क्रमश: AB और BC हैं।

∴ भुजाओं में सम्बन्ध :
(AC)² = (AB)² + (BC)²
यदि दो भुजाओं की माप ज्ञात हो, तो तीसरी भुजा की माप ज्ञात कर सकते हैं।
→ त्रिकोणमिति (Trigonometry): त्रिकोणमिति गणित की वह शाखा है, जिसके अन्तर्गत एक त्रिभुज की भुजाओं और कोणों के बीच के सम्बन्धों का अध्ययन किया जाता है।
→ त्रिकोणमितीय अनुपात (Trigonometric Ratios): एक समकोण त्रिभुज में किसी न्यून कोण के सापेक्ष भुजाओं के अनुपात का अध्ययन त्रिकोणमितीय अनुपात कहलाता है।
→ त्रिकोणमितीय सर्वसमिकाएँ (Trigonometric Identities): एक कोण के त्रिकोणमितीय अनुपातों से सम्बन्धित समीकरण को त्रिकोणमितीय सर्वसमिका कहते हैं। जबकि यह सम्बन्धित कोण (कोणों) के सभी मानों के लिए सत्य होता है।
→ पूरक कोण (Complimentary angles): यदि दो कोणों का योग 90° हो, तो उन कोणों को परस्पर पूरक कोण कहते हैं।
→ sin θ : प्रतीक sin θ का प्रयोग कोण θ, के sinθ के संक्षिप्त रूप में किया गया है।
→ cos θ : प्रतीक cos θ का प्रयोग कोण θ, के cosinθ के संक्षिप्त रूप में किया गया है।
→ tan θ : प्रतीक tan θ का प्रयोग कोण θ के tangent के संक्षिप्त रूप में किया गया है।
→ cot θ : प्रतीक cot θ का प्रयोग कोण θ के, cotangent के संक्षिप्त रूप में किया गया है।
→ sec θ : प्रतीक sec θ का प्रयोग कोण θ के, secant के संक्षिप्त रूप में किया गया है।
→ cosec θ : प्रतीक cosec θ का प्रयोग कोण θ, के cosecant के संक्षिप्त रूप में किया गया है।

त्रिकोणमितीय अनुपात :
समकोण त्रिभुज ABC की भुजाओं के कुछ अनुपातों का उसके न्यूनकोणों के सापेक्ष अध्ययन को त्रिकोणमितीय अनुपात कहते हैं।
समकोण त्रिभुज ABC में ∠B समकोण और न्यूनकोण A के सापेक्ष त्रिकोणमितीय अनुपातों को निम्नांकित प्रकार से परिभाषित कर सकते हैं-


टिप्पणी : cosec A, sec A और cot A अनुपातों sin A, cos A और tan A के क्रमशः व्युत्क्रम हैं।

इसलिए, समकोण त्रिभुज के एक न्यूनकोण के त्रिकोणमितीय अनुपात, त्रिभुज के कोण और उसकी भुजाओं की लम्बाई के बीच के सम्बन्ध को व्यक्त करते हैं।

त्रिकोणमितीय अनुपातों में पारस्परिक सम्बन्ध (Relations among Trigonometric Ratios) :
(i) sin और cosec त्रिकोणमितीय अनुपात परस्पर व्युत्क्रम (Reciprocal) हैं।
(ii) cos और sec त्रिकोणमितीय अनुपात परस्पर व्युत्क्रम हैं।
(iii) tan और cot त्रिकोणमितीय अनुपात, परस्पर व्युत्क्रम हैं।

इन्हें निम्न प्रकार से भी व्यक्त किया जा सकता है :

sin² θ + cos² θ =1
sin² θ = 1 – cos² θ
cos² θ = 1 – sin² θ
sec² = 1 + tan² θ
sec² θ – tan² θ = 1
tan² θ = sec²2 θ – 1
cosec² θ = 1 – cot² θ
cosec² θ – 1 = cot² θ
cosec² θ – cot² θ = 1
अग्रांकित सारणी की सहायता से प्रत्येक त्रिकोणमितीय अनुपात को दूसरे त्रिकोणमितीय अनुपातों में परिवर्तित किया जा सकता है :

कुछ विशिष्ट कोणों के त्रिकोणमितीय अनुपात :
0° तथा 90° के त्रिकोणमितीय अनुपात: यदि समकोण ΔABC में कर्ण AC तथा ΔABC’ में कर्ण AC’ बराबर लम्बाई के हैं। दोनों त्रिभुजों से हम देखते हैं कि θ का मान ज्यों-ज्यों बढ़ाते जाते हैं, त्यों-त्यों उसकी सम्मुख भुजा की लम्बाई बढ़ती जाती है। इसके विपरीत θ का मान कम करने पर उसकी सम्मुख भुजा की लम्बाई कम होती जाती है।

यदि θ का मान घटते घटते शून्य हो जाए, तो उस स्थिति में θ की सम्मुख भुजा BC शून्य और आधार भुजा AB = AC हो जाती है तथा बिन्दु C बिन्दु B के ठीक ऊपर होगा।
∴ sin 0° = \(\frac{B C}{A C}=\frac{0}{A C}\) = 0
तथा cos 0° = \(\frac{A B}{A C}=\frac{A C}{A C}\) = 1 (∵ AB = AC)
tan 0° = \(\frac{\sin 0^{\circ}}{\cos 0^{\circ}}=\frac{0}{1}\) = 0

sin 0° = 0
cos 0° = 1
tan 0° = 0
विलोमतः
cosec 0° = अपरिभाषित
sec 0° = 1
cot 0° = अपरिभाषित
यदि समकोण ΔABC में 6 का मान बढ़ाने पर ∠θ के सामने की भुजा BC की लम्बाई बढ़ती है और आधार भुजा घटती है। θ = 90° की स्थिति में BC भुजा, कर्ण AC के बराबर हो जाती है और बिन्दु के ठीक ऊपर होता है और भुजा AB शून्य हो जाती है।
sin 90° = \(\frac{B C}{A C}=\frac{A C}{A C}=1\)
(∵ 90° पर BC = AC)
तथा cos 90° = \(\frac{A B}{A C}=\frac{0}{A C}\)
tan 90° = \(\frac{\sin 90^{\circ}}{\cos 90^{\circ}}=\frac{1}{0}\) = ∞ (अपरिभाषित)
sin 90° = 1
cos 90° = 0
tan 90° = अपरिभाषित
cosec 90° = 1
विलोमत: sec 90° = अपरिभाषित
cot 90° = 0
30° और 60° के त्रिकोणमितीय अनुपात : एक समबाहु त्रिभुज ABC लेते हैं। जिसका प्रत्येक कोण 60° का होता है।
∴ ∠A = ∠B = ∠C = 60°
शीर्ष A से भुजा BC पर लम्ब AD डालते हैं।

अर्थात् AD ⊥ BC
ΔABD ≅ ΔACD, (RHS सर्वांगसमता नियम से)
∴ BD = DC
∠BAD = ∠CAD (CPCT)
अत: ΔABD एक समकोण Δ है, जिसका कोण D समकोण है।
जहाँ ∠BAD = 30° और ∠ABD = 60°
माना AB = BC = CA = 2a
तब BD = \(\frac{1}{2}\)BC = \(\frac{1}{2}\) × 2a = a
समकोण ΔABD में,
AD² = AB² – BD²
= (2a)² – (a)²
= 4a² – a² = 3a²
AD = a\(\sqrt{3}\)
अब sin 30° = \(\frac{B D}{A B}=\frac{a}{2 a}=\frac{1}{2}\)
cosec 30° = \(\frac{1}{\sin 30^{\circ}}=\frac{1}{\frac{1}{2}}=2\)

45° के त्रिकोणमितीय अनुपात : समकोण ΔABC में, जिसका ∠B = 90° है। यदि एक ∠A = 45°, तो दूसरा ∠B = 45° का होगा।

अर्थात्
∠A = ∠C = 45°
∴ AB = BC
माना AB = BC = a
पाइथागोरस प्रमेय से,
AC² = AB² + BC² = a² + a²
⇒ AC² = 2a²
∴ AC = \(\sqrt{2}\)a.

विशेष कोणों के त्रिकोणमितीय अनुपातों की सारणी (Tables of Trigonometric Ratios of Particular Angles)

पूरक कोणों के त्रिकोणमितीय अनुपात (Trignometrical Ratios of Complementary Angles) :
यदि दो कोणों का योग 90° के बराबर हो, तो वे दोनों कोण एक-दूसरे के पूरक कहलाते हैं। एक समकोण ΔABC में ∠B समकोण है। इसलिए
∠A + ∠C = 90° ⇒ ∠C = 90° – A
अर्थात् ∠A व ∠C एक-दूसरे के पूरक हैं।

∠A के लिए त्रिकोणमितीय अनुपात :
कर्ण = AC, लम्ब = BC, आधार = AB

अब ∠C = (90° – A) के लिए त्रिकोणमितीय अनुपात :
कर्ण = AC, आधार = BC, लम्ब = AB
sin ( 90° – A) = लम्ब / कर्ण = \(\frac{A B}{A C}\)
cos (90° – A) = \(\frac{B C}{A C}\)
tan (90° – A) = \(\frac{A B}{B C}\)
cot (90° – A ) = \(\frac{B C}{A B}\)
sec (90° – A) = \(\frac{A C}{B C}\)
cosec (90° – A) = \(\frac{A C}{A B}\)
∠A और ∠C के त्रिकोणमितीय अनुपातों की तुलना करने पर,
sin ( 90° – A) = \(\frac{A B}{A C}\) = cos A
cos (90° – A) = \(\frac{B C}{A C}\) = sin A
tan (90° – A) = \(\frac{A B}{B C}\) = cot A
cot (90° – A ) = \(\frac{B C}{A B}\) = tan A
sec (90° – A) = \(\frac{A C}{B C}\) = cosec A
cosec (90° – A) = \(\frac{A C}{A B}\) = sec A

टिप्पणी : जब हम कोण को बदलेंगे, तो sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ भी परिवर्तित हो जायेंगे। याद रखने के लिए ‘co’ को जोड़िए यदि यह नहीं है तो ‘co’ को हटाइए।

त्रिकोणमितीय सर्वसमिकाएँ :
एक समकोण ΔABC में ∠B समकोण है। पाइथागोरस प्रमेय से,
AB² + BC² = AC² …..(1)

समीकरण (1) के प्रत्येक पदों को AC² से विभाजित करने पर,
\(\frac{A B^2}{A C^2}+\frac{B C^2}{A C^2}=\frac{A C^2}{A C^2}\)
या \(\left(\frac{A B}{A C}\right)^2+\left(\frac{B C}{A C}\right)^2=1\)
या (cos A)² + (sin A)² = 1
अर्थात् cos² A + sin² A = 1, जहाँ 0° ≤ A ≤ 90°
sin² A + cos² A = 1 …..(2)
अब समीकरण (1) को AB² से विभाजित करने पर,
\(\frac{A B^2}{A B^2}+\frac{B C^2}{A B^2}=\frac{A C^2}{A B^2}\)
या \(\left(\frac{A B}{A B}\right)^2+\left(\frac{B C}{A B}\right)^2=\left(\frac{A C}{A B}\right)^2\)
अर्थात् 1 + tan² A = sec² A …..(3)
अब समीकरण (1) को BC2 से विभाजित करने पर,
\(\frac{A B^2}{B C^2}+\frac{B C^2}{B C^2}=\frac{A C^2}{B C^2}\)
⇒ \(\left(\frac{A B}{B C}\right)^2+\left(\frac{B C}{B C}\right)^2=\left(\frac{A C}{B C}\right)^2\)
⇒ cot² A + 1 = cosece² A ….( 4 )
इन सर्वसमिकाओं का प्रयोग करके हम प्रत्येक त्रिकोणमितीय अनुपात को, दूसरे त्रिकोणमितीय अनुपातों के पदों में व्यक्त कर सकते हैं,
sin² θ + cos² θ = 1
⇒ sin² θ = 1 – cos² θ
या cos² θ = 1 – sin² θ
1 + tan² θ = sec² θ
⇒ tan² = sec² θ – 1
या sec² θ – tan² θ = 1
cot² θ + 1 = cosec² θ
⇒ cot² θ = cosec² θ – 1
या cosec² θ – cot² θ = 1

Students must go through these JAC Class 10 Science Notes Chapter 13 Magnetic Effects of Electric Current to get a clear insight into all the important concepts.

JAC Board Class 10 Science Notes Chapter 13 Magnetic Effects of Electric Current

→ Magnet: A magnet is a piece of magnetic material, occuring naturally or made artificially by magnetizing iron or steel, which attracts pieces of magnetic substances such as iron, nickel and cobalt.

→ Magnetic field: The region surrounding a magnet in which the force of attraction and repulsion due to that magnet can be detected (using magnet or magnetic substances) is called the magnetic field.

[Each point in this field has a particular strength. The field at each point also has a definite direction.]

→ Magnetic field lines : The lines along which the iron filings align / arrange themselves due to force acting on them in the magnetic field of a bar magnet are called magnetic field lines.

Note : The path followed by the north pole of a compass needle (magnetic needle) placed in a region of magnetic field, such as that of a bar magnet, is called a magnetic field line. It shows how the magnetic force changes from point to point.

→ Characteristics of magnetic field lines:

• The magnetic field lines emerge from north pole and merge at the south pole outside the magnet, while inside the magnet the direction of field lines is from its south pole to its north pole.
  Thus, the magnetic field lines are closed and continuous curve.
• The magnetic field lines are crowded near the pole where the magnetic field is strong and are far apart near the middle of the magnet and far from the magnet where the magnetic field is weak.
• The magnetic field lines never intersect each other because if they do so, there would be two directions of magnetic field at that point which is absurd.
• In case the field lines are parallel and equidistant, these represent a uniform magnetic field.
  Important note: The relative strength of the magnetic field is shown by the degree of closeness of the field lines.

→ Oersted’s observations : When a magnetic needle is placed near a conducting wire carrying a current, the magnetic needle deflects. The deflection is observed in the opposite direction, when the direction of the current is reversed. This shows that when an electric current passes through a conducting wire a magnetic field is produced around it.

→ Magnetic field due to a current through a straight conductor: The magnetic field lines due to a current through a straight conductor form concentric circles around the conductor. The strength of this magnetic field is proportional to the current through the conductor and inversely proportional to the distance of the given point from the conductor.

→ Right-hand thumb rule: Imagine that you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your fingers of right hand wrap around the conductor in the direction of the field lines of the magnetic field.

→ Magnetic field due to a current through a circular loop: The magnetic field lines due to a current through a circular loop form concentric circles. These circles become larger and larger as we move away from the wire and at the centre of the circular loop, the arcs of these large circles appear as straight lines. The magnetic field at the centre of the circular loop is proportional to the current through the loop and inversely proportional to the radius of the circular loop.

→ Solenoid: A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder is called a solenoid. The magnetic field due to a current in a solenoid is similar to that of a bar magnet. The magnetic field lines inside a solenoid are straight lines parallel to each other. The magnetic field inside a solenoid is uniform.

→ Electromagnet: An electromagnet is a magnet consisting of a long coil of insulated copper wire wound around a soft iron core in the form of a rod. It is magnetised only when an electric current is passed through the coil.

→ Force on a current- carrying conductor placed in a magnetic field: When a conductor carrying an electric current is placed in a magnetic field, magnetic force acts on it. This force is perpendicular to the direction of the magnetic field as well as the direction of the current and is proportional to the electric current, magnetic field of the magnet, the length of the conductor inside the magnetic field and on the angle between the directions of the magnetic field and the current. It is maximum when this angle is 90°. The direction of this force can be determined using Fleming’s left-hand rule.

→ Fleming’s left-hand rule: Stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular. If the first (fore) finger points in the direction of magnetic field and second (middle) finger points in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

→ Electric motor: It converts electrical energy into mechanical energy. In this case, a current-carrying coil kept in a strong magnetic field experiences a force. As a result, the coil starts rotating.

→ Electromagnetic induction: The process, by which a changing magnetic field in a conductor induces a current in another conductor is called electromagnetic induction.
OR
An electric current produced in a closed circuit by a changing magnetic field is called an induced current. This phenomenon is called electromagnetic induction.

→ Induced potential difference (or electromotive force): Because of the rate of change of number of magnetic field lines linked with the loop during the relative motion of a magnet and loop or due to a changing current in a nearby conductor, an electric potential difference is induced in the coil. It is called induced potential difference (or electromotive force).

The electromotive force induced in a loop is proportional to the rate of change of number of magnetic field lines associated with the loop and the number of turns of a loop.

→ Fleming’s right-hand rule : Stretch the thumb, forefinger and middle finger of right hand so that they are perpendicular to each other. If forefinger indicates the direction of the magnetic field and thumb shows the direction of motion of conductor, then the middle finger will show the direction of induced current.

→ Electric generator: It converts mechanical energy into electrical energy. Its working is based on electromagnetic induction. The generator by which a unidirectional current (DC) can be obtained is called a DC generator and that by which an alternating current (AC) is obtained is called an AC generator.

→ Direct Current (DC) and Alternating Current (AC) : If a current flows only in one direction, it is called a direct current (DC). DC is obtained with a battery and DC generator.

The current whose direction changes periodically with time is called an alternating current (AC). AC is obtained with an AC generator.

→ Domestic electric circuits: In India, the AC voltage used for domestic purposes is 220 V and its frequency is 50 Hz. Three wires, namely live, neutral and earthing wire enter our house through a main fuse passing through an electric meter at the meter-board. The potential difference between the live and neutral wire is 220 V.

The earthing wire is connected to a copper plate placed in a deep pit dug near the house. This wire is connected to the metallic body of the appliances, so that one does not experience shock.

Two types of electric lines are available for domestic usage. One is of 5 A and other is of 15 A rating. The entire wiring of the house is done in parallel connections.

→ Fuse : An electric fuse is an important component of all domestic circuits. It is used to avoid incidents such as electric shock, fire, damage to an electric appliance due to (a) short-circuiting or (b) overloading (drawing a current beyond a specified limit) in a circuit. A fuse is a most important safety device. It consists of a short, thin, tin-plated copper wire having low melting point. It melts and breaks the circuit if the current exceeds a safe value.

A fuse is always connected in series with the live wire and in the beginning of the circuit.