CLASS X · STRUCTURAL FORMULAE & NOMENCLATURE · PRACTICE SHEET
Section A — Drawing Structural Formulae
Q1. Draw the electron dot structure and structural formula of methane (CH₄).
Show Answer
Methane has one carbon atom forming four single covalent bonds with four hydrogen atoms.
\[
\text{H} – \underset{\displaystyle \big| }{\overset{\displaystyle \text{H}}{\text{C}}} – \text{H} \quad \text{with a fourth H below the C}
\]
Structural formula: CH₄, with C at the centre bonded to 4 H atoms by 4 single bonds (tetrahedral arrangement)
Q2. Write the structural formula of ethane (C₂H₆) and identify the number of covalent bonds present.
Show Answer
\[
\text{H}_3\text{C} – \text{CH}_3
\]
Ethane contains 1 C–C bond and 6 C–H bonds, making a total of 7 covalent bonds.
Q3. Draw the structural formula of ethene (C₂H₄) and ethyne (C₂H₂). State the type of bond between the two carbon atoms in each.
Show Answer
Ethene:
\[
\text{H}_2\text{C} = \text{CH}_2
\]
— carbon atoms joined by a double bond.
Ethyne:
\[
\text{HC} \equiv \text{CH}
\]
— carbon atoms joined by a triple bond.
Q4. Draw the structural formula of propane, and show all the possible structural isomers of butane (C₄H₁₀).
Show Answer
Propane:
\[
\text{CH}_3 – \text{CH}_2 – \text{CH}_3
\]
Butane has two structural isomers:
Q6. Write the IUPAC names of the following compounds: (a) CH₃–CH₂–CH₂–CH₃ (b) CH₃–CH₂–OH (c) CH₃–CHO (d) CH₃–CO–CH₃
Show Answer
(a) Butane
(b) Ethanol
(c) Ethanal
(d) Propanone
Q7. Name the functional groups present in the following compounds and give the general formula of each homologous series: (a) alcohols (b) aldehydes (c) ketones (d) carboxylic acids
Show Answer
(a) Alcohol — functional group –OH; general formula \( \text{C}_n\text{H}_{2n+1}\text{OH} \)
(b) Aldehyde — functional group –CHO; general formula \( \text{C}_n\text{H}_{2n+1}\text{CHO} \)
(c) Ketone — functional group \(>\!\text{C}=\text{O}\); general formula \( \text{C}_n\text{H}_{2n+1}\text{CO}\text{C}_m\text{H}_{2m+1} \)
(d) Carboxylic acid — functional group –COOH; general formula \( \text{C}_n\text{H}_{2n+1}\text{COOH} \)
Q8. Write the structural formula for: (a) propanal (b) pentan-1-ol (c) but-2-ene (d) propyne
Q9. Differentiate between saturated and unsaturated hydrocarbons with one structural example each, and state the general formula for each class (alkanes, alkenes, alkynes).
Show Answer
Saturated hydrocarbons (alkanes) contain only single bonds between carbon atoms.
General formula: \( \text{C}_n\text{H}_{2n+2} \) e.g. ethane, \( \text{CH}_3-\text{CH}_3 \)
Unsaturated hydrocarbons contain at least one double or triple bond.
Alkenes: \( \text{C}_n\text{H}_{2n} \) e.g. ethene, \( \text{H}_2\text{C}=\text{CH}_2 \)
Alkynes: \( \text{C}_n\text{H}_{2n-2} \) e.g. ethyne, \( \text{HC}\equiv\text{CH} \)
Q10. Draw all possible isomers of pentane (C₅H₁₂) and give their IUPAC names.
Show Answer
n-pentane:
\[
\text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{CH}_2 – \text{CH}_3
\]
2-methylbutane:
\[
\text{CH}_3 – \underset{\displaystyle \text{CH}_3}{\text{CH}} – \text{CH}_2 – \text{CH}_3
\]
2,2-dimethylpropane (neopentane):
\[
\text{CH}_3 – \underset{\displaystyle \text{CH}_3}{\underset{\displaystyle |}{\text{C}}} – \text{CH}_3, \text{ with a second } \text{CH}_3 \text{ also on the central C}
\]
Instructions:
Name each of the following carbon compounds using the correct IUPAC nomenclature.
Important: Students must write the names themselves. The answer key is hidden below each set.
सरल भाषा में व्याख्या — दैनिक जीवन के उदाहरणों के साथ
1.1हमें मात्रकों की आवश्यकता क्यों है?
कल्पना कीजिए कि आप किसी से कहते हैं, “मेज 5 लंबी है।” पाँच क्या? पाँच इंच? पाँच फीट? या पाँच मीटर? बिना किसी मात्रक के, संख्या “5” का कोई अर्थ नहीं रह जाता। एक मात्रक (Unit) सर्वसम्मत अंतरराष्ट्रीय मानक होता है, जिसके साथ हम भौतिक राशियों की तुलना करते हैं, ताकि हर व्यक्ति एक ही माप को एक ही अर्थ में समझ सके।
इसे ऐसे समझें
यदि आप कहें “मैं 5 दौड़ा” और यह न बताएं कि किलोमीटर या मील, तो कोई नहीं समझ पाएगा कि आप वास्तव में कितनी दूर दौड़े। मात्रक ही वह साधन है जो संख्या को वास्तविक अर्थ प्रदान करता है।
भौतिकी में केवल कुछ ही मूल राशियां (Fundamental/Base Quantities) होती हैं — जैसे लंबाई, द्रव्यमान और समय — जिन्हें और छोटे घटकों में नहीं तोड़ा जा सकता। शेष सभी भौतिक राशियां (जैसे चाल, बल, ऊर्जा, दाब) इन्हीं मूल राशियों के संयोजन से बनती हैं। संयुक्त राशियों के मात्रकों को व्युत्पन्न मात्रक (Derived Units) कहा जाता है।
1.2मात्रकों की अंतर्राष्ट्रीय प्रणाली (SI)
प्राचीन समय में, अलग-अलग देश अलग-अलग प्रणालियों का उपयोग करते थे — जैसे CGS (सेन्टीमीटर-ग्राम-सेकंड), FPS (फुट-पाउंड-सेकंड, ब्रिटेन व अमरीका में प्रचलित), तथा MKS (मीटर-किलोग्राम-सेकंड)। इससे विभिन्न देशों के वैज्ञानिकों के बीच शोध साझा करते समय भ्रम पैदा होता था।
इसलिए, संपूर्ण विश्व ने एक सर्वमान्य प्रणाली अपनाई: SI प्रणाली (Système International d’Unités — फ्रांसीसी भाषा में “मात्रकों की अंतर्राष्ट्रीय प्रणाली”)। यह 7 मूल मात्रकों पर आधारित है, जिनमें से प्रत्येक को एक स्थिर नियतांक (Physical Constant) से परिभाषित किया गया है ताकि दुनिया में कहीं भी कोई भी वैज्ञानिक इन्हें पूरी शुद्धता से पुनरुत्पादित कर सके।
7 मूल मात्रक (Base Units)
मूल राशि
मात्रक का नाम
प्रतीक
परिभाषा का आधार
लंबाई (Length)
मीटर (metre)
m
निर्वात में प्रकाश की चाल
द्रव्यमान (Mass)
किलोग्राम (kilogram)
kg
प्लांक नियतांक (Planck constant)
समय (Time)
सेकंड (second)
s
सीज़ियम परमाणु की विकिरण आवृत्ति
विद्युत धारा (Electric Current)
एम्पियर (ampere)
A
मूल विद्युत आवेश (e)
तापमान (Temperature)
केल्विन (kelvin)
K
बोल्ट्ज़मान नियतांक (Boltzmann constant)
पदार्थ की मात्रा (Amount of Substance)
मोल (mole)
mol
आवोगाद्रो नियतांक (Avogadro constant)
ज्योति तीव्रता (Luminous Intensity)
कैंडेला (candela)
cd
प्रकाश की दीप्त दक्षता
केवल धातु की छड़ को “1 मीटर” का मानक क्यों नहीं माना गया?
क्योंकि धातु की छड़ें समय के साथ घिस सकती हैं, उनमें ज़ंग लग सकता है या तापमान से उनकी लंबाई बदल सकती है। प्रकृति के मूलभूत नियतांकों (जैसे प्रकाश की चाल) पर आधारित परिभाषाएं कभी नहीं बदलतीं — ब्रह्मांड में कहीं भी सही उपकरणों के साथ इन्हें सटीक रूप से पुनः मापा जा सकता है।
इन 7 मूल मात्रकों के अतिरिक्त कोण मापने के लिए दो पूरक मात्रक भी हैं: समतल कोण के लिए रेडियन (radian – rad) और घन कोण के लिए स्टेरेडियन (steradian – sr)। ये दोनों “विमाहीन” (Dimensionless) राशियां हैं — अर्थात ये केवल दो समान राशियों के अनुपात हैं (जैसे चाप की लंबाई / त्रिज्या), इसलिए इनकी कोई भौतिक विमा नहीं होती।
प्रश्न: “घन कोण” (Solid Angle) क्या होता है?
उत्तर: एक साधारण समतल कोण यह मापता है कि कागज पर दो रेखाओं के बीच का क्षेत्र कितना चौड़ा है। घन कोण 3D (त्रिविम) स्थान में यह मापता है कि कोई शंकु (cone) कितना फैला हुआ है — जैसे आपकी दृष्टि सीमा में कोई वस्तु कितना स्थान घेरती है। इसे स्टेरेडियन में मापा जाता है।
1.3सार्थक अंक (Significant Figures)
प्रत्येक मापन में कुछ न कुछ अनिश्चितता होती है — कोई भी मापक यंत्र शत-प्रतिशत परिशुद्ध नहीं होता। सार्थक अंक हमें यह बताते हैं कि किसी मान को कितनी यथार्थता के साथ मापा गया है, ताकि हम उपकरण की क्षमता से अधिक यथार्थता का दावा न करें।
दैनिक जीवन का उदाहरण
यदि आपकी रसोई का तराजू केवल निकटतम ग्राम तक ही माप सकता है, तो यह कहना कि “मैंने 250.437 ग्राम चीनी मापी है” गलत और भ्रामक है। सार्थक अंक हमें मापन की वास्तविक परिशुद्धता के प्रति ईमानदार रखते हैं।
सार्थक अंक गिनने के सरल नियम
सभी गैर-शून्य (Non-zero) अंक सार्थक होते हैं। 247 में 3 सार्थक अंक हैं।
दो गैर-शून्य अंकों के बीच आने वाले सभी शून्य सार्थक होते हैं। 1005 में 4 सार्थक अंक हैं।
प्रथम गैर-शून्य अंक से पहले आने वाले प्रारंभिक शून्य सार्थक नहीं होते। 0.0025 में केवल 2 सार्थक अंक हैं (2 और 5)।
दशमलव बिंदु के बिना अंत में आने वाले शून्य सार्थक नहीं होते। 1200 में केवल 2 सार्थक अंक हैं (जब तक कि इसे वैज्ञानिक संकेतन में न लिखा जाए)।
दशमलव बिंदु के बाद अंत में आने वाले शून्य हमेशा सार्थक होते हैं। 3.500 में 4 सार्थक अंक हैं — क्योंकि इन शून्यों को जानबूझकर मापा और दर्ज किया गया है।
वैज्ञानिक संकेतन (Scientific Notation) भ्रम को कैसे दूर करता है?
1200 लिखना अस्पष्ट है — क्या इसमें 2, 3 या 4 सार्थक अंक हैं? परंतु 1.200 × 10³ लिखने से यह पूरी तरह स्पष्ट हो जाता है कि इसमें ठीक 4 सार्थक अंक हैं।
गणितीय गणनाओं के नियम
गुणा/भाग: अंतिम उत्तर में केवल उतने ही सार्थक अंक होने चाहिए जितने कि सबसे कम सार्थक अंक वाली दी गई राशि में हैं।
योग/घटाव: अंतिम उत्तर में दशमलव के बाद केवल उतने ही स्थान होने चाहिए जितने कि सबसे कम दशमलव स्थान वाली दी गई राशि में हैं।
हल किया गया उदाहरण
द्रव्यमान = 4.237 g, आयतन = 2.51 cm³। कैलकुलेटर से भाग देने पर घनत्व 1.68804780876 g/cm³ प्राप्त होता है — लेकिन चूंकि आयतन (2.51) में केवल 3 सार्थक अंक हैं, इसलिए अंतिम उत्तर को भी 3 सार्थक अंकों तक पूर्णांकित (Round off) करना होगा:
अंक “5” को पूर्णांकित करने का नियम (Rounding off rule for 5)
यदि छोड़ा जाने वाला अंक ठीक 5 हो और उसके बाद कोई अंक न हो, तो पूर्ववर्ती अंक को निकटतम सम अंक (Even digit) में बदल दिया जाता है। उदाहरण के लिए 2.745 → 2.74 (क्योंकि 4 पहले से सम है), परंतु 2.735 → 2.74 (3 विषम था, इसलिए बढ़ाकर सम 4 कर दिया गया)।
1.4भौतिक राशियों की विमाएं (Dimensions)
किसी भी भौतिक राशि को मूल राशियों के संयोजन के रूप में व्यक्त किया जा सकता है — मुख्य रूप से द्रव्यमान [M], लंबाई [L], और समय [T] (तथा आवश्यकतानुसार धारा, ताप आदि)। इस संयोजन को उस राशि की विमा (Dimension) कहते हैं।
विमाओं की तुलना किसी पकवान की सामग्री से करें
जिस प्रकार कोई रेसिपी यह बताती है कि “इस डिश में आटा, चीनी और अंडे चाहिए” बिना सटीक मात्रा बताए, उसी प्रकार विमीय सूत्र यह बताता है कि “यह राशि द्रव्यमान, लंबाई और समय से बनी है” बिना उसके आगे की संख्या बताए।
उदाहरण के लिए, चाल “प्रति इकाई समय में तय की गई दूरी” है, इसलिए इसकी विमा लंबाई को समय से भाग देने पर प्राप्त होती है:
जब हम किसी भौतिक राशि को उसकी विमाओं के पदों में व्यक्त करते हैं, तो इसे विमीय सूत्र (Dimensional Formula) कहते हैं। जब इसे समीकरण के रूप में लिखा जाता है, तो यह विमीय समीकरण (Dimensional Equation) कहलाता है।
भौतिक राशि
विमीय सूत्र
आयतन (Volume)
\([\text{M}^0\text{L}^3\text{T}^0]\)
चाल / वेग (Speed/Velocity)
\([\text{M}^0\text{L}^1\text{T}^{-1}]\)
बल (Force)
\([\text{M}^1\text{L}^1\text{T}^{-2}]\)
द्रव्यमान घनत्व (Mass Density)
\([\text{M}^1\text{L}^{-3}\text{T}^0]\)
1.6विमीय विश्लेषण और इसके उपयोग
विमीय विश्लेषण का मुख्य आधार विमाओं की समांगता का सिद्धांत (Principle of Homogeneity of Dimensions) है: “हम केवल उन्हीं भौतिक राशियों को जोड़ या घटा सकते हैं जिनकी विमाएं समान हों।” आप “5 किग्रा” में “3 मीटर” नहीं जोड़ सकते — यह भौतिक रूप से निरर्थक है।
उपयोग 1: समीकरण की सत्यता की जांच करना
मुख्य नियम
यदि किसी समीकरण के दोनों पक्षों के प्रत्येक पद की विमाएं समान हैं, तो समीकरण विमीय रूप से सही (Consistent) है। परंतु ध्यान दें: विमीय रूप से सही समीकरण का पूरी तरह सही होना आवश्यक नहीं है — लेकिन यदि विमाएं समान नहीं हैं, तो समीकरण निश्चित रूप से गलत है।
हल किया गया उदाहरण — \(\frac{1}{2}mv^2 = mgh\) की जांच
वामावर्त पक्ष (LHS): \([\text{M}][\text{L}\text{T}^{-1}]^2 = [\text{M}\text{L}^2\text{T}^{-2}]\)
दक्षिणावर्त पक्ष (RHS): \([\text{M}][\text{L}\text{T}^{-2}][\text{L}] = [\text{M}\text{L}^2\text{T}^{-2}]\)
चूंकि दोनों पक्षों की विमाएं समान हैं, इसलिए यह समीकरण विमीय रूप से सत्य है (यह गिरती हुई वस्तु के लिए ऊर्जा संरक्षण का समीकरण है)।
उपयोग 2: विभिन्न भौतिक राशियों के बीच संबंध (सूत्र) स्थापित करना
यदि हमें यह अनुमान हो कि कोई भौतिक राशि किन-किन अन्य राशियों पर निर्भर करती है, तो विमीय विश्लेषण के उपयोग से हम उनके बीच का सूत्र व्युत्पन्न कर सकते हैं — हालांकि यह सूत्र में आने वाले आंकिक नियतांक (Numerical Constant) का मान नहीं बता सकता।
हल किया गया उदाहरण — सरल लोलक का आवर्तकाल
मान लीजिए कि आवर्तकाल \(T\) लोलक की लंबाई \(l\), द्रव्यमान \(m\), और गुरुत्वीय त्वरण \(g\) पर निर्भर करता है:
\[ T = k\, l^x g^y m^z \]
दोनों पक्षों की विमाओं की तुलना करने पर हमें \(x = \frac{1}{2}\), \(y = -\frac{1}{2}\), तथा \(z = 0\) प्राप्त होता है — अर्थात लोलक का द्रव्यमान आवर्तकाल को बिल्कुल प्रभावित नहीं करता! इससे हमें निम्नलिखित सूत्र प्राप्त होता है:
\[ T = k\sqrt{\frac{l}{g}} \]
विमीय विश्लेषण यह नहीं बता सकता कि नियतांक \(k = 2\pi\) है — इसके लिए प्रयोगों या सैद्धांतिक निगमन की आवश्यकता होती है। परंतु इसने सूत्र की संपूर्ण संरचना का सटीक अनुमान लगा लिया।
विमीय विश्लेषण की सीमाएं (Limitations)
यह सूत्र में आने वाले बीजीय नियतांकों (जैसे 2, ½, या π) का मान ज्ञात नहीं कर सकता।
यह उन विभिन्न राशियों के बीच भेद नहीं कर सकता जिनकी विमाएं समान होती हैं (उदाहरण के लिए: कार्य और बल-आघूर्ण दोनों का विमीय सूत्र समान है, परंतु वे भौतिक रूप से भिन्न हैं)।
यह केवल तभी कार्य करता है जब आपको पहले से ज्ञात हो कि कौन-कौन सी राशियां प्रक्रिया में शामिल हैं।
त्रिकोणमितीय, चरघातांकी (exponential) और लघुगणकीय (logarithmic) पदों वाले सूत्रों का निगमन इससे नहीं किया जा सकता।
संक्षेप में (Quick Recap)
मात्रक ही संख्याओं को भौतिक अर्थ प्रदान करते हैं; SI प्रणाली वैश्विक रूप से स्वीकृत प्रणाली है जिसमें 7 मूल मात्रक हैं।
सार्थक अंक यह दर्शाते हैं कि मापन कितना यथार्थ है — मापक यंत्र की क्षमता से अधिक यथार्थता नहीं लिखनी चाहिए।
गुणा/भाग करते समय न्यूनतम सार्थक अंक तथा योग/घटाव करते समय न्यूनतम दशमलव स्थानों का ध्यान रखें।
विमाएं यह बताती हैं कि कोई राशि द्रव्यमान, लंबाई, समय आदि के किस संयोजन से बनी है।
केवल समान विमाओं वाली राशियों को ही आपस में जोड़ा या घटाया जा सकता है।
विमीय विश्लेषण गलत समीकरणों को पकड़ने और सूत्रों का रूप निर्धारित करने में सहायक है — परंतु यह आंकिक नियतांक का मान ज्ञात नहीं कर सकता।
Chapter 6: How Forces Affect Motion — Detailed Notes | Class 9 Science
Class 9 · Science · NCERT Exploration
Chapter 6: How Forces Affect Motion
Complete Study Notes Newton’s Three Laws, Friction & Momentum Exam-Ready Format
1. The Concept of Force
A force can: (i) make an object move from rest, (ii) change the speed or direction of a moving object, and (iii) change the shape of an object (e.g., kicking a ball, striking a ball with a bat, squeezing a lemon).
Force is a Vector Quantity
Like position, displacement, velocity, and acceleration, force needs both magnitude and direction to be fully specified (e.g., friction acts opposite to motion; gravity acts downward; buoyant force acts upward). SI unit: newton (N) — written with a small “n” as a word, but capital “N” as a symbol.
Exam Tip
Common slip: writing “Newton” for the unit name or “n” for the symbol. Correct: unit name = newton (lowercase), symbol = N (uppercase) — this rule applies to any unit named after a person (e.g., kelvin/K, pascal/Pa).
1.1 Measuring the Magnitude of a Force
A spring balance measures the force with which you pull on the spring inside it — it can measure weight (gravitational force) or force in general.
Threads of Curiosity — Scale of Forces
Everyday smallest perceivable forces: order of millinewtons ($10^{-3}$ N, e.g., a light touch). Scientists can measure forces as small as yoctonewtons ($10^{-24}$ N) in specialised experiments.
2. Balanced and Unbalanced Forces
Usually, more than one force acts on an object simultaneously (e.g., pushing a box: applied force + friction; a floating ball: gravity + buoyant force).
Balanced Forces
Equal in magnitude, opposite in direction → net force = 0. Object’s motion does not change (e.g., tug-of-war with equal pulling teams — rope doesn’t move).
Unbalanced Forces
Net force ≠ 0 → causes a change in the object’s state of motion. The object moves/accelerates in the direction of the larger force.
2.1 Calculating Net Force
6 N10 N
Same direction → Net force = sum of magnitudes, along that direction. Opposite directions → Net force = difference of magnitudes, along the larger force’s direction.
Solved Example 6.1 — Net Force on a Block
Two forces, 10 N and 6 N, act on a block. Find the net force in each case:
(a) Both acting right: $F_{net} = 10 + 6 = 16\ \text{N}$, towards the right.
(b) 10 N right, 6 N left: $F_{net} = 10 – 6 = 4\ \text{N}$, towards the right (direction of larger force).
(c) 6 N right, 10 N left: $F_{net} = 10 – 6 = 4\ \text{N}$, towards the left.
Exam Tip
Key line for definitions: forces equal in magnitude and opposite in direction are called balanced forces; the rope in tug-of-war “does not move if forces are balanced, but moves if forces are unbalanced” — a favourite fill-in-the-blank/reasoning line.
Beyond the Basics
Forces that act at an angle to each other (not parallel/opposite), and equal-and-opposite forces applied to two ends of an extended object (causing rotation — e.g., turning a handlebar or a tap), are studied in higher grades.
3. The Force of Friction
Friction is the force that opposes relative motion (or attempted motion) between two surfaces in contact. It always acts opposite to the direction of motion.
Note — Multiple Forces, One Net Effect
Multiple forces may act on an object (applied force, friction, gravity/weight, normal force), but its motion depends only on the net force. Weight (downward) and normal force (upward, perpendicular to surface) are balanced for an object on a horizontal surface — they don’t affect horizontal motion.
A stack of coins launched by a stretched rubber band travels different distances on different surfaces (wooden table < laminated table < polished marble/tile) even though launched with the same force each time.
Conclusion: smoother surfaces → smaller force of friction → object decelerates more slowly → travels farther before stopping. A spring balance pulling a block confirms this directly: a smaller spring-balance reading (force needed to just start motion) = smaller friction on that surface.
Exam Tip — “What If” ReasoningIf friction disappeared entirely: once set in motion, objects would continue moving forever at constant velocity (per Newton’s First Law) — nothing would ever naturally come to rest by sliding to a stop. This is a classic HOTS/thought-experiment question.
4. Newton’s First Law of Motion (Law of Inertia)
Statement
An object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless a net force acts upon it.
If net force = 0 → acceleration = 0 → object cannot begin to move or change its velocity.
Galileo Galilei
Through thought experiments, argued (17th century) that if a body moves on a horizontal plane with all impediments (friction) removed, it will continue moving indefinitely — challenging the old belief that force is needed to sustain motion.
Isaac Newton
Introduced the term inertia — the tendency of objects to resist a change in their state of rest or uniform motion. Framed the First Law using this idea; presented all three laws of motion in 1687 (Principia).
Exam Tip — Position-Time & Velocity-Time Graphs (No Net Force)At rest: position-time graph = horizontal line at a fixed position; velocity-time graph = horizontal line at $v=0$. Constant velocity: position-time graph = straight sloped line; velocity-time graph = horizontal line at some non-zero $v$.
This is one of the most frequently asked graph-sketching questions in this chapter.
Solved Example 6.2
A person pushes a moving box forward with a force exactly equal to friction. Will the box stop or keep moving? Answer: The two forces are equal and opposite → balanced → net force = 0 → by Newton’s First Law, the box continues moving with constant velocity (it does NOT stop, and does NOT speed up).
5. Newton’s Second Law of Motion
A force produces acceleration. Experiments (pulling a cart with varying weights via a pulley system) show:
For a fixed mass, acceleration increases as net force increases (larger force → larger acceleration)
For a fixed force, acceleration decreases as mass increases (larger mass → smaller acceleration)
Statement
When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of acceleration is directly proportional to net force and inversely proportional to mass.
Newton’s Second Law
$$a = \frac{F}{m} \qquad \text{or} \qquad F = ma$$
Definition of the Newton (N)
If $m = 1\ \text{kg}$ and $a = 1\ \text{m s}^{-2}$: $F = 1\ \text{kg} \times 1\ \text{m s}^{-2} = 1\ \text{kg m s}^{-2} = 1\ \text{N}$ One newton is the force that produces an acceleration of $1\ \text{m s}^{-2}$ on a mass of $1\ \text{kg}$.
Gravitational Force / Weight
$$F = mg \qquad g \approx 9.8\ \text{m s}^{-2}\ (\text{take}\ 10\ \text{m s}^{-2}\ \text{for quick estimates})$$
$g$ does NOT depend on the mass of the object.
Threads of Curiosity — Real vs. Ideal
In real experiments, doubling force doesn’t exactly double acceleration (and vice versa for mass) due to measurement errors and friction between the cart’s wheels and the surface — a good explanation to cite for “why did your experimental ratio not match theory” type questions.
Beyond the Basics — Momentum FormMomentum = mass × velocity (direction same as velocity). The more complete Second Law: the rate of change of momentum of an object is proportional to the net force and occurs in the direction of the net force. This form applies even when mass is not constant.
5.1 Newton’s Second Law in Everyday Life
Catching a fast ball: pulling hands back while catching increases the time over which velocity drops to zero → reduces acceleration → reduces force on hands (less injury)
Airbags: increase the time of impact during a collision → reduce deceleration → reduce force on the passenger
Cracking a coconut: very short stopping time on hitting the ground → very large force → shell breaks
Quick Recall: Longer stopping/impact time → smaller force (catching, airbags, landing mats). Shorter stopping time → larger force (coconut cracking, bullet penetration).
Solved Numericals — Newton’s Second Law
Example 6.4 — Weightlifter’s Force
Barbell: 10 kg on each side + 10 kg bar = 30 kg total.
$$F = mg = 30 \times 9.8 = 294\ \text{N (downward, due to gravity)}$$
To hold it steady, the weightlifter applies an equal and opposite force of 294 N upward.
Example 6.5 — Pushing a Block Against Friction
Block mass = 25 kg; max friction = 50 N. (i) Applied force = 50 N → balances friction exactly → net force = 0 → block stays stationary. (ii) Applied force = 55 N → $F_{net} = 55 – 50 = 5\ \text{N}$
$$a = \frac{F}{m} = \frac{5}{25} = 0.2\ \text{m s}^{-2}$$
$$s = ut + \frac{1}{2}at^2 = (0)(2) + \frac{1}{2}(0.2)(2)^2 = 0.4\ \text{m (forward)}$$
Example 6.6 — Force from a Velocity-Time Graph
Sports car, mass 1500 kg. Graph: 0–5 s (0 → 10 m/s), 5–10 s (constant 10 m/s), 10–15 s (10 → 0 m/s). 0–5 s: $a = \frac{v-u}{t} = \frac{10-0}{5} = 2\ \text{m s}^{-2}$; $F = ma = 1500 \times 2 = 3000\ \text{N}$ (towards east) 5–10 s: constant velocity → $a = 0$ → $F = 0$ N 10–15 s: $a = \frac{0-10}{5} = -2\ \text{m s}^{-2}$; $F = 1500 \times (-2) = -3000\ \text{N}$ (negative sign → force acts towards west, opposite to motion)
6. Newton’s Third Law of Motion
Statement
Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first object.
Exam Tip — Critical Distinction
Action-reaction force pairs act on two different objects — so they never cancel each other (they cannot “balance” since they act on different bodies). This is distinct from balanced forces (which act on the same object).
6.1 Everyday Applications
Paddle → waterWater → paddle & canoe
Canoeing: paddle pushes water backward; water pushes paddle (and canoe) forward with equal force. Pushing harder → larger forward force → higher canoe velocity.
Walking/running: feet push the ground backward; the ground pushes feet forward (via friction) — here friction helps motion rather than opposing it
Climbing a tree: legs push the trunk down; friction pushes the person up — harder on smooth (low-friction) trunks
Rocket launch: engine expels gas downward; gas exerts equal-opposite force on rocket upward; when this exceeds the rocket’s weight, it lifts off
Chandrayaan-3’s Vikram lander: fired its engine in the direction of motion (forward) to produce a backward reaction force, slowing it down for a soft lunar landing
Exam Tip — Grooves & Treads
Grooves on footwear soles and treads on vehicle tyres increase friction between the surfaces — necessary for the reaction force (via friction) that pushes us/vehicles forward. This is why it’s hard to walk on wet polished floors/ice, and risky to drive on wet/snowy roads (friction, hence reaction force, is too low).
6.2 Newton’s Third Law Applies to All Force Types
Not just contact forces — also applies to non-contact forces: two magnets repel/attract each other equally; two charged balloons apply equal electrostatic forces; the Earth and a fruit apply equal gravitational forces on each other.
Note — Equal Forces, Unequal Accelerations
Forces in a third-law pair are always equal in magnitude, but they generally do NOT produce equal accelerations, because $a = F/m$ — if the masses of the two objects differ, their accelerations differ too.
Example 6.7 — Why Doesn’t the Earth Move Towards the Fruit?
The Earth and a falling fruit exert equal and opposite gravitational forces on each other. But the Earth’s mass is astronomically larger than the fruit’s, so $a_{Earth} = F/m_{Earth}$ is extremely tiny — undetectably small — while the fruit visibly accelerates towards Earth.
Example 6.8 — Gun Recoil
0.1 kg bullet fired from a 5 kg gun; force = 2 N.
By Newton’s Third Law, recoil force on gun = 2 N (equal, opposite).
$$a_{gun} = \frac{2\ \text{N}}{5\ \text{kg}} = 0.4\ \text{m s}^{-2} \qquad a_{bullet} = \frac{2\ \text{N}}{0.1\ \text{kg}} = 20\ \text{m s}^{-2}$$
Equal forces, but very different accelerations because the masses differ hugely.
7. Forces on a System of Objects
For two objects connected by a string (e.g., two boxes, masses $m_1$ and $m_2$, pulled by external force $F$), treat them as a single system: internal forces (tension $T$) cancel out within the system; only the external force $F$ matters.
Acceleration of a Connected System
$$a = \frac{F}{\text{mass of system}} = \frac{F}{m_1 + m_2}$$
Exam Tip
The system also has external gravitational force $(m_1+m_2)g$ downward, balanced by normal force $(N_1+N_2)$ upward from the ground — these don’t affect horizontal acceleration. This “treat connected objects as one system” trick massively simplifies numerical problems on connected bodies/systems.
8. Quick Summary Table — All Three Laws
Table 8.1 — Newton’s Laws at a Glance
Law
Statement (essence)
Key formula
Real-life cue
First (Inertia)
No change in motion without net force
$F_{net}=0 \Rightarrow a=0$
Object slides to rest only due to friction
Second
Net force produces acceleration, proportional to F, inverse to m
$F=ma$
Catching a ball softly (increase time, reduce force)
Third
Every action has an equal & opposite reaction (on a different body)
$F_{12} = -F_{21}$
Rowing a canoe, rocket launch, walking
9. Exam Question Bank
A. MCQ / Assertion–Reason (1 mark each)
A table moved at constant velocity by force $F$ across a floor — frictional force = $F$ (equal and opposite, since velocity is constant → net force = 0).[1]
Two identical spring balances connected and pulled from both ends show the same reading — confirms Newton’s Third Law experimentally.[1]
B. Short Answer (2–3 marks)
Why does a canoe move forward when the canoeist pushes water backward with the paddle? Why does it move faster with a harder push?[3]
Explain why airbags reduce injury in a car crash, using Newton’s Second Law.[3]
Distinguish between balanced and unbalanced forces with one example each.[2]
Why do treads on tyres and grooves on shoe soles help us move rather than slip?[2]
C. Numerical Problems
A bullet of mass 50 g moving at 100 m/s penetrates 50 cm into a wooden block and stops. Estimate the stopping force (assume constant deceleration).[3]
A footballer kicks a 0.4 kg ball to a speed of 108 km/h with a force of 800 N. Find the time of contact between foot and ball.[3]
An object of mass 2 kg moving at constant 10 m/s enters a rough patch where friction (7 N) plus an additional opposing force (3 N) act. Find the distance travelled before it stops.[3]
A 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N. Find the initial accelerations of the bullet and the gun.[3]
D. Long Answer / HOTS (4–5 marks)
A tractor pulls a harrow of mass $m_1$ with force $F$ giving acceleration $a_1$, and separately pulls a trolley of mass $m_2$ with the same force $F$ giving acceleration $a_2$. If the tractor now pulls both together (trolley + harrow) with the same force $F$, derive the resulting acceleration in terms of $a_1$ and $a_2$.[5]
Explain, using Newton’s third law, why a bar magnet brought near a compass causes the compass needle to move but the bar magnet appears stationary — even though the forces are equal and opposite.[4]
A sailor jumps forward from a small boat onto the shore. Explain, using Newton’s laws, whether the boat moves, and in which direction.[3]
Explain why a landing mat or sand bed is used in a high jump event, linking your answer to the concept of impact time and force.[3]
At a Glance — Chapter Summary
Force is a vector — needs magnitude AND direction; SI unit = newton (N); measured using a spring balance.
Balanced forces → net force = 0 → no change in motion. Unbalanced forces → net force ≠ 0 → causes acceleration.
Friction opposes relative motion; depends on the nature of the surfaces in contact.
Newton’s First Law: An object at rest stays at rest, an object in motion continues with constant velocity, unless acted on by a net force (Law of Inertia).
Newton’s Third Law: Every action has an equal and opposite reaction, acting on a different object — hence these paired forces never balance each other.
Connected objects can be treated as a single system: $a = F/(m_1+m_2)$ using only external forces.
Real-life applications: catching a ball, airbags, rocket propulsion, walking/rowing (friction/reaction helps motion), tyre treads and shoe grooves.
Chapter 8: Journey Inside the Atom — Detailed Notes | Class 9 Science
Class 9 · Science · NCERT Exploration
Chapter 8: Journey Inside the Atom
Complete Study Notes Atomic Models, Subatomic Particles & Isotopes Exam-Ready Format
1. The Ancient Origins of Atomic Theory
Over 2,000 years ago, thinkers in both ancient India and ancient Greece independently asked: What is everything made up of?
Acharya Kanada (India)
Proposed that matter (dravya) divided repeatedly reaches a stage of smallest, indivisible particles called parmanus, recorded in the Vaisesika Sutras. Parmanus combine to form dyads (pairs) and triads (groups of three), building up the entire material universe.
Leucippus & Democritus (Greece)
Proposed similar indivisible particles called atomos (Greek for “indivisible”) — the origin of the modern word “atom”.
Exam Tip
Remember: both ideas were philosophical/imaginary, not based on experiments. The first scientific, experiment-based atomic theory came from John Dalton in 1808. This distinction (imaginary vs experimental) is a favourite 1-mark question.
Dalton’s Atomic Theory (1808)
All matter is composed of indivisible particles called atoms — the fundamental building blocks of matter that cannot be broken into smaller parts. This became the starting point for modern atomic theory, though we now know atoms ARE divisible.
Following Dalton, scientists sought answers to: What are atoms made of? What would they look like? What makes atoms of one element different from another?
2. Evolution of Atomic Models
Discovery of radioactivity (certain elements emitting invisible energy/particles called radiation) in the late 19th century proved that atoms are not indivisible — they must contain smaller particles.
2.1 Thomson’s Model (1897) — “Plum Pudding”
The Discovery of the Electron
J. J. Thomson studied conduction of electricity through gases at very low pressure using a cathode ray tube (two electrodes in a glass tube with high voltage). He observed rays moving from cathode (–) to anode (+), called cathode rays. Studying their behaviour in electric/magnetic fields, he concluded these are streams of negatively charged particles — much lighter than atoms — later named electrons.
Key finding: the nature of cathode rays was independent of the cathode material and the gas used — proving electrons are a fundamental component of all atoms. Charge of an electron = $-1.602 \times 10^{-19}\,\text{C}$, taken as $-1$ by convention.
Since atoms are neutral overall, Thomson had to explain where the positive charge sits. He proposed the atom as a sphere of positive charge with electrons embedded in it — like a pudding studded with plums, or seeds in a watermelon.
2.2 Testing Thomson’s Model — The Gold Foil Experiment (1911)
Geiger–Marsden Experiment (under Rutherford)
A beam of alpha (α) particles (positively charged; later shown to be helium nuclei — 2 protons + 2 neutrons) was fired at an extremely thin gold foil.
Expected (per Thomson’s model): particles pass straight through or deflect only slightly, since positive charge is spread evenly.
Observed: most particles passed straight through undeflected; some were sharply deflected; a few bounced straight back. This deflection is called scattering — hence also called the α-ray scattering experiment.
Exam Tip — Classic Reasoning QuestionWhy did most α-particles pass through undeflected, while a few bounced back? Most of the atom is empty space (→ undeflected passage); the positive charge and mass are concentrated in a tiny, dense region (→ occasional sharp deflection/bounce-back when an α-particle happens to approach this dense region head-on).
2.3 Rutherford’s Nuclear (Planetary) Model
Based on the gold foil results, Rutherford proposed:
Most of an atom is empty space
All positive charge and most mass is concentrated in a tiny, dense central region — the nucleus
Electrons revolve around the nucleus like planets around the Sun — the planetary model
Scale of the Atom
Diameter of atom $\approx 10^{-10}\,\text{m}$; diameter of nucleus $\approx 10^{-15}\,\text{m}$ — the nucleus is about $10^{5}$ (one lakh) times smaller than the whole atom. If an atom were the size of a cricket ground (~100 m across), the nucleus would be a tiny pepper grain at its centre!
Limitation of Rutherford’s Model
A revolving (accelerating) electron should continuously lose energy, spiral inward, and eventually crash into the nucleus — meaning atoms should collapse. But atoms are stable in reality. This is the single biggest weakness of Rutherford’s model, and it’s exactly what Bohr’s model fixed. This “why doesn’t the electron fall into the nucleus” question is extremely common.
Discovery of the Proton
Rutherford showed the nucleus carries positive charge from particles called protons — much heavier than electrons, with charge equal and opposite to an electron’s. For a neutral atom: number of protons = number of electrons.
2.4 Bohr’s Model (1913) — Fixed Energy Levels
Niels Bohr proposed that electrons:
Move only in fixed circular paths called stationary states, orbits, or shells — not randomly
Each shell has a definite energy — hence called energy levels, labelled K, L, M, N… (or $n = 1, 2, 3, 4…$)
Do not lose energy while moving in a fixed shell (this is the key postulate that fixes Rutherford’s stability problem)
K-shell ($n=1$, closest to nucleus) has the least energy; energy increases with distance from the nucleus
An electron changes shells only by absorbing or releasing energy exactly equal to the difference between the two shells’ energies
Each shell holds only a limited number of electrons
Quick Recall: Dalton (indivisible ball) → Thomson (plum pudding — + sphere with e⁻) → Rutherford (nucleus + empty space, but unstable) → Bohr (fixed energy shells, stable) → Modern (electron cloud/quantum model).
Dalton (1808)
Solid indivisible sphere
Thomson (1897)
Plum pudding model
Rutherford (1911)
Nucleus + empty space
Bohr (1913)
Fixed energy shells
Modern Model
Quantum electron cloud
Threads of Curiosity — Why K, L, M, N?
The naming came from physicist Charles Barkla’s early X-ray experiments — he called the first observed X-ray line “K”, leaving room (unused, as it turned out) for a possible earlier series. Bohr adopted the same notation for atomic shells.
3. Discovery of the Neutron & the Three Subatomic Particles
A puzzle remained: a helium atom (2 protons) has about 4× the mass of a hydrogen atom (1 proton) — not 2×. This suggested another mass-contributing particle existed.
James Chadwick (1932)
Discovered the neutron — mass nearly equal to a proton’s, but electrically neutral (no charge). Found in the nucleus of all atoms except ordinary hydrogen ($^1_1\text{H}$). Neutrons also help bind the nucleus together by countering the mutual repulsion between protons (the nuclear force) — this is why heavier atoms need proportionally more neutrons.
Table 3.1 — The Three Subatomic Particles
Particle
Symbol
Relative Charge
Location
Discovered by
Electron
$e^-$
−1
Revolves around nucleus
J. J. Thomson (1897)
Proton
$p^+$
+1
Inside nucleus
Ernest Rutherford (1911)
Neutron
$n^0$
0
Inside nucleus
James Chadwick (1932)
4. Symbols of Elements
Dalton (1803) introduced pictorial symbols. Berzelius (1813) proposed deriving symbols from Latin names — giving us modern alphabetic symbols. Today, IUPAC approves all element names/symbols internationally.
Rules for Writing Symbols
Often the first letter, or first two letters, of the element’s name
First letter always capital; second letter (if any) always lowercase — e.g., Aluminium = Al (not AL), Cobalt = Co (not CO)
Some symbols use a letter other than the second letter — e.g., Chlorine = Cl, Zinc = Zn
Some come from Latin/Greek/German names — Iron = Fe (Latin ferrum), Mercury = Hg (Greek hydrargyros), Tungsten = W (German wolfram)
Exam Tip
Common mistake: writing AL for aluminium or CO for cobalt (which is actually carbon monoxide!). Always: first letter capital, second letter small.
5. Atomic Number, Mass Number & Notation
Definitions
Atomic Number $(Z)$ = Number of protons in the nucleus $=$ Number of electrons (in a neutral atom)
Mass Number $(A)$ = Number of protons $+$ Number of neutrons (protons and neutrons together = nucleons)
Standard notation for an atom:
$$\ ^{A}_{Z}\text{Symbol} \quad\quad \text{e.g., for Carbon:} \quad\ ^{12}_{6}\text{C}$$
Table 5.1 — First 18 Elements: Composition & Electron Distribution
Element
Symbol
Z
Protons
Neutrons
Electrons
K
L
M
Hydrogen
H
1
1
0
1
1
–
–
Helium
He
2
2
2
2
2
–
–
Lithium
Li
3
3
4
3
2
1
–
Beryllium
Be
4
4
5
4
2
2
–
Boron
B
5
5
6
5
2
3
–
Carbon
C
6
6
6
6
2
4
–
Nitrogen
N
7
7
7
7
2
5
–
Oxygen
O
8
8
8
8
2
6
–
Fluorine
F
9
9
10
9
2
7
–
Neon
Ne
10
10
10
10
2
8
–
Sodium
Na
11
11
12
11
2
8
1
Magnesium
Mg
12
12
12
12
2
8
2
Aluminium
Al
13
13
14
13
2
8
3
Silicon
Si
14
14
14
14
2
8
4
Phosphorus
P
15
15
16
15
2
8
5
Sulfur
S
16
16
16
16
2
8
6
Chlorine
Cl
17
17
18
17
2
8
7
Argon
Ar
18
18
22
18
2
8
8
5.1 Bohr–Bury Rules for Electron Distribution
Maximum electrons in a shell $= 2n^2$ (where $n$ = shell number): K-shell ($n=1$) → 2; L-shell ($n=2$) → 8; M-shell ($n=3$) → 18
Maximum electrons in the outermost shell = 8 (except when it’s the only shell, which holds max 2)
Shells fill stepwise from innermost outward (K → L → M → N…); a shell fills completely before the next one starts
Solved ExampleQ: Write the electronic configuration of Magnesium (Z = 12). A: 12 electrons to distribute: K-shell takes 2 (full) → 10 remain; L-shell takes 8 (full) → 2 remain; M-shell takes the remaining 2. Configuration: 2, 8, 2
6. Valency — Combining Capacity
Key TermsValence shell = outermost shell containing electrons. Valence electrons = electrons in that shell. Octet = 8 electrons in the outermost shell (a stable, largely unreactive configuration; for the first shell, 2 electrons = stable, as in helium).
Valency = number of electrons an atom gains, loses, or shares to complete its octet (achieve stability).
Fewer than 4 valence electrons
Tends to lose electrons. E.g., Sodium (2, 8, 1) loses 1 electron → valency 1.
More than 4 valence electrons
Tends to gain electrons. E.g., Oxygen (2, 6) gains 2 electrons → valency 2.
Exam Tip
Elements with a complete octet already (like Neon, Argon) are stable and generally do not lose/gain/share electrons — this is why noble gases are largely unreactive.
7. Isotopes and Isobars
7.1 Isotopes
Isotopes are atoms of the same element (same atomic number $Z$, same number of protons/electrons) but with different numbers of neutrons, hence different mass numbers $A$.
Table 7.1 — Isotopes of Hydrogen
Isotope
Notation
Protons
Neutrons
Electrons
Natural abundance
Protium
$^1_1\text{H}$
1
0
1
~99.98%
Deuterium
$^2_1\text{H}$
1
1
1
~0.015%
Tritium
$^3_1\text{H}$
1
2
1
traces
Exam Tip
Isotopes have identical chemical properties (same number of electrons/valence electrons) but different physical properties (e.g., boiling/melting points), because chemical behaviour depends on valence electrons, while physical properties depend on mass.
Applications of Isotopes
$^{235}_{92}\text{U}$ — fuel in nuclear reactors (electricity generation)
$^{60}_{27}\text{Co}$ — radiotherapy for cancer treatment
$^{131}_{53}\text{I}$ — treatment of goitre and thyroid cancer
$^{14}_{6}\text{C}$ — radiocarbon dating of fossils/artefacts in archaeology and geology
Average (Weighted) Atomic Mass
Since isotopes don’t occur in equal proportions, the atomic mass of an element is calculated as a weighted average, not a simple average.
Solved Example — Chlorine
Chlorine has two isotopes: $^{35}\text{Cl}$ (75% abundance) and $^{37}\text{Cl}$ (25% abundance).
$$\text{Average atomic mass} = \left(35 \times \frac{75}{100}\right) + \left(37 \times \frac{25}{100}\right)$$
$$= \frac{105}{4} + \frac{37}{4} = \frac{142}{4} = 35.5\ \text{u}$$
Interpretation: No single chlorine atom weighs 35.5 u — but in 1 million chlorine atoms, about 7.5 lakh would be $^{35}\text{Cl}$ and 2.5 lakh would be $^{37}\text{Cl}$, giving this weighted average.
7.2 Isobars
Isobars are atoms of different elements with the same mass number but different atomic numbers.
Table 7.2 — Example: Isobars with Mass Number 40
Element
Atomic Number (Z)
Mass Number (A)
Argon
18
40
Potassium
19
40
Calcium
20
40
Do Not Confuse: Isotopes → SAME element, SAME protons, DIFFERENT neutrons/mass number. Isobars → DIFFERENT elements, DIFFERENT protons, SAME mass number.
8. Beyond Bohr — The Modern View
Bohr’s model, too, was eventually found incomplete. Electrons do not follow fixed, well-defined circular paths. The modern quantum mechanical model describes electrons as existing in “electron clouds” — regions around the nucleus where an electron is most likely to be found, rather than an exact, predictable path. (Detailed treatment is reserved for higher grades.)
9. Scientists to Remember
J. J. Thomson
Discovered the electron via cathode ray studies; Nobel Prize in Physics, 1906; headed the Cavendish Laboratory, Cambridge, and mentored Ernest Rutherford.
Ernest Rutherford
Born in New Zealand; known as the “Father of Nuclear Physics”; discovered the atomic nucleus and the proton; Nobel Prize in Chemistry, 1908; his portrait appears on New Zealand’s $100 banknote.
Niels Bohr
Proposed fixed energy levels/shells to explain atomic stability; Nobel Prize in Physics, 1922.
James Chadwick
Discovered the neutron in 1932 while working under Rutherford at the Cavendish Laboratory; Nobel Prize in Physics, 1935.
Homi Jehangir Bhabha
“Father of the Indian nuclear programme”; established the Tata Institute of Fundamental Research (TIFR) and the Bhabha Atomic Research Centre (BARC) for peaceful uses of atomic energy — electricity generation, agriculture, medical treatment.
10. Exam Question Bank
A. MCQ / Assertion–Reason (1 mark each)
The gold foil experiment showed most alpha particles pass undeflected, proving that most of the atom is empty space (NOT evidence of neutrons — neutrons were discovered separately by Chadwick in 1932).[1]
Assertion: Electrons do not fall into the nucleus in Bohr’s model. Reason: Electrons in a stationary state (fixed shell) do not lose energy. — Both true, R correctly explains A.[1]
The maximum number of electrons in the M-shell is $2n^2 = 2(3)^2 = 18$.[1]
B. Short Answer (2–3 marks)
State the postulates of Bohr’s model of the atom that explain the stability of atoms.[3]
Differentiate between isotopes and isobars with one example each.[3]
Why do atoms of the same element have the same chemical properties even when they are different isotopes?[2]
Explain why the gold foil experiment’s results could not be explained by Thomson’s model.[3]
C. Numerical Problems
An atom has atomic number 26 and 56 nucleons. Find the number of electrons, protons, and neutrons.[2]
An atom has 18 neutrons and atomic number 17. Find its mass number.[2]
Bromine has two isotopes: $^{79}_{35}\text{Br}$ (49.7%) and $^{81}_{35}\text{Br}$ (50.3%). Calculate its average atomic mass.[3]
An atom $^{23}A$ has 11 electrons. Find the number of neutrons in it.[2]
D. Long Answer / HOTS (4–5 marks)
Trace the historical development of atomic models from Dalton to the modern quantum mechanical model, highlighting the key experimental evidence that led to each revision.[5]
Explain, with reasoning, why Rutherford’s model failed to explain atomic stability, and how Bohr’s model resolved this issue.[4]
For an element with mass number 35 and 18 neutrons: (i) find protons and electrons, (ii) find atomic number, (iii) identify the element, (iv) write its electronic configuration, (v) state its valency.[5]
Explain why isotopes of an element have the same valency but the weighted average atomic mass of the element is often a non-whole number.[4]
At a Glance — Chapter Summary
Atomic theory origins: Acharya Kanada’s parmanu and Democritus’s atomos — imaginative, not experimental. Dalton (1808) gave the first scientific atomic theory.
Model evolution: Dalton (indivisible sphere) → Thomson (plum pudding — electrons in positive sphere) → Rutherford (nucleus, mostly empty space, but unstable) → Bohr (fixed energy shells, stable) → Modern (quantum/electron cloud model).
Gold foil experiment (Geiger–Marsden, under Rutherford) → discovered the nucleus via α-particle scattering.
Three subatomic particles: electron ($e^-$, −1, Thomson), proton ($p^+$, +1, Rutherford), neutron ($n^0$, 0, Chadwick).
Atomic number ($Z$) = number of protons; Mass number ($A$) = protons + neutrons (nucleons).
Bohr–Bury rule: max electrons per shell = $2n^2$; outermost shell max = 8.
Valency = electrons lost/gained/shared to complete the octet.
Isotopes: same element, same $Z$, different $A$ (different neutrons) — same chemical, different physical properties.
Isobars: different elements, same $A$, different $Z$.
Average atomic mass = weighted average based on natural isotopic abundance.
Chapter 3: Tissues in Action — Detailed Notes | Class 9 Science
Class 9 · Science · NCERT Exploration
Chapter 3: Tissues in Action
Complete Study Notes Plant & Animal Tissues Exam-Ready Format
1. Introduction — The Idea of a Tissue
A tissue is a group of cells that are similar in structure and work together to perform a specific function. The formation of tissues creates division of labour in the body, which increases efficiency and allows complex life processes to occur.
Hierarchy of Organisation
Cell → Tissue → Organ → Organ System → Organism
In unicellular organisms (e.g., Amoeba), a single cell performs all life functions. In multicellular organisms (plants and animals), different groups of specialised cells carry out different functions — this is why tissues exist.
Exam Tip
A very common 1-mark/2-mark question: “Define tissue and give one example each from a plant and an animal.” Answer: Tissue = group of similar cells performing a specific function. Plant example: xylem (conducts water). Animal example: muscle tissue (movement).
Why Are Plant and Animal Tissues Different?
Table 1.1 — Key structural/functional differences
Basis
Plants
Animals
Movement
Fixed in one place; need support to stay upright
Generally capable of movement (except a few like sponges)
Cell wall
Present — gives rigidity and strength
Absent — allows change of shape, aiding locomotion
Mode of nutrition
Autotrophic — tissues help utilise solar energy via photosynthesis
Heterotrophic — tissues help digest food from external sources
Growth pattern
Localised growth (only at meristems) — growth continues throughout life
Generalised growth — occurs throughout the body, but growth stops after a point
2. Tissues for Growth in Plants
Plants grow in three distinct ways:
Increase in length — height of stem and depth of roots
Increase in girth — thickness of the stem
Regrowth — after cutting of branches or grazing by animals
All three types of growth depend on actively dividing cells, collectively called meristematic tissue. There are three types, based on location and function.
2.1 Apical Meristem — Growth in Length
Key Experiment: Activity 3.1 (Onion Root Growth)
Two onion bulbs are placed in water-filled jars (Jar A and Jar B). Root growth is measured daily. On Day 3, the root tips of Jar B are cut by about 1 cm.
Observation: Roots in Jar A continue growing in length steadily. Roots in Jar B stop growing after their tips are removed.
Conclusion: Roots grow only from their tips, which contain cells that divide continuously (confirmed earlier by observing mitosis in onion root tips).
This growth zone at the tips of roots and shoots is called the apical meristem. It is responsible for increase in length of the plant.
Exam Tip — Diagram-Based Question
Be ready to label a sapling diagram showing shoot apical meristem (tip of shoot) and root apical meristem (tip of root). This is a frequently asked 2-mark diagram question.
2.2 Lateral Meristem — Growth in Girth
Located along the circumference of the stem (in a ring), this meristem produces new cells inside and outside in a concentric manner, increasing the stem’s diameter. This is called the lateral meristem (also linked to formation of annual growth rings seen in a cut tree trunk).
Application: Age of a Tree
Counting annual rings lets scientists estimate a tree’s age and the climatic conditions (favourable/unfavourable years) it experienced — wide rings = good growth year; narrow rings = poor growth year.
2.3 Intercalary Meristem — Regrowth After Cutting
Located at the base of the internode, or just above a node (the point where branches/leaves arise). When the tip of a stem is cut, growth in length stops, but new branches sprout from the nodes — this is why hedges become bushier after trimming and grass regrows after mowing/grazing.
Large, prominent nucleus and dense cytoplasm rich in organelles
Vacuoles generally absent
Tightly packed with little/no intercellular space
These features allow rapid and continuous cell division.
Differentiation
Cells that lose the ability to divide undergo structural/functional change to become specialised — this process is called differentiation. Meristematic tissue becomes permanent tissue through differentiation.
3. Permanent Tissues in Plants
Permanent tissues are classified as:
Simple — composed of only one type of cell (parenchyma, collenchyma, sclerenchyma)
Complex — composed of more than one type of cell working together (xylem, phloem)
3.1 Protective Tissue — Epidermis
The epidermis is the outermost layer of the plant body — a single, tightly packed layer of flat, rectangular cells. It is covered by a waxy layer called the cuticle, which reduces water loss. In roots, epidermal projections form root hairs that increase surface area for water/mineral absorption. In leaves, the epidermis contains pores called stomata, which allow gaseous exchange and transpiration (evaporation of water vapour), creating a transpiration pull that helps move water up through the xylem.
HOTS QuestionWhy is a thick cuticle advantageous for a desert plant but disadvantageous for an underwater plant? — A thick cuticle minimises water loss by transpiration (vital in dry conditions), but an aquatic plant needs gas exchange directly through its surface, so a thick waterproof cuticle would hinder that exchange.
Exam Tip
Classic assertion-reason style question: fresh green twigs bend (collenchyma provides flexibility) but dry/woody twigs snap (sclerenchyma is rigid due to lignin, cannot bend).
Companion cells regulate loading/unloading of sugars in sieve tubes
Frequently Confused
Xylem = water/minerals, mostly dead cells (except xylem parenchyma). Phloem = food, mostly living cells. If a question says “food transport tissue malfunctioning” → answer is Phloem, not Xylem.
3.4 Plant Tissue Systems
All plant tissues are organised into three larger tissue systems:
Dermal tissue system — outer covering; protects and reduces water loss (epidermis)
Ground tissue system — main body between dermal and vascular tissue; includes parenchyma, collenchyma, sclerenchyma
Vascular tissue system — conducting tissues: xylem and phloem
4. Animal Tissues
Animal tissues are broadly of four types: Epithelial, Connective, Muscular, and Nervous.
4.1 Epithelial Tissue
Forms the outer covering of the body (skin) and lines internal organs (mouth, lungs, blood vessels, intestine). Cells are closely packed with minimal intercellular space, preventing entry of germs and reducing water loss.
Table 4.1 — Types of Epithelial Tissue (structure–function relationship)
Function
Structure
Location
Exchange (diffusion of gases/liquids)
Single layer of thin, flat cells (squamous)
Lining of blood vessels, lungs
Protection
Many layers; outer cells flat and tightly packed (stratified)
Skin, mouth, oesophagus
Secretion
Cuboidal/columnar cells specialised for producing & releasing substances
Salivary glands, sweat glands, stomach lining
Sensory function
Specialised receptor cells with hair-like cilia
Nostrils, taste buds, inner ear
Absorption
Single layer of tall, pillar-like (columnar) cells, often with hair-like structures
Lining of small intestine
Exam TipWhy is epithelium lining the lungs/blood vessels only one cell thick? — To allow rapid/efficient diffusion of gases and liquids across the tissue; a thicker layer would slow down exchange.
4.2 Connective Tissue
Connects and supports other tissues/organs. Composition varies in the matrix (the substance between cells) — from fluid (blood) to hard (bone).
Platelets — help in blood clotting at injury sites
4.3 Muscular Tissue
Table 4.3 — Three Types of Muscle
Type
Cell shape
Nuclei
Striations
Control
Location
Skeletal (striated)
Long, cylindrical, unbranched
Multinucleate
Present (striated)
Voluntary
Attached to skeleton (limbs, etc.)
Smooth (unstriated)
Spindle-shaped
Single nucleus
Absent
Involuntary
Stomach, intestines (slow, continuous movement)
Cardiac
Cylindrical, branched
Single nucleus
Faint striations
Involuntary
Only in the heart
Exam Tip — Assertion/Reason styleAssertion: Cardiac muscle can contract continuously without fatigue. Reason: Cardiac muscle cells have abundant mitochondria and a rich blood supply, sustaining continuous energy production. (Both true and R correctly explains A.)
4.4 Nervous Tissue
Forms the body’s control and coordination network. Made up of neurons (nerve cells), specialised to receive, process, and transmit messages.
Structure of a Neuron (label carefully for diagram questions)
Cell body — contains nucleus; controls cell activities
Dendrites — receive signals from other neurons
Axon — long fibre carrying messages away from the cell body, ending in axon terminals that transmit signals to other cells
5. The Musculoskeletal System
Made up of bones, muscles, joints, cartilage, tendons, and ligaments. It helps the body stand upright, move, maintain posture, and protect delicate organs — functioning under control of the nervous system. Muscles pull on bones via tendons; muscle contraction → tendon transmits force → movement at a joint.
Quick Fact
On average, the adult human skeleton makes up about 12–15% of body weight (varies with age, gender, body composition).
5.1 Types of Joints
Table 5.1 — Joint types, movement, and location
Joint Type
Movement Allowed
Example Location
Ball and socket
Free movement — forward, backward, sideways, circular
Shoulder (rounded arm bone fits shoulder bone hollow)
Hinge
Movement in one direction only (like a door hinge)
Elbow, knee
Pivot
Side-to-side rotation (like a doorknob)
Joint between skull and backbone (neck)
Fixed
No movement
Skull bones (protect brain)
Common Mistake to AvoidAssertion: In a hinge joint, movement occurs primarily in one plane. Reason: The bone ends are shaped to allow sliding in all directions. — This Reason is FALSE; hinge joints restrict movement to one plane precisely because the bone ends do NOT allow sliding in all directions (that describes a ball-and-socket joint).
5.2 Skeletal System — Backbone & Rib Cage
Vertebral column (backbone) — series of small bones called vertebrae; supports the body, allows upright posture. Cartilage discs between vertebrae cushion and allow flexibility while protecting the spinal cord.
Rib cage — 12 pairs of ribs; attached to the spine (back) and sternum/breast bone (front) via flexible cartilage. This flexibility allows the rib cage to expand/contract during breathing.
6. Important Named Concepts & Scientists
Totipotency (F. C. Steward, 1958)
Demonstrated that single cells from carrot phloem could regenerate an entire plant. Phloem cells first dedifferentiate (regain ability to divide) forming an unspecialised cell mass, which then divides and redifferentiates into roots, shoot, and a complete plant. This ability of a mature cell to give rise to a whole organism is called totipotency, and such cells are totipotent cells — similar to how a zygote develops into a full organism.
Crown Gall Disease
Caused by the bacterium Agrobacterium tumefaciens; produces tumour-like swellings on stems due to uncontrolled cell division. Scientists studied how this bacterium transfers genetic material into plant cells — this knowledge is now used as a tool in plant tissue culture and genetic engineering to introduce useful genes into crops.
B. G. L. Swamy
Renowned Indian botanist known for contributions to plant morphology and anatomy; authored Hasuru Honnu (Kannada), which won the Kendra Sahitya Akademi Award (1978).
Sipra Guha Mukherjee & S. C. Maheshwari
Achieved a breakthrough in plant tissue culture — developed a complete plant through anther culture using an artificial nutrient medium under controlled conditions, contributing significantly to crop improvement.
7. Exam Question Bank
A. Multiple Choice / Assertion-Reason (1 mark each)
Meristematic tissues can divide repeatedly because their cells have thin walls, dense cytoplasm, and a large prominent nucleus (NOT because they have thick walls or large vacuoles).[1]
If a plant cannot transport food from leaves to roots, the malfunctioning tissue is Phloem.[1]
Epithelial tissue lining internal organs is usually one/few cells thick to allow quick exchange of materials by diffusion.[1]
Bending knees and ankles involves a Hinge joint.[1]
B. Short Answer (2–3 marks)
Distinguish between xylem and phloem on the basis of function and cell nature (living/dead).[3]
Why does a fresh green twig bend without breaking, while a dry twig snaps? Name the tissues responsible.[2]
Differentiate between a tendon and a ligament with one example/function of each.[2]
Explain why cardiac muscle can work continuously throughout life without fatigue.[2]
What is meant by differentiation in the context of meristematic tissue?[2]
C. Long Answer / HOTS (4–5 marks)
An elephant severely debarks a tree to feed on its bark. (a) Which function(s) of the tree are hampered? (b) Which tissue would be further affected if damage continues beneath the bark? (c) What function would be hampered if tissue beneath the bark is severely damaged? — Hint: bark removal damages phloem (just under the bark), disrupting food transport; further damage into the wood affects xylem (water transport) and the lateral meristem (girth growth).[5]
Explain the totipotency experiment of F. C. Steward using carrot phloem cells. What does this reveal about the potential of mature plant cells?[5]
With reference to Sohan’s sugarcane cutting experiment (Type A vs Type B), identify the likely difference between the cuttings (presence of a node/bud with meristematic tissue) and explain why only Type B could sprout.[4]
Explain the three types of meristematic tissues with their location and function. How do they together account for a plant’s overall growth in length, girth, and branching?[5]
Describe the structure and function of a neuron, and explain how nervous tissue coordinates both voluntary and involuntary muscle activity.[4]
D. Numerical / Graph-Based
Using the teak tree data (age vs. diameter vs. annual rings), plot the graph and explain the relationship between diameter and number of annual rings formed. Identify the tissue responsible for girth and its location.[5]
At a Glance — Chapter Summary
Tissue = group of similar cells performing specific functions; enables division of labour.
