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Students can use Class 8 Malayalam Adisthana Padavali Notes Pdf and കുരുവിയും കാട്ടുതീയും Kuruviyum Kaattu Theeyum Summary in Malayalam to grasp the key points of a lengthy text.

Class 8 Malayalam Kuruviyum Kaattu Theeyum Summary

Kuruviyum Kaattu Theeyum Summary in Malayalam

കുരുവിയും കാട്ടുതീയും Summary in Malayalam

എഴുത്തുകാരനെ പരിചയപ്പെടാം

ആനന്ദിന്റെ കഥകൾ – സച്ചിദാനന്ദ്
തൂലികാനാമം – ആനന്ദ്

പ്രശസ്തനായ ഒരു മലയാള നോവലിസ്റ്റും എഴുത്തുകാരനുമാണ് ആനന്ദ് എന്നറിയപ്പെടുന്ന പി. സച്ചിദാനന്ദൻ. 1936 ൽ ഇരിങ്ങാലക്കുടയിലാണ് ജനിച്ചത്. തിരുവനന്തപുരം എൻജിനീയറിങ്ങ് കോളേജിൽ നിന്ന് സിവിൽ എൻജിനീ യറിങ്ങിൽ ബിരുദം. നാലുകൊല്ലത്തോളം പട്ടാളത്തിൽ സേവനമനുഷ്ഠി ച്ചിട്ടുണ്ട്. ന്യൂഡെൽഹിയിൽ സെൻട്രൽ വാട്ടർ കമ്മീഷനിൽ പ്ലാനിങ്ങ് ഡയറ ക്ടറായി വിരമിച്ചു. ശില്പ കലയിലും തത്പരനായ ആനന്ദിന്റെ പല നോവലു കളിലും മുഖച്ചിത്രമായി അദ്ദേഹം നിർമിച്ച ശില്പങ്ങളുടെ ഫോട്ടോയാണ് ഉപയോഗിച്ചിട്ടുള്ളത്. 2016 ലെ കൊച്ചിൻ മുസിരിസ് ബിനലയിൽ അദ്ദേഹം ശിൽപ്പങ്ങൾ പ്രദർശിപ്പിച്ചിരുന്നു.

വീടും തടവും, ജൈവമനുഷ്യൻ ഇവ കേരള സാഹിത്യ അക്കാദമി അവാർഡും മരുഭൂമികൾ ഉണ്ടാകുന്നത് വയലാർ അവാർഡും ഗോവർദ്ധനന്റെ യാത്രകൾ 1997ലെ കേന്ദ്ര സാഹിത്യ അക്കാദമി അവാർഡും നേടി. മഹാശ്വേതാദേവിയുടെ ‘കവി ബന്ദ്യഘടിഗായിയുടെ ജീവിതവും മരണവും’ എന്ന കൃതിയുടെ മലയാള വിവർത്തനത്തിന് 2012ൽ കേന്ദ്രസാഹിത്യ അക്കാദമി പുരസ്കാരം ലഭിച്ചു. 2019 ലെ എഴുത്തച്ചൻ പുരസ്കാരം ലഭിച്ചു.

വ്യവസ്ഥകളിലും ശീലങ്ങളിലും ക്രമപ്പെട്ടുപോയ മനുഷ്യരുടെ അകമേനിന്നു പുറപ്പെടുന്ന ഒച്ചയാണ് ആനന്ദിന്റെ കഥകൾ. കരച്ചിലോ, വിലാപങ്ങളോ അല്ല; സമൂഹത്തിന്റെയും ചരിത്രത്തിന്റെയും രാഷ്ട്രീയ ത്തിന്റെയും ജീർണ്ണസത്തകളിലേക്കുള്ള ശ്രദ്ധ ക്ഷണിക്കലാണ് ആ ശബ്ദം. ആ ഒച്ചകൾ നമുക്കു ചുറ്റും പ്രതിധ്വനിക്കുന്നു. സമകാലീനതയുടെ ആത്മകഥകളെന്നു വിശേഷിപ്പിക്കാ വുന്നവയാണ് ആനന്ദിന്റെ ശ്രദ്ധേയങ്ങളായ രചനകൾ.

പാഠസംഗ്രഹം

‘കുരുവിയും കാട്ടുതീയും’ ഏറെ സമകാലീന പ്രസക്തിയുള്ള ആഴമുള്ള സാമൂഹിക ചിന്ത കാഴ്ചവയ്ക്കുന്ന കഥയാണ് പ്രകൃതിയിലെ ദുരന്തമായി കാട്ടുതീ പടർന്നപ്പോൾ മൃഗങ്ങളും പക്ഷികളും ജീവനുവേണ്ടി ഓടി രക്ഷപ്പെടുന്നു. എന്നാൽ ഒരു കൊച്ചുകുരുവി മാത്രം, സ്വല്പമെങ്കിലും പ്രതിരോധിക്കാൻ ശ്രമിക്കുന്നു. തന്റെ ചെറുചുണ്ടിൽ വെള്ളം എടുത്ത് തീയിൽ തളിച്ച് തീ കെടുത്താൻ കുരുവി കഠിന ശ്രമം തുടരുന്നു. വലിയ കാട്ടുതീ അണയ്ക്കാൻ ചെറിയ ചുണ്ടിൽ വെള്ളം തേവുന്ന കുരുവിയെ കണ്ടു മഴദേവൻ പരിഹസിക്കുമ്പോൾ, കുരുവി മറുപടി പറയുന്നു: ‘ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു.’ മഴമേഘത്തിന് ‘തീയണയ്ക്കാൻ’ കഴിവുണ്ടായിട്ടും പ്രവർത്തിക്കാത്തിനെ വ്യംഗ്യമായി വിമർശിക്കുന്നു കുരുവി.

ഈ വാചകം കഥയുടെ ആത്മാവാണ്. മനുഷ്യൻ ശക്തരായും ശേഷിയുള്ളവരായും ഇരിക്കെ ഭൂമിയിലെ നശീകരണങ്ങളോടും ദുരന്തങ്ങളോടും ഉള്ള അവന്റെ മനോഭാവത്തെ ചോദ്യം ചെയ്യുകയാണ് ഈ കുരുവിയുടെ വാക്കുകളിലൂടെ കഥാകാരൻ. ഇവിടെ സാമാന്യേന ശേഷി കുറഞ്ഞ കൊച്ചുകുരുവി പോലുള്ള ജീവിയും സ്വന്തം കഴിവിനൊത്ത് വല്ലതും ചെയ്യുകയാണ്. അതിനുവേണ്ടിയാണ് കുരുവി ശ്രമിക്കുന്നത് പല നിർണായക സന്ദർഭങ്ങളിലും മനുഷ്യർ നമ്മൾ ഒരാൾ വിചാരിച്ചാൽ ഒന്നും സാധിക്കുകയില്ലെന്ന് പറഞ്ഞു മൗനദർശിയായി നിൽക്കാറുണ്ട്. ചെറുശേഷിയുള്ളതായിരുന്നാലും തന്റെ പ്രവർത്തി മറ്റുള്ളവരിൽ ഉത്തരവാദിത്ത ബോധം വളർത്തുമെന്നും മൃഗങ്ങൾ തന്നെപ്പോലെ ചിന്തിച്ചാൽ കാട്ടുതീ അണയ്ക്കാം എന്നുമുള്ള ബോധ്യമാകുരുവിക്ക് ഉണ്ടായിരിക്കണം. (നമ്മുടെ സ്വാതന്ത്ര്യ സമര ചരിത്രത്തിലെ ദണ്ഡിയാത്ര ഓർമ്മിക്കുക.) ഒരാളുടെ ചെറിയ പ്രവർത്തനം പോലും വലിയ പ്രചാരണങ്ങൾക്ക് വഴിവെക്കാം എന്നതിന്റെ മാതൃക ചിത്രമാണ് ഈ കുരുവി. അനാസ്ഥയുടെയും ഉപേക്ഷയുടെയും ആധുനിക സമൂഹ മനോഭാവത്തിന് എതിരായ പ്രതിരോധം തന്നെയാണിത്.

ഇന്നത്തെ കാലത്ത് പരിസ്ഥിതി പ്രശ്നങ്ങൾ, സാമൂഹിക അനീതികൾ, യുദ്ധങ്ങൾ, സാംസ്കാരിക മൂല്യച്ചുതി തുടങ്ങിയ പ്രശ്നങ്ങൾ മനുഷ്യന്റെ മുന്നിലുണ്ട്. അവയെ തടയാൻ പുതിയ തലമുറ മൗനപരമായ മനോഭാവം സ്വീകരിക്കുമ്പോൾ, അങ്ങേയറ്റത്തെ അനാസ്ഥയും നിരുത്തരവാദപരവും, പുരോഗമന കാഴ്ചപ്പാടുകളുടെ മറവിൽ നടത്തുന്ന ചൂഷണവും ഭൂമിയെ ആകെ ഒരു കാട്ടുതീ പോലെ പടർന്നുപിടിക്കുന്നത് കഥാകാരൻ ഈ കഥയിൽ ചേർത്ത് വച്ചിരിക്കുന്ന സത്യമാണ്.

‘കുരുവിയും കാട്ടുതീയും’ എന്ന കഥയിലൂടെ ആനന്ദ് സമകാലീന സമൂഹത്തിന് മുന്നിൽ ആധുനിക ചോദ്യങ്ങൾ ഉന്നയിക്കുന്നു: ‘നിനക്ക് എന്ത് ചെയ്യാനാകും,’, ‘നിനക്ക് അതിന് മനസ്സുണ്ടോ?’ അതിനുള്ള ഉത്തരം കഥാകൃത്ത് നല്കുന്നില്ല; മറിച്ച്, അത് വായനക്കാരനോട് തന്നെ ആവശ്യപ്പെടുകയാണ്. ചെറുതെങ്കിലും ഒരു കുരുവിയുടെ നീക്കം, വലിയൊരു മാറ്റത്തിന് തുടക്കമാകാമെന്നത് ഈ കഥയുടെ ഉദാത്തമായ സന്ദേശമാണ്.

പുതിയ പദങ്ങൾ

പ്രാണവായു = പ്രാണൻ നിലനിർത്തുന്ന വായു (ഓക്സിജൻ)
ജീവജലം = ജീവദായകമായ ജലം
അന്നം = ആഹാരം, ഭക്ഷ്യവസ്തു
മടയത്തം = വിഡ്ഢിത്തം / വിഫലമായ ശ്രമം
ആളിക്കത്തുന്ന = വളരെ വേഗത്തിൽ കത്തുന്ന വേഗത്തിൽ വ്യാപിക്കുന്ന

Practicing with Std 8 Malayalam Adisthana Padavali Notes Unit 1 Chapter 1 കുരുവിയും കാട്ടുതീയും Kuruviyum Kaattu Theeyum Notes Questions and Answers Pdf improves language skills.

കുരുവിയും കാട്ടുതീയും Question Answer Notes Std 8 Malayalam Adisthana Padavali Chapter 1

Class 8 Malayalam Adisthana Padavali Unit 1 Chapter 1 Notes Question Answer Kuruviyum Kaattu Theeyum

Class 8 Malayalam Kuruviyum Kaattu Theeyum Notes Questions and Answers

പാഠപുസ്തകത്തിലെ ചോദ്യങ്ങളും ഉത്തരങ്ങളും

കഥപറയാം

Question 1.
കഥ മൗനമായി വായിക്കൂ. സംഘമായി തിരിഞ്ഞ് ഒരാൾ ഒരുവാക്യം എന്ന ക്രമത്തിൽ ആശയം ചോർന്നുപോകാതെ പറയൂ.
Answer:
കൂട്ടുകാരെ മുകളിൽ പറഞ്ഞതുപോലെ ചെയ്യുമല്ലോ…

അഭിപ്രായ കുറിപ്പ്

Question 1.
കാട്ടുതീ ഒത്തിരി പക്ഷികളുടെയും മൃഗങ്ങളുടെയും ജീവനെടുക്കുന്നത് കാണുന്നില്ലേ? എന്ന് കുരുവി മഴ ദേവനോട് ചോദിക്കുന്നു എന്നാൽ മഴ ദേവൻ എന്താണ് ചെയ്തത് രണ്ടുപേരുടെയും മനോഭാവങ്ങളിൽ എന്ത് വ്യത്യാസമാണുള്ളത്? നിങ്ങൾ ആരുടെ ഭാഗത്ത് നിൽക്കുന്നു? നിങ്ങളുടെ അഭിപ്രായം എഴുതുക.
Answer:
കുരുവി തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നപ്പോൾ മഴദേവൻ അവളെ പരിഹസിക്കുന്നു. ഇത് രണ്ടുവിധമുള്ള മനോഭാവങ്ങൾ പ്രതിപാദിക്കുന്നു:

മഴദേവൻ: പ്രതീക്ഷയില്ലാത്ത, സംഭവിച്ചുകൊണ്ടിരിക്കുന്ന സാഹചര്യങ്ങൾ അഥവാ സംഭവങ്ങൾ മാറ്റാൻ കഴിയില്ലെന്ന് കരുതുന്ന, മറ്റുള്ളവരുടെ ശ്രമത്തെ പരിഹസിക്കുന്ന, വിഫലം എന്ന് കരുതുന്ന, നോക്കിക്കൊണ്ടിരിക്കാൻ മാത്രം തയ്യാറുള്ള മനോഭാവത്തോട് കൂടിയവരെ പ്രതിനിധീകരിക്കുന്നു.

എന്നാൽ കുരുവി: ധൈര്യവും ഉത്തരവാദിത്വബോധവും നിറഞ്ഞവൾ ആണ്. വലിയത് ചെയ്തില്ലെങ്കിലും കഴിയുന്നതുമാത്രം ചെയ്യുന്നവൾ.സഹജീവികൾക്കായി കരുതുന്നവൾ. സ്വതന്ത്രമായി നീതിനിമിത്തം പ്രവർത്തിക്കുന്നവൾ. കുഞ്ഞിക്കിളിയുടെ പ്രവൃത്തി മാതൃകാപരമാണ്. അതുകൊണ്ടു തന്നെ ഞാൻ കുഞ്ഞിക്കിളിയുടെ ഭാഗത്തു നിൽക്കുന്നു. പ്രതീക്ഷയും കരുണയും ഉള്ള ഈ ചെറിയ പക്ഷിയുടെ മനോഭാവമാണ് യഥാർത്ഥത്തിൽ മനുഷ്യനാകാനുള്ള ആദ്യപടി. മാറ്റങ്ങൾ ചെറുതായിരുന്നാലും, അവ തുടങ്ങേണ്ടത് നമ്മിൽ കൂടെയാണ്. നോക്കിക്കൊണ്ടിരിക്കുന്നതിലല്ല, പ്രവർത്തനത്തിലായിരിക്കണം നമ്മുടെ പങ്ക്.

കഥയെഴുതൂ, പതിപ്പാക്കു

Question 1.
ഈ കഥയിലെ കുരുവിയും മഴ ദേവനും തമ്മിലുള്ള സംഭാഷണം ശ്രദ്ധിച് കുരുവിയും കാടും തമ്മിലുള്ള സംഭാഷണം ആയിരുന്നുവെങ്കിൽ കഥാഗതിയിൽ എന്തു മാറ്റമാണ് സംഭവിക്കുക സങ്കൽപ്പിച്ച് കഥയെഴുതി പതിപ്പാക്കും…
കുരുവിയും കാടും തമ്മിലുള്ള സംഭാഷനത്തിൽ സംഭവിക്കുന്ന മാറ്റം.
Answer:
സങ്കല്പ കഥ:
• കാട്ടുതീ വലിയ ശബ്ദത്തോടെ കാട്ടിലാകെ പടരുമ്പോൾ, കുരുവി തന്റെ കുഞ്ഞി ചുണ്ടുകളിൽ പുഴയിൽ നിന്ന് വെള്ളം കൊണ്ടുപോയി തീ കെടുത്താൻ ഒരു വിഫല ശ്രമം നടത്തുകയാണ് അതു കണ്ട് കാട് കുരുവിയോട് പറഞ്ഞു:

കാട്: കുരുവി തീ പടർന്നു പിടിക്കുന്നിടത്ത് നീ എന്താണ് ചെയ്യുന്നത്? നിന്റെ കുഞ്ഞി ചിറകുകൾക്ക് പൊള്ളലേൽക്കും മുമ്പ് ഇവിടെ നിന്നും ദൂരേക്ക് പറന്നു പോകൂ…

കുരുവി: ഞാൻ വസിക്കുന്ന കാട് ആണ് നീ. നിന്നെ ഈ അവസ്ഥയിൽ ഉപേക്ഷിച്ച് ഞാൻ എങ്ങനെ പോകും? (അതും പറഞ്ഞ് കുരുവി വീണ്ടും പുഴയിലേക്ക് ചെന്ന് വെള്ളവുമായി വന്നു)

കാട്: ‘നിന്റെ സ്നേഹം ഞാൻ അറിയുന്നു കുരുവി പക്ഷേ നിനക്ക് ഈ തീ അണയ്ക്കുവാൻ ആവുകയില്ല നീ നിന്റെ വലിപ്പവും കഴിവും മനസ്സിലാക്കണം. ഞാൻ നിന്റെ മാത്രമല്ല മറ്റു ജീവജാലങ്ങളുടെയും കാടാണ്. പക്ഷേ എല്ലാവരും സ്വന്തം ജീവന് വേണ്ടി പരക്കം പായുന്നത് നീ കാണുന്നില്ലേ…?
നിന്റെ കുഞ്ഞി ചുണ്ടിലെ രണ്ടിറ്റു വെള്ളം കൊണ്ട് ഈ തീ അണയുകയില്ല. സ്വന്തം ജീവനെങ്കിലും രക്ഷപ്പെടുത്തി കൊള്ളൂ ഇനിയും ഒരു കാട് ദൂരെയെങ്ങാനും കാണും.

കുരുവി: ‘നിന്റെ വേദന എനിക്ക് മനസ്സിലാകുന്നു. നിന്റെ സ്നേഹവും. തീ നിന്നെ തിന്നുമ്പോൾ ഞാനെന്തെങ്കിലും ചെയ്യേണ്ടതല്ലേ? എനിക്കാവുന്നതു ഞാൻ ചെയ്യുന്നു. നിന്റെ വിയോഗം നോക്കി എനിക്ക് മൗനമായി ഇരിക്കാൻ കഴിയില്ല. ഞാൻ പ്രത്യാശിക്കുകയാണ്. എന്റെ പ്രവർത്തി മറ്റുള്ളവരെ ചിന്തിപ്പിച്ചിരുന്നുവെങ്കിൽ അവരും എന്നോട് ചേർന്ന് നിന്നെ സംരക്ഷിക്കാൻ പ്രയത്നിച്ചിരുന്നു എങ്കിൽ… ഇവിടന്നങ്ങോട്ട് ഈ കാടിനെ സംരക്ഷിക്കാൻ, തീ പടരുന്നത് തടഞ്ഞു നിർത്തുവാൻ കഴിയുമായിരിക്കാം. എന്റെ ശ്രമം മറ്റുള്ളവർക്കും പ്രചോദനമാകുമെങ്കിൽ…

കാട്: ഞാൻ നിന്നിൽ അഭിമാനിക്കുന്നു…. നിനക്ക് വാസസ്ഥലമായതിലും നിന്നെ ഊട്ടിയതിലും.

കുരുവി : നീ നിന്റെ ഉത്തരവാദിത്വം നിർവഹിച്ചിരിക്കുന്നു. ഞാനും നിന്നെപ്പോലെ…… (മഴപെയ്തതായും തീ അണഞ്ഞതായും സങ്കൽപ്പിക്കാം. മറ്റു മൃഗങ്ങൾ ഒത്തുചേർന്ന് തീ അണച്ചതായും സങ്കൽപ്പിക്കാം കഥ കൂട്ടുകാരുടെ ആശയത്തിനൊത്ത് അവസാനിപ്പിക്കാം).

ആശയവ്യത്യാസം

Question 1.
ഇത് പലതവണ ആവർത്തിച്ചു
ഇത് പലതവണ ആവർത്തിക്കും
ഇത് പലതവണ ആവർത്തിക്കുന്നു
അടിവരയിട്ട് പദങ്ങളുടെ പ്രയോഗം വാക്യങ്ങളുടെ ആശയതലത്തിൽ ഉണ്ടാക്കുന്ന മാറ്റം എന്ത് അതിന്റെ കാരണം ചർച്ച ചെയ്യുക ഇത്തരം പദങ്ങൾ കൂടുതൽ കണ്ടെത്തി എഴുതുക.
Answer:
പദപ്രയോഗങ്ങളുടെ വ്യത്യാസവും അർത്ഥ വ്യത്യാസങ്ങളും

വാക്യങ്ങൾ:

1. ഇത് പലതവണ ആവർത്തിച്ചു → കഴിഞ്ഞിട്ടുള്ള പ്രവൃത്തി (ഭൂതകാലം)
2. ഇത് പലതവണ ആവർത്തിക്കും → വരാനിരിക്കുന്ന പ്രവൃത്തി (ഭാവികാലം)
3. ഇത് പലതവണ ആവർത്തിക്കുന്നു → ഇപ്പോൾ നടക്കുന്നതായ പ്രവൃത്തി (വർത്തമാനകാലം)

പദപ്രയോഗം	കാലം	ആശയം
ആവർത്തിച്ചു	ഭൂതകാലം (കഴിഞ്ഞത്)	ഒരു കാര്യം വളരെവട്ടം കഴിഞ്ഞുപോയിരിക്കുന്നു
ആവർത്തിക്കുന്നു	വർത്തമാനകാലം(ഇപ്പോൾ നടക്കുകയാണ്)	പ്രവർത്തനം ഇപ്പോഴും തുടരുന്നു.
ആവർത്തിക്കും	ഭാവികാലം നടക്കാൻ പോകുന്നു	ഇപ്പോഴും പ്രതീക്ഷയുണ്ട്.

ഈ വ്യത്യാസങ്ങൾ വാചകത്തിന്റെ തീവ്രതയും സമയബോധവും മാറ്റുന്നു. ഉദാഹരണത്തിന്, ‘അവൻ പഠിച്ചു’ എന്ന് പറയുമ്പോൾ പഠനം കഴിഞ്ഞതായി സൂചനയുണ്ട്, പക്ഷേ ‘പഠിക്കുന്നു’ എന്നത് ഇപ്പോഴും പ്രവർത്തനം നടക്കുകയാണെന്നത് വ്യക്തമാക്കുന്നു. ഈ ചെറിയ വ്യത്യാസങ്ങൾ കഥയിലെ സ്ഥിതിഗതികൾ വായനക്കാരന് കൂടുതൽ വ്യക്തമായി അനുഭവപ്പെടാൻ സഹായിക്കുന്നു.

ഉദാഹരണം

തുടർപ്രവർത്തനങ്ങൾ

Question 1.
തീയണക്കാൻ ശ്രമിച്ച കുരുവിയെ പരിഹസിച്ചതാര്?
Answer:
തീയണയ്ക്കാൻ ശ്രമിച്ച് കുരുവിയെ പരിഹസിച്ചത് മഴദേവൻ ആയിരുന്നു. കുരുവിയുടെ ചെറിയ ശ്രമം കാണുമ്പോൾ മഴ ദേവൻ അത് ഫലപ്രദമല്ലെന്ന് പരിഹസിക്കുന്നു. എന്നാൽ ചെറിയ സഹായം പോലും ചെയ്യാൻ മഴദേവൻ തയ്യാറാവുന്നുമില്ല.

Question 2.
“ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു”-
ഈ വാക്യം ആരോട് പറഞ്ഞു? ഈ വാക്യത്തിൽ പ്രതിഫലിക്കുന്ന മാനസികാവസ്ഥയെ വിശകലനം ചെയ്യുക. സന്ദർഭം വ്യക്തമാക്കുക.
Answer:
ഈ വാക്യം കുരുവിയാണ് പറയുന്നത്. കാട്ടുതീ പടർന്നപ്പോൾ കുരുവി തന്റെ കുഞ്ഞി ചുണ്ടിൽ വെള്ളം കൊണ്ടുവന്ന് തീയിൽ തളിച്ച് തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നു. അതുകണ്ട് മഴദേവൻ പരിഹസിക്കുമ്പോൾ കുരുവി ഉത്തരമായി ഈ വാക്യം പറയുന്നു തന്റെ ചെറിയ ശ്രമം ചോദ്യം ചെയ്യുന്ന മഴദേവനോട് പറയുന്നു. ‘ഞാൻ എനിക്കാവുന്നത് ചെയ്യുന്നു.’
ഇവിടെ തെളിയുന്ന മാനസികാവസ്ഥ എന്തെന്നാൽ

ഈ വാക്യം പ്രതിനിധീകരിക്കുന്നത് ഒരു വ്യക്തിയുടെ ഉത്തരവാദിത്വബോധവും ധൈര്യവുമാണ്. കുരുവി തന്റെ കഴിവിനു എന്ത് ചെയ്യാനാവുമോ അതാണ് ചെയ്യുന്നത്. മറ്റുള്ളവരുടെ പരിഹാസം അതിന്റെ ചാരുത ഇല്ലാതാക്കുന്നില്ല.

ഇത് പാഠത്തിലെ മുഖ്യ സന്ദേശമായ ‘നമ്മിൽ ഒരാൾ ഒരു ചെറിയ ശ്രമം നടത്തിയാൽ പോലും വലിയ മാറ്റത്തിന് തുടക്കമാകാം’ എന്നതിനെ പ്രതിഫലിപ്പിക്കുന്നു.

Question 3.
ആനന്ദിന്റെ എഴുത്തിൽ പ്രകടമാകുന്ന സമകാലീനതയുടെ ആത്മാവിനെ’ കുറിച്ച് പ്രതിപാദിക്കുക.
Answer:
ആനന്ദിന്റെ എഴുത്തിൽ ‘സമകാലീനതയുടെ ആത്മാവ്’ എന്നത് ആധുനിക സാമൂഹിക പ്രശ്നങ്ങളെ നേരിട്ട് അഭിമുഖീകരിക്കുന്നു (പരിസ്ഥിതി നാശം, മനുഷ്യത്വമില്ലായ്മ, യുദ്ധ ഭീഷണി, അനാസ്ഥ, നിരുത്തരവാദിത്വം). ഏകാന്തമായെങ്കിലും പ്രതിരോധം നടത്താനുള്ള വ്യക്തിയുടെ നീക്കം (കുരുവി പ്രതിനിധാനം ചെയ്യുന്ന രീതിയിൽ). ഒരു വ്യക്തിയുടെ പ്രചോദനാത്മകമായ പ്രവൃത്തികൾ വലിയ മാറ്റത്തിന് തുടക്കമാകാമെന്ന സന്ദേശം. ആനന്ദിന്റെ കഥകൾ വായനക്കാരെ ചിന്തിപ്പിക്കുന്നു, മറുപടി നിർബന്ധിക്കുന്നില്ല – മറിച്ച് വായനക്കാരൻ സ്വയം ഉത്തരമെത്തേണ്ട അവസ്ഥയിലേക്കാണ് എഴുത്ത് നയിക്കുന്നത്.

Question 4.
കുരുവിയുടെ പ്രയത്നം കണ്ട് മറ്റു മൃഗങ്ങളും കുരുവിയോടൊപ്പം ചേർന്നു എന്നിരിക്കട്ടെ കഥയുടെ അവസാനം മാറ്റി എഴുതൂ.
Answer:
കുരുവിയുടെ കൃപയും പ്രതിബദ്ധതയും കണ്ട് കാട്ടിലുള്ള മറ്റു മൃഗങ്ങൾക്ക് അവരുടെ പ്രവർത്തിയിൽ ലജ്ജ തോന്നുന്നു. മൃഗങ്ങളെല്ലാവരും വെള്ളം കൊണ്ട് തീ അണയ്ക്കാൻ ശ്രമിക്കുന്നു. ഒടുവിൽ, എല്ലാ ജീവികളുടെയും ചേർന്ന പരിശ്രമം കാട്ടുതീ അണയ്ക്കാൻ കാരണമാകുന്നു. അവരുടെ കാടിന്റെ അവർ വീണ്ടെടുത്തു. ഒരുമയുടെ ബലം അവർക്ക് മനസിലായി.
മഴദേവനും ആ ശ്രമം കാണുമ്പോൾ നാണംകെട്ട് ക്ഷമ ചോദിക്കുന്നു.
സന്ദേശം: ‘നല്ലത് ചെയ്യുന്ന ഒരാളുടെ ശ്രമം മറ്റുള്ളവരെയും ഉണർത്തുന്നു.’

അധിക അറിവിലേക്ക്

സമാന ആശയം വരുന്ന കവികളുടെയും മഹത് വ്യക്തികളുടെയും വാക്കുകൾ
‘ചെറുതെങ്കിലും തനിക്കാവുന്നത് ചെയ്യുക’ എന്ന സന്ദേശം മലയാളം സാഹിത്യത്തിലും ലോകസാഹിത്യത്തിലും പലവിധരൂപങ്ങളിലായി പ്രതിഫലിച്ചിട്ടുണ്ട്. മനുഷ്യൻ തന്റെ പരിധിയിൽ നല്ലതൊരിക്കൽ ചെയ്താൽ തന്നെ അതിന് വലിയ പ്രഭാവമുണ്ടാകാമെന്ന ചിന്ത ഉൾക്കുന്ന വചനങ്ങളും കവിതാശകലങ്ങളും ചുവടെ പങ്കുവെക്കുന്നു;

മഹത് വചനങ്ങൾ (മഹത്വമുള്ള ഉദ്ധരണികൾ).
ഗാന്ധിജി
• “Be the change you wish to see in the world.”
(നിങ്ങൾ ലോകത്തിൽ കാണാൻ ആഗ്രഹിക്കുന്ന മാറ്റം തന്നെയാവുക.)
വലിയ വ്യതിയാനങ്ങൾ ചെറുതായെങ്കിലും വ്യക്തി തലത്തിൽ തുടങ്ങണം എന്ന സന്ദേശം.

മദർ തെരേസ:
• “Not all of us can do great things. But we can do small things with great love.”
(എല്ലാവർക്കും വലിയ കാര്യങ്ങൾ ചെയ്യാൻ കഴിയില്ല. എന്നാൽ വലിയ സ്നേഹത്തോടെ ചെറുകാര്യങ്ങൾ ചെയ്യാം.)
കുരുവിയുടെ ചിന്തയോടും പ്രവർത്തിയോടും ഏറെ സാമ്യമുണ്ട്.

സ്വാമി വിവേകാനന്ദൻ:
• “Each soul is potentially divine. Serve man, serve God.”
ഓരോ മനുഷ്യനും ദിവ്യതയുള്ളവൻ. സേവനത്തിലൂടെ ലോകം മാറാം.

മലയാള കവിതാശകലങ്ങൾ:

1. നന്മ ചെയ്യുവാൻ പുറപ്പെടുക,
നിന്റെ ഉള്ളമറിഞ്ഞു നീ ഭരിക്ക…’
♦ കുമാരനാശാൻ – ‘ചിന്താവിഷ്ടയായ സീത’:
ഭവനത്തിന്റെ അകത്തു നിന്ന് തന്നെ നന്മയുടെ വിത്തിടാൻ കൃത്യമായ ഒരു ആഹ്വാനം.

2. ‘ഞാൻ മനുഷ്യനെന്നതോർത്തു
ഞാൻ മനുഷ്യരെ സ്നേഹിച്ചു…’
♦ വള്ളത്തോൾ – ‘ജാതിവെറിക്കെതിരെ’:
ഒരാളുടെ മനസ്സിനുള്ളിലെ മാറ്റം സമൂഹത്തിന് മികച്ചതാവാൻ ഇടയാക്കുന്നു.

3. ‘ഞാൻ നിശ്ശബ്ദമായി നിന്നു
ചെറുതായെങ്കിലും തണൽ
നല്കാൻ ശ്രമിച്ചു…’
♦ സുഗതകുമാരി – ‘മൂലവൃക്ഷം’:

4. വലിയൊരു ലോകം
മുഴുവൻ നന്നാവാൻ
ചെറിയൊരു സൂത്രം
ചെവിയിലോതാം ഞാൻ
‘സ്വയം നന്നാവുക’
♦ കുഞ്ഞുണ്ണി മാഷ്
കുരുവിയുടെ നിലപാടിന്റെ കാവ്യപരമായ രൂപമാണ് ഇത്

5. ‘കർമണേവാധികാരസ്തേ മാ ഫലേഷു കദാചന’
ഭഗവത് ഗീത (ചാപ്റ്റർ 2, ശ്ലോകം 47):
(ഫലമിഛിക്കാതെ കർമ്മം ചെയ്യുക.)
കുരുവി ഫലത്തെക്കുറിച്ച് ചിന്തിക്കാതെ ദൈനംദിനം ചിന്താപരമായ കർമ്മം ചെയ്യുന്നു.

മറ്റു ഭാഷകളിൽ നിന്ന് കവിതാശകലങ്ങൾ
• “If I can stop one heart from breaking,
I shall not live in vain.”
♦ എമിലി ഡിക്കിൻസൺ:
ഒരാൾക്ക് പോലും ആശ്വാസം നൽകാനാകുന്നത് ജീവിതം ഫലപ്രദമാക്കിയതായി കാണുന്നു.

കിളിമൊഴി

(കിളിമൊഴി – പക്ഷികൾക്ക് വേണ്ടി 35 ഭാഷണങ്ങൾ) ഡോ. സലിം അലി.
പരിഭാഷ എസ്. ശാന്തി

നിരീക്ഷണക്കുറിപ്പ്

Question 1.
പക്ഷിനിരീക്ഷണത്തെക്കുറിച്ചുള്ള ഡോ. സാലിം അലിയുടെ നിർദേശങ്ങൾ ശ്രദ്ധിച്ചല്ലോ. അവ കൂടി പരിഗണിച്ചുകൊണ്ട് നിങ്ങൾക്കിഷ്ടപ്പെട്ട ഒരു പക്ഷിയെ നിരീക്ഷിച്ച് കുറിപ്പ് തയ്യാറാക്കുക.
Answer:
ഡോക്ടർ സലിം അലിയുടെ നിർദ്ദേശങ്ങൾ അനുസരിച്ച് ഒരു പക്ഷിയെ ഞാൻ നിരീക്ഷിച്ചെന്നു കണക്കാക്കി തയ്യാറാക്കിയ മാതൃക കുറിപ്പാണ്. നിങ്ങൾക്ക് ഇഷ്ടമുള്ള പക്ഷിയെ ഇവിടെ ചേർത്ത് ഈ മാതൃകയുടെ അടിസ്ഥാനത്തിൽ കൂട്ടുകാരും കുറുപ്പ് തയ്യാറാക്കണേ…

പക്ഷിനിരീക്ഷണ കുറിപ്പ്

പക്ഷിയുടെ പേര്: കുഞ്ചി തത്ത (Common Green Bee eater)

1. അവലോകനം ചെയ്ത സ്ഥലം
പാലക്കാട് ജില്ലയിലെ ഒരു കൃഷി ഭൂമിയിലായിരുന്നു ഈ പക്ഷിയെ കണ്ടത്. രാവിലെ ഏകദേശം 8 മണിയായിരുന്നു സമയം. വാനിൽ വെളിച്ചവും ചുറ്റും നിശബ്ദതയും ഉണ്ടായിരുന്നു.

2. ദൃശ്യവിവരണം
വലുപ്പം: ഏകദേശം ഒരു ചെറുതത്തയുടെ വലിപ്പം (16 – 18 സെ.മീ.).
നിറം: പച്ച നിറം ആധാരമായി കാണപ്പെടുന്നു. തലയിൽ ചെറിയ വെള്ള ചുവപ്പ് നിറമുള്ള പാലിളക്കം. കഴുത്ത് ചുറ്റി കറുത്ത വര.
ശരീരരൂപം: അതിസുന്ദരമായി നീളമുള്ള ചിറകും വാൽപ്പക്ഷ ത്തേയും കാണാം. രണ്ടു വാടികൾ കുറച്ച് നീളത്തിൽ നീളിയി രിക്കുന്നു.
കൊക്ക്: കുറച്ച് നീളമുള്ളതും വളഞ്ഞതും കറുത്ത നിറത്തി ലുള്ളതും.
കണ്ണുകൾ: ചെറുതായി ചുവപ്പു നിറം പതിച്ച കാഴ്ച.
കാൽ: ചെറുതും കറുത്തതും.
ഇരിപ്പിടം: വരമ്പത്ത് നാട്ടിയ ഒരു തൂണിന്മേൽ ഇരിക്കുകയായിരുന്നു.
കണ്ടത്: ഒറ്റയ്ക്കായിരുന്നു പക്ഷി.
നടപ്പുകൾ: വീണ്ടും വീണ്ടും വരമ്പത്ത് തൂണിൽ നിന്ന് പറന്നു ചെറു കീടങ്ങൾ പിടിച്ച് തിരികെ തൂണിലേയ്ക്ക് വരികയായിരുന്നു.
ശബ്ദം: ചെറുതായി ചിര് ചിര്’ പോലുള്ള ശബ്ദം ഉരിയുന്നു.

4. ആവാസ സ്ഥലം
ഈ
പക്ഷിയെ കൃഷിഭൂമികൾ, തുറന്ന ശുദ്ധവായുവുള്ള ഇടങ്ങൾ, വയൽ പാടങ്ങൾ, മരക്കൊമ്പുകൾ എന്നിവിടങ്ങളിൽ സാധാരണ യായി കാണാം.

5. നിരീക്ഷണ അഭിപ്രായം
കുഞ്ചി തത്തയെ നിരീക്ഷിക്കുന്നത് വളരെ രസകരമായ അനുഭവമാണ്. നാട്ടുവേലി തത്ത എന്നും ഇതിനെ പേരുണ്ട്. അതിന്റെ തെളിഞ്ഞ പച്ച നിറം പച്ചപ്പ് നിറഞ്ഞ പശ്ചാത്തലത്തിൽ ആകർഷകമായി നിറഞ്ഞുനിൽക്കുന്നു. തത്സമയം തന്നെ പറന്ന് കീടങ്ങൾ പിടിക്കുന്ന അതിന്റെ കൃത്യത ഉറ്റുനോക്കിയാൽ അതിന്റെ മനോഹാരിതയെ ഏറെ മികവോടെ കാണാം. ശാന്തമായ ഒരു രാവിലെ ഈ പക്ഷിയെ കാണാനായതിൽ ഞാൻ സന്തോഷവാനാണ്.

പക്ഷി നിരീക്ഷണം

വ്യവസ്ഥാധിഷ്ഠിതമായ പക്ഷിനിരീക്ഷണത്തിന് ഇന്ത്യയിൽ അടിസ്ഥാനമിട്ട ആളാണ് സാലിം അലി (സാലിം മുഇസുദ്ദീൻ അബ്ദുൾ അലി, നവംബർ 12, 1896 – ജൂലൈ 27, 1987) അദ്ദേഹത്തിന്റെ നിരീക്ഷണങ്ങൾ, ഭാരതത്തിലെ ജനങ്ങളിൽ പക്ഷിനിരീക്ഷണത്തിനും, പ്രകൃതി സ്നേഹത്തിനും അടിത്തറയിട്ടു. പക്ഷിനിരീക്ഷണ ശാസ്ത്രത്തെക്കുറിച്ചും പക്ഷികളെക്കുറിച്ചും സലിം എഴുതിയ ഗ്രന്ഥങ്ങൾ വിജ്ഞാനപ്രദവും പ്രസിദ്ധവുമാണ്. ഇവയിൽ കേരളത്തിലെ പക്ഷികളെ പറ്റിയെഴുതിയ ഗ്രന്ഥവും ഉൾപ്പെടും. ഒരു കുരുവിയുടെ പതനം’ അദ്ദേഹത്തിന്റെ ആത്മകഥയാണ്. പക്ഷിമനുഷ്യൻ എന്നും ഇദ്ദേഹം അറിയപ്പെടുന്നു

Class 8 Malayalam Adisthana Padavali Notes Unit 1 കനിവും കരുതലും

താഴെ കാണുന്ന ചിത്രങ്ങൾ ‘കനിവും കരുതലും’ എന്ന അധ്യായത്തിന്റെ ഭാഗമായി ഉൾപ്പെടുത്തിയതാണ്. ഇവ തമ്മിൽ ഉള്ള തിരിച്ചറിയലുകളും പ്രതിപാദ്യവും മനസ്സിലാക്കുന്ന പാഠഭാഗങ്ങളാണ് ഈ യൂണിറ്റിൽ പഠിക്കാൻ ഉള്ളത്. കുരുവിയും കാട്ടുതീയും, കൊച്ചു ദേവദാരു, പെരുമഴയത്ത് എന്നീ കഥകളുടെ ആകെ തുകയാണ് ഈ ചിത്രങ്ങൾ.

ചിത്രം 1: കാനനസൗന്ദര്യം – ചിത്രകാരൻ: മുരളി നാഗപ്പുഴ
(ഒ.എൻ.വി കുറുപ്പിന്റെ ഭൂമിക്ക് ഒരു ചരമഗീതം എന്ന പുസ്തകത്തിന്റെ കവർ ചിത്രം കൂടിയാണിത്)

മുരളി നാഗപ്പുഴയുടെ ചിത്രങ്ങൾ പച്ചപ്പും സാധാരണവും ഗ്രാമീണതയും അതിനോടൊപ്പം നിറങ്ങൾ ചാർത്തിയ കൗതുകങ്ങളും തിങ്ങി നിൽക്കുന്നുണ്ടായിരിക്കും അതുകൊണ്ടുതന്നെ അന്താരാഷ്ട്രതലത്തിൽ അവാർഡുകൾ നേടിയ ചിത്രങ്ങളാണ് മുരളി നാഗപ്പുഴയുടേത്.

ഈ ചിത്രം ശാന്തവും പച്ചപ്പും നിറഞ്ഞ ഒരു വനപ്രദേശം കാണിക്കുന്നു. കുട്ടികൾ വനത്തിൽ ശാസ്ത്രീയമായി പഠിക്കുകയും ആനന്ദിക്കുകയും ചെയ്യുന്നു.ആകാശത്ത് പറക്കുന്ന പക്ഷികൾ, തഴുകുന്ന മരങ്ങൾ, നിറഞ്ഞ ചെടികൾ എന്നിവ പ്രകൃതിയുടെ സമൃദ്ധിയും നിമിഷികതയും ചൂണ്ടിക്കാട്ടുന്നു. പ്രകൃതിയോട് സ്നേഹത്തോടെ സമീപിക്കുക, അതിൽ നിന്ന് പഠിക്കുക എന്നതാണ് ഈ ചിത്രത്തിന്റെ പ്രധാന താത്പര്യം.

ചിത്രം 2:- ചിത്രകാരൻ: സബീനബി
ആഗോളതാപനത്തെക്കുറിച്ചുള്ള ഒരു അന്താരാഷ്ട്ര കാർട്ടൂൺ (റഷ്യൻ എക്കോളജിക്കൽ മൂവ്മെന്റ് സംഘടിപ്പിച്ച) മത്സരത്തിൽ നാമനിർദ്ദേശം ചെയ്യപ്പെട്ട ആർട്ടിസ്റ്റ് സിബി ഷിബുവിന്റെ ചിത്രം.

ഈ ചിത്രത്തിൽ വരൾച്ച വിണ്ടുപോയ ഭൂമിക്ക് നടുവിലൂടെ രണ്ട് ആളുകൾ നടന്നു പോവുകയാണ്, അവരുടെ കൈയിൽ വിലയേറിയ ഒരു ജലത്തുള്ളിയുണ്ട്. തിളങ്ങുന്ന ആ ഒറ്റ തുള്ളി വെള്ളം, ഒരൊറ്റ വെള്ളതുള്ളിയെ ഏറ്റവും വിലപ്പെട്ടതായി കണക്കാക്കേണ്ടി വരുന്ന ആ നിമിഷം ഓർത്ത് നോക്കൂ. പ്രകൃതിയോട് അനാദരവ് കാട്ടുമ്പോൾ അതിന്റെ പ്രത്യാഘാതം എങ്ങനെയായിരിക്കും എന്നതിന്റെ മുന്നറിയിപ്പാണ് ഈ ചിത്രം.

ആദ്യ ചിത്രം പ്രകൃതിയോട് സ്നേഹപരമായ അനുഭാവവും സംരക്ഷണവും നൽകുമ്പോൾ രണ്ടാം ചിത്രം പ്രകൃതിദുരന്തത്തിന്റെ ദാരുണ ഫലങ്ങളെയും കാണിക്കുന്നു. ഈ രണ്ട് ചിത്രങ്ങളിലൂടെ മനുഷ്യന് പ്രകൃതിയോടുള്ള ഉത്തരവാദിത്വം വ്യക്തമാക്കുന്നു.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 6 Multiples and Factors Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 6 Solutions Multiples and Factors

Class 6 Kerala Syllabus Maths Solutions Chapter 6 Multiples and Factors Questions and Answers

Multiples and Factors Class 6 Questions and Answers Kerala Syllabus

Multiples of Multiples (Page No. 85)

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 8
(ii) Multiples of 10
(iii) Multiples of 12
Answer:
(i) The factors of 8 are 2 and 4
8 is a multiple of 2 and 4
Therefore, the multiples of 8 are also the multiples of 2 and 4

(ii) The factors of 10 are 2 and 5
10 is a multiple of 2 and 5
Therefore, the multiples of 10 are also the multiples of 2 and 5

(iii) The factors of 12 are 2, 3, 4, and 6
12 is a multiple of 2, 3, 4, and 6
Therefore, the multiples of 12 are also the multiples of 2, 3, 4, and 6.

Question 2.
Check whether each of the statements below is true or false. For true statements, explain why they are so. For the false statements, give an example in which it is not true.
(i) All multiples of 20 are multiples of 10
(ii) All multiples of 10 are multiples of 2
(iii) All multiples of 15 are multiples of 5
(iv) All multiples of 15 are multiples of 3
(v) All multiples of 5 are multiples of 15
(vi) All multiples of 3 are multiples of 15
Answer:
(i) True
Since 20 = 2 × 10,
So every multiple of 20 is a multiple of 10.

(ii) True
Since 10 = 2 × 5,
So every multiple of 10 is a multiple of 2.
Or
Since 10 is an even number, and every multiple of 10 ends in 0, it is divisible by 2
That is, every multiple of 10 is also a multiple of 2.

(iii) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 5.

(iv) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 3.

(v) False
Not all multiples of 5 are divisible by 15.
E.g.: 10 is a multiple of 5 but not of 15.

(vi) False
Not all multiples of 3 are divisible by 15.
E.g.: 6 is a multiple of 3, but not a multiple of 15.

Primary Factors (Page No. 88)

Question 1.
Can you write the numbers below as a product of primes?
(i) 24
(ii) 35
(iii) 36
(iv) 60
(v) 100
Answer:
(i) 24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
Therefore 24 = 2 × 2 × 2 × 3

(ii) 35 = 5 × 7

(iii) 36 = 2 × 18
= 2 × 2 × 9
= 2 × 2 × 3 × 3
Therefore 36 = 2 × 2 × 3 × 3

(iv) 60 = 2 × 30
= 2 × 2 × 15
= 2 × 2 × 3 × 5
Therefore 60 = 2 × 2 × 3 × 5

(v) 100 = 2 × 50
= 2 × 2 × 25
= 2 × 2 × 5 × 5
Therefore 100 = 2 × 2 × 5 × 5

Textbook Page No. 89

Question 1.
Write each of the numbers below as a product of primes.
(i) 72
(ii) 105
(iii) 144
(iv) 330
(v) 900
Answer:
(i) 72 = 12 × 6
12 = 2 × 2 × 3
6 = 2 × 3
Therefore, 72 = 2 × 2 × 2 × 3 × 3

(ii) 105 = 21 × 5
21 = 3 × 7
5 = 5
Therefore 105 = 3 × 5 × 7

(iii) 144 = 12 × 12
12 = 2 × 2 × 3
Therefore 144 = 2 × 2 × 3 × 2 × 2 × 3

(iv) 330 = 10 × 33
10 = 2 × 5
33 = 3 × 11
Therefore, 330 = 2 × 3 × 5 × 11

(v) 900 = 30 × 30
30 = 2 × 3 × 5
Therefore, 900 = 2 × 3 × 5 × 2 × 3 × 5

All Factors (Page No. 89)

Question 1.
Find all the factors of the number below:
(i) 35
(ii) 77
(iii) 26
(iv) 51
(v) 95
Answer:
(i) 35 = 1 × 35 = 5 × 7
Factors of 35 are: 1, 5, 7, and 35

(ii) 77 = 1 × 77 = 11 × 7
Factors of 77 are: 1, 7, 11, and 77

(iii) 26 = 1 × 26 = 2 × 13
Factors of 26 are: 1, 2, 13, and 26

(iv) 51 = 1 × 51 = 3 × 17
Factors of 51 are: 1, 3, 17, and 51

(v) 95 = 1 × 95 = 5 × 19
Factors of 95 are: 1, 5, 19, and 95

Textbook Page No. 90

Question 1.
Write each of the numbers below as a product of three primes and find all its factors:
(i) 66
(ii) 70
(iii) 105
(iv) 110
(v) 130
Answer:
(i) 66 is the product of three prime numbers.
66 = 2 × 3 × 11
Factors of 66 are:
1
2, 3, 11
2 × 3 = 6
2 × 11 = 22
3 × 11 = 33
Therefore, the factors are: 1, 2, 3, 6, 11, 22, 33, and 66

(ii) 70 as the product of three prime numbers.
70 = 2 × 5 × 7
Factors of 70 are:
1
2, 5, 7
2 × 5 = 10
2 × 7 = 14
5 × 7 = 35
Therefore, the factors are: 1, 2, 5, 7, 10, 14, 35, and 70

(iii) 105 is the product of three prime numbers.
105 = 3 × 5 × 7
Factors of 105 are:
1
3, 5, 7
3 × 5 = 15
3 × 7 = 21
5 × 7 = 35
Therefore, the factors are: 1, 3, 5, 7, 15, 21, 35, and 105

(iv) 110 as the product of three prime numbers.
110 = 2 × 5 × 11
Factors of 110 are:
1
2, 5, 11
2 × 5 = 10
2 × 11 = 22
5 × 11 = 55
Therefore, the factors are: 1, 2, 5, 10, 11, 22, 55, and 110

(v) 130 is the product of three prime numbers.
130 = 2 × 5 × 13
Factors of 130 are:
1
2, 5, 13
2 × 5 = 10
2 × 13 = 26
5 × 13 = 65
Therefore, the factors are: 1, 2, 5, 10, 13, 26, 65, and 130

Prime Numbers (Page No. 92)

Question 1.
Find all primes less than 100. Find the primes that differ by 2 among these.
Answer:
Prime numbers less than 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers that differ by 2 are:
(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

Question 2.
Can the product of two natural numbers be a prime?
Answer:
The product of two natural numbers can only be a prime number if one of the numbers is 1 and the other is a prime.
If both numbers are greater than one, their product will always have more than two factors, so the result cannot be a prime.

Question 3.
Can the sum of two prime numbers be prime?
Answer:
Yes, sometimes, but not always.
If the sum of two prime numbers is a prime only when one of the numbers is 2 (the only even prime).
If 2 is added to an odd prime, the result is strange and may be prime.
But if two odd primes are added, the result is even and will never be a prime (except 2 + 2 = 4, which is not a prime number).

Class 6 Maths Chapter 6 Kerala Syllabus Multiples and Factors Questions and Answers

Class 6 Maths Multiples and Factors Questions and Answers

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 15
(ii) Multiples of 21
(iii) Multiples of 33
Answer:
(i) The factors of 15 are 3 and 5.
15 is a multiple of 3 and 5.
Therefore, the multiples of 15 are also the multiples of 3 and 5.

(ii) The factors of 21 are 3 and 7.
21 is a multiple of 3 and 7.
Therefore, the multiples of 21 are also the multiples of 3 and 7.

(iii) The factors of are 3 and 11.
33 is a multiple of 3 and 11.
Therefore, the multiples of 33 are also the multiples of 3 and 11.

Question 2.
Write the numbers below as a product of primes?
(i) 18
(ii) 40
(iii) 150
(iv) 210
(v) 300
Answer:
(i) 18 = 2 × 9 = 2 × 3 × 3
Therefore 18 = 2 × 3 × 3

(ii) 40 = 2 × 20
= 2 × 2 × 10
= 2 × 2 × 2 × 5
Therefore 40 = 2 × 2 × 2 × 5

(iii) 150 = 2 × 75
= 2 × 3 × 25
= 2 × 3 × 5 × 5
Therefore 150 = 2 × 3 × 5 × 5

(iv) 210 = 2 × 105
= 2 × 3 × 35
= 2 × 3 × 5 × 7
Therefore 210 = 2 × 3 × 5 × 7

(v) 300 = 2 × 150
= 2 × 2 × 75
= 2 × 2 × 3 × 25
= 2 × 2 × 3 × 5 × 5
Therefore 300 = 2 × 2 × 3 × 5 × 5

Question 3.
Write the following numbers as the product of three prime numbers, and find all the factors of them.
(i) 174
(ii) 385
(iii) 182
Answer:
(i) 174 is the product of three prime numbers.
174 = 2 × 3 × 29
Factors of 174 are:
1
2, 3, 29
2 × 3 = 6
2 × 29 = 39
3 × 29 = 78
Therefore, the factors are: 1, 2, 3, 6, 29, 58, 87, 174

(ii) 385 is the product of three prime numbers.
385 = 5 × 7 × 11
Factors of 385 are:
1
5, 7, 11
5 × 7 = 35
5 × 11 = 55
7 × 11 = 77
Therefore, the factors are: 1, 5, 7, 11, 35, 55, 77, 88, and 385.

(iii) 182 as the product of three prime numbers.
182 = 2 × 7 × 13
Factors of 182 are:
1
2, 7, 13
2 × 7 = 14
2 × 13 = 26
7 × 13 = 91
Therefore, the factors are: 1, 2, 7, 13, 14, 26, 91, and 182.

Class 6 Maths Chapter 6 Notes Kerala Syllabus Multiples and Factors

→ The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…

→ All multiples of the multiple of a number are also multiples of that number.

→ All multiples of a number are also multiples of any of its factors.

→ A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

→ Any composite number can be written as a product of primes.

→ The only even number among the prime numbers is 2.

Multiples and factors are fundamental concepts in mathematics that help us understand the relationships between numbers. A multiple of a number is the result of multiplying that number by any natural number, while a factor is a number that divides another number exactly, without leaving a remainder.

For example, 4 is a multiple of 2, and 2 is a factor of 4. Learning about multiples and factors is essential for solving problems involving divisibility, simplifying fractions, finding the greatest common factor (GCF), the least common multiple (LCM), and much more. In this chapter, we discuss multiples of multiples, primary factors, and prime numbers.

Multiples of Multiples
The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…
For example:
The multiples of 2 are the numbers 2, 4, 6,… obtained by multiplying the natural number by 2.
The multiples of 4 are the numbers 4, 8, 12,… obtained by multiplying the natural number by 4.
Here, all the multiples of 4 can be written as multiples of 2 also:
1 × 4 = 4
2 × 4 = 8
3 × 4 = 12
4 × 4 = 16
5 × 4 = 20
……………

2 × 2 = 4
4 × 2 = 8
6 × 2 = 12
8 × 2 = 16
10 × 2 = 20
……………….

Similarly, if 6 is a multiple of 2 and 3.
So all multiples of 6 can be written as multiples of 2 and 3.

1 × 6 = 6
2 × 6 = 12
3 × 6 = 18
4 × 6 = 24
5 × 6 = 30
………………..

3 × 2 = 6
6 × 2 = 12
9 × 2 = 18
12 × 2 = 24
15 × 2 = 30
………………….

2 × 3 = 6
4 × 3 = 12
6 × 3 = 18
8 × 3 = 24
10 × 3 = 30
………………..

In general, all multiples of the multiple of a number are also multiples of that number.

The multiples of 15 can be written as the multiples of what numbers?
Answer:
15 is a multiple of 3 and 5.
So the multiples of 10 can be written as the multiples of 2 and 5.

We have seen that multiples can also be put in terms of factors.
For example:
4 is the multiple of 2 can also be written as 2 is a factor of 4.
6 is the multiple of 2 and 3 can also be written as 2 and 3 are two factors of 6.
12 is a multiple of 3, and 4 can also be written as 3 and 4 are factors of 12.

In general, we can say that all multiples of a number are also multiples of any of its factors.

14 is a multiple of 2 and 7. Express it in the form of factors.
Answer:
14 is a multiple of 2 and 7.
Since, 2 × 7 = 14
So 2 and 7 are two factors of 14.

70 is a multiple of 2, 5, and 7. Express it in the form of factors.
Answer:
70 is a multiple of 2, 5, and 7.
Since, 2 × 5 × 7 = 70
So 2, 5, and 7 are the factors of 70.

If 2, 3, and 7 are factors of a number. Then what is that number?
Answer:
2, 3, and 7 are factors of a number.
That means 2 × 3 × 7 = 42
Therefore, the number is 42, and multiples of 42 are 2, 3, and 7.

Primary Factors
Any number can be written as the product of its factors in different ways.
For example, consider the number 70,
1 × 70 = 70
2 × 35 = 70
5 × 14 = 70
10 × 7 = 70
We can write 70 as a product of three factors, without using 1:
That is 70 = 2 × 5 × 7
The only factors of each of the numbers 2, 5, and 7 are 1 and the number itself.
For any number 1, the number itself is are factor.

The numbers below 20 with factors 1 and itself are: 1, 2, 3, 5, 7, 11, 13, 17, 19.
Such numbers, excluding 1, are said to be prime numbers.

A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

Numbers greater than 1, which are not primes, are called composite numbers.
For example, 4 is a composite number.
Since 4 = 2 × 2

A composite number can be written as the product of a prime number.

When a number is written as the product of two factors and any one of them is not a prime, then that factor can be written as the product of two factors. This can continue till all factors are prime.

Write 48 as the product of primes.
Answer:
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
Therefore 48 = 2 × 2 × 2 × 2 × 3

Write the following as the product of primes.
(i) 30
(ii) 45
(iii) 64
Answer:
(i) 30 = 2 × 15 = 2 × 3 × 5
Therefore 30 = 2 × 3 × 5

(ii) 45 = 5 × 9 = 5 × 3 × 3
Therefore 45 = 5 × 3 × 3

(iii) 64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
Therefore 64 = 2 × 2 × 2 × 2 × 2 × 2

Product of Primes:
Once we write two numbers as a product of primes, it is easy to write the product of these numbers also as a product of primes.
For example: 12 × 24 can be split like this:
12 = 2 × 2 × 3
24 = 2 × 2 × 2 × 3
Split 12 × 24 as shown below:
288 = 12 × 24
= (2 × 2 × 3) × (2 × 2 × 2 × 3)
= 2 × 2 × 3 × 2 × 2 × 2 × 3

To split a number into a product of primes, we first split it into the product of any two factors, then split each of these factors into a product of primes, and finally put these prime factors together.
For example: Split 140 into a product of primes,
140 = 14 × 10
Next, write 14 and 10 as products of primes
14 = 2 × 7
10 = 2 × 5
We can write 140 like this;
140 = 14 × 10
= (2 × 7) × (2 × 5)
= 2 × 2 × 5 × 7

Split the following numbers into a product of primes.
(i) 420
(ii) 180
(iii) 336
Answer:
(i) 420 = 15 × 28
15 = 3 × 5
28 = 2 × 2 × 7
Therefore, 420 = 2 × 2 × 3 × 5 × 7

(ii) 180 = 12 × 15
12 = 2 × 2 × 3
15 = 3 × 5
Therefore 180 = 2 × 2 × 3 × 3 × 5
(iii) 336 = 14 × 24
14 = 2 × 7
24 = 2 × 2 × 2 × 3
Therefore, 336 = 2 × 2 × 2 × 2 × 3 × 7

All Factors
If we know the prime factors of a number, we can find all its factors.
For example, the factors of 6 are: 1, 2, 3, and 6.

Write the prime factors of 15. And what are its other factors?
Answer:
Prime factors of 15 are: 3 and 5
Factors of 15 are: 1, 3, 5, and 15

Find all the factors of the numbers given below:
(i) 42
(ii) 54
(iii) 63
Answer:
(i) 42 = 1 × 42
= 2 × 21
= 3 × 14
= 6 × 7
Therefore, factors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42.

(ii) 54 = 1 × 54
= 2 × 27
= 3 × 18
= 6 × 9
Therefore, factors of 54 are: 1, 2, 3, 6, 9, 18, 27, and 54.

(iii) 63 = 1 × 63
= 3 × 21
= 7 × 9
Therefore, factors of 63 are: 1, 3, 7, 9, 21, and 63.

Product of Three Prime Numbers:
Now let’s look at the product of three different primes.

Write 42 as the product of three prime numbers and find all the factors?
Answer:
42 as the product of three prime numbers,
42 = 2 × 3 × 7
Factors of 42 are:
1
2, 3, 7
2 × 3 = 6
2 × 7 = 14
3 × 7 = 21
Therefore, the factors are: 1, 2, 3, 6, 7, 14, 21, and 42

Write the following numbers as the product of three prime numbers, and find all the factors of it?
(i) 102
(ii) 154
(iii) 195
Answer:
(i) 102 is the product of three prime numbers.
102 = 2 × 3 × 17
Factors of 102 are:
1
2, 3, 17
2 × 3 = 6
2 × 17 = 34
3 × 17 = 51
Therefore, the factors are: 1, 2, 3, 6, 17, 34, 51, and 102

(ii) 154 is the product of three prime numbers.
154 = 2 × 7 × 11
Factors of 154 are:
1
2, 7, 11
2 × 7 = 14
2 × 11 = 22
7 × 11 = 77
Therefore, the factors are: 1, 2, 7, 11, 14, 22, 77, and 154.

(iii) 195 as the product of three prime numbers.
195 = 3 × 5 × 13
Factors of 195 are:
1
3, 5, 13
1 × 5 = 15
3 × 13 = 39
5 × 13 = 65
Therefore, the factors are: 1, 3, 5, 13, 15, 39, 65, and 195

Prime Numbers
The only even number among the prime numbers, 2, 3, 5, 7, 11,… is 2.
All primes afterwards are odd numbers. But not all odd numbers are primes;
For example: 9 = 3 × 3, 15 = 3 × 5,……. They are not prime numbers.

There is no definite pattern for the odd primes.
For example, after 3, 5, 7 are consecutive primes that differ by 2, the next prime is not 9 (which is not a prime), but 11. Thus, the difference between 7 and 11 is 4. Similarly, after the prime 31, the next prime is 37, and their difference is 6; the prime after 89 is 97, with a difference of 8. But even as such consecutive primes drift further apart, there are consecutive primes like 41 and 43 or 71 and 73 in between, which are only 2 apart. There is a technique to list all primes less than a specified number. That is, first write all numbers up to 50 in rows and columns like this:

Strike off 1 from this. Then strike off all multiples of 2, except 2:

Keep 3 and strike off all multiples of 3:

Strike all multiples of 5, except 5 itself.
If we remove the multiples of 7 other than itself also, we can see that there are no multiples, except themselves, of the other numbers that remain:

Now the numbers not struck off are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
These are the prime numbers less than 50.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 3 Pressure Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 3 Pressure Question Answer Notes

Class 8 Basic Science Chapter 3 Notes Kerala Syllabus Pressure Question Answer

Pressure Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
Give reasons.
• Heavy vehicles have more tyres.
• Some passengers experience nosebleeds when airplanes fly at high altitudes.
• The size of the bubbles rising from the bottom of an aquarium increases gradually.
• The foundation of buildings is made wider.
Answer:
• Heavy vehicles are provided with more tyres to increase the surface area of contact with the surface of the road so that it reduce the pressure on each tyre and on the surface of the road and ensures smooth motion.
• As the altitude increases the atmospheric pressure decreases. This lower atmospheric pressure leads to nosebleeds in some passengers.
• At the bottom of the aquarium, the pressure is high. For a bubble coming up the pressure experienced is less. Hence the volume of bubble increases.
• As the foundation is made wider, the pressure exerted by the building on the ground decreases because surface area of contact between building and ground increases. Thus in turn makes the building stronger and safer.

Question 2.
Which of the bottles below shows the shape of rising bubbles correctly? Explain why.

Answer:
Bottle C shows the correct shape.

Liquid pressure is greater at the bottom. For a bubble coming up pressure decreases. Hence the size of the bubbles increases as it moves from bottom to top.

Question 3.
Observe the picture below. Which of the following jars containing the same liquid, will experience the highest pressure at the bottom? Explain why.

Answer:
All the three jars containing the same liquid will experience the same pressure. Liquid pressure depends on the depth and the density of the liquid. Since the liquids are same liquid pressure is not dependent on density here. Liquid pressure is same here as the depth of the liquid (or height of liquid column) is same in all the jars.

Question 4.
Cooking is faster in a pressure cooker than in an open vessel. Explain why.
Answer:
When we cook food in a pressure cooker, the pressure inside it increases as the temperature increases. The boiling point of water is raised by this increased pressure inside it. This allows food to be cooked at a higher temperature and makes cooking faster than in an open vessel.

Basic Science Class 8 Chapter 3 Question Answer Kerala Syllabus

Question 1.
Let’s now have a look at the figure given below, in which the same weight is applied in two different ways.

A brick is placed vertically and then horizontally on a sponge (Fig(a)). In both cases, does the weight of the brick remain the same? Are the compressions on the sponge equal?
Answer:
The weight of the brick remain the same in both cases. The compressions on the sponge is not equal.

Question 2.
The weight of the child remains the same whether they stand or lie on the mattress (Fig.(b)) as shown above. Yet, the mattress is compressed more while standing. What could be the reason for this?
Answer:
The mattress is compressed more while standing because the surface area in contact is less in this case.
Activity

Fix a single nail on the cardboard as shown in the figure and place a balloon on top of it. Place a small weight on the balloon. Observe what happens to the balloon. Now, nail multiple pins very close to each other as shown in the second figure. Place the same weight on top of the balloon. Observe the result.

Observation
When weight is placed on the balloon, placed on a single nail it bursts. When the same weight is placed on the top of the balloon, placed on the top of multiple pins it will not burst.

Question 3.
Record the observations from the above activity in the table below.

Answer:

Question 4.
Analyze the table.
Answer:
Even when the same force is applied in each case, the results vary depending on the change in surface area in contact.

Question 5.
What is pressure?
Answer:
Pressure is the force acting normally per unit area.

Question 6.

Who will win the race? Write your opinion.
Answer:
Duck will win the race. Hen cannot move at the same speed as that of duck in mud.

Question 7.
What could be the reason?
Answer:
Webbed feet of duck makes more surface area of contact with the mud. Thus the pressure experienced on its feet is less and it can walk over mud easily. But the hen’s feet make less surface area of contact with the mud and hence experience more pressure which makes it difficult for it to move over the mud.

Question 8.
See the figure of two concrete slabs placed in sand.

Each of them is made of cubes weighing 18 N fixed together. The area of one side of the cube is 0.36 m²


Referring to the figures, complete the following table.
Answer:

Measurements	Slab 1	Slab 2
Force experienced on the sand	18 N	18 N
Surface area of contact	0.36 × 4 = 1.44 m2	0.36 m2
Force experienced per unit area	18/1.44 = 12.5 N/m2	18/0.36 = 50 N/m2

The weight of the slab acts normally on the sand. This is called thrust.
The force acting normally on a surface is called thrust. The force acting normally per unit area is pressure.
If pressure is denoted by P, thrust by F and area by A, pressure = \(\frac{\text { Thrust }}{\text { Area }}\) P = \(\frac{F}{A}\)
Unit of pressure = \(\frac{\text { Unit of thrust }}{\text { Unit of area }}\)
The SI unit of pressure is pascal. 1 pascal = 1N/m²

Question 9.
Now, find out the pressure exerted by the above slabs, on the sand.
Answer:
Pressure exerted by slab 1 = 12.5 N/ m² or 12.5 Pa
Pressure exerted by slab 2 = 50 N/ m² or 50 Pa

Question 10.
A concrete block weighing 100 N is placed on sand in three different ways as shown in the figure below.

a) In which position will the highest pressure be exerted on the sand?
Answer:
Higher pressure will be exerted on the sand when the brick is placed vertically (position 3)

b) Which position will cause the concrete block to sink deeper into the sand? Why?
Answer:
The third position (vertical) will cause the concrete block to sink deeper into the sand. It is because the surface area of contact between the concrete block and sand is less here.

When thrust remains constant, the pressure is inversely proportional to the surface area. This means, when the surface area increases, the pressure decreases and vice versa.

Question 11.
Can you explain the reason why a lorry would not get stuck in the mud if a wooden plank is put.
Answer:
When a wooden plank is put over mud, the surface area of contact between the lorry and the mud increases, so the pressure experienced on the mud becomes less, so that the lorry will not get stuck in mud.

Question 12.
Try to lift your school bag by tying it with a thin twine. How do you feel?
Answer:
It is very difficult to lift a school bag by tying it with a thin twine because pressure felt on the hands is very high.

Question 13.
What if you lift it using a wide strap as shown in the figure? Can you explain the difference by relating its area to pressure?

Answer:
When the bag is lifted using a wide strap the surface area of contact between the bag and the hand is more, so the pressure felt on the hands is less.

But when thin strap is used, the surface area of contact between the bag and the hand is less, so the pressure felt on the hands is more. This makes it difficult to lift the bag.

Question 14.
Similarly, explain the following situations in the figure. Suggest more such situations.

Answer:

• The edge of a knife is made thinner – Thin edges have less surface area of contact which helps to exert more pressure on the objects to be cut.
• The wheels of bulldozers are connected by wide chains -Wide chains make more surface area of contact with the mud to exert less pressure over it. This helps to prevent the wheels getting stuck in the mud and ensures its smooth motion.
• A person lying on a bed of nails will not bleed- As the bed of nails make large surface area of contact with the person, the pressure exerted on the person is less. So person will not bleed.
• The basement of dams is made wider -Basements are made wider to increase the surface area of contact betw een walls and basement and thus to withstand the high pressure exerted by water stored in dams.

Question 15.
Do only solids exert pressure?
Answer:
No

Question 16.
Can’t liquids and gases exert pressure?
Answer:
Yes, Liquids and gases also exert pressure.
Activity
Tie a polythene cover tightly around your hand as shown in the figure. Dip your hand into a bucket of water.

Observation: The cover sticks to your hand.
Inference: This happens due to the pressure exerted by the water on the polythene cover.
Thus we can understand that, just like solids, liquids can also exert pressure.

The normal force exerted by a liquid is called thrust. The thrust that a liquid exerts per unit area is called liquid pressure. Liquids exert pressure on all sides of the container.

Question 17.
If the force of water jetting out is greater, wouldn’t the pressure also be.greater?
Answer:
Yes. When the force of water jetting out is greater the pressure also will be greater.

Question 18.
What happens to the water stream when the water level decreases?
Answer:
When the water level decreases the force of water jetting out becomes less and hence the pressure also will be lesser.

Question 19.
What can be inferred from this?
Answer:
As the force decreases due to decrease in water level pressure also decreases and pressure depends on the depth of the liquid.

Pressure increases as the depth in a liquid increases.

Activity
Attach a syringe on a board as shown in the figure below.

Attach one end of the I.V. set tube to the hub of the syringe. Attach the piston of the syringe to the bottom end of a pen barrel. Arrange it so that the pen barrel deflects when the piston moves. Now fix a thin stick to the top end of pen barrel as a pointer. Attach the other end of the I.V. set to the plastic bottle. Fill the plastic bottle with water and place it at a higher level than the syringe.

Question 20.
Raise the bottle. The force applied by the piston will change according to the pressure of the water. Now’ you can see the thin stick deflecting according to the force exerted in the piston, right?
Answer:
Yes. We can see the thin stick deflecting according to the force exerted in the piston.

Question 21.
Raise the bottle further. What do you observe? What is the reason?
Answer:
The thin stick deflects more. When the bottle is raised further, height of the liquid column increases and pressure of water increases and the force applied by the piston also increases.

Question 22.
Repeat the experiment using salt solution instead of water. Why does the thin stick deflect more?
Answer:
Water and saline water have different densities. Density of saline water is higher than that of water. When density of a liquid increases, pressure exerted by it increases. So the force exerted on the piston by saline water is more. That is why the piston moves more, causing greater deflection of the thin stick.

The density of a liquid influences its pressure. If the density of a liquid mercases, the pressure also increases.

The instrument used to measure liquid pressure is called a manometer.

Question 23.
Do you see any difference in the level of water in the tube?
Answer:
Yes . When slight pressure is applied the level of water in tube rises up.
Apply more pressure. You can see a difference in water levels also.

Question 24.
Now take different solutions in the beaker and measure the pressure at different levels. Write your observations in the table below.

Answer:

Position of the funnel	Difference in the water levels of U-tube/pressure
On the surface	In water	In saline water	In kerosene
Midway in the beaker	12 cm	12 cm	12 cm
At the bottom of the beaker	18 cm	20 cm	16 cm
Position of the funnel	24 cm	26 cm	20 cm

Question 25.
Analyse the table.
a) In which position is the pressure highest?
Answer:
Pressure is highest at the bottom of the beaker.

b) Which liquid exerts the highest pressure?
Answer:
Saline water exerts the highest pressure.

Question 26.
Fill a deep vessel with water. Dip a straw into the water and blow gently. Do bubbles come up?

Answer:
Yes, bubbles come up.

Question 27.
What happens to their size when they reach the surface?
Answer:
Size of the bubble increases.

Question 28.
Explain their change in size.
Answer:
The increase in size of bubble when it reach the surface is due to the decrease in pressure.

Question 29.
Explain why dams are built with a wider base.
Answer:
Dams are built with a wider base to withstand the pressure exerted by the water stored in it. When there is a wider base, the surface area of contact between water and the base of the dam is more and so pressure exerted decreases.

Question 30.
Can you think of more situations in daily life where liquid pressure is experienced?
Answer:

• Water tanks at home are placed at a height to exert liquid pressure to the pipelines in the building.
• Oil tankers are made of thick metals to withstand the pressure of huge volume of oil.
• When a bucket of water is carried on head pressure is felt on head.

Factors influencing gas pressure
Question 31.
Inflate a balloon at its maximum. What happens when the balloon is inflated further? Will it eventually burst?
Answer:
The balloon bursts as it fails to withstand the air pressure inside.

Question 32.
Find the change in the number of gas particles, in figure shown below, when more air is filled. What can be inferred from this?

Answer:
When more air is filled, the number of gas particles increases and the gas pressure also increases.

Gas pressure depends on the number of particles. As the number of particles increases, gas pressure also increases.

Question 33.
Fill the same quantity of air into two balloons of different sizes. Now the number of gas particles in both . balloons is equal, right? Which of the balloons has more pressure inside? What can be inferred from this?
Answer:
As the same quantity of air is filled into two balloons the number of gas particles in both balloons is equal. But the balloon with smaller size has more pressure as it has less volume. When the volume is less, the gas particles has less space to move around. This leads to the frequent collision with the walls of the balloon and hence more pressure is exerted.

Gas pressure depends on its volume.

Question 34.
Leave an inflated balloon in the sunlight. What do you observe? What is the change in the pressure when the air inside gets heated up?
Answer:
When the air inside gets heated up, the pressure inside the balloon increases and it bursts after some time.

Gas pressure depends on its temperature.

When we cook food in a pressure cooker, the pressure inside increases as the temperature increases.

Question 35.
List the factors influencing gas pressure.
Answer:

• Number of particles
• Volume
• Temperature

Question 36.
Is atmospheric pressure the same everywhere on the Earth? Observe the figure.

Which book will experience the highest pressure? Why?
Answer:
No, the atmospheric pressure is different at different places on earth. The book at the bottom experiences the highest pressure as it experiences the atmospheric pressure and the combined weight of all the books stacked above.

Question 37.
Similarly, we know that atmospheric pressure is the weight of air column experienced per unit area. So, what is the change that occurs in atmospheric pressure as altitude increases? Note your inferences.
Answer:
Atmospheric pressure decreases as height from the Earth’s surface increases.

Atmospheric pressure at sea level is known as standard atmospheric pressure. It is defined as the weight of a mercury column that is 0.76 m high, with a unit cross sectional area, exerting a pressure of 1 atm. The instrument used to measure atmospheric pressure is a barometer.

Question 38.
Take an aluminum can and fill it with hot water. Pour out the hot water and immediately seal the can. Immerse the can in cold water. What do you observe?

When the can cools down, the pressure inside the can decreases. The external atmospheric pressure is strong enough to crush the can.

Question 39.
Calculate the force exerted by atmospheric pressure on a table surface of area 1 m². Atmospheric pressure is 101325 Pascal.
Answer:
Force = Pressure × Area
= 1 Pa × 1 m²
= 101325 N
This is equivalent to the weight of an object with a mass of 10339 kg. If such a large force is exerted on the table, it doesn’t collapse. It is because the pressure is applied from all sides on it and so the forces get balanced.

Gas pressure is exerted in all directions, similar to that of liquids.

Activity
Fill a glass with water and cover it with a cardboard. Hold the cardboard with your hand and turn the glass upside down. Remove your hands slowly.

Question 40.
Does the water fall?
Answer:
The force due to atmospheric pressure can hold the entire weight of the water in the glass. So the water does not fall down.
The equilibrium of pressure between gases and liquids is a common phenomenon occurring in nature.

Question 41.
What happens to the candle flame?
Answer:
The candle flame goes out.

Question 42.
What happens to the air pressure inside the glass?
Answer:
Air pressure inside the glass decreases.

Question 43.
What is the change in the water level?
Answer:
Water level inside the glass rises.
The water level remains stable when the pressure inside the glass and the atmospheric pressure outside the glass are in equilibrium.

Question 44.
Gently pull back the piston of a syringe and seal the opening with your hand. Then pull the piston all the way back and release. What do you see?
Answer:
The piston will go back. When the syringe opening is sealed and the piston is pulled back and released, the piston will move into the syringe barrel. This movement is due to the difference in pressure between the outside and inside of the syringe. The higher outside pressure forces the piston back into the syringe barrel.

Question 45.
Why does the piston go back? Discuss the reason.
Answer:
The piston move back into the syringe barrel, because the pressure inside the syringe is less than the pressure outside. So the higher pressure outside forces the piston back into ehe syring barrel.

Question 46.
Read the following situations and explain them based on atmospheric pressure.
• Rubber suckers stick to smooth surfaces.
• Mountain climbers often experience nosebleeds at high altitudes.
• Holes are made on the injection bottle with a needle during a drip injection.
• Passengers travelling uphill in vehicles on a ghat road experience ear pain.
Answer:
• When a rubber sucker is pressed on a smooth surface, it pushes the air out. This creates low pressure inside. The higher air pressure outside pushes it onto the smooth surface making, it stick.
• Mountain climbers often experience nosebleeds at high altitudes because as altitude increases atmospheric pressure decreases.
• For the smooth flow of liquid there should be air flow. In the absence of a hole, air cannot enter the bottle and the flow stops. Making a hole lets air into the injection bottle and keeps the pressure balanced, ensuring continuous flow.
• As the altitude increases the atmospheric pressure decreases. But the pressure inside the ears remain the same for a while and this difference in pressure causes a feeling of pain in the ears.

Class 8 Basic Science Chapter 3 Question Answer Extended Activities

Question 1.
Conduct a study on the topic ‘Rain and Pressure’ and prepare a project report.
Answer:
An example of a project report is shown below.
Title: Rain and pressure

Introduction:
Rainfall is one of the most important phenomena. It plays a major role in agriculture, environment and many more day to day activities. One of the important factors that influence weather patterns is pressure. Here we explore the relationship between pressure and rain.

Main points to be included in the content

• The continuous movement of water through the phases of evaporation, condensation, precipitation, and collecting is known as the water cycle.
• Water vapour, which rises into the atmosphere, is created when the Sun heats water in lakes, rivers, and seas.
• In regions of low pressure, the air is generally less dense and rising.
• As warm air rises it cools down.
• When air cools, the vapour in it turns back into small liquid droplets or ice crystals to form clouds. This process is called condensation.
• Precipitation is the term for the water that returns to the earth as rain, snow, or hail when the clouds becomes heavy.
• The cycle then restarts when the water gets collected in lakes, rivers, or underground, with atmospheric pressure being a major factor in causing rain to fall and water vapour to rise.
• When the air pressure is high it means the air is generally more dense and sinking.
• Sinking air warms up.
• Warm air can hold more water vapour without condensing into droplets.
• So high pressure areas are usually associated with sunny weather and clear skies.
  (You can also refer to weather reports in newspapers and include points from it, add flowcharts and diagrams, various weather study reports)

Conclusion
Pressure and rainfall are interrelated.

Question 2.
Attach a pipe to each end of a plastic bottle, ensuring it airtight. One pipe should be longer and the other shorter. Place one end of the pipe in a bucket filled with water and placed at a highest level. Place the shorter pipe in another vessel. Squeeze and release the plastic bottle two or three times. Does the water from the bucket flow into the lower vessel? State the reason for this.
Answer:
Yes, the water from the bucket will flow into the lower vessel. When the bottle is squeezed, the pressure inside it increases. This higher pressure pushes air into the water through the longer pipe and forces water to rise up through it and then water is pushed down the longer pipe and into the shorter pipe, which then flows into the lower vessel due to gravity.

Pressure Class 8 Notes

Class 8 Basic Science Pressure Notes Kerala Syllabus

• The force acting normally on a surface is called thrust. The force acting normally per unit area is pressure.
• If pressure is denoted by P, thrust by F and area by A, Pressure = \(\frac{\text { Thrust }}{\text { Area }}\) P = \(\frac{\mathrm{F}}{\mathrm{~A}}\)
• Unit of pressure = \(\frac{\text { Unit of thrust }}{\text { Unit of area }}\). The SI unit of pressure is pascal. 1 pascal = 1N/m²
• When thrust remains constant, the pressure is inversely proportional to the surface area. This means, when the surface area increases, the pressure decreases and vice versa.
• The normal force exerted by a liquid is called thrust. The thrust that a liquid exerts per unit area is called liquid pressure.
• The factors affecting liquid pressure are
  • Height of the liquid column (h)-As height of the liquid column increases, liquid pressure increases.
  • Density of the liquid (d)-As density of liquid increases, liquid pressure increases.
• The instrument used to measure liquid pressure is called a manometer.
• The force exerted by a gas normally on a unit area is gas pressure.
• Factors influencing gas pressure.
  • Number of particles-As number of particles increases, gas pressure increases.
  • Volume-As volume increases, gas pressure decreases.
  • Temperature-As temperature increases, gas pressure increases.
• Atmospheric air exerts pressure on objects.
• The weight of air column per unit area on the earth’s surface is atmospheric pressure. The unit of atmospheric pressure is ‘bar’.
• Atmospheric pressure at sea level is known as standard atmospheric pressure. The instrument used to measure atmospheric pressure is a barometer.
• Gas pressure is exerted in all directions, similar to that of liquids.

INTRODUCTION

The concept of pressure is important in science and it is an inevitable part of our day to day life. It explains simple daily life situations like using a sharp knife to cut better, making use of syringes, air pumps to applications like making the base of dams wider and weather predictions. Not only solids, but liquids and gases also exert pressure. This chapter deals with topics like pressure and its measurement, liquid pressure, gas pressure and atmospheric pressure.

PRESSURE AND ITS MEASUREMENT
We have learnt about different types of forces. The earth exerts an attractive force on all objects. That is the weight of the object.

ACTIVITY TO FIND OUT THE FACTORS LIQUID PRESSURE DEPENDS ON
Make a hole at the bottom of a plastic bottle. Close the hole and fill the bottle with water. Open the hole and observe the water jetting out.

Repeat the experiment using bottles of different shapes and sizes.

MAKING A MANOMETER

Fix a plastic tube on a board in a U shape. Fill it with water. Connect a funnel to one end of the tube as shown in the figure. Make a diaphragm with a balloon on the mouth of the funnel. Attach a scale to the board. Now the manometer is ready.
Apply slight pressure with your finger on the balloon.

GAS PRESSURE
Just like solids and liquids, gases also have the ability to exert pressure. As we know, pressure is measured when air is filled into vehicle tyres.

Pressure is the force exerted by the particles of a gas on unit area of a surface. This is due to the collision of gas particles.

The force exerted by a gas normally on a unit area is gas pressure.

ATMOSPHERIC PRESSURE
The atmosphere is a blanket of air surrounding the Earth. The density of air decreases as we go higher.

ACTIVITY TO UNDERSTAND IF ATMOSPHERE CAN EXERT PRESSURE
Place a scale on the table as shown in the figure. Drop an ice cream ball on the scale from a small height. The scale falls down. Place an A4 size paper on the scale. Drop the ball again from the same height.

Now, the scale does not fall down. It is because of the weight of the air above the paper. The atmospheric pressure counteracts the force exerted by the ball.

Atmospheric air exerts pressure on objects.

From the above activity, we understood the influence of atmospheric pressure.

The weight of air column per unit area on the earth’s surface is atmospheric pressure. The unit of atmospheric pressure is ‘bar’.

PRESSURE BALANCE
Activity
Pour water in a flat container. Place a lit candle in it. Cover the candle with a glass.

Reviewing SCERT Class 8 Basic Science Solutions and Kerala Syllabus Class 8 Basic Science Chapter 2 Motion and Force Question Answer Notes Pdf can uncover gaps in understanding.

Class 8 Basic Science Chapter 2 Motion and Force Question Answer Notes

Class 8 Basic Science Chapter 2 Notes Kerala Syllabus Motion and Force Question Answer

Motion and Force Class 8 Questions and Answers Notes

Let’s Assess

Question 1.
A car starts from A and reaches B, which is 75 m away. The uniform speed of the car is 25 m/s. Another car with a uniform speed of 30 m/s starts from A and reaches B through C. Which car will reach the destination first?

Answer:
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)

Car 1
Distance travelled_from A to B = 75 m
Speed of car 1 = 25 m/s
Time taken bv Car 1 = \(\frac{\text { Distance }}{\text { Speed }}\) = 75 ÷ 25 = 3 S

Car 2
Distance travelled from A to B through C = AC + CB = 50 + 70 = 120 m
Speed of car 2 = 30 m/s
Time taken by Car 2 = \(\frac{\text { Distance }}{\text { Speed }}\) = 120 ÷ 30 = 4 s
So Car 1 will reach the destination first because it takes less time.

Question 2.
A bus starts from C and reaches D in 7 s. If the uniform speed of the bus is 50 m/s, find the distance from C to D.
Answer:
Speed of the bus = 50 m/s
Time taken = 7 s
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
Distance travelled from C to D = Speed × Time = 50 × 7 = 350 m

Question 3.
How long will it take to hear thunder from 12000 m away? (The speed of sound is 340 m/s).
Answer:
Speed of sound = 340 m/s
Distance travelled by thunder = 12000 m
Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)
Time taken to hear thunder \(\frac{\text { Distance }}{\text { Speed }}\) = 12000 ÷ 340 = 35.29 s

Question 4.
Complete the puzzle given below:

Answer:

• Starts moving a stationary object
• Changes the direction of motion
• Changes the speed of a moving object
• Changes the shape of an object
• Changes the size of an object

Basic Science Class 8 Chapter 2 Question Answer Kerala Syllabus

Question 1.
Analyze the picture given above. What are the situations shown?
Answer:

• Child sitting on a moving giant wheel
• Person riding a horse
• Log of wood in a moving lorry
• Twig in the beak of a flying bird
• Earth in the solar system
• A book on a table

Question 2.
Tabulate the above situations that the objects change their position relative to the surroundings and those do not.

Answer:

Objects changing position relative to surroundings	Objects not changing position relative to surroundings
• Child sitting on a moving giant wheel • Person riding a horse • Log of wood in a moving lorry • Twig in the beak of a flying bird • Earth in the solar system	• A book on a table

Objects that change the position with respect to their surroundings are considered to be in motion, while those that do not change position are considered to be stationary.

Question 3.
The above situations are given in the table below. Complete the table suitably by putting ✓ marks.

Answer:

From the table, we can infer that to determine whether an object is moving, we need to refer to another object. The object that is used as a reference is called the reference object.

The object taken to determine the state of motion or the state of rest of a body is called the reference object.

If an object changes its position relative to the reference object, it is said to be in motion, and if it does not change position, it is said to be stationary. Moving objects undergo a change in position.

Question 4.
The picture of a 400 m track is given below.

What will be the length of the path travelled by an athlete completing two rounds on this track?
Answer:
400 + 400 = 800 m

Distance is the length of path travelled by an object. The SI unit of distance is metre.

Odometre

An odometre records distance travelled by a vehicle in kilometre.

Question 5.
Since 1 km = 1000 m, calculate how far a vehicle has travelled so far, according to its odometer given above.
Answer:
1,44,969 km = 14,49,69,000 m

Question 6.
Figure 2.5 shows a railway track from A to F. Note down the distance traveled by the train upon reaching each place.

Answer:

Place	Distance traveled
At A	0 km
While reaching B	60 km
While reaching C	110 km
While reaching D	150 km
While reaching E	250 km
While reaching F	320 km

Question 7.
Bus A travelled 75 m in 5 seconds. Bus B travelled 112 m in 7 seconds. Which bus was faster?
Answer:
Speed of Bus A = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{75 m}{5 s}\) = 15 m/s
Speed of Bus B = \(\frac{\text { distance travelled }}{\text { time taken }}\) = \(\frac{112 m}{7 s}\) = 16 m/s
Bus B was faster.
Speedometer

Speedometre is the device that shows the speed of a vehicle.

Question 8.
The details of 3 children who participated in a 400 m running race in school sports meet are given in the table below. Can we find the fastest athlete among them.

Answer:

Child	Distance	Time	Speed
A	400 m	180 s	\(\frac{400}{180}\) = 2.22 m/s
B	400 m	120 s	\(\frac{400}{120}\) = 3.33 m/s
C	400 m	400 s	\(\frac{400}{400}\) = 1 m/s

Here, speed is the distance travelled in one second. The distance travelled in one second will not always be the same.

Question 9.
Look at the clock dial shown.
a) Complete the table based on the movement of the tip of the second hand (P in the figure). You can measure the distance between different points using a thread.


Answer:

Change in position	Distance (cm)	Time (s)
From A to C	10	10
From C to E	10	10
From E to G	10	10

Change in position	Distance (cm)	Time (s)
From A to D	15	15
From D to G	15	15
From G to J	15	15

b) What is the distance travelled in every 10 seconds?
Answer:
10 cm

c) What is the distance travelled in every 15 seconds?
Answer:
15 cm

If an object covers equal distances in equal time intervals, it is said to have uniform speed.

If an object travels equal distance in equal intervals of time, it is in uniform speed.

In our daily life, all movements are not in uniform speed.
Birds flying, humans walking, a ball rolling, etc., do not move with uniform speed. They move with nonuniform speed.

If an object travels unequal distance in equal intervals of time, it is in non-uniform speed.

Question 10.
We see vehicles ranging from slow-moving bicycles to high-speed cars on the roads. Road accidents due to excessive speed and carelessness are daily news. What can we do to avoid such road accidents? Discuss
Answer:
Road accidents caused by excessive speed and carelessness happen every day. To avoid such accidents, we : should always drive carefully and follow traffic rules. We must not rush or speed while driving. Wearing seat belts, paying attention while crossing the road, and avoiding distractions like using mobile phones while driving are also important. Being alert and responsible while on the road helps keep everyone safe.

Question 11.
Examine the figures.


Identify whether the force applied in each case is a contact or a non-contact force.

Answer:

Situation	Force	Through Contact / Non-Contact
Mango falling downward	Earth’s Gravitational Force	Through Non-Contact
Leaves swaying in the wind	Force applied by wind	Through Contact
Magnet attracting a pin	Magnetic force	Through Non-Contact
Hammering a nail	Muscular force	Through Contact
Swimming	Muscular force	Through Contact

The force experienced when objects come into contact with each other is called contact force. The force effected when there is no contact with the object is called non-contact force.

The standard unit of force is called the newton. It can be represented by the letter N. The capital letter is used because it is the first letter of the scientist’s name.

Question 12.
Where do we make use of friction in our life?
Answer:
Friction helps us walk, hold objects, drive, write, and stop vehicles. It provides grip and prevents slipping in everyday life.

Question 13.
Rub your palms together. Do you feel the warmth?
Answer:
When you rub your palms together, you can feel warmth. This happens because rubbing creates friction, which generates heat. Friction is the force that opposes the motion of two surfaces sliding over each other. So, the more you rub your palms, the more heat is produced, making your palms warm.

The same thing happens when striking a matchstick and lighting a lighter.

Advantages of friction

• Helps to hold objects firmly.
• Helps us in walking.
• Helps vehicles to move without slipping.
• Slows the fall and keeps the parachute open.

Question 14.
Is friction always beneficial?
Answer:
No, friction is not always beneficial. It has both advantages and disadvantages depending on the situation. Friction is helpful when we need to hold on or stop slipping. But in machines, too much friction is a problem because it wastes energy and makes parts wear out.

Disadvantages of friction

• Surfaces in contact wears out.
• Obstructs smooth movement of machine parts.
• Continuous friction between the bones during movement causes knee wear.

Question 15.
List some examples of lubricants.
Answer:

• Oil (such as motor oil, cooking oil)
• Grease (used in bicycles, machines)
• Wax (like bees wax)
• Soap solutions (used in some machinery)
• Graphite is a solid lubricant. It is commonly used as a lubricant between machine parts at high temperature.

Ball bearings seen in connection with tyres in vehicles and between hubs and the axles of ceiling fans and bicycles are used to reduce friction with the axle.

Airplanes and boats are made in special shapes to reduce friction. Smooth, streamlined designs help them move easily through air and water, making travel faster and easier by decreasing the resistance they face.

This method of reducing friction by changing the shape is called streamlining.

Question 16.
Write the various ways to reduce friction?
Answer:

• Use lubricants like oil, grease, or wax.
• Make objects streamlined.
• Use ball bearings in machines and vehicles.

Class 8 Basic Science Chapter 2 Question Answer Extended Activities

Question 1.
Prepare and present a seminar paper on the ways to reduce road accidents.
Answer:
Hints
Main Ways to Reduce Road Accidents
1. Follow Traffic Rules

• Obey speed limits
• Use signals and helmets

2. Improve Road Conditions

• Maintain roads properly
• Install clear signs

3. Raise Awareness

• Educate people about road safety
• Conduct awareness campaigns

4. Enforce Laws Strictly

• Penalties for Violations
• Regular checks for drunk driving

5. Use Technology

• Speed cameras
• Traffic lights and CCTV

6. Encourage Public Transport
• Less traffic, safer roads

7. Pedestrian Safety

• Crosswalks and footpaths
• Educate pedestrians

Question 2.
Prepare a science article on friction in daily life.
Answer:

• Friction is a force that opposes the motion of two surfaces sliding against each other. It is a common and important force we experience every day. Friction helps us in many activities, but sometimes it can also cause problems.
• Friction occurs when two surfaces come into contact and resist sliding. It acts in the opposite direction to movement. The amount of friction depends on the nature of the surfaces and how hard they press together.

Uses of Friction in Daily Life

• Walking and Running: Friction between shoes and the ground prevents slipping.
• Writing: Friction between pen and paper helps in writing.
• Driving: Friction between tyres and road helps in moving the vehicle without slipping.
• Climbing: Friction helps climbers grip rocks or walls.

Problems Caused by Friction

• Friction causes wear and tear of objects, like shoes or machine parts.
• It can slow down moving objects and waste energy.

How to Reduce Friction

• Use lubricants like oil or grease on machines.
• Smoothen surfaces to decrease friction.
• Use wheels and ball bearings.

Motion and Force Class 8 Notes

Class 8 Basic Science Motion and Force Notes Kerala Syllabus

• If an object changes its position relative to the reference object, it is said to be in motion, and if it does not change position, it is said to be stationary. Moving objects undergo a change in position.
• Distance is the length of path travelled by an object. The SI unit of distance is metre.
• An odometre records distance travelled by a vehicle in kilometre and a speedometre is the device that shows the speed of a vehicle in km/h.
• Speed is the distance travelled by an object in unit time. The SI unit of speed is m/s.
• If an object travels equal distance in equal intervals of time, it is in uniform speed.
• If an object travels unequal distance in equal intervals of time, it is in non-uniform speed.
• Force is a push or pull that changes the shape, size, volume, state of rest or state of motion of a body. The standard unit of force is called the newton.

INTRODUCTION

Understanding how objects move and interact is fundamental for the study of physics. This chapter introduces key concepts related to motion, including the difference between being at rest and in motion, and how distance and speed describe an object’s movement. We explore the forces that cause changes in motion, distinguishing between contact and non-contact forces, and examine the role of friction, a force that both opposes motion and its numerous practical applications in daily life. By grasping these fundamental ideas, we can better understand motion and force seen around us and learn how to control them to make our daily activities easier and safer.

STATE OF MOTION AND STATE OF REST
In our daily life we always engage in some activities. We move in various situations like running, playing and walking. A bird flying, a car running, leaves swaying in the wind – all these are different forms of motion.

SPEED

A photo finish picture of the 100 m race in the Olympics is given above. The athlete who completed 100 m in the least time would be the winner.

Experiment

Take a tall glass jar. Mark the top and bottom as A and B, respectively. Fill the glass jar with glycerin and drop a stone from the top. Start the stopwatch when the stone reaches A. Stop the watch when the stone reaches B.
Time taken for the stone to reach B = 5 s
Distance travelled = 30 cm
Distance travelled bv the stone in one second = \(\frac{30}{5}\) = 6 cm/s

Speed is the distance travelled by an object in unit time. The SI unit of speed is m/s. Speed = \(\frac{\text { distance travelled }}{\text { time taken }}\)

Although the SI unit of speed is m/s, the speed of vehicles is usually expressed in km/h.

Relation between km/h and m/s.
1 km/h = \(\frac{1000 m}{3600 s}\) = \(\frac{5}{18}\) m/s

FORCE

• Pulling a trolley
• Pushing a car
• Pushing a wall
• Pulling a table

We can see a push or a pull in all these situations. This is force. It is not only applied during pushing and pulling. Force can be applied in many ways, such as pressing, lifting, twisting, or even when objects attract each other (like gravity). Any action that causes a change in the motion or shape of an object involves applying force.

• Catching a ball
• Rolling dough to chapati
• Kicking a ball
• Hitting a cork

Here, we can see that force is applied to change the shape and direction of the object and to stop a moving object.

Force is a push or pull that changes the shape, size, volume, state of rest or state of motion of a body.

In all the situations shown above, force is applied through direct contact with the object. It is not necessary to have a direct contact in all the cases.

FRICTIONAL FORCE

You must have noticed that a ball rolling freely on the floor gradually comes to rest. The force responsible for this is called friction.

Let’s try an activity.
Take a wooden block and make one side smooth leaving the other rough. Slide its smooth surface down the inclined plane Then try to slide its rough side down the inclined plane.

Observation
More force is effected against the movement on a rough surface. This is frictional force. You can see that the speed decreases with the increase in friction.

When a surface moves or tries to move over another surface, a parallel force is produced between them against their relative motion. This is frictional force.

Activity

Slide a heavy box on a rough surface. Then move the same box on a trolley with tyres.

Observation

• The box moves faster on the trolley with tyres.
• The frictional force decreases in this case because rolling friction (or rolling resistance) is less than sliding friction.

Using tyres reduces the frictional force, making it easier and quicker to move the box.

More friction occurs when sliding. This is called sliding friction. When using a trolley, the tyre rolls. This is rolling friction. Sliding friction is greater than rolling friction.

WAYS TO REDUCE FRICTION
To understand the ways to reduce friction, we need to know the factors that influence it.
Activity

Take a wooden block, an ice block, and a rubber piece of the same mass. Allow them to slide down an inclined plane. The area in contact with the inclined surface should be equal in all cases.

Observations

• The rubber piece will slide slowly because of higher friction.
• The wooden block will slide faster than rubber but slower than ice.
• The ice block will slide the fastest because it has the least friction.

Nature of the surfaces in contact influence the frictional force
Experiment
Take some water in a glass jar and drop a round stone and a sharp-edged stone of equal mass into the water.

Observation
The round stone has a smooth shape, so it moves easily through water. It faces less friction and falls faster.
The sharp-edged stone has an uneven, rough shape. It faces more friction and falls slower.
From this, we can understand that the shape of the object affects friction.

Factors that affect friction

• Nature of surfaces in contact:
  Rough surfaces create more friction than smooth surfaces.
• Area of contact

Larger contact areas can increase friction.
We have seen that friction occurs when our palms are rubbed together. If oil is applied on the hands and rubbed together, they slip quickly. From this, it can be assumed thar friction is reduced when oil is applied.

Substances that help to reduce friction between contacting surfaces are lubricants.

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Class 8 Malayalam Adisthana Padavali Notes Pdf

8th Standard Malayalam Adisthana Padavali Question Answer Textbook Solutions Notes

SCERT Class 8 Malayalam Adisthana Padavali Solutions

Unit 1 കനിവും കരുതലും

• Chapter 1 കുരുവിയും കാട്ടുതീയും Notes – Kuruviyum Kaattu Theeyum Question Answer
• Chapter 2 കൊച്ചുദേവദാരു Notes – Kochu Devadaru Question Answer
• Chapter 3 പെരുമഴയത്ത് Notes – Perumazhayath Question Answer

Unit 2 ആയുരാരോഗ്യസൗഖ്യം

• Chapter 4 പ്രത്യാശയുടെ കിരണങ്ങൾ Notes – Prathyashayude Kiranangal Question Answer
• Chapter 5 ഒപ്പം മിടിക്കുന്നത് Notes – Oppam Midikkunnath Question Answer
• Chapter 6 സ്നേഹപൂർവം, അമ്മ Notes – Snehapoorvam Amma Question Answer

Unit 3 ഹൃദയംതൊടുന്ന നക്ഷത്രങ്ങൾ

• Chapter 7 അമ്മമ്മ Notes – Ammamma Question Answer
• Chapter 8 തോട്ടക്കാരി Notes – Thottakkari Question Answer
• Chapter 9 ഭഗീരഥപ്രയത്നം Notes – Bhagiratha Prayatnam Question Answer

Adisthana Padavali Class 8 Notes Chapters Summary

• Chapter 1 Kuruviyum Kaattu Theeyum Summary
• Chapter 2 Kochu Devadaru Summary
• Chapter 3 Perumazhayath Summary
• Chapter 4 Prathyashayude Kiranangal Summary
• Chapter 5 Oppam Midikkunnath Summary
• Chapter 6 Snehapoorvam Amma Summary
• Chapter 7 Ammamma Summary
• Chapter 8 Thottakkari Summary
• Chapter 9 Bhagiratha Prayatnam Summary

SCERT Class 8 Malayalam Adisthana Padavali Solutions (Old Syllabus)

Unit 1 പിന്നെയും പൂക്കുമീ ചില്ലകൾ

• Chapter 1 പുതുവർഷം Notes (Puthuvarsham)
• Chapter 2 ആ വാഴവെട്ട് Notes (Aa Vazha Vettu)
• Chapter 3 എണ്ണനിറച്ച കരണ്ടി Notes (Enna Niracha Karandi)

Unit 2 കണ്ണുവേണമിരുപുറമെപ്പോഴും

• Chapter 4 രണ്ടു മത്സ്യങ്ങൾ Notes (Randu Malsyangal)
• Chapter 5 കിട്ടും പണമെങ്കിലിപ്പോൾ Notes (Kittum Panam Enkil Ippol)
• Chapter 6 തേൻകനി Notes (Thenkani)

Unit 3 ഓർമ്മകൾക്കെന്തു സുഗന്ധം!

• Chapter 7 കുപ്പായം Notes (Kuppayam)
• Chapter 8 ബഷീർ എന്ന ബല്യ ഒന്ന് Notes (Basheer Enna Balya Onnu)
• Chapter 9 നനയാത്ത മഴ Notes (Nanayatha Mazha)

Malayalam Adisthana Padavali Class 8 Question Paper Pdf

• Class 8 Malayalam Adisthana Padavali Annual Exam Question Paper 2023-24
• Class 8 Malayalam Adisthana Padavali First Term Question Paper 2022-23
• Class 8 Malayalam Adisthana Padavali Second Term Question Paper 2022-23
• Class 8 Malayalam Adisthana Padavali Annual Exam Question Paper 2022-23
• Std 8 Malayalam Adisthana Padavali Annual Exam Question Paper 2021-22

Class 8 Adisthana Padavali Notes Chapters Summary (Old Syllabus)

• Chapter 1 Puthuvarsham Summary
• Chapter 2 Aa Vazha Vettu Summary
• Chapter 3 Enna Niracha Karandi Summary
• Chapter 4 Randu Malsyangal Summary
• Chapter 5 Kittum Panam Enkil Ippol Summary
• Chapter 6 Thenkani Summary
• Chapter 7 Kuppayam Summary
• Chapter 8 Basheer Enna Balya Onnu Summary
• Chapter 9 Nanayatha Mazha Summary

We hope the given 8th Class Std Malayalam Adisthana Padavali Notes, 8th Standard Malayalam Adisthana Padavali Notes will help you. If you have any queries regarding Adisthana Padavali Class 8 Notes, Malayalam Notes Class 8 Kerala Syllabus Adisthana Padavali drop a comment below and we will get back to you at the earliest.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 4 Arithmetic of Parts Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 4 Solutions Arithmetic of Parts

Class 6 Kerala Syllabus Maths Solutions Chapter 4 Arithmetic of Parts Questions and Answers

Arithmetic of Parts Class 6 Questions and Answers Kerala Syllabus

Intext Questions (Page No. 50)

In each of the pictures below, write the parts of each colour and the total coloured parts as fractions. Write the sum of the fractions obtained from each picture in the lowest terms.

Question 1.

Answer:
The fraction of circle coloured in blue = \(\frac {3}{8}\)
The fraction of circle coloured in green = \(\frac {1}{8}\)
The circle is divided into 8 equal parts.
Sum of fractions is \(\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\)

Question 2.

Answer:
The fraction of circle coloured in orange = \(\frac {3}{8}\)
The fraction of circle coloured in brown = \(\frac {3}{8}\)
The circle is divided into 8 equal parts.
Sum of fractions is \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)

Question 3.

Answer:
The fraction coloured in yellow = \(\frac {1}{10}\)
The fraction coloured in red = \(\frac {3}{10}\)
The ribbon is divided into 10 equal parts.
Sum of fractions is \(\frac{1}{10}+\frac{3}{10}=\frac{4}{10}=\frac{2}{5}\)

Question 4.

Answer:
The fraction coloured in yellow = \(\frac {5}{16}\)
The fraction coloured in green = \(\frac {7}{16}\)
The ribbon is divided into 10 equal parts.
Sum of fractions is \(\frac{5}{16}+\frac{7}{16}=\frac{12}{16}=\frac{3}{4}\)

Addition of Fractions (Page No. 56)

Question 1.
In each pair of pictures below, find the fraction of the circle we get by cutting up the coloured pieces of both circles and putting them together:

Answer:

Question 2.
Calculate the sums given below:
(i) \(\frac{1}{4}+\frac{1}{8}\)
(ii) \(\frac{3}{4}+\frac{1}{6}\)
(iii) \(\frac{1}{3}+\frac{2}{5}\)
(iv) \(\frac{1}{2}+\frac{2}{5}\)
(v) \(\frac{2}{3}+\frac{1}{5}\)
Answer:

Question 3.
There are two taps to fill a tank with water. If the first tap alone is opened, the tank would fill up in 10 minutes. If the second tap alone is opened, it would take 15 minutes to fill up the tank.
(i) If the first tap alone is opened, what fraction of the tank would be filled in one minute?
(ii) If the second tap alone is opened, what fraction of the tank would be filled in one minute?
(iii) If both the taps are opened, what fraction of the tank would be filled in one minute?
(iv) If both the taps are opened, how much time would it take for the tank to be filled up?
Answer:
(i) The first tap fills the whole tank in 10 minutes.
Therefore, in 1 minute it fills upto \(\frac {1}{10}\) of the tank.

(ii) The second tap fills the whole tank in 15 minutes.
Therefore, in 1 minute it fills upto \(\frac {1}{15}\) of the tank.

(iii) If both taps are opened,
First tap fills \(\frac {1}{10}\)
Second tap fills \(\frac {1}{15}\)
So together they fill = \(\frac{1}{10}+\frac{1}{15}=\frac{1 \times 3}{10 \times 3}+\frac{1 \times 2}{15 \times 2}\) = \(\frac{3}{30}+\frac{2}{30}=\frac{5}{30}=\frac{1}{6}\)
That means both taps together fill \(\frac {1}{6}\) of the tank in 1 minute.

(iv) The time taken to fill the tank when both taps are opened.
\(1 \div \frac{1}{6}=\frac{1}{\frac{1}{6}}=1 \times \frac{6}{1}\) = 6 minutes

Some Other Sums (Page No. 57)

Question 1.
For each fraction given below, can you mentally calculate the fraction to be added to make it 1?
(i) \(\frac {2}{7}\)
(ii) \(\frac {4}{7}\)
(iii) \(\frac {3}{8}\)
(iv) \(\frac {3}{10}\)
Answer:
(i) \(\frac {5}{7}\) is added with \(\frac {2}{7}\) to get 1.
That is, \(\frac{2}{7}+\frac{5}{7}=\frac{7}{7}\) = 1

(ii) \(\frac {3}{7}\) is added with \(\frac {4}{7}\) to get 1.
That is, \(\frac{3}{7}+\frac{4}{7}=\frac{7}{7}\) = 1

(iii) \(\frac {5}{8}\) is added with \(\frac {3}{8}\) to get 1.
That is, \(\frac{3}{8}+\frac{5}{8}=\frac{8}{8}\) = 1

(iv) \(\frac {7}{10}\) is added with \(\frac {3}{10}\) to get 1.
That is, \(\frac{7}{10}+\frac{3}{10}=\frac{10}{10}\) = 1

Textbook Page No. 60

Question 1.
Calculate the sums below:
(i) \(\frac{5}{6}+\frac{1}{3}\)
(ii) \(\frac{7}{8}+\frac{1}{4}\)
(iii) \(\frac{5}{6}+\frac{1}{4}\)
(iv) \(\frac{5}{8}+\frac{3}{4}\)
(v) \(2 \frac{1}{3}+3 \frac{1}{2}\)
Answer:

Question 2.
One jar contains one and a half litres of milk, and another contains two and three-quarters litres of milk. How much milk is in both jars together?
Answer:
Milk contained in first jar = 1\(\frac {1}{2}\) litres
Second jar contained in second jar = 2\(\frac {3}{4}\) litres
Milk contained in both jars is,

Question 3.
Two strings of lengths one and a half metres are joined end to end. What is the total length?
Answer:
Lengths of strings = 1\(\frac {1}{2}\)
Two strings are joined together,
\(1 \frac{1}{2}+1 \frac{1}{2}=(1+1)+\left(\frac{1}{2}+\frac{1}{2}\right)\)
= 2 + \(\frac {2}{2}\)
= 2 + 1
= 3 metres

Question 4.
Ahirath bought one and a half kilograms of beans and three-quarters kilograms of yams. What is the total weight?
Answer:
Quantity of beans Ahirath bought = 1\(\frac {1}{2}\) kg
Quantity of yam Ahirath bought = \(\frac {3}{4}\) kg

Removing Parts (Page No. 61)

Question 1.
(i) \(\frac{1}{2}-\frac{1}{8}\)
Answer:

(ii) \(\frac{3}{4}-\frac{1}{8}\)
Answer:

(iii) \(\frac{1}{3}-\frac{1}{5}\)
Answer:

(iv) \(\frac{2}{5}-\frac{1}{3}\)
Answer:

(v) \(\frac{2}{3}-\frac{1}{5}\)
Answer:

Textbook Page No. 64

Question 1.
Natasha drew a circle and coloured \(\frac {5}{12}\) of it. What fraction of the circle remains to be coloured?
Answer:
Coloured portion of the circle = \(\frac {5}{12}\)
Portion of the circle to be coloured is,
\(1-\frac{5}{12}=\frac{12}{12}-\frac{5}{12}\)
= \(\frac{12-5}{12}\)
= \(\frac {7}{12}\)

Question 2.
A bucket can hold 10 litres of water, and it contains 3\(\frac {3}{4}\) litres. How much more is needed to fill it?
Answer:
Total quantity of water that can be held in the bucket = 10 litres
Quantity of water contained in the bucket = 3\(\frac {3}{4}\) litres
The quantity of water needed to fill the bucket is,
10 – 3\(\frac {3}{4}\) = \(\left(9 \frac{1}{4}+\frac{3}{4}\right)-\left(3 \frac{3}{4}\right)\)
= \(\left(9+\frac{1}{4}+\frac{3}{4}\right)-\left(3+\frac{3}{4}\right)\)
= (9 – 3) + \(\left(\frac{1}{4}+\frac{3}{4}-\frac{3}{4}\right)\)
= 6 + \(\frac {1}{4}\)
= 6\(\frac {1}{4}\) litres
OR
If \(\frac {1}{4}\) added to 3\(\frac {3}{4}\), we get 4
And if 6 is added to 4 will results 10
That means, 6 + \(\frac {1}{4}\) = 6\(\frac {1}{4}\) litres

Question 3.
From a string, one and three-quarters of a metre long, a piece half a metre long is cut off. What is the length of the remaining piece?
Answer:
Length of the original string = 1\(\frac {3}{4}\) metre
Piece cut off = \(\frac {1}{2}\) metre
Length of the remaining piece is

Question 4.
A panchayat constructed a new road, 14\(\frac {3}{4}\) kilometres long last year. This year, a road 16\(\frac {1}{4}\) kilometres long was constructed. How much more was constructed this year?
Answer:
Road constructed last year = 14\(\frac {3}{4}\) km
Road constructed this year = 16\(\frac {1}{4}\) km
The more roads constructed this year are,

Question 5.
Ashadev bought 20 metres of string. He cut off a piece 5\(\frac {3}{4}\) metres long first, and then a piece 6\(\frac {1}{2}\) metres long later. What is the length of the string left?
Answer:
Total length of the string = 20 metres
First cut piece = 5\(\frac {3}{4}\) metre
Second cut piece = 6\(\frac {1}{2}\) metres
Total length of the string cut is,

Length of the remaining string is,

Question 6.
The milk society got 75\(\frac {1}{4}\) litres in the morning and 55\(\frac {1}{2}\) litres in the evening. Of this 85\(\frac {3}{4}\) litres of milk is sold. How much milk is left?
Answer:
Milk received in the morning = 75\(\frac {1}{4}\) litres
Milk received in the evening = 55\(\frac {1}{2}\) litres
Milk sold is = 85\(\frac {3}{4}\) litres
Remaining milk is,

Large and Small (Page No. 66)

Question 1.
Find the larger and smaller of each pair of fractions below and write this using the < or > symbol:
(i) \(\frac{2}{5}, \frac{3}{5}\)
(ii) \(\frac{2}{5}, \frac{2}{3}\)
(iii) \(\frac{2}{5}, \frac{3}{4}\)
(iv) \(\frac{3}{7}, \frac{2}{9}\)
(v) \(\frac{2}{7}, \frac{3}{8}\)
(vi) \(\frac{4}{9}, \frac{3}{8}\)
Answer:
(i) Here, the denominators are the same.
So compare the numerators.
That is 2 < 3,
Therefore \(\frac{2}{5}<\frac{3}{5}\)

(ii) Here, the numerators are the same.
So the smaller denominator is larger.
Therefore \(\frac{2}{3}>\frac{2}{5}\)

(iii) Convert the fractions into forms with the same denominators
\(\frac{2}{5}=\frac{2 \times 4}{5 \times 4}=\frac{8}{20}\) and \(\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)
Therefore \(\frac{2}{5}<\frac{3}{4}\)

(iv) Convert the fractions into forms with the same denominators
\(\frac{3}{7}=\frac{3 \times 9}{7 \times 9}=\frac{27}{63}\) and \(\frac{2}{9}=\frac{2 \times 7}{9 \times 7}=\frac{14}{63}\)
Therefore \(\frac{3}{7}>\frac{2}{9}\)

(v) Convert the fractions into forms with the same denominators
\(\frac{2}{7}=\frac{2 \times 8}{7 \times 8}=\frac{16}{56}\) and \(\frac{3}{8}=\frac{3 \times 7}{8 \times 7}=\frac{21}{56}\)
Therefore \(\frac{2}{7}<\frac{3}{8}\)

(vi) Convert the fractions into forms with the same denominators
\(\frac{4}{9}=\frac{4 \times 8}{9 \times 8}=\frac{32}{72}\) and \(\frac{3}{8}=\frac{3 \times 9}{8 \times 9}=\frac{27}{72}\)
Therefore \(\frac{4}{9}>\frac{3}{8}\)

Question 2.
Arrange each triple of fractions below from the smallest to the largest and write it using the < symbol:
(i) \(\frac{2}{5}, \frac{3}{4}, \frac{3}{5}\)
(ii) \(\frac{3}{7}, \frac{2}{9}, \frac{2}{7}\)
(iii) \(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}\)
Answer:
(i) Convert the fractions into forms with the same denominators
\(\frac{2}{5}=\frac{2 \times 4}{5 \times 4}=\frac{8}{20}, \quad \frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}, \quad \frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
\(\frac{8}{20}<\frac{12}{20}<\frac{15}{20}\)
Therefore \(\frac{2}{5}<\frac{3}{5}<\frac{3}{4}\)

(ii) Convert the fractions into forms with the same denominators
\(\frac{3}{7}=\frac{3 \times 9}{7 \times 9}=\frac{27}{63}, \quad \frac{2}{9}=\frac{2 \times 7}{9 \times 7}=\frac{14}{63}, \frac{2}{7}=\frac{2 \times 9}{7 \times 9}=\frac{18}{63}\)
\(\frac{14}{63}<\frac{18}{63}<\frac{27}{63}\)
Therefore \(\frac{2}{9}<\frac{2}{7}<\frac{3}{7}\)

(iii) Convert the fractions into forms with the same denominators
\(\frac{1}{2}=\frac{1 \times 3}{2 \times 3}=\frac{3}{6}, \quad \frac{1}{3}=\frac{1 \times 2}{3 \times 2}=\frac{2}{6} \quad, \frac{2}{3}=\frac{2 \times 2}{3 \times 2}=\frac{4}{6}\)
\(\frac{2}{6}<\frac{3}{6}<\frac{4}{6}\)
Therefore \(\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\)

Class 6 Maths Chapter 4 Kerala Syllabus Arithmetic of Parts Questions and Answers

Class 6 Maths Arithmetic of Parts Questions and Answers

Question 1.
Compare \(\frac {5}{8}\) and \(\frac {3}{4}\)?
Answer:
Convert the fractions into forms with the same denominators
\(\frac{5}{8}<\frac{6}{8}\)
Therefore \(\frac{5}{8}<\frac{3}{4}\)

Question 2.
Arrange from smallest to largest: \(\frac{1}{6}, \frac{1}{2}, \frac{5}{12}\)
Answer:
Convert the fractions into forms with the same denominators
\(\frac{1}{6}=\frac{1 \times 2}{6 \times 2}=\frac{2}{12}, \quad \frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12} \quad, \frac{5}{12}\)
\(\frac{2}{12}<\frac{5}{12}<\frac{6}{12}\)
Therefore \(\frac{1}{6}<\frac{5}{12}<\frac{1}{2}\)

Question 3.
Calculate: \(\frac{3}{5}+\frac{1}{2}\)
Answer:

Question 4.
A bottle contains 2\(\frac {1}{4}\) litres of juice, and another bottle contains 1\(\frac {2}{3}\) litres. What is the total amount of juice?
Answer:
Juice in first bottle = 2\(\frac {1}{4}\) litres
Juice in second bottle = 1\(\frac {2}{3}\) litres
Total amount of the juice is;

Question 5.
A ribbon is 2\(\frac {1}{2}\) metres long. Another ribbon is 3\(\frac {1}{3}\) metres long. What is their total length?
Answer:
First ribbon = 2\(\frac {1}{2}\) metres
Second ribbon = 3\(\frac {1}{3}\) metres
Total length is:

Question 6.
Reshma had 12\(\frac {1}{2}\) metres of ribbon. She used 4\(\frac {3}{4}\) metres for a project. How much ribbon is left?
Answer:
Total ribbon = 12\(\frac {1}{2}\) metres
Ribbon used = 4\(\frac {3}{4}\) metres
The remaining ribbon is,

Question 7.
A container holds 18 litres of oil. 6\(\frac {2}{3}\) litres are used in the morning and 5\(\frac {1}{6}\) litres in the evening. How much oil is left?
Answer:
Total quantity of oil held in the container = 18 litres
Oil used is,

The quantity of oil remaining is,

Question 8.
A pipe is 9\(\frac {1}{2}\) metres long. A piece 3\(\frac {3}{4}\) metres is cut from it. How much pipe remains?
Answer:
Length of the original pipe = 9\(\frac {1}{2}\) metres
Length of the cut piece = 3\(\frac {3}{4}\) metres
Length of the remaining pipe is,

Question 9.
A recipe needs 1\(\frac {1}{4}\) cups of sugar. Aliya only has \(\frac {3}{4}\) cup. How much more does she need?
Answer:
Required sugar = 1\(\frac {1}{4}\) cups
Available sugar = \(\frac {3}{4}\) cups
Sugar needed is,

Question 10.
Meera walked 2\(\frac {2}{5}\) kilometre in the morning and 3\(\frac {3}{5}\) kilometre in the evening. How much did she walk in total?
Answer:
Meera walked in the morning = 2\(\frac {2}{5}\) km
Walked in the evening = 3\(\frac {3}{5}\) km
Total she walked:

Class 6 Maths Chapter 4 Notes Kerala Syllabus Arithmetic of Parts

→ To find the sum of two fractional measures put together, we must see them as a set of the same equal pieces.

→ Convert both fractional measures to forms with the same denominator.

→ Increasing the numerator alone makes a fraction larger.

→ Of two fractions with the same denominator, the one with the larger numerator is the larger and the one with the smaller numerator is the smaller.

→ If the denominator alone of a fraction is increased, the fraction becomes smaller.

→ Of two fractions with the same numerator, the one with the larger denominator is smaller and the one with the smaller denominator is larger.

→ To compare two fractions with the same denominator, we need only compare the numerator; to compare two fractions with the same numerator, we need only compare the denominators.

In our everyday life, we often come across situations where we need to work with parts of a whole, like dividing a chocolate bar between friends, measuring ingredients for a recipe, or cutting ribbon into equal pieces. This idea of working with parts is what we study in this chapter, Arithmetic of Parts. This chapter mainly deals with understanding fractions – numbers that represent parts of a whole, dividing object quantities into equal parts, performing operations such as addition, subtraction, multiplication, and division with fractions, comparing and converting fractions, especially simplifying them to their lowest terms. This chapter helps build a strong foundation in dealing with fractions and their applications in real-life situations.

Joining Parts
A circle is divided into 4 equal parts, and two of them are joined together to get half of the circle.

The quarter circle joined to a quarter circle makes half a circle.
That means, two quarters make a half.
We can write like this: \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Suppose a circle is divided into eight equal parts ,and two of them are joined together.
2 of 8 equal parts is \(\frac {2}{8}\); also, \(\frac{2}{8}=\frac{1 \times 2}{4 \times 2}=\frac{1}{4}\)
So we have \(\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\)

Question 1.
Join \(\frac {1}{8}\) of a circle and \(\frac {3}{8}\) of the circle, what fraction of the circle would we get?
Answer:
We took 1 + 3 = 4 parts of 8 equal parts, that is \(\frac {4}{8}\)
Reducing the numerator and denominator,
\(\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\)

Question 2.
Look at the picture given below;

What sum of fractions of the ribbon do we get from this?
Answer:
The ribbon is divided into 8 equal parts
The fraction of the ribbon coloured in red = \(\frac {1}{8}\)
The fraction of the ribbon coloured in green = \(\frac {5}{8}\)
Sum of the fraction is, \(\frac{1}{8}+\frac{5}{8}=\frac{6}{8}=\frac{3}{4}\)

Addition of Fractions
If a circle is cut into four equal pieces and two of them joined together, we get half a circle:

If one more piece is joined to this, we get three-quarters of a circle.

That is half and a quarter make three quarters, \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

Question 1.
What sum do we get from this picture below?

Answer:
Here, the circle is divided into 8 equal pieces
The pieces coloured in red = 1
The pieces coloured in yellow = 4
Fraction of the coloured portion = \(\frac{1}{8}+\frac{4}{8}=\frac{5}{8}\)
The lowest form of \(\frac{4}{8}=\frac{1}{2}\)
Therefore the sum is \(\frac{1}{8}+\frac{1}{2}=\frac{5}{8}\)

Question 2.
Look at the picture below.

\(\frac {1}{4}\) of a circle and \(\frac {3}{8}\) of another circle of the same size are cut out, and these pieces are put together. What fraction of the full circle is this?
Answer:
In the first figure, the circle is divided into 4 equal parts, and 1 is coloured.
Here one one-fourth of the circle can be considered as two one-eighths joined together.
That is \(\frac{1}{4}=\frac{2}{8}\)
In the second figure, the circle is divided into 8 equal parts, and 3 is coloured.
That is \(\frac {3}{8}\)

Therefore, the fraction of the full circle is \(\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\)

Question 3.
Two ribbons of length \(\frac {3}{10}\) metre and \(\frac {2}{5}\) metre are joined end to end. What is the total length?
Answer:
The first ribbon is 3 of 10 equal parts of a 1 metre long ribbon:

The second ribbon is 2 of 5 equal parts of a 1 metre long ribbon:

Considering \(\frac {2}{5}\) metre as 4 of 10 equal parts of a metre:

The sum of fraction is \(\frac{3}{10}+\frac{2}{5}=\frac{3}{10}+\frac{4}{10}=\frac{7}{10}\)
Therefore, the total length is \(\frac {7}{10}\) metre.

To find the sum of two fractional measures put together, we must see them as a set of the same equal pieces.
Convert both fractional measures to forms with the same denominator.

Question 4.
Two ribbons of length \(\frac {1}{2}\) metre and \(\frac {2}{5}\) metre are joined end to end. What is the total length?
Answer:
For the same denominators we want for both must be a multiple of 2 and 5.
That is \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) and \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)
Total length of the ribbon is \(\frac{1}{2}+\frac{2}{5}=\frac{5}{10}+\frac{4}{10}=\frac{9}{10}\)

Some Other Sums
Of the two fractions added, one is some parts of one divided into equal parts taken together, and the other is the remaining parts taken together. When both these are taken together, we get all the parts or the whole.
\(\frac{1}{2}+\frac{1}{2}\) = 1
\(\frac{1}{3}+\frac{2}{3}\) = 1
\(\frac{1}{4}+\frac{3}{4}\) = 1
\(\frac{1}{5}+\frac{4}{5}\) = 1
\(\frac{2}{5}+\frac{3}{5}\) = 1

Question 1.
What fraction to be added to get 1?
(i) \(\frac {3}{7}\)
(ii) \(\frac {5}{6}\)
(iii) \(\frac {5}{9}\)
Answer:
(i) \(\frac{3}{7}+\frac{4}{7}=\frac{7}{7}\) = 1
(ii) \(\frac{5}{6}+\frac{1}{6}=\frac{6}{6}\) = 1
(iii) \(\frac{5}{9}+\frac{4}{9}=\frac{9}{9}\) = 1

Question 2.
Draw two circles of the same size and colour half of one and two-thirds of the other. If we cut out these pieces, can we put them together?

What if we cut them like this?

Answer:
Figure 1
The first circle is divided into 2 equal parts, and 1 is considered. In the second circle, it is divided into 3 equal parts, and 2 is considered from it.
If we join these two circles together, we get,

Figure 2
If we join these two circles together, we get,

Removing Parts
Here we can do this subtraction just as we did the addition of fractions:
If a half metre piece is cut off from a three-quarter metre long ribbon, then the length of the remaining portion is;

Answer:
Length of the ribbon is 1 metre.
Here, a half metre piece is cut off from a three-quarter metre long ribbon,

If one-third of a metre is removed from half a metre;
Answer:

Therefore, the remaining portion is \(\frac {1}{6}\) metres.

We have seen some pairs of fractions that add up to 1.
We can rewrite them as subtractions:

Question 3.
From one liter of milk, a quarter liter was used up. How much milk is left?
Answer:
The remaining portion of milk is,
\(1-\frac{1}{4}=\frac{4}{4}-\frac{1}{4}\)
= \(\frac{4-1}{4}\)
= \(\frac {3}{4}\) litres

Question 4.
From two and a half kilograms of yams, one and a quarter kilograms are cut off. What is the weight of the remaining piece?
Answer:
Weight of yam = 2\(\frac {1}{2}\)
The weight cut off from the yam = 1\(\frac {1}{4}\)
Weight of the remaining piece

Large and Small
Case 1: To find the largest fraction with the same denominator,
Increasing the numerator alone makes a fraction larger.
For example:
\(\frac{1}{7}<\frac{2}{7}<\frac{3}{7}<\frac{4}{7}<\frac{5}{7}<\frac{6}{7}\)

Which is larger, \(\frac {2}{5}\) or \(\frac {3}{5}\)?
Answer:
\(\frac {2}{5}\) is 2 parts of 5 equal parts taken together, and \(\frac {3}{5}\) is 3 parts of 5 equal parts taken together.

Therefore \(\frac {2}{5}\) is less than the number \(\frac {3}{5}\).
We can write like this:
\(\frac {2}{5}\) < \(\frac {3}{5}\) Or \(\frac {3}{5}\) is greater than \(\frac {2}{5}\).
That is, \(\frac {3}{5}\) > \(\frac {2}{5}\)

In general, we can say “Of two fractions with the same denominator, the one with a larger numerator is the larger and the one with a smaller numerator is the smaller.”

Case 2
If the denominator of a fraction is increased, the fraction becomes smaller.
For example:
\(\frac{7}{2}>\frac{7}{3}>\frac{7}{4}>\frac{7}{5}>\frac{7}{6}\)

Which is larger \(\frac {3}{4}\) and \(\frac {3}{5}\)?
Answer:
4 equal parts are larger than each of 5 equal parts.

3 of the first kind of pieces taken together is larger than 3 of the second kind of pieces.

This means \(\frac {3}{4}\) > \(\frac {3}{5}\)
In general, of two fractions with the same numerator, the one with the larger denominator is smaller and the one with the smaller denominator is larger.

In conclusion,
To compare two fractions with the same denominator, we need only compare the numerators
To compare two fractions with the same numerator, we need only compare the denominators.
If both the numerator and denominator are different, convert the fractions into forms with the same denominators.

Which of \(\frac {1}{2}\) and \(\frac {3}{5}\) are larger?
Answer:
\(\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) and \(\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)
That is, \(\frac{5}{10}<\frac{6}{10}\)
Therefore \(\frac{1}{2}<\frac{3}{5}\)

Which is larger, \(\frac {3}{5}\) or \(\frac {7}{5}\)?
Answer:
\(\frac {3}{5}\) < \(\frac {7}{5}\)

Which is larger, \(\frac {4}{5}\) or \(\frac {4}{6}\)?
Answer:
\(\frac {4}{5}\) > \(\frac {4}{6}\)

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 3 Volume Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 3 Solutions Volume

Class 6 Kerala Syllabus Maths Solutions Chapter 3 Volume Questions and Answers

Volume Class 6 Questions and Answers Kerala Syllabus

Size as Number (Page No. 37)

Question 1.
All blocks shown below are made up of cubes of side 1 centimetre. Calculate the violume of each.


Answer:
Figure 1
Length = 4 small cube = 4 cm
Width = 4 small cube = 4 cm
Height = 4 small cube = 4 cm
Therefore volume = 4 × 4 × 4 = 64 cubic cm

Figure 2
Length = 2 small cube = 2 cm
Width = 2 small cube = 2 cm
Height = 3 small cube = 3 cm
Therefore volume = 2 × 2 × 3 = 12 cubic cm

Figure 3
Length = 4 small cube = 4 cm
Width = 3 small cube = 3 cm
Height = 3 small cube = 3 cm
Therefore volume = 4 × 3 × 3 = 36 cubic cm

Figure 4
Length = 7 small cube = 7 cm
Width = 2 small cube = 2 cm
Height = 3 small cube = 3 cm
Therefore volume = 7 × 2 × 3 = 42 cubic cm

Figure 5
Length = 3 small cube = 3 cm
Width = 2 small cube = 2 cm
Height = 2 small cube = 2 cm
Therefore volume = 3 × 2 × 2 = 12 cubic cm

Intext Questions (Page No. 40)

Question 1.
Calculate the volume of each of the rectangular blocks shown below:

Answer:
Figure 1
Length = 7 small cube = 7 cm
Width = 4 small cube = 4 cm
Height = 1 small cube = 1 cm
Therefore volume = length × width × height
= 7 × 4 × 1
= 28 cubic cm

Figure 2
Length = 6 small cube = 6 cm
Width = 3 small cube = 3 cm
Height = 3 small cube = 3 cm
Therefore volume = length × width × height
= 6 × 3 × 3
= 54 cubic cm

Figure 3
Length = 5 small cube = 5 cm
Width = 5 small cube = 5 cm
Height = 5 small cube = 5 cm
Therefore volume = length × width × height
= 5 × 5 × 5
= 125 cubic cm

Figure 4
Length = 5 small cube = 5 cm
Width = 4 small cube = 4 cm
Height = 5 small cube = 5 cm
Therefore volume = length × width × height
= 5 × 4 × 5
= 100 cubic cm

Volume Calculation (Page No. 40)

Question 1.
The length, width, and height of a brick are 21 centimetres, 15 centimetres, and 7 centimetres. What is its volume?
Answer:
Length of a brick = 21 cm
Width = 15 cm
Height = 7 cm
Volume = 21 × 15 × 7
= 315 × 7
= 2205 cubic cm

Question 2.
An iron cube is of side 8 centimetres. What is its volume? 1 cubic centimetre of iron weighs 8 grams. What is the weight of this cube?
Answer:
Volume = 8 × 8 × 8 = 512 cubic cm.
Weight of this cube = 512 × 8 = 4096 cubic cm

Volume and Length (Page No. 41)

Question 1.
The table shows the measurements of some rectangular blocks. Calculate the missing measures.

Answer:

New Shapes (Page No. 42)

Question 1.
Calculate the volumes of the shapes shown below. All lengths are in centimetres.

Answer:
Figure 1

(i) 8 × 4 × 2 = 64 cubic cm.
(ii) 8 × 4 × 2 = 64 cubic cm.
(iii) 20 × 4 × 2 = 160 cubic cm.
(iv) 4 × 4 × 2 = 32 cubic cm.
Total volume = 64 + 64 + 160 + 32 = 320 cubic cm.

Figure 2

(i) 16 × 3 × 4 = 192 cubic cm.
(ii) 16 × 3 × 4 = 192 cubic cm.
(iii) 4 × 3 × 4 = 48 cubic cm.
Total volume = 192 + 192 + 48 = 432 cubic cm.

Figure 3

(i) 16 × 3 × 4 = 192 cubic cm.
(ii) 11 × 3 × 4 = 132 cubic cm.
Total volume = 192 + 132 = 324 cubic cm

Large Measures (Page No. 43)

Question 1.
A truck is loaded with sand, 4 metres long, 2 metres wide, and 1 metre high. The price of 1 cubic metre of sand is 1000 rupees. What is the price of this truckload?
Answer:
Volume of the truck = 4 × 2 × 1 = 8 cubic metre
The price of 1 cubic metre of sand = 1000 rupees.
The price of 8 cubic metre sand = 8 × 1000 = 8000 rupees

Question 2.
What is the volume in cubic centimetres of a platform 6 metres long, 1 metre wide, and 50 centimetres high?
Answer:
Volume of the platform = 6 m × 1 m × 50 cm
= 600 × 100 × 50
= 3,000,000 cubic cm

Question 3.
What is the volume of a piece of wood that is 5 metres long, 1 metre wide, and 25 centimetres high? The price of 1 cubic metre of wood is 60000 rupees. What is the price of this piece of wood?
Answer:
Volume of apiece of wood = 5 m × 1 m × 25 cm
= 500 cm × 100 cm × 25 cm
= 1250000 cubic cm
= 1250000 ÷ 1000000
= 1 cubic metre 250000 cubic cm
= 1\(\frac {1}{4}\) cubic metre
The price of 1 cubic metre of wood = 60000 rupees.
The price of \(\frac {1}{4}\) cubic metre of wood = 15000 rupees.
Therefore the total price = 60000 + 15000 = 75000 rupees

Capacity & Liquid Measures (Page No. 45)

Question 1.
The inner sides of a cubical box are of length 4 centimetres. What is its capacity? How many cubes of side 2 centimetres can be stacked inside it?
Answer:
Inner length of the cubical box = 4 cm
Capacity of the box = 4 × 4 × 4 = 64 cubic cm.
Volume of one small cube = 2 × 2 × 2 = 8 cubic cm.
Number of small cubes that can fit inside = 64 ÷ 8 = 8 cubes

Question 2.
The inner sides of a rectangular tank are 70 centimetres, 80 centimetres, and 90 centimetres. How many litres of water can it contain?
Answer:
Capacity of the water tank = 70 × 80 × 90
= 504000 cubic cm
= 504 litres (Since 1000 cubic cm = 1 liter)

Question 3.
The length and width of a rectangular box are 90 centimetres and 40 centimetres. It contains 180 litres of water. How high is the water level?
Answer:
1 litre = 1000 cubic cm.
180 litres = 180 × 1000 = 180,000 cubic cm.
Volume = length × width × height
180,000 = 90 × 40 × height
height = 180,000 ÷ (90 × 40) = 50 cm

Question 4.
The inner length, width, and height of a tank are 80 centimetres, 60 centimetres, and 50 centimetres, and it contains water 15 centimetres high. How much more water is needed to fill it?
Answer:
Inner dimensions of the tank:
Length = 80 cm
Width = 60 cm
Height = 50 cm
Current water height = 15 cm
Capacity of the tank,
Total volume = 80 × 60 × 50 = 240,000 cubic cm.
Capacity = 240 litres
Volume of the water already in the tank = 80 × 60 × 15 = 72,000 cubic cm
Current Capacity = 72 litres
Remaining water needed = 240 – 72 = 168 litres
Or
Now the water is at 15 cm height, and the remaining height is 35 cm.
The water can fill in the remaining portion = 80 × 60 × 35
= 168,000 millilitres
= 168 litres

Question 5.
The panchayat decided to make a rectangular pond. The length, width, and depth were decided to be 20 metres, 15 metres, and 2 metres. How many litres of water are needed to fill this pond to a height of one and a half metres?
Answer:
Length = 20 m = 2000 cm
Width = 15 m = 1500 cm
Height = 1\(\frac {1}{2}\) m = 150 cm
Volume = 2000 × 1500 × 150 = 450,000,000 cubic cm.
= 450 cubic metres
= 450000 litre
Or
Volume = 20 × 15 × 1\(\frac {1}{2}\)
= 300 × 1\(\frac {1}{2}\)
= 300 + 150
= 450 cubic metres
= 450000 litre

Question 6.
The inner length and width of an aquarium are 60 centimetres and 30 centimetres. It is half-filled with water. When a stone is immersed in it, the water level rises by 10 centimetres. What is the volume of the stone?
Answer:
When a stone is immersed in the aquarium, the water level rises by 10 centimetres
Volume of the water roses = length × width × 10 cm
= 60 × 30 × 10
= 18000 cubic metres.
Volume of the stone = 18000 cubic cm.

Question 7.
A rectangular iron block has a length of 20 centimetres, a width of 10 centimetres, and a height of 5 centimetres. It is melted and recast into a cube. What is the length of a side of this cube?
Answer:
Volume = 20 × 10 × 5 = 1000 cubic cm
The side of the cube = 10 cm
10 × 10 × 10 = 1000 cubic cm

Question 8.
A tank 2 metres long and 1 metre wide is to contain 10000 litres of water. What should be the height of the tank?
Answer:
1 cubic metre = 1000 litres
10,000 litres = 10 cubic metres
Volume = length × width × height
10 = 2 × 1 × height
height = 10 ÷ (2 × 1)
= 10 ÷ 2
= 5 metres

Question 9.
From the four corners of a square piece of paper of side 12 centimetres, small squares of side 1 centimetre are cut off. The edges of this are bent up and joined to form a container of height 1 centimetre. What is the capacity of this container? If squares of side 2 centimetres are cut off. What would be the capacity?
Answer:

Original side of the square = 12 cm
After cutting 1 cm from each corner, the length and width were reduced by 2 cm (1 cm from each end)
New length = 12 – 2 = 10 cm
New width = 12 – 2 = 10 cm
Height = 1 cm
Capacity = 10 × 10 × 1 = 100 cubic cm.
If 2 cm is cut off from each side,
New length = 12 – 4 = 8 cm
New width = 12 – 4 = 8 cm
Height = 2 cm
Volume = 8 × 8 × 2 = 128 cubic cm.

Class 6 Maths Chapter 3 Kerala Syllabus Volume Questions and Answers

Class 6 Maths Volume Questions and Answers

Question 1.
Which of the following rectangular blocks has a volume of 30 cubic centimetres?
(a) 3 cm, 4 cm, 5 cm
(b) 5 cm, 3 cm, 2 cm
(c) 4 cm, 7 cm, 2 cm
(d) 10 cm, 2 cm, 2 cm
Answer:
(b) 5 cm, 3 cm, 2 cm
Volume = 5 × 3 × 2 = 30 cubic cm.

Question 2.
Which of the following is correct for the volume of a rectangular block?
(a) Volume is the product of its length and height.
(b) Unit of volume is the square centimetre.
(c) If the length of a rectangular block is doubled, then its volume is also doubled.
(d) 100 cubic centimetres is equal to 1 cubic metre.
Answer:
(a) False.
Reason: volume = length × width × height
(b) False.
Reason: unit of volume is cubic centimetre or cubic metre.
(c) If the length of a rectangular block is doubled, then its volume is also doubled.
(d) False.
Reason: 1000 cubic centimetres is 1 litre
100 cubic centimetres is 100 millilitres.

Question 3.
Match the following:
(a) 1000 cubic centimetre – 10 litre
(b) 1000000 cubic centimetre – 1 millilitre
(c) 1 cubic centimetre – 1 cubic metre
(d) 10000 millilitre – 1 litre
Ans:
(a) 1000 cubic centimetre – 1 litre
(b) 1000000 cubic centimetre – 1 cubic metre
(c) 1 cubic centimetre – 1 millilitre
(d) 10000 millilitre – 10 litre

Question 4.
Find the volume of the rectangular block whose length is 12 centimetres, width is 6 centimetres, and height is 4 centimetres.
Answer:
Volume = 12 × 6 × 4 = 288 cubic centimetres

Question 5.
A rectangular block of length 50 centimetres and width 20 centimetres has a volume of 3000 cubic centimetres. What is the height of the rectangular block?
Answer:
Volume = length × width × height
3000 = 50 × 20 × height
height = 3000 ÷ (50 × 20)
= 3000 ÷ 1000
= 3 cm

Question 6.
A wooden block of length 15 metres, width 4 metres, and height 2 metres. What is its volume?
Answer:
Volume of the wodden block = 15 × 4 × 2 = 120 cubic metres

Question 7.
A tank 4 metres long and 2 metres wide is to contain 16000 litres of water. What should be the height of the tank?
Answer:
Volume = 16000 litre = 16 cubic metre
Volume = length × width × height
16 = 4 × 2 × height
height = 16 ÷ (4 × 2)
= 16 ÷ 8
= 2 metres

Question 8.
Complete the following given below:
(a) 180 cubic metre = _____________ litre
(b) 3000 cubic centimetre = _____________ litre
(c) 2000 litre = _____________ cubic metre
(d) 15 cubic centimetre = _____________ millilitre
(e) 75 cubic metre = _____________ litre
Answer:
(a) 180 cubic metre = 180,000 litre
(b) 3000 cubic centimetre = 3 litre
(c) 2000 litres = 2 cubic metres
(d) 15 cubic centimetre = 15 millilitre
(e) 75 cubic metre = 75,000 litre

Class 6 Maths Chapter 3 Notes Kerala Syllabus Volume

→ To compare the size of two rectangular blocks, we must consider their length, width, and height.

→ The volume of a rectangular block can be calculated by multiplying its length, width, and height.

→ If the measurements are in centimetres, then the volume will be in cubic centimetres.

→ If measurements are in metres, then the volume should be in cubic metres.

→ The volume of a rectangular block is the product of its length, width, and height.

→ For a rectangular block,

• If the length is doubled, the volume becomes twice.
• If the length and width are doubled, the volume becomes four times.
• If the length, width, and height are all doubled, the volume becomes eight times.

→ The volume of a rectangular block is 1 cubic metre if its length is 1 metre, width is 1 metre, and height is 1 metre.

→ 1 cubic metre = 100 × 100 × 100 = 1000000 cubic centimetre

→ The volume is called the capacity of the box or a vessel.

→ The capacity of a rectangular box or a vessel is calculated by multiplying its inner length, width, and height.

→ 1 cubic centimeter = 1 millilitre

→ 1000 cubic centimeter = 1 litre

→ 1 cubic metre =1000 litre

→ 1 cubic metre = 1000000 cubic centimetre

In our everyday life, we often come across objects that occupy spaces – like a water bottle, a box, a tank, or even a cupboard. The amount of space an object takes up is called its volume. Volume helps us understand how much space is inside a three-dimensional object or how much it can hold. In this chapter, we will learn how to measure the volume of different solid shapes, such as cubes, cuboids. We will also explore the formulas used to calculate volume, and understand the units in which volume is measured, like cubic centimetres or liters.

Large and Small & Rectangular Blocks
In the previous classes, we have already learned how to compare the lengths of objects, to find their perimeter. The size of an object can be determined by measuring its length of the objects. Measuring only the length of a rectangular-shaped object is not sufficient to determine its size. Instead, we need to measure both its length and breadth to find its area. To compare the size of two rectangular blocks, we must consider their length, width, and height.

Size of a Rectangular Block & Size as a Number
Look at the rectangular block given below.

The length, width, and height of the rectangular block are 10 cm, 5 cm, and 10 cm, respectively. Also, consider that it is made by stacking smaller blocks, each having length, width, and height as 1 cm, 1 cm, 1 cm.

Let us find out how many small rectangular blocks are contained within the large rectangular block. In the bottom row, there are 10 numbers in length and 5 numbers in width. So the total number of small rectangular blocks is 50.

The height of the larger rectangular block is 10 cm, so we can place 10 smaller blocks along the height. If we stack 50 such columns, each with 10 blocks in height, we get a total of 50 × 10 = 500 blocks. So the size of the larger rectangular block is the same as the size of the small 500 rectangular blocks, each measuring 1 cm in length, 1 cm in width, and 1 cm in height.

We learned that the area of a rectangle with a length of 1 cm and a width of 1 cm is 1 square centimeter. Similarly, the volume of a rectangular block with a length of 1 cm, a width of 1 cm, and a height of 1 cm is 1 cubic centimeter. Therefore, the volume of the above rectangular block is 10 × 5 × 10 = 500 cubic centimeters. That means the volume of a rectangular block can be calculated by multiplying its length, width, and height. If the measurements are in centimeters, then the volume will be in cubic centimeters. If measurements are in metres, then the volume should be in cubic metres.

Question 1.
Find the volume of the given rectangular block.

Answer:
Length of the rectangular block = 20 cm
Width =10 cm
Height = 8 cm
Therefore, Volume = 20 × 10 × 8 = 1600 cubic cm

Question 2.
How many small cubes are used to make this large cube? If one small block is removed from each corner of the large block, how many would be left?

Answer:
The large cube is made by stacking small cubes.
Length = 4 small cubes
Width = 4 small cubes
Height = 4 small cubes
Total number of small cubes = 4 × 4 × 4 = 64
A cube has 8 corners. Removing 1 cube from each corner means a total of 8 cubes are removed.
Therefore the cubes left = 64 – 8 = 56
The number of cubes remaining after removing corners is 56.

Question 3.
All sides of the large cube are painted. How many small cubes would have no paint at all?

Answer:
Total number of small cubes = 3 × 3 × 3 = 27
Only the cubes that are completely inside the large cube will have no paint on them.
The top layer and bottom layer have 9 cubes each, and all have at least one side painted.
In the middle layer, only the outer edge cubes have paint on any of their sides, which means 8 cubes have at least one side painted.
So the total number of cubes with at least one side painted is = 9 + 9 + 8 = 26
Therefore, the small cube with no paint at all = 27 – 26 = 1 cube

Volume Calculation
How do we calculate its volume?

For calculating the volume, we must find out how many cubes of side 1 cm we need to make it.

30 small cubes are needed to make it.
So the volume of the cube is 30 cubic cm.
Or
Volume of the cube = 5 × 3 × 2 = 30 cubic cm.
The volume of a rectangular block is the product of its length, width, and height.

Question 1.
The dimensions of a rectangular block are given below. Find its volume.
(i) Length = 10 cm, Width = 7 cm, Height = 4 cm.
(ii) Length = 5 cm, Width = 5 cm, Height = 5 cm.
(iii) Length = 8 cm, Width = 4 cm, Height = 6 cm.
(iv) Length = 20 cm, Width = 10 cm, Height = 10 cm.
(v) Length = 40 cm, Width = 10 cm, Height = 10 cm.
Answer:
(i) 10 × 7 × 4 = 280 cubic cm
(ii) 5 × 5 × 5 = 125 cubic cm
(iii) 8 × 4 × 6 = 192 cubic cm
(iv) 20 × 10 × 10 = 2000 cubic cm
(v) 40 × 10 × 10 = 4000 cubic cm

Volume and Length
Volume can be calculated when the length, width, and height are given. If any three of these four terms are given, we can find the fourth term.

Question 1.
The volume of a rectangular block is 160 cubic centimetres. Its height and width are 5 centimetres and 4 centimetres respectively. What is its length?
Answer:
Volume = length × breadth × height
160 = length × 5 × 4
160 = length × 20
To find the length, divide 160 by 20
Length = 160 ÷ 20 = 8 cm

Question 2.
A rectangular block of length 23 centimetres and height 6 centimetres has a volume of 1380 cubic centimetres. What is its width?
Answer:
Volume = length × breadth × height
1380 = 23 × width × 6
1380 = width × 138
Width = 1380 ÷ 138 = 10 cm

New Shapes
We can make shapes other than rectangular blocks by stacking cubes.

Question 1.
It is made by stacking cubes of side 1 centimetre. Calculate its volume?

Answer:
1st row (bottom row) = 9 × 9 = 81
2nd row = 7 × 9 = 63
3rd row = 5 × 9 = 45
4th row = 3 × 9 = 27
Total = 81 + 63 + 45 + 27 = 216
Therefore, volume = 216 cubic cm.
Or
If the length and width are 9 in every row, and the height is 4.
Then total number of blocks = 9 × 9 × 4 = 324
Now consider the missing blocks.
1st row (bottom row) = 2 × 9 = 18
2nd row = 4 × 9 = 36
3rd row = 6 × 9 = 54
Total missing blocks = 18 + 36 + 54 = 108
Total number of blocks = 324 – 108 = 216
Therefore, volume = 216 cubic cm.

Question 2.
It is made by stacking square blocks. The bottom block is of side 9 centimetres. As we move up, the sides decrease by 2 centimetres at each step. All blocks are of height 1 centimetre. What is the volume of this figure?

Answer:
The blocks are arranged in 5 rows.
Volume of the first square block = 9 × 9 × 1 = 81 cubic cm.
Volume of the second square block = 7 × 7 × 1 = 49 cubic cm.
Volume of the third square block = 5 × 5 × 1 = 25 cubic cm.
Volume of the fourth square block = 3 × 3 × 1 = 9 cubic cm.
Volume of the fifth (top) square block = 1 × 1 × 1 = 1 cubic cm.
Toatal volume of the square blocks = 81 + 49 + 25 + 9 + 1 = 165 cubic cm.

Question 3.
What is the volume of a rectangular block of length 4 centimetres, width 3 centimetres, and height 1 centimetre? If the length, width, and height are doubled, what happens to the volume?
Answer:
Length of the rectangular block = 4 cm
Width = 3 cm
Height = 1 cm
Volume of the rectangular block = 4 × 3 × 1 = 12 cubic cm.
If the length, width, and height are doubled.
Length = 8 cm
Width = 6 cm
Height = 2 cm
Volume = 8 × 6 × 2 = 96 cubic cm.

For a rectangular block,
If the length is doubled, the volume becomes twice.
If the length and width are doubled, the volume becomes four times.
If the length, width, and height are all doubled, the volume becomes eight times.

Large Measures
The volume of a rectangular block is 1 cubic centimetre if its length is 1 centimetre, its width is 1 centimetre, and its height is 1 centimetre.
The volume of a rectangular block is 1 cubic metre if its length is 1 metre, width is h metres, and height is 1 metre.
1 cubic metre = 100 × 100 × 100 = 1000000 cubic centimetre
Then, the volume of a rectangular block, if its length is 5 metres, width 3 metres, and height 2 metres, is 5 × 3 × 2 = 30 cubic metres
If it is converted into cubic centimetres, 30 × 1000000 = 30000000 cubic cm

Question 1.
What is the volume of a platform if its length is 12 metres, width is 8 metres, and height is 75 centimetres?
Answer:
Here, the measurements are both in centimetres and metres.
Convert every value into centimetres.
1200 cm × 800 cm × 75 cm = 72,000,000 cubic cm
If converting it into cubic metres, 72,000,000 ÷ 1000000 = 72 cubic metres

Capacity & Liquid Measures
Capacity refers to the amount a box or a vessel can hold. So it is closely related to volume. That is, the volume is called the capacity of the box or a vessel. The capacity of a rectangular box or a vessel is calculated by multiplying its inner length, width, and height. While considering the thickness of a rectangular box or vessel, the inner and outer dimensions are different.

For this rectangular box, the outer dimensions are greater than the inner dimensions. For calculating the capacity, we have to consider the inner dimensions.

Question 1.
Calculate the capacity when its inner dimensions are of length 20 centimetres, width 10 centimetres, and height 5 centimetres.
Answer:
Capacity = 20 × 10 × 5 = 1000 cubic cm

If we are discussing the quantity of water contained in a rectangular box, it can be expressed in millilitres or litres.
Consider a rectangular box with dimensions of length 1 centimetre, width 1 centimetre, and height 1 centimetre.
Its capacity is calculated as = 1 × 1 × 1 = 1 cubic cm. This is equal to 1 millilitre

If we combine 1000 such boxes, their total capacity becomes 1000 cubic cm, which is equal to 1 litre.
1000 millilitre = 1 litre
Therefore, the capacity of the above-mentioned rectangular box is 1000 cubic cm = 1 litre.

Consider a rectangular box with dimensions of length 10 centimetres, width 10 centimetres, and height 10 centimetres.
Its capacity is calculated as = 10 × 10 × 10 = 1000 cubic cm. This is equal to 1 litre. These relationships can be written like this:

• 1 cubic centimeter = 1 millilitre
• 1000 cubic centimeter = 1 litre
• 1 cubic metre = 1000 litre
• 1 cubic metre = 1000000 cubic centimetre

Question 2.
What is the capacity of a box whose inner length, width, and height are 15 centimetres, 10 centimetres, and 8 centimetres?
Answer:
Capacity = 15 × 10 × 8
= 1200 cubic cm.
= 1000 cubic cm + 200 cubic cm
= 1 litre + 200 millilitre
= 1 litre 200 millilitre

Question 3.
A vessel is filled with water. If a cube of side 1 centimetre is immersed in it, how many cubic centimetres of water would overflow? What if 20 such cubes are immersed?
Answer:
If a cube of side 1 centimetre is immersed in a vessel filled with water, the amount of water that overflows will be equal to the volume (or capacity) of the cube.
Volume of the cube = 1 cm × 1 cm × 1 cm = 1 cubic centimeter
Since 1 cubic centimetre = 1 millilitre
1 millilitre of water will overflow.
If 20 such cubes are immersed.
Each cube displaces 1 millilitre of water.
So, 20 × 1 = 20 millilitres of water will overflow.

Question 4.
A water tank of length 4 metres, width 3 metres, and height 2 metres. How many litres of water does it contain?
Answer:
Capacity of the water tank = 4 × 3 × 2 = 24 cubic metre
Since 1 cubic metre = 1000 litre
24 cubic metre = 24000 litre
Therefore, the tank can hold 24,000 litres of water.

Question 5.
A rectangular tank of length 8 metres and width 6 metres. It contains 96000 litres of water. Calculate the height of the water in the tank?
Answer:
1 cubic metre = 1000 litres
96000 litres = \(\frac {96000}{10000}\) = 96 cubic metres
Volume = length × width × height
96 = 8 × 6 × height
height = 96 ÷ 48 = 2 metre

Question 6.
A swimming pool is 25 metres long, 10 metres wide, and 2 metres deep. It is half-filled. How many litres of water does it contain now?
25 × 10 × 1 = 250 cubic metres = 250000 litres
Suppose the water level is increased by 1 centimetre. How many more litres of water does it contain now?
Answer:
If it is half-filled
Full depth = 2 metre, Half depth = 1 metre
Volume = 25 × 10 × 1 = 250 cubic metres
250 cubic metres = 250 × 1000 = 250,000 litres
If the water level rises by 1 cm
Additional volume = 2500 × 1000 × 1 = 2500000 cubic cm
2500000 cubic cm = \(\frac {2500000}{1000}\) = 2500 litres

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 2 One Fraction Many Forms Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 2 Solutions One Fraction Many Forms

Class 6 Kerala Syllabus Maths Solutions Chapter 2 One Fraction Many Forms Questions and Answers

One Fraction Many Forms Class 6 Questions and Answers Kerala Syllabus

Numerator and Denominator (Page No. 24)

Question 1.
Find the fractions specified below:
(i) The form of \(\frac {1}{2}\) with denominator 24
(ii) The form of \(\frac {1}{2}\) with numerator 24
(iii) The form of \(\frac {1}{3}\) with denominator 24
(iv) The form of \(\frac {1}{3}\) with numerator 24
(v) The form of \(\frac {1}{4}\) with numerator 100
Answer:
(i) \(\frac{1}{2}=\frac{12}{24}\)
(ii) \(\frac{1}{2}=\frac{24}{48}\)
(iii) \(\frac{1}{3}=\frac{8}{24}\)
(iv) \(\frac{1}{3}=\frac{24}{72}\)
(v) \(\frac{1}{4}=\frac{100}{400}\)

Question 2.
Write three different forms of \(\frac {1}{4}\).
Answer:
Three different forms of \(\frac {1}{4}\) is
\(\frac{1}{4}=\frac{2}{8}=\frac{20}{80}=\frac{50}{200}\)

Question 3.
For each pair of fractions below, find three forms with the same denominator:
(i) \(\frac{1}{2}, \frac{1}{3}\)
(ii) \(\frac{1}{2}, \frac{1}{4}\)
(iii) \(\frac{1}{3}, \frac{1}{4}\)
Answer:

Question 4.
Does \(\frac {1}{3}\) have another form with a denominator of 10, 100, or 1000? Give reasons.
Answer:
No, \(\frac {1}{3}\) does not have another form with a denominator 10, 100, or 1000, because none of these numbers is divisible by 3.

Textbook Page No. 26

Question 1.
Now, can’t you fill up the following table by multiplying the numerator and denominator by a number?

Answer:

Textbook Page No. 28

Question 1.
Write each of the following fractions in lowest terms:
(i) \(\frac {32}{64}\)
(ii) \(\frac {27}{81}\)
(iii) \(\frac {30}{45}\)
(iv) \(\frac {12}{21}\)
(v) \(\frac {45}{54}\)
Answer:
(i) \(\frac{32}{64}=\frac{16}{32}=\frac{8}{16}=\frac{4}{8}=\frac{2}{4}=\frac{1}{2}\)
The largest common factor of 32 and 64 is 32
So divide 32 and 64 by 32, we get \(\frac{32}{64}=\frac{1}{2}\)

(ii) \(\frac {27}{81}\)
3, 9, 27 are the common factors of 27 and 81
\(\frac{27}{81}=\frac{9 \times 3}{27 \times 3} ; \quad \frac{27}{81}=\frac{3 \times 9}{9 \times 9} ; \quad \frac{27}{81}=\frac{1 \times 27}{3 \times 27}\)
So divide it by the largest common factor, 27, to get \(\frac {1}{3}\)
Therefore, the lowest form is \(\frac {1}{3}\).

(iii) \(\frac {30}{45}\)
Common factors are 5 and 15
So divide it by 15
\(\frac{30}{45}=\frac{2}{3}\)

(iv) \(\frac {12}{21}\)
3 is the common factor of 12 and 21
\(\frac{12}{21}=\frac{3 \times 4}{3 \times 7}=\frac{4}{7}\)

(v) \(\frac {45}{54}\)
3, 9 are the common factors of 45 and 54.
\(\frac{45}{54}=\frac{15 \times 3}{18 \times 3}=\frac{15}{18}, \quad \frac{15}{18}=\frac{5 \times 3}{6 \times 3}=\frac{5}{6}\)
First, removing the factor 9 from it will result in \(\frac {5}{6}\)
\(\frac{45}{54}=\frac{9 \times 5}{9 \times 6}=\frac{5}{6}\)

Fraction as Division (Page No. 32)

Question 1.
20 litres of water are used to fill 8 identical bottles. How many litres of water are there in each bottle?
Answer:
20 ÷ 8 = \(\frac {20}{8}\)
\(\frac {16}{8}\) = 2
The remaining 4 is divided into 8 equal parts, that is,
\(\frac{4}{8}=\frac{1}{2}\)
Therefore the quantity of water contained in one bottle is 2\(\frac {1}{2}\) liters
Or \(\frac{20}{8}=\frac{10}{4}=\frac{5}{2}=2 \frac{1}{2}\)

Question 2.
A rope of length 140 centimetres is cut into 16 equal pieces. What is the length of each piece?
Answer:
140 ÷ 16 = \(\frac {140}{16}\)
= \(\frac {70}{8}\)
= \(\frac {35}{4}\)
= 8\(\frac {3}{4}\) metre

Question 3.
If 215 kilograms of rice is divided equally among 15 people, how many kilograms of rice would each get?
Answer:
215 ÷ 15 = \(\frac{215}{15}=\frac{43 \times 5}{3 \times 5}=\frac{43}{3}\)
If 42 is divided by 3 we get 14
The remainder 1 is divided into 3 parts, we get \(\frac {1}{3}\)
The kilogram of rice each get 14\(\frac {1}{3}\) kg

Class 6 Maths Chapter 2 Kerala Syllabus One Fraction Many Forms Questions and Answers

Class 6 Maths One Fraction Many Forms Questions and Answers

Question 1.
Which is the correct statement given below:
(a) The form of \(\frac {1}{3}\) with denominator 9 is \(\frac {2}{9}\)
(b) The form of \(\frac {1}{3}\) with numerator 9 is \(\frac {9}{27}\)
(c) The form of \(\frac {1}{3}\) with denominator 6 is \(\frac {4}{6}\)
(d) The form of \(\frac {1}{3}\) with numerator 2 is \(\frac {2}{6}\)
Answer:
(b) The form of \(\frac {1}{3}\) with numerator 9 is \(\frac {9}{27}\)
(d) The form of \(\frac {1}{3}\) with numerator 2 is \(\frac {2}{6}\)

Question 2.
Which is the wrong statement given below:
(a) Another form of \(\frac {1}{4}\) is \(\frac {5}{20}\)
(b) Another form of \(\frac {3}{7}\) is \(\frac {9}{21}\)
(c) Another form of \(\frac {5}{8}\) is \(\frac {30}{40}\)
(d) Another form of \(\frac {9}{10}\) is \(\frac {27}{30}\)
Answer:
(c) Another form of \(\frac {5}{8}\) is \(\frac {30}{40}\)
Multiplying 8 by 5 we get 40, and multiplying 5 by 5 we get 25.
So the another form of \(\frac {5}{8}\) is \(\frac {25}{40}\)

Question 3.
Match the following:

Answer:

Question 4.
The form of \(\frac {3}{5}\) with denominator 25 is _____________
Answer:
\(\frac{3}{5}=\frac{15}{25}\) (3 × 5 = 15, 5 × 5 = 25)

Question 5.
The form of \(\frac {1}{8}\) with numerator 10 is _____________
Answer:
\(\frac{1}{8}=\frac{10}{80}\) (1 × 10 = 10, 8 × 10 = 80)

Question 6.
Write 3 different forms of \(\frac {3}{10}\)?
Answer:
\(\frac{3}{10}=\frac{6}{20}=\frac{12}{40}=\frac{24}{80}\)

Question 7.
Write 3 different forms of the pair \(\frac {2}{5}\), \(\frac {1}{6}\) with same denominator.
Answer:
\(\frac {2}{5}\), \(\frac {1}{6}\)
Denominator 30: \(\frac{12}{30}, \frac{5}{30}\)
Denominator 60: \(\frac{24}{60}, \frac{10}{60}\)
Denominator 120: \(\frac{48}{120}, \frac{20}{120}\)

Question 8.
If 10 litres of milk are filled into 3 identical bottles. How many litres of milk are contained in one bottle?
Answer:
Milk contained in one bottle is 10 ÷ 3 = \(\frac {10}{3}\) = 3\(\frac {1}{3}\) litres

Question 9.
Dividing 12 by which number is equal to 24 divided by 18.
Answer:
24 ÷ 18 = 12 ÷ 9 (\(\frac{24}{18}=\frac{12}{9}\))

Question 10.
Write the lowest term of \(\frac {60}{90}\).
Answer:
Lowest term is \(\frac{60}{90}=\frac{2 \times 30}{3 \times 30}=\frac{2}{3}\)

Class 6 Maths Chapter 2 Notes Kerala Syllabus One Fraction Many Forms

→ Multiplying the numerator and denominator of a fraction by the same natural number gives a form of the same fraction.

→ If the numerator and denominator of a fraction have a common factor, then dividing them by this common factor gives a form of this fraction.

→ The form of a fraction in the lowest terms is obtained by removing all common factors of the numerator and denominator by division.

→ If we remove common factors of the numerator and denominator of a fraction, then we get another form of the same fraction.

→ In writing divisions with remainders as fractions, we can remove common factors of the numerator and denominator.

Fractions are numerical representations used to describe parts of a whole or the ratio between quantities. A fraction consists of two components: the numerator (the top number), which shows how many parts are being considered, and the denominator (the bottom number), which indicates the total number of equal parts the whole is divided into. Fractions are commonly used in everyday life to divide objects or quantities evenly, express proportions, and make comparisons. In this chapter, we will explore the concept of fractions in detail and learn how a single fraction can be expressed in various forms.

Numerator and Denominator

→ A portion of two equal parts is called a half. It is also known as one by two portion and is written as \(\frac {1}{2}\).

→ If an object is divided into 4 equal parts and 2 parts are taken, or it is divided into 6 equal parts and 3 parts are taken, or even if it is divided into 8 equal parts and 4 parts are taken. In all these cases, the portion taken is the same as \(\frac {1}{2}\) of the whole.

→ If an object is divided into 3 equal parts, then 1 part is \(\frac {1}{3}\) (one-third) and 2 parts are \(\frac {2}{3}\) (two-thirds) of the whole.

→ If an object is divided into 4 equal parts, then 1 part represents \(\frac {1}{4}\) (one-fourth or a quarter), and 3 parts represent \(\frac {3}{4}\) (three-fourths or three-quarters) of the whole.

→ Like this, fractions are the numbers used to represent parts of an object being considered.

→ If 2 objects are divided into 3 equal parts, then one part will be \(\frac {2}{3}\) of it.

→ If 3 objects are divided into 2 equal parts, each part will be \(\frac {3}{2}\), which is equal to one and a half.
\(\frac{3}{2}=1+\frac{1}{2}=1 \frac{1}{2}\)

Now let’s check the different forms of a fraction in detail:

→ Different forms of the fraction \(\frac {1}{2}\) are \(\frac{2}{4}, \frac{3}{6}, \frac{4}{8}, \frac{5}{10}\)

→ Similarly different forms of \(\frac {1}{3}\) are \(\frac{2}{6}, \frac{3}{9}, \frac{4}{12}, \frac{5}{15}\)

→ If an object is divided into 100 equal parts and we take 50 of them or divide it into 50 equal parts and we take 25 of them will be half \(\frac {1}{2}\). So \(\frac{1}{2}, \frac{25}{50}, \frac{50}{100}\) are all different forms of a fraction.

→ Similarly, we can see the different forms of the fraction \(\frac {1}{3}\) as \(\frac{10}{30}, \frac{20}{60}, \frac{100}{300}\)

→ For the fraction \(\frac {1}{2}\), its numerator is 1 and denominator is 2.

→ For the fraction \(\frac {2}{3}\), its numerator is 2 and denominator is 3.

To get the different forms of a fraction when the numerator is 1, we can write the numerator as the number used to multiply the denominator.
Another form of \(\frac {1}{2}\)
If we multiply 2 by 8, we get 16
So another form of \(\frac {1}{2}\) is \(\frac {8}{16}\).

Another form of \(\frac {1}{5}\)
If we multiply 5 by 10, we get 50
Numerator = 10, Denominator = 50
So another form of \(\frac {1}{5}\) is \(\frac {10}{50}\).

The form of \(\frac {1}{7}\) with denominator 21
If we multiply 7 by 3, we get 21
The numerator of another form is 3
So the another form is \(\frac {3}{21}\)

The form of \(\frac {1}{4}\) with denominator 10 is \(\frac {10}{40}\)
Because the denominator is multiplied by 10 is its numerator.

Write the different forms of \(\frac {1}{7}\)
Write the numerator as the number multiplied by 7
7 × 4 = 28 → \(\frac {4}{28}\)
7 × 6 = 42 → \(\frac {6}{42}\)
7 × 10 = 70 → \(\frac {10}{70}\)

Complete the following given below.

Question 1.
The form of \(\frac {1}{6}\) with denominator 12 is _____________
Answer:
The denominator of \(\frac {1}{6}\) is multiplied by 2, we get 12.
So its numerator is 12
Therefore the form is \(\frac {2}{12}\)

Question 2.
The form of \(\frac {1}{10}\) with denominator 40 is _____________
Answer:
The denominator of \(\frac {1}{10}\) is multiplied by 4, we get 40.
So its numerator is 4
Therefore the form is \(\frac {4}{40}\)

Question 3.
The form of \(\frac {1}{8}\) with numerator 4 is _____________
Answer:
The denominator of \(\frac {1}{8}\) is multiplied by 4, we get 32.
So its numerator is 4
That is \(\frac{1}{8}=\frac{4}{32}\)
(Since the numerator will be the number used to multiply the denominator)

Question 4.
The form of \(\frac {1}{12}\) with numerator 5 is _____________
Answer:
The denominator of \(\frac {1}{12}\) is multiplied by 5, we get 60.
So its numerator is 5
Therefore the form is \(\frac{1}{12}=\frac{5}{60}\)

Question 5.
The form of \(\frac {1}{5}\) with denominator 50 is _____________
Answer:
The denominator of \(\frac {1}{5}\) is multiplied by 10, we get 50.
So its numerator is 10
Therefore the form is \(\frac{1}{5}=\frac{10}{50}\)

Question 6.
The form of \(\frac {1}{7}\) with numerator 20 is _____________
Answer:
The denominator of \(\frac {1}{7}\) is multiplied by 20, we get 140.
So its numerator is 20
Therefore the form is \(\frac{1}{7}=\frac{20}{140}\)

Question 7.
Any 3 different forms of \(\frac {1}{4}\) is _____________
Answer:
Different forms of \(\frac {1}{4}\) is,

Question 8.
Any 3 different forms of \(\frac {1}{8}\) is _____________
Answer:

Different forms of fractions:
So far, we have discussed the different forms of fractions with a numerator of 1.
To find the different forms of fractions in this form, we can take the numerator as the number that is used to multiply the denominator.
\(\frac{1}{4}=\frac{1 \times 10}{4 \times 10}=\frac{10}{40}\)

How do we find the different forms of \(\frac {2}{5}\)?
For this, we have to multiply both the numerator and denominator by the same number.
In the fraction \(\frac {2}{5}\), multiply 5 by 2, we get 10.
Here, we multiply the denominator by 2, so we need to multiply the numerator also by 2.
That is, \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)
Another form, \(\frac{2}{5}=\frac{2 \times 5}{5 \times 5}=\frac{10}{25}\)
That means instead of taking 2 from 5 equal parts, take 4 from 10 equal parts (\(\frac{2}{5}=\frac{4}{10}\))
Similarly, instead of taking 2 from 5 equal parts, take 10 from 25 equal parts (\(\frac{2}{5}=\frac{10}{25}\))

Now check the different forms of \(\frac {3}{7}\).
\(\frac{3}{7}=\frac{6}{14}\) (multiply 7 by 2, we get 14, and multiply 3 by 2, we get 6)
\(\frac{3}{7}=\frac{30}{70}\) (multiply 7 by 10 we get 70, and multiply 3 by 10 we get 30)

Complete the following given below:

Question 1.
The form of \(\frac {2}{3}\) with denominator 15 is _____________
Answer:
Multiply 3 by 5 to get the denominator as 15
So, multiply 2 by 5, and we get the numerator as 10
\(\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}\)

Question 2.
The form of \(\frac {3}{4}\) with denominator 100 is _____________
Answer:
Multiply 4 by 25 to get the denominator as 100
So, multiply 3 by 25, and  we get the numerator as 75
\(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\)

Question 3.
The form of \(\frac {5}{8}\) with numerator 25 is _____________
Answer:
Multiply 5 by 5 to get the numerator as 25
So, multiply 8 by 5, and we get the denominator as 40
\(\frac{5}{8}=\frac{5 \times 5}{8 \times 5}=\frac{25}{40}\)

Question 4.
The form of \(\frac {7}{10}\) with numerator 70 is _____________
Answer:
Multiply 7 by 10 to get the numerator as 70
So, multiply 10 by 10, we get the denominator as 100
\(\frac{7}{10}=\frac{7 \times 10}{10 \times 10}=\frac{70}{100}\)

Question 5.
The 3 forms of \(\frac {3}{8}\) is _____________
Answer:
Different forms is,

Question 6.
The 3 forms of \(\frac {5}{7}\) is _____________
Answer:
Different forms is,

Till now, we have discussed the fractions with a numerator of 1 and different numbers separately.
In general, we can say this: “Multiplying the numerator and denominator of a fraction by the same natural number gives a form of the same fraction.”
For example: \(\frac{1}{5}=\frac{1 \times 6}{5 \times 6}=\frac{6}{30}, \quad \frac{2}{5}=\frac{2 \times 6}{5 \times 6}=\frac{12}{30}\)

Lowest terms:
We can form the different forms of a fraction by multiplying the numerator and denominator by the same number.

Let’s look at another form of \(\frac {20}{25}\);
We can divide 20 and 25 by their factor 5.
Divide 20 by 5, and we get 4
Divide 25 by 5, we get 5
Therefore \(\frac{20}{25}=\frac{4}{5}\)
Now 1 is the only number that can divide 4 and 5.
So the lowest form of \(\frac {20}{25}\) is \(\frac {4}{5}\).

Now let’s look at another form of \(\frac {30}{36}\);
Divide 30 and 36 by their common factor 2, we get
\(\frac{30}{36}=\frac{15}{18}\)
Now, divide 15 and 18 by their common factor 3, and we get
\(\frac{15}{18}=\frac{5}{6}\) (15 ÷ 3 = 5, 18 ÷ 3 = 6)
There is no common factor for 5 and 6.
So the lowest form of \(\frac {30}{36}\) is \(\frac {5}{6}\).
(Instead of dividing the numerator and denominator of the fraction \(\frac {30}{36}\) by 2 and 3, it is enough to divide both 30 and 36 by their common factor 6)
\(\frac{30}{36}=\frac{15}{18}=\frac{5}{6}\)

Like this, find another form of \(\frac {40}{50}\) by removing their common factor;
Divide both 40 and 50 by their common factor 5,
40 ÷ 5 = 8, 50 ÷ 5 = 10
\(\frac{40}{50}=\frac{8}{10}\)
2 is the factor of 8 and 10.
So, divide both 8 and 10 by 2, we get
8 ÷ 2 = 4, 10 ÷ 2 = 5
\(\frac{8}{10}=\frac{4}{5}\)
There is no common factor for 4 and 5, so the lowest form of \(\frac {40}{50}\) is \(\frac {4}{5}\).
(When the numerator and denominator of \(\frac {40}{50}\) is divided by 10 we get \(\frac {4}{5}\))
By removing the common factor of the numerator and denominator of any fraction, we get the lowest form of this fraction.
In general, if the numerator and denominator of a fraction have a common factor, then dividing them by this common factor gives a form of this fraction.

Write each of the following fractions in lowest terms:
(i) \(\frac {21}{28}\)
Answer:
7 is a factor of 21 and 28
Divide 21 by 7 we get 3
Divide 28 by 7 we get 4
\(\frac{21}{28}=\frac{3}{4}\) (3 and 4 do not have any common factor)

(ii) \(\frac {35}{50}\)
Answer:
5 is a factor of 35 and 50
35 ÷ 5 = 7, 50 ÷ 5 = 10
\(\frac{35}{50}=\frac{7}{10}\)

(iii) \(\frac {6}{60}\)
Answer:
2 is the factor of 6 and 60
So, \(\frac{6}{60}=\frac{3}{30}\)
3 is the factor of 3 and 30
\(\frac{3}{30}=\frac{1}{10}\)

(iv) \(\frac {40}{70}\)
Answer:
5 is a factor of 40 and 70
So, \(\frac{40}{70}=\frac{8}{14}\)
2 is a factor of 8 and 14
\(\frac{8}{14}=\frac{4}{7}\)
4 and 7 do not have any common factor.
So the lowest form is \(\frac {4}{7}\)
(Also 40 and 70 can be divided by 10 to get \(\frac {4}{7}\))

(v) \(\frac {7}{70}\)
Answer:
7 is the factor of 7 and 70
7 ÷ 7 = 1, 70 ÷ 10 = 7
\(\frac{7}{70}=\frac{1}{10}\)

Fraction as Division
In class 5, we have seen that the fraction \(\frac {2}{3}\) can be expressed in two different ways.

• If one is divided into 3 equal parts and 2 fares are taken from it.
• If two is divided into 3 equal parts and 1 is removed from it.

Any division can be written as a fraction like this.
For example 8 ÷ 4 = \(\frac {8}{4}\)
The lowest form is
\(\frac{8}{4}=\frac{2}{1}\) = 2
10 ÷ 5 = \(\frac {10}{5}\) = 2
24 ÷ 6 = \(\frac {24}{6}\) = 4
We can write like this.
\(\frac {4}{1}\) means 4 itself.
Like this, any natural number with a denominator of 1 can be expressed as a fraction.
\(\frac {5}{1}\) = 5, \(\frac {50}{1}\) = 50, \(\frac {100}{1}\) = 100
The remaining divisions can also be written as fractions.

If a 3 metre ribbon is divided equally among 2 people. Then what is the length of the ribbon each will get?
Answer:
One will get 3 ÷ 2.
Each person will get 1 metre of ribbon and 1 metre as a remainder.
Divide the remainder of 1 metre equally among the 2 people.
Each will get \(\frac {1}{2}\) metre.
So one person will get, 1 metre + \(\frac {1}{2}\) metre = 1\(\frac {1}{2}\) metre.

Like this, what is 8 ÷ 3?
8 ÷ 3 = \(\frac {8}{3}\)
Divide 6 into 3 equal parts, we get 2
Divide the remaining 2 into 3 equal parts, that is \(\frac {2}{3}\)
So, \(\frac{8}{3}=2 \frac{2}{3}\)

\(\frac {15}{4}\)
Divide 12 by 4, and we get 3
Divide the remaining 3 into 4 equal parts, we get \(\frac {3}{4}\)
So, \(\frac{15}{4}=3 \frac{3}{4}\)

If 14 is divided into 4 parts
\(\frac{14}{4}=\frac{7}{2}\)
That means dividing 14 into 4 parts is the same as dividing 7 into 2 parts.
\(\frac{14}{4}=3 \frac{2}{4}=3 \frac{1}{2}\)
\(\frac{7}{2}=3 \frac{1}{2}\)

Write the following in the form of fractions:
15 ÷ 3 = \(\frac {15}{3}\) = 5
21 ÷ 7 = \(\frac {21}{7}\) = 3
24 ÷ 4 = \(\frac {24}{4}\) = 6
32 ÷ 16 = \(\frac {32}{16}\) = 2
7 ÷ 3 = \(\frac {7}{3}\)
If 7 is divided into 3 parts, we get 2, and the remaining 1 is divided into 3 parts
\(\frac{7}{3}=2 \frac{1}{3}\)

25 ÷ 7 = \(\frac {25}{7}\)
If 21 is divided into 7 equal parts, we get 3, and the remaining 4 is divided into 7 parts
\(\frac{25}{7}=3 \frac{4}{7}\)

Answer the following questions:

Question 1.
If a 12 metre long ribbon is divided among 5 people. How long will a person get?
Answer:
12 ÷ 5 = \(\frac {12}{5}\)
If 10 is divided into 5 equal parts, we get 2, and the remaining 2 is divided into 5 parts, that is \(\frac {2}{5}\)
One person will get 2\(\frac {2}{5}\)

Question 2.
If a 37 metre iron rod is divided into 5 equal parts. What is the length of the rod in metres?
Answer:
37 ÷ 5 = \(\frac {37}{5}\)
If 35 is divided into 5 equal parts, we get 7, and the remaining 2 is divided into 5 parts, that is \(\frac {2}{5}\)
The length of one rod is 7\(\frac {2}{5}\) metre

Question 3.
If 18 kg of sugar is packed equally in 7 identical packets. What is the quantity of sugar in one packet?
Answer:
18 ÷ 7 = \(\frac {18}{7}\)
If 14 is divided into 7 equal parts, we get 2, and the remaining 4 is divided into 7 parts, that is \(\frac {4}{7}\)
Sugar in one packet is 2\(\frac {4}{7}\)

Question 4.
If 20 litres of coconut oil are filled in 6 bottles, then how many liters are in each bottle contains.
Answer:
20 ÷ 6 = \(\frac {20}{6}\)
If 18 is divided into 6 equal parts, we get 3, and the remaining 2 is divided into 6 parts, that is \(\frac{2}{6}=\frac{1}{3}\).
One bottle contain 3\(\frac {1}{3}\)
Or dividing 20 into 6 parts is the same as dividing 10 into 3 parts.
\(\frac{20}{6}=\frac{10}{3}, \quad \frac{10}{3}=3 \frac{1}{3}\)

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 1 Angles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 1 Solutions Angles

Class 6 Kerala Syllabus Maths Solutions Chapter 1 Angles Questions and Answers

Angles Class 6 Questions and Answers Kerala Syllabus

Drawing Angles (Page 13)

Question 1.
Measure and mark each angle below:


Answer:
(i) 60°
(ii) 135°
(iii) 60°
(iv) 130°

Question 2.
Draw the pictures below in the notebook:

Answer:
(i)

• Draw PQ of length 4 cm.
• From P measure 50° (right measure).
• From Q measure 50° (left).
• Join the two lines.

(ii)

• Make a line of 5 cm.
• From the right end, measure 90° and make a line.
• From the left end, draw a line making a 30° angle.
• Meet the two lines to get the required figure.

(iii)

• Draw a line of length 5 cm.
• From one end, draw a line of 3 cm at an angle of 60°.
• From the other end, draw a line of 3 cm, making 120°.
• Join the angles of the two lines.

(iv)

• Draw a line (dotted) of 6 cm.
• From both ends, draw a line to make a 20° angle on both sides.
• Join the lines together to get the required figure.

Circle Division (Pages 17 & 18)

Question 1.
In each of the pictures below, calculate what fractions of the circle are the yellow and green parts:


Answer:
The measure of an angle inside a circle is 360°
Figure-1
Angle measure of the yellow portion is 20°
Total fractions of the circle is 360° ÷ 20 = 18
The fraction of yellow portion is \(\frac {1}{18}\)
The green portion is \(\frac {17}{18}\)

Figure-2
Angle measure of the yellow portion is 24°
Total fractions of the circle is 360° ÷ 24 = 15
The fraction of yellow portion is \(\frac {1}{15}\)
The green portion is \(\frac {14}{15}\)

Figure-3
Angle measure of the yellow portion is 54°
But for 18° its fraction is, \(\frac {1}{20}\). So for 54° it is \(\frac {3}{20}\).
Therefore the fraction of yellow portion is \(\frac {3}{20}\)
The green portion is \(\frac {17}{20}\)

Figure-4
The angle measure of the yellow portion is 80°
Total fractions of the circle when it is 40°;
360° ÷ 40 = 9. So for 40° it is \(\frac {1}{9}\)
The yellow portion is 80°.
The fraction of yellow portion is \(\frac {2}{9}\)
The green portion is \(\frac {7}{9}\)

Figure-5
Angle measure of the yellow portion is 108°
If it is 36° its fraction is \(\frac {1}{10}\).
That means 36 × 3 = 108
So for 108° it is \(\frac {3}{10}\)
Therefore the fraction of yellow portion is \(\frac {3}{10}\)
The green portion is \(\frac {7}{10}\)

Figure-6
Angle measure of the yellow portion is 150°
If it is 30° its fraction is, \(\frac {1}{12}\).
So for 150° it is \(\frac {5}{12}\).
Therefore the fraction of yellow portion is \(\frac {5}{12}\)
The green portion is \(\frac {7}{12}\)

Question 2.
Mark each of the fractions below as part of a circle and colour the pictures.
(i) \(\frac {3}{8}\)
(ii) \(\frac {2}{5}\)
(iii) \(\frac {4}{9}\)
(iv) \(\frac {5}{12}\)
(v) \(\frac {5}{24}\)
Answer:
(i) \(\frac {3}{8}\)

Angle inside a circle is 360°.
360 is divided into 8 equal parts.
\(\frac {360}{8}\) = 45°
Each part measures 45°.
Angle of shaded region = 3 × 45 = 135°

(ii) \(\frac {2}{5}\)

\(\frac {360}{5}\) = 72°
Each part measures 72°.
Shaded angle = 72 × 2 = 144°

(iii) \(\frac {4}{9}\)

\(\frac {360}{9}\) = 40°
Each part measures 40°.
Shaded angle = 4 × 40° = 160°

(iv) \(\frac {5}{12}\)

\(\frac {360}{12}\) = 30°
Each part measures 30°.
Shaded angle = 30 × 5 = 150°

(v) \(\frac {5}{24}\)

\(\frac {360}{24}\) = 15°
Each part measures 15°.
Shaded angle = 15 × 5 = 75°

Question 3.
Draw the pictures below:

Answer:
Figure-1
Draw a circle and mark 72° in the centre of the circle, and complete the pattern. From the corner-1, draw a line to the corners 3 and 4 similarly, from corner-5 draw a line to the corners 2 and 3, and from corner-2 draw lines to the corners 4 and 5. And colour the picture.

Figure-2
To draw the second figure, make some changes to the above second figure.

Figure-3
Draw an 8-sided figure and draw lines inside the figure to get a square.

Figure-4
Draw it like this.

Class 6 Maths Chapter 1 Kerala Syllabus Angles Questions and Answers

Class 6 Maths Angles Questions and Answers

Question 1.
Draw the angles on the lines that are marked below.

Answer:
Place the point which the bottom line and the perpendicular line joins together in the protractor at the end of the line where the angle should be drawn. For drawing the figures 4 and 5 place the protractor downwards and mark the angles and complete the angles.

Question 2.
Measure and mark each angle below:

Answer:
(i) 50°
(ii) 130°
(iii) 90°
(iv) 105°
(v) 103°

Question 3.
From the angles given below, without measuring it classify them into angles right angle, less than and greater than right angles.

Answer:
Less than right angle: a, d, h
Greater than right angle: b, e, f, g
Right angle: c

Question 4.
Measure the angles below.

Answer:
(a) 120°
(b) 60°
(c) 58°
(d) 125°
(e) 90°
(f) 22°

Question 5.
Divide the circle into 6 equal parts and draw different figures.
Answer:

Question 6.
Draw a picture with 5 and 8 sides.
Answer:
Draw a circle and mark a horizontal line (radius), and mark a 72° angle on it.
Again, mark a 72° angle with each line.

Draw a circle and mark a horizontal line (radius), and mark a 45° angle on it.
Again, mark a 45° angle with each line.

Question 7.
Measure all the angles.

Answer:
(a) 1 – 90°, 2 – 90°, 3 – 130°, 4 – 50°
(b) 1 – 150°, 2 – 120°, 3 – 60°, 4 – 30°

Question 8.
Draw these pictures with the same measurements.

Answer:

Question 9.
Calculate what fraction of the circle is the shaded portion.

Answer:
Figure-1
Angle measure of the shaded portion is 36°
Total fractions of the circle is 360° ÷ 36 = 10
The fraction of shaded portion is \(\frac {1}{10}\)

Figure-2
Angle measure of the shaded portion is 120°
Total fractions of the circle is 360° ÷ 120 = 3
The fraction of shaded portion is \(\frac {1}{3}\)

Class 6 Maths Chapter 1 Notes Kerala Syllabus Angles

→ A figure formed by two lines meeting at a point is called an angle.

→ A protractor is a simple measuring instrument that is used to measure angles.

→ It is in the shape of a semicircle with an inner scale and an outer scale and with markings from 0° to 180° on it.

→ The angle at a square corner is 90°. It is also called a right angle.

→ The measure of a circle is 360°.

An angle is formed when two lines meet at a point. The space between these lines is called an angle. Angles are measured in degrees or radians, and they are grouped based on their size. Angles are very useful in our daily lives. For example, engineers use angles to build houses, bridges, and buildings. Athletes use angles in sports to improve their movements. Carpenters use angles to make things like doors, tables, and chairs. In this chapter, we will learn how to measure angles using a protractor, and how to divide a circle into equal parts.

When Lines Join
Remember how we drew various shapes joining lines straight up and slanted in the section Line Math of the lesson Lines and Circles in the class 5 textbook. Here we draw different patterns using the set square.

If we draw two lines upward of the same length on the two ends of a line and join the top ends using another line, the lengths of the lines in the top and bottom are of the same length.

Instead of drawing the straight-up lines, two slanted lines can be drawn using the set square. We know that the length of the top and bottom lines is are same.

We learn more about the topic straight up and slanted in this unit.
A figure formed by two lines meeting at a point is called an angle.

For example: Arrange these angles based on their size.

Write the numbers of the cones in ascending order.
The angle with less spread out is 4, and the more spread out is 6.
Therefore, the smallest angle is 4 and the largest angle is 6.
Thu,s the arrangement is of the form, 4, 1, 3, 2, 5, 6.

Protractor:
A protractor is a simple measuring instrument that is used to measure angles. It is in the shape of a semicircle with an inner scale and an outer scale, and with markings from 0° to 180° on it.

Considering the inner scale, starting with the bottom line marked 0, there are other lines upward; and as they move up, the angles between them and the bottom line become larger and larger. The numbers at their ends show the sizes of these angles.

How do you measure these angles using a protractor?

Place the protractor at the corner of each angle, as shown below:

Here we can see that the angle on the second figure is 60°.
In the first figure, the slant line of this angle passes between 50 and 60 on the protractor.
We can see that the small lines in the protractor divide the gaps between the multiples of 10 into ten equal parts. Each of them shows a difference of 1°.
Here, in the first figure, the slant line goes through the fifth (slightly larger) line between 50 and 60. That means the angle is 55°.

The angle at a square corner is 90°. It is also called a right angle.

A line making an angle of 90° with another line is said to be perpendicular to the first line.

For example, from the following figures, find out the perpendicular lines.

The lines in Figures 1, 6, and 8 are perpendiculars.
The other figures the angles does not form 90°; therefore, the lines are not perpendicular.

Drawing Angles
Draw an angle of 40° using a protractor.
First, draw a horizontal line. Then place the protractor at its left end as shown below and mark a point at the number 40 in it.

Now remove the protractor and join this point and the left endpoint of the first line to get a 40° angle:

For example: Draw an angle measure of 80°.
First, draw a horizontal line. Then, place the point where the bottom line and the perpendicular line join together in the protractor at the end of the line where the angle should be drawn.

Then mark the point 80° and join this point.

Circle Division
A protractor is a semicircle with 180° measure. Then for a circle, its measure is 360°.
That is, the measure of the angle around the centre of a circle is 360°.
In a circle, 360 equally spaced radii are drawn, and then the angle between any two nearby radii is 1°.
Here, the circle is divided into 360 equal parts.

What if we divide a circle into 4 equal parts?
Its one part is, 360° ÷ 4 = 90°

What if we draw radii 10° apart?
We get 36 radii, which divide the circle into 36 equal parts.

For example, divide a circle into 8 equal parts.
360° ÷ 8 = 45°
Here we get 8 radii, which are 45° apart.

The angle measures of some other portions are;