Class 9

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Std 9 Chemistry Chapter 1 Notes Solutions Malayalam Medium ആറ്റത്തിൻ്റെ ഘടന

Kerala Syllabus 9th Standard Chemistry Chapter 1 Notes Solutions Malayalam Medium ആറ്റത്തിൻ്റെ ഘടന

Class 9 Chemistry Chapter 1 Notes Malayalam Medium Let Us Assess Answers

Question 1.
കാഥോഡ് രശ്മികളുമായി ബന്ധപ്പെട്ട ചില പരീക്ഷണങ്ങളുടെ നിരീക്ഷണങ്ങൾ നൽകിയിരിക്കുന്നു. ഓരോ നിരീക്ഷണത്തിന്റെയും അനുമാനം എഴുതുക.
a) കാഥോഡ് രശ്മികളുടെ പാതയിൽ വച്ച നേർത്ത ഇതളുകളുള്ള ചക്രം കറങ്ങുന്നു.
b) കാഥോഡ് രശ്മികളുടെ പാതയിൽ ഒരു വസ്തു വച്ചാൽ നിഴൽ ഉണ്ടാകുന്നു.
c) കാഥോഡ് രശ്മികളുടെ പാതയ്ക്ക് ലംബമായി ഒരു വൈദ്യുത മണ്ഡലം പ്രയോഗിക്കുമ്പോൾ അത് പോസിറ്റീവ് പ്ലേറ്റിനടുത്തേക്ക് വ്യതിചലിക്കുന്നു.
Answer:
a) കാഥോഡ് രശ്മികളുടെ പാതയിൽ നേർത്ത ഇതളുകളുള്ള ചക്രം കറങ്ങുന്നു. – ഇതിൽ നിന്നും കാഥോഡ് രശ്മികളിലെ കണങ്ങൾക്ക് മാസ് ഉണ്ടെന്നു മനസ്സിലാക്കാം.

b) കാഥോഡ് രശ്മികളുടെ പാതയിൽ ഒരു വസ്തു വച്ചാൽ നിഴൽ ഉണ്ടാകുന്നു. – ഇതിൽനിന്നും കാഥോഡ് രശ്മികൾ നേർരേഖയിലാണ് സഞ്ചരിക്കുന്നതെന്ന് ബോധ്യപ്പെടുന്നു.

c) കാഥോഡ് രശ്മികളുടെ പാതയ്ക്ക് ലംബമായി ഒരു വൈദ്യുത മണ്ഡലം പ്രയോഗിക്കുമ്പോൾ അത് പോസിറ്റീവ് പ്ലേറ്റിനടുത്തേക്ക് വ്യതിചലിക്കുന്നു. – ഇതിൽ നിന്നും കാഥോഡ് രശ്മികൾക്ക് നെഗറ്റീവ് ചാർജ് ഉണ്ടെന്നു മനസ്സിലാക്കാം.

Question 2.
ഒരു ആറ്റത്തിന്റെ അറ്റോമിക നമ്പർ 16-ഉം മാസ് നമ്പർ 32-ഉം ആണ്.
a) ഈ ആറ്റത്തിൽ എത്ര ഇലക്ട്രോൺ, പ്രോട്ടോൺ, ന്യൂട്രോൺ എന്നിവ അടങ്ങിയിരിക്കുന്നു?
b) ഈ ആറ്റത്തിന്റെ’ഇലക്ട്രോൺ വിന്യാസം എഴുതുക.
c) ഇതിന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം ചിത്രീകരിക്കുക.
Answer:
a) അറ്റോമിക നമ്പർ = 16
മാസ് നമ്പർ = 32
പ്രോട്ടോണുകളുടെ എണ്ണം = അറ്റോമിക നമ്പർ = 16
ഇലക്ട്രോണുകളുടെ എണ്ണം = 16
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = 32 – 16 = 16

b) ഇലക്ട്രോൺ വിന്യാസം = 2, 8, 6

c) ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം

Question 3.
ഒരു ആറ്റത്തിലെ K, L, M എന്നീ ഷെല്ലുകളിൽ ഇലക്ട്രോണുകൾ ഉണ്ട്.
a) ഈ ഷെല്ലുകളിൽ ഏറ്റവും ഊർജം കൂടിയ ഷെൽ ഏത്?
b) M ഷെല്ലിൽ 3 ഇലക്ട്രോണുകൾ മാത്രമേ ഉള്ളൂവെങ്കിൽ ഈ ആറ്റത്തിന്റെ അറ്റോമിക നമ്പർ എഴു
തുക.
c) ഈ ആറ്റത്തിലെ ഇലക്ട്രോണുകളുടെ എണ്ണമെത്രയാണ്?
d) ഈ ആറ്റത്തിന്റെ ന്യൂക്ലിയസിൽ 16 ന്യൂട്രോണുകളാണുള്ളതെങ്കിൽ അതിന്റെ മാസ് നമ്പർ എത്ര യാണ്?
Answer:
a) M ഷെൽ

b) പൊതുവേ താഴ്ന്ന ഊർജനിലയിൽ ഉള്ള ഒരു ഓർബിറ്റിൽ ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോ ണുകൾ നിറഞ്ഞതിനുശേഷം മാത്രമേ അടുത്ത ഊർജനിലയിലുള്ള ഓർബിറ്റിൽ ഇലക്ട്രോൺ പൂരണം നടക്കുകയുള്ളൂ.
K ഷെല്ലിൽ ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകളുടെ എണ്ണം 2 ഉം L ഷെല്ലിൽ ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകളുടെ എണ്ണം 8 ഉം ആണ്. അങ്ങനെയെങ്കിൽ,
അറ്റോമിക നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം = ഇലക്ട്രോണുകളുടെ എണ്ണം = 2 + 8 + 3 = 13

c) ആകെ ഇലക്ട്രോണുകളുടെ എണ്ണം = 13

d) ന്യൂട്രോണുകളുടെ എണ്ണം = 16
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം = 16 + 13 = 29

Question 4.
ഒരു ആറ്റത്തിന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം ചിത്രീകരിച്ചിരിക്കുന്നു.

a) ഈ ആറ്റത്തിന്റെ മാസ് നമ്പർ എത്ര?
b) ഇതിന്റെ ഇലക്ട്രോൺ വിന്യാസം എഴുതുക.
Answer:
ചിത്രത്തിൽ നിന്ന്
a) പ്രോട്ടോണുകളുടെ എണ്ണം = 13
ന്യൂട്രോണുകളുടെ എണ്ണം = 14
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം = 13 + 14 = 27

b) ഇലക്ട്രോണുകളുടെ എണ്ണം = 13
ഇലക്ട്രോൺ വിന്യാസം = 2, 8, 3

Question 5.
ചില മൂലകങ്ങളുടെ പ്രതീകങ്ങൾ നൽകിയിരിക്കുന്നു.
Answer:
\({ }_{12}^{24} \mathrm{Mg}\), \({ }_{6}^{12} \mathrm{C}\), \({ }_{7}^{15} \mathrm{N}\), \({ }_{6}^{14} \mathrm{C}\), \({ }_{11}^{24} \mathrm{Na}\)
a) ഇവയിൽ നിന്നും ഒരു ജോഡി ഐസോടോപ്പുകൾ തിരഞ്ഞെടുത്തെഴുതുക. ഈ ജോഡി തിരഞ്ഞ
ടുക്കാനുള്ള കാരണം എഴുതുക.
b) തന്നിരിക്കുന്ന മൂലകങ്ങളിൽ നിന്നും ഒരു ജോഡി ഐസോബാറുകൾ തിരഞ്ഞെടുക്കുക.
Answer:
a) ഐസോടോപ്പ് ജോഡി = \({ }_{6}^{12} \mathrm{C}\), \({ }_{6}^{14} \mathrm{C}\)
കാരണം – ഒരേ അറ്റോമിക നമ്പറും വ്യത്യസ്ത മാസ് നമ്പറുമുള്ള ഒരേ മൂലകത്തിന്റെ വ്യത്യസ്ത ആറ്റങ്ങ ളാണ് ഐസോടോപ്പുകൾ.

b) \({ }_{12}^{24} \mathrm{Mg}\), \({ }_{11}^{24} \mathrm{Na}\)

Question 6.
A, B കോളങ്ങൾ അനുയോജ്യമായ രീതിയിൽ ചേർത്തെഴുതുക.

A	B
പ്ലം പുഡിങ് മാതൃക	ജെയിംസ് ചാഡ്വിക്
സൗരയൂഥ മാതൃക	ഗോൾഡ്സ്റ്റൈൻ
കനാൽ രശ്മികൾ	ജെ. ജെ. തോംസൺ
ന്യൂട്രോൺ	റഥർഫോർഡ്

Answer:

A	B
പ്ലം പുഡിങ് മാതൃക	ജെ. ജെ. തോംസൺ
സൗരയൂഥ മാതൃക	റഥർഫോർഡ്
കനാൽ രശ്മികൾ	ഗോൾഡ്സ്റ്റൈൻ
ന്യൂട്രോൺ	ജെയിംസ് ചാഡ്വിക്

Question 7.
ഒരു മൂലകത്തിന്റെ അറ്റോമിക നമ്പറും മാസ് നമ്പറും യഥാക്രമം 15, 31 എന്നിങ്ങനെയാണ്.
a) ഈ ആറ്റത്തിലെ ബാഹ്യതമ ഇലക്ട്രോണുകളുടെ എണ്ണം എത്ര?
b) ഇതിൽ എത്ര ന്യൂട്രോണുകൾ അടങ്ങിയിരിക്കുന്നു?
c) ഈ മൂലകത്തിന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം ചിത്രീകരിക്കുക.
Answer:
a) അറ്റോമിക നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം = ഇലക്ട്രോണുകളുടെ എണ്ണം = 15
ഇലക്ട്രോൺ വിന്യാസം = 2, 8, 5
ബാഹ്യതമ ഇലക്ട്രോണുകളുടെ എണ്ണം = 5

b) മാസ് നമ്പർ = അറ്റോമിക നമ്പർ + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – അറ്റോമിക നമ്പർ = 31 – 15 = 16

c) ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം

Question 8.
ഫോസിലുകളുടെ കാലപ്പഴക്കം നിർണയിക്കാൻ ഒരു മൂലകത്തിന്റെ ഐസോടോപ്പ് ഉപയോഗി ക്കുന്നു.
a) ഈ ഐസോടോപ്പ് ഏത്?
b) ഈ മൂലകത്തിന്റെ മറ്റ് രണ്ട് പ്രധാന ഐസോടോപ്പുകൾ ഏതൊക്കെ?
c) ഓരോ ഐസോടോപ്പിലുമുള്ള ന്യൂട്രോണുകളുടെ എണ്ണം എഴുതുക.
Answer:
a) \({ }_6^{14} \mathrm{C}\)

b) \({ }_6^{12} \mathrm{C}\), \({ }_6^{13} \mathrm{C}\)

c) \({ }_6^{14} \mathrm{C}\) ഐസോടോപ്പിൽ ഉള്ള ന്യൂട്രോണുകളുടെ എണ്ണം = 8
\({ }_6^{13} \mathrm{C}\) ഐസോടോപ്പിൽ ഉള്ള ന്യൂട്രോണുകളുടെ എണ്ണം = 7
\({ }_6^{12} \mathrm{C}\) ഐസോടോപ്പിൽ ഉള്ള ന്യൂട്രോണുകളുടെ എണ്ണം = 6

തുടർപ്രവർത്തനങ്ങൾ

Question 1.
ആറ്റം ചരിത്രവുമായി ബന്ധപ്പെട്ട ശാസ്ത്രജ്ഞരെക്കുറിച്ചും അവരുടെ സംഭാവനകളെക്കുറിച്ചും ഒരു പ്രസന്റേഷൻ തയ്യാറാക്കി ക്ലാസ്സിൽ അവതരിപ്പിക്കുക.
Answer:

• ഡെമോക്രിറ്റസ്: ആറ്റം സിദ്ധാന്തത്തിന്റെ തുടക്കം; എല്ലാ വസ്തുക്കളും വളരെ ചെറുതും വിഭജിക്ക പ്പെടാത്തതുമായ കണങ്ങൾ കൊണ്ടാണ് നിർമ്മിതമെന്ന സിദ്ധാന്തം.
• ജോൺ ഡാൽട്ടൻ: (1809) ആധുനിക ആറ്റം സിദ്ധാന്തം;(ആറ്റങ്ങൾ വിഭജിക്കപ്പെടാത്ത കണങ്ങളാണ്, ഓരോ ഘടകത്തിനും തന്മാത്രകളുണ്ട്, ‘രാസപ്രവർത്തനങ്ങളിൽ ആറ്റങ്ങൾ പുനഃസംഘടന ചെയ്യ പ്പെടുന്നു).
• ജെ.ജെ. തോംസൺ: ഇലക്ട്രോണിന്റെ കണ്ടെത്തൽ;(ആറ്റത്തിന്റെ ഉള്ളിൽ നെഗറ്റീവ് ചാർജുള്ള കണങ്ങൾ ഉണ്ടെന്നു കണ്ടെത്തി) (1897).
• ഏണസ്റ്റ് റഥർഫോർഡ്: (1901) ആറ്റത്തിന്റെ ന്യൂക്ലിയർ മോഡൽ;( ആറ്റത്തിന്റെ മദ്ധ്യഭാഗത്ത് പോസിറ്റീവ് ചാർജുള്ള ന്യൂക്ലിയസ് ഉണ്ട്, ഇലക്ട്രോണുകൾ പുറത്ത് ചുറ്റിനടക്കുന്നു.)
• നീൽസ് ബോർ: ബോർ മോഡൽ; ഇലക്ട്രോണുകൾ പ്രത്യേക ഊർജതലങ്ങളിലായാണ് ചുറ്റുന്നത്, ഈ ഊർജതലങ്ങൾക്കിടയിൽ ഇലക്ട്രോണുകൾ മാറുമ്പോൾ ഊർജം പരാമാവധി കൈമാറ്റം ചെയ്യപ്പെടുന്നു.

Question 2.
വിവിധ സബ് ആറ്റോമിക കണങ്ങളുടെ കണ്ടുപിടിത്തത്തിലേക്ക് നയിച്ച പ്രധാന സംഭവങ്ങൾ എഴുതി ടൈംലൈൻ ചാർട്ട് തയ്യാറാക്കുക.
Answer:

Question 3.
ഐസോടോപ്പുകളെക്കുറിച്ച് മനസ്സിലാക്കിയല്ലോ. റേഡിയോ ഐസോടോപ്പുകൾക്ക് കൂടുതൽ ഉദാഹരണങ്ങൾ കണ്ടെത്തുക. ഓരോ റേഡിയോ ഐസോടോപ്പിന്റെയും ഉപയോഗത്തെ ക്കുറിച്ച് ലേഖനം തയ്യാറാക്കി ശാസ്ത്രമാസികയിൽ പ്രസിദ്ധീകരിക്കുക. വേർഡ് പ്രോസസറിന്റെ സഹായത്തോടെ ഈ പ്രവർത്തനം ചെയ്യാമല്ലോ.
Answer:

Question 4.
നിങ്ങൾക്ക് റഥർഫോർഡുമായി ഒരു അഭിമുഖം നടത്താൻ അവസരം ലഭിക്കുകയാണെങ്കിൽ അതിന് ആവശ്യമായ ചോദ്യാവലി തയ്യാറാക്കുക.
Answer:

• നിങ്ങൾ പ്രതിപാദിച്ച ആറ്റം ഘടനയുടെ അടിസ്ഥാനത്തിൽ ആധുനിക രാസ ശാസ്ത്രം എങ്ങനെ വികസി ച്ചുവെന്ന് നിങ്ങൾ കരുതുന്നു?
• വളർന്നു വരുന്ന തലമുറയ്ക്ക് ഗവേഷണത്തിൽ അഭിരുചി ഉണ്ടാകാൻ എന്തെല്ലാം കാര്യങ്ങൾ നിങ്ങൾക്ക് നിർദേശിക്കാൻ കഴിയും?
• ആറ്റം ഘടനയെ കുറിച്ചുള്ള ഇപ്പോഴത്തെ ഗവേഷണങ്ങളോട് നിങ്ങൾക്ക് എന്താണ് പറയാനുള്ളത്?
• നിങ്ങളുടെ ഗവേഷണ ജീവിതത്തിലെ ഏറ്റവും അവിസ്മരണീയമായ അനുഭവം എന്തായിരുന്നു?
• ആറ്റം ഘടനയിലെ നിങ്ങളുടെ സിദ്ധാന്തങ്ങൾ ശാസ്ത്രലോകം സ്വീകരിക്കുന്നതിനായി വന്ന പ്രചാരങ്ങളും ആശങ്കകളും എങ്ങനെ മറികടന്നു?

9th Class Chemistry Notes Pdf Malayalam Medium Chapter 1

Question 1.
പരിചയമുള്ള പദാർഥങ്ങളിൽ അടങ്ങിയിരിക്കുന്ന ആറ്റങ്ങൾ ഏതൊക്കെയെന്ന് തിരിച്ചറിയാമോ? പട്ടിക വിശകലനം ചെയ്യുക.
വിവിധ പദാർഥങ്ങളിലെ തന്മാത്രകൾ എങ്ങനെയെല്ലാം വ്യത്യാസപ്പെട്ടിരിക്കുന്നു?

Answer:

• തന്മാത്രയിലടങ്ങിയിരിക്കുന്ന ഘടക മൂലകങ്ങൾ
• ഘടക മൂലക ആറ്റങ്ങളുടെ എണ്ണത്തിന്റെ അനുപാതം

Question 2.
ആറ്റങ്ങളിൽ അടങ്ങിയിരിക്കുന്ന പ്രധാന കണങ്ങൾ എന്തൊക്കെയാണ്?
Answer:

• ഇലക്ട്രോൺ
• പ്രോട്ടോൺ
• ന്യൂട്രോൺ
  ഇവ സബ്അറ്റോമിക കണങ്ങൾ എന്നറിയപ്പെടുന്നു

Question 3.
ഇലക്ട്രോണിന് മാസുണ്ടെന്ന് തെളിയിച്ചതെങ്ങനെ?
Answer:
കാഥോഡ് രശ്മികളുടെ പാതയിൽ നേർത്ത ഇതളുകളുള്ള ചക്രം (Paddle wheel) വെച്ചാൽ അത് കറങ്ങുന്നു. ഇതിൽ നിന്നും കാഥോഡ് രശ്മികളിലെ കണങ്ങൾക്ക് മാസ് ഉണ്ടെന്നു മനസ്സിലാക്കാം.

Question 4.
കാഥോഡ് രശ്മികളുടെ പാതയിൽ ഒരു അതാര്യ വസ്തു വച്ചാൽ നിഴൽ ഉണ്ടാകുന്നു. ഇതിൽ നിന്ന് എന്ത് മനസ്സിലാക്കാം?
Answer:
കാഥോഡ് രശ്മികളുടെ പാതയിൽ അതാര്യ വസ്തുക്കൾ വെച്ചാൽ നിഴൽ ഉണ്ടാകുന്നു. ഇതിൽനിന്നും കാഥോഡ് രശ്മികൾ നേർരേഖയിലാണ് സഞ്ചരിക്കുന്നതെന്ന് ബോധ്യപ്പെട്ടു.

Question 5.
സബ്അറ്റോമിക കണങ്ങളായ ഇലക്ട്രോൺ, പ്രോട്ടോൺ, ന്യൂട്രോൺ എന്നിവയുടെ ചില സവി ശേഷതകൾ പട്ടികയിൽ കൊടുത്തിരിക്കുന്നു. വിട്ടുപോയ ഭാഗം പൂരിപ്പിച്ച് സയൻസ് ഡയറിയിൽ രേഖപ്പെടുത്തുക.

Answer:

Question 6.
ചില പ്രസ്താവനകൾ നൽകിയിരിക്കുന്നു. ഇവയിൽ ജെ. ജെ. തോംസണുമായി ബന്ധപ്പെട്ട പ്രസ്താവനകൾ ഏതെല്ലാം?
Answer:
a) ഓർബിറ്റ് എന്ന ആശയം മുന്നോട്ടുവച്ചു.
b) ഡിസ്ചാർജ് ട്യൂബ് പരീക്ഷണങ്ങൾ നടത്തി.
c) ന്യൂട്രോണിനെ കണ്ടെത്തി.
d) ഇലക്ട്രോണിനെ കണ്ടെത്തി.
e) പ്ലം പുഡിങ് മാതൃക മുന്നോട്ടുവച്ചു.
Answer:
b) ഡിസ്ചാർജ് ട്യൂബ് പരീക്ഷണങ്ങൾ നടത്തി.
d) ഇലക്ട്രോണിനെ കണ്ടെത്തി.

Question 7.
ആറ്റം ഘടനയെക്കുറിച്ച് ഗവേഷണം നടത്തിയ ശാസ്ത്രജ്ഞർ, ‘അവരുടെ സംഭാവനകൾ എന്നിവയെ സംബന്ധിച്ച് ചോദ്യാവലി തയ്യാറാക്കി ക്ലാസിൽ ഒരു ക്വിസ് മത്സരം സംഘടിപ്പിക്കുക.
Answer:
നിങ്ങളുടെ റഫറൻസിനായുള്ള ചില ചോദ്യങ്ങൾ

• ഡിസ്ചാർജ് ട്യൂബുകളും വാക്വം ട്യൂബുകളും ആരാണ് വികസിപ്പിച്ചത്?
  – ഹെൻറിച്ച് ഗീസ്ലർ
• ഇലക്ട്രോണുകൾ കണ്ടെത്തിയത് ആരാണ്?
  – ജെ.ജെ. തോംസൺ
• ഇലക്ട്രോണുകളുടെ അനുപാതം കണ്ടെത്തിയ ശാസ്ത്രജ്ഞന്റെ പേര്.
  – ജെ.ജെ. തോംസൺ
• കനാൽ രശ്മികൾ കണ്ടെത്തിയത് ആരാണ്?
  – ഒയ്ഗൻ ഗോൾഡ്സ്റ്റൈൻ
• കാഥോഡ് രശ്മികളുടെ കണികകളുടെ ചാർജ് എന്താണ്?
  – നെഗറ്റീവ്
• ഇലക്ട്രോണിന്റെ ചാർജും മാസും കണ്ടെത്തിയത് ആരാണ്?
  – റോബർട്ട് മില്ലിക്കൺ
• റേഡിയോ ആക്റ്റീവത കണ്ടുപിടിച്ച ശാസ്ത്രജ്ഞൻ?
  – ഹെൻറി
  ബെക്വറൽ
• വൈദ്യുത കാന്തിക മണ്ഡലങ്ങളിൽ വ്യതിചലിക്കാത്ത കണികകൾ ഏതാണ്?
  – ന്യൂട്രോൺ

Question 8.
ആറ്റത്തിന്റെ ന്യൂക്ലിയസിലെ കണങ്ങൾ ഏതൊക്കെയാണ്?
Answer:
പ്രോട്ടോണുകളും ന്യൂട്രോണുകളും.
ഒരാറ്റത്തിലെ പ്രോട്ടോണുകളുടെയും ന്യൂട്രോണുകളുടെയും ആകെ എണ്ണത്തെ മാസ് നമ്പർ എന്ന് പറയുന്നു. ഇതിനെ A എന്ന അക്ഷരം ഉപയോഗിച്ച് സൂചിപ്പിക്കാം.

Question 9.
2 പ്രോട്ടോണുകളും, 2 ന്യൂട്രോണുകളും ഉള്ള ഒരു ആറ്റത്തിന്റെ മാസ് നമ്പർ എത്രയായിരിക്കും?
Answer:
പ്രോട്ടോണുകളുടെ എണ്ണം = 2
ന്യൂട്രോണുകളുടെ എണ്ണം = 2
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം = 2 + 2 = 4

മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം = അറ്റോമിക നമ്പർ + ന്യൂട്രോണുകളുടെ എണ്ണം

ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം
= മാസ് നമ്പർ – അറ്റോമിക നമ്പർ
= (A – Z)

ഒരു ആറ്റത്തെ പ്രതീകം ഉപയോഗിച്ച് പ്രതിനിധാനം ചെയ്യുമ്പോൾ പ്രതീകത്തിന്റെ ഇടതുവശത്ത് മുകളിലും താഴെ യുമായി യഥാക്രമം മാസ് നമ്പറും അറ്റോമിക നമ്പറും എഴുതുന്നു.
ഉദാ : \({ }_{17}^{35} \mathrm{Cl}\), \({ }_{20}^{40} \mathrm{Ca}\)

Question 10.
ക്ലോറിൻ, കാൽസ്യം എന്നീ ആറ്റങ്ങളിലെ പ്രോട്ടോണുകൾ, ഇലക്ട്രോണുകൾ, ന്യൂട്രോണുകൾ എന്നിവയുടെ എണ്ണം കണ്ടെത്തുക.

Answer:
a) ക്ലോറിൻ സ്കൂ\({ }_{17}^{35} \mathrm{Cl}\)
അറ്റോമിക നമ്പർ, Z = 17
മാസ് നമ്പർ, A = 35
പ്രോട്ടോണുകളുടെ എണ്ണം = 17
ഇലക്ട്രോണുകളുടെ എണ്ണം = 17
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
∴ ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 35 – 17 = 18

b) കാൽസ്യം \({ }_{20}^{40} \mathrm{Ca}\)
അറ്റോമിക നമ്പർ, Z = 20
മാസ് നമ്പർ, A = 40
പ്രോട്ടോണുകളുടെ എണ്ണം = 20
ഇലക്ട്രോണുകളുടെ എണ്ണം = 20
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
∴ ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 40 – 20 = 20

Question 11.
ചുവടെ നൽകിയിരിക്കുന്ന പട്ടിക പൂർത്തിയാക്കി സയൻസ് ഡയറിയിൽ രേഖപ്പെടുത്തുക.

Answer:
അറ്റോമിക നമ്പർ (Z) = പ്രോട്ടോണുകളുടെ എണ്ണം = ഇലക്ട്രോണുകളുടെ എണ്ണം
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
= അറ്റോമിക നമ്പർ + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം
= മാസ് നമ്പർ – അറ്റോമിക നമ്പർ
= (A – Z)

a) \({ }_1^1 \mathrm{H}\)
അറ്റോമിക നമ്പർ, Z = 1
മാസ് നമ്പർ, A = 1
പ്രോട്ടോണുകളുടെ എണ്ണം = 1
ഇലക്ട്രോണുകളുടെ എണ്ണം = 1
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 1 – 1 – 0

b) \({ }_3^7 \mathrm{Li}\)
അറ്റോമിക നമ്പർ, Z= 3
മാസ് നമ്പർ, A = 7
പ്രോട്ടോണുകളുടെ എണ്ണം = 3
ഇലക്ട്രോണുകളുടെ എണ്ണം = 3
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z – 7 – 3 = 4

c) \({ }_8^16 \mathrm{O}\)
അറ്റോമിക നമ്പർ, Z = 8
മാസ് നമ്പർ, A = 16
പ്രോട്ടോണുകളുടെ എണ്ണം = 8
ഇലക്ട്രോണുകളുടെ എണ്ണം = 8
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z – 16 – 8 = 8

d) \({ }_11^23 \mathrm{Na}\)
അറ്റോമിക നമ്പർ, Z = 11
മാസ് നമ്പർ, A = 23
പ്രോട്ടോണുകളുടെ എണ്ണം = 11
ഇലക്ട്രോണുകളുടെ എണ്ണം = 11
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 23 – 11 – 12

e) \({ }_10^20 \mathrm{Ne}\)
അറ്റോമിക നമ്പർ, 2 = 10
മാസ് നമ്പർ, A = 20
പ്രോട്ടോണുകളുടെ എണ്ണം = 10
ഇലക്ട്രോണുകളുടെ എണ്ണം = 10
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 20 – 10 = 10

f) \({ }_22^48 \mathrm{Ne}\)
അറ്റോമിക നമ്പർ, Z = 22
മാസ് നമ്പർ, A = 48
പ്രോട്ടോണുകളുടെ എണ്ണം = 22
ഇലക്ട്രോണുകളുടെ എണ്ണം = 22

മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം = A – Z = 48 – 22 = 26

g) \({ }_92^235 \mathrm{U}\)
അറ്റോമിക നമ്പർ Z = 2
മാസ് നമ്പർ A = 235
പ്രോട്ടോണുകളുടെ എണ്ണം = 92
ഇലക്ട്രോണുകളുടെ എണ്ണം = 92
മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം ( ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം – A – Z – 235 – 92 – 143

h) \({ }_90^232 \mathrm{Th}\)
അറ്റോമിക നമ്പർ Z = 90
മാസ് നമ്പർ, A = 32
പ്രോട്ടോണുകളുടെ എണ്ണം = 90
ഇലക്ട്രോണുകളുടെ എണ്ണം = 90
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പാട്ടോണുകളുടെ എണ്ണം = A – Z – 232 – 90 = 142

i) \({ }_30^65 \mathrm{Th}\)
അറ്റോമിക നമ്പർ Z = 30
മാസ് നമ്പർ, A = 65
പ്രോട്ടോണുകളുടെ എണ്ണം = 30
ഇലക്ട്രോണുകളുടെ എണ്ണം = 30
മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + തുടകളുടെ എണ്ണം
ന്യൂട്രോണുകളുടെ എണ്ണം = മാസ് നമ്പർ – പ്രോട്ടോണുകളുടെ എണ്ണം – A – Z = 65 – 30 = 35

Question 12.
ബോർ ആറ്റം മാതൃക അനുസരിച്ച് ഇലക്ട്രോൺ എവിടെയാണ് കാണപ്പെടുന്നത്?
Answer:
ഓർബിറ്റിൽ

Question 13.
1, 2, 3, 4 എന്നീ ഊർജനിലകൾക്ക് യഥാക്രമം ഏതെല്ലാം പ്രതീകങ്ങളാണ് നൽകിയിരിക്കുന്നത്?
Answer:

n	ഊർജനില
1	K
2	L
3	M
4	N

ഒരു ആറ്റത്തിലെ ഇലക്ട്രോണുകൾ വിവിധ ഓർബിറ്റുകളിൽ ക്രമീകരിക്കപ്പെടുന്നത് ചില നിയമങ്ങൾ അനുസരിച്ചാണ്.

ഏതൊരു ഓർബിറ്റിലും ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകളുടെ എണ്ണം 21 ആണ് (n = ഓർബിറ്റ് നമ്പർ).

ഓർബിറ്റ് നമ്പർ (n)	പേര്	ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകളുടെ എണ്ണം (2n2)
1	K	2 × l2 = 2
2	L	2 × 22 = 8
3	M	2 × 32 = 18
4	N	2 × 42 = 35
5	O	2 × 52 = 50

പൊതുവേ താഴ്ന്ന ഊർജനിലയിൽ ഉള്ള ഒരു ഓർബിറ്റിൽ ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകൾ നിറഞ്ഞതിനുശേഷം മാത്രമേ അടുത്ത ഊർജനിലയിലുള്ള ഓർബിറ്റിൽ ഇലക്ട്രോൺ പൂരണം നടക്കു കയുള്ളൂ.

ഏതൊരു ആറ്റത്തിന്റെയും ബാഹ്യ ഓർബിറ്റിൽ ഉൾക്കൊള്ളാവുന്ന പരമാവധി ഇലക്ട്രോണുകളുടെ എണ്ണം 8 ആയിരിക്കും.

ഒരു ആറ്റത്തിന്റെ ഓർബിറ്റുകളിൽ ഇലക്ട്രോണുകൾ നിറയുന്നത് രേഖപ്പെടുത്തുന്നതാണ് ഇലക്ട്രോൺ വിന്യാസം.

Question 14.
ചില മൂലകങ്ങളുടെ ഇലക്ട്രോൺ വിന്യാസം എഴുതി നോക്കാം. പട്ടിക പൂർത്തിയാക്കി സയൻസ് ഡയറിയിൽ രേഖപ്പെടുത്തുക,

Answer:

1 മുതൽ 18 വരെ അറ്റോമിക നമ്പറുള്ള മൂലകങ്ങളുടെ ഇലക്ട്രോൺ വിന്യാസം മാത്രമേ ഈ രീതിയനുസരിച്ച് കൃത്യമായി എഴുതാൻ കഴിയൂ. അറ്റോമിക നമ്പർ 18-ൽ കൂടുതലുള്ള മൂലകങ്ങളുടെ ഇലക്ട്രോൺ വിന്യാസം എഴുതുന്ന രീതി ഉയർന്ന ക്ലാസുകളിൽ പരിചയപ്പെടാം.

Question 15.
\({ }_{13}^{27} \mathrm{Al}\)-ന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം ചിത്രീകരിക്കുക.
Answer:
അലുമിനിയത്തിന്റെ അറ്റോമിക നമ്പർ, Z = 13
അലുമിനിയത്തിന്റെ മാസ് നമ്പർ, A = 27
അലുമിനിയത്തിലെ ന്യൂട്രോണുകളുടെ എണ്ണം = A – 2 = 27 – 13 = 14
അലുമിനിയത്തിന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം

Question 16.
ഒരു ആറ്റത്തിന്റെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം നൽകിയിരിക്കുന്നു.

ചിത്രം വിശകലനം ചെയ്ത് ചുവടെ കൊടുത്തിരിക്കുന്നവ കണ്ടെത്തുക.
i) അറ്റോമിക നമ്പർ
ii) മാസ് നമ്പര
ii) പ്രോട്ടോണുകളുടെ എണ്ണം
iv) ന്യൂട്രോണുകളുടെ എണ്ണം
v) ഇലക്ട്രോൺ വിന്യാസം
Answer:
i) അറ്റോമിക നമ്പർ = 18
ii) മാസ് നമ്പർ = പ്രോട്ടോണുകളുടെ എണ്ണം + ന്യൂട്രോണുകളുടെ എണ്ണം = 18 + 22 = 40
iii) പ്രോട്ടോണുകളുടെ എണ്ണം = 18
iv) ന്യൂട്രോണുകളുടെ എണ്ണം = 22
v) ഇലക്ട്രോൺ വിന്യാസം = 28, 8

Question 17.
1 മുതൽ 18 വരെ അറ്റോമിക നമ്പറുള്ള മൂലക ആറ്റങ്ങളുടെ ഇലക്ട്രോൺ വിന്യാസം എഴുതി അവയുടെ ഷെൽ ഇലക്ട്രോൺ വിന്യാസം സയൻസ് ഡയറിയിൽ ചിത്രീകരിക്കുക.
Answer:

Question 18.
ഒരു മൂലകം ഏതാണെന്ന് നിശ്ചയിക്കുന്നത് അതിലെ ഏതു സബ്അറ്റോമിക കണങ്ങളുടെ എണ്ണമാണ്? (പ്രോട്ടോൺ/ന്യൂട്രോൺ)
Answer:
പ്രോട്ടോൺ

Question 19.
താഴെ കൊടുത്തിരിക്കുന്ന ചിത്രം നോക്കൂ.

Answer:

Question 20.
ഈ ആറ്റങ്ങളുടെ അറ്റോമിക നമ്പർ എത്രയാണ്?
Answer:
1

Question 21.
അറ്റോമിക നമ്പർ 1 ഉള്ള മൂലകം ഏതാണ്?
Answer:
ഹൈഡ്രജൻ
എങ്കിൽ ഇവ മൂന്നും ഹൈഡ്രജൻ ആറ്റങ്ങൾ ആണല്ലോ.

Question 22.
ഈ ആറ്റങ്ങൾ തമ്മിൽ ഏത് കണത്തിന്റെ എണ്ണത്തിലാണ് വ്യത്യാസം?
Answer:
ന്യൂട്രോൺ

Question 23.
ഇവയുടെ മാസ് നമ്പർ ഒരുപോലെയാണോ?
Answer:
ആല്ല

Question 24.
ഇവയിൽ ന്യൂക്ലിയസിൽ ന്യൂട്രോൺ ഇല്ലാത്ത ആറ്റമേത്?
Answer:
പ്രോട്ടിയം

Question 25.
ഈ ആറ്റങ്ങൾ ഹൈഡ്രജന്റെ ഐസോടോപ്പുകൾ ആണ്. എങ്കിൽ ഐസോടോപ്പുകൾ എന്നാൽ എന്താണെന്ന് എഴുതാമോ?
Answer:
ഒരേ അറ്റോമിക നമ്പറും വ്യത്യസ്ത മാസ് നമ്പറുമുള്ള ഒരേ മൂലകത്തിന്റെ വ്യത്യസ്ത ആറ്റങ്ങളാണ് ഐസോ ടോപ്പുകൾ,

ഒരേ അറ്റോമിക നമ്പറും വ്യത്യസ്ത മാസ് നമ്പറുമുള്ള ഒരേ മൂലകത്തിന്റെ വ്യത്യസ്ത ആറ്റങ്ങളാണ് ഐസോടോപ്പുകൾ,

ഐസോടോപ്പുകൾ ഒരേ രാസസ്വഭാവം കാണിക്കുന്നു. എന്നാൽ ഭൗതിക സ്വഭാവങ്ങളിൽ ചെറിയ വ്യത്യാസ ങ്ങൾ കാണിക്കുന്നു.

ഘനജലം (Heavy water) ഹൈഡ്രജന്റെ ഐസോടോപ്പായ ഡ്യൂറ്റീരിയത്തിന്റെ ഓക്സൈഡാണ്. ഘനജലം ആണവ നിലയങ്ങളിൽ ഉപയോഗിക്കുന്നു.

Question 26.
ഹൈഡ്രജന് മാത്രമാണോ ഐസോടോപ്പുകൾ ഉള്ളതെന്നു നോക്കാം. താഴെ കൊടുത്തിരിക്കുന്ന ചിത്രം നോക്കൂ.

Answer:

\({ }_6^{12} \mathrm{C}\), \({ }_6^{13} \mathrm{C}\), \({ }_6^{14} \mathrm{C}\) എന്നിവ കാർബണിന്റെ പ്രകൃതിദത്ത ഐസോടോപ്പുകളാണ്.
\({ }_6^{12} \mathrm{C}\) ആണ് ഏറ്റവും സ്ഥിരതയുള്ളതും ലഭ്യത കൂടിയതുമായ കാർബൺ ഐസോടോപ്പ് കാർബണിനും ഐസോടോപ്പുകളുണ്ടെന്ന് മനസ്സിലായല്ലോ.

കാർബണിന്റെ ആകെ ഐസോടോപ്പുകളിൽ ഏകദേശം 1.1% മാത്രമാണ് C. ഇത് സസ്യങ്ങളിലും ജന്തുക്കളിലും നടക്കുന്ന ജീവൽ പ്രവർത്തനങ്ങളെക്കുറിച്ച് പഠിക്കാൻ ഉപയോഗിക്കുന്നു. ‘C ഒരു റേഡിയോ ആക്റ്റീവ് ഐസോടോപ്പ് ആണ്. ഇത് ഫോസിലുകളുടെ കാലപ്പഴക്കം നിർണയിക്കുന്നതിന് ഉപയോഗിക്കുന്നു. ഹൈഡ്രജന്റെ ഐസോടോപ്പുകൾക്കുമാത്രമേ പ്രത്യേക പേരുകൾ നൽകിയിട്ടുള്ളു എന്ന് ശ്രദ്ധിക്കുമല്ലോ. മറ്റ് ചില ഐസോടോപ്പുകളും അവയുടെ ഉപയോഗങ്ങളും പട്ടികയിൽ നൽകിയിരിക്കുന്നു.

ഐസോടോപ്പ്	ഉപയോഗം
അയോഡിൻ -131	തൈറോയ്ഡ് ഗ്രന്ഥിയുടെ പ്രവർത്തനങ്ങളുടെ പഠനത്തിനും ചികിത്സയ്ക്കും
യുറേനിയം – 235	ആണവ നിലയങ്ങളിൽ ഇന്ധനം
കൊബാൾട്ട് – 60	കാൻസർ ചികിത്സക്ക്
സോഡിയം – 24	വ്യാവസായിക പൈപ്പ് ലൈനുകളിലെ ചോർച്ച കണ്ടെത്തൽ
അയൺ – 59	അനീമിയ നിർണയിക്കൽ

Question 27.
ആർഗൺ (Ar), പൊട്ടാസ്യം (K), കാൽസ്യം (Ca) എന്നീ ആറ്റങ്ങളുടെ ഓർബിറ്റ് ഇലക്ട്രോൺ വിന്യാസം നൽകിയിരിക്കുന്നത് ശ്രദ്ധിക്കൂ.

ചിത്രം വിശകലനം ചെയ്ത് പട്ടിക പൂർത്തിയാക്കി സയൻസ് ഡയറിയിൽ രേഖപ്പെടുത്തുക.

Answer:

Question 28.
ഈ മൂലകങ്ങളുടെ മാസ് നമ്പറിന്റെ പ്രത്യേകത എന്താണ്?
Answer:
മാസ് നമ്പർ തുല്യമാണ്.

Question 29.
അറ്റോമിക നമ്പർ തുല്യമാണോ?
Answer:
തുല്യമല്ല
ഈ ആറ്റങ്ങൾ ഐസോബാറുകൾ എന്നറിയപ്പെടുന്നു.
ഒരേ മാസ് നമ്പറും വ്യത്യസ്ത അറ്റോമിക നമ്പറുമുള്ള ആറ്റങ്ങളാണ് ഐസോബാറുകൾ.
ഇവ ന്യൂക്ലിയസിലെ ആകെ കണങ്ങളുടെ എണ്ണം (പ്രോട്ടോൺ + ന്യൂട്രോൺ) തുല്യമായ വ്യത്യസ്ത മൂലക ആറ്റ ങ്ങളായിരിക്കും.
ന്യൂട്രോണുകളുടെ എണ്ണം തുല്യമായ ആറ്റങ്ങൾ ഐസോടോണുകൾ എന്നറിയപ്പെടുന്നു.
ഉദാ: \({ }_7^{15} \mathrm{C}\), \({ }_7^{14} \mathrm{C}\)

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 5 Reproductive Health Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 5 Notes Questions and Answers Reproductive Health

Std 9 Biology Chapter 5 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 5 Let Us Assess Answers Reproductive Health

Question 1.
What is the duration of full-term pregnancy in humans?
a) 200 – 210 days
b) 210 – 220 days
c) 270-280 days
d) 2 8 0 – 2 9 0 days
Answer:
c) 270-280 days

Question 2.
Choose the one which is used to observe the growth of the foetus?
a) Ultrasound scan
b) Stethoscope
c) ECG
d) Thermometer
Answer:
a) Ultrasound scan

Question 3.
The fluid filled sac that surrounds and protects the foetus is ………..
a) Amnion
b) Placenta
c) Uterus
d) Ovary
Answer:
a) Amnion

Question 4.
Implantation means
a) Deposition of sperm in the vagina
b) Blastocyst attaches to the endometrium and grows
c) The fusion of sperm and the ovum
d) Surgical removal of the baby
Answer:
b) Blastocyst attaches to the endometrium and grows

Question 5.
The illustration given below shows a surgical procedure that men can adopt for contraception.

(a) Identify the contraceptive method in males.
(b) How is contraception possible through this surgery?
Answer:
(a) Vasectomy is the surgical contraceptive method in males
(b) Vasectomy is done by cutting and sealing the vas deferens in males. This prevents pregnancy by stopping sperm from leaving the body.

Question 6.
Which of the following organisms reproduce by external fertilisation?
a) Amphibians
b) Reptiles
c) Birds
d) Mammals
Answer:
a) Amphibians

Question 7.
“First breast milk, First defence”. These are the words in the poster released by the Department of Women and Child Development, Government of Kerala. What is your response to the phrases in the poster?
Answer:
The phrase in the poster is true regarding the first breast milk. Colostrum is light yellow-coloured milk produced after giving birth. This is rich in antibodies that protect the baby from infection, diarrhoea, respiratory diseases, and allergies. Thus, the colostrum, the first breast milk, prepares the baby for its first defence.

Question 8.
What are the factors that reduce fertility in both men and women?
Answer:
Factors that reduce fertility in both men and women are:

• Exposure to toxins and pollution
• Smoking
• Drug abuse
• Alcohol consumption
• Sexually transmitted infections
• Inflammation of the reproductive organs

Question 9.
Describe the physical changes of the foetus during each trimester of pregnancy.
Answer:

First trimester	Second trimester	Third trimester
The heartbeat starts	Hair growth begins on the head and the body	Lungs attains complete growth
Formation of limbs, fingers and toes	The foetus starts moving	Body size increases
Sex organs and organ systems are formed	Eyelids open, eyelashes are formed	Gaining of body weight

Question 10.
What are the chief ways to avoid Sexually Transmitted Infections (STIs)?
Answer:
Methods to avoid Sexually Transmitted Infections (STIs)

• Practice safe sex: Use polyurethane condoms or other barrier methods each time having sex
• Get vaccinated: Some STIs, such as HPV and Hepatitis B, can be prevented with vaccines.
• Ensure genital hygiene.
• Avoid having sex with multiple partners

Extended Activities

Question 1.
Prepare a poster on adolescent health and food habits and display in the class.
Answer:

Question 2.
Prepare the list of vaccines to be administered to newborns with the help of imaging software and post them in the social media.
Answer:

Age	Vaccine
At birth	BCG, OPV (zero dose), Hepatitis B
6 Weeks	OPV-1, Pentavalent-1, IPV-1, RVV-1
10 W eeks	OPV-2, Pentavalent-2, RVV-2
14 Weeks	OPV-3, Pentavalent-3, IPV-2, RVV-3
9-12 W eeks	Measles & Rubella (MR)-l
16-24 W eeks	MR-2. Diphtheria, Pertussis & Tetanus (DPT)-Booster-l, OPV-Booster
5-6 years	DPT-Booster-2
10 years	Tetanus & adult diphtheria (Td)                                                                             ‘
16 years	Td

Question 3.
Prepare a manuscript on the importance of motherhood by including pictures and descriptions.
Answer:
Key Points for the Manuscript on the Importance of Motherhood

The Nurturing Instinct:

• Biological basis: Mothers are biologically equipped to nurture their offspring with instincts and physical attributes that facilitate caregiving.
• Emotional bond: The mother-child bond is one of the most profound emotional connections, forged through shared experiences, love, and empathy.

The Shaping of Lives:

• Early development: A mother’s love and care are essential for a child’s cognitive, emotional, and social development.
• Role modelling: Mothers serve as powerful role models, influencing their children’s values, beliefs, and behaviours.
• Resilience: The challenges of motherhood can strengthen a woman’s resilience, character, and determination.

The Impact on Families and Communities:

• Family dynamics: The presence of a mother in a family can significantly impact its dynamics, creating a harmonious and nurturing environment.
• Community well-being: Strong maternal figures contribute to the overall well-being of a community, influencing education, healthcare, and social cohesion.

The Challenges and Rewards:

• Balancing act: Motherhood often involves balancing the demands of work, family, and personal life, requiring effective time management and prioritisation skills.
• Unconditional love: The unconditional love that mothers offer their children is one of the greatest rewards of motherhood, transcending boundaries and challenges.

Additional Considerations:

• Cultural perspectives: Explore different cultural perspectives on motherhood to understand its global significance.
• Challenges faced by specific groups: Discuss the unique challenges faced by single mothers, working mothers, or mothers in disadvantaged circumstances.
• Supporting mothers: Highlight the importance of supporting mothers and promoting maternal well-being through policies, programs, and community initiatives.

Reproductive Health Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What can be the reasons for the decrease in the maternal and infant mortality rate in Kerala?
Answer:

• High literacy rate of women
• Best Maternal and Child Health Care Centers
• High quality public health projects
• Early intervention and preventive measures

Question 2.
What all can be done to ensure gender justice? Based on the hints given below discuss and summarise the ideas and prepare a note.

• Equal opportunities in leadership, decision making and positions
• Opportunity to travel anywhere at any time with freedom and security
• Equal right to education for boys and girls
• Equal wages for equal work
• Shared responsibilities for men and women in family care and household chores

Answer:
Gender justice aim to eliminate discrimination and promote equal rights for all genders.

• It promote policies that encourage women’s participation in leadership roles across all sectors, including politics, business, and community organisations.
• Implement mentorship programs and training initiatives to empower women and equip them with the skills necessary for leadership positions.
• Create safe environments that allow individuals, regardless of gender, to travel freely without fear of harassment or violence.
• Ensure that both boys and girls have equal access to quality education, addressing barriers such as poverty, cultural norms, and discrimination.
• Ensure that individuals receive fair compensation regardless of gender.
• Encourage shares responsibilities between men and women, which supports work-life balance.

Question 3.
What are the main changes and their causes of adolescence? List them.
Answer:

• Rapid physical growth occurs due to hormonal changes
• Development of Secondary Sexual Characteristics including breast development in girls, facial hair and voice deepening in boys, and the beginning of menstruation in girls.
• Increase in emotional changes leading to mood changes.

Question 4.
Based on the indicators, analyse the given illustration and formulate inferences.

Indicators:

• The change in uterus during the menstrual cycle
• Menstrual hygiene

Answer:
Menstrual cycle is a monthly process occurring in women’s body. It prepares body for pregnancy. During menstruation, the lining of the uterus (endometrium) sheds, and blood, tissue, and fluid flow out of the vagina. The menstrual cycle is divided into four phases: Menstrual phase, Follicular phase, Ovulation, and Luteal phase.

Menstrual phase is the stage when you have your period.
During follicular phase, the pituitary gland releases follicle-stimulating hormone (FSH), which stimulates the ovaries to produce follicles. Each follicle contains an egg.

During Ovulation, the mature egg is released from the ovary. It travels down the fallopian tube towards the uterus. After ovulation, the follicle that released the egg turns into a corpus luteum. The corpus luteum produces progesterone, which helps to thicken the lining of the uterus in preparation for a possible pregnancy. If the egg is not fertilised, the corpus luteum breaks down, progesterone levels drop, and the lining of the uterus sheds, starting the menstrual cycle again.

Hygiene is important during menstruation.
It involves:

• Choose suitable menstrual products
• Change pads every 4-8 hours
• Wash hands before and after changing menstrual products.
• Dispose used products in a hygienic way
• Shower regularly
• Stay hydrated

Question 5.
Isn’t the zygote a single cell? How does it become a multicellular baby?
Answer:
The zygote undergoes repeated cell division and forms morula. Morula transforms into a structure filled with fluid and cells called Blastocyst. The blastocyst attaches to a lining of uterus called endometrium and grows there. This process is called Implantation. Further, the cells begin to differentiate and form specific organs and systems, eventually developing into a multicellular baby.

Question 6.
Only one of the sperms that reaches the ovum fuses with it. Why? Find it out.
Answer:
Only one sperm fertilises an ovum to ensure the correct number of chromosomes in the offspring. If multiple sperms fertilize an ovum, it may lead to abnormal number of chromosomes.

Question 7.
Prepare a flowchart including the various steps from fertilisation to implantation.
Answer:

Question 8.
How does the foetus get nutrients and how is waste removed from its body?
Answer:
The foetus receives nutrients and eliminates waste through the placenta and umblical cord.

Question 9.
What are the changes that occur in the body during pregnancy?
Answer:

• Gains body weight
• The thickness of the inner lining of the uterus increases.
• Menstruation stops temporarily.
• Changes in skin
• Fatigue and tiredness

Question 10.
What all things have to be taken care of, to ensure the physical and mental health during pregnancy?
Answer:

• A healthy diet
• Regular checkups
• Moderate exercises
• Enough rest
• Stray hydrated
• Avoid harmful substances like alcohol, tobacco, etc.
• Ensure emotional well-being
• Attend prenatal classes

Question 11.
Discuss with adults, the food items to be included in the diet of a pregnant woman, and those to be regulated. Complete the table.

Food items to be included	Food items to be regulated

Answer:

Food items to be included	Food items to be regulated
Dairy Products Legumes Whole Grains Fruits and Vegetables Lean Protein	Caffeine Raw or undercooked fish Unpasteurized Dairy Products Processed Meats Alcohol

Question 12.
Organise a seminar in the class based on the topic ‘The challenges posed by alcohol, drugs and smoking’.
Sub topics

• Reproductive health
• Other health issues
• Social and economic issues.

Answer:
Seminar: The Challenges Posed by Alcohol, Drugs, and Smoking

Introduction:
Substance abuse, including alcohol, drugs, and smoking, poses significant challenges to individuals, families, and societies. These substances have far-reaching consequences not only on physical health but also on mental well-being, relationships, and economic stability. For young people, understanding these risks is crucial to making informed choices. This seminar aims to shed light on the challenges posed by alcohol, drugs, and smoking by exploring their impact on reproductive health, overall health, and broader social and economic issues.

Reproductive Health:
In women, alcohol consumption, smoking, and drug use can cause fertility problems, leading to complications in conception. These substances also have a negative impact during pregnancy, resulting in miscarriages, preterm births, or fetal alcohol syndrome, a condition that severely impairs a child’s development.

In men, drugs and smoking are known to decrease sperm quality and lead to erectile dysfunction. Moreover, substance abuse often leads to high-risk behaviours, increasing the likelihood of contracting sexually transmitted diseases (STDs). Therefore, substance use not only affects the individual’s reproductive capabilities but also poses long-term health risks for future generations.

Other Health Issues:
Alcohol is a leading cause of liver diseases, such as cirrhosis, and contributes to heart disease and strokes. Chronic alcohol use also damages the brain, leading to mental health problems like depression, anxiety, and even alcohol-induced dementia. Drug abuse, particularly with narcotics, can cause permanent brain damage, addiction, and overdose, which can be fatal. Smoking is responsible for a wide range of diseases, including chronic obstructive pulmonary disease (COPD), lung cancer, and heart disease.

Social and Economic Issues:
Substance abuse often leads to strained relationships, broken families, and social isolation, with many addicts experiencing stigma and shame. It can increase criminal behavior, contributing to violence, domestic abuse, and illegal activities. The economic impact of substance abuse is staggering, with productivity loss due to absenteeism or poor performance at work, leading to unemployment. Additionally, health care systems bear a significant financial burden due to the treatment of substance-related illnesses, rehabilitation programs, and public health initiatives aimed at prevention.

Conclusion:
The challenges posed by alcohol, drugs, and smoking extend far beyond the individual, affecting families, communities, and entire economies. Understanding the profound impact these substances have on reproductive health, general health, and socio-economic well-being is crucial for making better lifestyle choices.

Question 13.
You have understood the importance of care and nutrition during pregnancy and parturition. How should the nutrition and care of a newborn baby be?
Answer:
Proper nutrition and care are essential for the healthy growth and development of a newborn. Breastfeeding, hygiene, safe sleep, regular health check-ups, vaccinations, and emotional bonding are key aspects. Equally important is the postnatal care of the mother to ensure her well-being during this critical period.

Question 14.
What are the main vaccines given to newborns? What is the time schedule for vaccination? Visit the nearest health centre, check the National Immunisation Schedule and complete the table. Prepare a chart and display it in the class.
Find out the disease that is prevented by each vaccine.

Answer:

National Immunisation Schedule
At Birth	BCG, OPV (zero dose). Hepatitis B
6 Weeks	OPV-1, Pentavalent-1, IPV-1. RVV-1
10 Weeks	OPV-2, Pentavalent-2, RVV-2
14 Weeks	OPV-3, Pentavalent-3, IPV-2, RVV-3
9-12 Weeks	MR-1
16-24 Weeks	MR-2, DPT-Booster-1, OPV-Booster
5-6 years	DPT-Booster-2
10 years	Td
16 years	Td

 

Vaccine	Disease
Bacillus Calmcttc-Guerin (BCG)	Tuberculosis (TB)
Oral Polio Vaccine (OPV)	Polio
Rotavirus Vaccine (RVV)	Rotavirus (diarrhoea and vomiting in young children and babies)
Hepatitis B vaccine	Hepatitis B
Pentavalent	Diphtheria, Tetanus. Pertussis (whooping cough), Hepatitis B. and Haemophilus influenzae type B (Hib)
MR vaccine	Measles & Rubella Infections
DPT Vaccine	Diphtheria, Tetanus, and Pertassis (whooping cough)
Td Vaccine	Tetanus & adult diphtheria

Question 15.
What are the benefits of mother who breastfeeds?
Answer:
Health Benefits of the mother due to breastfeeding:

• Lowers the Risk of Certain Diseases like breast cancer, ovarian cancer, and even heart disease later in life.
• Breastfeeding creates a close connection between the mother and the baby, helping them bond better.
• Breastfeeding releases hormones that help the mother feel relaxed and calm, which can reduce stress and anxiety.
• Breastfeeding can delay the return of the mother’s menstrual cycle, which can act as a natural form of birth control, though it’s not completely reliable.

Question 16.
How about conducting a study on creating awareness among the public in your area about antenatal and postnatal care and the intervention of health workers? Prepare a report after interviewing health workers and the public by including the given topics.

• Antenatal care
• Diet, Treatment
• Intervention of health workers
• Home birth
• Vaccines

Answer:
Some sample interview questions for health workers and the public are given below
For Health Workers:

• What are the most important aspects of antenatal care?
• How do you assist mothers during pregnancy and childbirth?
• What challenges do you face in providing care, especially in rural areas?

For the Public:

• How often do you visit health workers during pregnancy?
• What kind of advice have you received about diet and vaccinations?
• Do you prefer home births or hospital births, and why?

Sample of the report after conducting the interview:
Health workers emphasized the importance of regular antenatal check-ups to monitor the health of both the mother and the baby. They highlighted the need for early detection of complications, such as high blood pressure, gestational diabetes, and anaemia. These visits are also important for educating mothers on nutrition, exercise, and general well-being. They provide medical support and guidance during pregnancy.

They offer regular monitoring, educate about healthy practices, and intervene in cases of complications. Midwives and community health workers also provide emotional support and are often the first point of contact in rural areas. Health workers recommend a balanced diet rich in proteins, iron, calcium, and folic acid. They also prescribe supplements when necessary to ensure that the mother and baby receive adequate nutrition.

Some members of the public were aware of antenatal care and its importance, but others, especially in rural areas, lacked knowledge or access to regular check-ups. Many rely on traditional methods and only visit a healthcare facility if there is a problem. Many women in urban areas are more aware of the importance of a nutritious diet during pregnancy, but in rural areas, women may not have access to the necessary foods or supplements. Some rely on traditional remedies and diets that may lack the needed nutrients.

Question 17.
Analysing the statements given below, discuss your opinions and form inferences.
Frequent pregnancy may affect the health of the mother and children.
Increase in population creates adverse effects in the environment and in the utilisation of resources.
In some countries where the birth rate is lesser, extra time and financial assistance for child care are given.
Answer:
Population growth and decline are global issues with wide-ranging implications for health, society, and the environment. While high birth rates strain resources and increase environmental damage, low birth rates threaten economic growth and social stability. Both challenges require comprehensive strategies, including access to contraception, education on family planning, and supportive policies for parents. Promoting reproductive health and creating policies that support families can lead to healthier populations and more sustainable development paths for nations across the globe.

Question 18.
Interview a doctor to clear your doubts about sexually transmitted diseases. Collect more information and prepare and display a poster on the pathogens causing the above diseases, their transmission and prevention.
Answer:
Some sample interview questions to ask Doctor about STIs:

• What are the most common sexually transmitted diseases (STDs) you encounter in your practice?
• Can you explain the different pathogens responsible for these STDs (bacteria, viruses, fungi, etc.)?
• How are these diseases typically transmitted?
• What are the most effective methods of prevention against STDs?
• What are some common misconceptions about STDs that patients often have?
• How important is early detection and treatment in managing STDs?
• Could you explain the role of vaccines, like the HPV vaccine, in preventing certain STDs?
• What are the long-term health effects of untreated STDs?

Information gathered through interview:

Disease	Causative organism	Pathogen
Acquired immunodeficiency syndrome (AIDS)	Virus	Human Immunodeficiency Virus (HIV)
Chlamydiosis	Bacteria	Chlamydia trachomatis
Syphilis	Bacteria	Treponema pallidum
Gonorrhea	Bacteria	Neisseria gonorrhoeae
Genital Herpes	Virus	Herpes simplex virus (HSV)
Genital warts	Virus	Human Papilloma Virus (HPV)
Hepatitis B	Virus	Hepatitis B virus (HBV)
Trichomoniasis	Protozoan parasite	Trichomonas vagina!is
Candidiasis	Fungi	Candida albicans

Transmission of STIs:

• Unprotected sexual contact (vaginal, anal, oral)
• Mother-to-child transmission during childbirth (e.g., HIV, syphilis)
• Shared needles (e.g., HIV)

Prevention of STIs:

• Condom use during sexual intercourse
• Vaccination (HPV and Hepatitis B vaccines)
• Regular screenings and early detection
• Avoid sharing needles

 

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 8 Organic Chemistry Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 8 Solutions Organic Chemistry

Kerala Syllabus Std 9 Chemistry Chapter 8 Organic Chemistry Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 8 Let Us Assess Answers Organic Chemistry

Question 1.
The structural formula of a hydrocarbon is given below.

a. Write its condensed formula.
b. Write its molecular formula.
c. Draw the structure of the first compound of the homologous series to which this hydrocarbon belongs.
d. Write the IUPAC name of this compound.
Answer:
a.

b. C₄H₈

c. Ethene (C₂H₄) is the first compound.

d. Butene

Question 2.
C₂H₆, C₃H₈, ……………., C₅H₁₂ belong to the same homologous series.
a. Write the molecular formula of the missing compound.
b. Write the name of the homologous series to which these compounds belong.
c. Write the structural formula of C₂H₆.
Answer:
a. C₄H₁₀
b. Homologous series of alkanes
c. C₂H₆ is

Question 3.
Molecular formulas of some hydrocarbons are given below.
C₃H₈, C₄H₈, C₄H₁₀, C₃H₆
a. Which among these are alkanes?
b. What is the general formula of alkenes?
c. Write the molecular formula of alkyne having 4 carbon atoms.
Answer:
a. Alkanes – C₃H₈, C₄H₁₀
b. C_nH_2n
c. C₄H₆

Question 4.
a. Write the molecular formula of the missing compounds in the homologous series.

b. To which homologous series does category C belong?
c. Write the general formula of category A.
Answer:

b. Homologous series of alkynes
c. C_nH_2n

Question 5.
The molecular formulas of a few hydrocarbons are given below.
C₂H₄, C₂H₂, C₂H₆, C₃H₄, C₃H₈
a. Which among them belongs to the alkene group?
b. Which category does C₂H₂ belong to?
c. Which are the hydrocarbons having the general formula C_nH_2n+2?
Answer:
a. C₂H₄
b. Alkynes
c. C₂H₆ and C₃H₈

Question 6.
Two hints regarding a hydrocarbon are given below.
• It has 3 carbon atoms.
• The general formula of the category to which this hydrocarbon belongs is C_nH_2n+2.
a. Write the molecular formula and IUPAC name of this compound.
b. Draw the structure of this compound.
c. Write the molecular formula of the hydrocarbon having the same number of carbon atoms and having a double bond.
Answer:
a. Molecular formula – C₃H₈
IUPAC name – Propane

c. C₃ H₆

Question 7.
Hints about a cyclic compound are given below.
It has 6 carbon atoms and 12 hydrogen atoms.
a. Draw the structure of this compound.
b. Write the structural formula of the open chain hydrocarbon having the same molecular formula.
c. Write the molecular formula of the alkane having the same number of carbon atoms.
Answer:

c. C₆H₁₄

Question 8.
A chain having carbon atoms is given below.

a. Complete the structure by adding hydrogen atoms to each carbon atom. Write its IUPAC name also.
b. Write the molecular formula of this compound.
c. Draw the structure of a cyclic compound having the same molecular formula.
d. Write the IUPAC name of this cyclic compound.
Answer:
a. IUPAC Name – Pentene

b. C₅H₁₀

d. Cyclopentane

Question 9.
The molecular formula of an alicyclic compound is C₄H₈.
a. Write the structural formula of this compound.
b. Write the structural formula of the open-chain hydrocarbon having the same molecular formula.
Answer:
a. Alicyclic C₄H₈

b. Open chain C₄H₈

Question 10.
The structure of an alicyclic compound is given below.

a. What is its molecular formula?
b. Write its IUPAC name.
c. Write the structural formula of an open chain hydrocarbon having the same molecular formula.
Answer:
a. C₄H₆
b. Cyclobutene
c. Butyne

Question 11.
The molecular formula of a hydrocarbon is C₃H₆.
a. Write the structural formula of this compound.
b. To which category does it belong?
(Alkane, alkene, alkyne)
c. Draw the structure of an alicyclic compound having the molecular formula C₃H₆.
d. Write the IUPAC name of the compound.
Answer:

b. Alkene
c. Alicyclic C₃H₆

Question 12.
a. Write the molecular formula of naphthalene.
b. Draw the structure of naphthalene.
Answer:
a. C₁₀H₈
b. Naphthalene

Question 13.
What is the method used to separate the components from petroleum?
Butane, the main component of LPG, is an alkane. It has four carbon atoms. Write the structural formula of butane.
Answer:
a. Fractional distillation
b. Butane

Question 14.
a. Which among the following gases does not cause global warming? (Methane, carbon dioxide, nitrogen, nitrous oxide)
b. Write two ways to prevent global warming.
Answer:
a. Nitrogen

b. (i) Use public transportation whenever possible.
(ii) Reduce, Reuse, Recycle: This practice helps minimise waste and reduces the demand for new products, which often require energy-intensive manufacturing processes.
(iii) Reduce carbon emissions.

Extended Activities

Question 1.
Prepare and exhibit the ball and stick model of alkane, alkene and alkyne compounds having 4 carbon atoms.
Answer:
Hints
Alkane – Butane – C₄H₁₀
Alkene – Butene – C₄H₈
Alkyne – Butyne – C₄H₆

Question 2.
Present a seminar on the topic of “Global warming and climate change”.
Answer:
Hints: Include points like factors causing global warming, global warming and climate, consequences, and remedies.

Question 3.
Prepare and present a paper on the topic “The importance of Organic Chemistry”.
Answer:
Hints: Introduction – Define organic chemistry and its scope; body – Discuss the role of organic compounds in living organisms, explore the vast applications of organic chemistry in various industries, discuss the environmental implications of organic chemistry; conclusion – summarise the key points and emphasise the ongoing importance of organic chemistry.

Question 4.
The structural formula of a few organic compounds is given below.
i. CH₃-CH₂-OH
ii. CH₃-CH₂-CH₂-OH
a. Write the molecular formulae of these compounds.
b. Are they homologous? Substantiate your answer.
Answer:
a. i. – C₂H₅OH or C₂H₆O
ii. – C₃H₇OH or C₃H₈O

b. Yes, they are homologous. Ethanol (C₂H₆O) and propanol (C₃H₈O) differ by one -CH₂ unit, which is a key feature of homologous series.

Question 5.
Construct and exhibit a model of cyclic compounds having 6 carbon atoms.
Answer:
Hints

Organic Chemistry Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Complete the table with the help of the periodic table.

Symbol	…………………………..
Atomic number	……………………………
Electronic configuration	………………………….
Valency	……………………………

Answer:

Symbol	C
Atomic number	6
Electronic configuration	2, 4
Valency	4

Carbon compounds are referred to as organic compounds due to their abundance and diversity
Carbon contains 4 electrons in its outermost shell and a valency of 4. This allows carbon to make covalent bonds in various ways.

Question 2.
The structure of an organic compound is given in the figure. What are the constituent atoms in this compound?

Answer:
Carbon and hydrogens.

Question 3.
What is the peculiarity of the bond between carbon and hydrogen?
Answer:
It is a single covalent bond, and in this bond, both carbon and hydrogen share electrons to achieve a stable electron configuration.

Question 4.
Do you know any other compounds that contain only carbon and hydrogen?
Answer:
Yes, there are a lot of other compounds that contain only carbon and hydrogen.

Question 5.
Look at the compounds given below. Complete the valency of carbon using hydrogen atoms.

Answer:

Question 6.
What are the peculiarities of the bond between carbon atoms?
Answer:
Three types of covalent bonds are present between these carbon atoms that are single, double and triple covalent bonds.
Hydrocarbons are compounds that contain only carbon and hydrogen

Question 7.
Analyse the structure of the given hydrocarbon.

a) Molecular formula?
b) What is the type of covalent bond in this compound?
Answer:
a) Molecular formula – CH₄
b) Single covalent bond

Question 8.
The structure of another hydrocarbon with two carbon atoms and having only a single bond between them is illustrated.

What is the molecular formula of this compound?
Answer:
C₂H₆ – ethane.

The structural formula of this compound can also be represented as CH₃—CH₂. Such a representation is known as a condensed formula.

Question 9.
Complete the table 8.2.

Answer:

The open-chain hydrocarbons having only single bonds between the carbon atoms are called alkanes.
n alkanes, since all the four valencies of each carbon atom are satisfied by single bonds, they are also known as saturated hydrocarbons.

Question 10.
Analyse Table 8.2 and answer the given questions.
a) How many hydrogen atoms are present in an alkane having one carbon atom? ……………….
b) How many hydrogen atoms are present in an alkane having 2 carbon atoms? ………………..
c) What if there are 3 or 4 carbon atoms? …………………
d) What number is added to twice the number of carbon atoms to get the number of hydrogen
atoms? ………………….
e) If an alkane contains an ‘n’ number of carbon atoms, how many hydrogen atoms will be
there? (2n, 2n+2,2n-2) ………………..
f) Write the general formula of alkanes ………………
Answer:
a) 4
b) 6
c) For 3 carbon atoms – 8 H atoms
For 4 carbon atoms – 10 H atoms
d) 2 is added
e) 2n+2
f) C_nH_2n+2

Question 11.
The structure of a hydrocarbon having a double bond between two carbon atoms is given

a) What is the molecular formula of the compound given above?
b) Write its condensed formula.
Answer:
a) C₂H₄
b)

Question 12.
Represent the structure of such hydrocarbons having a double bond between any of the two carbon atoms by adding more carbon atoms.
Answer:

Question 13.
Complete the table given below.

Answer:

Hydrocarbons having atleast one double bond between any two carbon atoms are called alkenes.

Question 14.
Analyse table 8.3
a) What is the relation between the number of carbon atoms and the number of hydrogen atoms in alkenes?
b) If an alkene contains an ‘n’ number of carbon atoms, how many hydrogen atoms will be there?
c) Write the general formula of alkenes.
Answer:
a) In alkenes, the number of hydrogen atoms is always twice the number of carbon atoms.
b) 2n
c) C_nH_2n

Question 15.
Look at the structure of a hydrocarbon having a triple bond between two carbon atoms.

a) Write the molecular formula of the compound given above.,
b) Write its condensed formula.
Answer:
a) C₂H₂
b) CH≡CH

Question 16.
Complete the table given below.

Hydrocarbons with atleast one triple bond between any two carbon atoms are called alkynes.
Answer:

Question 17.
Analyse table 8.4
a) In these, are the number of hydrogen atoms twice that of carbon atoms?
b) What number is subtracted from twice the number of carbon atoms to get the number of
hydrogen atoms in each of these?
c) If an alkyne has an ‘n’ number of carbon atoms, how many hydrogen atoms will be there? (2n+2, 2n, 2n-2)
d) Write the general formula of alkynes.
Answer:
a) No
b) 2
c) 2n-2
d) C_nH_2n-2

Question 18.
Analyse the molecular formula of the. hydrocarbons given below and classify them as alkane, alkene and alkyne.
C₅H₁₀, C₃H₄, C₂H₄, C₅H₁₂, C₆H₁₂, C₇H₁₂, C₉H₂₀, C₃H₈, C₄H₆, C₄H₈.
Answer

Alkane	Alkane	Alkane
C5H12, C9H20, C3H8	C5H10, C2H4, C6H12, C4H8	C3H4, C7H12, C4H6

Question 19.
Analyse the molecular formulae C₂H₆ and C₃H₈.
a) To which category do they belong?
(Alkane, alkene, alkyne)
b) What is the general formula of this category?
c) What is the difference between C₂H₆ and C₃H₈ in the number of carbon and hydrogen atoms? Let us examine.

Compound	Number of carbon atoms	Number of hydrogen atoms
C3H8 C2H6	3 2	8 6
Difference in the number of carbon and hydrogen atoms	1	……………………………..

d) Have you understood that the difference in the number of carbon and hydrogen atoms in these is -CH₂-?
e) Do C₄H₁₀ and C₅H₁₂, which belong to this category, have the same difference?
Answer:
a) Alkane
b) C_nH_{2n + 2}
c) There is a difference by -CH₂– group.

Compound	Number of carbon atoms	Number of hydrogen atoms
C3H8 C2H6	3 2	8 6
Difference in the number of carbon and hydrogen atoms	1	2

d) Yes
e) Yes, C₄H₁₀ and C₅H₁₂ also have a difference by -CH₂ group.

Question 20.
Alkene: C₆H₆ and C₄H₈
a) What is the difference in the number of carbon and hydrogen atoms in these alkenes?
b) The general formula of alkene ………………..
Answer:
a) There is a difference in 1 carbon and 2 hydrogens, i.e., a -CH₂– group.
b) C_nH_2n

Question 21.
Alkyne: C₂H₂ and C₃H₄
a) What is the difference in the number of carbon and hydrogen atoms in these alkynes?
b) The general formula of alkyne ……………..
Answer:
a) There is a difference in 1 carbon and 2 hydrogens, i.e., a -CH₂– group.
b) C_nH_2n-2
A homologous series is a series of organic compounds that can be represented by a general formula and have a difference of -CH₂– group between two successive members.

The characteristics of a homologous series are:
(i) The members can be represented by a general formula.
(ii) Successive members differ by a -CH2- group.
(iii) Members show similarity in chemical properties.
(iv) There is a regular gradation in their physical properties.

Question 22.
The molecular formulas of some hydrocarbons are given below.
C₂H₄, C₂H₆, C₃H₄, C₃H₈ .
a. Which of these compounds belong to the same homologous series?
b. Write the general formula of this homologous series.
Answer:
a. C₂H₆ and C₃H₈
b. C_nH_2n+2

Question 23.
The molecular formula of hydrocarbons that belong to the same homologous series are given below.
C₂H₂, C₃H₄, C₄H₆, A, B
a. Write the molecular formula of the compounds A and B.
b. To which category do these compounds belong?
(Alkane, alkene, alkyne)
c. What is their general formula?
d. Draw the structure of compound A.
Answer:
a. A – C₅H₈ and B – C₆H₁₀
b. Alkyne
c. C_nH_2n-2
d. C₅H₈

Question 24.
Write the IUPAC names of alkanes having 4 to 10 carbon atoms.
Answer:
C₄H₁₀ = Butane, C₅H₁₂ = Pentane, C₆H₁₄ = Hexane, C₇H₁₆ = Heptane, C₈H₁₈ = Octane, C₉H₂₀ = Nonane, C₁₀H₂₂ Decane.

Question 25.
Which is the suffix added here?
Answer:
“ene”.

Question 26.
Write the IUPAC names of alkenes having 4 to 10 carbon atoms.
Answer:
C₄H₈ = Butene, C₅H₁₀ = Pentene, C₆H₁₂ = Hexene, C₇H₁₄ = Heptene, C₈H₁₆ = Octene, C₉H₁₈ = Nonene, C₁₀H₂₀ = Decene.

Question 27.
Write the IUPAC names of alkynes having 4 to 10 carbon atoms.
Answer:
C₄H₆ = Butyne, C₅H₈ = Pentyne, C₆H₁₀= Hexyne, C₇H₁₂ = Heptyne, C₈H₁₆ = Octyne, C₉H₁₆ = Nonyne, C₁₀H₁₈ = Decyne

Question 28.
Notice the structure of a few carbon compounds given below.

What is the peculiarity of the structures of these carbon compounds?
Answer:
These carbon compounds are cyclic.
Carbon atoms can combine with one another to form cyclic compounds. Cyclic hydrocarbon compounds can be classified into two groups, namely alicyclic compounds and aromatic compounds.

Question 29.
The structure and IUPAC name of some alicyclic hydrocarbons are given. Write their molecular formula.

Answer:
Cyclobutane – C₄H₈, Cyclobutene – C₄H₆, Cyclopentene – C₅H₈, Cyclohexane – C₆H₁₂.

Question 30.
How are alicyclic compounds named?
Answer:
Alicyclic compounds can be named by adding the term ‘Cyclo-‘ as a prefix along with the IUPAC name of the hydrocarbon.

Question 31.
Write down the molecular formula of benzene.
Answer:
C₆H₆
Naphthalene is another aromatic hydrocarbon with a characteristic odour and a white crystalline form. Two benzene rings are fused together to form its structure. It is the main ingredient in mothballs. The structure of naphthalene is

Question 32.
Write the molecular formula of naphthalene.
Answer:
C₁₀H₈

Question 33.
The structural formula of a hydrocarbon is given below.
CH₂=CH-CH₂-CH₃
a. Write the molecular formula of this compound.
b. Draw the structure of a cyclic compound with the same molecular formula.
Answer:
C₄H₈
b. Cyclobutane

Question 34.
The structures of two organic compounds are given below. Compare these.

Answer:

	Compound A	Compound B
Molecular formula	C6H6	C6H12
IUPAC Name	Benzene	Cyclohexane
Aliphatic/ Aromatic	Aromatic	Alicyclic

Question 35.
You have seen that the number of carbon compounds is very high. List the reasons for this.
Answer:

• The valency of carbon is four. So, it has the ability to form four covalent bonds with other atoms or carbon itself.
• Ability for catenation.
• Can form single, double and triple covalent bonds.
• Can form an open chain and cyclic or ring compounds.

Question 36.
Which form of coal has the highest carbon content?
Answer:
Anthracite (94%).

Question 37.
Which form of coal has the lowest carbon content?
Answer:
Peat (57%).

Question 38.
What are the consequences of global warming? What are the ways to effectively counter global warming?
Answer:
Consequences

• Global warming is causing increasingly frequent and intense heatwaves, which threaten ecosystems and
• humans.
• Sea levels are rising faster as ice caps and glaciers melt, threatening coastal regions.
• Hurricanes, droughts, and floods are becoming more severe due to global warming, inflicting extensive damage and death.
• Many plant and animal species are losing habitat, endangering, and extinction due to climate change.

Ways to reduce

• Reduce the use of fossil fuels.
• Afforestation.
• Control deforestation.

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 1 To Life Processes Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 1 Notes Questions and Answers To Life Processes

Std 9 Biology Chapter 1 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 1 Let Us Assess Answers To Life Processes

Question 1.
Compare the outer membrane of raw egg and boiled egg using the indicators given below

• Permeability
• Possibility of osmosis
• Possibility of active transport

Answer:

Raw egg	Boiled egg
Permeability is more	Permeability is less
Possibility of osmosis is more	Possibility of osmosis is less
Active transport is present	Active transport is not present

Question 2.
Given below is an answer written by a child to the question ‘How is oxygen released in photosynthesis? ‘Evaluate and comment on it. ‘Carbon dioxide and water are the raw materials for photosynthesis. Both these breakdown and oxygen is released’
Answer:
The oxygen released during photosynthesis is from the water. The plants will absorb water as well as carbon dioxide during photosynthesis. Later these water molecules split and produce hydrogen and oxygen. Then this hydrogen is utilized for the dark phase of photosynthesis and is converted into sugar (Glucose). The oxygen is then released into the atmosphere whereas the sugar molecules are stored for energy.

Question 3.
Though photosynthesis is ultimately anabolism, it also involves catabolism’. Analyze the statement.
Answer:
Photosynthesis is an anabolic process during which plants use energy from sunlight to convert carbon dioxide gas and water into sugar molecules. But during this anabolic process, some breaking down reactions have also happened. An example of a catabolic reaction during photosynthesis is the process of food digestion, where different enzymes break down food particles so they can be absorbed by the small intestine.

Extended Activities

Question 1.
There are so many people who have dedicated their lives for environmental activities. Collect information about them and prepare an album.
Answer:
(Hints: Given below is a list of some environmentalists and their achievements as a model for the preparation of your album).
Greta Thunberg: A Swedish climate activist who is famous for starting the Fridays for Future movement. Thunberg began by protesting in front of the Swedish parliament in 2018, and her lone strike inspired millions of students around the world to walk out of school and demand action on climate change.

Vandana Shiva: An Indian environmental activist, physicist, and author who has written extensively about the dangers of genetically modified organisms (GMOs) and the importance of biodiversity. Shiva is the founder of Navdanya, an organization that promotes organic farming and seed saving.

Question 2.
Complete the table given below by observing plants in the surroundings.

Plants	Value added products	Consumption
Coconut tree	Coconut oil	For cooking
	Medicine

Answer:

Plants	Value added products	Consumption
Coconut tree	Coconut oil	For cooking
Aloe vera	Medicine	Treating skin problems, wounds, cancers, diabetes, gastrointestinal problems, etc.
Rubber	Reclaimed rubber, rubber mats	Used in waterproof materials and works as an electrical insulator.
T urmeric	Cosmetics, Food	Ingredient in dietary supplements, cosmetics, flavouring for foods
Coffee plant	Coffee powder. Bio-oil	For cooking. To produce essential oils from coffee leaves
Palm trees	Toddy, Wine, Honey, Jaggery	To produce fresh juice (sweet toddy), fermented drinks (toddy, wine, and arak).

To Life Processes Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Complete the illustration given below to gain an understanding of metabolism.


Answer:

Question 2.
Find examples of enzymes and hormones.
Answer:

Enzymes	Hormones
1. Salivary amylase in saliva	1. Sex hormone – Testosterone, Estrogen, Progesterone
2. Pepsin in the gastric juice	2. Thyroxine
3. Sucrase	3. Calcitonin
4. Lipase	4. Thymosin

Question 3.
Many factors required for metabolism are obtained from their external environment. Which are they? List out.
Answer:
Air, water, temperature, etc.

Question 4.
Is there internal environment like external environment? Note down your guess.
Answer:
Yes, there is internal environment, like external environment. In animals the fluid found in the Ispace between the cells (extracellular fluid) serves as the internal environment. The internal environment of plants consists of cell walls and their components, extracellular fluid and air sacs between cells.

Question 5.
Why plasma membrane is known as selectively permeable membrane? Find out.
Answer:
The plasma membrane is called selectively permeable membrane, because, it allows only certaint molecules to pass in and out.

Question 6.
What happens to raisins placed in fresh water and why? Find out.
Answer:
The raisins will bulge, because high amount of water enter into it. This is because of the movement of water molecules from higher concentration to lower concentration.

Question 7.
Will diffusion take place through and without semi permeable membrane? Find out.
Answer:
Yes, diffusion will takes place through and without semi permeable membrane.
Ex: Diffusion of CO₂ and O₂.

Question 8.
Complete the work sheet by including processes involved in the exchange of materials.

The nature of flow of molecule	Name of the process
from a region of higher to lower concentration
From a region of low er to higher concentration
Applicable to water only
Energy required
Energy not required
Carrier protein not required
Carrier protein required

Answer:

The nature of flow of molecule	Name of the process
from a region of higher to lower concentration	Diffusion
From a region of low er to higher concentration	Active transport
Applicable to water only	Osmosis
Energy required	Active transport
Energy not required	Facilitated diffusion
Carrier protein not required	Diffusion
Carrier protein required	Active transport

Question 9.
Nutrients are essential for metabolism. How do animals get them? What about plants?
Answer:
Animals get their nutrients from their surroundings, and they depend on other organisms for their food. They are heterotrophic organisms. But Plants follow an autotrophic mode of nutrition. They can prepare their own food by photosynthesis.

Question 10.
Photosynthesis is the process by which plants make food. List the components required for photosynthesis.
Answer:

• Chlorophyll
• Sunlight
• Water
• Carbon dioxide

Question 11.
Observe the illustration and discuss the parts involved in photosynthesis based on the indicators and make a note.

Indicators:

• Structure of chloroplast.
• Position of chlorophyll.
• Thylakoid, grana and stroma.

Answer:
There are three types of plastids in plant cells: chloroplast, chromoplast, and leucoplast. Chloroplast is a green-coloured plastid mainly responsible for photosynthesis due to the presence of chlorophyll pigments.

Chromoplasts are plastids that contain carotenoid pigments. They lack chlorophyll. Carotenoid pigments are responsible for different colours like yellow, orange, and red colour imparted to fruits, flowers, old leaves, roots, etc. Chromoplasts may develop from green chloroplasts.
Leucoplasts are colourless plastids that generally occur in non-green plant cells near the nucleus and store starch, proteins, and oils.

Anatomically, leaves consist of an outer protective group of cells called epidermal cells, both upper and lower epidermis. The lower epidermis contains numerous openings called stomata for water and gaseous exchange. Mesophyll cells form the ground tissue, contains two types of cells: elongated palisade mesophyll and spherical spongy mesophyll. Numerous green-coloured, dot-like structures are seen in the mesophyll cells, the chioroplasts.

Chioroplast is a double-membrane bound cell organelle, consisting of an outer and an inner membrane. The fluid part inside the inner membrane is the stroma, in which the dark reaction of photosynthesis takes place. Thylakoids are the numerous membranous sacs arranged like stacks of coins within the stroma. The group of thylakoids are termed as grana. Each granum is interconnected by a bridge-like structure called the stroma lamellae. The light reaction of photosynthesis occurs inside the grana (Thylakoid).

Question 12.
Complete the given table by analyzing illustration and the informations given.

Photosynthesis
Hints	Light phase	Dark phase
Place where reaction takes place
Reactions
Products

Answer:

Photosynthesis
Hints	Light phase	Dark phase
Place where the reaction takes place	Grana	Stroma
Reactions	Water splits into hydrogen and oxygen.	Glucose is formed by combining, hydrogen and carbon dioxide.
Products	Hydrogen, Oxygen, ATP	Glucose, Water

Question 13.
Complete the illustration by including the reactants and products of photosynthesis.

Answer:

Question 14.
Is sunlight itself is required for photosynthesis? Can photosynthesis take place under the light of an LED bulb? Find out.
Answer:
Yes, photosynthesis can take place in artificial light if the plant is exposed to the correct wavelength of light. Photosynthesis is the natural process by which plants use chlorophyll to absorb atmospheric carbon dioxide and convert it to sugar in the presence of sunlight

Question 15.
Many substances are produced when starch undergoes metabolism. Observe the pictures given below and complete the illustration.


Answer:

Question 16.
You know about nutrition and nutrients. List the nutrients.
Answer:

• Carbohydrates
• Minerals
• Water
• Proteins
• Vitamins

Question 17.
Plants grow in water as well as on land. Who are the producers in the ocean and other water bodies?
Answer:
Aquatic plants include larger plants, or macrophytes, and microscopic algae, or phytoplankton. These are the primary producers in the ocean and other water bodies.

Question 18.
What are the steps to be taken to prevent ocean pollution? Discuss.
Answer:

• Reduce plastic production & waste
• Improve wastewater systems Use eco-friendly products
• Reduce chemical pollution
• Manage oil spills
• Beach & river cleanups
• Monitoring & measuring progress.

Question 19.
Are food and oxygen the only things that plants provide? Arrive at inferences by analyzing the illustration and the description.

Answer:

• The service rendered by plants for the sustenance of the living world is unique. Plants serve as the cheapest, most effective, and natural means for the purification of air.
• Mangroves are found where back waters meet the sea. Kerala has 43 species of mangrove plants that grow in salt water.
• Plants are the foundation stones of the biosphere. Depletion of plants will ultimately affect the survival of life itself.
• The concept of sustainable development is formulated by considering plants as well.
• Nature conservation should become part and parcel of everyone’s life. Nature should be used wisely and preserved for generations to come.

The nature and humans are receiving countless services from plants. Most plant parts have economic importance in one way or the other. A variety of value-added products are made from them. The processing and marketing of these resources open doors to a wide range of job opportunities. The service rendered by plants for the sustenance of the living world is unique.

Plants serve as the cheapest, effective, and natural means for the purification of air. By absorbing carbon dioxide from the atmosphere and releasing oxygen, plants provide invaluable service to the living world. Plants also have a major role in the mitigation of natural disasters. Mangrove forests help in controlling Tsunami to some extent. Bamboo forests, reed, vetiver, lemongrass, etc. protect the riverbanks from collapsing during floods. Trees and bushes in mountains and hills prevent soil erosion and landslide.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 4 Malayalam Medium ഗുണനസമവാക്യങ്ങൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 4 Solutions Malayalam Medium ഗുണനസമവാക്യങ്ങൾ

Class 9 Maths Chapter 4 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 4 Malayalam Medium Textual Questions and Answers

Question 1.
ചുവടെയുള്ള കണക്കുകൾ മനസ്സിൽത്തന്നെ ചെയ്ത് ഉത്തരം കണ്ടുപിടിക്കുക.
(i) 71 × 91
(ii) 42 × 62
(iii) 10\(\frac{1}{2}\) × 6\(\frac{1}{2}\)
(iv) 9.5 × 3.5
(v) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
Answer:
(i) 71 x 91 = (70 + 1)(90 + 1)
= (70 × 90) + 70 + 90 + 1
= 6300 + 160 + 1
= 6461

(ii) 42 × 62 = (40 + 2)(60 + 2)
= (40 × 60) + 2(40 + 60) + 4
= 2400 + 200 + 4
= 2604

(iii) 110\(\frac{1}{2}\) × 6\(\frac{1}{2}\) = (10 + \(\frac{1}{2}\)) × (6 + \(\frac{1}{2}\))
= (10 × 6) + \(\frac{1}{2}\)(10 + 6) + \(\frac{1}{4}\)
= 60 + 8 + \(\frac{1}{4}\)
= 68\(\frac{1}{4}\)

(iv) 9.5 × 3.5 = (9 + 0.5)(3 + 0.5)
= (9 × 3) + 0.5(9 + 3) + 0.25
= 27 + 6 + 0.25
= 33.25

(v) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)

Question 2.
രണ്ട് എണ്ണൽ സംഖ്യകളുടെ ഗുണനഫലം 1400 ഉം, തുക 81 ഉം ആണ്. ഇവ ഓരോന്നിന്റെയും തൊട്ടടുത്ത രണ്ട് എണ്ണൽ സംഖ്യകളുടെ ഗുണനഫലം എന്താണ്?
Answer:
ഒരു സംഖ്യ x എന്നും മറ്റേ സംഖ്യ y എന്നും എടുത്താൽ,
xy = 1400
x + y = 81
തൊട്ടടുത്ത എണ്ണൽ സംഖ്യകൾ x + 1, y + 1
∴ (x + 1)(y + 1) = xy + x + y + 1
= 1400 + 81 + 1
= 1482

Question 3.
രണ്ട് ഒറ്റസംഖ്യകളുടെ ഗുണനഫലം 621 ഉം, തുക 50 ഉം ആണ്. ഈ ഓരോ ഒറ്റസംഖ്യയുടെയും തൊട്ടടുത്ത രണ്ട് ഒറ്റസംഖ്യകളുടെ ഗുണനഫലം എന്താണ്?
Answer:
ഒരു സംഖ്യ x എന്നും മറ്റേ സംഖ്യ y എന്നും എടുത്താൽ,
xy = 621
x + y = 50
തൊട്ടടുത്ത ഒറ്റസംഖ്യകൾ x + 2, y + 2
∴ (x + 2)(y + 2) = xy + 2(x + y) + 4
= 621 + 100 + 4
= 725

Question 4.
ചുവടെപ്പറയുന്ന ഓരോ കാര്യവും, പല സംഖ്യകളെടുത്തു പരിശോധിക്കുക. അവയിൽനിന്ന് പൊതുവായ ഒരു തത്വം ഊഹിക്കുക. ഊഹിച്ചത് ശരിയാണെന്ന് ബീജഗണിതം ഉപയോഗിച്ച് തെളിയിക്കുക.
i. 3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, 2 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, ഗുണനഫലത്തെ 3 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം.
ii. 4 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, 2 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, ഗുണനഫലത്തെ 4 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം.
iii. അടുത്തടുത്ത ആറ് എണ്ണൽസംഖ്യകളിൽ, അറ്റത്തുള്ള രണ്ടു സംഖ്യകളുടെയും, നടുക്കുള്ള രണ്ടു സംഖ്യകളുടെയും ഗുണനഫലം തമ്മിലുള്ള വ്യത്യാസം.
Answer:
i. 3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യകൾ,
1, 4, 7, 10,…
3 കൊണ്ട് ഹരിച്ചാൽ 2 മിച്ചം വരുന്ന സംഖ്യകൾ, 2, 5, 8, 11,…
സംഖ്യകൾ 4, 11 എന്നിങ്ങനെ എടുത്താൽ,
4 × 11 = 44

ഗുണനഫലത്തെ 3 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2
സംഖ്യകൾ 7, 8 എന്നിങ്ങനെ എടുത്താൽ,
7 × 8 = 56

ഗുണനഫലത്തെ 3 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2
3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും,
2 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, ഗുണനഫലത്തെ 3 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം 2 ആണ്.
ബീജഗണിതം ഉപയോഗിച്ചാൽ,

3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യയെ 31 + 1 എന്നും 3 കൊണ്ട് ഹരിച്ചാൽ 2 മിച്ചം വരുന്ന സംഖ്യയെ 3m + 2 എന്നും എടുത്താൽ,
ഇവയുടെ ഗുണനഫലം,
(3n+1)(3m + 2) = 9mn + 6n + 3m + 2
= 3(3mn + 2n + m) + 2
∴ ഗുണനഫലത്തെ 3 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2

ii. 4 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യകൾ, 1, 5, 9, 13,…
4 കൊണ്ട് ഹരിച്ചാൽ 2 മിച്ചം വരുന്ന സംഖ്യകൾ, 2, 6, 10, 14,…
സംഖ്യകൾ 5, 10 എന്നിങ്ങനെ എടുത്താൽ,
5 × 10 = 50

ഗുണനഫലത്തെ 4 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2
സംഖ്യകൾ 9, 6 എന്നിങ്ങനെ എടുത്താൽ,
9 × 6 = 54

ഗുണനഫലത്തെ 4 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2
ഗുണനസമവാക്യങ്ങൾ
4 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും,
2 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും, ഗുണനഫലത്തെ 4 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം 2 ആണ്.
ബീജഗണിതം ഉപയോഗിച്ചാൽ,

4 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യയെ 41 + 1 എന്നും 4 കൊണ്ട് ഹരിച്ചാൽ 2 മിച്ചം വരുന്ന സംഖ്യയെ 4m + 2 എന്നും എടുത്താൽ,
ഇവയുടെ ഗുണനഫലം,
(4n + 1)(4m + 2) = 16mn + 8n + 4m + 2
= 4(4mn + 2n + m) + 2
ഗുണനഫലത്തെ 4 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം = 2

iii. അടുത്തടുത്തുള്ള 6 എണ്ണൽ സംഖ്യകൾ 1, 2, 3, 4, 5, 6
അറ്റത്തുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം = 1 × 6 = 6
നടുക്കുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം = 3 × 4 = 12
∴ വ്യത്യാസം
സംഖ്യകൾ 5, 6, 7, 8, 9, 10 എന്നിങ്ങനെ എടുത്താൽ,

അറ്റത്തുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം = 5 × 10 = 50
നടുക്കുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം = 7 × 8 = 56
∴ വ്യത്യാസം = 6
ബീജഗണിതം ഉപയോഗിച്ചാൽ,

അടുത്തടുത്തുള്ള ഏതു 6 എണ്ണൽ സംഖ്യകൾ
x, x + 1, x + 2, x + 3, x + 4, x + 5 എന്നെ ടുക്യക.
അറ്റത്തുള്ള സംഖ്യകളുടെ ഗുണനഫലം = x(x + 5) = x² + 5x
നടുക്കുള്ള സംഖ്യകളുടെ ഗുണനഫലം (x + 2)(x + 3) = x² + 5x + 6
∴ വ്യത്യാസം = 6.

Question 5.
36 × 28 എന്ന ഗുണനഫലം കണ്ടുപിടിക്കാനുള്ള ഒരു രീതി ചുവടെ കാണിച്ചിരിക്കുന്നു.

i. മറ്റു ചില രണ്ടക്ക സംഖ്യകളിൽ ഈ രീതി പരീക്ഷിക്കുക.
ii. ഇത് ശരിയാകാനുള്ള കാരണം ബീജഗണിതരീതിയിൽ വിശദീകരിക്കുക.
(രണ്ടക്കസംഖ്യകളെയെല്ലാം 10m + 1 എന്ന ബീജഗണിതരൂപത്തിൽ എഴുതാമെന്ന് ഏഴാം ക്ലാസ്സിൽ കണ്ടത് ഓർക്കുക.)
Answer:
i.

ii. രണ്ടക്ക സംഖ്യകളെയെല്ലാം 10m + n എന്ന ബീജഗണിതരൂപത്തിൽ എഴുതാം
10m + n എന്ന സംഖ്യയിൽ ആദ്യത്തെ അക്കം m രണ്ടാമത്തെ അക്കം
10p + q എന്ന സംഖ്യയിൽ ആദ്യത്തെ അക്കം p രണ്ടാമത്തെ അക്കം മു
(10m + n)(10p + q) 100mp + 10(mq + np) + nq

Question 6.
ചുവടെ കാണിച്ചിരിക്കുന്നതുപോലെ സംഖ്യകൾ എഴുതുക:

i. കലണ്ടറിൽ ചെയ്തതുപോലെ നാലു സംഖ്യകളുള്ള ഒരു സമചതുരം അടയാളപ്പെടുത്തി, സംഖ്യകൾ കോണോടുകോൺ ഗുണിച്ചു വ്യത്യാസം കണ്ടുപിടിക്കുക. ഏതു സമചതുര ത്തിലെ നാലു സംഖ്യകളെടുത്താലും ഒരേ വ്യത്യാസമാണോ കിട്ടുന്നത്?
ii. ഇത് എന്തുകൊണ്ടാണെന്ന് ബീജഗണിതം ഉപയോഗിച്ച് വിശദീകരിക്കുക.
Answer:

കോണോടുകോൺ വരുന്ന സംഖ്യകളുടെ ഗുണനഫലം,
7 × 13 = 91

12 × 8 = 96
വ്യത്യാസം = 96 – 91 = 5

കോണോടുകോൺ വരുന്ന സംഖ്യകളുടെ ഗുണനഫലം,
9 × 3 = 27
4 × 8 = 32
വ്യത്യാസം = 32 – 27 = 5

i. ടേബിളിൽ ഉള്ള നമ്പറുകൾ എഴുതിയിരിക്കുന്നത് ഇങ്ങനെയാണ്;

n	n + 1
n + 5	n + 6

കോണോടുകോൺ വരുന്ന സംഖ്യകളുടെ ഗുണനഫലം,
n × (n + 6) = n² + 6n
(n + 1) × (n + 5) = n² + 6n + 5
വ്യത്യാസം = 5

Question 7.
ഗുണിതപ്പട്ടികയിൽ നാലു സംഖ്യകളുള്ള സമചതുരത്തിനു പകരം, ഒൻപതു സംഖ്യകളുള്ള ഒരു സമചതുരമെടുത്തു, നാലു മൂലകളിലുമുള്ള സംഖ്യകൾ മാത്രം അടയാളപ്പെടുത്തുക:

i. കോണോടുകോൺ തുകകളുടെ വ്യത്യാസം എന്താണ്?
ii. ഇങ്ങനെയുള്ള സമചതുരങ്ങളിലെല്ലാം വ്യത്യാസം 4 തന്നെ കിട്ടുന്നത് എന്തുകൊണ്ടാണെന്ന് ബീജഗണിതം ഉപയോഗിച്ച് വിശദീകരിക്കുക.
iii. പതിനാറ് സംഖ്യകളുടെ സമചതുരമെടുത്താലോ?
Answer:
കോണോടുകോൺ വരുന്ന സംഖ്യകൾ കൂട്ടിയാൽ,
6 + 20 = 26
10 + 12 = 22
വ്യത്യാസം = 4

ii. ടേബിളിൽ ഉള്ള നമ്പറുകൾ എഴുതിയിരിക്കുന്നത് ഇങ്ങനെയാണ്;

n	n + 2	n + 4
n + 3	n + 6	n + 9
n + 6	n + 10	n + 14

നാലു മൂലകളിലുമുള്ള സംഖ്യകൾ കോണോടുകോൺ കൂട്ടിയാൽ
n+ (n + 14) = 2n + 14
n+ 4 + (n + 6) = 2n + 10
വ്യത്യാസം = 4

iii.

കോണോടുകോൺ വരുന്ന സംഖ്യകൾ കൂട്ടിയാൽ,
6 + 30 = 36
15 + 12 = 27
വ്യത്യാസം = 9

കോണോടുകോൺ വരുന്ന സംഖ്യകൾ കൂട്ടിയാൽ,
n + n + 24 = 2n + 24
n + 6 + n + 9 = 2n + 15
വ്യത്യാസം = 9

Question 8.
ഒരു ചതുരത്തിന്റെ ചുറ്റളവ് 40 സെന്റിമീറ്ററും, പരപ്പളവ് 70 ചതുരശ്ര സെന്റിമീറ്ററുമാണ്. നീളവും വീതിയും ഇതിനേക്കാൾ 3 സെന്റിമീറ്റർ കുറവായ ചതുരത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.
Answer:
ചുറ്റളവ് = 40 സെന്റിമീറ്റർ
പരപ്പളവ് = 70 ചതുരശ്ര സെന്റിമീറ്റർ
നീളം x എന്നും വീതി y എന്നും എടുത്താൽ,
2(x + y) = 40 ⇒ x + y = 20
xy = 70
നീളവും വീതിയും ഇതിനേക്കാൾ 3 സെന്റിമീറ്റർ കുറവായ ചതുരത്തിന്റെ പരപ്പളവ് = (x – 3)(y – 3)
= xy – 3x – 3y + 9
= xy − 3(x + y) + 9
= 70 – 3 × 20 + 9
= 19 ചതുരശ്ര സെന്റിമീറ്റർ

Question 9.
ഒരു ചതുരത്തിന്റെ നീളവും വീതിയും ഒരു മീറ്റർ വീതം കുറച്ചാൽ, പരപ്പളവ് 741 ചതുരശ്ര മീറ്ററാകും; ഒരു മീറ്റർ വീതം കൂട്ടിയാൽ 861 ചതുരശ്ര മീറ്ററും.
i. ചതുരത്തിന്റെ പരപ്പളവ് എത്രയാണ്?
ii. ചുറ്റളവ് എത്രയാണ്?
iii. നീളവും വീതിയും എത്രയാണ്?
Answer:
നീളം x എന്നും വീതി y എന്നും എടുത്താൽ,
(x + 1)(y + 1) = 861 ⇒ xy + x + y + 1 = 861 ….(1)
(x − 1)(y-1) = 741⇒ xy – (x + y) + 1 = 741 ….(2)
(1) + (2) = = 2xy + 2 = 1602
⇒ xy = 800
(1) – (2) ⇒ 2(x + y) = 120
⇒ x + y = 60
xy = 800 & x + y = 60
∴ (x − y)² = (x + y)² – 4xy
= 60² – (4 × 800)
=400
= 20²
⇒ x – y = 20
x = 40
y = 20
i. ചതുരത്തിന്റെ പരപ്പളവ് = xy = 800 ചതുരശ്ര മീറ്റർ
ii. ചുറ്റളവ് = 2(x + y) = 120
iii. നീളം x = 40 മീറ്റർ
വീതി, y = 20 മീറ്റർ

Question 10.
രണ്ടു സംഖ്യകൾ ഓരോന്നിനോടും ഒന്ന് കൂട്ടി ഗുണിച്ചപ്പോൾ 1271 ഉം, ഒന്ന് കുറച്ചു ഗുണിച്ചപ്പോൾ 1131 ഉം കിട്ടി.
i. സംഖ്യകളുടെ ഗുണനഫലം എത്രയാണ്?
ii. സംഖ്യകളുടെ തുക എത്രയാണ്?
iii. സംഖ്യകൾ ഏതൊക്കെയാണ്?
Answer:
സംഖ്യകൾ x, y എന്നെടുത്താൽ,
(x + 1)(y + 1) = 1271 xy + x + y + 1 = 1271 ….(1)
(x – 1)(y – 1) = 1131 ⇒ xy – (x + y) + 1 = 1131 ….(2)
(1) + (2) ⇒ 2xy + 2 = 2402
⇒ xy = 1200
(1) – (2) = 2(x + y) = 140
⇒ x + y = 70
xy = 1200 & x + y = 70
∴ (x – y)² = (x + y)² – 4xy
= 70² – (4 × 1200) = 100 = 10²
⇒ x – y = 10
x = 40
y = 30
i. സംഖ്യകളുടെ ഗുണനഫലം, xy
ii. സംഖ്യകളുടെ തുക, x + y = 70
iii. സംഖ്യകൾ, x = 40 y = 30

Question 11.
രണ്ടു ഒറ്റസംഖ്യകളിൽ ഓരോന്നിന്റെയും തൊട്ടുമുമ്പിലുള്ള ഒറ്റ സംഖ്യകളുടെ ഗുണനഫലം 285 തൊട്ടുപുറകിലുള്ള ഒറ്റസംഖ്യകളുടെ ഗുണനഫലം 165 ഉം ആണ്. സംഖ്യകൾ എന്തൊക്കെയാണ്?
Answer:
സംഖ്യകൾ x, y എന്നെടുത്താൽ,
(x + 2)(y + 2) = 285 ⇒ xy + 2(x + y) + 4 = 285 ….(1)
(x – 2)(y – 2) = 165 ⇒ xy – 2(x + y) + 4 = 165….(2)
(1) + (2) ⇒ 2xy + 8 = 450
⇒ xy = 221

(1) – (2) ⇒ 4(x + y) = 120
⇒ x + y = 30
xy = 221 & x + y = 30
.. (x − y)² = (x + y)² – 4xy
= 30² – (4 × 221) = 16 = 4²
⇒ x – y = 4
x = 17
y = 13
സംഖ്യകൾ = 17, 13

Question 12.
ചുവടെയുള്ള കണക്കുകൾ ചെയ്തുനോക്കൂ.
i) 52 × 19
ii) 101 × 48
iii) 97 × 102
iv) 9\(\frac{3}{4}\) × 20\(\frac{1}{2}\)
Answer:
i) 52 × 19 = (50 + 2)(20 – 1)
= (50 × 20) – (50 × 1) + (2 × 20) -(2 × 1)
= 1000-50 + 40 – 2
= 988

ii) 101 × 48 = (100 + 1)(50 – 2)
= (100 × 50) – (100 × 2) + (1 × 50) -(1 × 2)
= 5000 – 200 + 50 – 2
= 4848

iii) 97 × 102 = (100 – 3)(100 + 2)
= (100 × 100) + (100 × 2) – (3 × 100) -(3 × 2)
= 10000 + 200 – 300-6
= 9894

iv) 9\(\frac{3}{4}\) × 20\(\frac{1}{2}\) = (10 – \(\frac{1}{4}\))(20 + \(\frac{1}{2}\))
= (10 × 20) + (10 × \(\frac{1}{2}\)) – (\(\frac{1}{4}\) × 20) – (\(\frac{1}{4}\) × \(\frac{1}{2}\))
= 200 + 5 – 5 – \(\frac{1}{8}\)
= 199\(\frac{7}{8}\)

(x + 1 )(y – 1) = xy — x + y – 1
(x – 1)(y + 1) = xy + x – y – 1

Question 13.
രണ്ടു സംഖ്യകളുടെ ഗുണനഫലം 713 ഉം, വ്യത്യാസം 8 ഉം ആണ്
i. വലിയ സംഖ്യയോട് 1 കൂട്ടിയതും, ചെറിയ സംഖ്യയിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം എന്താണ്?
ii. വലിയ സംഖ്യയിൽ നിന്ന് 1 കുറച്ചതും, ചെറിയ സംഖ്യയോട് 1 കൂട്ടിയതും തമ്മിലുള്ള ഗുണനഫലം എന്താണ്?
Answer:
വലിയ സംഖ്യ x എന്നും ചെറിയ സംഖ്യ y എന്നും എടുത്താൽ,
xy = 713
x – y = 8
i. വലിയ സംഖ്യയോട് 1 കൂട്ടിയതും, ചെറിയ സംഖ്യയിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം,
(x + 1)(y − 1) = xy−x + y − 1 = xy − (x − y) − 1
= 713 – 8 – 1
= 704

ii. വലിയ സംഖ്യയിൽ നിന്ന് 1 കുറച്ചതും, ചെറിയ സംഖ്യയോട് 1 കൂട്ടിയതും തമ്മിലുള്ള ഗുണനഫലം,
(x − 1)(y + 1) = = xy + xy – 1 = xy + (x – y) –
1
= 713 + 8 – 1
= 720

Question 14.
രണ്ടു സംഖ്യകളിൽ വലുതിനോട് 1 കൂട്ടിയതും, ചെറുതിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം, സംഖ്യകളുടെ ഗുണനഫലത്തേക്കാൾ 5 കുറവാണ്. വലുതിൽ നിന്ന് 1 കുറച്ചു ചെറുതിനോട് 1 കൂട്ടി ഗുണിച്ചാൽ ഗുണനഫലം സംഖ്യകളുടെ ഗുണനഫലത്തേക്കാൾ എത്ര കൂടും?
Answer:
വലിയ സംഖ്യ x എന്നും ചെറിയ സംഖ്യ y എന്നും എടുത്താൽ,
വലുതിനോട് 1 കൂട്ടിയതും, ചെറുതിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം, സംഖ്യകളുടെ ഗുണനഫലത്തേക്കാൾ 5 കുറവാണ്,
(x + 1)(y – 1) = xy – 5
-x + y – 1 = xy – 5
⇒ x – y = 4

വലുതിൽ നിന്ന് 1 കുറച്ചു ചെറുതിനോട് 1 കൂട്ടി ഗുണിച്ചാൽ ഗുണനഫലം,
(x – 1)(y + 1) = xy + (x – y) – 1
= xy + 4 = 1
= xy + 3
വലുതിൽ നിന്ന് 1 കുറച്ചു ചെറുതിനോട് 1 കൂട്ടി ഗുണിച്ചാൽ ഗുണനഫലം സംഖ്യകളുടെ ഗുണനഫലത്തേക്കാൾ 3 കൂടുതലാണ്.

Question 15.
രണ്ടു സംഖ്യകളിൽ, വലുതിനോട് 1 കൂട്ടിയതും, ചെറുതിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം 540; വലുതിൽ നിന്ന് 1 കുറച്ചതും ചെറുതിനോട് ഗുണനഫലം 560.
i. സംഖ്യകളുടെ ഗുണനഫലം എന്താണ്?
ii. സംഖ്യകൾ തമ്മിലുള്ള വ്യത്യാസം എന്താണ്?
iii. സംഖ്യകൾ എന്തൊക്കെയാണ്?
Answer:
വലിയ സംഖ്യ x എന്നും ചെറിയ സംഖ്യ y എന്നും എടുത്താൽ,
വലുതിനോട് 1 കൂട്ടിയതും, ചെറുതിൽ നിന്ന് 1 കുറച്ചതും തമ്മിലുള്ള ഗുണനഫലം 540
⇒ (x + 1)(y – 1) = 540
xy – x + y – 1 = 540 …….(1)

വലുതിൽനിന്ന് 1 കുറച്ചതും ചെറുതിനോട് 1 കൂട്ടിയതും തമ്മിലുള്ള ഗുണനഫലം 560
⇒ (x – 1)(y + 1) = 560
⇒ (x – 1) (y + 1) = 560
⇒ xy + (x – y) – 1 = 560 …..(2)

⇒ (1) + (2) ⇒ 2xy – 2 = 1100
⇒ xy = 551
സംഖ്യകളുടെ ഗുണനഫലം = 551

i. (2) – (1)⇒ 2(x − y) = 20
⇒ x – y = 10
സംഖ്യകൾ തമ്മിലുള്ള വ്യത്യാസം = 10

ii. (x + y)² = (x − y)² + 4xy
= 10² + 4 × 551
= 2304
x + y = 48
∴ x = 29
y = 19
സംഖ്യകൾ 29, 19

Question 16.
ഒരു ചതുരത്തിന്റെ നീളം 3 മീറ്റർ കൂട്ടുകയും, വീതി 2 മീറ്റർ കുറയ്ക്കുകയും ചെയ്താൽ പരപ്പളവ് 10 ചതുരശ്ര മീറ്റർ കുറയും. നീളം 2 മീറ്റർ കുറയ്ക്കുകയും വീതി 3മീറ്റർ കൂട്ടുകയും ചെയ്താൽ പരപ്പളവ് 30 ചതുരശ്ര മീറ്റർ കൂടും. ചതുരത്തിന്റെ നീളവും വീതിയും കണ്ടുപിടിക്കുക
Answer:
നീളം x എന്നും വീതി y എന്നും എടുത്താൽ,
നീളം 3 മീറ്റർ കൂട്ടുകയും, വീതി 2 മീറ്റർ കുറയ്ക്കുകയും ചെയ്താൽ പരപ്പളവ് 10 ചതുരശ്ര മീറ്റർ കുറയും
(x + 3)(y – 2) = xy – 10
⇒ xy – 2x + 3y – 6 = xy – 10
⇒ 2x – 3y = 4 ….(1)

നീളം 2 മീറ്റർ കുറയ്ക്കുകയും വീതി 3മീറ്റർ കൂട്ടുകയും ചെയ്താൽ പരപ്പളവ് 30 ചതുരശ്ര മീറ്റർ കൂടും
(x − 2)(y + 3) = xy + 30
⇒ xy + 3x – 2y – 6 = xy + 30
⇒ 3x – 2y = 36 …(2)

(1), (2) ഇവയിൽ നിന്ന്,
x = 20
y = 12
നീളം = 20 മീറ്റർ
വീതി = 12 മീറ്റർ

Class 9 Maths Chapter 4 Malayalam Medium Intext Questions and Answers

Question 1.
26 സെന്റീമീറ്റർ നീളവും 15 സെന്റീമീറ്റർ വീതിയുമുള്ള ചതുരത്തിന്റെ പരപ്പളവ് എത്രയാണ്?
Answer:

26 x 15
സാധാരണ പരപ്പളവ് കണ്ടുപിടിക്കുന്നത്:
പരപ്പളവ്= നീളം × വീതി
= 26 × 15

ഇത് നേരിട്ടു ഗുണിക്കാതെ കണ്ടുപിടിക്കാം.
26 × 15 = (20 + 6) × (10 + 5)
= (20 × 10) + (20 × 5) + (6 × 10) + (6 × 5)
=200 + 100 + 60 + 30
= 390 ച. സെ.മീ

ഇത് എണ്ണൽ സംഖ്യകളുടെ കാര്യത്തിൽ മാത്രമല്ല;
ഉദാഹരണത്തിന്,
6\(\frac{1}{2}\) × 8\(\frac{1}{3}\) = (6 + \(\frac{1}{2}\)) × (8 + \(\frac{1}{3}\))
= (6 × 8) + (6 × \(\frac{1}{3}\)) + (\(\frac{1}{2}\) × 8) + (\(\frac{1}{2}\) × \(\frac{1}{3}\))
= 48 + 2 + 4 + \(\frac{1}{6}\)
= 54\(\frac{1}{6}\)

Question 2.
(x + 1)(y + 1) = xy + x + y + 1
Answer:
ഉദാഹരണത്തിന്,
31 × 51 = (30 + 1)(50 + 1)
= (30 × 50) + 30 + 50 + 1
= 1500 + 80 + 1
= 1581

Question 3.
(x + \(\frac{1}{2}\))(y + \(\frac{1}{2}\)) = xy + \(\frac{1}{2}\)(x + y) + \(\frac{1}{4}\)
Answer:
ഉദാഹരണത്തിന്,
6\(\frac{1}{2}\) × 8\(\frac{1}{2}\) = (6 + \(\frac{1}{2}\)) × (8 + \(\frac{1}{2}\))
= (6 × 8) + \(\frac{1}{2}\)(6 + 8) + \(\frac{1}{4}\)
= 48 + 7 + \(\frac{1}{4}\)
= 55\(\frac{1}{4}\)

Question 4.
ഏതു രണ്ട് ഒറ്റസംഖ്യകളുടെയും ഗുണനഫലം ഒറ്റസംഖ്യ തന്നെയാണെന്ന് തെളിയിക്കുക.
Answer: ഒറ്റസംഖ്യകളുടെ പൊതുരൂപം,
2n + 1, n = 0, 1, 2, 3, ….
അതുകൊണ്ട് ഒരു ഒറ്റസംഖ്യയെ 2n + 1 എന്നും മറ്റേതിനെ 2m + 1 എന്നും എടുക്കുക.
ഇവയുടെ ഗുണനഫലം;
(2n + 1)(2m + 1) = 4mn + (2n + 2m) + 1
= 2(2mn + n + m) + 1
∴ ഗുണനഫലം ഒറ്റസംഖ്യയാണ്.

Question 5.
3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന രണ്ട് സംഖ്യകൾ തമ്മിൽ ഗുണിച്ചാൽ കിട്ടുന്ന ഗുണനഫലവും 3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യതന്നെയെന്ന് തെളിയിക്കുക.
Answer:
3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യകളുടെ പൊതുരൂപം,
3n + 1, n = 0, 1, 2, 3, ……………..
അതുകൊണ്ട് ഒരു സംഖ്യയെ 3n + 1 എന്നും മറ്റേതിനെ 3m + 1 എന്നും എടുക്കുക.
ഇവയുടെ ഗുണനഫലം;
(3n+1)(3m + 1) = 9mn + (3n+ 3m) + 1
= 3(3mn + n + m) + 1
∴ ഗുണനഫലം 3 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യയാണ്.

Question 6.
അടുത്തടുത്തുള്ള ഏതു നാല് എണ്ണൽ സംഖ്യകളെടുത്താലും, അറ്റത്തുള്ള സംഖ്യകളുടെ ഗുണനഫലവും, നടുക്കുള്ള രണ്ടു സംഖ്യകളുടെ ഗുണനഫലവും തമ്മിലുള്ള വ്യത്യാസം രണ്ട് ആണ് എന്ന് തെളിയിക്കുക.
Answer:
അടുത്തടുത്തുള്ള ഏതു നാല് എണ്ണൽ സംഖ്യകൾ x, x + 1, x + 2 x + 3
എന്നെടുക്കുക.
അറ്റത്തുള്ള സംഖ്യകളുടെ ഗുണനഫലം= x(x + 3) = x² + 3x
നടുക്കുള്ള സംഖ്യകളുടെ ഗുണനഫലം = (x + 1)(x+2) = x² + 3x + 2
∴ വ്യത്യാസം = 2.

Question 7.
ചുവടെയുള്ള കണക്കുകൾ ചെയ്തുനോക്കൂ.
(i) 38 × 49
(ii) 47 × 99
(iii) 29 × 46
(iv) 9\(\frac{1}{2}\) × 19\(\frac{1}{2}\)
Answer:
(i) 38 × 49 = (40 – 2)(50 – 1)
= (40 × 50) – (40 × 1) – (2 × 50) + (2 × 1)
= 2000 -40- 100 + 2
= 1862

(ii) 47 × 99 = (50 – 3)(100 – 1)
= (50 × 100) – (50 × 1) – (3 × 100) + (3 × 1)
= 5000- 50 – 300 + 3
= 4653

(iii) 29 × 46 = (30 – 1)(50 – 4)
= (30 × 50) – (30 × 4) – (1 × 50) + (1 × 4)
= 1500 – 120 – 50 + 4
= 1334

(iv) 9\(\frac{1}{2}\) × 19\(\frac{1}{2}\)
= (10 – \(\frac{1}{2}\)) (20 – \(\frac{1}{2}\))
= (10 × 20) – (10 × 1) – (\(\frac{1}{2}\) × 20) + (\(\frac{1}{2}\) × \(\frac{1}{2}\))
= 200 – 5 – 10 + \(\frac{1}{4}\)
= 185\(\frac{1}{4}\)

Multiplication Identities Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
50 × 40 = 2000 ആകുന്നു.
a) 51 × 41 =
b) 52 × 42 =
c) 49 × 39 =
d) 48 × 38 =
Answer:
(a) 51 × 41 = 50 × 40 + (50 + 40) + 1²
= 2000 + 90 + 1
= 2091

(b) 52 × 42 = 50 × 40 + 2(50 + 40) + 2²
= 2000 + 180 + 4
= 2184

(c) 49 × 39 = 50 × 40 – (50 + 40) + 1²
= 2000 – 90 + 1
= 1911

(d) 48 × 38 = 50 × 40 – 2(50 + 40) + 2²
= 2000 – 180 + 4
= 1824

Question 2.
രണ്ടു എണ്ണൽ സംഖ്യകളുടെ ഗുണനഫലം 300 ഉം തുക 35 ഉം ആണ്.
a) ഇവ ഓരോന്നിന്റെയും തൊട്ടടുത്ത രണ്ടു എണ്ണൽ സംഖ്യകളുടെ ഗുണനഫലം എന്താണ്? b) ഇവ ഓരോന്നിന്റെയും തൊട്ടുപുറകിലുള്ള രണ്ടു എണ്ണൽ സംഖ്യകളുടെ ഗുണനഫലം
എന്താണ്?
Answer: സംഖ്യകൾ x,y ആയാൽ,
xy = 300
x + y = 35
(a) (x + 1)(y + 1) = xy + (x + y) + 1;
= 300 + 35 + 1
= 336

(b) (x – 1)(y – 1) = xy – (x + y) + 1
= 300 – 35 + 1
= 266

Question 3.
രണ്ടു ഒറ്റസംഖ്യകളുടെ ഗുണനഫലം 899 ഉം, തുക 60 ഉം ആണ്. ഈ ഓരോ ഒറ്റ സംഖ്യകളുടെയും തൊട്ടടുത്തുള്ള രണ്ടു ഒറ്റസംഖ്യകളുടെ ഗുണനഫലം എന്താണ്?
Answer:
ഒറ്റസംഖ്യകൾ x, y ആയാൽ,
xy = 899
x + y = 60
(x + 2 )(y + 2) = xy + 2(x + y) + z²
= 899 + 2(60) + 4
= 1023

Question 4.
5 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും 2 മിച്ചം വരുന്ന ഒരു സംഖ്യയുടെയും ഗുണനഫലത്തെ 5 കൊണ്ട് ഹരിച്ചാലുള്ള മിച്ചം എത്രയാണ്?
Answer:
5 കൊണ്ട് ഹരിച്ചാൽ 1 മിച്ചം വരുന്ന സംഖ്യ 5m + 1 എന്നും, 5 കൊണ്ട് ഹരിച്ചാൽ 2 മിച്ചം വരുന്ന സംഖ്യ 5n + 2 എന്നും എടുത്താൽ, അവയുടെ ഗുണനഫലം
(5m + 1)(5n + 2) = 25mn + 10m + 5n + 2
= 5(5mn + 2m + n) + 2
അതായത് ഗുണനഫലത്തെ 5 കൊണ്ട് ഹരിച്ചാലുള്ള ശിഷ്ടം 2 ആകുന്നു.

Question 5.
(a) 8\(\frac{1}{2}\) × 4\(\frac{1}{2}\)
(b) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
Answer:
(a) 8\(\frac{1}{2}\) × 4\(\frac{1}{2}\)
= 8 × 4 + \(\frac{1}{2}\)(8 + 4) + (\(\frac{1}{2}\))²
= 32 + 6 + \(\frac{1}{4}\)
= 38\(\frac{1}{4}\)

(b) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
= 10 × 6 + \(\frac{1}{4}\)(10 + 6) + (\(\frac{1}{4}\))²
= 60 + 4 + \(\frac{1}{16}\)
= 64\(\frac{1}{16}\)

Question 6.
രണ്ടു എണ്ണൽ സംഖ്യകളിൽ ഓരോന്നിന്റെയും തൊട്ടുമുൻപിലുള്ള സംഖ്യകളുടെ ഗുണനഫലം 2201 ഉം തൊട്ടുപുറകിലുള്ള സംഖ്യകളുടെ ഗുണനഫലം 2001 ഉം ആകുന്നു. സംഖ്യകളുടെ ഗുണനഫലം എത്രയാണ്?
Answer:
സംഖ്യകൾ x,y ആയാൽ,
(x + 1)(y + 1) = 2201
(x − 1)(y − 1) = 2001
xy + (x + y) + 1 = 2201 … (1)
xy – (x + y) + 1 = 2001 ……… (2)
(1) + (2) ⇒ 2xy + 2 = 4202
2xy = 4200
xy = \(\frac{4200}{2}\) = 2100

Question 7.
ഒരു ചതുരത്തിന്റെ നീളവും വീതിയും ഒരു മീറ്റർ വീതം കുറച്ചാൽ, പരപ്പളവ് 240 ചതുരശ്ര മീറ്ററാകും, ഒരു മീറ്റർ വീതം കൂട്ടിയാൽ 306 ചതുരശ്രമീറ്ററും.
i. ചതുരത്തിന്റെ പരപ്പളവ് എത്രയാണ്?
ii. ചുറ്റളവ് എത്രയാണ്?
iii. നീളവും വീതിയും എത്രയാണ്?
Answer:
നീളവും വീതിയും x, y എന്നെടുത്താൽ,
(x + 1) (y + 1) = 306 … (1)
(x – 1) (y – 1) = 240 … (2)

i. ചതുരത്തിന്റെ പരപ്പളവ്,
xy = \(\frac{240+306-2}{2}\) = 272

ii. ചുറ്റളവ്= 2(x + y) = 306 – 240 = 66

iii. 2(x + y) = 66
x + y = 3 = 33 …(3)
(x − y)² = (x + y)² − 4xy
(3) + (4)
= 33² – 4 × 27² = 1 …(4)
= 2x = 34
⇒ x= 17
(3) — (4) → 2y = 32
⇒ y = 16

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 3 Laws of Motion Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 3 Notes Solutions Laws of Motion

SCERT Class 9 Physics Chapter 3 Notes Solutions Kerala Syllabus Laws of Motion Questions and Answers

Class 9 Physics Chapter 3 Let Us Assess Answers Laws of Motion

Question 1.
A body of mass 5 kg travelling with a velocity 144 km/h comes to rest in 4 s. Calculate its
a) initial momentum
b) final momentum
c) change in momentum
d) rate of change of momentum
Answer:
Mass, m = 5 kg
Initial velocity,
u = 144 km/h
= \(\frac{144 \times 5}{18}\) m/s (1km/h = \(\frac{5}{18}\) m/s) = 40 m/s

Final velocity, v = 0 (since it comes to rest)
Time, t = 4 s

a) Initial momentum, P_initial = m × u = 0 = 5 × 40 = 200 kg m/s
b) Final momentum, P_final = m × v = 0
Since the body comes to rest, the final momentum is 0.

c) Change in momentum, Δ p = P_final – P_initial = 0 – 200 = -200 kg m/s

d) Rate of change of momentum = \(\frac{\Delta p}{\Delta t}\) = \(\frac{-200}{4}\) = -50 kg m/s²

Question 2.
A hockey ball of mass 200 g hits a hockey stick with a speed 20 m/s and returns with the same speed through the same path. What is the change in momentum?
Answer:
Mass of the hockey ball. m = 200 g = 0.2 kg
Initial speed of the hockey ball (u) = 20 m/s
Final speed of the hockey ball (v) = 20 m/s (since it returns with the same speed)
Initial momentum, P_initial = m × u = 0.2 × 20 = 4 kg m/s
Final momentum, P_final = m × v = 0.2 × 20 = 4 kg m/s
Change in momentum, Δ p = P_final – P_initial = 4 – 4 = 0

Question 3.
What is the rate of change of momentum of a loaded truck of mass 10,000 kg, if its velocity changes from 15 m/s to 12 m/s in 4 s?
Answer:

Question 4.
Which of the following does not belong to the group?
Answer:
Speed. (Speed is a scalar quantity)

Question 5.
A cup is covered with a cardboard and a coin is kept on the cardboard.
a) What happens to the coin when the cardboard is struck off suddenly?
b) Why does it happen?
Answer:
a) The coin falls in the cup.
b) When the cardboard is struck off suddenly, it is thrown off because of the force applied on it. But the coin doesn’t get force. It continues in its state of rest due to its inertia of rest.

Question 6.
To clean a carpet, we hold it up and hit it with a stick. The dust falls off. Give reason.
Answer:
When a carpet is hung and hit with a stick, only the carpet gets an external unbalanced force. The dust particles do not get an external unbalanced force. The dust particles fall off due to its inertia of rest.

Question 7.
When a horse pulls the cart, the horse cart moves forward. The cart in turn pulls the horse with an equal and opposite force. But the horse and the cart go ahead. Explain.
Answer:
The horse’s feet exert force on the floor. At this time, an opposite force exerted on the legs by the floor helps the cart to move forward.

Question 8.
The velocity-time graph of a body of mass 250 g moving on a surface is given. Calculate the force of friction exerted by the surface.

Answer:
Mass, m = 250 g = 0.25 kg
From the graph,
Initial velocity, u=0
Final velocity, v = 6 m/s
Time, t = 12 s
Force, F = m × a
= m × \(\frac{v-u}{t}\)
= 0.25 × \(\frac{6-0}{12}\)
= 0.125 N

Question 9.
A body of mass 500 g moves with a velocity of 40 m/s. On applying a force for 4 s. the velocity changes to 80 m/s. Calculate the force applied.
Answer:
Mass, m = 500 g = 0.5 kg
Initial velocity, u = 40 m/s
Final velocity, v = 80 m/s
Time, t = 4s
Force, F = m × a
= m × \(\frac{v-u}{t}\)
= 0.5 × \(\frac{80-40}{4}\)
= 5 N

Question 10.
A person of mass 50 kg runs with a velocity of 8 m/s and makes a long jump. Another person of mass 60 kg makes the jump with a velocity 7 m/s. Compare their momenta. 50 kg
Answer:
Mass of first person, m, = 50 kg
Velocity of first person, v₁ = 8 m/s
Mass of second person, m₂ = 60 kg
Velocity of second person, v₂ = 7 m/s
Momentum of first person, p₁ = m₁ × v₁
=50 × 8
= 400 kg m/s
Momentum of second person, p₂ = m₂ × v₂
60 × 7 = 420 kg m/s
Comparing the two momenta, we can find that the second person has more momentum than the first person.

Question 11.
Calculate the force required to stop a vehicle of mass 14,000 kg by applying a retardation of 1.8 m/s².
Answer:
Mass, m = 14000 kg
Retardation, a = -1.8 m/s²
Force, F = m × a
= 14000 × (- 1.8)
= -25200 N

Question 12.
A force is applied on a body of mass 20 kg for 2s and its velocity changes from 10 m/s to 50 m/s. The same force is applied on another body of mass 10 kg moving with a velocity of 20 m/s for 2s in the direction of its motion. Calculate the final velocity.
Answer:
Mass of first body, m₁ = 20 kg
Initial velocity of first body, u₁ = 10 m/s
Final velocity of first body, v₁ = 50 m/s
Time, t₁ = 2s
Force on first body, F₁ = m × a = m × \(\frac{v_1-u_1}{t}\)
= 20 × \(\frac{50-10}{2}\) = 400 N
Given that force is same, ie., F₁ = F₂ = 400N
Mass of second body, m₂ = 10 kg
Initial velocity of second body, u₂ = 20 m/s
Time, t₂ = 2s
Force = m × \(\frac{v_2-u_2}{t}\)
400 = 10 × \(\frac{\mathrm{v}_2-20}{2}\)
v₂ – 20 = \(\frac{400 \times 2}{100}\)
v₂ = 80 + 20 = 100 m/s

Question 13.
A bullet of mass 20 g hits a wooden block with a velocity of 100 m/s and comes to rest after penetrating 4 cm.
a) What is the acceleration of the bullet?
b) What is the retardation of the bullet?
c) Calculate the force exerted by the bullet on the plank.
Answer:
Mass of bullet, m = 20g = 0.02 kg
Initial velocity, u = 100 m/s
Final velocity, v = 0
Distance penetrated by bullet, s = 4 cm = 0.04 m

a) From equation of motion, v² – u² = 2as
Acceleration, a = \(\frac{v^2-u^2}{2 s}\) = \(\frac{0-100^2}{2 \times 0.04}\) = – 125000 m/s²

b) Retardation of the bullet is 125000 m/s²
c) Force exerted by the bullet,
F = m × a
= 0.02 × (125000)
= – 2500 N

Question 14.
A graph showing the application of force on a body of mass 10 kg is given. The magnitude of force changes as indicated in the graph. (Frictional force need not be considered.)

a) What is the acceleration of the body when it is at 3 m?
b) Which are the instances when the body has uniform velocity?
c) Which are the instances when the body has uniform acceleration?
d) Which is the instance when the body has retardation?
Answer:
Mass of the body, m = 10 kg
a) Force at 3m, F = 8 N
Accelertion, a = \(\frac{F}{m}\)
= \(\frac{8}{10}\) = 0.8 N

b) Uniform velocity is at OA, DE and HI. (Here, force acting is zero. So, acceleration is zero)

c) Uniform acceleration is at BC and FG

d) Retardation is at CD and EF.

Question 15.
Which is the graph showing zero resultant force?

Answer:
Graph A

Question 16.
The figure shows forces applied on an object at rest. What is its acceleration? What is its displacement in 2 s?

Answer:
Mass of the object, m = 10 kg
Resultant force on the object, F = 16 – 7 = 9N
Acceleration, a = \(\frac{F}{m}\) = \(\frac{9}{10}\) = 0.9 m/s²
Initial velocity, u = 0 (since body is at rest)
Time, t = 2s
From equation of motion, S = ut + \(\frac{1}{2}\) at²
=0 × 2 + \(\frac{1}{2}\) × 0.9 × (2)²
= 1.8 m

Question 17.
Observe the figure.

A and B are two objects of masses 6 kg and 4 kg respectively. They are placed touching each other on a frictionless surface. Calculate the force exerted by object B on object A, when a force of 15 N is applied on them.
Answer:
Mass of A, m_A = 6 kg
Mass of B, m_B = 4 kg
Force, F = 15 N
Acceleration, a = \(\frac{F}{m_A+m_B}\)
= \(\frac{15}{6+4}\) = 1.5 m/s²
Force exerted by object B on object A,
F_BA = m_A × a
= 6 × 1.5
= 9 N

Class 9 Physics Chapter 3 Extended Activities Answers Laws of Motion

Question 1.
Prepare and present a seminar paper on how overload and overspeed of vehicles affect road safety.
Answer:
HINTS
Title: The Effects of Speeding and Overloaded Vehicles on Traffic Safety

Introduction:

• Begin by providing a brief definition of overload and overspeed in relation to automobiles.
• Stress how crucial road safety is to everyone’s health and safety.

Overloaded Automobiles:

• Describe the hazards connected to overloading
• Give situations or instances to highlight the risks associated with overloading.
• Discuss the significance of adhering to weight restrictions imposed by authorities.

Overspeed Vehicles:

• Talk about the dangers of exceeding the speed limit
• Emphasize how crucial it is to follow speed limits for everyone’s protection.

Effect on Traffic Safety:

• Provide evidence or statistics to back up your claims, if any are available.
• Stress the value of safe driving practices in order to avert collisions and save lives.

Conclusion:

• Recap the key ideas covered in the seminar paper.
• End with a request that everyone make road safety their top priority.

Question 2.
Write a report on how the concept of impulse can be used to explain the working of shock absorber in vehicles and discs in the spinal cord. Present it in the Science Club.
Answer:
HINTS
Title: Explaining Shock Absorbers in Vehicles and Discs in the Spinal Cord Using Impulse

Introduction:

• Define impulse as the change in momentum of an object.
• Explain how vehicle shock absorbers and spinal cord discs use impulse principles to perform properly.

Shock absorbers for vehicles:

• Define shock absorbers and their function in vehicles.
• Relate this process to the concept of impulse.

Discs in the Spinal Cord:

• Discuss the spinal cord’s role in signal transmission between the brain and the body.
• Relate spinal disc function to impulse

Conclusion:

• Summarize major points from the report.
• Emphasize the role of shock absorbers in automobiles and spinal discs in ensuring smooth and safe mobility.
• Emphasize the importance of impulse in both mechanisms, demonstrating how the concept can be utilized in a variety of real-world situations.

Question 3.
Present a seminar on some real life situations in which the concepts related to force are utilised.
Answer:
HINTS
Title: Practical Uses of Forces

Introduction:

• Begin by briefly defining force as a push or pull applied to an item and discussing its significance in day- to-day activities.
• Stress that practically everything we do, from walking to utilizing technology, involves forces.

Forces in Motion:

• Talk about situations in daily life where forces are at work to cause motion
• To demonstrate these points and describe how various forces, such as gravity and friction, affect motion, use demonstrations or movies.

Technological Forces:

• Describe the forces used to run different technology devices and how the engineers designed them to utilise a variety of forces.

Forces in Nature:

• Talk about the ways where natural occurrences are influenced by forces including tension, air resistance, and gravity.

Conclusion:

• Recap the key ideas covered in the seminar.
• Summarise more on how forces are everywhere and vital to life.

Laws of Motion Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
The rope moves only in one direction even though both the teams are applying force. Why?
Answer:
The Force F1 is greater than the force F2.

Question 2.
Is the force applied by both teams the same?
Answer:
No

Question 3.
Which team applied more force?
Answer:
The team on the left side applied more force.

Question 4.
Wasn’t it the excess force that caused the motion?
Answer:
Yes, it was the excess force that caused the motion.

Question 5.
What will be the resultant force if 100 N force is applied on an object in the east direction and 150 N force in the west direction?
Answer:
The resultant force = 100 N+(-150 N)=-50 N. The force applied in the east direction is positive. The resultant force is 50 N towards the west.

Question 6.
Complete the table by analysing the following figures.

Figure	Force F1	Force F2	Resultant Force(N)
1
2	+150	-150	0
3
4
5
6

Answer:

Figure	Force F1	Force F2	Resultant Force(N)
1	+150	0	+150 (To the right)
2	+150	-150	0
3	+150	-120	+30 (To the right)
4	0	-300	-300 (To the left)
5	0	-200	-200 (To the left)
6	+200	-200	0

Question 7.
In which of these situations is the resultant force zero?
Answer:
In fig. 2 and fig. 6

Question 8.
Which are the situations where the resultant force is not zero?
Answer:
In fig. 1, fig. 3, fig. 4, fig. 5

Question 9.
Which are the situations where there is no motion?
Answer:
In fig. 2 and fig. 6

Question 10.
In the tug of war shown in the figure below, is the resultant force experienced on the rope balanced or unbalanced?

Answer:
Unbalanced

Activity
To study if all forces cause motion

Fix a pulley each at the both ends of a wooden plank of length about 1.2 m and breadth 10 cm. Keep this plank on a table. Place a toy car in the middle of the plank as shown in the above figure. Hang pans of equal mass on strings attached to the two ends of the toy car. Place 200 g weight each in both the pans.

Question 11.
Does the toy car move?
Answer:
No

Question 12.
Are these forces balanced or unbalanced?
Answer:
Balanced

Question 13.
Add 50 g more to any one of the pans. What do you observe?
Answer:
The toy car moves in the direction in which the extra 50 g is added.

Question 14.
Are the forces balanced or unbalanced in this case?
Answer:
Unbalanced

Question 15.
When the toy car is moving, if 50 g more is added to the pan in the direction in which the car moves, what change can be observed in the motion of the car?
Answer:
Its speed increases.

Question 16.
When the car is moving, a mass of 200 g more is added to the pan attached to the string in the opposite direction of the motion of the car. What is the change observed?
Answer:
It moves in the opposite direction.

Question 17.
Are the forces balanced or unbalanced in the above situation?
Answer:
Unbalanced

Question 18.
What do you understand from these activities?
Answer:
The body moves in the direction of the resultant force.

Question 19.
Does the body move in the direction of resultant force? (moves / does not move)
Answer:
The body moves in the direction of the resultant force.

Question 20.
When does the speed of the car increase?
(when the magnitude of the resultant force increases/decreases)
Answer:
When the magnitude of resultant force increases.

Question 21.
Was the force that moved the car applied from inside the car or from outside?
Answer:
The force that moved the car was applied from outside.

Question 22.
In which situation does the direction of motion change?
Answer:
When the direction of the resultant force changes.
Note: In all the above cases, the force was given externally. Hence, all of them are external forces. An external force can be balanced or unbalanced.

Observe the figure

Question 23.
Can a vehicle move if pushed from inside?
Answer:
No

Question 24.
Isn’t it an internal force?
Answer:
Yes

Question 25.
All internal forces are………………
(balanced/unbalanced)
Answer:
balanced

Question 26.
Complete the chart and redraw it in the science diary

Answer:

Activity
Galileo’s Observations

A wiring channel is used for doing this experiment. The end C of the wiring channel is gradually lowered to horizontal level as shown in the figures.

Question 27.
What do you observe, if in each case, a marble is rolled from the end A in the wiring channel?
Answer:
In figures (a), (b) and (c), the marble that is rolled from A moves towards point C. It does not reach C. In fig (d), the marble moves beyond point C.

Question 28.
Does the distance travelled by the marble increase or decrease in each situation?
Answer:
increases

Question 29.
When did the marble travel the longest distance?
Answer:
fig.(d). The marble covers more distances as the slope decreases.
When the marble falls down it has a tendency to reach the original height. This tendency makes the marble move longer distances as the slope decreases.

Question 30.
Why did the marble come to rest after traversing some distance?
Answer:
The marble come to rest after traversing some distance due to friction.

Question 31.
What would have happened if the force of friction was absent?
Answer:
It will continue in its state of uniform motion.

Question 32.
What would have happened if no external force was applied to the marble?
Answer:
It would have continued in its state of uniform motion.

Question 33.
Write your inference from the above observation.
Answer:
If an unbalanced external force does not act on a body that is in motion, it will continue in its state of uniform motion.

Question 34.
What is the importance of Newton’s first law of motion?
Answer:
The first law of motion helps us to define the physical quantities like inertia and force

Question 35.
Passengers standing in a bus tend to fall backwards when the bus at rest moves forward suddenly. Why?
Answer:
Before the bus moved forward, the passengers and the bus were stationary. When the bus moves forward suddenly, the passengers tend to fall backwards because of the tendency to continue in the state of rest. This tendency is the inertia of rest.

Question 36.
Why do the passengers standing in a bus tend to fall forward when the moving bus stops suddenly?
Answer:
The passengers and the bus were moving. When the bus stops suddenly, the passengers tend to fall. forward because of the tendency to continue in the state of motion. This tendency is the inertia of motion.

Activity

Place a paper on a table. Keep a closed flat bottomed bottle filled with water over the paper. Quickly pull the paper horizontally.

Question 37.
What happened to the bottle?
Answer:
The bottle continues in the state of rest.

Question 38.
Name the inertia possessed by the bottle.
Answer:
The bottle possesses inertia of rest.

Activity

Place a glass filled with water on a desk. Slowly move it forward and gradually increase its speed. Stop it suddenly.

Question 39.
What do you observe? Name the inertia possessed by the water.
Answer:
The water in the cup splashes forward. The water possesses inertia of motion.

Activity

Stack some carrom coins, one above the other, as shown in the figure. Place a plastic cup filled with water above it. Using a long plastic scale, quickly strike out the carrom coins one by one from the bottom.

Question 40.
Does the cup possess inertia? Which type?
Answer:
Yes, the cup possesses inertia. The cup possesses inertia of rest.

Question 41.
Write down the following statements related to inertia in the table appropriately. Expand the table by including more examples for inertia of rest and inertia of motion.

• On shaking the branch of a mango tree, the mangoes get detached and fall down.
• A participant in long jump competition, runs some distance and then jumps.
• Travelling in a car. without wearing seat belt is dangerous.

Inertia of rest	Inertia of motion
When a bus moves forward suddenly, the standing passengers tend to fall backward.	A ball set rolling on a horizontal floor keeps moving forward.

Answer:

Inertia of rest

Inertia of motion

When a bus moves forward suddenly, the standing passengers tend to fall backwards. On shaking the branch of a mango tree, the mangoes get detached and fall down. Put a coin on a cardboard placed on the top of a glass. Strike off the cardboard quickly. The coin will fall into the cup due to its inertia of rest. When a carpet is hung and tapped with a stick, the dust particles fall down due to its inertia of rest.

A participant in long jump competition, runs some distance and then jumps. Travelling in a car without wearing seat belt is dangerous. A ball set rolling on a horizontal floor keeps moving forward If we try to jump out of a moving bus, we will fall forward due to inertia of motion. To avoid this, we have to step forward a little further. Sitting on a moving train, if we throw an object upwards, it falls back into our hands.

Activity

Place a paper on a table. Take two identical flat bottomed bottles. Fill one of them with sand. Place the bottles vertically on the paper. Quickly pull the paper horizontally.

Question 42.
Which bottle does not topple over?
Answer:
The bottle filled with sand does not topple over.

Question 43.
Which bottle has a higher mass?
Answer:
The bottle filled with sand has a higher mass.

Question 44.
Which bottle possesses more inertia?
Answer:
The bottle filled with sand possesses more inertia.

Question 45.
Based on the above observations, what is the relation between mass and inertia?
Answer:
As mass increases, inertia increases.

Question 46.
Which one possesses greater inertia – an empty barrel or a tar filled barrel? Give reason.
Answer:
The tar-filled barrel possesses greater inertia. Since the tar-filled barrel has more mass due to. the added weight of the tar, it will have greater inertia compared to the empty barrel.

Question 47.
Why people run in a zig-zag manner to escape from an elephant attack?
Answer:
As the mass of an elephant is greater, its inertia is also greater. So elephant cannot turn and run easily. Because we have relatively less mass, we can turn and run more easily than an elephant.

Question 48.
From the following, write the situations where an unbalanced force is experienced.
a) Brakes are applied on a car moving with a velocity of 20 m/s.
b) A book is supported by the hand.
c) An artificial satellite travels with uniform speed.
Answer:
(a) and (c)

Question 49.
A force of 200 N is applied on a body in one direction and another force of 250 N in the opposite direction.
a) Calculate the resultant force.
b) If it moves, what will be the direction of motion?
Answer:
a) F₁ = + 200 N
F₂ = -250 N
Resultant force = +200 N + -250 N = -50 N
b) It moves in the direction of 250 N force.

Activity
Observe the given figure.

Question 50.
Pull back the balls in a Newton’s cradle and release them in the following order. Write down the observations.
Answer:
First ball alone – When the first ball alone hits the next ball, the momentum thus transferred by it reaches the last ball through the other balls and results in the last ball being tossed off.

First two balls – When the first two balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last two balls. Hence, the last two balls are tossed off.

First three balls – When the first three balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last three balls. Hence, the last three balls are tossed off.

First four balls – When the first four balls together hit the next ball, they form a moving system and the momentum of this system is transferred to the last four balls. Hence, the last four balls are tossed off.

Factors influencing momentum

Activity
In the Newton’s cradle, pull back one ball alone to a small distance and release it. The last ball moves out only a little. When the same ball is pulled back further and released, it hits with a greater velocity. Now we can see that the last ball covers a greater distance on moving out. It is due to the increase in the velocity of the first ball.

Question 51.
Here, which factor influenced the momentum of the ball?
Answer:
Velocity of the ball.
We have seen that when a system of two balls together hit the others in Newton’s cradle, the last two balls are tossed out. This is due to the increase in the mass of the hitting system.

Question 52.
In this case, what decided the momentum of the balls?
Answer:
Mass of the ball.
Momentum is a vector quantity.
The direction of momentum is the same as that of its velocity.
Unit of momentum = Unit of mass × Unit of velocity Kg m/s

Question 53.
Calculate the momentum of a body of mass 200 kg moving with a velocity 16 m/s. 16 m/s
Answer:
m = 200 kg
v = 16 m/s
p = m v = 200 kg × 16 m/s =3200 kg m/s

Question 54.
The momentum of a body moving with a velocity 20 m/s is 200 kg m/s. What is its mass?
Answer:
p = 200 kg m/s
v = 20m/s
p=mv
m = p/v
=200/20
= 10 kg

Question 55.
Calculate the momentum of a bullet of mass 60 g moving with a velocity 200 m/s. What is its momentum when it is at rest?
Answer:
m = 60 g = \(\frac{60 \mathrm{~g}}{1000}\) = 0.06 kg
v = 200 m/s
p = mv = 0.06 kg × 200 m/s = 12 kg m/s
p = 0 (at rest)

Question 56.
A body of mass 20 kg is at rest. When a force is applied on it for 5 s, its velocity changes to 30 m/s. Find the change in momentum of the body.
Answer:
Initial momentum = mu = 20 kg × 0 = 0
Final momentum = mv =20 kg × 30 = 600 kg m/s
Change in momentum = mv – mu = 600 kg m/s – 0 = 600 kg m/s

Question 57.
What will be its change in momentum in unit time or its rate of change of momentum?
Answer:
Rate of change of momentum = change of momentum/time
= 600 / 5 = 120 kg m/s²

Question 58.
A body of mass 100 kg starts from rest and acquires a velocity of 30 m/s in the fourth second. If so,
a) what is its initial momentum?
b) what is its final momentum?
c) what is the change in momentum?
d) what is the rate of change of momentum?
Answer:
m = 100 kg, u = 0,v = 30 m/s, t = 4s
a) Initial momentum = mu = 100 kg × 0 = 0

b) Final momentum = mv = 1oo kg × 30 = 3000 kg m/s

c) Change in momentum = mv – mu = 3000 kg m/s – 0
= 3000 kg m/s

d) Rate of change of momentum = (mv – mu)/t = 3000/4 = 750 kg m/s²

Question 59.
A body of mass 20 kg is at rest. The velocity at various instances when a force of varying magnitude is applied on it for a time interval of 5 s is given. Calculate the initial momentum, final momentum and the rate of change of momentum of the body in each case. Complete the table and find the relation between the rate of change of momentum and the force applied on them.

Answer:

Question 60.
A body of mass 12 kg is moving with an acceleration of 4 m/s2. Calculate the force applied.
Answer:
m = 12 kg a = 4 m/s² F = ?
F = ma = 12 × 4 = 48 N

Question 61.
A force of 40 N is applied on a body of mass 20 kg. Calculate the acceleration produced.
Answer:
m = 20 kg F = 40 N a = ?
a = F/m = 40 N / 20 kg =2 m/s²

Question 62.
A vehicle of mass 1000 kg is travelling with a velocity of 90 km/h. The vehicle comes to rest when brakes are applied for 5 s. Calculate the force applied.
Answer:
Initial velocity u = 90 km/h = 90 × 5/18 m/s = 25 m/s
Final velocity v=0
Mass m = 1000 kg
F = ma
= m (v-u)/t
= 1000 (0-25)/5 = -5000 N

Question 63.
There is negative sign for this force. Why?
Answer:
Force is a vector quantity. The negative sign indicates that the force applied in a direction opposite to the motion of the vehicle.

Question 64.
The velocity of a body of mass 10 kg changes from 6 m/s to 18 m/s in 4 s.
a) What is the rate of change of momentum?
b) What is the force applied?
c) Calculate the acceleration of the body.
d) What will be its velocity if this force is applied for 6 s?
Answer:
mass = 10 kg
Initial velocity u = 6m/s
Final velocity v = 18 m/s

a) Rate of change of momentum
= m (v – u)/t
=10 (18 – 6)/4 = 30 N

b)Force, F = Rate of change of momentum = 30 N

c) Acceleration, a = F/m = 30 N/ 10 kg = 3 m/s²

d)Final velocity v=u+at
= 6 m/s +3 m/s² × 6 s = 24 m/s

Question 65.
A shot of mass 7 kg rolled on level ground, with a velocity 2 m/s came to rest in 5 s.
a) Which force brought it to rest?
b) Calculate the magnitude of this force.
Answer:
m = 7 kg, u = 2 m/s, t = 5 s, v = 0
a) Frictional Force
b) F = ma
= m(v – u)/t
7(0 – 2)/5 N
= -2.8 N

Question 66.
Can you find out the peculiarities of forces applied in following situations?
hitting the ball with cricket bat
kicking the ball while playing football
Nailing the wall
Answer:
Large forces are applied here for a short interval of time. Such forces are impulsive forces. Impulsive force

Question 67.
A ball of mass 200 g is moving with a velocity of 30 m/s. A person catches the ball.
a) If the time taken to bring the ball to rest is as follows, what will be the force felt on the arm in each case?
i) 0.3 s
ii) 0.2 s
iii) 0.1 s
b) The magnitude of the force is negative. This indicates that the force is applied in the opposite direction of the object’s motion.
c) As the time taken to bring the object to rest decreases, the force felt on the hand increases.
Answer:
a) m = 200g = 200/1000 = 0.2 kg
u = 30 m/s, v = 0
(i) F m (v – u) / t
= 0.2 (0 – 30) / 0.3
=0.2 × – 30/0.3
=- 20N

(ii)F = m(v – u)/t
= 0.2 (0 – 30) / 0.2
=0.2 × – 30/0.2
= – 30N

(iii)F =m(v – u)/t
0.2 (0 – 30) / 0.1
=0.2 × – 30/0.1
= -60N

b) The magnitude of the force is negative. This indicates that the force is applied in the opposite direction of the object’s motion.
c)As the time taken to bring the object to rest., decreases, the force felt on the hand increases.

Question 68.
Based on the conclusions formulated, find the reason for the following statements.

a) Cricket players move their hands backwards along with the ball while catching a fast moving ball.
b) In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball.
c) A foam bed is placed in a pole vault pit.
d) Sponge or hay is kept between glass vessels while packing.
Answer:
a) Cricket players move their hands backwards along with the ball while catching a fast-moving ball to increase the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

b) In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball, increasing the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

c) A foam bed is placed ¡n a pole vault pit.

d) Sponge or hay is kept between glass vessels while packing.
Answer:
a) Cricket players move their hands backwards along with the ball while catching a fist-moving ball to increase the time taken to bring the moving balito rest. This helps to decrease the impact or force felt on the hand.

b)In the game of football, while the goalkeeper catches the ball coming into the goal post, he moves his hands backwards along with the ball, increasing the time taken to bring the moving ball to rest. This helps to decrease the impact or force felt on the hand.

c) A foam bed is placed in a pole vault pit to increase the time taken to bring the athlete to rest. This helps to decrease the impact or force felt on the body.

d) Sponge or hay is kept between glass vessels while packing to increase the colliding time between the glass containers and thus reduce the impact or force.
Answer:
F = ma
a = F/m
when F = 0 then
a = 0
That means there will be no acceleration if no force is applied. That is, an object in motion with no acceleration will continue in its motion in a straight line or an object at rest will continue in its state of rest. This is the first law of motion. That means that the second law of motion is on par with the first law of motion.

Activity
A straw is passed through a smooth plastic thread tied diagonally. Attach an inflated balloon in the straw with cello tape and release the air from the balloon.

Question 70.
What is the direction of airflow from the balloon?
Answer:
The air flows backwards. (In the opposite direction of motion of the balloon).

Question 71.
What is the direction of motion of the balloon?
Answer:
The balloon moves forward. (In the opposite direction of motion of the air flow).

Activity

A and B are two identical spring balances.

Fix one end of the spring balance B firmly to the grill of a window. Apply force of 40 N on the spring balance B using A.

Question 72.
What is the reading shown by each spring balance?
Answer:
40 N

Question 73.
Are the readings the same?
Answer:
Yes

Question 74.
Are forces in same direction or the opposite direction?
Answer:
The forces are in the opposite direction.

Question 75.
A car will not move if we push it sitting inside. But we can move the front seat by pushing it sitting on the back seat. How is it possible?
Answer:
While sitting on the back seat and pushing the front seat, we are actually outside the front seat. Hence, we are able to exert an unbalanced external force. But when we are pushing the car sitting inside, the same force that is exerted on the car is transferred through our body to the platform of the car. Thus, the forces become balanced. Hence, the car will not move. On pushing the car by standing on the road the car moves as an unbalanced external force is acting on it.

Question 76.
Which happens first – action or reaction?
Answer:
Action and reaction act simultaneously.

Question 77.
While rowing a boat the water is pushed back, but the boat moves forward.
Answer:
When we row a boat, we push the water backwards, and the water pushes the boat forward. The boat moves forward due to the force exerted by water.

Question 78.
When a rocket is launched, gases are produced in its chamber by the combustion of fuels. These gases which are at high pressure move in one direction at high speed. But the rocket is propelled in the opposite direction.
Answer:
The high-temperature and high-pressure gases are produced by burning the fuel inside the rocket’s combustion chamber. These gases are ejected at very high speed through the nozzle of the rocket engine. The exhaust gases will exert a high force on the rocket. This will be in the opposite direction to the outflow of gases. As a result of this force, a forward acceleration is created in the rocket, and it moves forward.

Question 79.
When a person jumps from a boat onto a shore, the boat moves backwards.
Answer:
When a person jumps from a boat onto a shore, action and reaction will be exerted on the legs and boat. As a result of the action and reaction, the person jumps forward, and the boat moves backwards.

Question 80.
In the diagram below, are the forces in both directions equal?

Answer:
Forces occur only in pairs.
F₁₂ is the force exerted by the first body on the second body.
F₂₁, is the force exerted by the second body on the first body. If so, according to Newton’s third law of motion, F₁₂ – F₂₁
Answer the following questions and justify them.

Question 81.
Action and reaction are equal and opposite. If so, will they cancel each other?
Answer:
No, they will not cancel each other because action and reaction are acting on different bodies.

Question 82.
If you are pushing a vehicle standing on ice, will the vehicle move?
Answer:
No, Ice-covered surfaces have less frictional force. So, reaction force will not be obtained. The vehicle will not move.

Question 83.
Based on the third law of motion, establish how an internal force becomes balanced force.
Answer:
When an internal force is applied, both action and reaction are acting on the same body. They will cancel each other. Internal forces are always balanced.

A thorough understanding of Kerala Syllabus 9th Standard Biology Textbook Solutions Chapter 4 Behind Movements Notes Questions and Answers English Medium can improve academic performance.

SCERT Class 9 Biology Chapter 4 Notes Questions and Answers Behind Movements

Std 9 Biology Chapter 4 Notes Pdf Kerala Syllabus English Medium Solutions Questions and Answers

Class 9 Biology Chapter 4 Let Us Assess Answers Behind Movements

Question 1.
Identify the plant movement mentioned in each of those given below.
(a) The pea plant twines around a support
(b) The coconut tree near the bank of a river grows leaning towards the river.
(c) The pollen tube grows towards the ovary.
(d) The leaf of Touch-me-not plant folds while touching.
Answer:
(a) Haptotropism
(b) Phototropism (Sunlight as stimulus promotes the growth of coconut tree towards river)
(c) Chemotropism
(d) Nastic Movement

Question 2.
Identify the disease mentioned in the statement given below.
In some people certain cells of the immune system may destroy the cartilages and synovial membrane.
Answer:
Rheumatoid arthritis

Question 3.
Identify the muscle from the peculiarities given below.

• Cells with single nucleus.
• Spindle shaped cells

Answer:
Smooth Muscle

Question 4.
Observe the joints denoted as X, Y, Z and choose the one which comes in the correct order.

Answer:
c) X-Gliding Joint, Y-Hinge joint, Z-Pivot joint

Question 5.
Disorders of the bones and muscles are given in column 1, and their causes are given in column 2. Analyse them and choose the option including the correct pairs.

Column I	Column II
P) Sprain	i. Destruction of cartilage by certain defense cells
Q) Osteoporosis	ii. Changes that occur in genes
R) Rheumatoid arthritis	iii. Stretching or breaking of ligaments
S) Muscular dystrophy	iv. Deficiency of protein, calcium and Vitamin D

(a) P – ii, Q – iv, R – i, S – iii
(b) P – iv, Q – iii, R – ii, S – i
(c) P – i, Q – ii, R – iii, S – iv
(d) P – iii, Q – iv, R – i, S – ii
Answer:
(d) P – iii, Q – iv, R – i, S – ii

Question 6.
Re-draw the diagram and answer the following questions.
(a) Identify the parts mentioned below and label them.
i) Fluid present between the bones
ii) The part seen at the tip of bones which reduces friction

(b) Identify the part labelled as ‘ X ‘ in the diagram and write its function.

Answer:

Extended Activities

Question 1.
Collect pictures and information related to the diversity of locomotion in the living world and display them in the class.
Answer:

Question 2.
Prepare posters indicating the importance of exercise using graphics software and display them in the notice board.
Answer:

Question 3.
Observe various organisms in your surroundings and record the diversity in their movements in your Science diary.
Answer:
Following examples illustrate the diverse ways in which organisms move, each adapted to their environment for survival:

• Sparrow: A sparrow flies by flapping its wings rapidly and then gliding before landing on a branch. This movement helps it travel from tree to tree with ease.
• Ants: Ants move in a line using their six legs, walking in a coordinated manner. They can often be seen carrying food back to their nest.
• Dog: A dog moves smoothly on all four legs, walking or running in short bursts when excited or chasing something. This movement allows it to cover ground quickly.
• Fish: Fish swim by moving their tails from side to side. This allows them to navigate smoothly through the water, adjusting speed and direction as needed.
• Touch-me-not Plant (Mimosa pudica): When touched, the leaves of the Touch-me-not plant close rapidly. This movement is a response to touch, helping the plant protect itself.

Behind Movements Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
What all types of movements can be observed in the organisms in the picture?

Answer:
Prey capturing by birds, Honey capturing and pollination by bees and butterflies, Snakes move by dragging their bodies along the ground, plant leaves move in response to the wind in a variety of ways, including bending, twisting, and vibrating, plant parts grow towards the light, roots grow towards water, etc.

Question 2.
Observe the illustration and note down the importance of movement in each organism.

Answer:

• Food acquisition in amoeba
• Respiration in human beings (Both Expiration and inspiration)
• Growth of plant stem towards light
• The tiger, as a predator, captures its prey (deer) for its survival.

Question 3.
List out other kinds of movements that you are familiar with.
Answer:

• Movement of an organism from one place to another. Examples include walking, running, swimming, flying, and crawling.
• The movement of cells within an organism, such as during development or immune response.
• The movement of an object around a fixed axis. Examples include the spinning of a top or the rotation of the Earth.

Question 4.
Analyse the illustration based on the indicators and note down the inferences regarding the diverse movements in the living world.

Indicators:

• Movements in plants
• Movements in animals
• Common movements
• Microscopic and macroscopic movements

Answer:
a) Movements in plants:

• Plants cannot move from place to place.
• E.g., Seed germination, Change in parts of the plants according to stimuli

b) Movements in animals:

• Animals can change in a part of the body or position of the body with respect to the surroundings.
• E.g., The movement of sperm, Peristalsis during digestion, heartbeat, blood circulation, walking, running, jumping, etc.

c) Common movements:

• Transport of substances through osmosis, diffusion, active transport, transport of gases and nutrients, expulsion of waste, etc.

d) Microscopic and macroscopic movements:

• Microorganisms also exhibit various kinds of movements.
• E.g., Body movement in bacteria using flagella, Pseudopodia (false feet) in amoeba for locomotion and engulfing food particles, and movement in paramecium using cilia.

There is a great diversity in movements among organisms.

Question 5.
Identify the means of movement of the organisms shown in the illustration and complete it.

Answer:
Organisms exhibit different kinds of movements for their survival. There are different means in organisms that support these movements.

Organisms	Means of movements
Paramecium	Move using cilia, which are short, hair-like structures that cover their entire cell body.
Euglena	Move using a flagellum, a long, whip-like structure that acts as a motor.
Fishes	Possess a streamlined body that helps them to swim in the water. They swim with the help of their fins and their tails.
Frogs	Exhibit Toad movement. The hind limbs of frogs are muscular and strong and help in hopping long distances. And the webbed feet also help them in swimming.
Whale	Whales use their flippers and tail fin to move in water.
Birds	Birds use a variety of means of movement, including flight, walking, swimming, diving, jumping, etc.

Question 6.
Expand the given table by including more organisms as given above.

Organisms	Means of movement
Bacteria
Amoeba
Hydra

Answer:

Organisms	Means of movement
Bacteria	Bacterial movement typically involves the use of flagella.
Amoeba	Move by using pseudopodia, or false feet, in a process called amoeboid movement.
Hydra	Somersaulting, Sliding, Amoeboid motion. Floating
Cockroach	Flying and Crawling
Snake	Slithering
Duck	Swimming and walking
Snail	Slithering

Question 7.
Haven’t you understood the means of movement in different organisms? How does movement occur in humans? What are the means involved? Discuss, collect more information and complete the illustration.

Answer:

• Flagellar Movement: Sperm travels through the female reproductive tract to meet the ovum/egg by whipping its tail is an example of flagellar movement.
• Ciliary movement: Fertilised egg travels through the fallopian tube and reaches the uterus for implantation as a result of wave-like movement exhibited by fine hair present in the fallopian tube (cilia).

Question 8.
The discovery of movements and locomotion in humans is caused by the functioning of muscles. Which characteristics of muscles help in movement? Write the answer based on the indicators given below:

• Characteristics of muscle tissues
• Proteins in muscle cells and their importance

Answer:
Different body movements and locomotion are made possible by specialised tissue called muscles. They are formed of muscle cells. Unlike other cells, muscle cells contain more microfilaments made of proteins, such as Actin and Myosin. These filaments act together in the contraction and relaxation of muscles, which enables body movements.

There are different types of muscles in the body. They are:

Skeletal Muscles: Cylindrical-shaped cells with striations, and they do not show branching (Unbranched). More number of nuclei are present in a single cell (Multinucleated). Movements of the skeletal muscles can be controlled by one’s will (Voluntary muscles). These muscles are found attached to the bones.

Smooth Muscles: Spindle-shaped cells with tapered ends present in internal organs like the stomach. They are non-striated, uni-nucleated (cells with a single nucleus) and involuntary muscles (movements that cannot be controlled by will).

Cardiac Muscles: Cardiac muscles are muscles of the heart. They are involuntary, multinucleated, striated, branched, cylindrical-shaped cells.

Question 9.
How does Actin and Myosin help in the contraction of muscles? Find out.
Answer:
Actin is a thin protein filament that provides sites where myosin can attach during contraction. Myosin is a thicker protein filament with heads that stick out. These heads bind to actin during muscle contraction. When a muscle gets a signal to contract, calcium ions are released, exposing the binding sites on Actin.

The Myosin heads then attach to these sites, forming a cross-bridge. After attaching, the Myosin heads pull the Actin filaments inward, making the muscle shorter, which causes the muscle to contract. After the pulling action, the cross-bridge is broken, releasing actin from myosin, and muscle cells regain their original size; thus, muscle relaxation occurs. This repeated cycle of Myosin pulling on actin makes muscles move in a body.

Question 10.
Analyse the illustration, do the given activity and complete the table.

Answer:

	Muscles attached to the bones	Muscles in the hollow internal organs	Muscles in the walls of the heart
Name of the muscle	Skeletal Muscle	Smooth Muscle	Cardiac Muscle
Shape of the cell	Cylindrical	Spindle	Cylindrical
Presence of striations	Striated	Non-striated	Striated
Branches	Unbranched	Unbranched	Branched
Control of the muscles according to one’s will.	It can be controlled according to one’s will	It cannot be controlled according to one’s will	It cannot be controlled according to one’s will

Question 11.
Some muscles in our body that can be controlled (voluntary muscles) and some others cannot be controlled (involuntary) according to our will. In which all parts of the body are each one of these found? Find out examples through discussion and note them down in the table provided.

Answer:

Voluntary muscles	Involuntary muscles
Muscles in the hands Muscles in the legs Muscles found in the neck and back	Muscles in the oesophagus Muscles in the stomach Muscles in the intestine

Question 12.
By folding and stretching your hands and analysing the illustration, understand the contraction and relaxation of muscles. Discuss on the basis of indicators and prepare notes.

• The part that connects muscles to bones.
• Muscles involved in the movement of hands.
• The importance of connecting the two tips of the muscles to two bones.
• The change that should occur to the two muscles in order to fold the hands.
• The changes that should occur to the two muscles in order to stretch the hands.

Answer:
The contraction and relaxation of muscles in hand movements rely on tendons, which connect muscles to bones and allow the transfer of force to create movement. Muscles must be attached to two different bones so that when they contract, they can pull on the bones and create movement at the joint. This connection enables the hand to bend or stretch.

The main muscles involved are the biceps (Flexor muscle) and triceps (extensor muscle). To bend the arm, the biceps contracts and the triceps relaxes, pulling the hand upwards. In contrast, to straighten the arm, the triceps contracts while the biceps relaxes, allowing the arm to extend. This coordinated action of muscles attached to two different bones ensures smooth hand movements.

Question 13.
Analysing the given illustration and developing an understanding of the two divisions in the human skeletal system. Label the parts and complete the illustration.

Answer:
The human skeleton is divided into two divisions, namely, the Axial skeleton and the appendicular skeleton.

In human beings, muscles are connected with the help of either bones or cartilage. Diversity of movements are made possible due to the combined action of muscles and bones. The human skeleton is the internal framework of the human body. It is composed of around 270 bones at birth and this total decreases to around 206 bones by adulthood after some bones get fused together. It can be broadly divided into two types:

• Axial skeleton ( 80 bones): It consists of the bones seen in the central axis of the body.
• Appendicular skeleton (126 bones): It consists of the bones which are connected to the central axis.

Structure Of Bone

About 18 % of the human body weight is constituted by bones. They provide structure, support and protection to the body. Each bone is covered by a membrane known as periosteum. Blood vessels, nerves and lymph vessels are also found in bones. The components that provide hardness and strength to the bones include calcium, phosphate, collagen proteins and salts. Osteoblast cells deposit minerals in the bones which makes them strong and firm and also helps in the growth and repair of the bones.

Cartilage

It is the connective tissue which is softer and more flexible than bones. Cartilage is present in elbows, knees, ankles, at the tip of ribs, between the vertebrae of the vertebral column, pinna of the ear, tip of the nose and also in the trachea. Cartilages present at the tip of the bones reduce friction in the joints. Blood vessels and nerves are absent in them. The growth of cartilage cells is slower than the rest of the cells because of the absence of blood vessels.

Diversity In Structural Framework

All organisms do not have a skeletal framework (endoskeleton) like that of human beings; it varies from organism to organism. Hydroskeleton, exoskeleton and endoskeleton are some of the types of structural frameworks of the body present in different organisms.

Structure Of Bone

About 18 % of the human body weight is constituted by bones. They provide structure, support and protection to the body. Each bone is covered by a membrane known as periosteum. Blood vessels, nerves and lymph vessels are also found in bones. The components that provide hardness and strength to the bones include calcium, phosphate, collagen proteins and salts. Osteoblast cells deposit minerals in the bones which makes them strong and firm and also helps in the growth and repair of the bones.

Cartilage

It is the connective tissue which is softer and more flexible than bones. Cartilage is present in elbows, knees, ankles, at the tip of ribs, between the vertebrae of the vertebral column, pinna of the ear, tip of the nose and also in the trachea. Cartilages present at the tip of the bones reduce friction in the joints. Blood vessels and nerves are absent in them. The growth of cartilage cells is slower than the rest of the cells because of the absence of blood vessels.

Diversity In Structural Framework

All organisms do not have a skeletal framework (endoskeleton) like that of human beings; it varies from organism to organism. Hydroskeleton, exoskeleton and endoskeleton are some of the types of structural frameworks of the body present in different organisms.

Hydroskeleton

• Fluid filled chambers are present in the body of the earth worm.
• Here, water is the means to maintain body structure and locomotion. This mechanism is commonly called hydroskeleton.
• Hydroskeleton helps in the movements of hydra and snail.

Exoskeleton

• They connect muscles in respective places and help in movement, locomotion and protection of the body.
• Hard shells present in crabs, mussels and oysters made up of calcium carbonate, outer covering of grasshoppers and cockroaches made up of chitin are examples of exoskeleton.

Endoskeleton

• It provides shape to the body, protects internal organs and helps in movement and locomotion.
• It is seen in vertebrates including human beings which is made up of a framework of cartilages and bones.

Joints

Joints are the connection between two or more bones. Connecting the bones with the help of joints makes movement easier. They help to rotate our shoulders, bend our knees and elbows, swivel our neck and more. Joints differ according to their functions.

Structure of a typical joint

A typical joint consists of the following parts:

• Ligaments – They are found connecting two bones.
• Capsule – It is seen inside the ligament which helps in the smooth movement of the bones.
• Cartilage – It is seen covering the tip of each bone, which reduces the friction between the bones.
• Synovial fluid – The fluid present between the two bones of a joint which also reduces the friction between the bones.
• Synovial cells – Synovial fluid is produced by these cells, which is present in the synovial membrane.

Different types of joints

Based on the functions, there are different types of joints:

• Ball and socket joint
• Hinge joint
• Pivot joint
• Gliding joint

Name	Ball and Socket joint	Hinge joint	Pivot joint	Gliding joint
Peculiarities	The ball-shaped surface of one round bone fits into the cup-like depression of another bone, allowing greater freedom of movement	Allows movement only in one axis.	Enables movement in different directions	Allows movement only in two axes
Position	Shoulder joint.	Ankle joint	Between the atlas and axis	Wrist joint

Body growth and Bone Development

Growth during childhood and adolescence is associated with the development of the skeletal system. Childhood and adolescence is a critical period for bone development. It is essential that calcium should deposit in the bones to ensure their hardness and strength. Calcium-rich foods (dairy, fish, leafy greens) and vitamin D (sunlight, eggs, fish) are essential. Protein (meat, beans) also supports bone growth. As age advances, the density of bones decreases, making them weaker and more prone to fracture. A balanced diet throughout life helps maintain bone density and reduce the risk of osteoporosis.

Muscle And Exercise

The active functioning of muscle cells can be ensured only through movement. Exercise helps to strengthen muscles and increase their efficiency.

Part of the body	The benefits of exercise
Lungs	Vital capacity increases, gaseous exchange becomes efficient, strengthens respiratory muscles, improves oxygen utilization, reduces risk of respiratory diseases, promotes alveoli health.
Hands and legs	Improved muscle strength. enhanced flexibility and range of motion, reduces the risk of sprains, strains, increases oxygen and nutrients, increases bone density improvement.
Muscles, bones	Increased muscle strength, enhanced muscle coordination, increased bone density, enhanced joint function, improved posture and balance.
Heart and blood vessels	Strengthening the heart muscles. lo ers blood pressure, improves blood vessel health, lowers cholesterol levels, reduces risk of blood clots, helps in weight management

Vital Capacity

The total volume of air exhaled forcefully after a deep inhalation is called vital capacity. It is the measurement of a person’s respiratory health. It’s typically 4.5 litres in men and 3 litres in women. A decrease in vital capacity may be an indication of pulmonary diseases.

Question 14.
Write the total number of bones in the human skeleton.
Answer:
206 bones

Question 15.
Mention the number of bones in the axial skeleton and appendicular skeleton in the human skeleton.
Answer:
Axial skeleton – 80 bones, Appendicular skeleton – 126 bones.

Question 16.
Is the number of bones same in children and adults? What is the reason? Find out.
Answer:
No, the number of bones is not the same in children and adults. Babies are born with approximately 300 bones while adults have only 206. The reason behind this is that the babies are born with bone made of cartilage which is flexible. As children grow, this cartilage gradually hardens and fuses together, forming the stronger, more rigid bones of adulthood.

Question 17.
Which substances are responsible for the hardness of bones?
Answer:
The hardness of bones is due to the presence of calcium, phosphate, collagen proteins and salts.

Question 18.
What is the function of osteoblast cells of bones?
Answer:
It helps to deposit minerals in the bones which makes them strong and firm and helps in growth and repair.

Question 19.
Muscles are connected either to bones or to cartilages. What is the difference between bones and cartilage?
Answer:

Bones	Cartilages
It is very strong and firm when compared to cartilage.	It is softer and more flexible than bones.
Blood vessels, nerves and lymph vessels are present.	Blood vessels and nerves are absent.
They grow at a faster rate due to the presence of blood vessels.	Their growth is slower than the rest of the cells due to the absence of blood vessels.

Question 20.
Do all living organisms have a skeletal framework like that of humans?
Answer:
No

Question 21.
Complete the given illustration regarding the diversity in the structural framework of organisms. (Fill the square boxes with their characteristics and pentagonal boxes with suitable examples).

Answer:

Question 22.
Are there parts of exoskeleton in organisms with endoskeleton? Discuss and find out.
Answer:
Yes, some organisms have both an endoskeleton and an exoskeleton, including tortoises and crocodiles.

Question 23.
Some organisms having exoskeleton shed their outer covering. Why? Find out.
Answer:
Organisms with exoskeletohs, such as insects, prawns and spiders, shed their outer covering because if the exoskeleton becomes damaged or injured, it provides an opportunity for the organism to repair the damage and regenerate lost parts. In some insects, they shed their outer covering to remove the parasites or other harmful organisms that have attached themselves to their exoskeleton.

Question 24.
Analyse the figure of the joints and complete the table.

Answer:
Based on the functions, there are different types of joints:

• Ball and socket joint
• Hinge joint
• Pivot joint
• Gliding joint

Name	Ball and Socket joint	Hinge joint	Pivot joint	Gliding joint
Peculiarities	The ball-shaped surface of one round bone fits into the cup-like depression of another bone, allowing greater freedom of movement	Allows movement only in one axis.	Enables movement in different directions	Allows movement only in two axes
Position	Shoulder joint.	Ankle joint	Between the atlas and axis	Wrist joint

Question 25.
Collect more information about joints in different parts of the body. Prepare a chart and exhibit it in the class.
Answer:

Types of joints	Location	Description
Immovable	Skull, Pelvis, Sternum	Bones are fused together and do not allow any movement
Slightly movable	Vertebral column	Bones are connected by cartilage and allow limited movement
Freely movable	Shoulders, Hips, Knees, Elbows, Wrists, Ankles	Bones are separated by a synovial cavity, allowing for a wide range of movements.

Question 26.
What are the other functions of bones? Discuss and expand the list.
Answer:
The following are the functions of bones:

• Formation of blood cells
• Enables body movement
• Bones protect vital organs, such as the brain, heart, lungs, and spinal cord.
• Bones help regulate the body’s pH balance by absorbing or releasing minerals.
• Provides a structural framework for the body, supporting organs and tissues.
• Bones can store heavy metals and other toxins, helping to remove them from the body.

Question 27.
How does the deficiency of vitamin D affect the body? Find out.
Answer:
Vitamin D is essential for calcium absorption which is vital for strong bones. A deficiency can lead to osteoporosis, a condition characterized by weak, brittle bones that are susceptible to fractures. In children, vitamin D deficiency can cause rickets, a disease characterized by soft, weak bones that can lead to bowed legs and other deformities.

Vitamin D plays a role in regulating the immune system. A deficiency can impair the immune response, making individuals more susceptible to infections. Vitamin D deficiency has been associated with an increased risk of heart disease, including high blood pressure and stroke.

Question 28.
Complete the following table on disorders of bones and muscles.

Answer:

Disease	Causes	Symptoms
Osteoporosis	Deficiency of protein, calcium and vitamin D	Back pain, frequent bone fractures, etc.
Rheumatoid arthritis	The immune system destroys cartilage and synovial membrane.	Severe pain and swelling of joints
Muscular dystrophy	Changes in genes	Weakening and degeneration of muscles.
Muscular dystrophy	Injury caused by stretching or breaking of ligaments.	Pain, swelling, bruises, difficulty in moving joints.

Question 29.
Why is rheumatoid arthritis more in women than men? Find out.
Answer:
Rheumatoid arthritis (RA) is more common in women largely due to hormonal factors. The hormone estrogen, which is present in higher levels in women, is thought to affect the immune system. Estrogen can influence how the immune system responds, and fluctuations in this hormone, especially during pregnancy or menopause, can increase the risk of developing RA or make the symptoms worse.

During pregnancy, estrogen levels rise, which may temporarily improve RA symptoms, but after childbirth, when estrogen levels drop, the symptoms can flare up. Similarly, during menopause, when estrogen levels decrease significantly, women are more likely to experience the onset or worsening of RA. These hormonal changes play a key role in why women are more affected by RA than men. Other than this, stronger immunity and genetic factors increase the risks of rheumatoid arthritis in women than in men.

Question 30.
Identify the situations in which the first aid measures shown in the pictures are used and complete the illustration.

Answer:
First aid measures are crucial in stabilising injuries before professional medical treatment.

Sling: A sling is used to support and immobilise an injured arm or shoulder. It is typically used when someone has a fracture, dislocation, or severe sprain in the arm, wrist, or shoulder area. The sling helps to keep the injured part stable and prevents further movement, reducing pain and promoting healing.

Splint: A splint is used to keep an injured limb, such as a leg or arm, immobile. It is commonly used in cases of a fracture, severe sprain, or dislocation to prevent further injury. The splint provides support and protection to the injured area, ensuring that it remains in a fixed position until professional medical help is available.

Bandage: A bandage is used to cover wounds, support injured limbs, or secure dressings in place. It is commonly applied in cases of cuts, scrapes, bleeding, or sprains to protect the injury, stop bleeding, and provide support. Bandages are also used to prevent infection in open wounds and keep dressings clean.

Question 31.
Organise an awareness class by a health expert about first aid measures in coordination with the Health Club. Get hands-on training on the first-aid measures given

• How to prepare a sling
• How to use a splint
• How to prevent blood loss when a wound occurs
• The use of bandage and band-aid
• First aid to be given when there is a spinal injury.

Answer:
First aid is the immediate help we can give to someone who is injured or sick before professional medical help arrives. Having basic first aid knowledge enables one to act quickly in emergencies and can make a big difference in someone’s life. Here are some general instructions to perform first aid in particular situations:

A sling is used to support an injured arm or shoulder. To make one, use a triangular cloth placed under the injured arm, with the ends tied around the neck. Ensure the elbow is well-supported and the arm is immobilised without cutting off circulation.

A splint is needed to stabilise a broken or injured limb. Place a firm object, like a stick, along the injured area and tie it in place with bandages or cloth strips. Ensure it immobilised the limb but isn’t too tight to affect circulation.

To stop bleeding, apply direct pressure on the wound with a clean cloth or bandage. If possible, elevate the injured area above heart level and continue applying pressure until the bleeding slows or stops.

Bandages and band-aids both protect wounds but differ in use. Bandages are larger cloth or gauze used to cover and support bigger injuries, like cuts or sprains, often with extra dressings. Bandaids are small adhesive strips for minor cuts or scrapes, covering the wound directly to prevent infection. Bandages provide more support, while band-aids are for quick use on smaller injuries.

For suspected spinal injuries, avoid moving the person. Keep the head, neck, and spine aligned, and stabilise them with rolled towels if possible. Call for emergency help and move the person only if necessary.

Question 32.
Analyse the figure, discuss and find out the various plant movements and the stimuli that cause movements in plants.

Answer:

Type of movement	Stimuli	Directional/Non-directional
Curling of tendrils around a support	Touch	Directional
Bending of stern in the presence of light	Light	Directional
Folding of leaves of touch-me-not plant	Touch	Non-directional
Drooping of leaves after sunset	Light	Non-directional

Question 33.
Are all the plant movements that you have listed related to the direction of stimulus? Discuss.
Answer:
No. All plant movements are not related to the direction of the stimulus. Based on the findings, plant movements can be classified into:

Question 34.
Analyse the illustration, identify how the movements of the shoot and root are related to the direction of stimulus and complete the table.

Answer:

Plant movements	Stimulus	Direction of movement of the shoot	Direction of movement of the root
Phototropism	Light	Towards light	Away from light
Geotropism	Gravity	Away from gravity	Towards gravity
Hydrotropism	Water	Away from water	Towards water

Question 35.
Two other tropic movements found in plants are given in illustration. Find out the characteristics of these and record them in ‘our science diary.

Answer:
Haptotropism: Directional movement of plant parts in response to stimulus touch or contact with a solid surface. E.g. Tendrils of climbers curl around the support they grow.

Chemotropism: Growth of plant parts towards or away from chemical stimulus. E.g. Germination and growth of pollen tube towards the egg/female gamete in flowering plants.

Question 36.
Have you ever touched a Touch-me-not plant? How is the movement of the leaves of a Touch-me-not plant? Which type of movement is this? What is the peculiarity of such movements?
Answer:

When we touch a touch-me-not plant, the leaves of the plant fold immediately in response to the stimulus touch. This type of plant movements are called nastic movements. They are non-directional. i.e., they do not depend on the direction of stimulus. It happens because of changes in water pressure inside the plant’s cells, helping the plant protect itself from possible harm, like being eaten by predators.

Question 37.
List out more examples of nastic movements
Answer:
Examples of nastic movements:

• Leaves of certain plants (prayer plant, oxalis) fold during the night and spread out during the daytime.
• Opening and closure of jasmine flowers during night and day, respectively.
• Leaves of Venus fly trap shut when an insect sits on it.

When preparing for exams, Kerala SCERT Class 9 Maths Solutions Chapter 6 Malayalam Medium സദൃശത്രികോണങ്ങൾ can save valuable time.

Kerala SCERT Class 9 Maths Chapter 6 Solutions Malayalam Medium സദൃശത്രികോണങ്ങൾ

Class 9 Maths Chapter 6 Kerala Syllabus Malayalam Medium

Class 9 Maths Chapter 6 Malayalam Medium Textual Questions and Answers

Question 1.
ഒരു ത്രികോണത്തിന്റെ ഒരു വശം 8 സെ.മീ. ഉം അതിലെ രണ്ട് കോണുകൾ 60° യും 70° യും ആണ്. കോണുകൾ മാറാതെ വശങ്ങൾ ഇതിന്റെ ഒന്നരമടങ്ങായ ത്രികോണം വരയ്ക്കുക.
Answer:
8 ന്റെ ഒന്നരമടങ്ങ് = 8 × 1.5 = 12
ഒരു വശം 12 സെ.മീ. ഉം അതിലെ രണ്ട് കോണുകൾ 60° യും 70° യും ആയ ത്രികോണം വരച്ചാൽ നമ്മളോടു ചോദിച്ച ത്രികോണം കിട്ടും. അതിനായി,
12.സെ.മീ. നീളമുള്ള ഒരു വര വരയ്ക്കുക.
വരയുടെ ഒരറ്റത്തുനിന്ന് 60° യും മറ്റേ അറ്റത്തുനിന്ന് 70° യും അളന്നെടുക്കുക.

Question 2.
മട്ടത്രികോണത്തിന്റെ മട്ടമൂലയിൽനിന്ന് കർണ്ണത്തിലേക്കു വരയ്ക്കുന്ന കർണ്ണത്തിനെ 2 സെന്റിമീറ്ററും, 3 സെന്റിമീറ്ററും നീളമുള്ള ഭാഗങ്ങളാക്കുന്നു.

i) ലംബം മുറിച്ചുണ്ടാകുന്ന രണ്ടു ചെറിയ മട്ടത്രികോണങ്ങൾക്കും ഒരേ കോണുകളാണെന്ന് തെളിയിക്കുക.
ii) ലംബത്തിന്റെ ഉയരം h എന്നെടുത്താൽ \(\frac{h}{2}=\frac{3}{h}\) എന്നു തെളിയിക്കുക
iii) വലിയ മട്ടത്രികോണത്തിന്റെ ലംബവശങ്ങൾ കണക്കാക്കുക.
iv) ഒരു മട്ടത്രികോണത്തിന്റെ മട്ടമൂലയിൽ നിന്നു കർണ്ണത്തിലേക്കു വരയ്ക്കുന്ന ലംബത്തിന്റെ നീളം h എന്നും, അത് കർണ്ണത്തെ മുറിക്കുന്ന ഭാഗങ്ങളുടെ നീളം a, b എന്നുമെടുത്താൽ h² = ab എന്നു തെളിയിക്കുക.
Answer:
i) തന്നിരിക്കുന്ന ത്രികോണത്തിനെ ABC എന്നും മട്ടമല്ലാത്ത ഒരു കോണിനെ x എന്നും എടുത്താൽ തന്നിരിക്കുന്ന ത്രികോണം താഴെ കാണിച്ചിരിക്കുന്നുതു പോലെ ആകും.

∆ADC പരിഗണിക്കുക.
∠A = X
∠ADC = 90° (AD ലംബമായതുകൊണ്ട്)
∠A + ∠ADC + ∠ACD = 180°
∠ACD = 180° – ∠A – ∠ADC
= 180° – x – 90°
= 90° – x

∆BDC പരിഗണിക്കുക.
∠BDC = 90° (AD ലംബമായതുകൊണ്ട്)
∠BCD + ∠ACD = 90° (∠ACB മട്ടമായതുകൊണ്ട്)
∠BCD = 90° – ∠ACD
= 90° – (90° – x)
= x

∠BDC + ∠BCD + ∠B = 180°
∠B = 180° – BDC – BCD
= 180°- 90° – x
= 90° – x
∴ ലംബം മുറിച്ചുണ്ടാകുന്ന രണ്ടു ചെറിയ മട്ടത്രികോണങ്ങൾക്കും ഒരേ കോണുകളാണ്.

ii)

x ന് എതിരെയുള്ള വശങ്ങൾ: h, 3.
\(\frac{∆ADC ലെ വശം}{∆BDC ലെ വശം}\) = \(\frac{h}{3}\)

90° – x ന് എതിരെയുള്ള വശങ്ങൾ: 2, h.
\(\frac{∆ADC ലെ വശം}{∆BDC ലെ വശം}\) = \(\frac{2}{h}\)
∴ \(\frac{h}{3}=\frac{2}{h}\)

iii) \(\frac{h}{3}=\frac{2}{h}\) ⇒ h² = 6
⇒ h = √66m.øl.
∆ADC പരിഗണിക്കുക.
AC² = AD² + DC²
= 2² + (√6)²
= 4 + 6
= 10
AC = √10 സെ.മീ.

∆BDC പരിഗണിക്കുക.
BC² = BD² + DC²
BC = 3² + (√6)²
= 9 + 6
= 15
= √15 സെ.മീ.

വലിയ ത്രികോണത്തിന്റെ ലംബ വശങ്ങളുടെ നീളം = √10 സെ.മി, √15 സെ.മി.

iv)

∆BDC യും ∆ADC യും പരിഗണിക്കുക.
ഒരേ കോണുകളുള്ള ത്രികോണങ്ങളിൽ, തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങളുടെ നീളം മാറുന്നത് ഒരേ തോതിലാണ്. ആയതിനാൽ,
\(\frac{\mathrm{h}}{\mathrm{~b}}=\frac{\mathrm{a}}{\mathrm{~h}}\)
h² = ab

Question 3.
വിലങ്ങനെയുള്ള ഒരു വരയുടെ രണ്ടറ്റത്തും ഒരേ വലുപ്പമുള്ള കോണുകൾ മുകളിലും താഴെയുമായി വരച്ച്, ചരിഞ്ഞ വരകളിലെ രണ്ടു ബിന്ദുക്കൾ യോജിപ്പിക്കുന്നു.

i) വിലങ്ങനെയുള്ള (നീല) വരയുടെ ഭാഗങ്ങളും, ചരിഞ്ഞ (ചുവന്ന) വരയുടെ ഭാഗങ്ങളും ഒരേ അംശബന്ധത്തിലാണെന്നു തെളിയിക്കുക.
ii) വിലങ്ങനെയുള്ള വരയുടെ രണ്ടറ്റത്തുമുള്ള ചരിഞ്ഞ പച്ച വരകൾ തമ്മിലുള്ള അംശ ബന്ധവും ഇതുതന്നെയാണെന്ന് തെളിയിക്കുക.
iii) ഇതുപയോഗിച്ച്, 6 സെന്റിമീറ്റർ നീളമുള്ള വരയെ 3 : 4 എന്ന അംശബന്ധത്തിൽ എങ്ങനെ ഭാഗിക്കും?
Answer:
i)

\(\frac{A M}{M B}=\frac{C M}{M D}\) എന്നാണ് നമുക്ക് തെളിയിക്കേണ്ടത്.

അതിനായി, ∆ACM ഉം ∆BDM ഉം പരിഗണിക്കുക.
∠A = <B
∠AMC = ∠BMD (എതിർകോണുകൾ)
∠C = ∠D
∠A ക്ക് എതിരെയുള്ള വശം = CM
∠B ക്ക് എതിരെയുള്ള വശം = MD
∠C ക്ക് എതിരെയുള്ള വശം = AM
∠D ക്ക് എതിരെയുള്ള വശം = BM
ഒരേ കോണുകളുള്ള രണ്ടു ത്രികോണങ്ങളുടെ തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങൾ
ജോടികളായെടുത്താൽ, മൂന്നു ജോടികളിലും

ഒരേ
സംഖ്യയായിരിക്കും. ആയതിനാൽ, \(\frac{C M}{M D}=\frac{A M}{M B}\)
⇒ \(\frac{A M}{M B}=\frac{C M}{M D}\)
⇒ AM: MB = CM:MD

ii) \(\frac{A C}{B D}=\frac{A M}{M B}\) എന്നാണ് നമുക്ക് തെളിയിക്കേണ്ടത്.
അതിനായി, ∆ACM ഉം ∆BDM ഉം പരിഗണിക്കുക.
∠A = ∠B
∠AMC = ∠BMD (എതിർകോണുകൾ)
∠C = ∠D
∠C ക്ക് എതിരെയുള്ള വശം = AM
∠D ക്ക് എതിരെയുള്ള വശം = MB
∠AMC ക്ക് എതിരെയുള്ള വശം = AC
∠BMD ക്ക് എതിരെയുള്ള വശം = BD
ഒരേ കോണുകളുള്ള രണ്ടു ത്രികോണങ്ങളുടെ തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങൾ
ജോടികളായെടുത്താൽ, മൂന്നു ജോടികളിലും

ഒരേ
സംഖ്യയായിരിക്കും. ആയതിനാൽ, \(\frac{A C}{B D}=\frac{A M}{M B}\)
⇒ AC : BD = AM : MB

iii) 6 സെ.മീ നീളമുള്ള ഒരു വര വിലങ്ങനെ വരയ്ക്കുക. വരയുടെ രണ്ടറ്റത്തും ഒരേ വലുപ്പമുള്ള കോണുകൾ മുകളിലും താഴെയുമായി വരയ്ക്കുക. മുകളിലെ ചെരിഞ്ഞ വരയിൽ 3 സെ.മീ അകലത്തിൽ ഒരു ബിന്ദു അടയാളപ്പെടുത്തുക. താഴെയുള്ള ചെരിഞ്ഞ വരയിൽ 4 സെ.മീ അകലത്തിൽ ഒരു ബിന്ദു അടയാളപ്പെടുത്തുക. ചരിഞ്ഞ വരകളിലെ രണ്ടു ബിന്ദുക്കൾ യോജിപ്പിക്കുക.

M എന്ന ബിന്ദു AB യെ 3:4 എന്ന അംശബന്ധത്തിൽ മുറിക്കുന്നു.

Question 4.
ചുവടെയുള്ള ചിത്രത്തിൽ ഒരു മട്ടത്രികോണത്തിലെ മട്ടമൂലയും, മൂന്നു വശങ്ങളിലെയും ഓരോ ബിന്ദുക്കളും മൂലകളായി ഒരു സമചതുരം വരച്ചിരിക്കുന്നു.

i) സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം കണക്കാക്കുക.
ii) വശങ്ങളുടെ നീളം 3, 4, 5 സെന്റിമീറ്ററായ മട്ടത്രികോണത്തിൽ ഇങ്ങനെ വരയ്ക്കുന്ന സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം എത്ര സെന്റിമീറ്ററാണ്?
Answer:
i)

∆APQ ഉം ∆ABC ഉം പരിഗണിക്കുക.
P = B = 900
∠PQA = ∠BCA CPO ഉം BC ഉം സമാന്തരവരകൾ)
∴ ∠A = ∠A
∠A ക്ക് എതിരെ ∆APD ൽ ഉള്ള വശം = PQ
∠A ക്ക് എതിരെ ∆ABC ൽ ഉള്ള വശം = BC
∠PQA ക്ക് എതിരെയുള്ള വശം = AP
∠BCA ക്ക് എതിരെയുള്ള വശം = AB
ഒരേ കോണുകളുള്ള രണ്ടു ത്രികോണങ്ങളുടെ തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങൾ
ജോടികളായെടുത്താൽ, മൂന്നു ജോടികളിലും

ഒരേ
സംഖ്യയായിരിക്കും. ആയതിനാൽ, \(\frac{P Q}{B C}=\frac{A P}{A B}\)
സമചതുരം BRQP യുടെ ഒരു വശത്തിന്റെ നീളം x എന്ന് എടുത്താൽ
\(\frac{x}{x+1}=\frac{2}{x+2}\)
x(x + 2) = 2(x + 1)
x² + 2x = 2x + 2
x² = 2
x = √2
സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = √2 സെ.മീ.

ii) AB = 4 സെ.മീ, BC = 3 സെ.മീ, AC = 5 സെ.മീ എന്ന് തന്നിരിക്കുന്നു. സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം x എന്നെടുത്താൽ,

\(\frac{P Q}{B C}=\frac{A P}{A B}\)
⇒ \(\frac{x}{3}=\frac{4-x}{4}\)
4x = 3(4 – x)
4x = 12 – 3x
7x = 12
x = \(\frac{12}{7}\)
സമചതുരത്തിന്റെ ഒരു വശത്തിന്റെ നീളം = \(\frac{12}{7}\) സെ.മീ.

Question 5.
ചിത്രത്തിലെ വലിയ മട്ടത്രികോണത്തിന്റെ പരപ്പളവ് കണക്കാക്കുക.

Answer:

BM = h എന്നെടുക്കുക.
h² = AM × MC
= 9 × 4
= 36
h = √36 = 6 സെ.മീ.

∆ABC യുടെ പരപ്പളവ് = \(\frac{1}{2}\) × AC × h
= \(\frac{1}{2}\) x 13 × 6
= 39 ചതു.സെ.മീ

Question 6.
3 മീറ്ററും 2 മീറ്ററും ഉയരമുള്ള രണ്ടു കമ്പുകൾ കുത്തനെ നിലത്തു നാട്ടി, ഒരോ കമ്പിന്റെയും മുകളറ്റത്തുനിന്ന് മറ്റെ കമ്പിന്റെ ചുവട്ടിലേക്ക് കയർ വലിച്ചു കെട്ടിയിരിക്കുന്നു:

i) കയറുകൾ പരസ്പരം മുറിച്ചുകടക്കുന്നത്, നിലത്തുനിന്ന് എത്ര ഉയരത്തിലാണ്?
ii) കമ്പുകൾ തമ്മിലുള്ള അകലം എത്രയായാലും ഈ ഉയരം മാറുന്നില്ല എന്നു തെളിയിക്കുക.
iii) കമ്പുകളുടെ നീളം a, b എന്നും, കയറുകൾ മുറിച്ചുകടക്കുന്ന സ്ഥാനത്തിന്റെ ഉയരം h എന്നുമെടുത്ത്, a, b, h ഇവ തമ്മിലുള്ള ബന്ധം കണ്ടുപിടിക്കുക.
Answer:
i)

∆ADB യും ∆FEB പരിഗണിക്കുക.
∠DAB = ∠EFB (രണ്ടും 90°
∠ABD = ∠FBE (രണ്ടും ഒന്നാണ്)
അതിനാൽ < ADB = < FEB
ഒരേ കോണുകളുള്ള രണ്ടു ത്രികോണങ്ങളുടെ തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങൾ
ജോടികളായെടുത്താൽ, മൂന്നു ജോടികളിലും

ഒരേ
സംഖ്യയായിരിക്കും. ആയതിനാൽ, \(\frac{3}{h}=\frac{x+y}{y}\)
\(\frac{h}{3}=\frac{y}{x+y}\) ….(1)

ഇതുപോലെ AABC യും AFEA യും പരിഗണിച്ചാൽ,
\(\frac{2}{h}=\frac{x+y}{x}\)
\(\frac{\mathrm{h}}{2}=\frac{\mathrm{x}}{\mathrm{x}+\mathrm{y}}\) ……(2)

(1) + (2) → \(\frac{h}{3}+\frac{h}{2}=\frac{y}{x+y}+\frac{x}{x+y}\)
h[latex]\frac{1}{3}+\frac{1}{2}[/latex] = 1
h × \(\frac{5}{2}\) = 1
h = \(\frac{6}{5}\)
∴ കയറുകൾ പരസ്പരം മുറിച്ചുകടക്കുന്നത്, നിലത്തുനിന്ന് \(\frac{6}{5}\) ഉയരത്തിലാണ്.

ii) നിലത്തുനിന്നുള്ള ഉയരം കമ്പുകളുടെ ഉയരത്തെ മാത്രം ആശ്രയിച്ചിരിക്കുന്നു. അതിനാൽ, കമ്പുകൾ തമ്മിലുള്ള അകലം എത്രയായാലും ഈ ഉയരം മാറുന്നില്ല.

iii)

∆ADB യും ∆FEB യും പരിഗണിക്കുക.
∠DAB = ∠EFB (രണ്ടും 909)
∠ADB = ∠FEB
∴ ∠ABD = ∠FBE
ഒരേ കോണുകളുള്ള രണ്ടു ത്രികോണങ്ങളുടെ തുല്യമായ കോണുകൾക്കെതിരെയുള്ള വശങ്ങൾ
ജോടികളായെടുത്താൽ, മൂന്നു ജോടികളിലും

ഒരേ
സംഖ്യയായിരിക്കും. ആയതിനാൽ, \(\)
\(\frac{h}{a}=\frac{y}{x+y}\) ………. (1)

ഇതുപോലെ ∆ABC യും ∆FEA യും പരിഗണിച്ചാൽ,
\(\frac{b}{h}=\frac{x+y}{x}\)
\(\frac{b}{h}=\frac{x}{x+y}\) …….. (2)

(1) + (2) \(\frac{h}{a}+\frac{h}{b}=\frac{y}{x+y}+\frac{x}{x+y}\)
h\(\left[\frac{1}{a}+\frac{1}{b}\right]\) = 1
\(\left[\frac{1}{a}+\frac{1}{b}\right]=\frac{1}{h}\)

Question 7.
ചിത്രത്തിൽ ABC എന്ന ത്രികോണത്തിലെ ∠A യുടെ സമഭാജിയാണ് AP:

i) ABP എന്ന ത്രികോണത്തിനും, CPO എന്ന ത്രികോണത്തിനും ഒരേ കോണുകളാണെന്നു തെളിയിക്കുക.
ii) \(\frac{B P}{P C}\) കണക്കാക്കുക.
iii) ഏതു ത്രികോണത്തിലും ഒരു കോണിന്റെ സമഭാജി എതിർവശത്തെ മുറിക്കുന്നത്, കോൺ ഉൾപ്പെടുന്ന വശങ്ങളുടെ അംശബന്ധത്തിലാണ് എന്നു തെളിയിക്കുക.
Answer:
i) ∠BAP = ∠CAP (∠A യുടെ സമഭാജിയാണ് AP)
∠APB = ∠CPQ … (1)
∆ACQ സമഭുജ ത്രികോണമായതുകൊണ്ട്, ∠CAP = ∠AQC = x
∆APB യും ∆CPQ യും പരിഗണിക്കുക.
∠BAP = ∠CQP
∠APB = ∠CPQ
∴ ∠ABP = ∠PCQ
∴ ABP എന്ന ത്രികോണത്തിനും, CPQ എന്ന ത്രികോണത്തിനും ഒരേ കോണുകളാണ്.

ii) ∆APB ലെയും ∆CPO ലെയും എല്ലാ കോണുകളും തുല്യമായതിനാൽ അവ സദൃശ
ത്രികോണങ്ങളാണ്. അതിനാൽ, \(\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{5}{3}\)

iii)

∠A യുടെ സമഭാജിയാണ് AP.
P യിൽ നിന്ന് AB ലേക്കും AC ലേക്കും ഓരോ ലംബങ്ങൾ വരയ്ക്കുക.
∆AQP യും ∆ARP യും പരിഗണിക്കുക.
∠QAP = ∠RAP (∠A യുടെ സമഭാജിയാണ് AP)
∠AQP = ∠ARP = 90°
<QPA = ∠RPA (രണ്ടു ത്രികോണങ്ങളിലെ രണ്ടു കോണുകൾ തുല്യമാണെങ്കിൽ മൂന്നാമത്തെ കോണുകളും തുല്യമായിരിക്കും.) (പൊതുവായ വശം)
AP = AP
∴ ∆AQP = ∆ARP
PQ = PR = h എന്നെടുത്താൽ,
∆ABP യുടെ പരപ്പളവ് = \(\frac{1}{2}\) × AB × h
∆ACP യുടെ പരപ്പളവ് = \(\frac{1}{2}\) × AC × h

പരപ്പളവുകളുടെ അംശബന്ധം = \(\frac{1}{2}\) × AB × h: \(\frac{1}{2}\) × AC × h
= AB: AC
BP: PC = ∆ABP യുടെ പരപ്പളവ് – ∆APC യുടെ പരപ്പളവ്
= \(\frac{1}{2}\)AB × h : \(\frac{1}{2}\) × AC × h
= AB: AC
∴ ഏതു ത്രികോണത്തിലും ഒരു കോണിന്റെ സമഭാജി എതിർവശത്തെ മുറിക്കുന്നത്, കോൺ ഉൾപ്പെടുന്ന വശങ്ങളുടെ അംശബന്ധത്തിലാണ്.

Question 8.
ചുവടെ വരച്ചിരിക്കുന്ന ത്രികോണത്തിന്റെ അതേ കോണുകളും, വശങ്ങളുടെ നീളം 1- മടങ്ങുമായ ത്രികോണം വരയ്ക്കുക.

Answer:
“കോണുകൾ മാറാതെ ഒരു ത്രികോണം ചെറുതോ വലുതോ ആക്കി മാറ്റാൻ, കോണുകൾ അളക്കണമെന്നില്ല; വശങ്ങൾ ഒരേ തോതിൽ മാറ്റിയാൽ മതി” എന്ന് നമുക്കറിയാം.
8 ന്റെ 1\(\frac{1}{4}\) മടങ്ങ് = 8 × 1\(\frac{1}{4}\) = 8 × \(\frac{5}{4}\) = 10 സെ.മീ
4 ന്റെ 1\(\frac{1}{4}\) മടങ്ങ് = 4 × 1\(\frac{1}{4}\) = 4 × \(\frac{5}{4}\) = 5 സെ.മീ

∆ABC വരയ്ക്കുക. ഇതിൽ AB = 8 സെ.മീ , BC = 6 സെ.മീ, AC = 4 സെ.മീ ആയിരിക്കും
AC യുടെ നീളം 1 സെ.മീ, AB യുടെ നീളം 2 സെ.മീ വീതം കൂട്ടി വരച്ച് D, E അടയാളപ്പെടുത്തുക.
∆ADE വരയ്ക്കുക.

Question 9.
ഒരു ചതുർഭുജത്തിന്റെ ചിത്രം ചുവടെയുണ്ട്.

i) ഇതേ കോണുകളും, വശങ്ങളുടെ നീളമെല്ലാം 1\(\frac{1}{2}\) മടങ്ങുമായ ചതുർഭുജം വരയ്ക്കുക.
ii) കോണുകൾ വ്യത്യസ്തവും, വശങ്ങളുടെയെല്ലാം നീളം ഇതിലെ വശങ്ങളുടെ 1\(\frac{1}{2}\) മടങ്ങുമായ ഒരു ചതുർഭുജം വരയ്ക്കുക.
Answer:
4 × 1\(\frac{1}{4}\) = 4 × \(\frac{3}{2}\) = 6 സെ.മീ
5 × 1\(\frac{1}{2}\) = 5 × \(\frac{3}{2}\) = 7\(\frac{1}{2}\) സെ.മീ
6 × 1\(\frac{1}{2}\) = 6 × \(\frac{3}{2}\) = 9 സെ.മീ
ചതുർഭുജം ABCD വരയ്ക്കുക. ഇതിൽ AB = 6 സെ.മീ, BC = 3 സെ.മീ, CD = 2 സെ.മീ, AD 4സെ.മീ ആയിരിക്കും
AD യുടെ നീളം 2 സെ.മീ, AC യുടെ നീളം 2.5 സെ.മീ, AB യുടെ നീളം 3 സെ.മീ വീതം കൂട്ടിവരച്ച് E, F, G അടയാളപ്പെടുത്തുക.
ചതുർഭുജം ∆GFE വരയ്ക്കുക.

ii) “വശങ്ങളുടെ നീളം മാറാതെ കോണുകൾ വ്യത്യസ്തമായ ചതുർഭുജം വരക്കുന്നതിന് അതിന്റെ വികർണ്ണത്തിന്റെ നീളം മാറ്റിയാൽ മതി”.

4 ന്റെ 1\(\frac{1}{2}\) മടങ്ങ് = 4 × 1\(\frac{1}{2}\) = 4 × \(\frac{3}{2}\) = 6 സെ.മീ
2 ന്റെ 1\(\frac{1}{2}\) മടങ്ങ് = 2 × 1\(\frac{1}{2}\) = 2 × \(\frac{3}{2}\) = 3 സെ.മീ
3 ന്റെ 1\(\frac{1}{2}\) മടങ്ങ് = 3 × 1\(\frac{1}{2}\) = 3 × \(\frac{3}{2}\) = \(\frac{9}{2}\) = 4.5 സെ.മീ
6 ന്റെ 1\(\frac{1}{2}\) മടങ്ങ് = 6 × 1\(\frac{1}{2}\) = 6 × 2\(\frac{3}{2}\) = 9 സെ.മീ
വികർണ്ണത്തിന്റെ നീളം = 8.5 സെ.മീ

Question 10.
ഒരു ത്രികോണത്തിന്റെ പരപ്പളവ് 6 ചതുരശ്രസെന്റിമീറ്റർ. ഈ ത്രികോണത്തിന്റെ ഓരോ വശത്തിന്റെയും നാല് മടങ്ങ് വശമായിട്ടുള്ള ത്രികോണത്തിന്റെ പരപ്പളവ് എത്രയാണ്? ഓരോ വശവും പകുതിയാണെങ്കിലോ?
Answer:
“ഒരു ത്രികോണത്തിന്റെ വശങ്ങളെല്ലാം ഒരേ തോതിൽ വലുതാക്കുകയോ ചെറുതാക്കുകയോ ചെയ്താൽ, പരപ്പളവ് മാറുന്ന തോത്, വശങ്ങൾ മാറുന്ന തോതിന്റെ വർഗമാണ്” എന്ന് നമുക്കറിയാം. വശങ്ങൾ 4 മടങ്ങാക്കിയാൽ:
വശങ്ങൾ മാറുന്ന തോത് = 4
പരപ്പളവ് മാറുന്ന തോത് = 4² = 16
ആദ്യത്തെ പരപ്പളവ് = 6 ച.സെ.മീ
രണ്ടാമത്തെ പരപ്പളവ് = 6 × 16 = 96 ച.സെ.മീ

വശങ്ങൾ പകുതിയാക്കിയാൽ:
വശങ്ങൾ മാറുന്ന തോത് = \(\frac{1}{2}\)
പരപ്പളവ് മാറുന്ന തോത് = (\(\frac{1}{2}\))² = \(\frac{1}{4}\)
ആദ്യത്തെ പരപ്പളവ് = 6 ച.സെ.മീ
രണ്ടാമത്തെ പരപ്പളവ് = 6 × \(\frac{1}{4}=\frac{3}{2}\) ച.സെ.മീ

പാഠപുസ്തകത്തിലെ ചോദ്യോത്തരങ്ങൾ

Question 11.
രണ്ടു മട്ടത്രികോണങ്ങളുടെ ലംബ വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാണെങ്കിൽ, അവയുടെ കർണ്ണങ്ങളിലെ മാറ്റവും ഇതേ തോതിലാണെന്നു തെളിയിക്കുക.
Answer:
രണ്ടു മട്ടത്രികോണങ്ങളുടെ ലംബ വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാണെന്ന് തന്നിരിക്കുന്നു. രണ്ടു മട്ടത്രികോണങ്ങളിലും ലംബവശങ്ങൾക്കിടയിലുള്ള കോൺ തുല്യമാണ് (രണ്ടും 90).
“രണ്ടു വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലും, അവയുടെ ഇടയിലെ കോണുകൾ തുല്യവും ആയ ത്രികോണങ്ങളിൽ മൂന്നാം വശങ്ങളിലെ മാറ്റവും ഇതേ തോതിലാണ്” എന്ന് നമുക്കറിയാം.
ഇവിടെ മൂന്നാമത്തെ വശം കർണ്ണമാണ്. ആയതിനാൽ, രണ്ടു മട്ടത്രികോണങ്ങളുടെ ലംബ വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാണെങ്കിൽ, അവയുടെ കർണ്ണങ്ങളിലെ മാറ്റവും ഇതേ തോതിലാണ്.

Question 12.
രണ്ടു മട്ടത്രികോണങ്ങളുടെ ഏതെങ്കിലും രണ്ടു വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാണെങ്കിൽ, മൂന്നാമത്തെ വശങ്ങളിലെ മാറ്റവും ഇതേ തോതിലാണെന്നു തെളിയിക്കുക.
Answer:

AC, PR എന്നീ വശങ്ങളിലെ മാറ്റവും BC, QR എന്നീ വശങ്ങളിലെ മാറ്റവും ഒരേ തോതിലാണെന്ന് സങ്കൽപ്പിക്കുക.
⇒ \(\frac{A C}{P R}=\frac{B C}{Q R}\)
\(\frac{A C}{P R}\) = k എന്നും \(\frac{B C}{Q R}\) = k എന്നും എടുത്താൽ,
AC = kPR എന്നും BC = kQR എന്നും കിട്ടും.
പൈഥാഗറസ് സിദ്ധാന്തമനുസരിച്ച്,
AB² = AC² – BC²
AB = \(\sqrt{\mathrm{AC}^2-\mathrm{BC}^2}\)
= \(\sqrt{(k P R)^2-(k Q R)^2}\)
= \(\sqrt{k^2\left(P R^2-Q R^2\right)}\)
= k\(\sqrt{P R^2-Q R^2}\)
= k\(\sqrt{P Q^2}\)
= KPQ
⇒ \(\frac{A B}{P Q}\) = k
⇒ രണ്ടു മട്ടത്രികോണങ്ങളുടെ ഏതെങ്കിലും രണ്ടു വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാണെങ്കിൽ, മൂന്നാമത്തെ വശങ്ങളിലെ മാറ്റവും ഇതേ തോതിലാണ്.

Question 13.
ഒരു ത്രികോണം വരച്ച്, അതിനുള്ളിൽ ഒരു കുത്തിടുക. ത്രികോണത്തിന്റെ മൂലകൾ ഈ കുത്തുമായി യോജിപ്പിച്ചു വരയ്ക്കുക. ഈ വരകളോരോന്നും അവയുടെ പകുതി കൂടി പുറത്തേക്ക് നീട്ടി, അറ്റങ്ങൾ യോജിപ്പിക്കുക:

ഇങ്ങനെ കിട്ടിയ വലിയ ത്രികോണത്തിന്റെ വശങ്ങളെല്ലാം, ആദ്യത്തെ ത്രികോണത്തിന്റെ വശങ്ങളുടെ ഒന്നര മടങ്ങാണെന്നു തെളിയിക്കുക.
Answer:
ചോദ്യത്തിൽ തന്നിരിക്കുന്ന വിവരങ്ങൾ എല്ലാം സംയോജിപ്പിച്ചാൽ നമുക്ക് താഴെ കാണിച്ചിരിക്കുന്ന ചിത്രം ലഭിക്കും.

∆BDC യും ∆QDR യും പരിഗണിക്കുക.
രണ്ടു ത്രികോണത്തിലും ∠D പൊതുവാണ്.
∆BDC ൽ ∠D ക്ക് ഇടത്തുള്ള വശം = BD = 2x
∆QDR ൽ ∠D ക്ക് ഇടത്തുള്ള വശം = QD = 3x
\(\frac{Q D}{B D}=\frac{2 x}{3 x}=\frac{3}{2}\)

∆BDC ൽ LD ക്ക് വലത്തുള്ള വശം = CD = 2y
∆QDR ൽ LD ക്ക് വലത്തുള്ള വശം = RD = 3y
\(\frac{R D}{C D}=\frac{2 y}{3 y}=\frac{3}{2}\)

രണ്ടു വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാകുകയും, അവയുടെയിടയിൽ ഒരേ കോൺ ആയിരിക്കുകയും ചെയ്താൽ അവ സദൃശ ത്രികോണങ്ങളാണെന്ന് നമുക്കറിയാം. അതിനാൽ, ABDC യും AQDR യും സദൃശ ത്രികോണങ്ങളാണ്. അതിനാൽ,
\(\frac{Q R}{B C}=\frac{3}{2}\) … (1)

ഇതുപോലെ, ACAD, ARPD സദൃശ ത്രികോണങ്ങളാണെന്ന് കാണിക്കാം. അതിനാൽ,
\(\frac{\mathrm{RP}}{\mathrm{CA}}=\frac{3}{2}\) … (2)

ഇതുപോലെ, AABD, APOD സദൃശ ത്രികോണങ്ങളാണെന്ന് കാണിക്കാം. അതിനാൽ,
\(\frac{P Q}{A B}=\frac{3}{2}\)… (3)
(1), (2), (3) = വലിയ ത്രികോണത്തിന്റെ വശങ്ങളെല്ലാം, ആദ്യത്തെ ത്രികോണത്തിന്റെ വശങ്ങളുടെ ഒന്നര മടങ്ങാണ്.

Question 14.
ഒരു ചതുർഭുജത്തിനകത്തെ ഒരു ബിന്ദുവും ചതുർഭുജത്തിന്റെ മൂലകളും യോജിപ്പിക്കുന്ന വരകൾ, ഒരേ തോതിൽ പുറത്തേക്കു നീട്ടുന്നു; ഈ വരകളുടെ അറ്റങ്ങൾ യോജിപ്പിച്ച് മറ്റൊരു ചതുർഭുജമുണ്ടാക്കുന്നു.

(i) വലിയ ചതുർഭുജത്തിന്റെ വശങ്ങൾ, ചെറിയ ചതുർഭുജത്തിന്റെ വശങ്ങളെ ഒരേ തോതിൽ വലുതാക്കിയതാണെന്നു തെളിയിക്കുക.
(ii) രണ്ടു ചതുർഭുജങ്ങൾക്കും ഒരേ കോണുകളാണെന്നു തെളിയിക്കുക.
Answer:
(i)

\(\frac{D S}{O D}=\frac{C R}{O C}=\frac{B Q}{O B}=\frac{A P}{O A}\) = k
⇒ OS = (1 + k)OD
OP = (1 + k)OA
OQ = (1 + k)OB
OR = (1 + k)OC
∆OAB യും ∆OPQ യും പരിഗണിക്കുക.
രണ്ടു ത്രികോണത്തിലും 20 പൊതുവാണ്.
∆OAB ൽ 20 ക്ക് ഇടത്തുള്ള വശം = OA
∆OPQ ൽ 20 ക്ക് ഇടത്തുള്ള വശം = OP = (1 + k)OA
\(\frac{O P}{O A}=\frac{(1+k) O A}{O A}\) = 1 + k

∆OAB ൽ 20 ക്ക് വലത്തുള്ള വശം = OB
∆OPQ ൽ 20 ക്ക് വലത്തുള്ള വശം OQ = (1 + k)OB
\(\frac{O Q}{O B}=\frac{(1+k) O B}{O B}\) = 1 + k

രണ്ടു വശങ്ങളിലെ മാറ്റം ഒരേ തോതിലാകുകയും, അവയുടെയിടയിൽ ഒരേ ആയിരിക്കുകയും ചെയ്താൽ അവ സദൃശ ത്രികോണങ്ങളാണെന്ന് നമുക്കറിയാം. അതിനാൽ,

∆OAB യും ∆OPQ യും സദൃശ ത്രികോണങ്ങളാണ്. … (1)
⇒ \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = 1 + k
ഇതുപോലെ,

∆OBC യും ∆OQR യും സദൃശ ത്രികോണങ്ങളാണ്. … (2)
⇒ \(\frac{B C}{Q R}\) = 1 + k

∆OCD യും ∆ORS യും സദൃശ ത്രികോണങ്ങളാണ്. … (3)
⇒ \(\frac{C D}{R S}\) = 1 + k

∆ODA യും ∆OSP യും സദൃശ ത്രികോണങ്ങളാണ്. … (4)
⇒ \(\frac{\mathrm{DA}}{\mathrm{SP}}\) = 1 + k
∴ വലിയ ചതുർഭുജത്തിന്റെ വശങ്ങൾ, ചെറിയ ചതുർഭുജത്തിന്റെ വശങ്ങളെ ഒരേ തോതിൽ വലുതാക്കിയതാണ്.

(ii) (1), (2), (3), (4) ⇒ രണ്ടു ചതുർഭുജങ്ങൾക്കും ഒരേ കോണുകളാണ്.

Similar Triangles Class 9 Extra Questions and Answers Malayalam Medium

Question 1.
ചിത്രത്തിൽ ∠Q = 90, QR = 5 സെ.മീ, SR = 3 സെ.മീ ആകുന്നു. QS ഉം PR ഉം പരസ്പരം ലംബമാണ്.

a) QS ന്റെ നീളമെത്ര?
b) PS ന്റെ നീളമെത്ര?
Answer:
a) ∆SQR മട്ടത്രികോണമാണ്.
QS² = QR² – SR²
= 5² – 3²
= 25 – 9
= 16
∴ QS = √16 = 4 സെ.മീ

b) PS × SR = QS²
PS × 3 = 4²
PS = \(\frac{16}{3}\) സെ.മീ

Question 2.
6 സെന്റീമീറ്റർ വശമുള്ള ഒരു സമപാർശ്വ ത്രികോണം വരയ്ക്കുക. ഇതിന്റെ ഒന്നരമടങ്ങ് വരുന്ന മറ്റൊരു ത്രികോണം വരയ്ക്കുക.
Answer:
6 ന്റെ ഒന്നരമടങ്ങ് = 6 × \(\frac{3}{2}\) = 9

Question 3.
ചിത്രത്തിൽ POR, QST മട്ട ത്രികോണങ്ങളാണ്. അങ്ങനെയാണെങ്കിൽ QR × QS = QP × QT എന്ന് തെളിയിക്കുക.

Answer:
∆PQR യും ∆QST യും സദൃശ ത്രികോണങ്ങളാണ്.
⇒ \(\frac{\mathrm{QP}}{\mathrm{QS}}=\frac{\mathrm{QR}}{\mathrm{QT}}\)
QP × QT = QR × QS

Question 4.
90 സെ.മീ ഉയരമുള്ള ഒരു ആൺകുട്ടി ഒരു വിളക്കുതൂണിന്റെ അടിയിൽ നിന്ന് 1.2 മീറ്റർ/സെക്കന്റ് വേഗതയിൽ നടക്കുന്നു. വിളക്ക്, ഭൂമിയിൽ നിന്ന് 3.6 m ഉയരത്തിലാണെങ്കിൽ സെക്കൻഡുകൾക്ക് ശേഷം കുട്ടിയുടെ നിഴലിന്റെ നീളം കണ്ടുപിടിക്കുക.
Answer:
ആൺകുട്ടിയുടെ ഉയരം = 90 സെ.മീ = 0.9 മീ
വിളക്കുതൂണിന്റെ ഉയരം = 3.6 മീ
ആൺകുട്ടിയുടെ വേഗം 1.2 മീറ്റർ/സെക്കന്റ്
4 സെക്കൻഡുകൾക്ക് ശേഷം വിളക്കുതൂണിൽ നിന്നും കുട്ടിയുടെ അകലം = 1.2 × 4 = 4.8 മീറ്റർ/സെക്കന്റ്

∆DAB ഉം ∆DEC ഉം സദൃശ ത്രികോണങ്ങളാണ്. അതിനാൽ
\(\frac{A B}{E C}=\frac{B D}{C D}\)
CD = x എന്നെടുത്താൽ,
\(\frac{3.6}{0.9}=\frac{4.8+x}{x}\)
4 = \(\frac{4.8+x}{x}\)
4x = 4.8 + x
3x = 4.8
x = \(\frac{4.8}{3}\) = 1.6 മീ

Question 5.
ചിത്രത്തിൽ ABCD ഒരു ചതുരമാണ്. BC = 24 സെ.മീ, DP = 10 സെ.മീ, CD = 15 സെ.മീ ആണെങ്കിൽ, AQ ഉം CQ ഉം കണ്ടെത്തുക.

Answer:
BC = 24 സെ.മീ ⇒ AD = 24 സെ.മീ (ABCD ചതുരമായതുകൊണ്ട്)
∆APD ളം ∆CPQ ളം സദ്യശ ത്രിേകാണആദഉാണ്
⇒ \(\frac{A D}{C Q}=\frac{D P}{C P}=\frac{A P}{P Q}\) അതായത്, \(\frac{24}{C Q}=\frac{10}{5}=\frac{A P}{P Q}\)
\(\frac{24}{C Q}=\frac{10}{5}\) ⇒ CQ = \(\frac{24}{2}\) = 12 സെ.മീ
\(\frac{10}{5}=\frac{A P}{P Q}\) ⇒ AP = 2 × PQ

∆APD ഒരു മട്ടത്രികോണമാണ്.
AP² = DA² + DP²
= 24² + 10²
= 576 + 100
= 676
AP = \(\sqrt{676}\)
= 26 സെ.മീ
⇒ 2 × PQ = 26
PQ = \(\frac{26}{2}\) = 13
AQ = AP + PQ
= 26 + 13
= 39 സെ.മീ

Question 6.
ചിത്രത്തിൽ ∠A = ∠P, ∠B = <Q, AB = 5 സെ.മീ BC = 4 സെ.മീ, AC = 26 സെ.മീ, PR = 6 സെ.മീ.

(a) PQ-ൻ്റെ ദൈർഘ്യം എന്താണ്?
(b) ∆ABC, ∆PQR എന്നിവയുടെ ചുറ്റളവുകളുടെ അനുപാതം എത്രയാണ്?
Answer:
(a) \(\frac{P R}{A C}=\frac{P Q}{A B} \Rightarrow \frac{6}{2}=\frac{P Q}{5} \Rightarrow 3=\frac{P Q}{5}\) ⇒ PQ = 15 സെ.മീ.
(b) ∆ABC യുടെ ചുറ്റളവ് = 5 + 4 + 2 = 11 സെ.മീ
∆POR ന്റെ ചുറ്റളവ് = 15 + 12 + 6 = 33 സെ.മീ
ചുറ്റളവുകളുടെ അനുപാതം 11:33 = 1: 3

Question 7.
ചിത്രത്തിൽ ∠B = ∠D = 90°, AB = 15 സെ.മീ, AD = 5.6 സെ.മീ ആണ്

a) ∠DAE = 40° ആണെങ്കിൽ, ∠AED, ∠BAC കണ്ടുപിടിക്കുക.
b) ∠C കണ്ടുപിടിക്കുക.
c) \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = …………
Answer:
(a) ∠DAE = 90 – 40 = 50°
∠BAC = 40°

(b) ∠C = 50°

(c) ∆ADE and ∆ABC are similar
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{15}{5}=\frac{3}{1}\)
So,
\(\frac{\mathrm{BC}}{\mathrm{DE}}\) is also \(\frac{3}{1}\) = 3

Question 8.
താഴെ തന്നിരിക്കുന്ന ത്രികോണത്തിന്റെ അതെ കോണുകളും വശങ്ങൾ 1 മടങ്ങായതുമായ ത്രികോണം വരക്കുക.

Answer:
6 ന്റെ ഒന്നരമടങ്ങ് = 6 × \(\frac{3}{2}\) = 9
7 ന്റെ ഒന്നരമടങ്ങ് = 7 × \(\frac{3}{2}\) = \(\frac{21}{2}\) = 10.5

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 2 Equations of Motion Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 2 Notes Solutions Equations of Motion

SCERT Class 9 Physics Chapter 2 Notes Solutions Kerala Syllabus Equations of Motion Questions and Answers

Class 9 Physics Chapter 2 Let Us Assess Answers Equations of Motion

Question 1.
A car starts from rest and moves with uniform acceleration. Calculate the acceleration of the car if it covers a distance of 200 m in 20 s.
Answer:
Initial speed u = 0
distance travelled s = 200 m
time taken t = 20 sec
acceleration a = ?
Using the second equation
s = ut + \(\frac{1}{2}\) at²
⇒ 200 = (0)t + \(\frac{1}{2}\) a(20)²
a = 200/200 = 1 m s^-2

Question 2.
If an object starts from rest and moves with an acceleration of 2 m/s², what will be the velocity of the object after 10 s?
Answer:
Given, a = 2 m/s², time = 10 s
Initial velocity u=0
According to the equation of motion:
v = u + at
=0 + 2 × 10
= 20 m/s

Question 3.
Three different graphs related to the motion of a vehicle are given below. Analyse the graphs and find the characteristics of the motion.

Answer:

• Graph 1 shows a vehicle moving with uniform acceleration.
• Graph 2 shows a vehicle moving with uniform deceleration.
• Graph 3 shows a vehicle moving with non-uniform acceleration.

Question 4.
Graph related to the motion of Car A and Car B is given.
a) Which car has more acceleration? Why?
b) Redraw the graph by changing the scale and compare the graphs.

Answer:
a) Car A

When the scale of a graph changes, it affects the appearance and readability of the graph, but the underlying data and relationships remain the same.

Question 5.
Observe the figure.

A child runs along a circular path of circumference 440 m at a constant speed. The radius of the circular path is 70 m. The time taken to run from A and reach A in clockwise direction through B is 80 s. Find the distance, displacement, speed and velocity in each case in the table.
Answer:

Question 6.
A train starts from rest and attains a speed of 72 km/h in 5 minutes. Find the acceleration and displacement of the train.
Answer:
Given,
v = 72 km/h = 20 m/s
(I km/hr = 1000/3600 (m/s))
Time (t) = 5 minutes = 300 s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{20-0}{300}\) = \(\frac{1}{15}\) = m/s
Displacement s= ut + \(\frac{1}{2}\) at²
= 0 × 300 + \(\frac{1}{2}\) × \(\frac{1}{15}\) 300²
= 3000 m

Question 7.
Analyse Table and draw the velocity-time graph.

X Time (s)	0	2	4	6	8	10
Y Velocity(m/s)	10	15	20	20	20	15

a) From the graph, find the time interval during which there is no acceleration.
b) Find the time interval during which deceleration occurs.
c) Find the displacement between the fourth second and the eighth second.
Answer:

a) 4 s to 8 s
b) 8 s to 10 s
c) Time t = 4s
Velocity = 20 m/s
Displacement = Velocity × Time = 20 × 4 = 80 m

Question 8.
Observe the graphs.

In which graph,
a) does the object have uniform acceleration?
b) does the object have uniform velocity?
c) does the object have acceleration and deceleration?
Answer:
a) Graph 2
b) Graph 1
c) Graph 3

Question 9.
The velocity-time graph of an object in straight line motion is given.

Identify the regions of the graph where the object is moving with:
a) acceleration.
b) uniform velocity.
c) deceleration.
Answer:
a) Region OA
b) Region AB
c) Region BC

Question 10.
It is not safe for pedestrians to wear dark coloured clothes at night and in conditions with dim light. The school authorities decided to choose a dark coloured uniform for your school. Record your response to the decision. Justify the answer with respect to road safety.
Answer:
Choosing a dark coloured uniform for our school raises significant safety concerns, particularly regarding pedestrian safety at night and in low- light conditions. This decision is problematic because: visibility issues, road safety. For the safety of all students, it would be better for the school to choose a uniform that includes lighter or reflective elements. This change would help ensure that students are more visible to drivers, particularly in low-light conditions, thereby reducing the risk of accidents and enhancing overall road safety.

Question 11.
In the figure given below, a child travels from P to S through Q and R and comes back straight to P.

Analyse the figure and complete the table given below.
Answer:

Starting from P	Speed	Velocity
On reaching Q	\(\frac{60}{10}\) = 6 m/s	\(\frac{60}{10}\) = 6 m/s
On reaching R	\(\frac{150}{25}\) = 6 m/s	\(\frac{125}{25}\) = 5 m/s
On reaching S	\(\frac{204}{34}\) = 6 m/s	\(\frac{136}{16}\) = 8.5 m/s
While returning to P	\(\frac{340}{50}\) = 6.8 m/s	\(\frac{0}{50}\) = 0 m/s

Question 12.
A stone is thrown vertically upwards with a velocity of 20 m/s. (a = -10 m/s²)
a) What is the maximum height that the stone has reached?
b) How far will this stone travel in 3 s after it is thrown?
c) How high will the stone be from the ground, 3 s after it is released?
Answer:
Given,
v = 0m/s; u 20 m/s; a = -10 m/s²
a) Using equation of motion, v² = u² + 2as
0 = 20² + 2(-10)h
0 = 400 – 20 h
20 h = 400
h = 400/20 = 20 m

b) s = ut + 1⁄2 at²
20 = (20×t) + 1⁄2 (-10 t²)
20 = 20 t – 5 t²
t² – 4t + 4 = 0
(t-2)² = 0
t = 2s
During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Total distance travelled by the stone in 3 s = 20 + 5 = 25 m.
c) During the first 2 seconds, it reaches the maximum height of 20 m. Then it falls for 1 second. Now the stone falls down, then u=0.
s = 1⁄2 × 10 × (1)² = 5 m
Therefore, the height from the ground after 3 seconds: 20 m – 5m = 15 m.

Question 13.
An object is moving with a speed of 40 m/s.
If it is given a deceleration of 8 m/s²,
a) how long will it take to come to rest?
b) what is the displacement of the object in this time?
Answer:
a) Here, v = 0 m/s
u = 40 m/s
a = -8 m/s
By using the equation, v = u + at
0 = 40 + (-8) t
= 40 – 8t
8t = 40
t = 40/8 = 5 s
So, it will take 5 seconds for the object to come to rest.

b)Displacement s = ut + \(\frac{1}{2}\) at²
= (40 × 5 ) + \(\frac{1}{2}\) (-8)5²
= 200 + \(\frac{1}{2}\) (-200)
=200 – 100
= 100 m

Question 14.
An object is moving with a velocity of 20 m/s. This object is given an acceleration of 5 m/s². What is the velocity when the displacement is 120 m?
Answer:
Given,
u = 20 m/s
a = 5 m/s²
s = 120 m
By using equation, v² = u² + 2as
v² = 20² + 2 × 5 × 120
v² = 400 + 1200
v² = 1600
v = √1600 = 40 m/s

Question 15.
A bullet travelling with a velocity of 60 m/s comes to rest after penetrating 2 cm into a wooden block. What is the acceleration of this bullet? How much is its deceleration?
Answer: Given,
Initial velocity (u) = 60 m/s
Final velocity (v) = 0m/s
Distance (s) = 2 cm = 0.02 m.
We know, v² = u² + 2as
0² = 60² + 2 × a × 0.02
0 = 3600 + 0.04 a
a = \(\frac{-3600}{0.04}\) = 90000 m/s²
The deceleration (since it is negative acceleration) is 90000 m/s.

Question 16.
Observe the table illustrating the motion of an object, select an appropriate scale and draw the graph. Interpret the graph and write down the answers to the questions given below.

Time (s)	0	5	10	15	20	25	30
Velocity (m/s)	20	25	30	30	30	25	20

a) Which is the time interval with no acceleration?
b) Which is the time interval with deceleration?
c) Calculate the displacement of this object in 30 s.
Answer:

a) 10s to 20s
b) 20s to 30s
c) Time t = 30 s
Velocity v = 20 m/s
Displacement-Area under velocity-time graph.
= \(\frac{1}{2}\) × (20+30) × 10 + (10×30) + \(\frac{1}{2}\) ×(20+30) × 10
= \(\frac{500}{2}\) + 300 + \(\frac{500}{2}\)
= 250 + 300 + 250
= 800 m

Question 17.
The velocity-time graph of the motion of Car A and Car B is given below.

a) Which car started first?
b) How much time did each car take to attain the same speed?
c) Which car possesses more acceleration?
d) Which car has more displacement?
Answer:
a) Car A
b) 14 s
c) Car B
d) Car A

Class 9 Physics Chapter 2 Extended Activities Answers Equations of Motion

Question 1.
The Science Club organises an awareness class on ‘Road accidents due to over speeding. Prepare the necessary slides for the presentation. (Hints: traffic rules, signboards, traffic rules to be followed by pedestrians, etc.)
Answer:
Hint:
Slide 1:
Title: Road Accidents Due to Over speeding Subtitle: Understanding the Risks and Prevention Measures
Introduction: Brief overview of the importance of road safety and the focus on over speeding.

Slide 2: Introduction
Road accidents are a major cause of injury and death.
Over speeding is a leading factor in many accidents.
Today, we’ll learn about how to stay safe on the road.

Slide 3: Importance of Traffic Rules
Traffic rules are designed to keep everyone safe.
Following these rules helps prevent accidents.

Slide 4: Common Traffic Signs
Stop Sign: Stop your vehicle completely. Speed Limit Sign: Do not exceed the speed indicated.
Pedestrian Crossing Sign: Slow down and be ready to stop for people crossing.

Slide 5: Traffic Rules for Pedestrians
Use pedestrian crossings.
Look both ways before crossing the road. Avoid crossing between parked cars. Walk on sidewalks whenever possible.

Slide 6: Tips for Safe Driving
Always stay within speed limits.
Keep a safe distance from the vehicle in front of you.
Stay alert and focused on the road.

Slide 7: Conclusion
Following traffic rules saves lives.
Over speeding is dangerous for everyone.
Stay safe and be responsible on the road.

Question 2.
An object is thrown upwards with a velocity of 30 m/s. It returns to the same position after some time. Draw the velocity-time graph of this object and display it in the class (Consider the deceleration as 10 m/s²).
Hint:
Initial velocity (u): 30 m/s (upwards).
Final velocity (v) when it comes back to the same
position: -30 m/s (downwards).
Deceleration (a): 10 m/s².
Time to reach the highest point (t): Calculate using V = u + at.
After finding the value, then take the correct scale and draw the graph similar to this.

Prepare a project on whether the safety measures implemented in your area are adequate to reduce road accidents.
Project planning can be done with the help of your teacher.
Bring relevant findings to the attention of the Road Safety Authority

(For more information, the services of the National Transportation Planning and Research Conor (NATPAC) and the Motor Vehicles Department can be availed.

The report should include:

• Introduction
• Hypothesis
• Objectives
• Methodology
• Analysis
• Results
• Conclusion
• Suggestions

Hint:
Introduction
Road accidents are a serious issue affecting our community. This project aims to evaluate whether the current safety measures in our area are sufficient to reduce road accidents.

Hypothesis
The existing safety measures in our area are adequate to significantly reduce road accidents.

Objectives

• To identify the current safety measures implemented in our area.
• To assess the effectiveness of these safety measures.
• To gather opinions from local residents and authorities.
• To make suggestions for improvement based on findings.

Methodology
Planning: Work with our teacher to create a project plan.

Data Collection:

• Conduct surveys with local residents.
• Interview local traffic authorities.
• Observe road conditions and safety measures (e.g., traffic signs, speed limits, pedestrian crossings).

Research: Use resources from NATPAC and the Motor Vehicles Department.
Analysis: Compare collected data with road safety standards.

Analysis

• Survey Results: Summarize the responses from local residents about their perception of road safety.
• Interviews: Key points from interviews with traffic authorities.
• Observations: Note the presence and condition of traffic signs, speed bumps, pedestrian crossings, etc.

Results

• Positive Findings: List of effective safety measures currently in place (e.g., well-marked pedestrian crossings, functional traffic lights).
• Areas of Concern: Identify any shortcomings (e.g., lack of speed bumps in critical areas, poor visibility of traffic signs).

Conclusion
Summarize whether the safety measures are adequate. If they are not, highlight the main issues identified.

Equations of Motion Class 9 Notes Questions and Answers Kerala Syllabus

Activity
Observe figure.

A,B,C and Dare four electric poles erected on the side of the road. The distance between any two adjacent poles is 40 m. A child starts walking from pole B, passes C and reaches D. After that, the child returns from D and reaches the pole C.

Question 1.
What is the total distance travelled by the child?
Answer:
120 m

Question 2.
What is the distance between the current position C and the initial position B of the child?
Answer:
40 m

Question 3.
If the child travels 40 m from B, which are the possible positions of the child?
Answer:
Child can reach at A or C.

Question 4.
In which direction should the child travel 40 m to reach C from B? (towards the east /towards the west)
Answer:
Towards the east
Here 40 m eastward from B to C is the displacement of the child. So, Displacement is the shortest distance.

Question 5.
Is displacement a vector or a scalar?
Answer:
Vector

Question 6.
If the child travels 40 m in the westward direction from B, the current position of the child is …………
Answer:
A
Note: If the displacement from B in the forward direction (towards the east) is considered as positive, the displacement in the backward direction (towards the west) should be considered as negative. (These can also be considered in the reverse order). Once the direction is determined, the positive and negative directions should not be changed thereafter. Here, B is the initial position and A is the final position. Hence the displacement is negative.

Question 7.
Complete table based on the child’s travel given above.
Answer:

Stages of path covered by the child	Distance covered	Displacement
Directly from B to C	40 m	40 m
Starts from B, reaches D and returns to C	120 m	40 m (from B to C)
Starts from B and reaches D	80 m	80 m (from B to D)
From B to A	40 m	40 m (from B to A)
Starts from B, reaches A and returns to B	80 m	Zero

Question 8.
The child moved from A to D and returned to A. What is the distance covered? What is the displacement? Aren’t the initial position and final position the same?
Answer:
The distance covered is 240 m and the displacement is zero because of initial position and final position are same.

Question 9.
Write down the situations in which the distance covered and displacement are equal.
Answer:

• When you travel directly from the start point to the end point.
• When there are no turns or curves in your path.
• When you move in a straight line without changing direction.

Question 10.
Two different paths taken by a child to move from position P through Q are depicted.

a) What is the distance covered in figure (a)? What is the displacement?
b) What is the distance covered as per the motion in figure (b)? What is the displacement?
c) In which situation is the magnitude of the distance and displacement equal?
Answer:
a) Distance is 100 m. Displacement is 80 m. (From P to R)
b) From P to Q, the distance is 100 m and displacement is also 100 m. (From P to S)
c) In the second case; i.e., Displacement and distance are equal only when an object moves in a straight line without changing direction.

Question 11.
Tabulate the differences between distance and displacement related to the path traversed by a person in Table.
Answer:

Distance	Displacement
Length of the path covered	Straight-lone distance
Can’t be zero	Can be zero
Scalar	Vector

Question 12.
The classrooms and some other locations of a school are depicted.

During the interval, a student from Class 9B went to the staff room. The child then proceeded to the statue of the Father of the Nation in the school garden and returned to the class via the school office. Complete Table based on the path followed by the child.
Answer:

Child’s path	Distance (m)	Displacement (m)
When the child reaches the corridor in front of the staff room from Class 9B	15 m	15 m
When the child reaches the garden near the statue of the Father of the Nation from 9B via the staff room.	85 m	60 m
When the child returns to Class 9B	180 m	0 m

A child travels from P to R via Q in 18 seconds, as shown in the figure.

Question 13.
What is the total distance covered by the child to reach R from P through Q?
Answer:
90 m

Question 14.
What is the speed of the child when the child travels from P to R through Q?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{90}{18}\)
= 5 m/s

Question 15.
What is the displacement of the child?
Answer:
72 m

Question 16.
Isn’t the child’s displacement of 72 m taking place in 18 s?
Answer:
Yes
Let’s find out the displacement in one second.
Displacement in one second = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{72 m}{18 s}\) = 4 m/s
The displacement in unit time is the velocity.

Question 17.
What is its direction?
(P → R/R → P/P → Q → R)
Answer:
P → R
So the direction of displacement and velocity are same.

Question 18.
Now, let’s consider the doubt raised by the child at the beginning of this lesson. A train of length 200 m travels with a velocity of 20 m/s. What is the time taken by this train to cross a straight bridge of length 800 m?
Answer:

Displacement = Length of the bridge + length of the rain
s = 800 m + 200 m = 1000 m
Velocity (v) = 20 m/s
Time (t) = ?
t = \(\frac{s}{v}\) = \(\frac{1000 m}{20 m/ s}\)
Time = 50 s

Question 19.
Calculate the speed and velocity of the child from P to Q in figure c. What inferences can you draw from this?
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
Velocity = \(\frac{\text { Displacement }}{\text { Time}}\)
= \(\frac{40}{8}\) = 5 m/s.
This shows that the speed and velocity is same when it travels through a straight line.

Question 20.
A vehicle travels along a straight line with a velocity of 25 m/s and covers a distance of 400 m. Calculate the time taken for this.
Answer:
Given,
Velocity = 25 m/s
Distance = 400m
Here Speed and Velocity is same (along a straight line)
So, Speed = \(\frac{\text { Distance }}{\text { Time}}\)
Time = \(\frac{\text { Distance }}{\text { Speed}}\)
= \(\frac{400}{25}\) = 16 s.

Question 21.
What is the displacement of an object moving with a velocity of 36 m/s in one minute?
Answer:
Given,
Velocity = 36 m/s
Time = 1 minute = 60 s
Displacement = Velocity × Time
=36 × 60 = 2160 m

Question 22.
Is the velocity of CarA always the same? Why?
Answer:
Yes. Because it travels equal distance in equal interval of time.

Question 23.
What about the velocity of Car B? Why?
Answer:
Not same. Because it travels unequal distance in equal interval of time.

Question 24.
Haven’t you noticed the mud sticking to the tyres of vehicles being thrown off when they rotate? Does the mud splash in the same direction every time?
Answer:
No, the mud does not splash in the same direction every time. It is thrown off tangentially due to the rotation of the tires.

Question 25.
Isn’t the direction of motion of an object moving along a circular path always changing?
Answer:
Yes, the direction of motion of an object moving along a circular path is always changing.

Question 26.
Is the velocity of Car C the same every second? Does the velocity change?
Answer:
Even though the magnitude of the speed does not change, the velocity changes because the direction changes.

Question 27.
Complete the table based on the information in figures (1), (2), (3).

Answer:

Question 28.
Classify the situations given below as uniform velocity and non-uniform velocity. Record it in the science diary.

• Motion of a stone dropped from a height
• When light travels through vacuum
• A bus starts from a bus stop and is moving forward
• A train travelling at a uniform speed in the same direction
• Swinging on a swing

Answer:

Uniform velocity	Non-uniform velocity
When light travels through vacuum A train travelling at a uniform speed in the same direction	Motion of a stone dropped from a height A bus starts from a bus stop and is moving forward Swinging on a swing

Question 29.
The velocity of a bus starting from a bus stop keeps changing. Will the change in velocity be the same in each second?
Answer:
No.

The data related to the straight line motion of the bus is given below. Analyse the information and answer the questions.

Question 30.
Imagine that you are sifting in a bus. When the bus starts and moves forward in a straight line, doesn’t the velocity change?
Answer:
Yes

Question 31.
When the bus travels from A to B, the velocity at A is……………
(initial velocity / final velocity)
Answer:
Initial velocity.

Question 32.
The velocity at B is……………(initial velocity / final velocity)
Answer:
Final velocity.

Question 33.
While considering the motion from B to C, velocity at B is……………………
Answer:
Initial velocity.

Question 34.
Complete the table using the data of the motion of the bus.

Answer:

Question 35.
A car is moving along a straight road with a velocity of 10 m/s. It is given an acceleration of 5 m/s². Calculate the velocity of the car after 2s.
Answer:
Initial velocity u = 10 m/s
Acceleration a = 5 m/s²
time t = 2s
Final velocity v = ?
a = \(\frac{\text { v-u }}{\text { t}}\)
V – u = at
V = u + at = 10 + 5 × 2 = 20 m/s
To calculate the final velocity, we can use the equation v = u + at

Question 36.
The velocity of an object changes from 4 m/s to 28 m/s in 4 s. Calculate the acceleration.
Answer: Given,
Initial velocity (u) = 4 m/s
Final velocity (y) = 28 m/s
Time (t) = 4s
Acceleration = \(\frac{\text { change in velocity }}{\text { t}}\)
a = \(\frac{\text { v-u }}{\text { t}}\)
a = \(\frac{28-4}{4}\) = 6m/s²

Question 37.
Have a look at the scene in an amusement park. List the instances in which acceleration occurs.
Answer:
Motion of a giant wheel
Roller Coasters
Swing Rides: When the swings begin to move, accelerate outward, decelerate back inward.

Question 38.
Find instances of acceleration in your daily life and record them in the science diary.
Answer:

• Motion of a coconut falling from a coconut tree
• A car speeding up
• A bicycle starting from rest
• An airplane taking off
• An elevator starting to ascend or descend

Question 39.
Can you think of some instances ¡n everyday life where the velocity decreases? Expand the list by giving more examples.
Answer:

• Train arrivingata station
• The upward motion of a stone thrown upwards
• A car approaching a red traffic light
• A ball rolling up hill
• A parachutist landing

Question 40.
Observe the figure, and complete the table.


Answer:

Question 41.
Should negative sign be given while writing the value of retardation?
Answer:
Yes, a negative sign should be given while writing the value of retardation, as it indicates a decrease in velocity.

Question 42.
An object starts from rest and attains a velocity of 10 m/s in 5 s.
a) What is its acceleration?
b) What is the acceleration if it comes to rest in 5 s? What is the retardation?
Answer:
Given,
a) Initial velocity (u)=0 m/s
Final velocity (v) = 10 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 10-0 }}{\text { 5}}\) = 2 m/s²

b) Here,
initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s
Time (t) = 5 s
Acceleration (a) = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 0-10 }}{\text { 5}}\) = -2 m/s²
Retardation is simply the magnitude ofthe negative acceleration.
Thus, retardation = 2 m/s²

Question 43.
A vehicle travelling at a speed of 5 m/s is brought to. rest ¡n 2 s by applying brakes. Calculate the retardation of the vehicle.
Answer:
Initial velocity u = 5 m/s
Final velocity v = 0
Time taken = 2 s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\)
= \(\frac{\text { 0-5 }}{\text {2}}\) = -2.5 m/s²
Thus, retardation = 2.5 m/s²

Question 44.
If the velocity of an object in the 2nd second is 40 m/s and 30 m/s in the 4th second, what is its acceleration? What is its retardation? What is its velocity at the 8th second?
Answer:
Velocity at the 2^nd second u = 40 m/s
Velocity at the 4^th second v = 30 m/s
Time (t) = 2s
Acceleration a = \(\frac{\text { v-u }}{\text { t}}\) = \(\frac{\text { 30 – 40 }}{\text {2}}\) = -5 m/s²
So retardation is 5 m/s².
To find velocity at 8^th second.
Initiai velocity (u) = 40 m/s
Time (t) = 6 s
So, a = \(\frac{\text { v-u }}{\text { t}}\)
V = u + at
= 40 + (-5 × 6)
=40 – 30 = 10 m/s.

Question 45.
Was the acceleration obtained as per Table 1 the same on each occasion?
Answer:
Yes

Question 46.
What about the acceleration obtained as per Table 2?
Answer:
Not same and also have negative sign.
You have learned about speed, velocity, and acceleration. Over speeding of vehicles can cause accidents. We must follow traffic rules to reduce accidents. However, accidents aren’t caused only by over speeding. Pedestrians should also follow traffic rules.

Question 47.
Which are the traffic rules for pedestrians to follow?
Answer:

• Pedestrians should walk along the right side of the road.
• Cross the road only at the zebra crossing, obeying the traffic signal.
• Use sidewalks
• Avoid crossing between parked cars

Question 48.
In addition to the signs mentioned above, collect more symbols for each category, prepare separate posters and display them on the school bulletin board.
Answer:
Hint: An example shown below.

Question 49.

What information can be gathered from the above graph?
Answer:
From 2020 to 2022 there is a significant spike in accidents. The highest number of accidents happened in 2022. From 2019 to 2020 the number of accidents decreased.

Question 50.
List down your findings.
Answer:
The number of road accidents is generally increasing, this might indicate issues with traffic management, increasing vehicle numbers, or lack of road safety measures. Significant drops in 2019 to 2020 could be attributed to the COVID-19 pandemic, which resulted in lockdowns and reduced road traffic.

Question 51.
In which year is the number of accidents the least?
Answer:
2020

Question 52.
How many accidents occurred in 2019?
Answer:
43000

Question 53.
What is the nature of the graph obtained?
Answer:
An inclined straight line

Question 54.
By what name is this graph known?
Answer:
Position-Time graph

Question 55.
From the shape of the graph obtained, what is the nature of the velocity of the object?
Answer:
Uniform velocity

Question 56.
What is the displacement of the object in 5 s?
Answer:
2.5 m

Question 57.
What is the time taken to travel 1.5 m?
Answer:
3 s
Let’s consider another situation.
Position-Time graph of a moving car.

Question 58.
How can we find the velocity of the car from A to B from above graph?
Answer:
Here,
Displacement=30 m, Time=6 s
Velocity = Displacement/Time = \(\frac{30}{6}\) = 5 m/s

Question 59.
What is the displacement of the car from A to B in the graph?
Answer:
30 m

Question 60.
What is the time taken by the car to travel from A to B?
Answer:
6 s

Question 61.
What is the position of the car at the sixth second in the graph?
Answer:
30 m

Question 62.
Which type of velocity does this car have?
Answer:
Uniform velocity

Question 63.
Find out the velocity of the car between 6 s and 8 s from the graph.
Answer:
Here,
Displacement = 10m
Time = 2 m
Velocity = Displacement/ Time
= \(\frac{10}{2}\) = 5 m/s

A velocity-time graph is a graph that plots velocity on the Y-axis and time on the X-axis.

Question 64.
From the graph, find the displacement of the vehicle between the fourth and tenth second.
Answer:
Velocity = Displacement/Time
Then, displacement = velocity × time
On the graph, it will be equal to AB × AD (Equal to the area of the rectangle ABCD)
Displacement = 40 × 6 = 240 m

Question 65.
Isn’t this equal to the area of the portion below BC on the graph?
Answer:
Yes.
Area = 40 × 6 = 240 m

Question 66.
Find the displacement during the first 4 s from graph.
Answer:
displacement Area of triangle AOB
= \(\frac{1}{2}\) × 40 × 4 = \(\frac{160}{2}\) = 80 m

Question 67.
What is the change in velocity in the first 4 s? What is the acceleration?
Answer:
Initial velocity (u) = 0 m/s
Final velocity (v) = 40 m/s
Change in velocity = v – u – 0 – 40 = -40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{-40}{4}\) = -10 m/s²

Question 68.
What is the acceleration of this vehicle between 4 s and 10 s?
Answer:
Initial velocity (u) = 40 m/s
Final velocity (v) = 40 m/s
Acceleration a = \(\frac{v-u}{t}\) = \(\frac{40-40}{6}\) = 0 m/s²

Question 69.
A body starts from rest and acquires a velocity of 20 m/s in 2 s and 40 m/s in 6 s. What is the displacement of the object during this time interval?
Answer:
Initial velocity, u = 20 m/s
Time t = t₂ – t₁ = 6s – 2s = 4 s
Final velocity, v = 40 m/s
Acceleration, a = \(\frac{v-u}{t}\)
= \(\frac{40 \mathrm{~m} / \mathrm{s}-20 \mathrm{~m} / \mathrm{s}}{4 \mathrm{~s}}\)
= 5 m/s²
Displacement s = ut + \(\frac{1}{2}\) at²
= (20 m/s × 4 s) + [\(\frac{1}{2}\) × 5 m/s² (4 s)²]
= 80 m + 40 m
=120 m

Question 70.
If the velocity of a car increases from 6 m/s to 16 m/s in 10 s.
a) calculate the acceleration of the car.
b) what is the displacement of the car during this time?
Answer:
a) u = 6 m/s
v = 16 m/s
t = 10 s
Acceleration a = \(\frac{v-u}{t}\)
= \(\frac{16 \mathrm{~m} / \mathrm{s}-6 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= \(\frac{10 \mathrm{~m} / \mathrm{s}}{10 \mathrm{~s}}\)
= 1 m/s²

b) s = ut + \(\frac{1}{2}\) at²
=(6 m/s × 10 s) + [\(\frac{1}{2}\) × 1 m/s² (10 s)²]
=60 m + 50 m
= 110 m

Question 71.
The velocity of a train that started from a railway station becomes 90 km/h in 10 minutes. Calculate the acceleration of the train.
Answer:

Question 72.
An object falls down from rest and moves with an acceleration 10 m/s, hits the ground with a velocity 20 m/s. From what height does the object fall?
Answer:
u = 0
a = 10 m/s²
v = 20 m/s
v² = u² + 2as
(20 m/s)² = 0² + 2 × 10 × s
400 = 20 × s
s = \(\frac{400}{20}\) = 20 m

Students rely on Kerala Syllabus 9th Standard Chemistry Textbook Solutions Chapter 7 Non Metals Notes Questions and Answers English Medium to help self-study at home.

Kerala SCERT Class 9 Chemistry Chapter 7 Solutions Non Metals

Kerala Syllabus Std 9 Chemistry Chapter 7 Non Metals Notes Solutions Questions and Answers

Class 9 Chemistry Chapter 7 Let Us Assess Answers Non Metals

Question 1.
Some gases are given in the following box. Choose suitable answers from the box and answer
the questions below.
Hydrogen, Chlorine, Oxygen, Nitrogen
a. Which gas is formed in the thermal decomposition of KMn04?
b. Which gas is used for the purification of water?
c. Which element is essential for the growth of plants?
d. Which is the inflammable gas obtained during the electrolysis of water?
Answer:
a. Oxygen.
b. Chlorine.
c. Nitrogen.
d. Hydrogen.

Question 2.
Find examples of allotropes of carbon and complete the table.

Crystalline allotropes	Amorphous allotropes

Answer:

Crystalline allotropes	Amorphous allotropes
Diamond, Graphite, Fullerenes, Graphene	Charcoal, Carbon Black, Soot, Lampblack

Question 3.
An experiment is given to identify chloride salt. Write down the observations and inferences.

Experiment	Observation	Inference
i. Add AgNO3 solution to the given salt solution.
ii. Add NH4OH solution to the precipitate formed

Answer:

Experiment	Observation	Inference
i. Add AgNO3 solution to the given salt solution.	A white precipitate forms.	The formation of this white precipitate indicates the presence of Cl– ions in the original salt solution.
ii. Add NH4OH solution to the precipitate formed	The white precipitate of AgCl dissolves in the NH4 OH solution, forming a colouriess solution.	This solubility of AgCl in NH4OH is a characteristic property of AgCl. It confirms the presence of Cl– ions in the original salt solution.

Question 4.
Find the suitable gas for each of the following situations. (Chlorine, nitrogen, CFC, oxygen)
• As an oxidising agent in rocket fuels.
• Depletion of ozone.
• To inflate the tyres of vehicles.
Answer:
As an oxidising agent in rocket fuels – Oxygen
Depletion of ozone – CFC, Chlorine
To inflate the tyres of vehicles – Nitrogen

Question 5.
Potassium permanganate is heated in a boiling tube.
a. A burning matchstick is inserted into the mouth of the boiling tube. Write down your observation.
b. Which gas is produced?
Answer:
a. Observation: When a burning matchstick is inserted into the mouth of the boiling tube containing heated potassium permanganate, the matchstick will reignite with a bright flame.

b. The gas produced during the decomposition of potassium permanganate is oxygen. The oxygen gas supports combustion, hence the reignite of the matchstick.

Question 6.
Write the answers to the following questions on the preparation of chlorine gas in the laboratory.
a. What are the chemicals required to prepare chlorine gas?
b. Chlorine gas is collected by passing it through water. Why?
c. Chlorine gas is passed through concentrated sulphuric acid. Why?
Answer:
a. Potassium permanganate (KMnO₄)

Concentrated hydrochloric acid (HCl)

b. To remove the HCl vapours.

c. Chlorine gas is passed through concentrated sulphuric acid to remove any water vapour that may have been carried over from the reaction mixture. Concentrated sulphuric acid is a dehydrating agent, which means it can absorb water from other substances.

Question 7.
The bleaching action of chlorine requires moisture. Give reason
Answer:
Chlorine gas reacts with water to form hypochlorous acid.

• Hypochlorous acid is the active bleaching agent.
• The reaction between chlorine gas and water requires moisture.
• Without moisture, chlorine gas will not have any bleaching action.

Question 8.
“Chemical fertilisers must be banned completely, and organic fertilisers must be promoted.” What is your opinion about this argument? Justify your answer.
Answer:
Reasons to Ban:

• Environmental Health: Chemical fertilisers can harm soil and water, so reducing them can help ecosystems.
• Sustainability: Organic fertilisers are often better for long-term soil health
• Human Health: Less chemical use means fewer harmful residues in food and water.

Reasons Against Ban:

• Food Production: Chemical fertilisers help increase crop yields, which is vital for feeding a growing population.
• Farmer Struggles: Farmers who rely on chemicals may have a hard time switching to organic
  methods without support.

Question 9.
a. What are the chemicals required for preparing hydrogen in the laboratory? b. How will you identify that the gas obtained is hydrogen?
Answer:
a. Zinc granules: A reactive metal that reacts with acids to produce hydrogen.
Dilute hydrochloric acid: The acid reacts with the zinc to produce hydrogen gas
Zn + 2HCl → ZnCl₂ + H₂

b. Hydrogen gas can be identified by a simple test known as the “pop” test.
Collect the gas produced in a test tube.
Bring a lighted splint near the mouth of the test tube.
Observe the reaction: If the gas is hydrogen, it will ignite with a “pop” sound

Question 10.
a. What are the merits of hydrogen when it is used as a fuel?
b. What are the isotopes of hydrogen?
c. Which among these is an isotope without a neutron?
Answer:
a. High calorific value.
Clean combustion.
Abundant availability.
Possibility of pollution is low.

b. Hydrogen has three isotopes:
Protium: The most common isotope, with one proton and no neutrons in its nucleus.
Deuterium: An, isotope with one proton and one neutron in its nucleus.
Tritium: An isotope with one proton and two neutrons in its nucleus.

c. Protium

Question 11.
a. What is meant by calorific value?
b. Some fuel is given in the box. Which has the highest calorific value?
Petrol, Coal, Ethanol, Hydrogen, Methanol
c. Hydrogen is not used as a domestic fuel. What are the reasons for this limitation?
Answer:
a. The calorific value of a fuel is the heat energy released from unit mass of the fuel on complete combustion.

b. Hydrogen.

c. Hydrogen is a gas that bums with an explosion.
It is difficult to store and distribute hydrogen.
Safety concerns.
High Cost.
Technological limitations.

Question 12.
One of the crystalline forms of carbon is diamond. Diamond does not conduct electricjty. Why?
Answer:
In diamond, each carbon atom is covalently bonded to four other carbon atoms surrounding it. This
strong bond is responsible for the hardness of the diamond. Due to the absence of free electrons in this crystalline structure, diamond does not conduct electricity.

Question 13.
a. Which allotrope of carbon is a conductor of electricity?
b. Give a reason for this.
c. Write down any two uses of this allotrope.
Answer:
a. Graphite

b. The reason for graphite’s electrical conductivity lies in its unique structure:
Layered Structure: Graphite has a layered structure, where each layer consists of carbon atoms arranged in hexagonal rings.

Free Electrons: Within each layer, one of the carbon atom’s valence electrons is delocalised and can move freely between the hexagonal rings. These delocalised electrons act as charge carriers, allowing graphite to conduct electricity.

c) Graphite is used as the “lead” in pencils.
Graphite is used as an electrode in various types of batteries, such as lithium-ion batteries.

Question 14.
Match the following

A	B
Diamond	Nanotechnology
Graphite	Making of ornaments
Graphene	Lubricants

Answer:

A	B
Diamond	Making of ornaments
Graphite	Lubricants
Graphene	Nanotechnology

Question 15.
a. Dilute hydrochloric acid and calcium carbonate are taken in a test tube. Which gas is produced? b. Complete the chemical equation of the reaction.
CaCO₃ + 2HCl → CaCl₂ + …………… + ……………
c. The gas obtained is passed through clear lime water. Write down the observation.
Answer:
a. Carbon dioxide (CO₂) gas is produced.

b. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

c. When the gas obtained r(carbon dioxide) is passed through clear lime water (calcium hydroxide solution), it turns the lime water milky.

Extended Activities

Question 1.
Write a short note on the bleaching property of chlorine and present it.
Answer:
Chlorine is a powerful oxidising agent, which means it can remove electrons from other substanCes. This property allows chlorine to act as a bleaching agent, breaking down coloured compounds and making them colourless.
When chlorine comes into contact with a coloured compound, it oxidises the coloured molecules, breaking down their chemical structure.
The oxidation process results in the loss of colour, as the coloured molecules are transformed into colourless products.

Applications of Chlorine Bleaching:
Textile Industry: Chlorine is used to bleach cotton, linen, and other fabrics.
Paper Industry: Chlorine is used in the pulp and paper industry to bleach wood.
Water Treatment: Chlorine is added to drinking water and swimming pools to disinfect and remove impurities.

Question 2.
Prepare a table showing the use of oxygen, nitrogen and hydrogen in daily life.
Answer:

Gas	Common use
Oxygen	Breathing, medical treatments, industrial processes (combustion, steelmaking), rocket fuel, oxyacetylene flames is used for welding.
Nitrogen	Fertilizer production, food packaging (to prevent spoilage), industrial gases (creating inert atmospheres), automotive airbags, In the production of ammonia.
Hydrogen	Fuel cells (generating electricity), rocket fuel, production of ammonia (fertilizers), hydrogenation of fats and oils, used as an reducing agent in metal production.

Question 3.
Visit any chemical fertiliser factory, prepare a project report and present it
Answer:
Guidelines for Preparing Your Chemical Fertiliser Factory Project Report
1. Introduction:

• Clearly state the purpose of your visit and the overall goal of the report.
• Provide a brief overview of the factory, including its name, location, and any relevant background information.

2. Factory Tour:

• Describe the physical layout and facilities of the factory.
• Discuss the production process, step-by-step, including any diagrams or flowcharts to illustrate the process

3. Raw Materials and Production:

• Identify the primary raw materials used in fertiliser production.
• Explain the chemical reactions involved in the manufacturing process.
• Discuss any specific equipment or machinery used.

4. Environmental Impact:

• Assess the potential environmental impacts of fertiliser production (Eg: water pollution, airpollution, soil contamination).
• Discuss any sustainability initiatives or measures taken by the factory to minimise
  environmental harm.

5. Safety Measures:

• Describe the safety precautions and regulations implemented at the factory.
• Discuss any potential hazards or risks associated with fertiliser production.

6. Technology and Innovation:

• Highlight any advanced technologies or innovative practices used by the factory.
• Discuss the potential benefits of these technologies.

7. Challenges and Future Outlook:

• Identify any challenges or obstacles faced by the factory.
• Discuss the factory’s future plans and goals.

8. Conclusion:

• Summarize the key findings and observations from your visit.
• Offer your personal reflections or insights.
• Provide any recommendations or suggestions for improvement.

Non Metals Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Examine the periodic table, find out some examples of metals, non metals and metalloids, and

Metals	Non metals	Metalloids

Answer:

Metals	Non metals	Metalloids
Sodium Potassium Magnesium Calcium Aluminium Iron Copper Gold	Carbon Nitrogen Oxygen Hydrogen Chlorine Sulphur Fluorine	Metalloids Boron Silicon Germanium Arsenic Antimony Tellurium

Hydrogen, carbon, oxygen, nitrogen and chlorine are some important non-metals. Let us examine the methods of their preparation and some of their physical and chemical properties.

Hydrogen is the most abundant element in the universe. It is the major component of the sun and other stars. Hydrogen is seen in the atmosphere in a very small quantity in free state.

Question 2.
Complete the given table and write a short note on hydrogen.
Answer:

Atomic Number	1
Mass number	1
No. of electrons	1
No. of protons	1
No. of neutrons	No neutrons
Solubility in water	Insoluble in water
Isotopes	Protium, Deuterium, Tritium

Question 3.
Take 5 mL of dilute hydrochloric acid in a test tube. Add some zinc granules to it. Record your observation.
a) Is the gas produced hydrogen? How do you confirm it? Insert a burning splint to the mouth of the test tube. What do you observe?
Answer:
Yes, the hydrogen gas bums with a pop sound. Hydrogen is a flammable gas.

b) Write the balanced chemical equation of the above reaction.
Answer:
Zn + 2HCl → ZnCl₂ + H₂
Hydrogen is industrially prepared by the electrolysis of water.

Question 4.
To which category do the above reactions belong?
(Combination reaction, displacement reaction, decomposition reaction)
Answer:
Combination reaction.

Question 5.
The graph given below shows some fuels and their calorific values.

a. Which fuel has the lowest calorific value?
Answer:
Carbon monoxide.

b. Which fuel has the highest calorific value?
Answer:
Hydrogen.

c. What is the product obtained when hydrogen burns in air?
Answer:
Water.
The calorific value of hydrogen is higher than that of other fuels.

The merits of hydrogen as a fuel:

• When hydrogen is used as a fuel, the possibility of pollution is very low since water is the only product formed.
• High calorific value

In spite of these merits, hydrogen is not used as a domestic fuel due to the below-given limitations:

• Hydrogen is a gas that bums with an explosion.
• It is difficult to store and distribute hydrogen.

If these limitations are overcome, hydrogen will become a universal fuel. The problems like the scarcity of fossil fuels and environmental pollution can thus be resolved.

Question 6.
Which compounds of hydrogen are known to you?
Answer:
Water, Acids, Alkalies, Carbohydrates, Oils, Sugar

Question 7.
Which properties of hydrogen are used in the following situations?
a. Balloons filled with hydrogen fly high in the air.
b. Used as fuel.
Answer:
a. Hydrogen has less density than air

b.

• High calorific value
• The possibility of pollution is very low.
• Ease in preparing.

Question 8.
Prepare a note on the possibility of using hydrogen as a fuel.
Answer:
Hydrogen as a fuel:
Hydrogen is a colourless, odourless, tasteless gas that is highly flammable. It is the most abundant element in the universe and is found in water, natural gas, and other sources. Hydrogen can be used as a fuel in a variety of ways, including in fuel cells, internal combustion engines, and gas turbines.

Advantages of Hydrogen as a Fuel

• Hydrogen is a clean fuel that produces no greenhouse gases when burned.
• Hydrogen has a high energy density, which means that it can be stored in a small space and used to generate a large amount of energy.
• Hydrogen can be used in a variety of engines and fuel cells, making it a versatile fuel option.
• Hydrogen can be produced from a variety of sources, including renewable energy sources such as wind, solar, and geothermal power.

Disadvantages of Hydrogen as a Fuel:

• Hydrogen is a flammable gas, which means that it must be stored and handled carefully.
• Hydrogen is currently more expensive to produce than other fuels, such as gasoline and diesel.
• Hydrogen infrastructure is not yet widely available, which can make it difficult to refuel hydrogen-powered vehicles.

Future of Hydrogen as a Fuel:
Despite the challenges, hydrogen has the potential to become a major source of energy in the future. As technology advances and the cost of producing hydrogen decreases, it is likely that hydrogen will become a more viable option for transportation and other applications.

• Almost all substances around us contain carbon.
• Carbon is found both in an elemental state and a combined state in nature.

Question 9.
Examine the periodic table and record the position, atomic number and electronic configuration of carbon in your science diary.
Answer:
Atomic number – 6
Electronic configuration – 2, 4
Period – 2
Group – 14

Question 10.
In diamond, each carbon atom is covalently bonded to how many surrounding carbon atoms?
Answer:
In diamond, each carbon atom is covalently bonded to four surrounding carbon atoms.

Question 11.
What is the valency of carbon?
Answer:
The valency of carbon is 4.

Question 12.
Are free electrons present in the diamond?
Answer:
No, free electrons are not present in diamonds.

Properties of diamond

• Hardness is very high.
• Do not conduct electricity.
• High thermal conductivity.
• High refractive index.

Question 13.
Find out the reason for using diamonds for making ornaments and cutting glasses.
Answer:
Diamonds are used for making ornaments and cutting glasses due to their unique properties:

• Hardness
• Durability
• Precision

Question 14.
Some uses of graphite are given below. Find out and write the characteristics and properties that facilitate it.
a) Used as an electrode in dry cell
b) Used to make pencil lead
c) Used as a lubricant
Answer:
a) Used as an electrode in dry cell
Property: Good electrical conductivity
Characteristic: Graphite’s layered structure allows for the easy movement of electrons, making it an excellent conductor of electricity. This property is essential for its use as an electrode in dry cells, where it facilitates the flow of electrical current.

b) Used to make pencil lead
Property: Soft and easily markable
Characteristic: Graphite’s layered structure allows the layers to slide over each other easily, making it soft and easily markable. This property is ideal for pencil lead, as it allows the graphite to leave a mark on paper with minimal pressure.

c) Used as a lubricant
Property: Low friction between layers.
Characteristic: The force of attraction between these layers is weak. Hence, one layer can slide over the other. So, graphite is used as a lubricant

Question 15.
List out and compare the properties of graphite and diamond
Answer:

Graphite	Diamond
Layered structure with hexagonal rings of carbon atoms	Three-dimensional network structure with tetrahedral arrangement of carbon atoms
Soft	Extremely hard
Good conductor	Insulator
High thermal conductivity	High thermal conductivity
Lubricant, pencils, electrodes	Cutting tools, jewellery, abrasives

Question 16.
What are the carbon compounds that you know?
Answer:
Carbon dioxide (CO₂)
Carbon monoxide (CO)
Carbonates (CO₃^2-)
Bicarbonates

Question 17.
Which of these compounds contain carbon and oxygen?
Answer:
Carbon dioxide (CO₂)
Carbon monoxide (CO₂)
Carbonates (CO₃^2-)
Bicarbonates

Question 18.
What is the major compound produced when carbon or carbon compounds burn in the air?
Answer:
Carbon dioxide (CO₂).
C + O₂ → CO₂
Preparation of carbon dioxide in the laboratory

Question 19.
What are the reactants used here?
Answer:
Calcium carbonate, Diluted HCl

Question 20.
Complete the chemical equation of the reaction.
Answer:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Question 21.
How can we identify that the gas formed here is CO₂?
Answer:
When the gas formed is passed through lime water (Ca(OH)₂), if it turns milky we can say that the gas is Carbon dioxide (CO₂).

Carbon dioxide gas is produced when calcium carbonate and dilute hydrochloric acid react with each other. When this gas is passed through lime water Ca(OH)2, it turns milky.

Uses of Carbon dioxide

• In fire extinguishers.
• To make soda water and soft drinks.
• In the manufacture of washing soda (Na₂CO₃.10H₂O), and baking soda (NaHCO₃).
• In the manufacture of chemical fertilisers like urea.
• In Carbogen (95% oxygen and 5% carbon dioxide), which is used for artificial respiration.
• Solid carbon dioxide is known as dry ice. It is used to create a smoky effect in stage shows etc.

Question 22.
Tick the properties of carbon dioxide from the following.
Answer:

• Coloured / colourless (√)
• Helps combustion / does not help combustion (√)
• Odourous / odourless (√)
• Denser than air (√) / lighter than air
• Aqueous solution – Acidic (√) / Basic

Question 23.
Match the columns A and B suitably.

A	B
Washing soda	NaHCO3
Carbon monoxide + Hydrogen	Na2CO3.10H2O
Baking soda	Producer gas
	Water gas

Answer:

A	B
Washing soda	Na2CO3.10H2O
Carbon monoxide + Hydrogen	Water gas
Baking soda	NaHCO3

 

• Nitrogen is the most abundant gas in the atmosphere.
• Nitrogen is present in all living beings.
• Nitrogen is an element that is essential for the growth of plants.

Question 24.
Under normal circumstances, atmospheric nitrogen is inert. Write down the reason.
Answer:
Inert behaviour refers to the tendency of a substance to resist chemical reactions. An inert substance is relatively unreactive and does not readily combine with other elements or compounds.

Reason:

• Triple Covalent Bond: Nitrogen atoms share three pairs of electrons, forming a very stable bond. This bond is much stronger than a single or double bond.
• High, Activation Energy: Breaking this triple bond requires a high activation energy, which is the minimum amount of energy needed to start a chemical reaction. This high energy barrier makes it difficult for nitrogen to react with other substances.

Question 25.
Which gas helps combustion?
Answer:
Oxygen.

Question 26.
Which gas helps to control the rate of combustion?
Answer:
Nitrogen.
Plants cannot absorb nitrogen directly from the atmosphere. It involves following steps

Step: 1
Through thunder and lightning: The triple bond in the nitrogen molecule is broken down during thunder and lightning, and nitrogen combines with atmospheric oxygen to form nitric oxide (NO). N₂ + O₂ → 2NO
The nitric oxide thus produced combines with more oxygen to form nitrogen dioxide (NO₂).
2NO + O₂ → 2NO₂
In the presence of oxygen, nitrogen dioxide dissolves in rainwater and reaches the soil as nitric acid (HNO₃).
4NO₂ + 2H₂O + O₂ → 4HNO₃
This nitric acid reacts with the minerals in the soil and forms nitrate salts which are absorbed by plants.

Step: 2
Through nitrogen fixation: Rhizobium bacteria in the roots of leguminous plants absorb atmospheric nitrogen and convert it into compounds that can be absorbed by the plants.

Question 27.
What are the methods by which we can supply the elements required for plant growth in large amounts?
Answer:
Use of organic fertilisers
Use of chemical fertilisers
Merits and demerits of chemical and organic fertilisers

Organic fertilisers	Chemical fertilisers
Eco friendly	Increase crop yield
Maintain the original nature of the soil	Easily available
Delay in making the nutrients available	Quick supply of the three nutrients (N, P, K)
Do not destroy microorganisms in the soil	Change the organic structure of the soil.
More quantity is required.	Increase the acidity of the soil.
Reduce the possibility of pollution.	Cause pollution.

Question 28.
Observe the picture of the preparation of oxygen in the laboratory.
a. Which is the reactant in this chemical reaction?
Answer:
Potassium permanganate

b. Heat the boiling tube containing potassium permanganate.
Insert a burning matchstick to the mouth of the boiling tube. What is the observation?
Answer:
The glowing matchstick flares up.

c. Complete the chemical equation of the reaction.
Answer:

d. Pick out from the following the physical properties suitable to oxygen. Write them down.
Answer:

Colour	Yes/ No √
Odour	Yes/ No √
Solubility in water	Soluble √ / Insoluble
Nature of combustion	Bums/ Helps to burn

Oxygen is manufactured by the electrolysis of water.

Question 29.
Haven’t you noticed that some metals lose lustre in the course of time due to their reaction with oxygen? Write the reason.
Answer:
Oxygen reacts with metals and non-metals to produce their oxides. Due to this, metals lose lustre in the course of time.

Question 30.
Prepare a note on the role of plants in maintaining the level of oxygen in the atmosphere. Present it in the class.
Answer:
Plants play a crucial role in maintaining the oxygen levels in our atmosphere. Through the process of photosynthesis, plants convert carbon dioxide into oxygen, a gas essential for the survival of most life on Earth.

Photosynthesis is a complex process that occurs in the chloroplasts of plant cells. It involves the use of sunlight, water, and carbon dioxide to produce glucose and oxygen. The chemical equation for photosynthesis is:
6CO₂ + 6H₂O + sunlight → C₆H₁₂O₆ + 6O₂

The oxygen produced by plants is released into the atmosphere, where it is used by animals, humans, and other organisms for respiration. During respiration, these organisms consume oxygen and release carbon dioxide back into the atmosphere. This creates a continuous cycle of oxygen and carbon dioxide exchange.

Question 31.
Conduct a seminar on ozone depletion and its solutions.
Answer:
Seminar: Ozone Depletion and Its Solutions

Introduction:
The ozone layer, a vital shield in Earth’s atmosphere, is facing a serious threat: depletion. This depletion has far-reaching consequences for human health, ecosystems, and the planet as a whole. In this seminar, we will explore the causes, effects, and potential solutions to ozone depletion.

Causes of Ozone Depletion:
Chlorofluorocarbons (CFCs): These man-made chemicals, once widely used in refrigerants, aerosols, and foam insulation, are the primary culprits in ozone depletion. When released into the atmosphere, CFCs rise to the stratosphere and react with ozone molecules, destroying them.

Effects of Ozone Depletion:
Increased UV Radiation: Ozone depletion allows more harmful ultraviolet (UV) radiation to reach Earth’s surface.
This can lead to:

• Skin cancer
• Cataracts
• Weakened immune systems
• Damage to plants and marine ecosystems

Climate Change: Ozone depletion can indirectly contribute to climate change by affecting atmospheric circulation and temperature patterns.

Solutions to Ozone Depletion:

• Raising awareness about the importance of the ozone layer and the consequences of its depletion.
• Encouraging people to choose products that are ozone-friendly.
• Strict regulations and enforcement are essential.
• Develop advanced monitoring techniques to track ozone levels.

Question 32.
Complete the table about chlorine after analysing the periodic table.
Answer:

Atomic number	17
Electronic configuration	2, 8, 7
Mass number/Atomic mass	35
No. of electrons	17
No. of neutrons	18
Valency	1
Name of the element family	Halogen

Question 33.
Analyse the given figure depicting chlorine preparation and write answers to the questions given below.

a. What are the chemicals required to prepare chlorine in the laboratory?
Answer:
Potassium permanganate (KMnO₄), Concentrated Hydrochloric acid (Con. HCl)

b. What are the products formed?
Answer:
Potassium chloride (KCl), Manganese chloride (MnCl₂), Water (H₂O), Chlorine gas (Cl₂).

Question 34.
The chlorine gas obtained is collected in the jar after passing it first through water and then through sulphuric acid. Why is it done so?
Answer:
A small amount of hydrogen chloride vapour is also released along with the chlorine gas. Hydrogen chloride vapours dissolve in water when passed through it. Sulphuric acid can absorb the water vapour in the chlorine gas. So it is passed through concentrated sulphuric acid.

PROPERTIES OF CHLORINE

• Greenish yellow in colour
• Pungent smelling.
• Denser than air.

Question 35.
Take two jars filled with chlorine gas. Take a coloured cloth and cut it into two pieces. Put a wet piece into one jar and the other dry piece of cloth into the second jar. What happens to the colour of the cloth in the two jars?
Answer:
The cloth in the first jar remains unchanged in colour. The colour of the wet cloth in the second jar disappears. This is because of the bleaching properties of chlorine. Chlorine bleaches coloured materials into colourless materials in the presence of moisture.
H₂O + Cl₂ → HCl + HOCl
HOCl → HCl + [O]
Water reacts with chlorine to form hydrochloric acid and hypochlorous acid. Hypochlorous acid is unstable. So, it decomposes to form hydrochloric acid and nascent oxygen. Nascent oxygen acts as a powerful oxidising agent to decolourise the coloured substances.

Question 36.
Some questions are given below. Find suitable answers from the box given. Write them down.
Nitrogen, Nitric oxide, Oxygen, Carbon, Carbon dioxide
a) Which gas is obtained when potassium permanganate is heated?
b) What is the product formed when nitrogen combines with oxygen at high temperatures?
c) Which reactant is used along with hydrogen for the manufacture of ammonia?
Answer:
a) Oxygen
b) Nitric oxide
c) Nitrogen