Class-9

lbg M PsfpGg kf7 M x:| j bL3{ klxrfg cWoog v08 ‘d sf] x'F’ kf7sf] P3f/fcF} g'R5]b Wofg lbP/ k9\\gx' f];\\ . cEof; v08 pQm cgR' 5b] af6 x:| j Osf/ / bL3{ Os{ f/ nfu]sf zAb 5fgL ;\"rL agfpgx' f];\\ . lbg M afpGg kf7 M j0f{ljGof; cEof; cEof; v08 ‘bO' { ldtsf] syf’ kf7af6 lj:do;\"rs lrGx-¤_ ko| fu] ePsf tLg, k|Zgjfrs lrGx-<_ k|of]u ePsf rf/ / p4/0f lrGx -æ Æ_ ko| f]u ePsf rf/j6f jfSo vfh] /] sfkLdf l6Kg'xf;] \\ . lbg M lqkGg kf7 M hLjgL nv] gsf] cEof; cWoog v08 !# cfF} lbgsf] cWoog v08df xfdLn] hLjgL nv] gnfO{ d'Vo tLg v08df af8F ]/ n]lvG5 eGg] s'/f cWoog u/s] f lyof}F . kml] / Ps k6s pQm s'/f cWoog ug'x{ f;] \\ . cEof; v08 tn lbOPsf] ljj/0fsf cfwf/df Pp6f hLjgL tof/ kfg{x' f];\\ . gfd M ljZj]Zj/k|;fb sfO] /fnf hGdldlt M lj=;+= !(&! ebf} @$, jgf/; dftf÷lktf M lbJof sf]O/fnf÷si[ 0fk;| fb sfO] /fnf lzIff M aL=Pn= ky| d k|sflzt gk] fnL efiffsf] /rgf M rGbj| bg -lj=;=+ !((@ ;fndf zf/bf klqsfdf_ kd| 'v s[ltx¿ M syf ;ª\\ux| -’bfi] fL r:df’ / Zj]te/} jL’_ pkGof; -‘;'lDgdf’, ‘g/G] b| bfO’, ‘tLg 3D' tL’, ‘lx6n/ / ox'bL’, ‘df]bL cfOg’, ‘afa,' cfdf / 5f/] f’ ;:+ d/0f ‘cfÇgf] syf’, ‘cfTdj[QfGt’, ‘h]n hgn{ ’, ‘kml] / ;G' b/Lhn’ kj| l[ Q / ljz]iftf M g]kfnL ;flxTosf k|yd dgf]j1} flgs syfsf/, of}g dgf]lj1fgsf] ke| fj, ;zQm pkGof;sf/, kfqx¿sf] of}gefjgfnfo{ ;zQm ¿kdf k:| tt' u/s] f, /fhgLlt / ;flxTo bj' } Ifq] df ljlzi6 ofu] bfg k¥' ofPsf, axd' v' L k|ltefsf wgL d[To' M lj=;=+ @)#( ;fpg ^ ut] lbg M rfj} Gg kf7 M sfn / kIf -cEof;_ cWoog v08 xfdLn] kfFrf,}F 5}6fF} / ;ftf}F lbgdf sfn / kIf;Fu ;DalGwt sx] L hfgsf/L lnof}F / cEof; klg u¥of}F . To;nfO{ kg' M Psk6s bfx] f]¥ofpg'xg' cg'/fw] 5 . --51--

cEof; v08 != sf]i7sdf lbOPsf wft' / ;ªs\\ ]tsf cfwf/df vfnL 7fpF e/L cEof; k'l:tsfdf ;fg'{xf];\\ . s_ kf| lKtn] lrq ======= . -agfpM ;fdfGoe\"t_ v_ pgLx¿ af9Ldf ======== .-k/\\ M k0\" f{ jt{dfg_ u_ xfdL /fd|f uLt ====== . -ufp M ck\"0f{ eljiot_\\ 3_ kz| fGt klg dfdf3/ ========= < -hf M cEo:t et\" _ ª=_ l/of xfdLlt/ =========. -kms{M ck0\" f{ jtd{ fg_ r_ afafn] ahf/af6 /fdf| ] pkxf/ ======= . -NofpM ;fdfGo eljiot_\\ 5_ du[ x¿ af/Llt/ ======= lg . -3d' \\ M ;fdfGo jt{dfg_ h_ xh/' afn] dnfO{ cfzLjfb{ ===== . -lbM k\"0f{ e\"t_ em_ lk|of / l/of ;uF } dfjn ========= . -lxF8\\ M c1ft et\" _ `_ o'jf Snan] /Qmbfg sfos{ d| ====== . -/fv\\ M cEo:t e\"t_ lbg M krkGg kf7 M jfSodf ko| fu] cWoog v08 jfSo lgdf0{ f ubf{ Wofg lbgk' g{] s'/fx¿ M s_ ljleGg zAb, pvfg, 6S' sf cflbsf] cy{ /fd|/L a‰' g,] lgkftsf] ko| fu] af/] Wofg lbg], v_ cy{ a'lem;s]kl5 pQm zAb, pvfg, 6'Ssf jf lgkftnfO{ 5f]6f5] l/tf] / cyk{ 0\" f{ jfSo lgdf{0f ug,]{ u_ 6S' sfnfO{ jfSodf ko| fu] ubf{ 6'Ssfsf] cGTodf cfpg] ‘g'’ k|ToonfO{ ;dflksf lso| fdf abNg], 3_ kf| ljlws÷ kfl/eflifs zAbnfO{ pQm zAb ko| f]u xg' ] ljifo;uF ;DalGwt jfSo lgdf0{ f ug,{] ª_ jfSo lgdf0{ f ul/;sk] l5 jfSodf ko| f]u ul/Psf] cz+ nfO{ /]vfª\\sg ug]{ . jfSo lgdf0{ fsf sx] L pbfx/0f M sfg vfg' M k9\\g] an] fdf s]6fs]6Ln] xNnf u/]/ sfg vfP. pGd\"ng M /f0ffsfndf w]/} g/fd|f sfd eP klg ;tLk|yf / bf;k|yf pGdn\" g h:tf /fdf| sfd klg eP . lg M d/] f] efO t sIffdf k|yd eP/ k'/:sf/ kfof] lg. xftd'v hf]8g\\ ' M sf]/f]gf ;ªs\\ d| 0fkl5 yk' }| dflg;nfO{ xftdv' hf]8\\g klg wf}wf} 5 . b'Mv kfO;\\ dª\\un,] cfÇg} 9ªu\\ n] M afafn] b]vfPsf] af6f] glx8F /] afª\\uf] afªu\\ f] lx8F ]sf] k|ldz af6f] gkfP/ ;dodf 3/ kU' g} g;Sbf bM' v kfO;\\ dª\\un] cfÇg} 9ªu\\ n] eg] h:t} eof] . lgjf/0fM ;/sf/n] ul/aL lgjf/0fsf nflu klxnf] /fd|f] of]hgf NofPsf] 5 . cEof; v08 cy{ :ki6 x'g] u/L jfSodf k|ofu] ug'x{ f;] \\ M cfe] m]n kg{', /fb] L, jGbgf, kf7F} fhf/] L, lgTo, cZnLn, cab{' , lkm:6], xfF;L dhfs ug,{' t,jfrf afWF g' lbg M 5kGg kf7 M j0f{ljGof; cWoog v08 tnsf] cg'R5]b cWoog ugx{' f];\\ . lxdfnsf] s~rg kfgLdf / e~Hofªsf] lr;f] atf;df c;fwf/0f zlQm ePsfd] f s'g} zªs\\ f 5g} . ;+;f/sf yk' |} ;flxTosf/x¿n] lxdfnL atf; / kfgLsf] kljqtfsf] JofVof u/s] f 5g\\ . xfdLsxfF eg] hn;|ft] ;DaGwL plrt cg;' Gwfg ePsf] / pkoQ' m of]hgf ags] f] kfObF g} . cgluGtL gbLgfnf k|of]uljxLg ag]/ v]/ uO/xs] f 5g\\ . clncln ko| f]u ePsf gbLgfnf klg ljleGg ;lGw / ;Demf}tfsf cfwf/df g]kfnleq g]kfnLn] k|of]u ug{ kfPsf 5}gg\\ . xfd|f ;|ft] af6 cGoqLn] nfe lnO/xs] f 5g\\ . xfd|f] /fhgLlts --52--

;:+ sf/ kl/jt{g xg' g;Sgfn] bz] df ljsf; eGbf ljgfz a9L eO/x5] . l;h{gfsf] dfud{ f eGbf Wj+zsf] dfud{ f ;dfh cufl8 al9/x]sf] 5 . ;dfhdf p208x¿sf] x'n 7\"nf] aGb} uO/xs] f] 5 . /f0ffsfn / k~rfotsfnsf lg/ª\\sz' rl/q u0ftGqsf zf;sdf klg b]lvg' b'Mvb\\ 5 . v'n]cfd e|i6frf/Lsf] ;+/If0f ug{ ;ªs\\ fr] gdfGg] zLif{ g]tT[ jn] bz] nfO{ bzsfF} k5fl8 ws]Ng] cj:yf l;hg{ f ePsf] 5 . o'jfk':tfn] gofF cleofg yfn]/ snªl\\ st /fhgLltnfO{ ;ª\\nf] agfpg ;ª\\3if{ ug{'kg]{ an] f cfO;Sof] . cEof; v08 dflysf] cg'R5]baf6 cfwf ª, `, 0f, g / d -ª,\\ ~, 0, G, D_ ko| f]u ePsf zAb 5fg]/ sfkLdf nV] g'xf];\\ . lbg M ;GtfpGg kf7 M kT| ofudg cWoog v08 ‘bO' { ldtsf] syf’ kf7df /xs] f zAbfy{ k9g\\ 'xf;] \\ . cEof; v08 tkfOFs{ f] cleefjs jf bfhe' fO, lbbLalxgLnfO{ pQm kf7sf] zAbfy{ ;f]Wg nufpgx' f];\\ / tkfOF{n] eGgx' f];\\ . of] cEof; k|Tos] kf7df ug{ ;Sgx' 'G5 lsgls efiffdf zAb / To;sf] cy{ hfGg' cToGt dxŒjk\"0f{ x'G5 . lbg M cG7fpGg kf7 M Jofs/0f cWoog v08 tnsf] cgR' 5]b cWoog ugx'{ f;] \\ . xfd|f kv' fn{ ] xfd|f] ;+:s[lt lgdf{0fdf 7n\" f] of]ubfg u/] . pgLx¿n] cfÇg} df}lns ;+:sl[ t lgdf0{ fdf a9L hf8] lbPsf 5g\\ . lrqsnf, dl\" t{snf, jf:t'snf Pjd\\ ;ªu\\ Lt / gT[ odf klg gk] fnL dfl} nstf emNsg] u/L ltgnfO{ lgdf{0f ul/Psf] 5 . pgLx¿ o; sfo{df /ftlbg v6y\\ ] . /fi6« / ;+:s[ltnfO{ dxfg\\ 7fGy] . cfh tL df}lnstf w]/} nfk] eO;s]5g\\ . ljleGg b/af/x¿, dlGb/x¿, d\"ltx{ ¿ tyf nfs] uLt / g[To o;sf kd| f0f x'g\\ . xfdLn] kv' fa{ f6 w]/} s/' f l;Sg ;S5f}F . kv' fx{ ¿ cfTdlge/{ lyP . xfdL lbglbg} k/lge{/ aGb} 5fF} . xfdLn] cfÇgf dxŒjk0\" f{ s'/f nTofP/ csfs{ f g/fdf| s'/fnfO{ cg';/0f ul//x]sf 5f}F . of] s|d lg/Gt/ rln/x]df xfdL k0\" ft{ M k/lge{/ aGg]5f}F . xfd|f k'/ftflŒjs ;Dkbf / ;+:s[lt nf]k x'bF } hfg5] g\\ . xfdL kfgL l/lQPsf] ufu|Lh:t} df}lnstf x/fPsf] bz] sf gful/s ag]sf xg' ]5f}F . t;y{ xfdL cfh} ;r]t ag/] kv' fs{ f bg] sf ¿kdf /xs] f ;:+ s[lt / ;Dkbf hf]ufpglt/ nfufF} . cEof; v08 dflysf] cg'R5]bdf sfnsf ljleGg kIfsf ls|ofkbx¿ ko| fu] ePsf jfSox¿ 5g\\ . dflysf ;a} jfSonfO{ ;fdfGo e\"tsfndf kl/jt{g u/L kg' nv]{ g ugx{' f];\\ . lbg M pgfG;f7L kf7 M JofVof cWoog v08 != afx|fF} lbgs]f cWoog v08df xfdLn] ;k;| ªu\\ JofVofaf/] cWoog u/]sf lyofF} . To;nfO{ k'gM Psk6s x]g{'xf;] \\ . @= tn lbOPsf s]xL s/' f cWoog u/L ;k|;ªu\\ JofVof / JofVof nv] gsf km/s yfxf kfpg'xf;] \\ . JofVof / ;k;| ª\\u JofVof nueu p:t} x'g\\ . JofVofdf ;k|;ªu\\ JofVofh:tf] kf7, nv] s, n]vssf ljz]iftf / sg' kfqn] s'g ;Gbe{df egs] f] xf] eGg] lsl;dsf cf}krfl/s s'/f nV] g' kb{}g . --53--

;k|;ªu\\ JofVofdf klxnf] cg'R5]bdf hg' h'g s/' f pNnv] ug{ elgPsf] 5 JofVofdf tL ;a} pNnv] ug'{ kb}g{ . To;dWo] n]vs / k|Zgn] ;fw] s] f] d\"n cfzonfO{ ;d6] /] JofVofsf] klxnf] cgR' 5]b tof/ ugk'{ 5{ . cGo kl| s|of ;a} Pp6} xf] . cEof; v08 != JofVof ug{x' f;] \\ M s_ :jf:Yo g} wg xf] . lbg M ;f7L kf7 M >'ltn]vg cEof; v08 cleefjssf] ;xofu] lnO{ ‘d sf] x’'F kf7sf] cf7fF} cgR' 5]b >'ltnv] g ug{x' f;] \\ . kl5 cleefjs;uF } a;/] slt z4' n]Vg'eP5 k/LIf0f ugx'{ f];\\ . --54--

Subject : Compulsory Mathematics Day-1 Activity  Sate is a collection of well-defined objects or numbers. Representation of sets.  Listing method N = {1, 2, 3, …..}  Description method N = {Counting Numbers}  Set builder method N = { x:x∈ N} Exercise  Define sets with an example.  If A = {2, 4, 6, 8, 10, 12} write the given set in i) Description method. ii) Set builder method. Day-2 Activity If M & N are two sets such that every element of M is in N, we say that M is a sub set of N. Ty per of sets.  Empty set  having no element.  Unit set  having only one element.  Finite set  having finite number of elements.  Infinite set  having uncountable number of elements. Exercise 1. Write two examples of sub sets. 2. Define unit set & finite set with two examples. Day-3 Activity  Intersection of sets:- common elements on set A & set B is denoted by A∩ B. ie A∩ B = {x: x ∈ A, and x ∈ B}  Union of sets:- All elements present is set A & set B is denoted by A∪ B = {x:x∈ A or x ∈ B} Exercise Let A = {1, 2, 3,4, 5} and B = { 2, 3, 56} & C = {4, 5, 6, 7} verify (A∩ B) nc = A∩ (B ∩ C) Day-4 Activity  Difference of sets :- Elements that belong to set A and do not belong to B. It Q is denoted by A - B. i.e. A - B = {x:x∈ A, x ∉ B}  Complement of a set 1 - All elements of universal set. Which dese not belong to the set a. It is denoted by A i.e. A {x: x ∈∪, x ∉ A}. Exercise If ∪ = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 4, 7, 8}, B = {4, 6, 8, 9} & C = { 3, 4, 5, 7} then find (ii) (A-C)∪ (C − A) i) A - (B-C) --55--

Day- 5 Activity The number of elements contain is the set is cardinality of the set if A ={Apple, Banana, Mango}∪ (A) = 3.  n.(A ∪ B) = n(A). + n (B) (A & B are disjoints sets).  n(A ∪ B) = n(A)th(B) − n(A ∩ B)(A&B are overlapping sets) Exercise a) If = n (A) =200, n(B) = 300 & n (A ∪ B) = 400 find n(A ∩ B), Also, the information in venn diagram. b) If n (X) = 25, n (Y) = 40 & n (X∩ Y) = 10 find .n(X∪ Y), Also show the given information in venn diagram. Day-6 Activity  ∩°(A) = ∩ (A) − ∩ (A ∩ B)  ∩°(B) = ∩ (B) − ∩ (A ∩ B)  ∩ (A ∩ B) = ∩° (A) + ∩° (B) +∩ (A ∩ B) Also study Activity - 5 Exercise a) In a group of 300 students who like football or volleyball, 204 like football game and 180 like volleyball game by drawing venn-diagram find: i) How many students like both game? ii) How many students like only football? b) If ∩°(A) =30, ∩°(B) =45 &∩ (A ∩ B) = 10 find ∩ (A ∩ B). Day-7 Activity Dear students, study example - 3 (page - 15) from your text book, Also study again parous activities. Exercise In a survey of 90 people, it was found that 51 people liked to eat orange and 59 liked to eat apple. If there were 3 people who didn't like to eat both fruits. i) How many people liked to eat only orange? ii) How many people liked to eat only apple? iii) Show the above information in venndragram. Day-8 Activity Study carefully example - 4 (page - 15) from your text book. Exercise In a survey, it was found that 80% liked oranges, 85% liked mangoes and 75% liked both, but 333 people liked none of them by drawing a venn diagram, find the number of people surveyed. --56--

Day-9 Activity (Profit & Loss) Profit = SP - CP OR profit % of cp. Loss = CP - SP OR Loss% of CP. Exercise 1) A Toy which was bought for Rs 100 & was sold at Rs 85 find loss amount. 2) An article bought at Rs. 500 and sold with Rs. 100 as profit find selling price & profit percentage. Day-10 Activity SP = CP + profit OR CP + profit % of CP Exercise ହ ସ 1) An article bought for Rs. 800 is sold for of the cost price. What is it selling price? 2) A man bought a camera at Rs. 2000 and make a profit of 20% on it. What is its selling price? Day-11 Activity SP = CP - Loss SP = CP - Loss% of CP Exercise 1) Rita bought a jacket for Rs. 5000 and sold it for Rs. 4525 find it loss amount. 2) A shopkeeper bought a computer for Rs. 55200 and sold at a loss of 5% what is the selling price of that computer? Day-12 Activity  CP =1S0P0×-1L0%0  CP =10S0-PP×r1o0fi0t % Exercise 1) A man sold a watch for Rs. 252 at 5% profit. What was the cost price? 2) If a shopkeeper sold a cupboard for Rs. 6400 at a loss of 20% find the cost price of the watch. Day-13 Activity Study example - 2 (page - 25) & example - 4 (a page - 26) from your text book. Exercise 1) If the cost price of 10 chairs is equal to the selling price of 16 chairs. Find the loss or gain percentage. 2) By selling a watch for Rs. 4500 a dealer losses 10%. At what price should be sell to gain 10% on it? --57--

Day-14 Activity Read example - 5 (page - 26) from your text book carefully and do the following. Exercise Shishir bought 4000 oranges at 70 paisa each. But 400 of them were rotton. He sold 2000 oranges at 90 paisa each. If he plans to make a profit of Rs. 200, at what rate must he sell the rest of the orange? Day-15 Activity Read examples - 6 & 7 (page - 27) from your text book. Carefully and do the following. Exercise 1. Salmav sold two computers for Rs. 20,000 each on one, he gains 20% and on the other he losses 20% find his loss or gain in whole transaction. 2. A man bought two books for Rs. 1040. He sold on at a loss of 15% and other at a profit of 36% then he found both books are sold for the same price. Find the cost price of each book. Exercise 1. Amisha sells a bag to Amir at a profit of 10%. Amir sells the same bag to Abhishekh at a profit of 20% if Abhishek has sold it for Rs. 3300 there by earning a profit of 25% find the cost price of a bag for Ashmita? 2. A person sold an article at a profit of 15% if he sold it for Rs. 40 less, his loss would have been 12% find the lost price of the article. Day-17 Activity (Commission & Tax) A person involved in selling goods or worked as a middle man in selling, buying & renting things gets the percentage of values of sales which is percentage of values of sales which is called commission. Commission Amount = Commission % of total sale. Exercise 1. A man gets a commission of 10% on the sales of amount of Rs. 25000. How much much does the agent will get as commission. Also find the amount receive by business man. Day-18 Activity Commission (%) = Commission Amount × 100% Total Sales Also study example (page - 33) Exercise 1. Rita receives Rs. 4500 commission by selling an article worth Rs. 90,000. What is the commission rate? 2. A man monthly salary of Rs 8000 plus 5% commission on his total sale of the month. If his total monthly sales is Rs. 1,25,000. Find commission amount & his total salary of that month. --58--

Day-19 Activity Study Activity of day 17 & 18 again. Exercise 1) An agent sold some carpets at the rate of Rs. 4500 each and received Rs. 9000 as commission at 10% find the total sales of that month. 2) Suppose, you are a sales manager of a company and you get Rs. 15000 monthly salary plus 1.5% commission on total sales. In a particular month you received the total wages as Rs. 3250, find the total sales of that months. Day-20 Activity (Discount)  Some reduction amount on the original price is a discount. Discount = Marked price - selling price.  Discount = discount% of marked price.  Discount % = Discount Amount ×100% Marked Price Exercise Find the discount amount & discount percent. a) Marked price = Rs. 4500, selling price = Rs. 3800 b) Marked price = Rs. 1300, selling price Rs. 1000. Day-21 Activity SP = MP = discount OR MP - dis% of MP. Exercise 1. Find selling price when an article marked at Rs. 2000 & given a discount of Rs. 200. 2. The marked price of an article is Rs. 350. The discount of 10% is given what is the price of an article after discount? Day-22 Activity  MP = SP×100 100–Discount% Exercise 1. After discount of 10% an article was sold for Rs. 900. What was MP of the article? 2. The cost price of an article was Rs. 4000. Find the marked price of the article. So that there will be a profit of 20% after allowing a discount of 20%. Day-23 Activity Study Activities (20021, 22) again, also study example - 5 ( page - 40) from your text book. --59--

Exercise A shopkeeper allows discount of 10% and still makes a profit of 15% on selling an article if the marked price an article is Rs. 15,000 find, i) Selling price ii) The cost price iii) Profit Day-24 Activity Kindly request you to study example - 4 (Page- 39) from your text book. Exercise 1) The marked price of a radio 40% above the lost price. When it was sold allowing 30% discount on it there was a loss of Rs. 100. What was the marked price of the radio? 2) A shopkeeper fixed the marked price of his TV to make a profit of 30%. Allowing 15% discount of on marked price, the TV was sold. What percent profit will be make? Day-25 Activity : (Taxation  If an individual's or group of individuals income exceeds the minimum level fixed by the government then the government impose then to pay certain percent of their excess earnings as income tax.  Tax Rates in Nepal (2017 /018) i) For Unmarried Tax Banding Tax Rates a) First Rs. 3,50,000 1% b) Next Rs. 1,00,000 15% c) NextRs. 4,50,001 - Rs. 25,00,000 25% d) Balance exceeding Rs. 25,00,000 35% Exercise  The monthly salary of the cabinet Ministers of Nepal in BS 2074 was Rs. 60,970 Find total annual tax paid an unmarried cabinet Minister. (For tax rates see the above tax Banding table) Day-26 Rates 1% Activity 15% Tax Barding for married (2017/18) 25% 35% Tax Banding a) First Rs.4,00,000 b) Next Rs. 1,00,000 c) Next Rs. 5,00,001 - Rs. 25,00,000 d) Balance exceeding Rs. 25,00,000 Exercise i) The monthly salary of the secretary of Nepal in 2073 B.S. was Rs. 37390. Find the annual tax paid by married secretary. --60--

Day-27 Activity Study Activity 26 again. Also study example - 2 (page - 47) from your text book and answer the given question.  On the ratio of their investment. Exercise Mr. Racheet a CEO of a bank has a salary Rs. 3,00,000 in a month. He is married calculate his annual tax liability. Day-28 Activity Dividend  Dividend refers to the protion of net profit of company to the share holders.  Bonus is distributed to the employees as well.  1 share = Rs. 100  Rs. 1000 = 10 shares Exercise A school has 5,00,000 shares of Rs. 100 per share. Mr. Gobinda has 5000 shares in the school. The school earned Rs. 1,50,000 net profit in a year and board of director dicided to distribute 50% cash dividend to share holders. How much does Mr. Govinda get as cash dividend? Day-29 Activity  Study example - 2(page - 54) from your text book.  Also study Activity - 28 again. Exercise A company makes a profit of Rs. 15000000 annually. It distributes the dividend to its 500 shareholders with equal number of shares, equally so that each shareholders received Rs.15000 as dividend find the dividend rate. Day-30 Activity Findly request you to study example: 3(page - 54) from your text book. Exercise A noodles factory is profit increased from 25% to 40% and made a profit of Rs. 1,00,00,000. The factory decides to distribute 4% of its increased 15% equally among 50 employees. Find the bonus that each employee will get. Day-31 Activity factorization  a2- b2 = (a+b) (a-b)  (a + b)2 = a2+2ab+b2  (a-b)2 = a2 - 2ab + b2 --61--

Exercise a) 9x2 - 16y2 b) 64p2 - 25q2 b) (4n - 7m)2 Factorize: Expand the following: a) (p + 2q)2 Day-32 Activity  a2 + b2 = (a+b)2 - 2ab OR = (a-b)2 + 2ab  a2 + a2b2 + b4 = (a2)2 + (b2)2 + a2b2 = (a2 + b2)2 - 2a2b2 + a2b2 = (a2+b2)2 - a2b2 = (a2 + b2)2 - (ab)2 = (a2 + b2 + ab) (a2 + b2 - ab) Also study example - 1 (page - 114) Exercise b) 4m4 - x4 c) 4x4 + 625 Factorize: a) x4 + 4 Activity Day-33 Factorize: x4 + 23x2 + 256 (x2)2 + (16)2 + 23x2 = (x2 + 16)2 - 2.x2.16 + 23x2 (x2 + 16)2 - 32x2 + 23x2 = (x2 + 16)2 - 9x2 (x2 + 16)2 - (3x)2 = (x2 + 16 - 3x) (x2 + 16 + 3x) x4 + 6x2 + 25 = = = Exercise 625a4 + 25a2 + 1 Factorize: a) b) Day-34 Activity �� + 1+ �� Factorize: �� �� ����� + 2 .�� . ��+ �����- 1 Here, ��� + ���� - (1)2 ��� + � ��� � = � − 1� + � + 1� = Exercise a) x2 - 3 + � b) 4x2 - 27 + � Factorize: �� ��� Activity Day-35 Factorize: x2 - 10x + 24 + 6y - 9y2 Here, x2 - 2.x.5 + (5)2 - 25 + 24 + 6y - 9y2 --62--

= (x-5)2 - 1+ 6y - 9y2 = (x-5)2 - (1-6y + 9y2) = (x-5)2 - ((1)� − 2.1.3y + (3y)�) = (x - 5)2 - (1 - 3y)2 = (x - 5 - 1 + 3y) (x - 5 + 1 - 3y) = (x + 3y - 6) (x - 3y - 4) Also study example - 8 (page 115) Exercise a) 4a2- 20a + 24 + 6y - 9y2 Factorize: b) (m + n + p)2 - (m - n + p)2 Day-36 Activity am× an = am+n Indices am ÷ an = am-n (am)n = amn Exercise a) (ab)n = an . bn b) 216��� Evaluate: c) 16 ���� d) 6 × 27���� ÷ 3��� 5���� × 15 ���� × 3����� Day-37 Activity x��������������===��√����x�����     �√x = x���  x�� = 1�x� Exercise a) ��4x�y� × ��2x�y� Simplify: b) �√128a� ÷ �√2��a�� c) ��������� ������� Day-38 Activity ���������� × ���������� × ���������� Simplify: ��×���� ��×���� ��×���� = × ×��×���� ��×���� = ��×���� ���� ���� ���� × ×���� ���� ���� --63--

= x���� ������������������� = x0 = 1# Exercise - �������������� �������� �������������� Simplify: a) × ����� � × b) ������� × ������� × ������� Day-39 Activity Kindly request you to study activities of day- 36,37,38 again & do the following. Exercise ������� ��� × ������� ��� × ������� ��� a) ���� � ���� � ���� � b) ����������� × ������ ����� × ����������� Day-40 Activity �� ����������������� × ������������ Simplify: �÷� = ������������������� × ��������������� = ��� ×������������������������ ����������� ���� ��� ��������� ���� . ������������������ � ����� = ��� = = �������� ������� = ��� a) ��� ��� 1. Exercise ����������� �������� Simplify: ��� × ������������ Activity Day-41  If xa = xb then a=b --64--  If mq=nq then m=n

Also study example 1 & 2 (page 121) Exercise - Solve: a) 32x+1 = 92x-1 b) 729 = 3x+4 � c) 25x+3 = �.�� Day-42 Activity Revise Activity of Day-41. Also study example - 3 (page - 121) Exercise Solve: a) 2x+3 + 2x = 36 b) 2x - 2x-2 = 6 1� c) 3x+2 + 3x+1 = � Day-43 Activity 5x-3 . 32x-8 = 225 Solve: ���������××��������=���2=22525 (45)x = 225 ×6561 ×125 Solution: (45)x = (45)5 or, or, or, x=5 Exercise 5x-2 . 32x-3 = 135 b) 2x+3 . 3x+4 = 18 a) Day-44 Activity ��� ��� ��� Prove that: ���������� ����� + =���������� ������� L.H.S. +�������� � = � +� � � � � = � � �� � � � = = � = = +� � = ��� ��� ������(�×����+�����)���������+(����×�) �� ��� (���)(���) ����������� �������� = RHS proved. ����� --65--

Exercise (3�� + 2+����)��������=����= �� Prove that: a) � ��� b) ������� Day-45 Activity Triangle Kindly request you to study the revision protion of triangle which you have already learn in previous class page (152 & 153) Exercise Read and write some axioms and pot postulates from page - 154 (Text book) Day-46 Activity 46  The sum of three angles of a triangle is 1800 (two right angle)  Study theorem - 1 page - 155 from your text book. Exercise In a triangle MNP, prove that ∡M + ∡N + ∡P = 1800 Day-47 Activity  The exterior angle of a triangle is equal to the sum of the two opposite interior angles.  Study theorem - 2 (page - 155) W Exercise X YZ From the given figure prove that: ∡ywx + ∡wxy = ∡wyz Day-48 A 380 B Activity 0 D Calculate the value of x 810 in the adjoining figure. CX Solution: Here ∡ B∡ O = ∡ODC[being alternate ∡K or, ∡ODC = 380 Now ∡COD + ∡OCD + ODC = 1800 [Sum of all angles of a triangle] or, 810 + x + 380 = 1800 or, 1190 + x = 1800 or, x = 610  x = 610 Also, study example 1 & 3 (page - 156) --66--

Exercise b) Find the value of x. a) Day-49 Activity (Isosceles triangle) Using the concept of activity of Days - 47 & 48 f (i) do the following Exercise. Exercise a) In ∆ABC, if 3∡A = 4∡B = 6∡C calculate ∡A, ∡B &∡C. b) In ∆ABC, the angle bisectors of ∡B and ∡C meet at O. If ∡a = 700 find ∡BOC. Day-50 Activity Congruency test of Triangles. i) Side, Angle side = SAS ii) Angle side Angle = ASA iii) Side side side = SSS iv) Right angle, hypogenous and any other sides = RHS. Exercise Read and write the congruency test of triangle from your text book page - 160. Day-51 Activity  Base angles of an isosceles triangle are equal.  Read and write theorem 3(page - 161) from you text book. P Exercise In an isosceles triangle PQR PQ = PR, Q DR Prove that ∡Q = ∡R Day-52 Activity  If two angles of a triangle are equal then the sides of opposite to them are also equal.  Study carefully & write theorem 4 (converse of theorem - 3) Page - 161 (from your text book) W Exercise In a isosceles triangle WXY, ∡X = ∡Y, them prove that WY = WX XP Y --67--

Day-53 Activity Study and write example 1&2 )page - 162) from your text book. Exercise b) E C 35o B Find the value of x & y D a) E A R y 140o y 0 CD A Day-54 Activity Study & write examples 3 & 4 (page - 163) from your text book then solve the following Exercise. Exercise A In the given figure AB = AC, BC = EC and∡DAE = 40°, prove that the 400 ∡ADEis an isosceles and calculate the size of the ∡ADB. BD EC Day-55 Activity  The bisector of the vertical angle of an isosceles triangle is perpendicular to the base and bisects the base.  Also study the theorem - 5 (Page - 166) from your text book. Exercise M In the given isosceles triangle MNO N PO in which PQ - PR, and ∡NMP = ∡OMP then prove that MP⊥ NO and NP = PO. Day-56 Activity Study example (page - 167) from your textbook. also revise activities 54&55 carefully. Exercise A D a) From the each of the following B 70o yo xC find the value of x & y. --68--

P S 35 b) In the adjoin figure, QS & RS are the x bisectors of ∡PQR &∡PRT respectively if RT ∡QSR = 35°find the value of ∡QPR. Q Day-57 Activity Study & write example 2 & 3 (page - 197& 168) from your text book. O P Exercise - 57 Q In the given figure, MN = OP and PN = OM prove that, OQ = NQ M Day-58 N Activity (Relation between sides and angles of a triangle)  The sum of two sides of a triangle is greater than the third side. Also study experiment verification -1 page - 172. Exercise Verify experimentally that in a triangle PQR, a) PQ+QR > PR b) PQ + PR > QR c) QR + PR > PQ Day-59 Activity Study carefully experimentally variation -2 (page - 172) Exercise In any triangle, the angle opposite to the longer side is greater than the angle opposite to the shorter side. Day-60 Activity  In a right angled triangle, the angle opposite to right angle is hypotenuse and remaining two side are base and perpendicular (base and perpendicular are recognize on the bases of reference angle)  h2 = p2 + b2 Exercise E a) are 3cm, 4cm & 5cm sides 300 300 of right angled triangle? Fx y G b) Find the value of x H A D 12cm 13cm x cm B C --69--

Subject: C. Mathematics Self-Evaluation Question Paper F.M.: 100 Class: Nine Time: 3:00hrs Group A (6X1=6) 1. a) If SP = Rs x & loss % = y% of an article. What is CP? b) Write the formula to calculate commission rate (%). 2. a) What are the factors of (a + 2b)� � b) Solve: 9 � = 3 3. a) Find the value of x. b) From the given figure, write the longest side. Group B (17X2=34) 4. a) If P={5,6,9,12,15} and Q= {1,9,15,17,19} find the value of P ∩ Q and Q − P b) If n(A) =10, and n(B) =12, what is maximum and minimum possible value of n(A ∪ B) 5. a) A clock bought for Rs 120 and sold at a profit of 20% . Find its selling price. b) If the cost price of 10 chairs is equal to the selling price of 16 chairs. Find the loss or gain percentage. 6 a) An agent gets Rs 150 commission on the sale of an article at Rs 1000. Find the rate of commission. b) Find the selling price if marked price is Rs 50,000 and discount is one fifth of marked price. c) The profit on one share is Rs 25. Find the total profit of the investment Rs 20,000 if Rs 1000 equal to 10 shares. 7. a) Factorize: a� + 64 b) Factorize: a� + a�b� + b� 8. a) Simplify: �√2��x�. �√32x�� b) Solve: 3��� + 3� = 4. 9. a) Find the value of x. b) Find the value of y. c) Find the value of z. --70--

10. a) In the adjoining figure, ∡ACB = 60°, ∡BAN = x° and ∡NBC = y° , If AB=AC then find the values of x and y. b) In ∆PQR, ∡P = 43°, ∡Q = 58° find the longest and shortest side. c) Find the value of x in the given diagram. Group 'C' (10X4=40) 11. A & B are two subsets of a universal set U in which, n(U)=160, n(A)=100, n(B)=60 & n(A∩ B) = 20, then i) Show the given information in Venn-diagram. ii) Find the value of n 12. Out of 100 students, 50 passed in Science, 71 in Mathematics, 10 failed in both subjects and 7 did not appear in an examination. Find the number of students who passed in both the subject by representing the information in Venn-diagram. 13. A shopkeeper sold a piece of Jacket for Rs 1082 at a loss of 10%. At what price should he sell to gain 10% on it? 14. Ram allows a discount of 10% and still makes a profit of 15% on selling an article. If the marked price of an article is Rs 1500, find profit amount. 15. A factory makes a profit of Rs 10,00,000 per month. At the end of the year 5% of its profit is divided among its 500 shareholders with equal number of shares. Find the dividend received by each. 16. Factorize:- 4a� − 20a + 24 + 6y − 9y� 17. Solve: 2���. 3��� = 18 18. In∆ABC the angle bisectors of ∡B and ∡C meet at O. If ∡A = 70°, �ind ∡BOC. 19. Prove that base angle of an isosceles triangle are equal. 20. Verify experimentally that, the side opposite to the greater angle of a triangle is greater than the side opposite to the smaller angle. Group 'D' (4X5=20) 21. Shishir bought 4000 oranges at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 Paisa each. If he plans to make a profit of Rs 200 at what rate must he sell the rest of oranges? --71--

22. Mr. Miraj a CEO of a bank has Rs 2,50,000 monthly salary. He is married. Calculate his annual tax liability. Tax Banding Tax Rates Upto 4,50,000 1% next 1,00,000 10% next 2,00,000 20% next 12,50,000 30% above 20 lakhs 36% 23. Simplify: m + (mn�)�/� + (m�n)�/� × �1 − ���/� ��/� 24. In the given figure, MN = OP and PN = OM prove that OQ=NQ. The End --72--

Subject : Science Day-1 Chapter: Measurement Activity-1 Study the following tips: The accurate quantity of a substance can only be known by measurement. Measurement can be expressed in figures and in standard measurement. We use different instruments like measuring tape or scale for the measurement of length, beam balance for measuring masses and clock for the measurement of time. Measurement : The process of comparing an unknown physical quantity with a known standard quantity of the same kind is called measurement. Those substances which are available in our surrounding can be measured are known as physical quantities. Some of the examples of physical quantities are the area, mass, time, length, temperature, etc. There are two types of physical quantities. They are discussed below. 1. Fundamental Physical Quantity 2. Derived Physical Quantity 1. Fundamental Physical Quantity: Those physical quantities, which can neither be derived from other quantities nor be further resolved into simpler ones, are called fundamental quantities. The units of fundamental quantities are called fundamental units or basic units. 2. Derived Physical Quantity : Those physical quantities, which depend on two or more fundamental quantities or power of the fundamental quantity, are called derived quantities. The units of derived quantities are called derived units Activity-2 Answer the following questions: 1. Define measurement. State the benefits of measurements. 2. Differentiate between fundamental and derived quantities. Day-2 Activity-1 Study the following tips: Unit : A unit is defined as a convention to define an amount of physical property in a specific system of units. Fundamental unit: The units of the fundamental quantities are called fundamental unit. Those units which are independent cannot be derived from other unit is called fundamental unit. Physical Quantity Unit Symbol mass kilogram kg length meter m time second s temperature Kelvin K electric current ampere A amount of matter mole mol luminous intensity candela Cd Derived units : The units which are composed of the fundamental units and are originated from the products and ratios of the fundamental units. Examples are: area, velocity, force etc. Velocity is a derived physical quantity which is derived from fundamental quantities of length and time Velocity = distance / time = m/s --73--

So the units of velocity is ms-1 For example, an area is a derived quantity. Since area= l × b, so the value of area depends on the value of length and breadth. Activity -2 Answer the following questions: 1. Define Unit. Why unit of pressure is derived unit? 2. Differentiate between fundamental and derived unit. Day-3 Activity-1 Study the following tips: The ancient people use hand-span, foot-span, finger width, palm length, the distance of a step, etc. as a units of measurements. This types of measurements are not accurate. To measure various quantities such as distance, height, width, weight, etc., a standard system of measurement is needed. Standard System Of units : CGS System: The base units for length, mass and time in this system are centimeter, gram and second respectively. This is French system. FPS System: The base units for length, mass and time in this system are foot, pound and second respectively. This is British System. MKS System: The base units in this system are meter, kilogram and second respectively. It is also called metric system International System (SI) of Units: The system of units which is agreed by the international convention of scientists held in France in 1960 is called SI units. It is based on seven base unit and the internationally accepted system at present. SI Base Quantities and Units Length − meter (m) Mass − kilogram (kg) Time − second (s)Electric current − ampere (A) Amount of substance − mole (mol) Luminous intensity − candela (cd) Temperature: Kelvin (K) Activity-2 Answer them: 1. What are the standard system of unit? Define all. 2. What is SI unit? Give example. Day-4 Chapter: Force Activity -2 Study the following tips: Rest and Motion : A body is said to be at rest if its position does not change with respect to time and observer. A body is said to be at motion if its position changes with respect to time and observer. Rest and motion are relative terms. A body may seem to be at rest with respect to one object but may appear to be in motion with respect to another object. Force : Force is an external agent that changes or tends to change the state of the body. It can change an object’s direction and velocity. Force can also change the shape of an object. The SI unit of force is Newton (N) and in CGS system unit is dyne. 1 Newton = 106dyne. Balanced and Unbalanced Forces : When balanced forces are applied to an object, there will be no net effective force acting on the object. Balanced forces do not cause a change in --74--

motion. Unbalanced forces acting on an object change its speed and/or direction of motion. It moves in the direction of the force with the highest magnitude. Activity -2 Answer them 1. Define force. Write SI and CGS unit of force. 2. What are balanced forces and unbalanced forces? Day-5 Activity-1 Study the following tips: Speed : The distance travelled by the body in a unit time is called speed. Speed is a scalar quantity. It has only magnitude but no direction. Its unit is m/s. Mathematically, Speed= Distance travelled/ Time taken Numerical If a motorcycle travels 4.8 km in 10 minutes, what distance does it travel in 1 second? Solution Here, Distance (S) = 4.8 km = 4800 m Time (t) = 10 min = 600 sec Now, Distance travelled in 1 second = s/t = 4800/600 =8m Thus the distance travelled in one second is 8m. Velocity : The total displacement covered by the body in unit time is called velocity. Velocity is a vector quantity. It has both magnitude and direction. Its unit is m/s Mathematically, Velocity = Displacement/ Time taken Activity -2 Answer them 1. Differentiate between speed and velocity. 2. Why speed is scalar quantity? Day-6 Activity-1 Study the following tips: Types of Velocity a. Uniform Velocity : A body is said to have uniform velocity if the body covers an equal distance in an equal interval of time in a straight line. The velocity is same throughout the motion. This velocity is also called constant velocity. b. Non uniform velocity : A body is said to have non uniform velocity if the body covers an unequal distance in an equal interval of time in a straight line. For a body moving with unequal velocity, we should calculate average velocity. Average Velocity = (Initial Velocity + Final velocity) / 2 Also, Average Velocity = Total displacement/ Time taken --75--

Acceleration : The rate of change of velocity with time is called acceleration. Its unit is m/s2. It is vector quantity. It may be positive or negative .Negative acceleration is called retardation. Mathematically, Acceleration = (Final Velocity – Initial Velocity) / Time taken a = (v –u) / t Activity-2 Answer the following questions: 1. What are the types of velocity? Define. 2. What is acceleration? Write its unit. Day-7 Activity-2 Equations of Motion : Consider a body moving in a straight line with uniform acceleration. Let, Displacement = s Initial velocity = u Final velocity = v Acceleration = a Time taken = t Relation between u, v, a, and t a = (v-u) / t Or, at = v-u ∴ v= u + at ........... (i) This is the first equation of motion. Relation between s, u, v and t. We have Average velocity = (v+u) / 2 Also, Average velocity = s / t As both equations are equal u + v = 1 2 4 Or, 2s = (u + v) × t ∴s =u + v × t....... (ii) This is the second equation of the motion. 2 Activity -2 Answer the following question A bus starts from rest. After 15 seconds it gains the acceleration of 19m/s2. Find the velocity of bus. Find the distance travelled by the above bus. Day-8 Activity-1 Study the following tips: Relation between s, u, a and t We already have, V= u + at ........(i) s= u+ v × t......(ii) 2 Putting value of v from equation (i) in (ii) Or, s = u + (u + at) ×t 2 Or, s = (2u × t + at × t ) --76--

∴ s = ut + 1 at2........(iii) This is the third equation of motion. 2 Relation between u, v, a and s We have, v = u + at..............(i) ∴s = u + v × t.......(ii) 2 Putting the value of t from (i) in the equation (ii), s = u + v × u -v 2 a or 2as = v2 – u2 Or, v2 = u2 + 2as..................(iv) This is the fourth equation of motion Activity-2 Answer them: 1. A car is moving with a velocity of 45km/hr. The driver presses the brake and the car comes to rest in 3 seconds. What is its retardation? How far does it move before coming to rest? 2. Practice all numerical from your exercise. Day-9 Activity-1 Study the following tips: Inertia : Amount of matter contained in the body is called mass. Inertia is the tendency of a body to maintain its state of rest or a uniform motion unless it is acted upon by some external force. It only depends on mass.i.e. Larger the mass large will be the inertia. Inertia of rest : Inertia of rest is the property of a body by virtue of which it remains or tends to remain in the state of rest unless it is acted by external forces. When a blanket is given a sudden jerk, the dust particles fall off. Inertia of motion : Inertia of motion is the property of a body by virtue of which it remains or tends to remain in the state of motion unless it is acted by external forces. A passenger in a moving bus jerks forward when the bus stops suddenly. Inertia of direction :Inertia of direction is the property of a body by virtue of which it maintains or tends to maintain its motion unless it is acted by external forces. When a running bus suddenly takes a turn, the passengers experience a jerk in the outward direction. Activity-2 Answer the following question 1. Define inertia along with its types. Day-10 Activity - 1 Study the following tips: Momentum : Impacts produced by objects depend on their mass and velocity. The momentum of an object is defined as the product of its mass and velocity. Mathematically, Momentum = mass × Velocity p = mv. --77--

It is a vector quantity so it has direction and magnitude. The SI unit of momentum is kg.m/s. Activity-2 Answer them 1. What is momentum? Write mathematical formula of momentum. 2. Differentiate between inertia and momentum. Day-11 Activity-1 Study the following tips: Newton’s law of Motion : After Galileo, Sir Isaac Newton made a detail and systematic study of motions of bodies and formulated the three laws of motion. These law are known as Newton’s law of motion. First Law of Motion Frictional force : The force that opposes relative motion is called friction. It arises between the surfaces in contact. Example: When we try to push a table and it does not move is because it is balanced by the frictional force. First Law of Motion A body continues to be in the state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force. First Law is also called the Law of Inertia. Examples: 1. When we shake mango tree, the mangoes fall down. 2. When a person jumps out of a moving vehicle, he falls forward due to the inertia of motion Interpretation of first law : According to the first law, a body continues to remain at rest if no unbalanced forces act on it. Similarly, the body will remain moving with uniform velocity in a straight line unless unbalanced forces act on it. Activity -2 Answer them 1. State Newton’s First Law. 2. Define frictional force. 3. Why do athletes run before taking a long jump? Day-12 Activity-1 Study the following tips: Second Law of Motion : The rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force. Let us assume a body of mass ‘m’ is moved after applying a force ‘F’. If ‘a’ is the acceleration produced in a body on a straight line, then according to Newton’s second law of motion; a∝ F ………….(i) when mass is constant a∝ 1 …………(ii) when force is constant m From above, ⇒F∝ ma ⇒F = kma For 1 unit of force on 1 kg mass with the acceleration of 1m/s2, the value of k = 1. Therefore, F = ma. --78--

Examples; 1. A cricket player while catching a ball moves his hands backwards. 2. A person falling on the cemented floor is injured more than a person falling on a sandy floor or mattress. Interpretation of second law : If a small body is pushed gently, a small acceleration is produced. If it is pushed harder, a larger acceleration is produced. If two bodies, one lighter and another heavier, are pulled by the same amount of force in the same direction separately, it is found that the heavier body has less acceleration than the lighter body. Activity-2 Answer the following questions: 1. State and prove Newton’s Second law of motion. 2. Why a moving truck takes a much longer time to stop than that taken by a car when brakes are pressed at the same time? Day-13 Activity-2 Study the following tips: Third Law of Motion : Newton’s 3rd law states that every action has an equal and opposite reaction. Action and reaction forces are equal, opposite and acting on different bodies. Examples: 1. While rowing a boat, a person pushes water backwards with the help of the oars. Due to the reaction offered by water, the boat moves forwards. 2. Birds, while flying, push air with their wings (action). The air exerts an equal and opposite force on the birds (reaction) and that force causes birds to move forward. Interpretation of third law: Third law implies that forces always occur in a pair and a single force is thus impossible. Activity-2 Answer the following questions 1. State Newton’s Third law of motion. 2. Why do a gun recoils after firing a bullet? Day-14 Machine Activity-1 Study the following tips: We use various tools to perform different types of work in our daily life. The tools or simple devices used for making our work easier, faster and more convenient are simple machines. Mechanical advantage : The simple machine requires force to do work. The resistive force to be overcome is called load and the force applied to overcome the load is called effort. The ratio of the load to the effort in a simple machine is called mechanical advantages of the machine. Mechanical Advantage (M.A.) = Load / Effort It has no unit.It is theaffected by friction. Velocity ratio : Velocity ratio of simple machine is the ratio of distance travelled by an effort to the distance travelled by the load in the machine. Velocity ratio (V.R.) = Distance travelled by effort / Distance travelled by Load --79--

As velocity ratio or ideal mechanical advantage is a simple ratio of two distances, it also does not have the unit. The friction is not involved in it. Activity-2 Answer them 1. Define simple machine and write the purpose of using them. 2. Differentiate between M.A. and V.R. Day-15 Activity-1 Study the following tips: Efficiency of the machine : If a machine overcomes a load ‘L’ and the distance travelled by the load is ‘Ld’, the work done by the load is L × Ld. It is also called output work or useful work. Therefore, Output work = L × Ld. Likewise, the effort applied to overcome the load is E and the distance covered by effort is Ed, the work done by effort is E × Ed. It is also called input work. Therefore, Input work = E × Ed. Efficiency(Ƞ) = Output work / Input work For ideal or perfect machine, work output will be equal to the work input. Ideal machinesare those imaginary machines which are frictionless. In practice, the work output of a machine is always less than work input due to the effect of friction. If the frictional force in the machine increases the efficiency decreases. Since machines are not frictionless in practice, the efficiency of a machine can never be 100%. Relation between M.A, V.R. and efficiency We know that, Efficiency(Ƞ) = Output work / Input work Or, Ƞ= work done by leofafdort×100 % work done bby Or, Efficiency(Ƞ) = LE××LE..dd×100% Or, Ƞ= M.A × 100% V.R Activity-2 Answer the following questions: 1. Show the relation between M.A., V.R. and efficiency. 2. Calculate the efficiency of lever having output work of 600 J and input work of 900 J? Day-16 Activity -1 Study the following tips: To increase efficiency and mechanical advantage of the machines, we have to reduce friction. Ways to reduce friction 1. By using lubricants. 2. By using ball bearings, wheels and rollers. 3. By making surface smooth. 4. By rolling instead of sliding. Principle of Simple Machine --80--

It states that, “If there is no friction in a simple machine, output work and input work are found equal in that machine.” Mathematically, Output work = Input Work Activity -2 Answer the following questions: 1. Write some methods to increase efficiency of a simple machine. 2. Why the value of M.A. is always lesser than V.R.? 3. State Principle of Simple Machine. Day-17 Activity-1 Study the following tips: Types of Simple Machines Lever : A lever is a rigid bar may be straight or bent which is capable of rotating fixed point called fulcrum. In a lever, effort distance and load distance are measured from fulcrum. The distance between fulcrum and load is load distance and distance between fulcrum and effort is effort distance. Numerical Efficiency of lever is 60%. If its M.A. is 3, calculate a. effort applied to lift a load of 1200N by using the lever. b. velocity ratio of the lever c. output work if load is turned 10cm high. d. input work . Solution: M.A = 3 Efficiency = 60% Load = 1200N M.A = Load / effort ∴ Effort = 400N Efficiency = M.A / V.R× 100% 60% = 3 / VR×100% ∴ V.R = 5 Output work = L× L.d = 1200 × 0.1 = 120 J Efficiency = output work / input work× 100% 60% = 120 / input work× 100% ∴Input work= 200 J Activity-2 Answer the following questons 1. Define lever. State its types. 2. Calculate M.A. , V.R. and efficiency of a lever with following data: Load = 600N, Load distance = 200cm, Effort = 800N, Effort distance = 0.8m. Day-18 Activity-1 Study the following tips: Pulley : A pulley is a metallic or wooden disc with a grooved rim. The rim rotates about a horizontal axis passing through its center. A pulley can be used in single fixed pulley, single movable pulley or combined form (block and tackle). A single fixed pulley makes our work easier by changing the direction only; mechanical advantage is not gained because value of effort and load distance is equal. But in single movable pulley, mechanical advantage is --81--

gained. In a block and tackle, MA and VR are directly proportional with the number of pulleys used. If the number of pulley increases in block and tackle the VR also increases. In the same way MA increases. Numerical A load of 800N is lifted using a blocjk and tackle having 5 pulleys. If the applied effort is 200N, calculate a. M.A b. V.R. c. Efficiency d. Output work if load is lifted 5meter high. Load (L) = 800N Effort (E) = 200NV.R = no. of pulley used = 5 M.A = L / E M.A. = 800/200 = 4 V.R = No. of pulleys = 5 Efficiency = M.A / V.R×100% = 80% Effort distance = 5×5 = 25m If load is lifted to 5m high, output work = L× L.d = 800×5 = 4000J Input work = Effort× effort distance = 200×25 = 5000J Activity-2 Answer them: 1. Write the types of pulleys with advantages. 2. A load of 555 N is lifted by the five pulley system up to 5 meters by applying an effort of 222 N. Calculate: MA, VR, efficiency, output work and input work. Day-19 Activity-1 Study the following tips: Inclined plane : Slanted surface which is used to lift the heavy load by applying less effort is called inclined plane. In this device, the length of slope (l) acts as effort distance and height of slope (h) acts as load distance. The value of length of slope is always greater than value of height of slope; therefore value of VR is always greater than 1. V.R. = l / h Numerical A load of 680N is lifted from base to tip by 500N force on an inclined plane whose length of slope is 12m and height of slope is 8m. Calculate: a. output work b. input work c. M.A. d. V.R. e. Efficiency. Solution: Load (l) = 680N Effort (E) = 500N Length of slope (l) = 12m Height of slope (h) = 8 m Now, Output work = load × height 680 ×8 = 5.44 ×103 J I nput work = effort × length = 500 × 12 = 6000J Mechanical advantage (M.A) = load / effort= 680 / 500 =1.36600500=1.36 Velocity ratio (V.R) = L/ H = 12/8 = 1.5 Efficiency = M.A / V.R × 100% = 90.6% --82--

Activity-2 Answer the following questions: 1. Define inclined plane. Why VR is always greater than 1 in inclined plane? 2. Calculate output work, input work, MA, VR and efficiency of inclined plane having following parameters. Load = 0.5 N Effort = 0.35 N length of the slope = 0.4m, height of the slope = 0.19m. Day-20 Activity -1 Study the following tips: Wheel and axle : Wheel and axle consists of two coaxial cylinders of different diameters. Some examples of wheel and axle are string roller, screw driver etc. In wheel and axle, effort is applied on big cylinder called wheel and load is overcome by small cylinder called axle. The circumference of big cylinder (wheel) is considered as effort distance and circumference of small cylinder (axle) is known as load distance. Therefore, the calculation of VR in wheel and axle can be done by following formula VR = 2 πR / 2πr Or, VR = R / r When the wheel an axle is in use, the points of load and effort vary continuously for 3600 on the circumference of the wheel and axle, hence it is considered as continuous lever. Illustration : The radius of the wheel is 40cm and the radius of the axle is 0.1m in a wheel and axle. If aload of 1200N is overcome up to 4m by applying an effort of 400N, calculate a. M.A. b. V.R. c. efficiency d. Output work e. Input Work Solution; Radius of wheel (R) = 40cm =0.4m Radius of axle (r) = 0.1m Load = 1200N Effort = 400N Mechanical advantage (M.A) = load/effort= 3 Velocity ratio (V.R) =R/r = 4 Efficiency =M.A/V.R× 100% = 75% V.R = effort distance/load distance= 4 ∴Effort distance = 4×4 = 16m Output work = L× L.d = 1200 × 4 = 4800J Input work = E × E.d = 400 × 16 = 6400J Activity-2 Answer the following questions: 1. Why wheel and axle is called continuous lever? 2. In a wheel and axle, the radius of the wheel is 17cm and that of the axle is 8cm. If the load of 777 N is overcome by using the effort 189 N on it, calculate MA, VR and efficiency of the machine. --83--

Day-21 Activity-1 Study the following tips: Moment : Moment of the force may be defined as the product of the force and its perpendicular distance of the line of action from the axis of rotation. Moment = f×d Law of moment : It states that “In the balanced condition the sum of the clockwise moment acting on it is equal to the sum of anti-clockwise moment on it”. Mathematically, Effort × Effort distance = Load × Load Distance Illustration A lever is shown in the given figure (refer page no 53, q.no i). Three weights, 60N, 30N and 10N are suspended on it. Now, calculate a. Clockwise moment. B. Anti-clockwise moment c. Can the lever be in the condition of balanced in this situation? d. What will be the location of 10N weight by keeping other loads unchanged to balance the lever? Solution, Load (L1) =60N Effort (E1) = 30N Distance between L1 and fulcrum (Ld1) = 40cm =0.4m Distance between E1 and fulcrum (Ed1) = 10cm = 0.1m Distance between E2 and fulcrum (Ed2) = 30cm = 0.3m Clockwise moment = E1 × Ed1 + E2 × E d2 =10 × 0.1 + 30×0.3 = 10Nm Anticlockwise moment = L1×Ld1=60 ×0.4 = 24Nm The lever will not be in balance because the clockwise moment and anti-clockwise moment are not equal. To balance the lever, Clockwise moment = anticlockwise moment E1 × Ed1 + E2 × E d2 = L1×Ld1 Ed1 = 1.5m Activity-2 Answer the following question: 1. Define moment and state law of moment Day-22 Chapter: Classification of Elements Activity-1 Study the following tips: A matter is anything that occupies space and has mass. Book, wood, pens, water etc. are the examples of matter. The substances which is composed of its unit particles of a similar kind are known as pure substances. The substances which are composed of unit particles of different nature are known as impure substances. Element : An element is the simplest form of a pure substance, which cannot be split up into two or more simpler substances by any chemical reaction. All the substances are made up of elements. Hydrogen, oxygen, carbon, gold etc. are the examples of elements. There are 118 elements altogether, among them 92 elements occurs naturally. --84--

Symbol : A symbol is defined as the abbreviation of the full name of an element. Using the English alphabet symbols of elements are expressed. There are three main ways of making symbols a. Using the first letter of the name of elements. S. N Elements Symbol 1Hydrogen H b. If the name of two or more elements begins with the same letter, another significant letter is taken with the first letter. S.N Elements Symbol 1 Helium He c. Using letters from the name of elements of their original root also make symbols. So, symbols are derived from Latin and some are from German. S.N Elements Name in other language Symbol 1 Copper Cuprum (Latin) Cu Activity-2 Answer the following questions: 1. Define element. 2. What is symbol of an element? Write some examples. Day-23 Activity-1 Study the following tips: Compound : Compound is a substance of atoms of two or more elements formed by the chemical combination in a definite proportion by weight. For example, water is a compound because it is made of more than one element- Hydrogen and Oxygen in the ration of 2:1. The molecular formula of water is H2O. Molecule : The smallest particle of an element or compound, which is capable of independent existence is called molecule. Some of the examples of molecules are H2O (water), CO2 (Carbon dioxide), etc. Atom : The smallest particle of an element, which takes part in chemical change is called an atom. Activity -2 Answer the following questions : 1. Define atom and molecule. 2. Differentiate between element and compound. Day-24 Activity-1 Study the following tips: Structure of the atom : Atom is made up of tiny particles. These tiny particles are called sub-atomic particles or elementary particles. They are electrons, neutrons, and protons. The protons and neutrons are located at the center of an atom called nucleus. Nucleus shows the positive charge due to the presence of protons. It helps to find the atomic weight. The electrons are revolving around the nucleus in their respective orbit or shell. Electrons have negative charge in them. --85--

In any atom, the number of negatively charged electrons in the orbits is equal to the number of positively charged protons in the nucleus. Thus, an atom is electrically neutral. Atomic number It is denoted by Z. Mathematically, Atomic number (Z) = No. of protons (p+) = No. of electrons (e-) Atomic weight It is also called mass number. It is denoted by A. Mathematically, Atomic weight (A) = No. of protons (p+) + No. of neutrons (n0) Activity-2 Answer the following questions: 1. Why an atom is electrically neutral in nature? 2. Define atomic number and atomic weight. 3. Practice to tell the first 20 elements of Modern periodic table serially. Day-25 Activity-1 Study the following tips: Electronic configuration : The systematic distribution of an electron in different shells of an atom is called electronic configuration. To explain the arrangement of electrons in different shells, Bohr and Bury purposed a scheme as given below: The maximum number of electrons in each shell is determined by 2n2 rule, where n is the number of the shell. For example: For the K shell, the maximum number of electrons will be 2n = 2.(1)2 = 2 × 1 = 2 2. The maximum number of electrons in the outermost orbit is 8 and in second last orbit is 18. 3. It is not necessary for an orbit to be completed before another begins. The energy level or shell nearer to the nucleus is called lower energy level and energy level away from the nucleus is called higher energy level. Shells : The path traced by the electrons around the nucleus are called orbits or shells. There are K, L, M, N, O, P AND Q. Sub-shells : A sub shell is the pathway in which an electron moves within a shell. They are designated by s, p, d and f. Shells Sub shells K ( n = 1) 1s L ( n=2) 2s and 2p M ( n=3) 3s, 3p and 3d N ( n = 4) 4s, 4p, 4d and 4f Activity -2 1. Define electronic configuration. 2. Write the formula to determine the number of electrons in a shell. 3. Define shell and sub shell. Name them too. --86--

Day-26 Activity-1 Study the following tips: Aufbau principle : This principle explains how the atoms are being arranged in orbital. According to this principle, the sub shells of lowest energy is filled first then higher energy level are filled so on. Illustration Potassium (K) The atomic number of potassium is 19. Its electronic configuration is given below. Shells K LMN No. of Electrons 2 8 8 1 Electronic Configuration: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 Activity-2 Answer the following questions: 1. Write principle of Aufbau principle. 2. Write complete electronic configuration of Helium, Sodium, Calcium & Fluorine. Day-27 Activity-1 Study the following tips: Valence shell and Valence electrons :The outermost orbit or shell of an atom is called valence shell and the number of electrons present in the valence shell of an atom is called valence electrons. They are far from the nucleus. The valence electrons determine the valency of an element. Valence electron takes part in the chemical reaction. From valence electron, we get various information. Some of the information are given below; 1. It gives information about the combining capacity of the element. 2. The valence electron of an element gives information about the position of the element in a periodic table. 3. The number of shells determines the period to which the element falls in the periodic table. Valency : Valency is defined as the combining capacity of element or radicle with the other element or radical to form a molecule or a compound. It is represented by numbers like1, 2, 3, 4, 5, and 6. Activity-2 Answer the following questions: 1. Define valency. 2. Define valence electron. What are the information we obtained from valence shell? --87--

Day-28 Activity-1 Study the following tips: Ways of calculating valencies of an element or a radical 1. Hydrogen, oxygen, and chlorine are taken as the standard elements to determine the valency of an element or radical. The number of hydrogen atoms or chlorine atoms or a double number of oxygen atoms with which one atom of an element combines is known as its valency. If an element does not combine with hydrogen then in such condition, the valency is determined by comparing it with that of chlorine or oxygen. For example, The valency of nitrogen (N) in ammonia (NH) is 3 because 1 atom of N combines with 3 atoms of H. The valency of sodium (Na) in sodium chloride (NaCl) is 1 because 1 atom of Na combines with 1 atom of Cl. 2. Valence electron also determines the valency of an element. For example, the valence electron of Silicon is four. So, the valency of silicon is 4. 3. Valency is also determined by the number of electrons lost, gained or shared by the element during molecule formation. For example, a sodium atom loses an electron and chlorine gain one electron to form a compound. So, the valency of sodium and chlorine is 1. Variable valency : We use Greek prefixes to describes different valancies like mono for one, di for two, tri for three, tetra for four, penta for five and hexa for six. For example, sulphur and magnesium are divalent as they have valency two. The name of elements with the lower valency ends with a suffix- ous and that with the higher valency ends with the suffix- ic.Some elements have changeable combining capacity. When an element shows two or more than two valencies, then it is called variable valency. For example, copper shows valency 1 in cuprous chloride (CuCl) and valency 2 in cupric chloride (CuCl). Activity-2 Answer the following questions: 1. What are the ways to calculate valency of an element? 2. Define variable valency. Day-29 Activity-1 Study the following tips: Radicals : A radical is an atom or a group of atoms of different elements having positive or negative charge and acts as a single unit throughout the chemical reactions. They do not exist freely. Radicals can be classified into two types; electropositive radicals and electro negative radicals. Electro-positive radicals: The radicals which have the capacity of losing electrons and carry positive charge are called electropositive radicals. E.g. Ca, Na etc. Electro-negative radicals: The radicals, which have the capacity of gaining electrons and carry negative charge are called electronegative radicals. E.g. O, N etc. Inert gases : The elements fall into the zero group of the periodic table are inert gases or noble gases. They are helium, neon, argon, krypton, xenon and radon. These elements are chemically inert because of their stable electronic configuration. Duplet and duplet rule : The arrangement of two electrons in the K-shell is said to be duplet. It is chemically inert because it has a complete number of electrons in its K-shell. --88--

Helium is a duplet as it has two electrons in K- shells. The tendency of an atom with single shell to attain two electrons in its K- shell is called duplet rule. Octet and octet rule : Except helium, the other inert gases have 8 electrons in their valence shell. Such arrange of a stable group of 8 electrons in their valence shell is said to be octet. Some atom of elements make 8 electrons in their valenceshell by gaining, losing or sharing electrons between the combining atoms during the formation of a molecule is called octet rule. Activity-2 Answer the following questions: 1. Define radicals. Define its types. 2. Define inert gases. What is the valency of inert gases and why? 3. Define octet and duplet. Day-30 Active-1 Study the following tips: Chemical Bonding : Chemical bonding is the force of attraction that binds atoms or group of atoms to form a stable chemical molecule or compound. For example, hydrochloric acid (HCL) is formed of one molecule of hydrogen and one molecule of chlorine. The chlorine and hydrogen are found combined together due chemical bond. Here we study about two types of bonding which are given below; Electrovalent bond : The chemical bond formed by the transfer of electrons from valence shell of an atom to the valence shell of another atom is called electrovalent bond and the valency is known as electrovalency. Electrovalent bond is formed in between two atoms in which one atom belongs to metal and another is non-metal. The metal atom loses its valence electron and non-metal atom gains an electron from metal. Compounds like magnesium chloride, calcium oxide, sodium oxide, magnesium oxide and potassium chloride are formed by the electrovalent bonding. Characteristics of electrovalent compounds 1. They have high boiling point and melting point. 2. They are found in solid state. 3. They conduct electricity in molten or aqueous solution. 4. They contains metal in their molecule. Activity-2 Answer the following questions: 1. Define chemical bonding. 2. Define electrovalent bond and write its characteristics. Day-31 Activity-1 Study the following tips: Covalent bond : The covalent bond is formed by the sharing of a pair of electrons between two atoms each contributing an equal number of electrons to the electron pair. A covalent bond is represented by line (-) between the bonded atoms. When one pair of electron is shared, a single covalent bond is formed. Similarly, when two and three pairs of electrons are shared, they form double and triple bond respectively. Compounds like methane, ammonia, water and oxygen are formed by the covalent bonding. --89--

Characteristics of covalent bond 1. They have low melting point and boiling point. 2. They do not conduct electricity. 3. They are water insoluble but soluble in organic solvents. 4. They are found in all three state i.e. solid, liquid and gaseous state. Molecular Formula : A molecule is not a single atom but it is formed by the combination of two or more atoms of the same or different types in fixed proportion by their weight. Molecular formula is defined as the symbolic representation of the molecule of a substance. By observing the molecular formula, we can determine the no. of atoms involved in the formation of the molecule. Activity-2 Answer the following questions: 1. Define covalent bond with characteristics. 2. Define molecular formula. 3. Differentiate between electrovalent bond and covalent bond. Day-32 Activity-1 Study the following tips: Molecular formula of a compound: The molecular formula of a compound is defined as the symbolic representation which shows the actual number of an atom of different elements present in one molecule of the compound. Information obtained from molecular formula; a. It represents the composition of each element present in the given compound. b. It gives information about the total number of atoms of different elements involved in the formation of the molecule. c. It shows the combining capacity of its constituent elements with hydrogen i.e. valency. d. We can calculate molecular weight of the substance by observing its molecular formula. e. It represents one molecule of the substance. Writing molecular formula: To write the molecular formula of a molecule, we should follow the following steps given below; Calcium Sulphate (compound) CaSO4(symbol of basic and acidic radicals) Ca valency = 2 SO4 valency = 2 (valency at right upper corners of the symbols) . valency are exchanged and compound radical is enclosed in bracket CaSO4 → (L.C.M. is taken to get molecular formula of calcium sulphate) Activity -2 Answer the following questions: 1. What are the information we get from molecular formula of a compound? 2. Write molecular formula of following compounds. Ammonium Chloride, Calcium Carbonate, Magnesium Bicarbonate, Silver Chloride, Hydrogen peroxide, Aluminium Hydroxide. 3. Differentiate between symbol and molecular formula. --90--

Day-33 Chapter: Chemical Reaction Activity-1 Study the following tips: In our daily life, we observe many changes around us. The changes occurred in all the substance. When changes takes place in substances, either energy is given or energy release from them, all the changes can be classified into physical change and chemical change. Physical change : Physical change is a temporary or reversible change in which no new substances are formed. Examples: Lightening of an electric bulb. Magnetizing an iron nail. Preparing a solution etc. Characteristics of physical changes 1. Physical changes are temporary a change. 2. They are reversible change. In these changes, no new substances are formed. In these changes, only physical properties are changed but chemical compositions are not changed. . Chemical change : A chemical change is a permanent and usually irreversible change in which new substances with different properties are formed. Characteristics of chemical change 1. Chemical changes are permanent changes. 2. They are usually irreversible changes. In this change, new substances are formed. Rusting of iron. Burning magnesium ribbon in air to form magnesium oxide are some examples of chemical change. Activity-2 Answer the following questions: 1. List any five changes you observe today. 2. Define physical change and write its characteristics 3. Define chemical change and write its characteristics. 4. Differentiate between physical change and chemical change. Day-34 Activity-1 Study the following tips: Chemical reaction: The process by which a chemical change takes place is called as a chemical reaction. Those substances that take part in a chemical reaction are called reactants. Those substances which are formed by a chemical reaction are called products. Word equation : The chemical reaction which is expressed in terms of full names of reactant and product molecules is called word equation. For example, Hydrogen reacts with oxygen to produce water. Chemical Reaction Hydrogen + oxygen→Water Hydrogen + oxygen→Water (Reactants) (Products) Activity -2 Answer the following questions: 1. Define chemical reaction. 2. Define reactant and product. 3. What is word equation? Write a word equation for, when nitrogen reacts with hydrogen, ammonia gas is formed. --91--

Day-35 Activity-1 Study the following tips: Chemical equations : The chemical reaction which is expressed by using themolecular formula of reactants and products is called chemical equation. The chemical equation is also called formula equation. Hydrogen + Oxygen → Water 2H2+ O2→ 2H2O Skeletal chemical equation : The chemical equation in which the total number of atoms of each element in reactant and product is not equal is called an unbalanced chemical equation or skeletal chemical equation. Some examples: Sodium + Chlorine → Sodium chloride [Word equation] Na + Cl → NaCl [Skeletal equation] Zinc + Hydrochloric acid →Zinc chloride + Hydrogen [Word equation] Zn + HCl → ZnCl2 + H2 [Skeletal equation] Drawbacks of skeletal chemical equation a. Skeletal chemical equation does not follow the law of conservation of mass. b. It does not tell about the ratio of reactant and product molecules. c. It does not give information about the total number of atoms of each element in reactants and products. d. It is essential to balance the chemical equation because a balanced chemical equation satisfies the law of conservation of mass in a chemical reaction. Activity -2 Answer the following questions: 1. Define chemical equation. 2. What is skeletal equation and what are its drawbacks? Day-36 Activity-1 Study the following tips: Balanced Chemical Equation : There are many unbalanced chemical equations which need to be balanced. Mostly, we balance the chemical equation by comparing the numbers of atoms in reactant and product molecules and then equalizing the both sides. Balanced chemical equation is a type of equation in which a total number of atoms in reactants and products are equal. It is based on the law of conservation of mass. For example; Hydrogen and oxygen react with eachother to produce water. Hydrogen + Oxygen→ Water [Word equation] H2 + O2 → H2O (Skeletal equation)..........(i) 2H2 +O2 → 2H2O (Balanced chemical equation).............(ii) In the above equation (ii), the total number of atoms in reactant and product are equal so it is a balanced chemical equation. Activity-2 Answer the following questions: 1. Define a balanced chemical equation. On what law it is based on? 2. Practice few reactions to balance chemical equation. --92--

Day-37 Activity-1 Study the following tips: Balancing a Chemical Equation Illustration: Ammonia gas is prepared by the reaction of ammonium chloride and calcium hydroxide. i. Ammonium chloride + calcium hydroxide → calcium chloride +Ammonia + Water [ Word Equation] ii. NH4Cl + Ca(OH)2→ CaCl2 + NH3 + H2O [Skeletal Equation] iii. Here NH4Cl is multiplied by 2 to equalize with Cl atoms. 2NH4Cl + Ca(OH)2→ CaCl2 + NH3 + H2O iv. Now, to equalize N atoms, NH3 is multiplied by 2. 2NH4Cl + Ca(OH)2→ CaCl2 + 2NH3 + H2O v. To equalize oxygen and hydrogen, H2O is multiplied by 2. 2NH4Cl + Ca(OH)2→ CaCl2 + NH3 + 2H2O [Balanced Chemical equation] In this way, we get balanced chemical equation. Activity-2 1. Change the following word equations into balanced equations. a. Hydrogen + Chlorine → Hydrogen Chloride. b. Potassium chlorate → Potassium Chloride + Oxygen c. Copper + Sulphuric acid → Copper sulpahte + Water + Sulphur dioxide d. Aluminium + Sulphuric Acid → Aluminium sulphate + Hydrogen e. Irop + Oxygen → Iron Oxide Day-38 Activity-1 Study the following tips: Some more practices: 1. When calcium carbonate is heated, it forms calcium oxide and carbon dioxide. Calcium carbonate + Heat→ Calcium oxide + Carbon dioxide. [Word Equation] CaCO3 + Heat → CaO + CO2 [Skeletal Equation] CaCO3 + Heat → CaO + CO2 [Balanced Equation] 2 . When potassium burns with oxygen, it forms potassium oxide. Potassium Oxygen→ Potasssium oxide [Word Equation] K + O2 → K2O [Skeletal Equation] 4K + O2 → 2K2O [Balanced Equation] 3 .When ethane is burnt in oxygen, carbon dioxide gas and water are produced. Ethane + Oxygen → Carbon dioxide + water [Word Equation] C2H6 + O2 → CO2 + H2 [Skeletal Equation] 2C2H6 + 7O2 → 4CO2 + 6H2O [Balanced Equation] Activity -2 1. Balance the following skeletal equations. a. K + O2→ K2O b. Hg + O2 → HgO c. Na + Cl2 → NaCl --93--

2. Fill in the gaps and balance the following equations: a. Ca(OH)2 + ………….. → CaCO3 + …………….. b. H2O2 → H2O + ……………….. Day-39 Activity-1 Study the following tips: Irreversible reaction and reversible chemical reactions : The chemical reaction that occurs only in one direction is called irreversible reaction. For example, when calcium carbonate is being strongly heated, it decomposes to calcium oxide and carbon dioxide. Calcium carbonate → Calcium oxide + Carbon dioxide CaCO3 → CaO + CO2 ↑ The chemical reaction that occurs both in forward and backward directions is called reversible reaction. Nitrogen+Hydrogen ⇋ Ammonia N2+H2 ⇋NH3 Information we get from balanced chemical equation: 1. Formula of the reactant and product. 2. Total number of atoms or molecules of reactants and products. 3. Type of chemical reaction. 4. Ratio of masses of reactants and products. 5. Name of the type of reactants and products. Limitations of Chemical reactions: 1. It cannot give the physical state of reactants and products. 2. It do not give information about rate of chemical reaction. 3. It do not tell about the time taken for reaction. 4. It do not tell about the concentration of reactants. 5. It do not tell how hazardous is the reaction is. Activity -2 Answer the following questions: 1. What are the limitations of chemical reactions? 2. Define reversible and irreversible chemical reaction with examples. Day-40 Activity-1 Study the following tips: Endothermic and Exothermic Reactions Exothermic Reactions : An exothermic reaction is a reaction where energy is released from the system into the surroundings. Usually we observe this energy as heat. C + O2→CO2 + heat Example: Endothermic Reactions : An endothermic reaction is a reaction where energy is absorbed from the system into the surroundings. CaCo3+ heat → Cao + Co2↓ Example: Catalyst : A catalyst is defined as a chemical substance that changes the rate of chemical reaction itself without undergoing any permanent chemical change during the course of chemical reaction. They are of two types; they are positive catalyst and negative catalyst. Positive Catalyst: The catalysts that increases the rate of chemical reaction is called a positive catalysts. --94--

Manganese dioxide acts as a positive catalyst in the decomposition of hydrogen peroxide. This is because it increases the decomposition rate of hydrogen peroxide into water and oxygen. 2H2O2 →MnO2 2H2O + O2↑ Negative Catalyst: The catalyst that decreases the rate of chemical reaction is called a negative catalyst. Glycerin acts as a negative catalyst in the decomposition of hydrogen peroxide. This is because it decreases the decomposition rate of hydrogen peroxide into water and oxygen. 2H2O2 →Glycerol 2H2O + O2↑ Characteristics of Catalyst a. The mass and chemical composition of catalyst is unchanged. b. Catalyst is generally used in small amount. c. Catalyst do not initiate a reaction. d. Specific catalyst are used for specific reaction only. Activity -2 Answer the following questions: 1. Differentiate between exothermic and endothermic reactions. 2. Define catalyst and write its characteristics. 3. What are the types of catalyst? Give examples. Day-41 Chapter: Classification of Plants and Animals Activity-1 Study the following tips: There are various kinds of living beings in our surroundings. The living beings may be plants, animals and even microorganisms. The process of systematic and scientific grouping of living organisms into various groups and subgroups on the basis of their similar and dissimilar characteristics is called classification of living beings. Taxonomy : It is the branch of biology that deals with identification, nomenclature and classification of living organisms. Father of taxonomy is Carolus Linnaeus. Importance of classification 1. It makes study of organisms easy and systematic. 2. It helps to know inter-relationship between different groups of organisms. 3. It helps to know the evolution time of organisms. 4. It helps to study about bio diversity. Hierarchy of classification Kingdom Phylum/ Division Class Order Family Genus Species Activity-2 Answer the following questions: 1. Who is father of taxonomy? 2. Define classification with its importance. --95--

Day-42 Activity-1 Study the following tips: Genus : It is a group of closely related two or more species. Species : It is a group of living beings which can interbreed and reproduce among themselves and keep their population alive. Nomenclature : The process of giving scientific names to plants and animals is called nomenclature. During nomenclature two names are given. One is genus and another is species. The system of giving two name to each organism is binomial nomenclature. During writing scientific names, first we have to write genus and then species. Genus is written by a word with first letter capital and species by all small letter. Both genus and species have to be underlined. Eg. Humans = Homosapiens Frog =Ranatigrina Bee = Apisdorsata Carlous Linnaeus classified organisms into two kingdom; Plant Kingdom and Animal Kingdom on the basis of mode of nutrition and ability of locomotion. Activity -2 Answer the following questions: 1. What is binomial Nomenclature? 2. Write scientific name of Mosquito, pea and mustard. Day-43 Activity-1 Study the following tips: Five Kingdom system of classification Robert H. Whitaker, an American taxonomist proposed a five kingdom classification. This classification is more accurate and scientific than two kingdom classification on basis of following facts. 1. Prokaryotic and Eukaryotic organisms are separated. 2. Unicellular and Multicellular organisms are separated. 3. Green plants and non-green plants are separated. 4. It is based on the evolution of organisms making a clear concept of living organisms. Five Kingdom; Monera, Protista, Plantae, Fungi and Animalia. Kingdom Monera Characteristics: a. They are most primitive prokaryotic organisms. b. They are unicellular and microscopic c. They can be autotrophic or heterotrophic , parasitic or saprophytic. Examples are all bacteria, blue green algae etc. Kingdom Protista Characteristics: a. They are eukaryotic unicellular organisms. b. They can be autotrophic or heterotrophic. c. They have well developed nucleus. Examples are Amoeba, Paramecium etc. --96--

Activity -2 Answer the following questions: 1. Who proposed five kingdom of classification? 2. What are five kingdom of classification? 3. Write characteristics of kingdom Protista. Day-44 Activity-1 Study the following tips: Kingdom Fungi Characteristics: a. They are heterotrophs. b. Chlorophyll is absent. c. They grow in moist places. Examples are Mushroom, Mucor etc. Kingdom Plantae Characteristics: a. They are eukaryotic multicellular organisms. b. They have presence of chlorophyll. c. They are either flowering or non flowering. Examples are Maize, Pea etc. Kingdom Animalia Characteristics: a. They are eukaryotic multicellular organisms. b. They do not have chlorophyll. c. They are heterotrophic. Examples are earthworm, fish etc. Kingdom Plantae : This Kingdom is classified into three divisions on the basis of plant body, stage of development and presence of vascular tissues. Division Algae Characteristics: a. They are autotrophs. b. Chlorophyll is present. c. They are green filamentous plants. Examples are Spirogyra, Volvox etc. Division Bryophyta Characteristics: a. Life cycle completes by the alternation of generation. b. They are the developed plants in comparison to thallophyta. c. The plant body is differentiated into stem and leaves and roots are not present. Examples are Moss, Liverworts. Activity-2 Answer the following quesions 1. Write characteristics of Fungi and Animalia kingdom. 2. Write characteristics of division bryophyte. --97--

Day-45 Activity-1 Study the following tips: Division Tracheophyta : This division is classified into three sub divisions. Sub division Pteridophyta Characteristics: a. The body is differentiated into roots, stem and leaves. b. Complex tissues xylem and phylum are well developed. c. They are mostly found and cool and damped places. Examples are Fern, Horsetail etc. Sub division Gymnosperm Characteristics: a. They are cone bearing plants. b. They grow in dry places. c. They bear naked seeds. Examples are Cycas, Pinus etc. Sub division Angiosperm Characteristics: a. They grow both in land and water. b. They bear developed flowers. c. Their seed are enclosed in true fruits. Angiosperm are further divided into two classes. Class Dicotyledon Characteristics: a. They are also grown in both water and land. b. They bear two cotyledons in their seeds. c. They have a tap root system. Examples are Soyabean, Mustard etc. Class Monocotyledon Characteristics: a. They have parallel venation in leaves. b. They bear single cotyledons in their seeds. c. They have fibrous root system. Examples are Maize, Banana etc. Activity -2 Answer the following questions: 1. Enlist the characteristics of Pteridophytes. 2. Mustard is dicotyledon, justify. 3. Bryophytes are called amphibian plant, justify. Activity-1 Day-46 Cycas Study the following tips: Kingdom : Plantae Classifications of Some Plants: Mustard Division : Tracheophyta Kingdom : Plantae Sub Division : Fern Division : Tracheophyta Kingdom : Plantae Sub Division : Angiosperm Division : Tracheophyta Sub Division : Pteridophyta --98--

Type : Fern Class : Dicotyledon Gymnosperm Type : Mustard Type : Cycas Kingdom Animalia : This Kingdom is classified into nine phylum. The first eight phylum are invertebrates while the ninth one is vertebrate. Phylum Porifera : - Porifera are the organism having pores. - Digestion occurs within the cell. - Reproduction occurs sexually and asexually. Examples are Sycon, Spongilla etc. Phylum Coelenterata - They have body cavity. - They are mostly aquatic found in fresh water. - They are characterized by their stinging cells. Examples are Hydra, Jellyfish etc. Phylum Platyhelminthes - They are soft bilaterally symmetrical animals - Digestive respiratory and circulatory system is absent. - They are hermaphrodites. Examples are Tapeworm, Liverfluke etc. Activity -2 Answer the following questions: 1. Classify the plants: Volvox, Soyabean, Bamboo, Pistia. 2. Write characters of phylum porifera and coelentrata. Day-47 Activity-1 Study the following tips: Phylum Nemathelminthes - They are round in shapes and found in fresh water and damp soil. - They are triploblastic animals. - Their alimentary canal, mouth and anus can be easily distinguished. Examples are Hookworm, Pin worms etc. Phylum Annelida - They have circular bands and their body is divided into different segments. - Respiration takes place through skin. - Excretion takes place through nephridia. Examples are Leech, earthworm etc Phylum Arthropoda - Body is divided into head thorax and abdomen. - Excretion takes place through malphigian tube. - Circulatory system consists of dorsal heart with arteries only. Examples are Butterfly, Prawn etc. Phylum Mollusca - They are mosltlyacquatic. - Respiration takes place through gills - Alimentary canal is well developed. Examples are Snail, Cuttle fish etc. --99--

Phylum Echinodermata - Surface body is covered with spines - Mouth lies on the ventral surface. - Locomotion takes place through tube feet. Examples are Sea cucumber, Starfish etc. Activity-2 1. Write characteristics of phylum arthropoda and nemathehelminthes. Day-48 Activity-1 Study the following tips: Phylum Chordata : - They have a dorsal tubular hollow nerve tube. - Their eyes are originated from brain. - They have gill slit in embryonic stage. - Chordata are further classified into four sub phylum on the basis of development of notochord as described below. Sub Phylum Hemichordata : The animals possess the notochord in the anterior region of their body. Example: Balanoglossus. Sub Phylum Urochordata : The notochord is present in the tail region of the larval stage only but it is not present in the adult stage. Example Polycarpa. Sub Phylum Cephalochordata: The animals have the notochord that extends from the anterior end to the posterior end of the body. Example - Branchostoma This three sub phylum are collectively called protochordata. Sub Phylum Vertebrata - They have a vertebral column. - They have brain enclosed in a cranium. - They have a postanal tail and gill slits. - This sub phylum is further divided into five classes: 1. Class Pisces - Their body is long and streamlined. - Their body is covered with waterproof smooth scales. - They breathe through gills. Example Katla, Seahorse etc. Activity-2 1. Write characteristics of Class pisces. Day-49 Activity-1 Study the following tips: Class Amphibia - They can live on land and water both. - Their skin is usually thin, soft and moist. - They have two pairs of pentadactyl limbs. Examples are Frog, Toad etc. 2. Class Reptilia - They move by crawling. --100--